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Learning Objectives
At the completion of this lesson you should be able to…
• Explain how a mutation in a single gene can control resistance to herbicides such as ALS inhibitors.
• Predict the impact of a specific mutation on gene expression and the eventual trait observed in a plant.
• Distinguish between the molecular and classical definitions of an allele.
• Describe the molecular basis of dominance or a lack of dominance between alleles.
This lesson describes how changes in the DNA sequence of a gene can alter the synthesis of a protein and thus influence traits such as herbicide resistance. In this lesson we will describe how changes in the gene can alter the gene expression process and influence traits in an organism. The specific example of ALS-inhibitor herbicide resistance is used to demonstrate the impact of genetic change on trait expression in a plant.
Sequence Determines Structure, Structure Determines Function
Transcription and translation are fundamental to the production of all proteins found in all living things. That is why discovering the sequence of a gene can allow geneticists to predict the amino acid sequence of the protein that gene encodes. Remember, a completely synthesized protein will be hundreds of amino acids long and thus an average gene will have two or three thousand nucleotides in its sequence. As more is learned about how amino acid sequence dictates protein function, the value of gene sequence information to biologists increases. In this lesson, we will apply our understanding of the gene expression process to understanding the genetic basis of resistance to ALS herbicides in plants.
The ALS enzyme is about 800 amino acids. The specific order of amino acids in the protein chain dictates how the molecule can be folded up into a secondary structure (Fig. 1).[1] This is because each amino acid has a different chemistry that influences its interaction with other amino acids within the protein chain. The protein’s secondary structure determines the function of the protein. We can relate this principle to functional objects that are more familiar to us. Why would someone need a set of wrenches? Every tool in the set is made of the same materials. However, the specific shape and dimensions of each tool allow it to perform a specific function. The ALS enzyme functions to catalyze reactions for amino acid synthesis in the chloroplast in a plant cell. Therefore, the plant ALS enzyme structure signals the cell to transport the enzyme to the chloroplast and then allows the enzyme to catalyze the formation of a bond between amino acid precursor molecules. Clearly it was important to assemble the amino acids properly when building the ALS enzyme to allow it to function properly.
ALS and Weed Control
The function of the ALS enzyme in plants is also an interesting issue in the area of weed control. Herbicides have been discovered with sulfonyl urea or imidazolinone chemistries that will bind to ALS enzymes in plants and disrupt their catalytic functions (Figure 2). Because the herbicides bind tightly to ALS and ALS is not found in high copies in the cell, the herbicides will effectively halt the synthesis of the Leucine, Isoleucine, and Valine amino acids (Leu, Ile, Val). How will plants respond to the shortage of these three amino acids? They quit growing because they cannot make three of the twenty amino acids.
The proteins that need these amino acids will be only partially synthesized. The partial proteins will not work because they lack the structure needed for their function. The plant cells cannot grow and divide and the plant’s growth and development are halted. They will not necessarily die but they will no longer be able to compete for light and other resources if nearby plants are not affected by the ALS inhibitors. If the nearby plant that is not affected by the ALS inhibitors is a crop species such as corn or soybean, a valuable post emergence weed control strategy can be implemented.
ALS Alleles
Post emergence weed control by ALS inhibitors works in corn and soybean genotypes with altered ALS genes. The altered genes code for a slightly modified ALS enzyme which is resistant to ALS inhibitors. The molecular basis of this alteration can be determined and help us understand what alleles of a gene are. Even without molecular analysis, geneticists discovered alleles of the ALS gene that would make plants resistant to the ALS herbicides. How were these alleles discovered?
Corn breeders knew that ALS resistance did not naturally occur in their breeding lines. This discovery was easy to make. They planted thousands of different lines in short rows in the field, sprayed the field with the herbicide and observed the reaction of the lines. All of the lines tested had significant stunting. Variation did not exist in their lines so they decided to try to induce variation by mutagenesis. Pollen was collected from plants that had been exposed to a chemical mutagen. The chemical induced mistakes in DNA replication during pollen formation. The pollen was viable but carried mutant alleles. This pollen was applied to the silks of other corn plants to produce thousands of seeds. This seed was planted and the field test for ALS resistance was repeated. This time some plants resistant to stunting from ALS were observed, tagged and self-pollinated. From this experiment, the IT (immi tolerant) allele was discovered. The IT allele appeared to be dominant over the normal ALS allele because plants with one copy of the IT allele per cell (heterozygotes) had the same level of resistance as plants that had two copies of the IT allele (homozygotes). Molecular analysis of these plants was done to reveal how the IT allele was different for the original version.
The IT Allele
The IT allele of the ALS gene has a change in the coding region sequence. This was discovered by making a gene library from the DNA of homozygous IT plants and selecting a clone of the IT gene. The gene clone was sequenced and compared to the normal susceptible ALS gene from corn. A single nucleotide substitution in the IT allele was responsible for endowing a corn plant with resistance to ALS herbicides. The nucleotide substitution was in the gene coding region so it changed a codon sequence in the mRNA, coding for a different amino acid in the protein. The amino acid substitution had a minor or negligible effect on the enzyme’s ability to perform its normal functions. However, the amino acid substitution changed the site at which the ALS herbicide binds to the enzyme. The ALS herbicide will not interact with the ALS enzyme encoded by the IT allele, giving the plant resistance to the herbicide. This is an example of how a small change in a gene can have a big impact on a plant’s phenotype in the right environment.
The IR Allele, Tissue Culture Selection for Mutations
ALS alleles that confer resistance to immidazolinone herbicides have also been selected from mutations that have occurred in tissue culture cells. DNA replication does not occur 100 percent without errors so it is not surprising that natural mutations occur in cells. A mutation, though, that gives a single cell in a corn plant resistance to ALS herbicides will not render the whole plant resistant. Geneticists have used the power of tissue culture to select for these mutations at the single cell level. Hundreds of thousands of corn cells can be grown in tissue culture plates that contain the herbicide. Most of the cells will quit growing and dividing when they take up the herbicide but some cells continue to grow and form clumps of undifferentiated cells called callus. Plants regenerated from these callus cells will often express the herbicide resistance as well. The IR allele of the ALS gene was selected in this way.
This allele also has an altered coding region sequence that blocks the ALS herbicides from binding. The ALS enzyme encoded by the IR allele, however, does not provide desired levels of herbicide resistance in a heterozygous plant. The molecular mechanisms causing this difference in the IR compared to IT allele are not clear. The implication of this difference in plant breeding are significant. The IT allele can be inherited from just one parent to give the ‘Clearfield’ or “Immi” hybrid herbicide resistance. Hybrids must be homozygous for the IR allele requiring both parents be homozygous IR inbreds (Figure 3).
Molecular Basis of Dominance
The occurrence of the IT, IR and normal alleles of the ALS enzyme and their impact on plant phenotype provides us with an opportunity to think about dominance and lack of dominance at the molecular level. The IT allele has complete dominance to the normal allele because there is apparently no phenotype difference between homozygous dominant and heterozygous IT plants when they are sprayed with an ALS inhibitor. This suggests that in the heterozygous plants, one copy of the IT allele per cell encodes enough resistant ALS protein to drive the synthesis of sufficient quantities of Val, Leu and Ile for normal growth in the corn plant. However, the ALS enzyme encoded by the normal allele in these heterozygotes can be inhibited by the herbicide and will not contribute to amino acid synthesis.
In contrast, the IR allele has a lack of dominance to the normal allele (IN). Heterozygous plants that have one copy of the IR and one copy of the normal allele per cell do not maintain normal growth after ALS inhibitor treatment. One copy of the IR allele per cell does not encode sufficient copies of a resistant ALS enzyme to support normal growth. Instead, two copies of the IR allele per cell are needed so that all of the ALS enzyme made will not be bound by the herbicide.
The classification of the plant phenotype for characterizing possible genotypes at the ALS locus may depend on environment. In some growing environments it has been noticed that heterozygous IT hybrids will show a slight growth reduction in the sprayed plants compared to unsprayed. The difference is minor because the plants usually grow out of this difference later on and yield reduction is not a factor. This observation does emphasize that phenotype is a result of genotype and environment. Therefore, scoring traits in the proper environment will be critical to discovering the nature of gene action.
How Universal is the Genetic Code?
The table of codons used by organisms to translate mRNA into proteins is shown on the bottom of the page (Table 1). As was mentioned earlier in this lesson, the genetic code needed to be cracked one time because all organisms used the same codons to encode amino acids. As scientists began to sequence the coding regions of genes from different organisms, they discovered something called a codon preference. When you look at the codon table, you can see that the genetic code is redundant. This means that more than one codon can encode the same amino acid. This is because there are 61 codons that code for the placement of 20 different amino acids. A codon will only work in coding if a tRNA with a complementary anticodon is also found in the same cell and has the appropriate amino acid to deliver. Therefore, there could be 61 different tRNAs, one to complement each codon. Each different tRNA needs to be encoded by a different gene. If that gene is not expressed in the cell, the tRNA will not be found and a codon that needs to be complemented by that tRNA will not be complemented. In this case, the codon will act like a stop codon. The ribosome will halt translation and the protein made will be a shorter version of the intended protein. Organisms would not benefit from this situation so there is a tight complementation between what tRNAs genes are present and expressed in an organism’s cells and what codons are used to encode a specific mRNA. In this way the genetic code will have a dialect. The language is universal but certain words are used preferentially.
Table 1. RNA Codon Table. The genetic code. The 20 common amino acids are listed in their three and one-letter formats. Stop codons are UAA, AUG and UGA
First Position
Second Position
Third Position
U
C
A
G
U
Phe (F)
Ser (S)
Tyr (Y)
Cys (C)
U
C
Leu (L)
Stop
Stop
A
Stop
Trp (W)
G
C
Leu (L)
Pro (P)
His (H)
Arg (R)
U
C
Gln (Q)
A
G
A
Ile (I)
Thr (T)
Asn (N)
Ser (S)
U
C
Lys (K)
Arg (R)
A
Met (M)
G
G
Val (V)
Ala (A)
Asp (D)
Gly (G)
U
C
Glu (E)
A
G
Scientists are not sure why codon preferences are a part of the gene expression process in organisms. It may provide another level for the organism to control the amounts and kinds of proteins made in its cells. Recent experiences in genetic engineering of plants and animals, however, has made codon preference an important consideration. For example, scientists have put genes from a soil bacterium called Bacillus thuringiensis (Bt) into corn plant cells in order to give the corn plant the ability to make a protein that is toxic to European corn borer, a common pest to corn producers. They found that the gene would be transcribed but the mRNA would not be translated to make the desired protein. One reason was codon usage. Some of the codons the bacteria use to encode amino acids are rarely used by corn. The corn plant either lacked the tRNA to complement the codon or make the tRNA at such low levels that there were not enough copies in the cell to accommodate translation of the Bt mRNA. Therefore, the genetic engineers needed to make synthetic coding regions that substituted codons preferred by corn for those preferred by bacteria. The end result was they were able to get higher levels of the Bt protein made once these changes were made in the gene. Codon preference thus makes the genetic engineering process more challenging.
Learning Activities
Watch this animation of the transcription process.
Watch this animation of the translation process.
1. Graphics used in this figure were taken from The RCSB PDB "Molecule of the Month": Inspiring a Molecular View of Biology D.S. Goodsell, S. Dutta, C. Zardecki, M. Voigt, H.M. Berman, S.K. Burley (2015) PLoS Biol 13(5): e1002140. doi: 10.1371/journal.pbio.1002140 | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.08%3A_Gene_Expression-_Applied_Example_%28Part_2%29.txt |
Learning Objectives
• Understand the concept of gene expression.
• Understand transcriptional regulation of gene expression.
• Understand epigenetic gene regulation.4
• Understand post-transcriptional regulation of gene expression (RNA level).
• Understand post-translational protein modification and regulation.
Introduction
Every cell in a plant contains the same genetic information, the same set of genes. Yet Therefore different sets of genes are required for the various functions of different cells or tissues, as well as for plant responses to environmental stimuli or stresses. This is achieved by regulating the activity of genes according to the physiological demands of a particular cell type, developmental stage, or environmental condition. This regulation of activity is known as gene expression.
The term expression can be used in different ways that are sometimes confusing. Typically, if a gene product is produced, the gene is considered “expressed”. However, it sometimes occurs that a transcript might be produced but not a protein, or that a protein is produced but it is in an inactive state. In such cases, although a gene product is produced, the biological activity encoded by that gene is not present. For the purposes of this section, the key point is how the biological activity encoded by a gene is regulated.
The expression of genes in specific plant cells, tissues, and organs and the timing of this expression require a precise level of regulation. Expression, or genetic function, can potentially be regulated at any of the steps from transcription, RNA processing, translation, through post-translational protein modification, as discussed in lesson 1. Regulation can be qualitative (i.e. gene expression is either “on” or “off”) or quantitative (i.e. expression levels can be modulated “up” or “down”). Fluctuations in the intensities of external stimuli coupled to changes that occur at the genomic level result in different developmental outcomes or physiological states. Regulation of gene expression at the level of transcription can be brought about through chromatin and histone modifications. Also, a gene sequence can be differentially spliced to produce mRNA products of variable lengths leading to new protein products with novel functions. Some genes do not encode proteins but short forms of RNAs with regulatory functions such as induction of flowering. Finally, proteins products can be subjected to modifications such as phosphorylation or dephosphorylation to alter their functions, or can be completely degraded to turn off a gene.
Transcriptional gene regulation
Since transcription is the first step in gene expression, it makes sense in terms of cellular economy, to regulate expression at this point, and in fact this is one of the most important regulatory points. We already described the involvement of RNA polymerase in the transcription process, but in fact there are other protein factors that are required. Proteins involved in transcriptional regulation are known as transcription factors. It is the interaction of these transcription factors with specific DNA sequences that regulate the process of gene transcription.
A. The concept of differential (regulated) gene expression
As just described, not every gene is expressed all the time. When a gene displays different levels of expression in different circumstances, this is known as differential expression. Circumstances that might apply include, but are not limited to, different plant tissues (root vs. leaf), different developmental stages (germination vs. reproductive development), or in response to different environmental stimuli (cold stress or pathogen attack).
The term differential expression can also be used to compare the expression of different genes. If two genes show different expression patterns (among plant tissues or in response to environmental stimuli), they are considered differentially expressed, whereas genes that showed very similar patterns of expression would be considered co-expressed.
B. Promoters
As mentioned, transcription is regulated through the interactions of proteins, transcription factors, with specific DNA sequences. Most regulatory DNA sequences governing gene transcription are located on the 5′ border of the transcribed region. This region is called the gene promoter.
Promoters contain a core, which is required for the binding of the “basal transcriptional machinery”, including RNA polymerase. The “TATA” box, with a consensus sequence TATAA, is located within the core promoter, usually 25-30 nucleotides upstream of the transcription initiation site. Promoters also contain regulatory sequences that determine when, where, and to what level genes are transcribed. Promoters can vary in length from a hundred to a few thousand nucleotides.
The full promoter sequences of different genes that are expressed in a similar manner may be different. However, such promoters often contain short sequence “motifs” that are similar, referred to as cis elements. Early work (Benfy and Chua, 1990) to understand the function of different promoter elements in regulating gene expression in plant cells and tissues revealed that various combinations of cis elements are able to be interpreted by the cell and control gene expression. Sometimes cis elements promote gene transcription and sometimes they function to restrict transcription of genes in particular cells and tissues.
Promoter analysis is facilitated through the use of reporter genes. The reporter gene produces effects that are easily identifiable and quantifiable, which can be used to determine the function of a regulatory region of another gene (promoter, promoter elements, or enhancers) in cells, tissues, or organs. Such analyses are critical to crop biotechnology where targeted expression of genes in particular tissues is often desirable. To test whether a promoter is effective in conferring the expression of a gene to a particular tissue, scientists fuse the putative promoter to a reporter gene and introduce the promoter-gene fusion into plants. An example of a reporter is the GUS gene, which encodes a GUS enzyme (beta-Glucuronidase) that, when expressed, produces a blue color upon addition of a substrate. Figure 1 shows an example of GUS expression in maize seed using two promoters that act either in the endosperm or the embryo.
C. Enhancers
Enhancers are DNA sequences that increase the rate of transcription of a gene when they are present, although alone they cannot cause transcription to occur. Enhancers are usually position and orientation independent. Although they are normally located upstream of the promoter, they can also be located on the 3’ region of the gene or even with the coding region. Enhancers can increase the transcription when added to genes that they are not normally associated with. This is a useful property for biotechnology, allowing promoters to be manipulated for increased levels of transcriptional regulation. Some enhancers function at all times in all cells and tissues and they are referred to as constitutive. Other enhancers function in specific tissues at specific developmental stages. Some are active only in response to environmental signals. The AACCA enhancer on the promoter of soybean β-conglycinin gene encoding a seed storage protein functions specifically in seeds. Enhancers are thought to interact with specific nuclear proteins involved in transcription. For example, enhancers might facilitate the binding of transcription factors and direct these factors along the DNA strand to the direction of the promoter. Alternatively, enhancers may facilitate changes in DNA structure such as modification of the chromatin structure.
The counterpart of an enhancer is a silencer. Silencers have all the properties just described for enhancers, except they function to dampen, or decrease, the levels of transcription controlled by a promoter.
D. Transcription factors
RNA polymerase binds the promoter at the TATA box and in cooperation with other proteins drives gene transcription. The proteins that interact with RNA polymerase to facilitate its binding to the promoter and to regulate its activity are known as transcription factors. Some transcription factors, known as “basal transcription factors” are fundamental to RNA polymerase binding and function, and are expressed in all living cells.
Other transcription factors bind to cis regulatory DNA sequences of promoters, enhancers or silencers. These proteins interact in complex ways with the basal transcription machinery, to regulate the activity of RNA polymerase, and therefore gene transcription. Thus, transcription factors regulate when or where individual genes are expressed, and to what level.
Transcription factors can function as either positive or negative regulators. That is, they can either function to induce (increase) gene transcription or to repress it. The consequence of many transcription factors depends on their interactions with other proteins. Two factors together may be required for gene activity, and exclusion of one of the factors in space and time offers a mechanism for differential gene expression. Some transcription factors might function as a positive regulator in one context but as a negative regulator in another context, depending on what other cis elements and/or transcription factors might be present.
Epigenetic regulation
A. Chromatin structure and histone modification
At the molecular level chromatin is a product of an ordered and tight packaging of the double stranded DNA around nucleosomes (a core of proteins named histones), and an association with additional proteins.
Transcriptional regulation often involves modification of chromatin structure mediated by post-translational regulation (changes in acetylation and methylation) of histones. Histone acetylation involves the addition of acetyl groups and when histones are heavily acetylated the DNA is less tightly associated with them. This often correlates with increased transcriptional activity of specific genes. The idea is that when DNA is loosely associated with histones, it is more accessible to transcription factors that require interaction with the DNA to initiate transcription. Consequently, histone deacetylation (removal of acetyl groups) by histone deacetylase enzymes (for example, FLD, p462) stabilizes nucleosomes and represses transcription. On the other hand, histone acetylation by histone ecetyltransferase destabilizes nucleosomes and promotes transcription.
As described in the text, another form of histone modification is the addition of methyl groups to histone proteins, which similarly regulates chromatin condensation or decondensation.
B. DNA methylation
The nucleotide bases of DNA can be modified by the attachment of methyl groups at various locations. The addition of these methyl groups happens after the DNA has been synthesized and is controlled by enzymes that add methyl moieties to specific regions of DNA. The most common modified nucleotide base is C5 methylcytosine (m5C) due the activity of the enzyme DNA (cytosine-5) methyltransferase (MET). MET recognizes the unmethylated newly-replicated DNA strand and incorporates a methyl group if the template strand was methylated (hemimethylated).
Methylation status is correlated with gene expression (Figure 5) as demonstrated by the low level of methylation in regions of the genome undergoing active transcription.
Studies have shown that altering a plant’s overall DNA methylation status can affect growth and development. For example, cold treatment (vernalization) induces flowering in biannuals such as Arabidopsis and winter wheat, and lowers the level of methylation of particular genes. Also, treating plants with the drug 5 azacytidine prevents methylation at the 5 position of cytosine and stimulates flowering. Recent studies also suggest a relationship between epigenetic gene regulation and heterosis, with hybrids showing higher global levels of transcription, higher histone acetylation levels and lower DNA methylation levels (reviewed in He et al., 2011).
RNA-level regulation
Regulation of gene expression occurs at many levels, including post-transcriptionally. From the standpoint of the expression level for a given gene, a critical factor is the level of fully processed, mature mRNA. When we consider the level of the mature form a particular transcript at any given time reflects the steady state balance between synthesis, processing and degradation. The Post-transcriptional regulation of RNA occurs through several mechanisms.
RNA stability and degradation
The control of mRNA degradation rate, or turnover, is an important regulatory mechanism in gene expression. As mentioned, the level of a particular transcript at any given time reflects the steady state balance between synthesis and degradation. Thus, even though a gene may be transcribed at a high rate, a high rate of RNA turnover could effectively shut off the gene. The stability of mRNA can be regulated globally, to affect all or most transcripts, or it can be very specific to a particular mRNA. Regulated mRNA turnover may be particularly important in plants, since plants cannot move to avoid harsh environmental conditions and stresses are known to induce specific changes in RNA stability. Other factors such as light and plant hormones are also known to regulate mRNA stability.
The degradation of mRNA is catalyzed by enzymes called ribonucleases. Ribonucleases include exonucleases, which degrade only from one end of the transcript, and endonucleases, which can attack the mRNA molecule internally.
Therefore, controlling ribonuclease access to the mRNA substrate is an important regulatory mechanism in gene expression. This control is achieved through various mechanisms including, controlling the amount of nucleases present, regulating the activity of the nuclease, sequestering the nuclease to particular cellular locations to restrict its access to RNA, and to control the stability of the mRNA by decreasing accessibility to nuclease.
Four regions of an mRNA are important for its overall stability. These include, the 5′ untranslated region and the cap, coding region, 3’ untranslated region, and the poly(A) tail.
The cap at the 5′ end protects the mRNA from exonucleases that degrade RNA from the 5′ to the 3′ direction.
The 3′ untranslated region may contain short sequences that influence mRNA stability. For example, the repetition of the sequence AUUUA in the 3′ untranslated region of many animal and plant genes is associated with mRNAs with short half-lives (the time it takes for half the RNA to be degraded, after transcription is stopped).
The poly(A) tail increases mRNA stability but is not sufficient alone. It only provides stability when bound by proteins such as poly(A) binding protein (PABP). Test-tube experiments have shown that, removing PABP from a stable mRNA destabilizes it, while adding back purified PABP restores stability.
Translational regulation
As described, for most genes it is the protein product that performs the biological function. As such, regulating the amount of protein production effectively regulates gene expression. There are several known mechanisms by which the process of translation is known to be regulated. A detailed consideration of these mechanisms is beyond the scope of this course, but in general it is important to bear in mind their importance. For example, under certain stress conditions, “normal” translation is halted and only mRNAs related to stress tolerance are selectively permitted to be translated into proteins. This selective translation is important to allow plants to quickly respond to stress and to conserve energy under stress conditions.
Translation of mRNAs can play an important role in determining their overall stability. Mutations that add premature stop codons often lead to rapid degradation of the mRNA. Ribonucleases (figure 9) or other factors involved in degradation may recognize the number or spacing of ribosomes on an mRNA, and degrade those that are not produced properly. This may help prevent synthesis of proteins with incorrect functions which will negatively impact cellular processes. Errors during transcription could add or omit nucleotides which would alter the proper codon sequence creating mutant proteins.
Protein-level regulation
After translation, proteins are subject to a variety of modifications that can regulate their activity. There are many different ways by which proteins can potentially be modified and thereby regulated, and only a few of the basic mechanisms will be considered here.
Common types of post-translational modifications
The first mechanism is covalent modification by the addition of various chemical groups. A wide variety of groups can be involved, including small organic groups (methylation, acetylation) lipids (myristoylation, farnesylation, palmitylation), carbohydrates (glycosylation, glucosylation), small proteins (ubiquitination, sumoylation) and inorganic molecules (phosphorylation, sulfation). Such covalent modifications are generally accomplished by the activity of enzymes specialized to perform these modifications. Many of these modifications are also reversible; that is these groups can be added to a protein and subsequently removed. Phosphorylation is a particularly noteworthy reversible modification that is common in the regulation of many proteins. Enzymes that add phosphate groups to other proteins are called protein kinases, and those that remove phosphates are called phosphatases. As such, protein kinases and phosphatases are central to many cellular regulatory systems.
A second common mechanism of protein modification is through proteolytic cleavage. Proteolysis occurs as part of the general turnover of cellular proteins, which is required to eliminate damaged proteins and recycle amino acids. Proteolysis is important for the processing of certain proteins, for example in the removal of the signal peptide of proteins targeted to specific cellular compartments. It can also occur in a highly specific manner whereby particular proteins are targeted for degradation or for cleavage at a specific site within the protein. Proteolytic modifications are non-reversible.
Proteins can undergo modification through complex formation. Such complexes can occur among proteins or between a protein and a cofactor.
Proteins can also be modified according to conditions of the cellular environment. The redox state can result in oxidation or reduction of proteins, particularly of sulfhydryl side groups. Cellular pH can affect the charge of ionizable side groups.
Common types of protein regulation
All of the above mentioned types of protein modification can alter protein conformations and thus have regulatory consequences on protein function or activity. Protein regulation is highly complex and there are a myriad of different ways by which this occurs. Again we will just briefly consider a few of the more common mechanisms.
One major way protein modification can regulate protein function is by altering their activity. This can be true for many types of proteins including, but not limited to, enzymes, transcription factors, signaling proteins, and structural proteins.
Cells are compartmentalized into several membrane bound organelles, including the nucleus, chloroplasts, mitochondria, peroxisomes, endoplasmic reticulum (ER), golgi, and vacuoles. Each of these compartments performs unique metabolic functions that require a set of proteins. Compartmentation is regulated for some proteins. For example, upon exposure to light the phytochrome protein moves into the nucleus where it affects the expression of light-regulated genes.
Proteolytic processing is involved in several important regulatory processes. Many proteins are synthesized in an inactive form that requires proteolytic cleavage for activation. The full-length translation product before processing is often called the precursor, or a preprotein. As mentioned, cells contain membrane-bound compartments. Since membranes are impermeable to most proteins, an active mechanism is necessary to move a protein across a membrane. For proper delivery to their organellar destinations, specific amino acid sequences, called target signals must be present in a protein (to serve as an “address”). For example, entry into the chloroplast is achieved by the presence of a target signal called the transit peptide. This is at the amino terminal end of the protein and is proteolytically cleaved during import.
Another example of proteolytic processing is seen in a plant defense response in members of the solanaceae (for example, tomato and potato). An 18-amino acid peptide hormone called systemin is secreted by plant cells that are damaged by insects or mechanical wounding. Systemin production by wounded cells is required to induce the synthesis of proteins involved in defense. Systemin induces defense responses in wounded cells, and throughout the plant. Analogous to animal peptide hormones (e.g., insulin), systemin is initially synthesized as a much larger (200 amino acids) precursor called pro-systemin. Pro-system is inactive; however, upon wounding it undergoes proteolytic cleavage to produce activated systemin.
Targeted protein degradation
The amount of protein present in a cell or tissue is determined by both its rate of synthesis and its rate of degradation. Therefore, protein degradation is an important mechanism by which the plant can regulate biological activity (i.e., a genetic function). For example, one way to shut down a metabolic pathway is by degrading one of the key enzymes controlling the rate of the entire pathway. Therefore, protein degradation is an essential component of gene regulation to meet cellular demands for growth, development, and defense.
Protein degradation must be carefully controlled to fine tune gene expression to allow plants to adapt to new environmental conditions. Often, cells will adopt several complex mechanisms for proteolytic degradation of proteins. Enzymes that cleave or degrade proteins are referred to as proteases. For example, plant vacuoles are rich in proteases that play a similar function in protein degradation as that of lysosomes in animal cells. Protease activity must be tightly regulated to prevent accidental degradation of essential proteins. Sequestering proteases in particular organelles like the vacuole separates them from other organelles and is one of to control their activity.
An important mechanism by which specific proteins are targeted for degradation is through ubiquitin-mediated proteasomal degradation. The proteasome is a large complex of multiple protein subunits that have a protease activity to degrade proteins. Proteins get marked for proteasomal degradation with a small protein called ubiquitin. Ubiquitin is covalently attached to specific proteins in response to environmental or developmental signals. This allows plants to quickly adapt to changing conditions by eliminating proteins whose functions are not advantageous under the new conditions. For example, the photoreceptor phytochrome mentioned above becomes targeted for degradation when light is no longer available. This allows plants to change their physiological functions going from daylight to night conditions. Protein degradation by the proteasome system is also an important regulatory mechanism for plant hormone signaling, for example the signaling of the defense hormone jasmonic acid, and growth hormone gibberellic acid.
Proteins have lifetimes that range from a few minutes to weeks or more. Cells continuously make proteins from and break them down to amino acids. One of the functions of protein degradation is to eliminate aberrant or damaged proteins which could harm the cell. The second function is to facilitate the recycling of amino acids. For example, most of the amino acids required for growth of the seedling are derived from the degradation of seed storage proteins. Conversely, in annual crop plants, many of the amino acids in seed storage proteins are derived from proteins degraded in leaves and other plant parts during senescence.
Lesson Summary
The expression of genes in specific plant cells, tissues, and organs and the timing of this expression require a precise level of regulation. A single promoter may not be sufficient to regulate the expression of such gene(s) in space and time. Therefore, coding regions with the same function may have different promoters, and such genes are referred to as differentially regulated. Most regulatory sequences governing gene expression are located on the 5’ border of the coding region. Transcription is often initiated between 20 and 60 nucleotides upstream of the ATG start site. Enhancers are usually position and orientation independent. Although they are normally located upstream of the promoter, they can also be located on the 3’ region of the gene or even with the coding region. Enhancers can increase the transcription when added to genes that they are not associated with. RNA polymerase binds the promoter at the TATA box and in cooperation with other proteins drives gene transcription. The proteins that interact with RNA polymerase bind to regulatory sequences upstream and downstream of the transcription site.
Transcription factors can play a regulatory role by determining where individual genes are expressed. Transcriptional regulation often involves modification of chromatin by changes in acetylation and methylation of histone. The nucleotide bases of DNA can be modified by the attachment of methyl groups at various locations. Methylation status is correlated with gene expression. Alternative splicing describes an alternative mechanism of pre-mRNA processing to generate mRNAs that have different combinations of exons. The control of mRNA degradation rate, or turnover, is an important regulatory mechanism in gene expression. RNA interference (RNAi) is a post-transcriptional process involving the degradation of mRNA initiated by the formation of double-stranded RNA (dsRNA) of the target mRNA. Translation of mRNAs can play an important role in determining their overall stability. Mutations that add premature stop codons often lead to rapid degradation of the mRNA. Protein degradation is an essential component of gene regulation to meet cellular demands for growth, development, and defense. The plant can alter the activity of a metabolic pathway, by degrading one of the key enzymes controlling the rate of the entire pathway.
Exercises
1a. Which of the following could alter gene regulation
1. Deleting a promoter
2. Altering fertilizer application rate
3. Raising the temperature of the greenhouse
4. Herbicide application
5. Reduce irrigation frequency
1b. The detection of a gene product, for example, RNA or protein at any one moment is a reflection of the “steady state” of the product. Describe the term “steady state” in this context.
2. You are studying expression of a gene responsible for resistance to a pathogen. You cloned the gene and made antibody to detect the protein it encodes by a procedure called Western blot analysis, and the mRNA by a procedure called reverse transcription PCR. Explain the type of regulation from the scenarios in the table.
Scenarios
Not treated with pathogen
Treated with pathogen
mRNA
Protein
mRNA
Protein
1
Absent
Absent
Present
Present
2
Present
Absent
Present
Present
3
Present
Present
Absent
Absent
4
Present
Present
Present
Absent
5
Present
Present
Present
Present
3. You are studying the expression of a gene controlling height in your crop species. You have cloned the gene and are interested in determining its expression in different parts of the plant. You use the RT-PCR procedure as in the previous problem. After the PCR part, you load products on a gel and observe the following pattern. Explain your results in the context of gene regulation. | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.09%3A_Regulation_of_Gene_Expression.txt |
Learning Objectives
• Understand how multiple genes often function together to control a trait
• Understand the concept of a biosynthetic pathway
• Understand the concept of a regulatory pathway
• Understand the concept of a network
• Understand how mutations in different genes of a pathway can cause the same phenotype
• Understand how mutations in different genes of a pathway might cause different phenotypes
Introduction
Genes do not function in isolation but rather suites of genes act in concert to perform biological functions. When different genes function in different sequential steps of a biological process, this is known as a genetic pathway. Perhaps the most conceptually intuitive type of pathway is a biosynthetic pathway, where a precursor molecule is chemically modified through a series of enzymatically catalyzed intermediate steps to produce a bioactive product. Since enzymes are (proteins), the genes that encode these enzymes are considered a genetic pathway. In a regulatory pathway, some type of stimulus leads to a change in the expression or activity of a particular gene product, which in turn acts to alter the expression or activity of another gene product or products, which in turn could regulate yet another level of activity.
Ultimately, these regulatory changes in expression or activity lead to a response to the stimulus, typically involving changes in gene expression. Again, the genes that encode the various types of regulatory molecules, mainly proteins, constitute a genetic pathway. Pathways are often targets in the biotechnological manipulation of traits, which requires an understanding of how pathways behave. Also note that many biosynthetic and regulatory systems are better described as networks because of often, pathways branch or converge and interact with other systems. However, for simplicity such complex networks will not be discussed here.
Biosynthetic pathways
Genes encode enzymes that catalyze steps in the synthesis of a compound. A biosynthetic pathway actually describes a process of converting a precursor substrate into a product. In addition to the precursor substrate and product, it includes the enzymes, the chemical reactions catalyzed by the enzymes and all the intermediate compounds. A pathway consists of a series of steps where a precursor molecule acts as a substrate for an enzyme, which catalyzes a chemical reaction to produce a product, which then serves as a substrate for a subsequent step.
We have already discussed the one gene, one enzyme hypothesis which was an important advance in the discovery of how genes encode proteins. This analysis also provides a nice example of the concept of a biosynthetic pathway. This hypothesis was discovered by analyzing auxotrophic neurospora mutants that were defective in the production of amino acids. That is, the biosynthetic pathways producing particular amino acids were disrupted by mutations. The text discusses the analysis of the arginine pathway. Mutants that disrupt this pathway cannot grow on minimal medium lacking arginine but can grow if arginine is supplied in the medium. The same fundamental pathway also produces arginine in plants.
Mutations that disrupt the function of particular enzymes in the pathway block the progression of the pathway at that corresponding step. For example, a mutation in the argF gene would disrupt the enzyme ornithine carbamoyltransferase (OTC). This would block step 6 in the pathway, the conversion of L-ornithine to L-citruline. As a result, the intermediate compound L-Ornithine will likely accumulate, whereas all the compounds that occur later in the pathway (i.e., L-Citruline, L-argininosuccinate and L-arginine) will be deficient.
The anthocyanin pathway
The anthocyanin pathway is a slightly more complex pathway, although still simple as far as these things go. It is of significant interest to horticulturists because anthocyanin pigments are responsible for many plant colors, particularly in flowers and fruits (think red wine). Anthocyanin pigments range from orange to red, purple, and blue. In addition to the attractive colors, anthocyanins provide, there is also considerable recent interest in anthocyanins for their dietary value, acting as strong antioxidants and possibly anti-cancer agents.
Effects of mutations at different steps in anthocyanin biosynthesis.
Of course, each enzyme in the anthocyanin pathway is encoded by a gene. As such, mutations can generate variation or disrupt the pathway. The maize anthocyanin pathway and examples of mutants are shown in Figure 2. The a2 mutant causes a deficiency in the anthocyanidin synthase enzyme, blocking the production of any pigmented compounds. The a1 and c2 mutants are similarly colorless. The bz2 mutant is deficient in an enzyme related to glutathione-S-transferase, which is required for transport of anthocyanidins into the vacuole, resulting in the bronze-colored pigment due to the more alkaline environment of the cytosol. The bz1 mutant (not shown) produces a similar phenotype. Mutation of the pr1 gene eliminates just one class of anthocyanins. The pr1 gene encodes flavanone 3′-hydroxylase, required for the production of cyanidin, which generates purple anthocyanin. In the mutant, only pelargonidin is synthesized, producing a red pigment.
Regulatory pathways
A regulatory pathway refers to a situation where one gene or gene product controls the expression or activity of another gene or gene product, which may in turn also perform a regulatory function. As we discussed, gene function can be regulated at many points from transcriptional regulation to post-translational regulation. We will examine a couple of specific examples of regulatory pathways but first let’s consider some of the general features of regulatory pathways. Understanding the logic of regulatory pathways will become important later when we seek to use biotechnology to manipulate traits by altering their regulation.
A simplistic but useful way to help understand the logic of regulatory pathways is to consider them as a series of switches with on and off states.
Positive vs. negative regulators
A key feature of regulatory pathways is that regulators might function either positively or negatively. A positive regulator is one that functions to activate the next component in the pathway, whereas a negative regulator would inhibit the next component. For example, a transcription factor that activated the transcription of a gene would be considered a positive regulator, while one that repressed the transcription of a gene would be a negative regulator. Another common type of regulator is a protein kinase. Protein kinases are proteins that add a phosphate group to another protein. Phosphorylation is a common type of post-translational modification that can regulate the activity of proteins; sometimes the phosphorylated protein becomes activated and sometimes it becomes repressed.
A positive regulatory step is represented with an arrow. For example, “protein A activates protein B” would be represented like:
$\text{A} \to \text{B}$
A negative regulatory step is represented with a bar. “Protein C represses protein D” would be represented like:
$\text{C} \dashv \text{D}$
Active vs. inactive states of regulators
A second point to consider in the logic of regulatory pathways is the activity state of each component. A useful way to think about this is that each component can have an “on” or an “off” state. Of course, in reality, many proteins can have intermediate levels of activity but for the sake of simplicity, we will just consider the on and off states.
The effect of a regulator being in the on or off state depends on whether it is a positive or negative regulator.
If a positive regulator is in the on state, it will function to activate the next factor. In the example of $\text{A} \to \text{B}$, if A is in the on or active state, it will function to switch B to the on or active state.
If a negative regulator is in the on state, it will function to switch the next factor to the off or inactive state. For $\text{C} \dashv \text{D}$, if C is in the on or active state, it will switch D to the off or inactive state.
Regulatory pathways sometimes contain many, sometimes few, steps. They can also contain a mixture of positive and negative regulators. Let us explore a few hypothetical examples to see how the pieces fit together to regulate plant responses to stimuli.
When a pathway is depicted, all the steps are shown. For the sake of simplicity, it is assumed that the default activity state of any given component is such that the preceding step functions to change it. We will begin with a simple example containing several positive regulatory steps that activate a cellular activity, Activity 1, in response to Stimulus 1. In the pathway shown below, the Stimulus 1 activates A. Therefore, we assume that in the absence of this stimulus, A is in the inactive or off state. Since A is inactive, it is not functioning to activate B, which is therefore also in the inactive state, and likewise for C. Ultimately Activity 1 does not occur in the absence of Stimulus 1 but does occur in response to the stimulus.
Stimulus 1 → A → B → C → Activity 1
We can use color coding to help visualize this. Light grey represents the OFF state and blue represents the ON state.
No Stimulus 1 A B C Activity 1 (OFF)
Stimulus 1 A B C Activity 1 (ON)
Or we can make a table to help keep track of the states of each component in the presence or absence of stimulus.
Stimulus 1 A B C Activity 1
NO OFF OFF OFF NO
YES ON ON ON YES
Now let us look at another example containing both positive and negative regulators.
Stimulus 2 → D → E $\dashv$ F → Activity 2
In this case, components D and E would be in the inactive states in the absence of stimulus, because they would not have been activated. But E functions as a negative regulator of F. Since E is inactive, it is not functioning to inhibit F, which then remains in the active state promote cellular Activity 2. When Stimulus 2 is present, D and E become activated, and E functions to inactivate F. Since F is off, Activity 2 does not occur. Thus, the net response to Stimulus 2 is the repression of Activity 2.
$\dashv$
$\dashv$
OR
Stimulus 2 D E $\dashv$ F Activity 2
NO OFF OFF ON YES
YES ON ON OFF NO
Now that we have considered regulatory pathways from a hypothetical perspective, let’s consider a couple real ones. As mentioned, regulatory pathways can take on many forms. They can consist of a series of transcription factors that regulate the expression of one another’s genes, ultimately resulting in the regulation of genes that effect some biological response. Here, we will look at a signal transduction pathway consisting of a series of post-translational modifications that occur response to a hormone stimulus and culminate in gene expression changes to effect a response.
Signal transduction pathway for hormone signaling
Brassinosteroids (BRs) are a class of plant steroid hormones that control many aspect of plant physiology. They are most noted for their role in promoting plant growth and mutants deficient in BR biosynthesis or signaling show dwarfism. BRs also promote a range of other responses, including yield and increased stress resistance.
BR hormones, of which brassinolide (BL) is considered the most active, are recognized by a receptor called BRI1. BRI1 is a type of receptor known as a receptor kinase, which spans the cell membrane. On the outside of the cell is a receptor domain which specifically recognizes and binds to BR. There is also a membrane-spanning domain and then a protein kinase domain inside the cell. Binding to BR outside the cell then activates the protein kinase domain inside the cell. Protein kinases are enzymes that function to add phosphate groups to other proteins. As previously discussed, phosphorylation can alter the activity of a protein. Activation of the BRI1 receptor kinase then results in the activation of a signaling pathway that ultimately regulates the activity of transcription factors inside the nucleus, changing the expression of genes that regulate growth and stress resistance.
The BR signal transduction pathway is well studied and is summarized in Figure 3. Do not be concerned with memorizing the names of all these factors, but just focus on understanding the regulatory logic. The overall regulatory relationships of the pathway are depicted in Figure 3. The main target of regulation is a pair of closely related transcription factors called BZR1 and BZR2. BZR1/2 are negatively regulated by a protein called BIN2 in the absence of BR. BIN2 is a protein kinase that phosphorylates BZR1/2, which results in BZR1/2’s exclusion from the nucleus and their proteolytic degradation. In the absence of BR, BIN2 contains a phosphate group that is required for its activity.
Binding of BR to the BRI1 receptor domain outside the cell results in the activation of the kinase domain inside the cell. Activated BRI1 then phosphorylates a protein called BSK. Phosphorylated BSK binds yet another protein called BSU1, which is a protein phosphatase. Protein phosphatases are enzymes that remove phosphate groups from other proteins. BSK binding activates BSU1 which then catalyzes the removal of the phosphate on BIN2, thereby inactivating BIN2. Thus, BSU1 is a negative regulator of BIN2. When BIN2 is inactivated, that allows BZR1/2 to accumulate, to enter the nucleus, and to regulate the expression of genes that effect the BR response, including promoting plant growth and yield.
Effects of mutation on different steps
Mutations can affect regulatory pathways in multiple ways. Most mutations are loss-of-function mutations where the function of the gene or gene product is partly or completely impaired. Considering the BR signaling pathway, the phenotypes of mutation will differ depending on whether the mutation affects a positive or negative regulator of the pathway. A loss-of-function mutation in the bri1 gene will block the perception of the hormone and therefore the pathway will not be activated; a dwarf phenotype will result. On the other hand, a loss-of-function in the bin2 gene will release BZR1/2 from regulation. The result will be constant BZR1/2 activity regardless of whether BR hormone is present. This might be expected to generate giant plants but in fact the effects of deregulating hormone responses are more complex—let us just call it unregulated growth.
Mutations can also be gain-of-function, where the gene product assumes a form that is constantly in the active state. Such mutations are typically dominant and are much less common than loss-of-function. Again, the outcome of a gain-of-function mutation on the BR signaling pathway will vary depending on whether it affects a positive or negative regulator of the pathway. A gain-of-function mutation in a positive regulator would cause deregulated hormone responses, whereas such a mutation in a negative regulator would permanently shut down the pathway and cause dwarfism.
Complementary gene action and genetic epistasis in the context of pathways
Considering that biological processes are controlled by multiple genes acting together, several genetic principles make more sense. First, it becomes clear how mutations in different genes can produce the same phenotypic effects. Since multiple genes are often required for a single biological process, disruption of different genes in the process might have the same net effect. From this follows the concept of complementary gene action. Mutant plants with the same phenotype are crossed together and the offspring is normal looking. The mutations in the parent plants likely affected different genes in the same pathway. Finally, the concept of epistasis can be best understood in the context of pathways. Note that the term epistasis is used differently by different geneticists. In quantitative genetics, epistasis refers to any gene interaction. Here, the term epistasis refers to a specific type of gene interaction where the action of one gene is masked by the action of another gene. In other words, if two mutations are combined in an individual, only one of the phenotypes is apparent. The mutant whose phenotype is apparent is said to be epistatic to the one that is masked.
Complementary gene action in a regulatory pathway
The concept of complementary gene action is no different in a regulatory pathway than in a biosynthetic pathway. Mutations in different genes that have similar roles in the pathway (e.g., are both positive regulators of the pathway) would be expected to cause similar mutant phenotypes. If such mutants in different genes were crossed together, the F1 progeny would be normal because both genes would be heterozygous and therefore a functional copy of both factors would be present.
Epistasis in a regulatory pathway
Epistasis can be quite complex, especially in regulatory pathways. Unlike complementary gene action, epistasis is not restricted to recessive loss-of-function mutations. The situation is further complicated by the presence of positive and negative regulators in a pathway. In contrast to a biosynthetic pathway, the rule of thumb for a regulatory pathway is that the more “downstream” gene is epistatic to the more “upstream” one. Let’s turn again to the BR signaling pathway and look at a couple of examples to understand why.
First consider loss-of-function mutations in the bri1 and bin2 genes, which in single mutants cause a dwarf phenotype and an unregulated growth phenotype, respectively. If the two mutations are combined into a double mutant plant, what do we expect? The bri1 mutations disrupt the receptor function required to initiate signaling through the pathway. However, BIN2 functions to inhibit the activity of pathway by repressing the function of BZR1/2. If BIN2 is rendered non-functional, then BZR1/2 will be active and promote the growth response. This will be true even if previous steps are blocked by the bri1 mutation. So the double mutant would show the bin2 unregulated growth phenotype making bin2 epistatic to bri1.
What about a gain-of-function mutation in BZR1 or BZR2 combined with a loss of function mutation in bri1? A gain-of-function mutation would render BZR1/2 constitutively active, regardless of the activity state of BIN2, resulting in the unregulated growth phenotype. Since BIN2 activity would be irrelevant, upstream activities would also be irrelevant. Thus, the double mutant would show unregulated growth and BZR1/2 would be epistatic to bri1.
Lesson Summary
We saw that genes do not act alone but function in concert with other genes to perform various biological functions such as the biosynthesis of a compound or the regulation of a response to a stimulus. When genes control a sequential series of steps, that is called a pathway. But pathways typically branch and intersect with other pathways to form networks. Mutations affect the activity of pathways in various ways, depending on the nature of the mutation (loss- vs. gain-of-function) and the function of the gene product (positive vs. negative effector).
When mutations affect different steps in a pathway, they can often have a similar overall effect on the pathway and therefore lead to similar phenotypic consequences. Such mutations will complement one another even though their phenotypes are identical. It is also possible that mutations in different steps of a pathway can cause different phenotypes. When such mutations are combined into a double mutant individual, the phenotype of one is often masked by the phenotype of the other epistatic mutation. When pathways are linear, the “upstream” mutation is generally epistatic in biosynthetic pathways and in regulatory pathways it is generally the “downstream” mutation that is epistatic. In branched pathways or networks, it is more difficult to generalize.
Complementary gene action is important for geneticists trying to identify all the steps in a pathway. This is the basis of the classic complementation test used to determine if mutations are allelic or affect independent genes. Epistasis is also useful to geneticists who want to understand whether mutations with different phenotypes affect the same pathway, and if so, to determine the order of gene action.
When biotechnology is used to manipulate a trait, it is really the activity of a pathway or network that is being targeted. The same genetic principles outlined in this section are used to design biotechnological strategies to alter pathway activities so as to produce the desired outcome on the trait of interest.
Activity 1
To answer the following questions, refer to the pathways illustrated in Figure 3. They are labeled slightly differently but a1 encodes DFR, which catalyzes the production of leucoanthocyanidins and a2 encodes AS or LDOX/ANS, which subsequently catalyzes the production of anthocyanidins. Assume that normal, wild-type maize kernels are the deep purple/black color as in Figure 3.
1. What would you predict happens to the levels of leucoanthocyanidins in an a2 mutant compared to normal?
1. no change
2. increase
3. decrease
4. can’t predict
Show Answer
Answer: b or d.
The expectation would be b, that leucoanthocyanidins would increase. Because the pathway is blocked at the subsequent step, leucoanthocyanidins cannot be converted to anthocyanidins. But since all the preceding enzymes remain functional, those reactions would be expected to continue resulting in the accumulation of leucoanthocyanidins.
d is also a correct response. Biosynthetic pathways do not always behave as expected. For example, there could be feedback inhibition of one or more enzymes by intermediate compounds. Bottom line, we won’t know for certain until the levels are measured experimentally.
1. What would you predict happens to the levels of pelargonidin in an pr1 mutant compared to normal?
1. no change
2. increase
3. decrease
4. can’t predict
Show Answer
Answer: b or d.
Again, the expectation would be b, that pelargonidin would increase. Normally, some fraction of the intermediate compounds are channeled through the pathway branch leading to cyaniding. Because the pathway leading to cyanidin is blocked it might be expected that all the intermediates would be converted to pelargonidin, causing an increased accumulation.
d is also a correct response. As discussed in problem 1, biosynthetic pathways do not always behave as expected. Feedback inhibition could be a factor or it may be possible that the enzymatic capacity of the pelargonidin branch of the pathway is already saturated. To increase flux through the pathway, it could be required to increase the levels or activities of one or more rate-limiting enzymes. Again, we won’t know for certain until the levels are measured experimentally.
Activity 2
Predict the outcomes of the following mutations on the signaling output, or plant response for the BR signaling pathway.
1. A gain-of-function mutation in the BZR2 gene
1. increased output
2. no change
3. decreased output
4. can’t tell
Show Answer
Answer: a. Since BZR2 is a positive regulator of BR signaling, increased output from the pathway would be expected from a gain-of-function mutation.
1. A loss-of-function mutation in the bsu1 gene. (Study the pathway and think about this one!)
1. increased output
2. no change
3. decreased output
4. can’t tell
Show Answer
Answer: c. The output from the pathway would decrease causing decreased plant growth. Even though BSU1 is a negative regulator, its target is BIN2 which is another negative regulator. Without BSU1 function, BIN2 will always be in the ON state and inhibit the activities of BZR1/2, therefore decreasing the response to BRs.
Activity 3
1. Predict the phenotype of a BSK gain-of-function.
1. unregulated growth
2. no effect
3. dwarf
4. can’t predict
Show Answer
Answer: a. BSK is a positive regulator of the pathway that functions to activate BSU1. Therefore a gain-of-function would cause it to positively activate the pathway, even in the absence of BR hormone, causing unregulated growth.
2. Now predict the phenotype of a double mutant between the BSK gain-of-function and a BIN2 gain-of-function.
1. unregulated growth
2. no effect
3. dwarf
4. can’t predict
Show Answer
Answer: c. The gain-of-function BIN2 mutation locks BIN2 in the ON state where it would function to inhibit the activity of BZR1/2. It would be impervious to regulation by the upstream BSU1, so even if BSU1 were continually active due to the gain-of-function BSK, the BIN2 mutant would inhibit growth and therefore be epistatic.
Activity 4
Now you try a couple. Fill in each of the following tables to determine whether the hypothetical Activity will be activated or repressed in response to each Stimulus. | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.10%3A_Genetic_Pathways.txt |
Learning Objectives
• Understand the importance of recombinant DNA technology.
• Learn isolation of DNA and its separation on an agarose gel.
• Understand restriction and ligase enzymes and their application in gene cloning.
• Understand vectors and their application in gene cloning and expression.
• Understand polymerase chain termination reaction (PCR).
Introduction
Recombinant DNA (rDNA) technology has resulted in breakthroughs in crop and animal biotechnology. The power of rDNA technology comes from our ability to study and modify gene function by manipulating genes and transform them into cells of plant and animals. To arrive at this several tools of molecular biology are used including, DNA isolation and analysis, molecular cloning, quantification of gene expression, determination of gene copy number, transformation of the appropriate host for replication or transfer into crop plants and analyses of transgenic plants.
Definition and background
Recombinant rDNA technology involves procedures for analyzing or combining DNA fragments from one or several organisms (Figure 1) including the introduction of the rDNA molecule into a cell for its replication, or integration into the genome of the target cell.
Advances in molecular biology in the early 1970s, including the success in creating, and transferring DNA molecules into cells, revolutionized both science and industry. The first genetically modified organisms were bacteria that made simple proteins of pharmaceutical interest, for example, insulin. As the technologies improved, other organisms including plants became amenable for improvement by rDNA technology. Table 1 provides important milestones in the development and application of rDNA technology.
Table 1.Events related to the development and application of recombinant DNA technology.
Event
Year
Mendel’s experiments published
1866
DNA discovered in cell
1869
Mutation of genes by x-rays
1927
One gene-one enzyme hypothesis
1941
DNA is identified as the genetic material
1944
Structure of DNA determined
1953
Ribosomes synthesize protein
1954
Function of mRNA proposed
1961
Genetic code determined
1961-64
Isolation of a restriction enzyme
1970
Recombinant DNA techniques developed
Early 1970s
Isolation of a single copy gene from higher eukaryote
1977
Rapid method of DNA sequencing developed
1977
Plant transformation
1983
Field testing of transformed plants
ca. 1986
Release of engineered plants to general public in the US
1995-96
Transformation of cells with rDNA produces organisms called bioengineered or genetically modified organisms (GMOs). The GMOs contain new traits from another organism. The first GMOs were Escherichia coli cells that were transformed with genes from human to produce various proteins for pharmaceutical purposes.
Isolation of DNA, restriction digestions and its separation on an agarose gel:
Preparation of DNA
For recombinant DNA procedures to work, a pure DNA sample must be obtained. The challenge is that plant and animal cells produce numerous other compounds that often act as contaminants and may inhibit cloning or sequencing of the DNA. Also, tissues and organs from the same plant, or different plants often contain different composition of metabolites, for example, proteins, lipids, and carbohydrates. These compounds must be separated from the DNA during isolation. To achieve this, scientists take advantage of the chemical and physical properties of different molecules inside the cell. For example, DNA is negatively charged, making it soluble in aqueous solution. However, the polar sugar phosphate groups of the DNA are repelled by non-polar solutions. Therefore, the final step in many DNA purifications protocols involves precipitation using alcohol. Other compounds, for example, proteins can also be easily separated from DNA by altering the concentration of salt in the extraction buffer.
Digestion of DNA with restriction endonucleases
Restriction endonucleases are a group of enzymes derived (primarily) from bacteria. Although there are several different types of restriction enzymes, those most useful for rDNA technology recognize specific short sequences in DNA and cleave the DNA at that site to produce cohesive (sticky) or blunt-ended fragments (Figure 2).
More than 500 different restriction enzymes have been identified and can be purchased commercially. Thus, one may ask the question, how often does a restriction enzyme cut within a genomic sequence? It is not possible to give an exact answer for this question. However, let us assume that there are 4 bases on any strand of DNA. This means the probability of detecting an A (adenine) at a particular location is 1/4. Now, since most restriction enzymes recognize specific sequences of 6 bases long, the probability of finding such a site is (1/4)6 = 1 site in every 4,096 base pairs (bp). Assuming you have isolated genomic DNA from maize, and you want to digest it with a restriction enzyme that cuts every 4,096 (4,100) bp, how many fragments will you obtain? To answer this question, you need to have an idea of the size of the genome of the plant you are working with. The approximate size of the maize genome is 2,500,000,000 bp. Thus, using an enzyme that cuts every 4,100 nucleotides one would expect to obtain 2,500,000,000 bp/4,100 bp, approximately 610, 000 fragments. Note that nucleotide distribution is not always random and thus frequency may be different for given DNA. Also, methylation of specific bases in genomic DNA can prevent cleavage at some site.
Separation of digested genomic DNA on agarose gel
The only physical features of nucleic acid fragments that are routinely used for their characterization are their size and nucleotide sequence. Molecular weight of DNA is most conveniently evaluated by electrophoresis in agarose gels. Agarose forms a gel by hydrogen bonding when cooled from the melted state. This gel, interwoven network of agarose chains, interferes with the movement of DNA through the gel (see Lesson on PCR and Gel Electrophoresis). Pore size, which affects rate of movement of DNA fragments of a given size, depends on the concentration of agarose. The gel is submerged in electrolyte solution; sample is loaded into wells on one end and current is applied to facilitate movement of DNA fragments. Since DNA is negatively charged it will migrate in the electrical field. Fragments separate according to size. The distance of the migration in each time is proportional to 1/log MW. Following gel electrophoresis DNA can be visualized by staining with ethidium bromide (see Lesson on PCR and Gel Electrophoresis) or other DNA stains.
Tools for gene (DNA) cloning
Polymerase Chain Reaction (PCR)
Polymerase chain reaction (PCR) is a method by which millions of copies of a DNA fragment are produced in a test tube in a matter of minutes or a few hours. The basic steps in PCR reaction were discussed in the Lesson on PCR and Gel Electrophoresis.
Once the sequence of a particular gene is known, it becomes possible to use PCR to isolate that gene from any DNA sample. Recall in Lesson on Gene Transcription you learned that at the genomic level a gene is made up of regulatory sequences, coding, and non-coding sequences. Thus, if the goal is to use PCR to isolate both regulatory and non-regulatory sequences of a gene, the approach would be to use DNA as the starting material.
In cloning genes by PCR, restriction enzyme sites are added at the 5′-end of the primers to facilitate cloning of PCR fragments. A few additional nucleotides (~6 nucleotides) added at the 5′-end of the restriction sites to facilitate restriction digestion of PCR products prior to cloning in a plasmid vector. Alternatively, PCR products may be cloned directly into a T-vector without restriction digestions in E. coli (read more about Promega cloning vector systems). After cloning into E. coli, the fragment is analyzed by sequencing and then sub-cloned into suitable vectors for expression studies.
Like all other biochemical processes, DNA synthesis by PCR is not a perfect process, and occasionally the polymerase enzyme will add an incorrect base to the growing DNA strand. In the context of DNA replication in a cell, the errors are corrected by the DNA polymerase, this is called “proofreading”. Commercially available polymerases may or may not have proofreading capability.
Another important consideration in PCR analysis is contamination.
Minor contamination of the starting material can have serious consequences. Recall that minute amounts of starting DNA can be amplified to millions of copies through PCR. If one inadvertently (or carelessly) mixed DNA from two different sources, the results will be confounding making it impossible to distinguish lines and may cost a laboratory time and money. Ensure that proper procedures are followed in preparing PCR assays. One common source of contamination in plant biology is the products from previous amplification processes. A completed PCR reaction will contain millions of copies of amplified fragments so that even a minute droplet or aerosol from a pipette tip will contain an enormous number of amplifiable molecules. It is always essential to run negative controls which will reveal the presence of contaminating DNA in your PCR assays.
Cloning vector definition and requirements
A cloning vector is a specialized DNA sequence that can enter a living cell and provide means for detection of its presence to a researcher by conferring a selectable property on the host cell (e.g., resistance to antibiotics), and possess means for self-replication. A vector must also possess easily distinguishable physical traits, such as size, or shape, to allow purification away from the host cell’s genome.
Ligase enzyme and gene cloning
Cutting and joining together of vector and DNA fragments from different origins results in rDNA. Recall that restriction endonucleases are used to cut DNA. To join DNA molecules together, an enzyme called DNA ligase is used. The enzyme DNA ligase is used to seal together restriction fragments by forming new phosphodiester bonds. The ligated vector and DNA fragment can now be transformed into a host cell for replication and expression.
The transformation of E. coli takes several steps. First, a gene of interest is inserted into a plasmid that contains a selectable marker usually encoding for resistance to an antibiotic. Second, the plasmid construct containing the gene of interest is transformed into bacterial cells by briefly exposing the mixture of ligated plasmid-DNA fragment (rDNA molecule) and bacterial cells to cold (0oC) and heat (37-42oC). The next step is to grow the transformed cells on selection media containing an antibiotic. Only the cells that have been transformed with the plasmid containing the gene of interest and the marker for resistance to the antibiotic will survive. In addition to using an antibiotic, plasmid vector systems that contain the lacZ gene encoding β-galactosidase allow for easier selection of positive colonies that may harbor the rDNA molecule of interest.
Types of cloning vectors
An example of a cloning vector is a plasmid (Figure 4), defined as an autonomously replicating extra chromosomal circular DNA which is faithfully passed on to progeny. Plasmids are double stranded circular DNA and range in size from about 1 kb – 200 kb. The most useful for cloning are 2 – 10 kb because smaller plasmids are easier to manipulate and usually produce higher copy numbers when grown in bacterial host cells.
Generally, no more than 10 kb is cloned into plasmids. For cloning large DNA fragments with high efficiency, a vector called bacteriophage lambda is used. Large chromosomal DNA fragments close to 23 kb are stable when introduced into a lambda phage vector.
mRNA as starting material for gene cloning
Recall in the Lesson on Transcription you learned about synthesis of RNA from DNA through the process of transcription. Transcription is an important step in gene expression. The mRNA produced can be isolated and “copied” back to DNA by a process called reverse-transcription (Figure 5). The first step of reverse transcription mimics a strategy used by retroviruses (e.g., HIV) that have RNA genomes. As part of their gene-transmission package, retroviruses also contain an enzyme called RNA-dependent DNA Polymerases, commonly referred to as reverse transcriptase. After infecting a host cell the retrovirus uses its reverse transcriptase to copy its single stranded RNA genome into a strand of complementary DNA (cDNA). The reverse transcriptase then synthesizes the second DNA strand from the first strand to make a double stranded-DNA copy which integrates into the host genome.
If only a gene’s coding sequence is required, isolating the gene from cDNAs would be the strategy. It is important that cDNAs are synthesized from tissues expressing the gene of interest. Thus, prior knowledge of where the gene is functional is important in constructing the cDNA molecules for cloning the gene of interest.
The cDNAs produced in vitro (Figure 4) can be used for PCR analysis, similar to chromosomal DNA. The combination of reverse-transcription and PCR (RT-PCR) is a valuable tool in gene cloning and quantification of mRNA.
Lesson Summary
Recombinant DNA technology has contributed significantly to development of agricultural biotechnology. Transformation of cells with rDNA produces organisms called bioengineered or genetically modified organisms. The tools for plant rDNA technology include, vectors, restriction enzyme, ligation enzymes, bacterial hosts, methods to isolate and multiply nucleic acids, methods to quantify nucleic acids, Agrobacterium as a vector to insert foreign DNA into plants.
Lesson Activities
Activity 1
Below is a hypothetical DNA fragment containing restriction sites for EcoRI and BamHI.
1. How many fragments will be produced after cutting the DNA with BamHI? What size will these fragments be in bp?
2. How many fragments will be produced after cutting the DNA with both BamHI and EcoRI at the same time? What size will these fragments be in bp?
3. How many fragments will be produced after cutting the DNA with EcoRI? What size will these fragments be in bp?
Activity 2
1. You need to clone an EcoRI fragment in the EcoRI site of an expression plasmid vector in E coli. The fragment is 1,809 bp long. There is a BamHI site in the fragment at 1,204 bp. In the plasmid vector, there is a single BamHI site at the promoter region for expression of the gene in E. coli. The BamHI site is 100 bp upstream of EcoRI cloning site.
Following cloning the EcoRI fragment you will get two classes of clones, only one of which will produce the protein. Describe the approach you will take to identify the correct class of clones for expression of the gene in E. coli. | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.11%3A_Recombinant_DNA_Technology.txt |
Learning Objectives
• Define genetic engineering.
• List and briefly explain the five basic steps in genetic engineering. Describe why each is necessary.
• Identify the fundamental differences between genetically engineered crops and non-genetically engineered crops.
• Explain the limitations to traditional breeding that are overcome by genetic engineering.
• Identify the approximate length of time required to obtain a marketable transgenic crop line (complete the entire crop genetic engineering process).
Introduction
The production of genetically engineered plants became possible after Bob Fraley and others succeeded to use Agrobacterium tumefaciens to transform plant cells with recombinant DNA in the early 1980s (Vasil, 2008a). Since this breakthrough in plant biotechnology, GM crops are now routinely developed and grown in many parts of the globe. Current statistics on adoption of genetically engineered crops in the U.S. can be found on the USDA Economic Research Service’s website.
Genetic engineering has been used successfully to develop novel genes of economic importance that can be used to improve the genetics of crop plants. Genetic engineering is the targeted addition of a foreign gene or genes into the genome of an organism. The genes may be isolated from one organism and transferred to another or may be genes of one species that are modified and reinserted into the same species. The new genes, commonly referred to as transgenes, are inserted into a plant by a process called transformation. The inserted gene holds information that will give the organism a trait (Figure 1).
Crop genetic improvement (plant breeding) is an important tool but has limitations. First, in conventional terms, genetic improvement can only be done between two plants that can sexually mate with each other. This limits the new traits that can be added to those that already exist in that species. Second, when plants are mated, (crossed), many traits are transferred along with the trait of interest including traits with undesirable effects on yield potential.
Genetic engineering, on the other hand, is not bound by these limitations. It physically removes the DNA from one organism and transfers the gene(s) for one or a few traits into another. Since crossing is not necessary, the ‘sexual’ barrier between species is overcome. Therefore, traits from any living organism can be transferred into a plant. This method is more specific in that a single trait can be added to a plant.
The overall process of genetic engineering. A basic explanation of the five steps for genetically engineering a crop is provided. The five steps are:
1. Locating an organism with a specific trait and extracting its DNA.
2. Cloning a gene that controls the trait.
3. Designing a gene to express in a specific way.
4. Transformation, inserting the gene into the cells of a crop plant.
5. Cross the transgene into an elite background.
Step 1: DNA Extraction
The process of genetic engineering requires the successful completion of a series of five steps and discoveries. To better understand each of these, the development of Bt maize will be used as an example.
Before the genetic engineering process can begin, a living organism that exhibits the desired trait must be discovered. The trait for Bt maize (resistance to European corn borer) was discovered around 100 years ago. Silkworm farmers in the Orient had noticed that populations of silkworms were dying. Scientists discovered that a naturally occurring soil bacteria was causing the silkworm deaths. These soil bacteria, called Bacillus thuringiensis, or Bt for short, produced a protein that was toxic to silkworms, the Bt protein.
Although the scientists did not know it, they had made one of the first discoveries necessary in the process of making Bt corn. The same Bt protein found to be toxic to silkworms is also toxic to European corn borer because both insects belong to the Lepidoptera order. The production of the Bt protein in the bacteria is controlled by the bacteria’s genes.
To be able to work with the gene responsible for making the Bt toxin, scientists must extract DNA from the Bt bacteria (Figure 2). This is accomplished by taking a sample of bacteria containing the gene of interest and taking it through a series of steps that separate the DNA from the other parts of a cell.
Step 2: Gene Cloning
The second step of the genetic engineering process is gene cloning. During DNA extraction, all the DNA from the organism is extracted at once. This means the sample of DNA extracted from the Bacillus thuringiensis bacteria will contain the gene for the Bt protein, but also all the other bacterium’s genes. Scientists use gene cloning to separate the single gene of interest from the rest of the DNA extracted (Figure 2).
The next stages of genetic engineering will involve further study and experimentation with this gene. To do that, a scientist needs to have thousands of exact copies of it. This copying is also done during the gene cloning step.
Step 3: Gene Design
Gene design relies upon another major discovery. This was the ‘One gene One enzyme’ Theory first proposed by George W. Beadle and Edward L. Tatum in the 1940’s. Discoveries made during their research laid the groundwork for the theory that a single gene stores the information that directs the cell in how to produce a single enzyme (protein). Therefore, there is a single gene that controls the production of the Bt protein. It is called the Bt gene.
Once a gene has been cloned (Figure 2), genetic engineers begin the third step, designing the gene to work once inside a different organism. This is done in a test tube by cutting the gene apart with restriction enzymes and replacing certain regions (Figure 3).
Scientists replaced the bacterial gene promoter with promoters turn on the Bt gene in selected parts of the plant or promoters that can always turn on the Bt gene in all tissues. As a result, the first Bt gene released was designed to produce a level of Bt protein lethal to European corn borer and to only produce the Bt protein in green tissues of the corn plant, (stems, leaves, etc.). Later, Bt genes were designed to produce the lethal level of protein in all tissues of a corn plant, (leaves, stems, tassel, ear, roots, etc.).
Plant transformation and tissue culture
The process of transformation involves the insertion of the desired transgene construct (Figure 5) into cells of the recipient plant species. In this process, scientists isolate tissue or cells from the cultivar they wish to transform and use one of several methods to insert the transgene into the tissue or cells. The transgene construct contains the following key features.
1. A promoter that acts to turn the gene on and off in the cell. The CaMV 35s promoter from the cauliflower mosaic virus (CaMV) is commonly used in genetic engineering. Other types of promoters, such as, the nopaline synthase promoter (NOS-Pro) also may be used to express transgenes in plant tissues.
2. A selectable marker that is used to select cells that successfully obtained the construct during the transformation process. In figure 4, the selectable marker in the construct is NPT II (Kanr) that controls resistance to the antibiotic kanamycin. The cells of the plant used for transformation will be grown on a media containing the antibiotic. Other selectable markers that have been used successfully in plants include genes controlling herbicide resistance.
3. A terminator sequence, such as the nopaline synthase (NOS) is included to mark the end of the transgene sequence for proper expression in plant cells.
Two commonly used transformation methods include Agrobacterium tumefaciens-mediated transformation and biolistics transformation (aka gene gun), commonly referred to as particle bombardment (Figure 5). The biolistics method involves the use of high pressure to propel tungsten or gold beads coated with DNA of the gene construct into plant cells.
Agrobacterium-mediated plant transformation
Crown galls are tumors of plants that arise at the site of infection by some species of the Agrobacterium. Agrobacteria do not enter the plant cells but transfer a DNA segment called T-DNA from their circular extra chromosomal tumor-inducing (Ti) plasmid into the genome of the host cells. Ti plasmids are maintained in Agrobacteria because a part of their T-DNA contains genes that encode unusual amino acids used by Agrobacterium. The T-DNA also encodes genes that affect host plant hormone physiology resulting in induced growth of the infected cells and tumor formation. Scientists took advantage of Agrobacterium’s ability to stably integrate its T-DNA into the plant genome for introducing rDNA into plant cells. They first removed the genes that cause tumor or crown gall disease in plants from the T-DNA and engineered the plasmid for replication in both Escherichia coli and Agrobacterium cells. The initial replication of the construct in E. coli is useful for verifying the presence of the cloned gene and increasing the quantity of construct DNA for subsequent uses, including sequencing and transformation into Agrobacterium.
The steps in Agrobacterium-mediated transformation of plants are described in Figure 6.
At present, very few host cells receive the construct during the transformation process. Each random insertion of the construct into the genome of plant cells is referred as an event. Useful events are rare because of the random nature of the transformation process. Selectable markers are very important because they allow the identification of the rare events (Figure 7). Scientists must screen many potential transformants to identify events that are useful for breeding.
From there, the new DNA may or may not be successfully inserted into a chromosome. The cells that do receive the new gene are called transgenic and are selected from those that are not transgenic (Figure 7). Many types of plant cells are totipotent meaning a single plant cell can develop into an entire plant. Therefore, each transgenic cell can then develop into an entire plant which has the transgene in every cell. The transgenic plants are grown to maturity in greenhouses and the seed they produce, which has inherited the transgene, is collected. The genetic engineer’s job is now complete. He/she will hand the transgenic seeds over to a plant breeder who is responsible for the final step.
Inheritance of a transgene in plants
Transformation is successful when a transgene is incorporated into one of the chromosomes. The cells that have only one copy of the transgene in their genomes are said to be hemizygous (hemi = half, zygous = zygote). Because the segregation in the progeny of a hemizygous plant is the same as for a heterozygous plant, the term heterozygous will be used in this course when referring to a plant that is not homozygous for the transgene. The trait will segregate in the progeny in the same manner as any other gene in the plant as illustrated below (Figure 8).
Step 5: Backcross Breeding
The fifth and final part of producing a genetically engineered crop is backcross breeding (Figure 9). Transgenic plants are crossed with elite breeding lines using traditional plant breeding methods to combine the desired traits of elite parents and the transgene into a single line. The offspring are repeatedly crossed back to the elite line to obtain a high-yielding transgenic line. The result will be a plant with a yield potential close to current hybrids that expresses the trait encoded by the new transgene.
The Process of Plant Genetic Engineering
The entire genetic engineering process is basically the same for any plant. The length of time required to complete all five steps from start to finish varies depending upon the gene, crop species, and available resources. It can take anywhere from 6-15+ years before a new transgenic hybrid is ready for release to be grown in production fields.
The tissue culture process of regenerating transgenic plants from callus may result in genetic variation that is not associated with the transgene. Also, the parent line used for transformation commonly is selected for the frequency with which useful events can be obtained and not its agronomic performance. Therefore, transgenes are incorporated into commercial cultivars by conventional breeding procedures, such as backcrossing.
Key Takeaways
Genetic engineering is the directed addition of foreign DNA (genes) into an organism.
Five basic steps in crop genetic engineering:
1. DNA extraction – DNA is extracted from an organism known to have the desired trait.
2. Gene cloning – The gene of interest is located and copied.
3. Gene modification – The gene is modified to express in a desired way by altering and replacing gene regions.
4. Transformation – The gene(s) are delivered into tissue culture cells, using one of several methods, where hopefully they will land in the nucleus and insert into a chromosome.
5. Backcross breeding – Transgenic lines are crossed with elite lines to make highyielding transgenic lines. | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.12%3A_Genetic_Engineering.txt |
Learning Objectives
• Outline the experimental approach Mendel used to propose the idea that genes exist, control traits, and are inherited in predictable ways.
• Compare the methods used by Mendel and Punnett to predict trait inheritance.
Introduction
In plant and animal genetics research, the decisions a scientist will make are based on a high level of confidence in the predictable inheritance of the genes that control the trait being studied. This confidence comes from a past discovery by a biologist named Gregor Mendel, who explained the inheritance of trait variation using the idea of monogenic traits.
Monogenic characters are controlled by the following biological principles:
• Living things have genes in their cells that encode the information to control a single trait. These genes are stable and passed on from cell to cell without changing.
• The genes are in pairs in somatic cells. When these cells divide to form gametes, the pair of genes is divided. One gene from the pair goes into a gamete.
• Male gametes (pollen) combine with female gametes (eggs) in the wheat flower pistil and fuse to form the next generation (zygote). Gamete union is random.
• The zygote, again, has two copies of each gene. As the zygote grows into a multicellular seed and the seed grows into a plant, the same two gene copies are found in every cell.
Let’s take a short genetics history lesson to understand their confidence.
Mendel’s Peas
In the mid 1800’s, an Austrian monk named Gregor Mendel (Figure 1) decided he should try to understand how inherited traits are controlled. He needed a model organism he could work with in his research facility, a small garden in the monastery, and a research plan. His plan was designed to test a hypothesis for the inheritance of trait variation.
Since Mendel could obtain different varieties of peas that differed in easy to observe traits such as flower color, seed color and seed shape, and he could grow these peas in his garden, he chose peas as the model organism for conducting his inheritance control study. A model is easy to work with and often what you learn from the model you can apply to other organisms.
The Hypothesis
While many biologists were interested in trait inheritance, at the time Mendel conducted his experiments none of the biologists had published evidence that inheritance could be predicted. Mendel made this bold statement. His hypothesis was that he could observe “mathematical” regularities in the appearance of a trait that was passed on from parents to their offspring. Mendel had the idea that mathematical regularities could be observed and could be used to explain the biology of inheritance!
The Plan
Mendel’s experimental plan was designed to test the hypothesis. He identified true breeding lines of peas by allowing them to self pollinate (which we will refer to as “selfing”) and examining their offspring. Pea plants have flowers that contain both male and female reproductive parts; if a pea flower is left undisturbed, the male and female gametes from the same flower will combine to produce seeds, the next generation. If the pea always made offspring like itself, Mendel had his true breeding line. He then made planned crosses between lines that differed by just one trait (monohybrid crosses). The controlled monohybrid cross was the first step in his experiment that allowed him to look for mathematical regularities in the data for three generations. Table 1 below shows the data from a series of these monohybrid cross experiments.
The Analysis
By summarizing his data in a single table, Mendel could look for those hypothesized math regularities. A regularity is a repeated observation.
Table 1. Mendel result from crossing peas.
Character
Cross and Phenotypes
F1
F2
F2 Ratio
Seed form
Round X Wrinkled
All round
5474 Round 1850 Wrinkled
2.96 to 1
Cotyledon color
Yellow X Green
All yellow
6022 Yellow 2001 Green
3.01 to 1
Seed coat color*
Gray X White
All gray
705 Gray 224 White
3.15 to 1
Pod form
Inflated X Constricted
All inflated
882 Inflated 299 Constricted
2.95 to 1
Pod color
Green X Yellow
All green
428 Green 152 Yellow
2.82 to 1
Flower position
Axial X Terminal
All axial
651 Axial 207 Terminal
3.14 to 1
Stem length
Tall X Short
All tall
787 Tall 277 Short
2.84 to 1
*Gray seed coat also had purple flowers; White seed coat had white flowers.
Table 1 demonstrates that Mendel was serious about the math. He generated large numbers of offspring that allowed him to observe mathematical ratios. From his table of data, we can see mathematical patterns appear with every monohybrid cross he made.
• F1: All the plants had the same phenotype as one of the parents.
• F2: Both phenotypes are present, the phenotype that was not expressed in the F1 appears again in the F2 but is always the least frequently produced. The average ratio is about 3:1 for the two phenotypes.
What was striking to Mendel was that every character in his study exhibited the same kind of mathematical pattern. This suggested that the same fundamental processes inside the plant’s reproductive cells were at work controlling the inheritance of each trait.
Now Mendel had the task of providing a description of the fundamental biology process controlling each of these traits. He needed to come up with ideas that no one had yet proposed to explain biology.
New Idea #1:
The traits expressed in the pea plant were controlled by some kind of particle. These hereditary particles are stable and passed on intact from parent to offspring through the sex cells. (NOTE: Sex cells or gametes were not a new idea, Mendel was aware that biologists knew sexually reproducing plants and animals needed to make gametes.) We now call these particulate factors genes and will use that term in the rest of this reading.
New Idea #2:
Genes are stable, and genes can have alternative versions (alleles).
New Idea #3:
Genes are in pairs in somatic cells and these paired genes separate during gamete formation. Each gamete will have one gene from the pair of genes. The segregating of the paired genes from the somatic cells of the parent into gametes is random. Because segregation is random, a parent that has two different alleles for a gene pair will make two kinds of gametes and makes these gametes at equal frequencies.
From Mendel’s ideas, we can see that in a situation in which there was a normal version of a gene (we can call it the R gene) and an alternate version (r), the plant could produce gametes with just the R gene or just the r gene.
New Idea #4:
Plant flowers are designed to allow male gametes (pollen) to combine randomly with the female gametes (egg). When the gametes randomly come together, they bring the genes they carry to the same zygote. This means plants could have the genotype RR, Rr, or rr in families that have both the R and r alleles.
New Idea #5:
Mendel proposed that the genes controlling a trait not only paired in somatic cells, they also interacted in controlling the traits of the plants. For the traits in his experiment, he proposed that one allele interacted with the other in a dominant fashion. That means a plant that is the genotype RR would have the same phenotype as an Rr plant. The R allele is dominant to the r allele.
Ideas and Data advance science
Those were Mendel’s new ideas; he used them to make sense of his experiment data and observations. Let’s think like Mendel and apply those ideas.
All the F1 were the same
Mendel’s new ideas could explain this observation. Since his parents were true breeding, he was always making a cross between homozygous parents. Homo means the same, so the parents had two copies of the same version of the gene.
Crossing RR X rr plants to produce Rr
Since the R is dominant to r, then the Rr offspring (named the F1) look the same (have the same phenotype) as the RR parent. Therefore, only one phenotype is observed in the F1. But the F1 genotype is different from either parent. It is heterozygous (two different alleles).
The F2: both traits appear in about a 3:1 ratio
Mendel could explain the reappearance of the recessive trait and the ratio by combining the idea of genes with the idea of random segregation. Mendel used simple algebra to explain this result.
First, he wrote out a mathematical expression to account for the gametes made in the male part of the F1 flower or in the female part.
½ R + ½ r = all the gametes made (Figure 2).
Next, he reasoned that if pollen randomly united with the egg to combine the genes in the gametes, then algebra could be used to predict the result by multiplying the gamete expressions.
(½ R + ½ r) X (½ R + ½ r) = all the F2 offspring made.
If we do the multiplication above, we get …
¼ RR + ¼ Rr + ¼ Rr + ¼ rr = ¼ RR + ½ Rr + ¼ rr = predicted fractions of F2 genotypes.
WAIT!
If this math is causing your brain to lose focus, you might be experiencing what Mendel’s contemporaries experienced when they read his published research paper. While many biologists were motivated to understand how the variation among animals and plants was controlled and inherited, it took biologists 30 years to recognize that Mendel’s new ideas to explain inheritance of traits in peas could be applied to inheritance of traits in other living organisms.
One possible explanation for this 30-year delay in appreciation is that it was difficult for biologists to understand how math could explain biology. One biologist that did understand what Mendel was describing was Punnett. Punnett decided to convert Mendel’s algebra into a more graphic representation of the process of gamete segregation and random union.
The Punnett Square
Math: (¼ RR + ½ Rr + ¼ rr).
Punnett designated the gametes made in the male and female parents with single letters (Figure 3). The diagram shows that when the gametes combine, the offspring (inside the squares) again have the genes in pairs in their cells. Accounting for the random union of gametes is accomplished with the four squares in the diagram. Two squares give the same Rr result, one the RR genotype and one rr. Both the algebra and diagram approaches provide the same prediction. Crossing an Rr with an Rr will produce three genotypes, RR, Rr and rr. They will be produced in a ratio based on the principle of segregation.
The genes controlling the monogenic traits behaved in predictable ways
Punnett’s diagram clarified for many biologists what Mendel was telling them in his published article. This was a challenging idea to understand because he was asking biologists to use something they could not see (genes) and explain something they could see (traits in peas or some other living organism).
Because Mendel recognized he was proposing a very different idea with the segregation principle, he was likely motivated to share the most convincing evidence possible. Mendel conducted additional experiments. One experiment was to test the hypothesis that there were two different kinds of F2 which expressed the dominant trait, and these two types were being made by the F1 in predictable fractions. How would Mendel show that F2 which had the same phenotype did not always have the same genotype?
Mendel tested the breeding behavior of the F2. Mendel harvested all the selfed seed produced by his F2 and grew progeny rows of F3. His segregation principle predicted that of the dominant F2, there should be two that are heterozygous for every one homozygote made (on average). The results of this experiment are summarized in Table 2. Did Mendel’s data support the hypothesis?
Table 2. Selfing dominant F2 to produce F3
F2 type
Mixed rows
True breeding
Ratio
Round seed
372
193
1.93 to 1
Yellow cotyledon
353
166
2.13 to 1
Gray seed coat
64
36
1.78 to 1
Inflated pod
71
29
2.45 to 1
Green pod
60
40
1.50 to 1
Axial flower
67
33
2.03 to 1
Tall plant
72
28
2.57 to 1
Average ratio heterozygote F2 to homozygote F2 was 2.06 to 1.
The data show that, if we select a sample of F2 with the dominant trait (Round seed or Yellow cotyledon), the principle of segregation predicts that there should be 2 heterozygotes for every 1 homozygotes.
Mendel’s data from rows of F3 that all came from F2 with the dominant trait supported his hypothesis. There were always two kinds of rows (true breeding and mixed) and the rows were in a 2:1 ratio. This fits with the principle of segregation.
By publishing these results in a scientific journal, Mendel allowed other scientists to learn from his work. This story reveals the real power of publishing research in the “permanent” scientific literature. The power of publication does not mean you were right with your science. The real power is that other scientists can find your paper, read it, think about your ideas, and then test them. In Mendel’s case, he was already dead when his fellow biologists discovered that his new ideas to explain the biology of peas were not only correct, but universal in their application.
Mendel’s Dihybrid Cross Experiments
Proper credit must be given to the idea of independent assortment. Gregor Mendel was the first to put this idea down on paper based on what he observed with his pea experiments. Furthermore, Mendel performed additional experiments to back up his ideas. Let’s examine his experiments with peas from the late 1800’s.
The outline below describes Mendel’s dihybrid cross experiments. The pattern observed in the results should look familiar!
The Experiment
• Parents: round seeds, yellow seeds (RRYY) x wrinkled seeds, green seeds (rryy).
• F1: All round and yellow seeds (RrYy).
• Selfing: F1 (RrYy x RrYy):
Mendel explained his results as follows:
The F1 plants have the genotype RrYy and can make four kinds of gametes RY, Ry, rY and ry.
Table 3. Gametes are in equal frequencies in the male and female parts of the plant.
Possible gametes
Male gametes
¼ RY + ¼ Ry + ¼ rY + ¼ ry
Female gametes
¼ RY + ¼ Ry + ¼ rY + ¼ ry
Table 4. Punnett square demonstrates the possible genotypes produced when the RrYy dihybrid is selfed.
RY
Ry
rY
ry
RY
RRYY
RRYy
RyYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
rY
RrYY
RrYy
RrYY
RrYy
ry
RrYy
Rryy
rrYy
rryy
Note that with both the Mendel algebra and Punnett square, the RRYY genotype occurs one time and the RrYy genotype occurs four times (Table 4). Mendel’s algebra and Punnett’s squares can be summarized to give the same results.
Selfing the F2 to produce F3
Table 5. Phenotype classes and their fractions in F2.
Phenotype Number Fraction
Round, yellow 315 9/16
Round, green 108 3/16
Wrinkled, yellow 101 3/16
Wrinkled, green 32 1/16
The easiest experiment to perform was to let the plants self-pollinate and then keep good records. After scoring his 556 F2 seeds (Table 5) he took the 315 that were round and yellow and planted them in one part of his garden. The plants that grew were allowed to self-pollinate. Of the 315 round and yellow seeds planted, 301 plants matured and produced seed. The seed produced was the F3 generation. At harvest, Mendel needed to exercise the utmost care. Each F2 plant was handled separately. The seeds from the plant were harvested and Mendel then scored the F3 seeds that came from the same F2 plant. This can be referred to as F2:3 data and the table below summarizes his complete experiment using all of the F2 phenotypes.
Mendel’s F2 data supported his principle of independent assortment. There were four different types of round yellow F2 based on the kinds of progeny they could produce or their breeding behaviors. Based on the F3 progeny produced, the F2 genotype was deduced. For example, if a round, yellow seed gave all round progeny it must have the genotype RR. If it gave both round and wrinkled it was Rr.
Furthermore, the numbers of F2 plants with each breeding behavior were in agreement with what was expected with independent assortment. There were four times as many round and yellow F2 that gave all four phenotypes of F3 seeds (138) compared to the round and yellow F2 that were true breeding (38). Overall, there were nine types of breeding behaviors demonstrated in the F2 demonstrating that there were nine F2 genotypes. In all cases, the fractions observed in the F2 agreed to the principle of independent assortment. Mendel’s well-planned experiment provided a convincing demonstration that genes behaved in this predictable manner.
The only thing better than performing an experiment that shows you were right about a new hypothesis is performing two experiments that show that you were right. That is what Gregor Mendel did! In his second experiment he crossed dihybrid F1 plants with homozygous recessive plants in a test cross. This type of cross is named because the geneticist wants to perform a cross that will test or reveal the genotype of an organism. Therefore, a test cross is usually made between an organism with a dominant trait and a partner with a recessive version of this trait. Mendel performed the RrYy x rryy testcross and the expected progeny are shown in the Punnett square below:
RrYy gametes: RY, Ry, rY, ry
rryy gametes: all ry
RY Ry rY ry
ry RrYy Rryy rrYy rryy
The observed result closely matched the expected. The testcross experiment provides additional support for the principle of independent assortment.
Mendel established a rigorous precedent for using carefully planned multi-generation experiments to reveal the principles that governed trait inheritance. The beauty of Mendel’s accomplishments is that both the principles and his experimental approach can be applied to understanding the genetic control and inheritance of traits in many kinds of organisms still today.
Lessson Summary
Mendel’s principles of segregation and independent assortment are valid explanations for genetic variation observed in many organisms. Alleles of a gene pair may interact in a dominant vs. recessive manner or show a lack of dominance. Even so, these principles can be used to predict the future…at least the potential outcome of specific crosses.
Video Example
Watch this video about Punnett Squares for more information | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.13%3A_Introduction_to_Mendelian_Genetics.txt |
Learning Objectives
At the completion of this lesson, you will be able to:
1. Contrast the inheritance of traits that are controlled by independent genes, linked genes and pleiotropic genes.
2. Demonstrate the physical basis of linkage by drawing the key events in meiosis.
3. Calculate map distances from 2-point test cross and F2 data.
4. Assemble linkage maps from map distance information.
Maize Genetics
Maize (Zea mays, corn) is not only a commercially important crop; it has also been a valuable organism used in basic genetic studies to understand genetic principles. Geneticists such as Dr. John Osterman, in the School of Biological Sciences at the University of Nebraska, use corn in basic genetic studies. Let’s examine one experiment.
Many seed traits in corn have been studied by geneticists. Color in the outside (aleurone) or inside (endosperm) of the seed can be important for end uses such as what color of corn chips you prefer. Starch development traits in the kernel influence what the corn seed can be used for (popcorn, sweet corn, waxy corn).
A line of corn that was true breeding for a red kernel color was crossed to a true breeding line with shrunken kernels but no red color (white or yellow). Even though the two parents differed by two traits, kernel color and kernel shape, the F1 offspring produced from the cross were all red with plump or normal kernels.
Parents: Red, plump x white, shrunken
F1 offspring: All red, plump
Which phenotypes would you say are dominant based upon this single result? Going a step further, what is your hypothesis for the genotypes of the parents and F1?
The next year these red plump seeds were planted in the crossing nursery near the white shrunken line. Dr. Osterman made a backcross or testcross between the F1 progeny and white, shrunken plants.
Thinking back on the genetics principles we have covered so far in this course, what phenotypes would you expect to observe in the testcross progeny?
If we use the principle of independent assortment and the information we already have, we would predict that there should be four types of seeds in testcross progeny.
We should get both parental combinations of red, plump and white, shrunken. We should also get two new combinations; red, shrunken and white, plump. Furthermore, the principle of independent assortment predicts the four combinations should be in a 1:1:1:1 ratio. (Make sure you understand why!)
What did Dr. Osterman actually observe? When he harvested the ear from the first plant, peeled back the husks, and examined the kernels, all he saw were red, plump and white, shrunken seeds. Only the parental combination of traits was found among the several hundred seeds on that ear!
Half the seeds had the dominant red and plump combination while the other half had the recessive traits. While this result does not fit what we expect based on the idea of independent traits, Dr. Osterman was not surprised. Obviously, another genetic hypothesis can explain this result. Let’s examine two alternative explanations.
Pleiotropy
Sometimes a single gene pair will control more than one trait. For example, barley that has good malting characteristics for brewing also tends to sprout prior to harvest. High sucrose lines of soybean have a sweeter flavor in soymilk and cause less gas production in the consumer compared to lines with normal sucrose levels. Perhaps a seed color gene in corn also controls whether the kernel is plump or shrunken. White kernels would always be shrunken, red would always be plump. When genes control more than one trait, these traits will always be inherited together. This explanation is consistent with what Dr. Osterman observed on the seeds of the testcross ear.
Linkage
The second explanation involves two genes but looks closely at their chromosome locations. Cytogeneticists have observed ten pairs of chromosomes in the somatic cells of corn. Corn has countless traits controlled by tens of thousands of genes. Since genes are on chromosomes, it makes logical sense that each chromosome is made of thousands of genes. When genes are on the same chromosome they will travel together as the chromosome is passed on to the gametes (Figure 3). If the gene for red vs. white kernel color is on the same corn chromosome as the gene for plump vs. shrunken, those genes will travel together during gamete formation. This could also explain why the red and plump gene alleles are inherited together. The same would be true of the white and shrunken alleles on the homologous chromosome in the F1. Therefore, we could explain the result by the hypothesis of linkage; a separate gene pair controls each trait but these genes are located near each other on the same chromosome.
More Data
Two genetic hypotheses can explain the testcross data. How can we determine which hypothesis is correct? Like all good geneticists, Dr. Osterman made an effort to generate large numbers of offspring in his studies. He had made several test crosses that summer and each cross generated ears with several hundred seeds. Upon careful examination of the seeds on all the ears, Dr. Osterman did observe a few red shrunken and white plump kernels. The new combinations were rare but their appearance in the testcross progeny allows us to eliminate one of the genetic hypotheses. Which hypothesis can be ruled out?
Linked Genes Tend to Stay Together
Because the red trait and plump trait are not always found together, one gene cannot be responsible for both traits, the occurrence of the recombinant red, shrunken and white, plump testcross offspring eliminates this as a reasonable hypothesis. Can we still use linkage as an explanation though? Yes, if we think through chromosome behavior during meiosis, we can see why genes on the chromosome tend to stay together but will not always be passed on together. Let’s think about meiosis and crossing over to understand these results.
Crossing Over
One of the key differences between meiosis and mitosis is the synapsis of homologous chromosomes during prophase I of meiosis. Synapsis is the process in which homologous chromosomes carefully pair. The pairing allows for an orderly first division to send one chromosome from each pair to separate cells. The close association of the homologous chromosomes also allows for crossing over between non-sister chromatids (Figure 4). During this process sections of the chromosomes break off and are exchanged between non-sister chromatids. When non-sister chromatids crossover, chromatids can be made that have a new combination of genes compared to the original combination on the chromosome. The original combination was inherited from the organism’s parents and is called the parental combination of genes. The new combination made is called the recombinant combination. In figure 4, a crossover occurs but the original or parental combination of CS (red and plump) and cs (white and shrunken) will stay together. Crossing over can cause new gene combinations to occur on a chromosome if the crossover occurs between the linked genes.
When a crossover occurs between genes, chromatids with both the parental combination and chromatids with a new combination will be made. We can see this in figure 5. Two of the chromatids are not involved in the crossing over. These chromatids will maintain the parental combination and when meiosis is complete, the two gametes made that have these chromosomes will be called parental gametes. The gametes made that have the other two chromosomes, those that went through crossing over and have the new gene combination, are called recombinant gametes (Figure 6).
When crossing over occurs between two non-sister chromatids, cells will make equal numbers of recombinant and parental gametes. In looking back at Dr. Osterman’s data though the number of plants that would have inherited recombinant type gametes was far below 50%. Why are parental combinations of linked genes made more frequently than new combinations? Understanding this requires us to imagine many cells going through gamete formation.
Making Lots of Gametes
Sexually reproducing organisms tend to make a large number of gametes to ensure reproductive success. Therefore, we need to think about crossing over events happening in many cells. Geneticists do not understand all the details of crossing over but in general, crossing over is a random process that occurs during prophase I. We also know that the occurrence of one crossover along a chromosome can reduce the chance of a second crossing over. As you look at figure 5, you can see that crossing over could occur at any point between where the C,c genes (the C,c locus) and the S,s locus reside on the chromosome causing recombinant gamete formation. If a crossover does not occur between the C,c and S,s loci (plural for locus), then only the parental gamete combination of CS and cs will be made. Considering the size of the chromosome and the relative positions of the C,c and S,s loci; how often would cells have crossovers occur between the genes compared to somewhere else along the chromosome? If crossing over is random, we would not expect a cross over to occur in very many cells between the C,c and S,s loci. Genes that are on the same chromosome tend to stay together at a rate that is proportional to how far apart they are on the chromosome. Since the C,c and S,s loci are close to each other, crossovers are rare. The ‘C’ gene tends to stay with the ‘S’ gene as it is passed from the red, plump line to the F1 and then to the testcross generation. The same is true of the ‘c’ and ‘s’ parental combination.
How does this discussion of crossing over help us understand what we see from our test cross result? The parental combinations of red, plump, and white shrunken were almost always made. The new combinations of red, shrunken and white, plump were rarely observed. It would take a recombinant gamete to make one of these rare combinations. While we never see the genes in the gametes or directly observe the genes on the chromosomes, the outcome suggests that these loci are linked close together on the chromosome. Making the connection between the results of crossing studies such as these has allowed geneticists to map genes or determine their relative positions on a chromosome. Let’s go through a mapping study to see how it is done.
Two-Point Testcross Mapping
The results of a testcross study different from Dr. Osterman’s is given below
• Red, Shrunken CCss X White, Plump ccSS
• F1: all Red, Plump (CcSs)
• F1 (CcSs) crossed with White Shrunken (ccss)
Table 1. Testcross progeny data
Phenotypes
# of progeny
Red, plump
120
Red, shrunken
3420
White, plump
3334
White, shrunken
126
Total
7000
Obviously, the geneticist who did this experiment made a lot of test crosses to generate these numbers. How can we use the numbers to map these genes? With this information we can answer one question. ‘What is the distance between the C,c and S,s loci?’ We cannot get out a fancy microscopic ruler and physically measure the distance on the corn chromosome. With this information all we can do is estimate the map unit distance. Map units are a measure of the tendency for crossovers to occur between two loci. Because genes that are farther apart will have a higher likelihood of crossovers, the higher the crossover frequency, the farther apart the genes are on the chromosome. Let’s apply this idea to our test cross data.
Test cross data allows us to indirectly measure the frequency of gametes made by an individual. All of the testcross progeny inherited a gamete with the recessive ‘c‘ and ‘s‘ alleles from the white, shrunken parent. Therefore, the alleles that the F1, dihybrid parent has passed on determine the traits in the seed (Table 1). We need to be able to measure how often crossovers occurred between the C,c and S,s loci when these dihybrids made gametes. From the data we have the following:
• Gametes that were passed to the F1 from the parent lines were Cs and cS.
• Gametes made from crossing over in the F1 (recombinant gametes) were CS and cs.
• Gametes with the original parent combination (parental gametes) were Cs and cS.
Therefore, the frequency of recombinant gametes was 246 (120 + 126) divided by the total gametes we have information on (7000) or 246/7000 = 3.5%
Map unit distance between the C,c and S,s loci = 3.5 Map units.
One map unit is equal to 1% recombinant gametes. Again, this is not a physical measurement. It is a relative measure of how often crossovers occurred between these loci. How reliable is this measurement? Let’s look at another data set (Table 2).
• Red, plump CCSS X White, shrunken ccss
• F1: all Red, Plump (CcSs or CS / cs)
• F1 (CS / cs) crossed with White Shrunken (ccss)
Table 2. Testcross progeny data from another data set
Phenotypes
# of Progeny
Red, plump
192
Red, shrunken
5
White, plump
3
White, shrunken
200
Total
400
Recombinant gamete frequency: 5 + 3 / 400 = 2%; 2 map units from C,c to S,s
Based on this result is the map unit distance measurement reliable? Yes, it is, to a certain degree. It is important to recognize the differences between the two test cross experiments.
Cis Versus Trans
The first difference to note is that the parental gametes are not always the same allele combinations, but they are always the most frequently produced gametes. In the smaller experiment above, the parents had either both dominant or both recessive traits together. Therefore, the F1 parent was making Cs and cS recombinant gametes. This dihybrid parent was in cis or coupling phase with respect to these genes (Figure 7).
In the larger experiment with 7000 testcross progeny the dihybrid F1 was in trans or repulsion phase for the C,c and S,s genes. In this case the Cs and cS are the parental gametes (Figure 8). Therefore, when genes are linked, they can be arranged in two ways in a dihybrid. This means that a CcSs plant can have its genotype in the cis arrangement with both dominant genes on one chromosome and both recessives on the other (written CS / cs) or can also be in trans (Cs / cS).
Map Distance Measurement
The first experiment gave us 3.5 map units and the latter gave 2 map units.
Why do our two experiments give us two different map unit distances? In both cases, we can see that the parental gene combinations have a strong tendency to stay together. Because the experiments were done with different population sizes, with different plants, and the difference in our map unit estimates is small, the difference could be attributed to chance. Environment may also influence the rate of crossing over to some degree. For practical purposes this gene mapping information is reliable enough to tell us that these two genes reside close to each other on the same chromosome. Once we have information on distances between other genes, we can improve our map.
Mapping Another Seed Trait Gene Pair
A third seed trait in corn is normal vs. waxy starch. Waxy starch has a different chemistry than normal starch and can be scored by staining with iodine. This gene is commercially important because some waxy corn is produced for specialty markets. Let’s look at the following test cross data (Table 3):
Parent: Red, Normal (CCWW) X White, waxy (ccww)
F1: Red, Normal (CcWw) X White, Waxy (ccww)
Table 3. Testcross data for another seed trait
Phenotypes
# of testcross progeny
Red, normal
2781
Red, waxy
759
White, normal
749
White, waxy:
2711
Total
7000
• Parental gametes made by the F1: CW and cw
• Recombinant gametes made by the F1: Cw and cW = 759 + 749 = 1508
• Map unit distance between the C,c and the W,w loci: 1508/7000 = 21.5 Map Units
It is clear from the testcross data (Table 3) that the genes at the C,c and W,w loci do not have as great a tendency to be inherited together compared to the genes at the C,c and S,s loci. This would be because the loci are farther apart, and our 21.5 map units is an indicator of the relative distance.
From the two distances calculated, how far apart are the W,w and S,s loci? Here is where we can try to use map unit distances to do gene mapping the same way we use miles to do road mapping. We can pencil our two possible maps (Figure 9).
In one map the C,c locus is in between the W,w and S,s loci. Using the 21.5 plus 3.5 map units we have calculated, the best estimate will be that 25 map units separate the W,w and S,s loci. The other map that fits our data is having the S,s locus in the middle with the W,w and S,s loci 18 map units apart (21.5 – 3.5). If the genes occupy a fixed position on the chromosome, only one map will be correct. How can we determine which is the correct map?
The Third Map Distance
A third map distance needs to be estimated in order to determine the linkage map of these three loci. We need to make a cross that gives us the opportunity to measure crossing over frequency between the S,s and W,w loci. One such cross is shown below.
• Parent: ssWW (shrunken, normal) X SSww (plump, waxy)
• F1: sW / Sw X sW / Sw
Test cross offspring:
Plump, normal:
630
Plump, waxy:
2824
Shrunken, normal:
2900
Shrunken, waxy:
646
Total:
7000
• Recombinant gametes: 646 + 630 = 1276
• Map distance S,s to W,w: 1276/7000 = 18/2 map units
Based on the third map distance our best map for all three loci is the second alternative (Figure 9). To map all three loci using this two-point test cross data, the geneticist needed to generate three different crosses. This is how gene maps in organisms such as corn have been generated. As new genes controlling traits are discovered, crosses can be made to test if these genes are independent or linked to other genes. Once linkage is determined, additional crosses can be analyzed to position the gene on the linkage map.
Map Distances from F2 Data
In our last test cross, the F1 dihybrid sW / Sw was crossed to a sw/sw, shrunken waxy plant. If the shrunken waxy line was not available, the geneticist could still obtain information to determine map distance by selfing the F1 to produce F2 offspring. Here is the procedure.
Results of Selfing the sW/Sw F1:
F2 Offspring
Number
Plump, normal
254
Plump, waxy
122
Shrunken, normal
120
Shrunken, waxy
4
Total F2s:
500
Step one: Focus on the double recessive (sw/sw).
How do we obtain recombinant gamete frequencies from this F2 phenotype information? The first thing to remember is that this is a self-cross rather than a testcross and an additional Punnett square is needed to depict how gametes produced these F2 (Figure 10). From the Punnett square we can see that four kinds of gametes are made in both the male and female in this cross. Therefore, the plump, normal seeds can be made nine different ways in our diagram and will consist of five different genotypes (SW/SW, SW/sW, sW/SW, sW/Sw and SW/sw). We cannot tell by observing the seed phenotype though, which of the genotypes it has. The plump, waxy and shrunken, normal F2 seeds will each consist of two different genotypes. The only F2 phenotypes that consist of a single genotype are the shrunken, waxy. They are all ssww (sw/sw). By focusing on how this phenotype was produced in the F2, we can estimate the frequency of recombinant gametes made by the selfed parent.
Step two: Go from phenotype frequency to gamete frequency.
Look at the Punnett square and think about how the shrunken waxy seeds were made. First, a sw gamete in the male part of the plant and a sw gamete in the female needed to be made. The parent had the genotype sW/Sw that means that these would be recombinant gametes in both the male and female. Once these gametes were made, they needed to come together at random during pollination to produce the seed. Mathematically, we can say that the frequency of the shrunken, waxy seeds is the frequency of the sw gamete made in the male times the frequency of the sw gamete made in the female. If we assume that crossovers occur in the pollen forming cells at the same frequency as the egg forming cells, then the frequency of the shrunken waxy seeds is the square of the frequency of sw. Make sure this last paragraph makes sense to you by walking through figure 10.
Step three: Take the square root.
Now we can use our seed phenotype frequency to estimate gamete frequency in the selfed parent. Four of the 500 seeds produced were shrunken waxy. That would give our square in the lower right a frequency of 0.008 (4/500). This frequency should be the sw gamete frequency squared (sw2). If we want to know the frequency of sw, we can take the square root of 0.008 and get about 0.09. Now we have the frequency of one gamete estimated. This would be the same for both the male and female gametes, where 9% of all gametes produced by the sW/Sw parent will be sw. As it turns out, we can calculate the frequency of the other three gametes from this information if we just think about how the gametes were made in meiosis.
Step four: Double the frequency, decide if parental or recombinant.
In our continuing example the sw gamete was made when a crossover occurred in the sW/Sw parent. Crossovers are a reciprocal exchange between non-sister chromatids so every time a sw gamete was made, a SW gamete was also made. Therefore, if the sw frequency is 0.09 then the SW frequency is also 0.09. These are our recombinant gametes, so the frequency of recombinant gametes is 2 X 0.09 or 0.18. We can tell from this frequency that these are the recombinant gametes, they are less frequently made than the parental gametes. The parental gametes would total 82% of the gametes, each parental gamete making up half (41%) of that total. Thus, 18 map units between the S,s and W,w loci is our map distance estimate from the data. Because we use only a subset of the information in this procedure (only the double recessive numbers) this estimate is more subject to chance than the test cross data. We can use more complex math to estimate gamete frequencies from all the F2 data but this square root method is a quick way to detect linkage and estimate map distance. In cases where geneticists cannot perform a testcross working with F2 data is a necessary part of gene mapping. Can you think of some instances where this would be true? | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.14%3A_Deviations_from_Mendelian_Genetics-_Linkage_%28Part_1%29.txt |
Learning Objectives
At the completion of this lesson, you will be able to:
1. Make predictions about inheritance using map unit distances and genetic markers.
2. Assemble maps from multiple-point linkage data.
3. Define the relationship between linkage maps, linkage groups and genome maps.
4. Describe how DNA or molecular markers are observed and used in gene mapping.
In this lesson you will learn to make predictions about inheritance using map unit distances and genetic markers, assemble maps from multiple-point linkage data, define the relationship between linkage maps, linkage groups and genome maps, and describe how DNA or molecular markers are observed and used in gene mapping.
Introduction
Why do geneticists map genes? This is a legitimate question after all this discussion. Gene mapping helps geneticists do two things, make better predictions and isolate genes. The following is an example of how gene mapping can help breeders make better selections.
Gene Maps and Selection
This example shows how gene mapping has a practical application. Soybeans have perfect flowers that self-pollinate. Tedious flower manipulation is required to make a hybrid seed. Soybean breeders would like to find ways to make hybrid seeds in commercial quantities. They have discovered a gene that causes a lack of pollen formation and makes plants male-sterile. Therefore, when bees visit these plants, they will make a cross-pollination and produce hybrid seed if they carry pollen from a nearby male-fertile plant. A hybrid seed production field could be set up if they could have all male-sterile plants in one row and male-fertile in the next. The genetic control of this trait is shown below:
Male-fertile plants: MM or Mm
Male-sterile: mm
Can you see the problem one would have in generating seed that is all male-sterile (genotype mm)? These plants will never be true breeding because they cannot self-pollinate (they have working female parts but no pollen). The only plants that can be self-pollinated to produce mm offspring are Mm plants. Progeny from selfed Mm plants will segregate 3 fertile to 1 sterile, making it impossible to obtain a pure collection of mm seeds to plant in a ‘female’ row for hybrid seed production. If the mm genotype could be picked out in seeds prior to planting, this would solve the problem. Unfortunately, the male sterile trait is impossible to select for in the seed but breeders discovered a trick they could use that took advantage of their knowledge of gene maps. A second trait in soybean that is easy to select for in seeds is controlled by a gene that was closely linked to the M,m male sterile gene. This trait is green vs. yellow seed coat. The seed coat trait is controlled as follows:
GG or Gg: green
gg: yellow
The G,g locus and M,m locus have been mapped and are about two map units apart. Therefore, we can take advantage of this linkage to select seeds from a cross that will tend to also be male sterile. This is how the process would work. The following cross is made:
GGMM (green, male-fertile) X ggmm (yellow, male sterile)
The cis dihybrid (GM/gm) will be male-fertile and can be self-pollinated. Because of linkage, the GM and gm parental gametes are made 98% of the time. Therefore, seeds that are gg and yellow are almost always going to be mm and male sterile. If the breeder sorts out the F2 seeds based on color, about one fourth of the seeds will be yellow. Based on 2-map unit distance, about 0.24 would be the expected frequency of ggmm out of all the F2 but 96% of the yellow F2 will be male sterile (0.24 out of 0.25). Therefore 96 out of 100 seeds planted in the ‘female’ row of selected yellow seeds will be male-sterile (mm). Harvesting seeds from these rows will provide a high percentage of hybrid seeds. Thus, the breeder took advantage of linkage to select for a trait that was easy to detect (yellow seeds) and obtain individuals with a trait that was impossible to select (male-sterile). The power of this prediction potential has driven much of the recent efforts in gene mapping in crop plants, livestock, and humans.
Table 1. Predicting the percentage of male-sterile F2 among yellow F2 seeds using a chart. Chart is in text form below for additional accessibility options.
0.49 big G, big M
0.01 big G, little m
0.01 little g, big M
0.49 little g, little m
0.49 big G, big M
0.01 big G, little M
0.01 big G, little m
0.01 little g, big M
0.0001
0.0049
0.49 little g, little m
0.0049
0.2401 (selected)
Three Point Test Cross: Multiple Point Gene Mapping
Gene mappers are motivated to map all of the tens of thousands of genes found on the chromosomes of plant or animals. Analyzing data from crosses to determine map distances for two genes at a time makes the process time-consuming and tedious. Therefore, geneticists will often attempt to map as many genes as possible from a single set of progeny. We will go through the simplest multiple point mapping example, a three-point testcross, to demonstrate this process.
We will use our three corn seed trait loci again and use data from one cross to map these three loci. The first step would be to obtain a trihybrid individual that is heterozygous at all three loci and then perform a testcross with this trihybrid.
Parents: CCssWW (CsW/CsW) X ccSSww (cSw/cSw)
Trihybrid: CcSsWs (CsW/cSw) X ccssww (csw/csw)
Table 2. Test cross offspring
Seed trait
Gamete from trihybrid
Number
Red, shrunken, normal
CsW
2777
White, plump, waxy
cSw
2708
Red, plump, waxy
CSw
116
White, shrunken, normal
csW
123
Red, shrunken, waxy
Csw
643
White, plump, normal
cSW
626
Red, plump, normal
CSW
4
White, shrunken, waxy
csw
3
Total number of progeny:
7000
We observe eight phenotypes of seeds in the testcross progeny because the trihybrid can make eight kinds of gametes. As we knew, these three genes are linked and so the uneven ratio of phenotypes reflects the combinations of two parental and six recombinant gametes. The parental gametes are still the most frequently produced type, and the six recombinant gametes are made when crossing over occurs between the loci. We can systematically account for these crossovers if we follow the following four steps. First, we will explain the steps and then we will show how to use them to map the three genes.
Step 1: Identify the parental gametes.
These are CsW and cSw, that combination which came from the parent generation and the combination made by the trihybrid at the highest frequency.
Step 2: Classify the recombinants.
If we observe the gene combination in each of the six recombinant gametes, we can ask ourselves if the gamete has a new combination for each pair of genes. For example, CSw has a new combination for the CS and the Cw compared to the parental gametes but the same combination for Sw as the cSw parental gamete. Therefore, these 116 gametes represent crossovers between C,c and S,s as well as C,c and W,s. This information is recorded (see below).
Step 3: Determine recombinant gamete frequency.
Once all six recombinant gametes have been classified, the total number of crossovers between the three loci can be added up and crossover percentage determined between each pair of loci. This number will reflect gene distance but one more step is needed to complete the process.
Step 4: Add in the double crossover gametes.
Two of the six recombinant gametes were made as a result of double crossovers between the two loci that are furthest apart. These crossovers have been added to the map distances between the middle locus and the two outside loci. Now we need to add these double crossovers to the outside loci distance.
Applying these four steps to our three-point test cross data would work as follows:
1. CsW and cSw
2. C,c to S,s: 116 (CSw) + 123 (csW) + 4(CSW) + 3(csw) = 246
C,c to W,w: 116 (CSw) + 123 (csW) + 643 (Csw) + 626 (cSW) = 1508
S,s to W,w: 643 (Csw) + 626 (cSW) + 4 (CSW) + 3 (csw) = 1276
3. 246 / 7000 = 3.5% recombinant gametes C,c to S,s 1508 / 7000 = 21.5% recombinant gametes.
C,c to W,w 1276 / 7000 = 18.2% recombinant gametes S,s to W,w.
4. From the information in step 3 we can see that the C,c and W,w loci are the farthest apart so the S,s locus must be between them. We can also see that we have underestimated the distance between the outside loci (3.5 C,c to S,s + 18.2 S,s to W,w = 21.7 map units not 21.5). While this difference is small, we can rectify this and double check our work by adding in the double crossovers. The CSW and csw gametes are made very rarely. That is because is takes two crossovers in the trihybrid’s chromosome to make them, not just one.
When two crossovers occur between the C,c and W,w loci the parental combinations of CW and cw are restored. Therefore, we did not count the CSW and csw gametes as representing crossovers when they actually represent two crossovers each (Figure 1). Thus, we should add up the double crossovers (3 + 4 = 7), multiply times two and add these 14 crossovers to the C,c to W,w distance (1508 + 14 = 1522 / 7000 = 21.7 map units). Now we have double checked our map distances and have our three-point map complete.
It can be intimidating to see all the data generated from a three-point test cross. This information, however, can be used systematically to save the geneticists time and map three genes in one experiment. The three point data also provides a more accurate measure of map distances compared to two-point data when genes are farther apart on a chromosome. This is because a third gene in between the more distant loci can account for double crossovers that would not be detectable in a two-point analysis. For this reason, map distances tend to be underestimated when genes are further apart as we saw to a modest extent in this example (21.5 two-point vs. 21.7 three-point).
How would a geneticist work with a four-point test cross? There would be sixteen phenotypes in the progeny as a result of the hybrid parent making two parental gametes and fourteen recombinant gametes. These recombinant gametes would be a result of single, double, and even triple crossovers that occurred in prophase I. While the data would be tedious to work with, one data set could be analyzed to reveal the map of four genes. It is not surprising that geneticists now have written computer programs which will perform these types of calculations, save time, and reduce the chance of calculator error. Geneticist’s mapping genes in economically important plants and animals will also generate mapping populations that are a result of crossing parents that have different alleles at hundreds or thousands of loci. Even though the development of computers and programs has become a part of modern gene mapping the process of indirectly measuring crossover frequency by observing the inheritance of trait combinations is the basis of generating these gene maps.
Linkage Maps, Linkage Groups
The organism that gene mapping was first performed in was fruit flies. Working with fruit flies gave the early gene mappers many advantages. First, the fruit fly geneticists observed a great of deal of genetic variation among fruit flies for body part traits (wing size, eye shape and color, leg bristle types etc.) that were easy to see if they had a low powered microscope. The phenotype variants arose naturally or could be induced by chemical or radiation mutagenesis. Additionally, short generation time, large offspring numbers and low rearing costs allowed them to complete informative linkage experiments in a short time. Finally, cytogeneticists knew that fruit flies had four pairs of chromosomes so all the genes would map into four linkage groups. One chromosome was very small, and few genes mapped to this chromosome. The X-chromosome and the largest autosome have many genes and more than 100 map units separate genes on opposite ends of the chromosome.
Let’s stop and think about how they determined this large map unit distance. If the A,a and W,w loci are 106 map units apart, how frequently will an AW / aw dihybrid make the gametes Aw and aW? Since these are the recombinant gametes and map distance is the frequency of recombinant gametes, we are tempted to say 106% but that is impossible. The parental gametes for any two loci will never be made less than 50% of the time, the percentage we observe when two loci are independently assorting. In fact, 50% is the correct answer here. Anytime genes that are on the same linkage group are 50 map units or more apart, they will behave as if they are independent of one another. That is because the genes are far enough apart to always allow at least one crossover to occur between them during prophase I. So how can we ever determine that A,a and W,w are 106 map units apart? Simply by determining the distances between these genes and other genes in between them that are less than 50 map units apart (Figure 3). Figure 3 shows that distances can be found between A,a and B,b, B,b and C,c, then C,c and W,w. Each of these distances is added to arrive at the 106 map units between A,a and W,w. Therefore, if the geneticist combines the mapping information from many linkage experiments or performs a multiple point linkage analysis on a progeny group segregating for many linked genes, they can deduce the larger map unit distances.
Limited Traits, Limited Linkage Maps
Many gene mappers do not enjoy the advantages that fruit fly geneticists have in mapping genes. Cattle for example have thirty linkage groups (2n = 60), they often have one offspring per cross, and it is difficult of observe hundreds of phenotype differences controlled by single genes among cattle. Soybean have 20 linkage groups but for many decades, soybean geneticists had compiled about 30 different linkage maps and had no idea as to which maps were really part of the same linkage group. Soybean geneticists needed to discover more genes that had clear phenotype effects and then perform crosses that compared the inheritance of these new genes with the genes that had already been mapped. In the 1970’s and 1980’s, a new type of genetic trait, molecular markers, was discovered that greatly accelerated the gene mapping process in all organisms. | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.15%3A_Deviations_from_Mendelian_Genetics-_Linkage_%28Part_2%29.txt |
Genetics – is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Heredity – Humans have always been aware that the characteristics of an individual plant or animal in a population could be passed down through the generations. Offspring look more like their parents.Humans also knew that some heritable characteristics (such as the size or color of fruit) varied between individuals, and that they could select or breed crops and animals for the most favorable traits. Knowledge of these hereditary properties has been of significant value in the history of human development. In the past, humans could only manipulate and select from naturally existing combinations of genes. More recently, with the discovery of the substance and nature of genetic material, DNA, we can now identify, clone, and create novel, better combinations of genes that will serve our goals. Understanding the mechanisms of genetics is fundamental to using it wisely and for the betterment of all.
Fig.1-1: Parent and offspring. Wolf’s Monkey. (Flickr-eclectic echos-CC:AND)
1.02: DNA is the Genetic Material
By the early 1900’s, biochemists had isolated hundreds of different chemicals from living cells. Which of these was the genetic material? Proteins seemed like promising candidates, since they were abundant, diverse, and complex molecules. However, a few key experiments demonstrated that DNA, rather than protein, is the genetic material.
Griffith’s Transformation Experiment (1928)
Microbiologists identified two strains of the bacterium Streptococcus pneumoniae. The R-strain produced rough colonies on a bacterial plate, while the other S-strain was smooth (Figure \(2\)). More importantly, the S-strain bacteria caused fatal infections when injected into mice, while the R-strain did not (top, Figure \(3\)). Neither did “heat-treated” S-strain cells. Griffith in 1929 noticed that upon mixing “heat-treated” S-strain cells together with some R-type bacteria (neither should kill the mice), the mice died and there were S-strain, pathogenic cells recoverable. Thus, some non-living component from the S-type strains contained genetic information that could be transferred to and transform the living R-type strain cells into S-type cells.
Avery, MacLeod and McCarty’s Experiment (1944)
What kind of molecule from within the S-type cells was responsible for the transformation? To answer this, researchers named Avery, MacLeod and McCarty separated the S-type cells into various components, such as proteins, polysaccharides, lipids, and nucleic acids. Only the nucleic acids from S-type cells were able to make the R-strains smooth and fatal. Furthermore, when cellular extracts of S-type cells were treated with DNase (an enzyme that digests DNA), the transformation ability was lost. The researchers therefore concluded that DNA was the genetic material, which in this case controlled the appearance (smooth or rough) and pathogenicity of the bacteria.
Hershey and Chase’s Experiment (1952)
Further evidence that DNA is the genetic material came from experiments conducted by Hershey and Chase. These researchers studied the transmission of genetic information in a virus called the T2 bacteriophage, which used Escherichia coli as its host bacterium (Figure \(4\)). Like all viruses, T2 hijacks the cellular machinery of its host to manufacture more viruses. The T2 phage itself only contains both protein and DNA, but no other class of potential genetic material.
To determine which of these two types of molecules contained the genetic blueprint for the virus, Hershey and Chase grew viral cultures in the presence of radioactive isotopes of either phosphorus (32P) or sulphur (35S). The phage incorporated these isotopes into their DNA and proteins, respectively (Fig 1.5). The researchers then infected E. coli with the radiolabeled viruses, and looked to see whether 32P or 35S entered the bacteria. After ensuring that all viruses had been removed from the surface of the cells, the researchers observed that infection with 32P labeled viruses (but not the 35S labeled viruses) resulted in radioactive bacteria. This demonstrated that DNA was the material that contained genetic instructions.
Meselson and Stahl experiment (1958)
From the complementary strands model of DNA, proposed by Watson and Crick in 1953, there were three straightforward possible mechanisms for DNA replication: (1) semi-conservative, (2) conservative, and (3) dispersive (Fig 1.6).
1. The semi-conservative model proposes the two strands of a DNA molecule separate during replication and then strand acts as a template for synthesis of a new, complementary strand.
2. The conservative model proposes that the entire DNA duplex acts as a single template for the synthesis of an entirely new duplex.
3. The dispersive model has the two strands of the double helix breaking into units that which are then replicated and reassembled, with the new duplexes containing alternating segments from one strand to the other.
Each of these three models makes a different prediction about the how DNA strands should be distributed following two rounds of replication. These predictions can be tested in the following experiment by following the nitrogen component in DNA in E. coli as it goes through several rounds of replication. Meselson and Stahl used different isotopes of Nitrogen, which is a major component in DNA. Nitrogen-14 (14N) is the most abundant natural isotope, while Nitrogen-15 (15N) is rare, but also denser. Neither is radioactive; each can be followed by a difference in density – “light” 14 vs “heavy”15 atomic weight in a CsCl density gradient ultra-centrifugation of DNA.
The experiment starts with E. coli grown for several generations on medium containing only 15N. It will have denser DNA. When extracted and separated in a CsCl density gradient tube, this “heavy” DNA will move to a position nearer the bottom of the tube in the more dense solution of CsCl (left side in Figure \(7\)). DNA extracted from E. coli grown on normal, 14 N containing medium will migrate more towards the less dense top of the tube.
If these E. coli cells are transferred to a medium containing only 14N, the “light” isotope, and grown for one generation, then their DNA will be composed of one-half 15N and one-half 14N. If the this DNA is extracted and applied to a CsCl gradient, the observed result is that one band appears at the point midway between the locations predicted for wholly 15N DNA and wholly 14N DNA (Figure \(7\)). This “single-band” observation is inconsistent with the predicted outcome from the conservative model of DNA replication (disproves this model), but is consistent with both that expected for the semi-conservative and dispersive models.
If the E. coli is permitted to go through another round of replication in the 14N medium, and the DNA extracted and separated on a CsCl gradient tube, then two bands were seen by Meselson and Shahl: one at the 14N-15N intermediate position and one at the wholly 14N position (Figure \(7\)). This result is inconsistent with the dispersive model (a single band between the 14N-15N position and the wholly 14 N position) and thus disproves this model. The two band observation is consistent with the semi-conservative model which predicts one wholly 14 N duplex and one 14N-15N duplex. Additional rounds of replication also support the semi-conservative model/hypothesis of DNA replication. Thus, the semi-conservative model is the currently accepted mechanism for DNA replication. Note however, that we now also know from more recent experiments that whole chromosomes, which can be millions of bases in length, are also semi-conservatively replicated.
These experiments, published in 1958, are a wonderful example of how science works. Researchers start with three clearly defined models (hypotheses). These models were tested, and two (conservative and dispersive) were found to be inconsistent with the observations and thus disproven. The third hypothesis, semi-conservative, was consistent with the observations and thereby supported and accepted as mechanism of DNA replication. Note, however, this is not “proof” of the model, just strong evidence for it; hypotheses are not “proven”, only disproven or supported.
RNA and protein
While DNA is the genetic material for the vast majority of organisms, there are some viruses that use RNA as their genetic material. These viruses can be either single or double stranded and include SARS, influenza, hepatitis C and polio, as well as the retroviruses like HIV-AIDS. Typically there is DNA used at some stage in their life cycle to replicate their RNA genome.
Also, the case of Prion infections agents transmit characteristics via only a protein (no nucleic acid present). Prions infect by transmitting a misfolded protein state from one aberrant protein molecule to a normally folded molecule. These agents are responsible for bovine spongiform encephalopathy (BSE, also known as "mad cow disease") in cattle and deer and Creutzfeldt–Jakob disease (CJD) in humans. All known prion diseases act by altering the structure of the brain or other neural tissue and all are currently untreatable and ultimately fatal. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/01%3A_Overview_DNA_and_Genes/1.01%3A_Overview.txt |
The experiments outlined in the previous sections proved that DNA was the genetic material, but very little was known about its structure at the time.
Chargaff’s Rules
When Watson and Crick set out in the 1940’s to determine the structure of DNA, it was already known that DNA is made up of a series four different types of molecules, called bases or nucleotides: adenine (A), cytosine (C), thymine (T), guanine (G). Watson and Crick also knew of Chargaff’s Rules, which were a set of observations about the relative amount of each nucleotide that was present in almost any extract of DNA. Chargaff had observed that for any given species, the abundance of A was the same as T, and G was the same as C. This was essential to Watson & Crick’s model.
The Double Helix
Using proportional metal models of the individual nucleotides, Watson and Crick deduced a structure for DNA that was consistent with Chargaff’s Rules and with x-ray crystallography data that was obtained (with some controversy) from another researcher named Rosalind Franklin. In Watson and Crick’s famous double helix, each of the two strands contains DNA bases connected through covalent bonds to a sugar-phosphate backbone (Fig 1.8, 1.9). Because one side of each sugar molecule is always connected to the opposite side of the next sugar molecule, each strand of DNA has polarity: these are called the 5’ (5-prime) end and the 3’ (3-prime) end, in accordance with the nomenclature of the carbons in the sugars. The two strands of the double helix run in anti-parallel (i.e. opposite) directions, with the 5’ end of one strand adjacent to the 3’ end of the other strand. The double helix has a right-handed twist, (rather than the left-handed twist that is often represented incorrectly in popular media). The DNA bases extend from the backbone towards the center of the helix, with a pair of bases from each strand forming hydrogen bonds that help to hold the two strands together. Under most conditions, the two strands are slightly offset, which creates a major groove on one face of the double helix, and a minor groove on the other. Because of the structure of the bases, A can only form hydrogen bonds with T, and G can only form hydrogen bonds with C (remember Chargaff’s Rules). Each strand is therefore said to be complementary to the other, and so each strand also contains enough information to act as a template for the synthesis of the other. This complementary redundancy is important in DNA replication and repair.
How can this molecule, DNA, contain the genetic material?
1.04: Genes are the Basic Units of Inheritance
Blending vs Particulate inheritance
The once prevalent (but now discredited) concept of blending inheritance proposed that some undefined essence, in its entirety, contained all of the heritable information for an individual. It was thought that mating combined the essences from each parent, much like the mixing of two colors of paint. Once blended together, the individual characteristics of the parents could not be separated again. However, Gregor Mendel (Fig 1.10) was one of the first to take a quantitative, scientific approach to the study of heredity.
He started with well-characterized strains, repeated his experiments many times, and kept careful records of his observations. Working with peas, Mendel showed that white-flowered plants could be produced by crossing two purple-flowered plants, but only if the purple-flowered plants themselves had at least one white-flowered parent (Fig 1.11). This was evidence that the genetic factor that produced white-flowers had not blended irreversibly with the factor for purple-flowers. Mendel’s observations disprove blending inheritance and favor an alternative concept, called particulate inheritance, in which heredity is the product of discrete factors that control independent traits. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/01%3A_Overview_DNA_and_Genes/1.03%3A__The_Structure_of_DNA.txt |
Beadle and Tatum: one gene, one enzyme hypothesis
Life depends on (bio)chemistry to supply energy and to produce the molecules to construct and regulate cells. In 1908, A. Garrod described “in born errors of metabolism” in humans using the congenital disorder, alkaptonuria (black urine disease), as an example of how “genetic defects” led to the lack of an enzyme in a biochemical pathway and caused a disease (phenotype). Over 40 years later, in 1941, Beadle and Tatum built on this connection between genes and metabolic pathways. Their research led to the “one gene, one enzyme (or protein)” hypothesis, which states that each of the enzymes that act in a biochemical pathway is encoded by a different gene. Although we now know of many exceptions to the “one gene, one enzyme (or protein)” principle, it is generally true that each different gene produces a protein that has a distinct catalytic, regulatory, or structural function.
Beadle and Tatum used the fungus Neurospora crassa (a mold) for their studies because it had practical advantages as a laboratory organism. They knew that Neurospora was prototrophic, meaning that it could synthesize its own amino acids when grown on minimal medium, which lacked most nutrients except for a few minerals, simple sugars, and one vitamin (biotin). They also knew that by exposing Neurospora spores to X-rays, they could randomly damage its DNA to create mutations in genes. Each different spore exposed to X-rays potentially contained a mutation in a different gene. After genetically screening many, many spores for growth, most appeared to still be prototrophic and still able to grow on minimal medium. However, some spores had mutations that changed them into auxotrophic strains that could no longer grow on minimal medium, but did grow on complete medium supplemented with nutrients (Figure \(12\)). In fact, some auxotrophic mutations could grow on minimal medium with only one, single nutrient supplied, such as arginine.
B&T’s 1 gene: 1 enzyme hypothesis led to Biochemical Pathway dissection using genetic screens and mutations
Beadle and Tatum’s experiments are important not only for its conceptual advances in understanding genes, but also because they demonstrate the utility of screening for genetic mutants to investigate a biological process – genetic analysis. Beadle and Tatum’s results were useful to investigate biological processes, specifically the metabolic pathways that produce amino acids. For example, Srb and Horowitz in 1944 tested the ability of the amino acids to rescue auxotrophic strains. They added one of each of the amino acids to minimal medium and recorded which of these restored growth to independent mutants. For example, if the progeny of a mutagenized spore could grow on minimal medium only when it was supplemented with arginine (Arg), then the auxotroph must bear a mutation in the Arg biosynthetic pathway and was called an “arginineless” strain (arg-).
Synthesis of even a relatively simple molecule such as arginine requires many steps, each with a different enzyme. Each enzyme works sequentially on a different intermediate in the pathway (Figure \(13\)). For arginine (Arg), two of the intermediates are ornithine (Orn) and citrulline (Cit). Thus, mutation of any one of the enzymes in this pathway could turn Neurospora into an Arg auxotroph (arg-). Srb and Horowitz extended their analysis of Arg auxotrophs by testing the intermediates of amino acid biosynthesis for the ability to restore growth of the mutants (Figure \(14\)).
They found that some of the Arg auxotrophs could be rescued only by Arg, while others could be rescued by either Arg or Cit, and still other mutants could be rescued by Arg, Cit, or Orn (Table \(1\)). Based on these results, they deduced the location of each mutation in the Arg biochemical pathway, (i.e. which gene was responsible for the metabolism of which intermediate).
Table \(1\): Ability of auxotrophic mutants of each of the three enzymes of the Arg biosynthetic pathways to grow on minimal medium (MM) supplemented with Arg or either of its precursors, Orn and Cit. Gene names refer to the labels used in Figure \(14\)
MM + Orn
MM + Cit
MM + Arg
gene A mutants
Yes
Yes
Yes
gene B mutants
No
Yes
Yes
gene C mutants
No
No
Yes | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/01%3A_Overview_DNA_and_Genes/1.05%3A_The_Function_of_Genes.txt |
The C-value of the nuclear genome
The complete set of DNA within the nucleus of any organism is called its nuclear genome and is measured as the C-value in units of either the number of base pairs or picograms of DNA. There is a general correlation between the nuclear DNA content of a genome (i.e. the C-value) and the physical size or complexity of an organism. Compare the size of E. coli and humans for example in Table \(2\).
Table \(2\): Measures of genome size in selected organisms. The DNA content (1C) is shown in millions of basepairs (Mb). For eukaryotes, the chromosome number is the chromosomes counted in a gamete (1N) from each organism. The average gene density is the mean number of non-coding bases (in bp) between genes in the genome.
DNA content (Mb, 1C)
Estimated gene number
Average gene density
Chromosome number (1N)
Homo sapiens
3,200
25,000
100,000
23
Mus musculus
2,600
25,000
100,000
20
Drosophila melanogaster
140
13,000
9,000
4
Arabidopsis thaliana
130
25,000
4,000
5
Caenorhabditis elegans
100
19,000
5,000
6
Saccharomyces cerevisiae
12
6,000
2,000
16
Escherichia coli
5
3,200
1,400
1
There are, however, many exceptions to this generalization, such as the human genome contains only 3.2 x 109 DNA bases, while the wheat genome contains 17 x 109 DNA bases, almost 6 times as much. The Marbled Lungfish (Protopterus aethiopicus – Figure \(16\)) contains ~133 x 109 DNA bases, (~45 times as much as a human) and a fresh water amoeboid, Polychaos dubium, which has as much as 670 x 109 bases (200x a human).
Fig 1.16: Marbled Lungfish. (Wikipedia-OpenCage-CC:AS)
The C-value paradox
This apparent paradox (called the C-value paradox) can be explained by the fact that not all nuclear DNA encodes genes – much of the DNA in larger genomes is non-gene coding. In fact, in many organisms, genes are separated from each other by long stretches of DNA that do not code for genes or any other genetic information. Much of this “non-gene” DNA consists of transposable elements of various types, which are an interesting class of self-replicating DNA elements discussed in more detail in a subsequent chapter. Other non-gene DNA includes short, highly repetitive sequences of various types.
Other genomes
Organelles such as mitochondria and chloroplasts also have their own genomes. These are, compared to the nuclear genome, relatively small and are also circular, like the prokaryotes from which they originated (Endosymbiont hypothesis).
1.07: Model Organisms Facilitate Genetic Advances
Model organisms
Many of the great advances in genetics were made using species that are not especially important from a medical, economic, or even ecological perspective. Geneticists, from Mendel onwards, have sought the best organisms for their experiments. Today, a small number of species are widely used as model organisms in genetics (Fig 1.17). All of these species have specific characteristics that make large number of them easy to grow and analyze in laboratories: (1) they are small, (2) fast growing with a short generation time, (3) produce lots of progeny from matings that can be easily controlled, (4) have small genomes (small C-value), and (5) are diploid (i.e. chromosomes are present in pairs).
The most commonly used model organism are:
• The prokaryote bacterium, Escherichia coli, is the simplest genetic model organism and is often used to clone DNA sequences from other model species.
• Yeast (Saccharomyces cerevisiae) is a good general model for the basic functions of eukaryotic cells.
• The roundworm, Caenorhabditis elegans is a useful model for the development of multicellular organisms, in part because it is transparent throughout its life cycle, and its cells undergo a well-characterized series of divisions to produce the adult body.
• The fruit fly (Drosophila melanogaster) has been studied longer, and probably in more detail, than any of the other genetic model organisms still in use, and is a useful model for studying development as well as physiology and even behavior.
• The mouse (Mus musculus) is the model organism most closely related to humans, however there are some practical difficulties working with mice, such as cost, slow reproductive time, and ethical considerations.
• The zebrafish (Danio rerio) has more recently been developed by researchers as a genetic model for vertebrates.Unlike mice, zebrafish embryos develop quickly and externally to their mothers, and are transparent, making it easier to study the development of internal structures and organs.
• Finally, a small weed, Arabidopsis thaliana, is the most widely studied plant genetic model organism. This provides knowledge that can be applied to other plant species, such as wheat, rice, and corn. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/01%3A_Overview_DNA_and_Genes/1.06%3A_The_Nuclear_Genome.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
1.1 How would the results of the cross in Figure 1.11 have been different if heredity worked through blending inheritance rather than particulate inheritance?
1.2 Imagine that astronauts provide you with living samples of multicellular organisms discovered on another planet. These organisms reproduce with a short generation time, but nothing else is known about their genetics.
a) How could you define laws of heredity for these organisms?
b) How could you determine what molecules within these organisms contained genetic information?
c) Would the mechanisms of genetic inheritance likely be similar for all organisms from this planet?
d) Would the mechanisms of genetic inheritance likely be similar to organisms from earth?
1.3 It is relatively easy to extract DNA and protein from cells; biochemists had been doing this since at least the 1800’s. Why then did Hershey and Chase need to use radioactivity to label DNA and proteins in their experiments?
1.4 Compare Watson and Crick’s discovery with Avery, MacLeod and McCarty’s discovery.
a) What did each discover, and what was the impact of these discoveries on biology?
b) How did Watson and Crick’s approach generally differ from Avery, MacLeod and McCarty’s?
c) Briefly research Rosalind Franklin on the internet. Why is her contribution to the structure of DNA controversial?
1.5 Starting with mice and R and S strains of S. pneumoniae, what experiments in additional to those shown in Figure 1.3 to demonstrate that DNA is the genetic material?
1.6 List the information that Watson and Crick used to deduce the structure of DNA.
1.7 Refer to Watson and Crick’
a) List the defining characteristics of the structure of a DNA molecule.
b) Which of these characteristics are most important to replication?
c) Which characteristics are most important to the Central Dogma?
1.8 Compare Figure 1.13 and Table 1.1. Which of the mutants (#1, #2, #3) shown in Figure 1.13 matches each of the phenotypes expected for mutations in genes A, B,C?
1.9 Refer to Table 1.2
a) What is the relationship between DNA content of a genome, number of genes, gene density, and chromosome number?
b) What feature of genomes explains the c-value paradox?
c) Do any of the numbers in Table 1.2 show a correlation with organismal complexity?
1.10 a) List the characteristics of an ideal model organism.
b) Which model organism can be used most efficiently to identify genes related to:
i) eye development
ii) skeletal development
iii) photosynthesis
iii) cell division
iv) cell differentiation
v) cancer
1.11 Refer to Figure 1.8
a) Identify the part of the DNA molecule that would be radioactively labeled in the manner used by Hershey & Chase
b) DNA helices that are rich in G-C base pairs are harder to separate (e.g. by heating) than A-T rich helices. Why?
1.S: Overview DNA and Genes (Summary)
• Mendel demonstrated that heredity involved discrete, heritable factors that affected specific traits.
• A gene can be defined abstractly as a unit of inheritance.
• The ability of DNA from bacteria and viruses to transfer genetic information into bacteria demonstrated that DNA is the genetic material.
• DNA is a double helix made of two anti-parallel strands of bases on a sugar-phosphate backbone.
• Specific bases on opposite strands pair through hydrogen bonding, ensuring complementarity of the strands.
• The Central Dogma explains how DNA dictates heritable traits.
• Not all DNA in an organism contains genes.
• Model organisms accelerate the use of genetics in basic and applied research in biology, agriculture and medicine.
Key Terms
blending inheritance
particulate inheritance
Mendel
gene
allele
trait
P, F1, F2
Griffith
Avery, MacLeod, & McCarty
Hershey and Chase
Meselson & Stahl
DNase
proteinase
35S
32P
bacteriophage
semi-conservative
conservative
dispersive
E. coli
Nitrogen-14
Nitrogen-15
heavy vs light
CsCl gradient
Beadle & Tatum
auxotroph
prototroph
metabolic pathway
Neurospora crassa
Chargaff’s Rules
Watson and Crick
DNA bases
sugar-phosphate backbone
anti-parallel
complementary
hydrogen bond
minor groove
major groove
adenine
cytosine
thymine
guanine
Central Dogma
transcription
reverse transcription
translation
RNA
prion
one-gene:one-enzyme
minimal medium
complete medium
arginine
genetic screen
nuclear genome
c-value paradox
model organism
Saccharomyces cerevisiae
Caenorhabditis elegans
Drosophila melanogaster
Mus musculus
Danio rerio
Arabidopsis thaliana
Escherichia coli | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/01%3A_Overview_DNA_and_Genes/1.E%3A_Overview_DNA_and_Genes_%28Exercises%29.txt |
DNA can be highly compacted
If stretched to its full length, the DNA molecule of the largest human chromosome would be 85mm. Yet during mitosis and meiosis, this DNA molecule is compacted into a chromosome approximately 5µm long. Although this compaction makes it easier to transport DNA within a dividing cell, it also makes DNA less accessible for other cellular functions such as DNA synthesis and transcription. Thus, chromosomes vary in how tightly DNA is packaged, depending on the stage of the cell cycle and also depending on the level of gene activity required in any particular region of the chromosome.
Levels of compaction
There are several different levels of structural organization in eukaryotic chromosomes, with each successive level contributing to the further compaction of DNA (Figure \(2\)). For more loosely compacted DNA, only the first few levels of organization may apply. Each level involves a specific set of proteins that associate with the DNA to compact it. First, proteins called the core histones act as spool around which DNA is coiled twice to form a structure called the nucleosome. Nucleosomes are formed at regular intervals along the DNA strand, giving the molecule the appearance of “beads on a string”. At the next level of organization, histone H1 helps to compact the DNA strand and its nucleosomes into a 30nm fibre. Subsequent levels of organization involve the addition of scaffold proteins that wind the 30nm fibre into coils, which are in turn wound around other scaffold proteins.
Chromatin Packaging Varies inside the Nucleus: Euchromatin & Heterochromatin
Chromosomes stain with some types of dyes, which is how they got their name (chromosome means “colored body”). Certain dyes stain some regions along a chromosome more intensely than others, giving some chromosomes a banded appearance. The material that makes up chromosomes, which we now know to be proteins and DNA, is called chromatin. Classically, there are two major types of chromatin, but these are more the ends of a continous and varied spectrum. Euchromatin is more loosely packed, and tends to contain genes that are being transcribed, when compared to the more densely compacted heterochromatin, which is rich in repetitive sequences and tends not to be transcribed. Heterochromatin sequences include short, highly-repetitive sequences called satellite DNA, which acquired their name because their bouyant density is distictly different from the main band of DNA following ultracentrifugation.
Morphological features of Chromosomes
Chromosomes also contain other distinctive features such as centromeres and telomeres. Both of these are usually heterochromatin. In most cases, each chromosome contains one centromere. These sequences are bound by centromeric proteins that link the centromere to microtubules that transport chromosomes during cell division. Under the microscope, centromeres of metaphase chromosomes can sometimes appear as constrictions in the body of the chromosome (Figure \(3\)) and are called primary (1°) constrictions. If a centromere is located near the middle of a chromosome, it is said to be metacentric, while an acrocentric centromere is closer to one end of a chromosome, and a telocentric chromsome is at, or near, the very end. In contrast, some species have a holocentric centromere, where no single centromere can be defined and the entire chromsome acts as the centromere. Telomeres are repetitive sequences near the ends of linear chromosomes, and are important in maintaining the length of the chromosomes during replication, and protecting the ends of the chromosomes from alterations.
It is essential to describe the similarity between chromosomes using appropriate terminology (Fig 2.4). Homologous chromosomes are typically pairs of similar, but non-identical, chromosomes in which one member of the pair comes from the male parent, and the other comes from the female parent. Homologs contain the same gene loci but not necessarily the same alleles. Non-homologous chromosomes contain different gene loci, and are usually distinguishable based on cytological features such as length, centromere position, and banding patterns.
An unreplicated chromosomes can undergo replication, to produce a replicated chromosome that has two sister chromatids, which are physically connected to each other at the centromere and remain joined until cell division. Because a pair of sister chromatids is produced by the replication of a single DNA molecule, their sequences are essentially identical (same alleles), differing only because of DNA replication errors. On the other hand, non-sister chromatids come from two separate, but homologous chromosomes, and therefore usually contain the same gene loci in the same order, but do not necessarily have identical DNA sequences (allelic differences).
The decondensed chromosomes are not randomly arranged in within the interphase nucleus. They often have specific locations within the nucleus and relative to one another (Figure \(5\))
Chromosome and DNA Replication
When the cell enters S-phase in the cell cycle (G1-S-G2-M) all the chromosomal DNA is replicated. This is done by enzymes called DNA polymerases. All DNA polymerases synthesize new strands by adding nucleotides to the 3'OH group present on the previous nucleotide. For this reason they are said to work in a 5' to 3' direction. DNA polymerases use a single strand of DNA as a template upon which it will synthesize the complementary sequence. This works fine for the middle of chromosomes - DNA-directed DNA polymerases travel along the original DNA strands making complementary strands (Figure \(6\)a).
DNA replication in both prokaryotes and eukaryotes begins at an Origin of Replication (Ori). Origins are specific sequences on specific positions on the chromosome. In E. coli, the OriC origin is ~245 bp in size. Chromosome replication begins with the binding of the DnaA initiator protein to an AT-rich 9-mer in OriC and melts the two strands. Then DnaC loader protein helps DnaB helicase protein extend the single stranded regions such that the DnaG primase can initiate the synthesis of an RNA primer, from which the DNA polymerases can begin DNA synthesis at the two replication forks. The forks continue in opposite directions until they meet another fork or the end of the chromosome (Figure \(7\)).
The ends of linear chromosomes present a problem – at each end one strand cannot be completely replicated because there is no primer to extend. Although the loss of such a small sequence would not be a problem, the continued rounds of replication would result in the continued loss of sequence from the chromosome end to a point were it would begin to loose essential gene sequences. Thus, this DNA must be replicated. Most eukaryotes solve the problem of synthesizing this unreplicated DNA with a specialized DNA polymerase called telomerase, in combination with a regular polymerase. Telomerases are RNA-directed DNA polymerases. They are a riboprotein, as they are composed of both protein and RNA. As Figure \(6\)b shows, these enzymes contain a small piece of RNA that serves as a portable and reusable template from which the complementary DNA is synthesized. The RNA in human telomerases uses the sequence 3-AAUCCC-5' as the template, and thus our telomere DNA has the complementary sequence 5'-TTAGGG-3' repeated over and over 1000’s of times. After the telomerase has made the first strand a primase synthesizes an RNA primer and a regular DNA polymerase can then make a complementary strand so that the telomere DNA will ultimately be double stranded to the original length (Figure \(8\)). Note: the number of repeats, and thus the size of the telomere, is not set. It fluctuates after each round of the cell cycle. Because there are many repeats at the end, this fluctuation maintains a length buffer – sometimes it’s longer, sometimes it’s shorter – but the average length will be maintained over the generations of cell replication.
In the absence of telomerase, as is the case in human somatic cells, repeated cell division leads to the “Hayflick limit”, where the telomeres shorten to a critical limit and then the cells enter a senescence phase of non-growth. The activation of telomerase expression permits a cell and its descendants to become immortal and bypass the Hayflick limit. This happens in cancer cells, which can form tumours as well as in cells in culture, such as HeLa cells, which can be propagated essentially indefinitely. HeLa cells have been kept in culture since 1951.
Eukaryote Chromosomes have Multiple Origins of Replication
In prokaryotes, with a small, simple, circular chromosome, only one origin of replication is needed to replicate the whole genome. For example, E. coli has a ~4.5 Mb genome (chromosome) that can be duplicated in ~40 minutes assuming a single origin, bi-directional replication, and a speed of ~1000 bases/second/fork for the polymerase.
However, in larger, more complicated eukaryotes, with multiple linear chromosomes, more than one origin of replication is required per chromosome to duplicate the whole chromosome set in the 8-hours of S-phase of the cell cycle. For example, the human diploid genome has 46 chromosomes (6 x 109 basepairs). The shortest chromosomes are ~50 Mbp long and so could not possibly be replicated from one origin. Additionally, the rate of replication fork movement is slower, only ~100 base/second. Thus, eukaryotes contain multiple origins of replication distributed over the length of each chromosome to enable the duplication of each chromosome within the observed time of S-phase (Fig 2.9). | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/02%3A_Chromosomes_Mitosis_and_Meiosis/2.01%3A_DNA_is_Packaged_into_Chromatin.txt |
Cell growth and division is essential to asexual reproduction and the development of multicellular organisms. Accordingly, the primary function of mitosis is to ensure that at division each daughter cell inherits identical genetic material, i.e. exactly one copy of each chromosome. To make this happen, replicated chromosomes condense (prophase), and are positioned near the middle of the dividing cell (metaphase), and then each of the sister chromatids from each chromosome migrates towards opposite poles of the dividing cell (anaphase), until the identical sets of unreplicated chromosomes are completely separated from each other within the two newly formed daughter cells (telophase) (Figures 2.10 and 2.11). This is followed by the division of the cytoplasm (cytokinesis) to complete the process. The movement of chromosomes occurs through microtubules that attach to the chromosomes at the centromeres.
2.03: Meiosis
Most eukaryotes replicate sexually - a cell from one individual joins with a cell from another to create the next generation. For this to be successful, the cells that fuse must contain half the number of chromosomes as in the adult organism. Otherwise, the number of chromosomes would double with each generation! The reduction in chromosome number is achieved by the process of meiosis. In meiosis, there are usually two steps, Meiosis I and II. In Meiosis I homologous chromosomes segregate, while in Meiosis II sister chromatids segregate. Most multicellular organisms use meiosis to produce gametes, the cells that fuse to make offspring. Some single celled eukaryotes such as yeast also use meiosis.
Meiosis I and II
Meiosis begins similarly to mitosis (a cell has replicated its chromosomes and grown large enough to divide), but requires two rounds of division (Figure \(8\)). In the first, known as meiosis I, the homologous chromosomes separate and segregate. During meiosis II the sister chromatids separate and segregate. Note how meosis I and II are both divided into prophase, metaphase, anaphase, and telophase. After two rounds of cytokinesis, four cells will be produced, each with a single copy of each chromosome.
Cells that will undergo the process of meiosis are called meiocytes and are diploid (2N). Meiosis is divided into two stages designated by the roman numerals I and II. Meiosis I is called a reductional division, because it reduces the number of chromosomes inherited by each of the daughter cells. Meiosis I is further divided into Prophase I, Metaphase I, Anaphase I, and Telophase I, which are roughly similar to the corresponding stages of mitosis, except that in Prophase I and Metaphase I, homologous chromosomes pair with each other, or synapse, and are called bivalents (Figs. 2.12). This is an important difference between mitosis and meiosis, because it affects the segregation of alleles, and also allows for recombination to occur through crossing-over, as described later. During Anaphase I, one member of each pair of homologous chromosomes migrates to each daughter cell (1N). Meiosis II resembles mitosis, with one sister chromatid from each chromosome separating to produce two daughter cells. Because Meiosis II, like mitosis, results in the segregation of sister chromatids, Meiosis II is called an equational division.
In meiosis I replicated, homologous chromosomes pair up, or synapse, during prophase I, lining up in the middle of the cell during metaphase I, and separating during anaphase I. For this to happen the homologous chromosomes need to be brought together while they condense during prophase I. These attachments are formed in two ways. Proteins bind to both homologous chromosomes along their entire length and form the synaptonemal complex (synapse means junction). These proteins hold the chromosomes in a transient structure called a bivalent. The proteins are released when the cell enters anaphase I.
Within the synaptonemal complex a second event, crossingover, occurs. These are places where DNA repair enzymes break the DNA two non-sister chromatids in similar locations and then covalently reattach non-sister chromatids together to create a crossover between non-sister chromatids. This reorganization of chromatids will persist for the remainder of meiosis and result in recombination of alleles in the gametes.
Crossovers function to hold homologous chromosomes together during meiosis I so they segregate successfully and they also cause the reshuffling of gene/allele combinations to create genetic diversity, which can have an important effect on evolution (see Chapter 7).
Stages of Prophase I
In meiosis, Prophase I is divided up into five visual stages, that are steps along a continuum of events. Leptotene, zygotene, pachytene, diplotene and diakinesis. From interphase, a cell enters leptotene as the nuclear material begins to condense into long visible threads (chromosomes). During Zygotene homologous chromosomes begin to pair up (synapse) and form an elaborate structure called the synaptonemal complex along their length. At pachytene homologous chromosomes are fully synapsed (two chromosomes and four chromatids) to form bivalents. Crossing over takes place in pachytene. After this, the pairing begins to loosen and individual chromatids become apparent in diplotene. This is when the consequences of each crossing over event can be seen as a chiasma (plural: chiasmata). Diakinesis follows as the chromosomes continue to condense and individualize. This is followed by metaphase I were the paired chromosomes orient on the metaphase plate in preparation for segregation (reductional).
Meiosis II and Gamete Maturation
At the completion of meiosis I there are two cells, each with one, replicated copy of each chromosome (1N). Because the number of chromosomes per cell has decreased (2->1), meiosis I is called a reductional cell division. In the second part of meiosis the chromosomes will once again be brought to the middle of the cell, but this time it is the sister chromatids that will segregate during anaphase.
After cytokinesis there will be four cells, each containing only one unreplicated chromosome of each type. Meiosis II resembles mitosis in that the number of chromosomes per cell is unchanged - both are equational cell divisions – but in meiosis II all cells won’t have the same genetic composition. There will be allelic differences among the gametes.
In animals and plants the cells produced by meiosis need to mature before they become functional gametes. In male animals the four products of meiosis are called spermatids. They grow tails and become functional sperm cells. In female animals the gametes are eggs. In order that each egg contains the maximum amount of nutrients only one of the four products of meiosis becomes an egg. The other three cells end up as tiny disposable cells called polar bodies. In plants the products of meiosis reproduce a few times using mitosis as they develop into functional male or female gametes.
2.04: The Cell Cycle and Changes in DNA Content
Four stages of a typical cell cycle
The life cycle of eukaryotic cells can generally be divided into four stages and a typical cell cycle is shown in Figure \(13\). When a cell is produced through fertilization or cell division, there is usually a lag before it undergoes DNA synthesis (replication). This lag period is called Gap 1 (G1), and ends with the onset of the DNA synthesis (S) phase, during which each chromosome is replicated. Following replication, there may be another lag, called Gap 2 (G2), before mitosis (M). Cells undergoing meiosis do not usually have a G2 phase. Interphase is as term used to include those phases of the cell cycle excluding mitosis and meiosis. Many variants of this generalized cell cycle also exist. Some cells never leave G1 phase, and are said to enter a permanent, non-dividing stage called G0. On the other hand, some cells undergo many rounds of DNA synthesis (S) without any mitosis or cell division, leading to endoreduplication. Understanding the control of the cell cycle is an active area of research, particularly because of the relationship between cell division and cancer.
Measures of DNA content and chromosome content
The amount of DNA within a cell changes following each of the following events: fertilization, DNA synthesis, mitosis, and meiosis (Fig 2.14). We use “c” to represent the DNA content in a cell, and “n” to represent the number of complete sets of chromosomes. In a gamete (i.e. sperm or egg), the amount of DNA is 1c, and the number of chromosomes is 1n. Upon fertilization, both the DNA content and the number of chromosomes doubles to 2c and 2n, respectively. Following DNA replication, the DNA content doubles again to 4c, but each pair of sister chromatids is still counted as a single chromosome (a replicated chromosome), so the number of chromosomes remains unchanged at 2n. If the cell undergoes mitosis, each daughter cell will return to 2c and 2n, because it will receive half of the DNA, and one of each pair of sister chromatids. In contrast, the 4 cells that come from meiosis of a 2n, 4c cell are each 1c and 1n, since each pair of sister chromatids, and each pair of homologous chromosomes, divides during meiosis. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/02%3A_Chromosomes_Mitosis_and_Meiosis/2.02%3A_Mitosis.txt |
Karyograms are images of real chromosomes
Each eukaryotic species has its nuclear genome divided among a number of chromosomes that is characteristic of that species. For example, a haploid human nucleus (i.e. sperm or egg) normally has 23 chromosomes (n=23), and a diploid human nucleus has 23 pairs of chromosomes (2n=46). A karyotype is the complete set of chromosomes of an individual. The cell was in metaphase so each of the 46 structures is a replicated chromosome even though it is hard to see the two sister chromatids for each chromosome at this resolution. As expected there are 46 chromosomes. Note that the chromosomes have different lengths. In fact, human chromosomes were named based upon this feature. Our largest chromosome is called 1, our next longest is 2, and so on. By convention the chromosomes are arranged into the pattern shown in Figure \(15\) and the resulting image is called a karyogram. A karyogram allows a geneticist to determine a person's karyotype - a written description of their chromosomes including anything out of the ordinary.
Figure \(15\): Karyogram of a normal human male karytype.(Wikipedia-NHGRI-PD)
Various stains and fluorescent dyes are used to produce characteristic banding patterns to distinguish all 23 chromosomes. The number of chromosomes varies between species, but there appears to be very little correlation between chromosome number and either the complexity of an organism or its total amount genomic DNA.
Autosomes and Sex Chromosomes
In the figure above note that most of the chromosomes are paired (same length, centromere location, and banding pattern). These chromosomes are called autosomes. However note that two of the chromosomes, the X and the Y do not look alike. These are sex chromosomes. In humans males have one of each while females have two X chromosomes. Autosomes are those chromosomes present in the same number in males and females while sex chromosomes are those that are not. When sex chromosomes were first discovered their function was unknown and the name X was used to indicate this mystery. The next ones were named Y, then Z, and then W.
The combination of sex chromosomes within a species is associated with either male or female individuals. In mammals, fruit flies, and some flowering plants embryos, those with two X chromosomes develop into females while those with an X and a Y become males. In birds, moths, and butterflies males are ZZ and females are ZW. Because sex chromosomes have arisen multiple times during evolution the molecular mechanism(s) through which they determine sex differs among those organisms. For example, although humans and Drosophila both have X and Y sex chromosomes, they have different mechanisms for determining sex .
In mammals, the sex chromosomes evolved just after the divergence of the monotreme lineage from the lineage that led to placental and marsupial mammals. Thus nearly every mammal species uses the same sex determination system. During embryogenesis the gonads will develop into either ovaries or testes. A gene present only on the Y chromosome called TDF encodes a protein that makes the gonads mature into testes. XX embryos do not have this gene and their gonads mature into ovaries instead (default). Once formed the testes produce sex hormones that direct the rest of the developing embryo to become male, while the ovaries make different sex hormones that promote female development. The testes and ovaries are also the organs where gametes (sperm or eggs) are produced.
How do the sex chromosome behave during meiosis? Well, in those individuals with two of the same chromosome (i.e. homogametic sexes: XX females and ZZ males) the chromosomes pair and segregate during meiosis I the same as autosomes do. During meiosis in XY males or ZW females (heterogametic sexes) the sex chromosomes pair with each other (Figure \(16\)). In mammals the consequence of this is that all egg cells will carry an X chromosome while the sperm cells will carry either an X or a Y chromosome. Half of the offspring will receive two X chromosomes and become female while half will receive an X and a Y and become male.
Figure \(16\): Meiosis in an XY mammal. The stages shown are anaphase I, anaphase II, and mature sperm. Note how half of the sperm contain Y chromosomes and half contain X chromosomes. (Original-Harrington-CC:AN)
Aneuploidy - Changes in Chromosome Number
Analysis of karyotypes can identify chromosomal abnormalities, including aneuploidy, which is the addition or subtraction of a chromosome from a pair of homologs. More specifically, the absence of one member of a pair of homologous chromosomes is called monosomy (only one remains). On the other hand, in a trisomy, there are three, rather than two (disomy), homologs of a particular chromosome. Different types of aneuploidy are sometimes represented symbolically; if 2n symbolizes the normal number of chromosomes in a cell, then 2n-1 indicates monosomy and 2n+1 represents trisomy. The addition or loss of a whole chromosome is a mutation, a change in the genotype of a cell or organism.
The most familiar human aneuploidy is trisomy-21 (i.e. three copies of chromosome 21), which is one cause of Down syndrome. Most (but not all) other human aneuploidies are lethal at an early stage of embryonic development. Note that aneuploidy usually affects only one set of homologs within a karyotype, and is therefore distinct from polyploidy, in which the entire chromosome set is duplicated (see below). Aneuploidy is almost always deleterious, whereas polyploidy appears to be beneficial in some organisms, particularly many species of food plants.
Aneuploidy can arise due to a non-disjunction event, which is the failure of at least one pair of chromosomes or chromatids to segregate during mitosis or meiosis. Non-disjunction will generate gametes with extra and missing chromosomes.
Chromosomal abnormalities
Structural defects in chromosomes are another type of abnormality that can be detected in karyotypes (Fig 2.17). These defects include deletions, duplications, and inversions, which all involve changes in a segment of a single chromosome. Insertions and translocations involve two non-homologous chromosomes. In an insertion, DNA from one chromosome is moved to a non-homologous chromosome in a unidirectional manner. In a translocation, the transfer of chromosomal segments is bidirectional and reciprocal – a reciprocal translocation.
Figure \(17\): Structural abberations in chromosomes.(Wikipedia-Zephyris-GFDL)
Structural defects affect only part of a chromosome (a subset of genes), and so tend to be less harmful than aneuploidy. In fact, there are many examples of ancient chromosomal rearrangements in the genomes of species including our own. Duplications of some small chromosomal segments, in particular, may have some evolutionary advantage by providing extra copies of some genes, which can then evolve in new, potentially beneficial, ways.
Chromosomal abnormalities arise in many different ways, some of which can be traced to rare errors in natural cellular processes such as DNA replication. Chromosome breakage also occurs infrequently as the result of physical damage (such as ionizing radiation), movement of some types of transposons, and other factors. During the repair of a broken chromosome, deletions, insertions, translocations and even inversions can be introduced. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/02%3A_Chromosomes_Mitosis_and_Meiosis/2.05%3A_Karyotypes_Describe_Chromosome_Number_and_Structure.txt |
If you refer back to Figure \(15\) you can see that humans, like most animals and most eukaryotic genetic model organisms, have two copies of each autosome. This situation is called diploidy. This means that most of their cells have two homologous copies of each chromosome. In contrast, many plant species and even a few animal species are polyploids. This means they have more than two chromosome sets, and so have more than two homologs of each chromosome in each cell.
When the nuclear content changes by a whole chromosome set we call it a change in ploidy. Gametes are haploid (1n) and thus most animals are diploid (2n), formed by the fusion of two haploid gametes. However, some species can exist as monoploid (1x), triploid (3x), tetraploid (4x), pentaploid (5x), hexaploid (6x), or higher.
Notation of ploidy
When describing polyploids, we use the letter “x” (not “n”) to define the level of ploidy. A diploid is 2x, because there are two basic sets of chromosomes, and a tetraploid is 4x, because it contains four chromosome sets. For clarity when discussing polyploids, geneticists will often combine the “x” notation with the “n” notation already defined previously in this chapter. Thus for both diploids and polyploids, “n” is the number of chromosomes in a gamete, and “2n” is the number of chromosomes following fertilization. For a diploid, therefore, n=x, and 2n=2x. But for a tetraploid, n=2x, and 2n=4x and for a hexaploid, n=3x, and 2n=6x.
Male Bees are Monoploid
Monoploids, with only one set, are usually inviable in most species, however, in many species of hymenoptera (bees, wasps, ants) the males are monoploid and develop from unfertilized eggs. These males don’t undergo meiosis for gametes; mitosis produces sperm. Females are diploid (from fertilized eggs) and produce eggs via meiosis. This is the basis for the haploid-diploid sex determination system (not the X/Y chromosome system). Female bees are diploid (2n=32) and are formed when an egg (n=16) is fertilized by a sperm (n=16). If an egg isn't fertilized it can still develop and the result is a n=16 male drone. Males are described as haploid (because they have the same number of chromosomes as a gamete) or monoploid (because they have only one chromosome set). Females produce eggs by meiosis while males produce sperm by mitosis. This form of sex determination produces more females – workers, which do the work (Figure 2-18) than males, who are only needed for reproduction.
Figure \(18\): A worker caste European honey bee, which is female and diploid. Male drones are haploid. (Wikipedia-J. Severns-PD)
Polyploids can be stable or sterile
Like diploids (2n=2x), stable polyploids generally have an even number of copies of each chromosome: tetraploid (2n=4x), hexaploid (2n=6x), and so on. The reason for this is clear from a consideration of meiosis. Remembering that the purpose of meiosis is to reduce the sum of the genetic material by half, meiosis can equally divide an even number of chromosome sets, but not an odd number. Thus, polyploids with an odd number of chromosomes (e.g. triploids, 2n=3x) tend to be sterile, even if they are otherwise healthy.
The mechanism of meiosis in stable polyploids is essentially the same as in diploids: during metaphase I, homologous chromosomes pair with each other. Depending on the species, all of the homologs may be aligned together at metaphase, or in multiple separate pairs. For example, in a tetraploid, some species may form tetravalents in which the four homologs from each chromosome align together, or alternatively, two pairs of homologs may form two bivalents. Note that because that mitosis does not involve any pairing of homologous chromosomes, mitosis is equally effective in diploids, even-number polyploids, and odd-number polyploids.
Many Crop Plants are Hexaploid or Octoploid
Polyploid plants tend to be larger and healthier than their diploid counterparts. The strawberries sold in grocery stores come from octoploid (8x) strains and are much larger than the strawberries formed by wild diploid strains. An example is bread wheat which is a hexaploid (6x) strain. This species is derived from the combination of three other wheat species, T. monococcum (chromosome sets = AA), T. searsii (BB), and T. tauschii (DD). Each of these chromosome sets has 7 chromosomes so the diploid species are 2n=2x=14 and bread wheat is 2n=6x=42 and has the chromosome sets AABBDD. Bread wheat is viable because each chromosome behaves independently during mitosis. The species is also fertile because during meiosis I the A chromosomes pair with the other A chromosomes, and so on. Thus, even in a polyploid, homologous chromosomes can segregate equally and gene balance can be maintained.
Bananas, Watermelons, and Other Seedless Plants are Triploid
The bananas found in grocery stores are a seedless variety called Cavendish. They are a triploid variety (chromosome sets = AAA) of a normally diploid species called Musa acuminata (AA). Cavendish plants are viable because mitosis can occur. However they are sterile because the chromosomes cannot pair properly during meiosis I. During prophase I there are three copies of each chromosome trying to “pair” with each other. Because proper chromosome segregation in meiosis fails, seeds cannot be made and the result is a fruit that is easier to eat because there are no seeds to spit out. Seedless watermelons (Figure \(19\)) have a similar explanation.
Figure \(19\): Seedless watermelon is triploid, with the white, aborted seeds within the flesh. (Flickr-Darwin Bell-CC:AN)
If triploids cannot make seeds, how do we obtain enough triploid individuals for cultivation? The answer depends on the plant species involved. In some cases, such as banana, it is possible to propagate the plant asexually; new progeny can simply be grown from cuttings from a triploid plant. On the other hand, seeds for seedless watermelon are produced sexually: a tetraploid watermelon plant is crossed with a diploid watermelon plant. Both the tetraploid and the diploid are fully fertile, and produce gametes with two (1n=2x) or one (1n=1x) sets of chromosomes, respectively. These gametes fuse to produce a zygote (2n=3x) that is able to develop normally into an adult plant through multiple rounds of mitosis, but is unable to compete normal meiosis or produce seeds.
Polyploids are often larger in size than their diploid relatives (Figure \(20\)). This feature is used extensively in food plants. For example, most strawberries you eat are not diploid, but octoploid (8x).
Polyploidy in animals is rare, essentially limited to lower forms, which often reproduce by parthenogenesis.
Credit: www.jamesandthegiantcorn.com/wp-content/uploads/2009/11/strawberries2.jpg | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/02%3A_Chromosomes_Mitosis_and_Meiosis/2.06%3A_Polyploidy_Arises_from_Changes_in_Whole_Sets_of_Chromosomes.txt |
Endoreduplication, is a special type of tissue-specific genome amplification that occurs in many types of plant cells and in specialized cells of some animals including humans. Endoreduplication does not affect the germline or gametes, so species with endoreduplication are not considered polyploids. Endoreduplication occurs when a cell undergoes extra rounds of DNA synthesis (S-phase) without any mitosis or cytokinesis to produce an endopolyploid cell. This produces multiple chromatids of each chromosome. Endopolyploidy seems to be associated with cells that are metabolically very active, and produce a lot of enzymes and other proteins in a short period of time. An example is the highly endoreduplicated salivary gland polytene chromosomes of D. melanogaster (Figure \(21\)) which can have over 1,000 chromatids that align together and form giant chromosomes that show a banding pattern that reflects the underlying DNA sequence and genes in that chromosome region. These chromosomes have been wonderful research models in genetics, since their relatively large, amplified size makes it easy to identify and study a wide variety of chromosome aberrations under the microscope.
Figure \(21\): Endoreduplicated chromosomes from a Drosophila salivary gland cell. The banding pattern is produced with fluorescent labels. (Flickr-Elissa Lei, Ph.D. @ NIH-CCA)
2.08: Gene Balance
Why do trisomies, duplications, and other chromosomal abnormalities that alter gene copy number often have a negative effect on the normal development or physiology of an organism? This is particularly intriguing because in many species, aneuploidy is detrimental or lethal, while polyploidy is tolerated or even beneficial. The answer probably differs in each case, but is probably related to the concept of gene balance, which can be summarized as follows: genes, and the proteins they produce, have evolved to function in complex metabolic and regulatory networks. Some of these networks function best when certain enzymes and regulators are present in specific ratios to each other. Increasing or decreasing the gene copy number for just one part of the network may throw the whole network out of balance, leading to increases or decreases of certain metabolites, which may be toxic in high concentrations or which may be limiting in other important processes in the cell. The activity of genes and metabolic networks is regulated in many different ways besides changes in gene copy number, so duplication of just a few genes will usually not be harmful. However, trisomy and large segmental duplications of chromosomes affect the dosage of so many genes that cellular networks are unable to compensate for the changes and an abnormal or lethal phenotype results.
2.09: Organellar Genomes
In eukaryotes, DNA and genes also exist outside of the chromosomes found in the nucleus. Both the chloroplast and mitochondrion have circular chromosomes (Figure \(22\)). These organellar genomes are often present in multiple copies within each organelle. In most sexually reproducing species, organellar chromosomes are inherited from only one parent, usually the one that produces the largest gamete. Thus, in mammals, angiosperms, and many other organisms, mitochondria and chloroplasts are inherited only through the mother (maternally).
These organelles are likely the remnants of prokaryotic endosymbionts that entered the cytoplasm of ancient progenitors of today’s eukaryotes (endosymbiont theory). These endosymbionts had their own, circular chromosomes, like most bacteria that exist today. Chloroplasts and mitochondria typically have circular chromosomes that behave more like bacterial chromosomes than eukaryotic chromosomes, i.e. these organellar genomes do not undergo mitosis or meiosis. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/02%3A_Chromosomes_Mitosis_and_Meiosis/2.07%3A_Endoreduplication.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
2.1 Define chromatin. What is the difference between DNA, chromatin and chromosomes?
2.2 Species A has n=4 chromosomes and Species B has n=6 chromosomes. Can you tell from this information which species has more DNA? Can you tell which species has more genes?
2.3 The answer to question 2 implies that not all DNA within a chromosome encodes genes. Can you name any examples of chromosomal regions that contain relatively few genes?
2.4 a) How many centromeres does a typical chromosome have?
b) What would happen if there was more than one centromere per chromosome?
c) What if a chromosome had zero centromeres?
2.5 For a diploid with 2n=16 chromosomes, how many chromosomes and chromatids are per cell present in the gamete, and zygote and immediately following G1, S, G2, mitosis, and meiosis?
2.6 Bread wheat (Triticum aestivum) is a hexaploid. Using the nomenclature presented in class, an ovum cell of wheat has n=21 chromosomes. How many chromosomes in a zygote of bread wheat?
2.7 For a given gene:
a) What is the maximum number of alleles that can exist in a 2n cell of a given diploid individual?
b) What is the maximum number of alleles that can exist in a 1n cell of a tetraploid individual?
c) What is the maximum number of alleles that can exist in a 2n cell of a tetraploid individual?
d) What is the maximum number of alleles that can exist in a population?
2.8 a) Why is aneuploidy more often lethal than polyploidy?
b) Which is more likely to disrupt gene balance: polyploidy or duplication?
2.9 For a diploid organism with 2n=4 chromosomes, draw a diagram of all of the possible configurations of chromosomes during normal anaphase I, with the maternally and paternally derived chromosomes labelled.
2.10 For a triploid organism with 2n=3x=6 chromosomes, draw a diagram of all of the possible configurations of chromosomes at anaphase I (it is not necessary label maternal and paternal chromosomes).
2.11 For a tetraploid organism with 2n=4x=8 chromosomes, draw all of the possible configurations of chromosomes during a normal metaphase.
2.12 A simple mnemonic for leptotene, zygotene, pachytene, diplotene, & diakinesis is Lame Zebras Pee Down Drains. Make another one yourself.
2.S: Chromosomes Mitosis and Meiosis (Summary)
• Chromosomes are complex and dynamic structures consisting of DNA and proteins (chromatin).
• The degree of chromatin compaction involves proteins and varies between heterochromatic and euchromatic regions and among stages of the cell cycle.
• Chromosomes can be distinguished cytologicaly based on their length, centromere position, and banding patterns when stained dyes or labeled with sequence-specific probes.
• Homologous chromosomes contain the same series of genes along their length, but not necessarily the same alleles.Sister chromatids initially contain the same alleles.
• Chromosomes are replicated by DNA polymerases and begin at an origin. Replication is bi-directional. Eukaryotes have multiple origins along each chromosome and have telomerase to replicate the ends.
• Mitosis reduces the c-number, but not the n-number.Meiosis reduces both c and n.
• Homologous chromosomes pair (sysnapse) with each other during meiosis, but not mitosis.
• Several types of structural defects in chromosomes occur naturally, and can affect cellular function and even evolution.
• Aneuploidy results from the addition or subtraction of one or more chromosomes from a group of homologs, and is usually deleterious to the cell.
• Polyploidy is the presence of more than two complete sets of chromosomes in a genome.Even-numbered multiple sets of chromosomes can be stably inherited in some species, especially plants.
• Endopolyploidy is tissue-specific type of polyploidy observed in some species, including diploids.
• Both aneuploidy and structural defects such as duplications can affect gene balance.
• Organelles also contain chromosomes, but these are much more like prokaryotic chromosomes than the nuclear chromosomes of eukaryotes.
Key Terms
chromosome
core histones
nucleosome
30nm fiber
histone H1
scaffold proteins
heterochromatin
euchromatin
satellite DNA
chromatid
centromere
metacentric
acrocentric
telocentric
holocentric
telomere
homologous
non-homologous
chromatid
sister chromatid
non-sister chromatid
interphase
mitosis
prophase
metaphase
anaphase
telophase
DNA polymerase
origin of replication
telomerase
riboprotein
Hayflick limit
HeLa cells
cytokinesis
meiosis
gametes
prophase (I, II)
metaphase (I, II)
anaphase (I, II)
telophase (I, II)
cytokinesis
meiocyte
bivalent
syanapse, pair up
synaptonemal complex
reductional division
equational division
leptotene
zygotene
pachytene
diplotene
diakinesis
crossing over
chiasma (chiasmata)
polar bodies
G1
G2
S
M
G0
interphase
n
c
replicated chromosome
karyotype/karyogram
autosome
sex-chromosome
homogametic
heterogametic
aneuploidy
monsomic
trisomic
Down syndrome
deletion
duplication
insertion
inversion
translocation
non-disjunction
chromosome breakage
polyploidy
x
monoploid
sterile
tetravalent
octoploid
hexaploid
triploid
endoreduplication
endopolyploidy
salivary gland chromosome
polytene
gene balance
cellular network
chloroplast
mitochondria
endosymbiont
endosymbiont theory
organellar chromosome
mtDNA | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/02%3A_Chromosomes_Mitosis_and_Meiosis/2.E%3A_Chromosomes_Mitosis_and_Meiosis_%28Exercises%29.txt |
Before Mendel, the basic rules of heredity were not understood. For example, it was known that green-seeded pea plants occasionally produced offspring that had yellow seeds; but were the hereditary factors that controlled seed color somehow changing from one generation to the next, or were certain factors disappearing and reappearing? And did the same factors that controlled seed color also control things like plant height?
• 3.1: Mendel’s First Law
Mendel’s First Law, also called The Law of Equal Segregation, states that during gamete formation, the two alleles at a gene locus segregate from each other; each gamete has an equal probability of containing either allele. More than one allele of a gene can be present in an individual since most eukaryotic organisms have at least two sets of homologous chromosomes. For organisms that are predominantly diploid, chromosomes exist as pairs, with one homolog inherited from each parent.
• 3.2: Relationships Between Genes, Genotypes and Phenotypes
A specific position along a chromosome is called a locus and each gene occupies a specific locus; each locus will have an allelic form. The complete set of alleles (at all loci of interest) in an individual is its genotype. The visible or detectable effect of these alleles on the structure or function of that individual is called its phenotype
• 3.3: Biochemical Basis of Dominance
For the majority of genes studied, the normal (i.e. wild-type) alleles are haplosufficient. So in diploids, even with a mutation that causes a complete loss of function in one allele, the other allele, a wild-type allele, will provide sufficient normal biochemical activity to yield a wild type phenotype and thus be dominant and dictate the heterozygote phenotype.
• 3.4: Crossing Techniques Used in Classical Genetics
Mendel also invented several testing and analysis techniques still used today. Classical genetics is the science of solving biological questions using controlled matings of model organisms. It began with Mendel in 1865 but did not take off until Thomas Morgan began working with fruit flies in 1908. Later, starting with Watson and Crick’s structure of DNA in 1953, classical genetics was joined by molecular genetics, the science of solving biological questions using DNA, RNA, and proteins isolated
• 3.5: Sex-Linkage- An Exception to Mendel’s First Law
In the previous chapter we introduced sex chromosomes and autosomes. For loci on autosomes, the alleles follow the normal Mendelian pattern of inheritance. However, for loci on the sex chromosomes this is mostly not true, because most of the loci on the typical X-chromosome are absent from the Y-chromosome, even though they act as a homologous pair during meiosis. Instead, they will follow a sex-linked pattern of inheritance.
• 3.6: Phenotypes May Not Be As Expected from the Genotype
The phenotypes described thus far have a nearly perfect correlation with their associated genotypes; in other words an individual with a particular genotype always has the expected phenotype. However, many phenotypes are not determined entirely by genotype alone. They are instead determined by an interaction between genotype and non-genetic, environmental factors.
• 3.7: Phenotypic Ratios May Not Be As Expected
For a variety of reasons, the phenotypic ratios observed from real crosses rarely match the exact ratios expected based on a Punnett Square or other prediction techniques. There are many possible explanations for deviations from expected ratios. Sometimes these deviations are due to sampling effects, in other words, the random selection of a non-representative subset of individuals for observation. On the other hand, it may be because certain genotypes have a less than 100% survival rate.
• 3.E: Genetic Analysis of Single Genes (Exercises)
• 3.S: Genetic Analysis of Single Genes (Summary)
Thumbnail: Pea plants were used by Gregor Mendel to discover some fundamental laws of genetics. (Flicker-Christian Guthier-CC:A)
03: Genetic Analysis of Single Genes
Character Traits Exist in Pairs that Segregate at Meiosis
Through careful study of patterns of inheritance, Mendel recognized that a single trait could exist in different versions, or alleles, even within an individual plant or animal. For example, he found two allelic forms of a gene for seed color: one allele gave green seeds, and the other gave yellow seeds. Mendel also observed that although different alleles could influence a single trait, they remained indivisible and could be inherited separately. This is the basis of Mendel’s First Law, also called The Law of Equal Segregation, which states: during gamete formation, the two alleles at a gene locus segregate from each other; each gamete has an equal probability of containing either allele.
Hetero-, Homo-, Hemizygosity
Mendel’s First Law is especially remarkable because he made his observations and conclusions (1865) without knowing about the relationships between genes, chromosomes, and DNA. We now know the reason why more than one allele of a gene can be present in an individual: most eukaryotic organisms have at least two sets of homologous chromosomes. For organisms that are predominantly diploid, such as humans or Mendel’s peas, chromosomes exist as pairs, with one homolog inherited from each parent. Diploid cells therefore contain two different alleles of each gene, with one allele on each member of a pair of homologous chromosomes. If both alleles of a particular gene are identical, the individual is said to be homozygous for that gene. On the other hand, if the alleles are different from each other, the genotype is heterozygous. In cases where there is only one copy of a gene present, for example if there is a deletion on the homologous chromosome, we use the term hemizygous.
Although a typical diploid individual can have at most two different alleles of a particular gene, many more than two different alleles can exist in a population of individuals. In a natural population the most common allelic form is usually called the wild-type allele. However, in many populations there can be multiple variants at the DNA sequence level that are visibly indistinguishable as all exhibit a normal, wild type appearance. There can also be various mutant alleles (in wild populations and in lab strains) that vary from wild type in their appearance, each with a different change at the DNA sequence level. Such collections of mutations are known as an allelic series. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/03%3A_Genetic_Analysis_of_Single_Genes/3.01%3A__Mendels_First_Law.txt |
Terminology
A specific position along a chromosome is called a locus. Each gene occupies a specific locus (so the terms locus and gene are often used interchangeably). Each locus will have an allelic form (allele). The complete set of alleles (at all loci of interest) in an individual is its genotype. Typically, when writing out a genotype, only the alleles at the locus (loci) of interest are considered – all the others are present and assumed to be wild type. The visible or detectable effect of these alleles on the structure or function of that individual is called its phenotype – what it looks like. The phenotype studied in any particular genetic experiment may range from simple, visible traits such as hair color, to more complex phenotypes including disease susceptibility or behavior. If two alleles are present in an individual, then various interactions between them may influence their expression in the phenotype.
Complete Dominance
Let us return to an example of a simple phenotype: flower color in Mendel’s peas. We have already said that one allele as a homozygote produces purple flowers, while the other allele as a homozygote produces white flowers (see Figures 1.8 and 3.3). But what about an individual that has one purple allele and one white allele; what is the phenotype of an individual whose genotype is heterozygous? This can only be determined by experimental observation. We know from observation that individuals heterozygous for the purple and white alleles of the flower color gene have purple flowers. Thus, the allele associated with purple color is therefore said to be dominant to the allele that produces the white color. The white allele, whose phenotype is masked by the purple allele in a heterozygote, is recessive to the purple allele.
To represent this relationship, often, a dominant allele will be represented by a capital letter (e.g. A) while a recessive allele will be represented in lower case (e.g. a). However, many different systems of genetic symbols are in use. The most common are shown in Table \(1\). Also note that genes and alleles are usually written in italics and chromosomes and proteins are not. For example, the white gene in Drosophila melanogaster on the X chromosome encodes a protein called WHITE.
Table \(1\): Examples of symbols used to represent genes and alleles.
Examples
Interpretation
A and a
Uppercase letters represent dominant alleles and lowercase letters indicate recessive alleles. Mendel invented this system but it is not commonly used because not all alleles show complete dominance and many genes have more than two alleles.
a+ and a1
Superscripts or subscripts are used to indicate alleles. For wild type alleles the symbol is a superscript +.
AA or A/A
Sometimes a forward slash is used to indicate that the two symbols are alleles of the same gene, but on homologous chromosomes.
Incomplete Dominance
Besides dominance and recessivity, other relationships can exist between alleles. In incomplete dominance (also called semi-dominance, Figure \(4\)), both alleles affect the trait additively, and the phenotype of the heterozygote is intermediate between either of the homozygotes. For example, alleles for color in carnation flowers (and many other species) exhibit incomplete dominance. Plants with an allele for red petals (A1) and an allele for white petals (A2) have pink petals. We say that the A1 and the A2 alleles show incomplete dominance because neither allele is completely dominant over the other.
Co-Dominance
Co-dominance is another type of allelic relationship, in which a heterozygous individual expresses the phenotype of both alleles simultaneously. An example of co-dominance is found within the ABO blood group of humans. The ABO gene has three common alleles which were named (for historical reasons) IA, IB, and i. People homozygous for IA or IB display only A or B type antigens, respectively, on the surface of their blood cells, and therefore have either type A or type B blood (Figure \(5\)). Heterozygous IAIB individuals have both A and B antigens on their cells, and so have type AB blood. Notice that the heterozygote expresses both alleles simultaneously, and is not some kind of novel intermediate between A and B. Co-dominance is therefore distinct from incomplete dominance, although they are sometimes confused.
Figure \(5\): Relationship between genotype and phenotype for three alleles of the human ABO gene. The IA and IB alleles show co-dominance. The IA allele is completely dominant to the i allele. The IB allele is completely dominant to the i allele. (Original-Deholos -CC:AN)
It is also important to note that the third allele, i, does not make either antigen and is recessive to the other alleles. IA/i or IB/i individuals display A or B antigens respectively. People homozygous for the i allele have type O blood. This is a useful reminder that different types of dominance relationships can exist, even for alleles of the same gene. Many types of molecular markers, which we will discuss in a later chapter, display a co-dominant relationship among alleles. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/03%3A_Genetic_Analysis_of_Single_Genes/3.02%3A_Relationships_Between_Genes_Genotypes_and_Phenotypes.txt |
Given that a heterozygote’s phenotype cannot simply be predicted from the phenotype of homozygotes, what does the type of dominance tell us about the biochemical nature of the gene product? How does dominance work at the biochemical level? There are several different biochemical mechanisms that may make one allele dominant to another.
For the majority of genes studied, the normal (i.e. wild-type) alleles are haplosufficient. So in diploids, even with a mutation that causes a complete loss of function in one allele, the other allele, a wild-type allele, will provide sufficient normal biochemical activity to yield a wild type phenotype and thus be dominant and dictate the heterozygote phenotype.
On the other hand, in some biochemical pathways, a single wild-type allele is not enough protein and may be haploinsufficient to produce enough biochemical activity to result in a normal phenotype, when heterozygous with a non-functioning mutant allele. In this case, the non-functional mutant allele will be dominant (or semi-dominant) to a wild-type allele.
Mutant alleles may also encode products that have new and/or different biochemical activities instead of, or in addition to, the normal ones. These novel activities could cause a new phenotype that would be dominantly expressed.
3.04: Crossing Techniques Used in Classical Genetics
Classical Genetics
Not only did Mendel solve the mystery of inheritance as units (genes), he also invented several testing and analysis techniques still used today. Classical genetics is the science of solving biological questions using controlled matings of model organisms. It began with Mendel in 1865 but did not take off until Thomas Morgan began working with fruit flies in 1908. Later, starting with Watson and Crick’s structure of DNA in 1953, classical genetics was joined by molecular genetics, the science of solving biological questions using DNA, RNA, and proteins isolated from organisms. The genetics of DNA cloning began in 1970 with the discovery of restriction enzymes.
True Breeding Lines
Geneticists make use of true breeding lines just as Mendel did (Figure \(6\)a). These are in-bred populations of plants or animals in which all parents and their offspring (over many generations) have the same phenotypes with respect to a particular trait. True breeding lines are useful, because they are typically assumed to be homozygous for the alleles that affect the trait of interest. When two individuals that are homozygous for the same alleles are crossed, all of their offspring will all also be homozygous. The continuation of such crosses constitutes a true breeding line or strain. A large variety of different strains, each with a different, true breeding character, can be collected and maintained for genetic research.
Monohybrid Crosses
A monohybrid cross is one in which both parents are heterozygous (or a hybrid) for a single (mono) trait. The trait might be petal colour in pea plants (Figure \(6\)b). Recall from chapter 1 that the generations in a cross are named P (parental), F1 (first filial), F2 (second filial), and so on.
Punnett Squares
Given the genotypes of any two parents, we can predict all of the possible genotypes of the offspring. Furthermore, if we also know the dominance relationships for all of the alleles, we can predict the phenotypes of the offspring. A convenient method for calculating the expected genotypic and phenotypic ratios from a cross was invented by Reginald Punnett. A Punnett square is a matrix in which all of the possible gametes produced by one parent are listed along one axis, and the gametes from the other parent are listed along the other axis. Each possible combination of gametes is listed at the intersection of each row and column. The F1 cross from Figure \(6\)b would be drawn as in Figure \(7\). Punnett squares can also be used to calculate the frequency of offspring. The frequency of each offspring is the frequency of the male gametes multiplied by the frequency of the female gamete.
A
a
A
AA
Aa
a
Aa
aa
Figure \(7\): A Punnett square showing a monohybrid cross. (Original-Deholos (Fireworks)-CC:AN)
Test Crosses
Knowing the genotypes of an individual is usually an important part of a genetic experiment. However, genotypes cannot be observed directly; they must be inferred based on phenotypes. Because of dominance, it is often not possible to distinguish between a heterozygote and a homozgyote based on phenotype alone (e.g. see the purple-flowered F2 plants in Figure \(6\)b). To determine the genotype of a specific individual, a test cross can be performed, in which the individual with an uncertain genotype is crossed with an individual that is homozygous recessive for all of the loci being tested.
For example, if you were given a pea plant with purple flowers it might be a homozygote (AA) or a heterozygote (Aa). You could cross this purple-flowered plant to a white-flowered plant as a tester, since you know the genotype of the tester is aa. Depending on the genotype of the purple-flowered parent (Figure \(8\)), you will observe different phenotypic ratios in the F1 generation. If the purple-flowered parent was a homozgyote, all of the F1 progeny will be purple. If the purple-flowered parent was a heterozygote, the F1 progeny should segregate purple-flowered and white-flowered plants in a 1:1 ratio.
A
A
a
Aa
Aa
a
Aa
Aa
A
a
a
Aa
aa
a
Aa
aa
Figure \(8\): Punnett Squares showing the two possible outcomes of a test cross. (Original-Deholos (Fireworks)-CC:AN) | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/03%3A_Genetic_Analysis_of_Single_Genes/3.03%3A__Biochemical_Basis_of_Dominance.txt |
In the previous chapter we introduced sex chromosomes and autosomes. For loci on autosomes, the alleles follow the normal Mendelian pattern of inheritance. However, for loci on the sex chromosomes this is mostly not true, because most of the loci on the typical X-chromosome are absent from the Y-chromosome, even though they act as a homologous pair during meiosis. Instead, they will follow a sex-linked pattern of inheritance.
X-Linked Genes: the white gene in Drosophila melanogaster
A well-studied sex-linked gene is the white gene on the X chromosome of Drosophila melanogaster. Normally flies have red eyes but flies with a mutant allele of this gene called white- (w-) have white eyes because the red pigments are absent. Because this mutation is recessive to the wild type w+ allele females that are heterozygous have normal red eyes. Female flies that are homozygous for the mutant allele have white eyes. Because there is no white gene on the Y chromosome, male flies can only be hemizygous for the wild type allele or the mutant allele.
A researcher may not know beforehand whether a novel mutation is sex-linked. The definitive method to test for sex-linkage is reciprocal crosses (Figure \(10\)). This means to cross a male and a female that have different phenotypes, and then conduct a second set of crosses, in which the phenotypes are reversed relative to the sex of the parents in the first cross. For example, if you were to set up reciprocal crosses with flies from pure-breeding w+ and w- strains the results would be as shown in Figure \(10\). Whenever reciprocal crosses give different results in the F1 and F2 and whenever the male and female offspring have different phenotypes the usual explanation is sex-linkage. Remember, if the locus were autosomal the F1 and F2 progeny would be different from either of these crosses.
A similar pattern of sex-linked inheritance is seen for X-chromosome loci in other species with an XX-XY sex chromosome system, including mammals and humans. The ZZ-ZW system is similar, but reversed (see below).
Sex Determination in animals.
There are various mechanisms for sex determination in animals. These include sex chromosomes, chromosome dosage, and environment.
For example in humans and other mammals XY embryos develop as males while XX embryos become females. This difference in development is due to the presence of only a single gene, the TDF-Y gene, on the Y-chromosome. Its presence and expression dictates that the sex of the individual will be male. Its absence results in a female phenotype.
Although Drosophila melanogaster also has an XX-XY sex chromosomes, its sex determination system uses a different method, that of X:Autosome (X:A) ratio. In this system it is the ratio of autosome chromosome sets (A) relative to the number of X-chromosomes (X) that determines the sex. Individuals with two autosome sets and two X-chromosomes (2A:2X) will develop as females, while those with only one X-chromosome (2A:1X) will develop as males. The presence/absence of the Y-chromosome and its genes are not significant.
In other species of animals the number of chromosome sets can determine sex. For example the haploid-diploid system is used in bees, ants, and wasps. Typically haploids are male and diploids are female.
In other species, the environment can determine an individuals sex. In alligators (and some other reptiles) the temperature of development dictates the sex, while in many reef fish, the population sex ratio can cause some individuals to change sex.
Dosage Compensation for Loci on Sex Chromosomes.
Mammals and Drosophila both have XX - XY sex determination systems. However, because these systems evolved independently they work differently with regard to compensating for the difference in gene dosage (and sex determination – see above). Remember, in most cases the sex chromosomes act as a homologous pair even though the Y-chromosome has lost most of the loci when compared to the X-chromosome. Typically, the X and the Y chromosomes were once similar but, for unclear reasons, the Y chromosomes have degenerated, slowly mutating and loosing its loci. In modern day mammals the Y chromosomes have very few genes left while the X chromosomes remain as they were. This is a general feature of all organisms that use chromosome based sex determination systems. Chromosomes found in both sexes (the X or the Z) have retained their genes while the chromosome found in only one sex (the Y or the W) have lost most of their genes. In either case there is a gene dosage difference between the sexes: e.g. XX females have two doses of X-chromosome genes while XY males only have one. This gene dosage needs to be compensated in a process called dosage compensation. There are two major mechanisms.
In Drosophila and many other insects, to make up for the males only having a single X chromosome the genes on it are expressed at twice the normal rate. This mechanism of dosage compensation restores a balance between proteins encoded by X-linked genes and those made by autosomal genes.
In mammals a different mechanism is used, called X-chromosome inactivation.
X-chromosome Inactivation in Mammals
In mammals the dosage compensation system operates in females, not males. In XX embryos one X in each cell is randomly chosen and marked for inactivation. From this point forward this chromosome will be inactive, hence its name Xinactive (Xi). The other X chromosome, the Xactive (Xa), is unaffected. The Xi is replicated during S phase and transmitted during mitosis the same as any other chromosome but most of its genes are never allowed to turn on. The chromosome appears as a condensed mass within interphase nuclei called the Barr body. With the inactivation of genes on one X-chromosome, females have the same number of functioning X-linked genes as males.
This random inactivation of one X-chromosome leads to a commonly observe phenomenon in cats. A familiar X-linked gene is the Orange gene (O) in cats. The OO allele encodes an enzyme that results in orange pigment for the hair. The OB allele causes the hairs to be black. The phenotypes of various genotypes of cats are shown in Figure \(11\). Note that the heterozygous females have an orange and black mottled phenotype known as tortoiseshell. This is due to patches of skin cells having different X-chromosomes inactivated. In each orange hair the Xi chromosome carrying the OB allele is inactivated. The OO allele on the Xa is functional and orange pigments are made. In black hairs the reverse is true, the Xi chromosome with the OO allele is inactive and the Xa chromosome with the OB allele is active. Because the inactivation decision happens early during embryogenesis, the cells continue to divide to make large patches on the adult cat skin where one or the other X is inactivated.
The Orange gene in cats is a good demonstration of how the mammalian dosage compensation system affects gene expression. However, most X-linked genes do not produce such dramatic mosaic phenotypes in heterozygous females. A more typical example is the F8 gene in humans. It makes Factor VIII blood clotting proteins in liver cells. If a male is hemizygous for a mutant allele the result is hemophilia type A. Females homozygous for mutant alleles will also have hemophilia. Heterozygous females, those people who are F8+/F8-, do not have hemophilia because even though half of their liver cells do not make Factor VIII (because the X with the F8+ allele is inactive) the other 50% can (Figure \(12\)). Because some of their liver cells are exporting Factor VIII proteins into the blood stream they have the ability to form blood clots throughout their bodies. The genetic mosaicism in the cells of their bodies does not produce a visible mosaic phenotype.
Other Sex-Linked Genes – Z-linked genes
One last example is a Z-linked gene that influences feather colour in turkeys. Turkeys are birds, which use the ZZ-ZW sex chromosome system. The E allele makes the feathers bronze and the e allele makes the feathers brown (Figure \(13\)). Only male turkeys can be heterozygous for this locus, because they have two Z chromosomes. They are also uniformly bronze because the E allele is completely dominant to the e allele and birds use a dosage compensation system similar to Drosophila and not mammals. Reciprocal crosses between turkeys from pure-breeding bronze and brown breeds would reveal that this gene is in fact Z-linked.
Mechanisms of Sex Determination Systems
Sex is a phenotype. Typically, in most species, there are multiple characteristics, in addition to sex organs, that distinguish male from female individuals (although some species are normally hermaphrodites where both sex organs are present in the same individual). The morphology and physiology of male and females is a phenotype just like hair or eye colour or wing shape. The sex of an organism is part of its phenotype and can be genetically (or environmentally) determined.
For each species, the genetic determination relies on one of several gene or chromosome based mechanisms. See Figure \(14\) for a summary. There are, for other species, also a variety of environmental mechanisms, too (rearing temperature, social interactions, parthenogenesis). Whatever the sex choice mechanism, however, there are two different means by which the cells of an organism carry out this decision: hormonal or cell-autonomous
Hormonal mechanism: With this system, used by mammals for example, including humans, the zygote initially develops into a sexually undifferentiated embryo that can become either sex. Then, depending on the sex choice of the genital ridge cells, they will grow and differentiate into male (testis) or female (ovary) gonads, which will then produce the appropriate hormones (e.g. testosterone or estrogen). This hormone will circulate throughout the body and cause all the other tissues to develop and differentiate accordingly, into a male or female phenotype for that individual. Thus, the circulating hormone “tells” all the cells and tissues what sex to be and which sexual phenotype to be.
A freemartin is a type of chimera found in cattle (and some other mammals). Externally it appears as a female but is infertile, and has masculinized behavior and non-functioning ovaries. The animal originates as a female (XX), but acquires male (XY) cells or tissues in utero by exchange of some cellular material from a male twin. The female reproductive development is altered by anti-Müllerian hormone from the male twin, acquired via vascular connections between placentas.
Cell-autonomous mechanism: With this system, used by many animals, including birds and insects, the zygote cell initially has a sex phenotype set at the cell level. All cells intrinsically know, individually, which sex they are and develop accordingly, giving the appropriate sexual characteristics and phenotype. Each cell is autonomous with respect to its sex; there are no sex hormone cues to determine the sex expressed. This autonomy can lead to sexual gynandromorphs, which are mosaics that display both male and female characteristics in a mosaic fasion, typically split down the midline of the organism. These rare individuals are thought to be the result of an improper sex chromosome segregation that occurs in a cell very early in development so that one half of the individual has cells with a male chromosome set while the other half has cells with a female set. If the species is sexually dimorphic (external morphology easily distinguishs males from females) they are easily visible and are even sometimes seen in the wild. See Figure \(15\) for a local example. A search on the internet will bring up many more examples.
While gynandromorphs are seen in cell-autonomous species, such as insects and birds, they are not seen in hormonally determined species, such as mammals, because all the cells display the same sex phenotype caused by the circulating sex hormones. Sexual gynandromorphs appear to be absent in reptiles, amphibians, and fish indicating that they don’t use a cell-autonomous mechanism. Nevertheless, there are genetic mosaic individuals in these groups but they do not appear to involve sex determined traits, which is required for a true gynandromorph. They often involve mosaicism of alleles at a single gene locus that affect external morphology (e.g. color).
• A gynandromorph is an organism that made up of mosaic tissues of male and female genotypes and displays both male and female characteristics.
• A mosaic is an organism or a tissue that contains two or more types of genetically different cells derived from the same zygote.
• A chimera is a single organism composed of genetically distinct cells derived from different zygotes. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/03%3A_Genetic_Analysis_of_Single_Genes/3.05%3A__Sex-Linkage-_An_Exception_to_Mendels_First_Law.txt |
Environmental Factors
The phenotypes described thus far have a nearly perfect correlation with their associated genotypes; in other words an individual with a particular genotype always has the expected phenotype. However, many phenotypes are not determined entirely by genotype alone. They are instead determined by an interaction between genotype and non-genetic, environmental factors and can be conceptualized in the following relationship:
$\mathrm{Genotype + Environment ⇒ Phenotype \hspace{30px} (G + E ⇒ P)}$
Or:
$\mathrm{Genotype + Environment + \overset{Genetics\: and\: Environment}{Interaction_{GE}} ⇒ Phenotype \hspace{30px} (G + E + I_{GE} ⇒ P)}$
This interaction is especially relevant in the study of economically important phenotypes, such as human diseases or agricultural productivity. For example, a particular genotype may pre-dispose an individual to cancer, but cancer may only develop if the individual is exposed to certain DNA-damaging chemicals. Therefore, not all individuals with the particular genotype will develop the cancer phenotype.
Penetrance and Expressivity
The terms penetrance and expressivity are also useful to describe the relationship between certain genotypes and their phenotypes.
• Penetrance is the proportion of individuals (usually expressed as a percentage) with a particular genotype that display a corresponding phenotype (Figure $16$). Because all pea plants that are homozygous for the allele for white flowers (e.g. aa in Figure $3$) actually have white flowers, this genotype is completely penetrant. In contrast, many human genetic diseases are incompletely penetrant, since not all individuals with the disease genotype actually develop symptoms associated with the disease.
• Expressivity describes the variability in mutant phenotypes observed in individuals with a particular phenotype (Figure $16$). Many human genetic diseases provide examples of broad expressivity, since individuals with the same genotypes may vary greatly in the severity of their symptoms. Incomplete penetrance and broad expressivity are due to random chance, non-genetic (environmental), and genetic factors (mutations in other genes).
3.07: Phenotypic Ratios May Not Be As Expected
Explanation
For a variety of reasons, the phenotypic ratios observed from real crosses rarely match the exact ratios expected based on a Punnett Square or other prediction techniques. There are many possible explanations for deviations from expected ratios. Sometimes these deviations are due to sampling effects, in other words, the random selection of a non-representative subset of individuals for observation. On the other hand, it may be because certain genotypes have a less than 100% survival rate. For example, Drosophila crosses sometimes give unexpected results because the more mutant alleles a zygote has the less likely it is to survive to become an adult. Genotypes that cause death for embryos or larvae are underrepresented when adult flies are counted.
The χTest For Goodness-of-fit
A statistical procedure called the chi-square (χ2) test can be used to help a geneticist decide whether the deviation between observed and expected ratios is due to sampling effects, or whether the difference is so large that some other explanation must be sought by re-examining the assumptions used to calculate the expected ratio. The procedure for performing a chi-square test is covered in the labs.
3.E: Genetic Analysis of Single Genes (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Q3.1
hat is the maximum number of alleles for a given locus in a normal gamete of a diploid species?
Q3.2
Wirey hair (W) is dominant to smooth hair (w) in dogs.
1. If you cross a homozygous, wirey-haired dog with a smooth-haired dog, what will be the genotype and phenotype of the F1 generation?
2. If two dogs from the F1 generation mated, what would be the most likely ratio of hair phenotypes among their progeny?
3. When two wirey-haired Ww dogs actually mated, they had alitter of three puppies, which all had smooth hair. How do you explain this observation?
4. Someone left a wirey-haired dog on your doorstep. Without extracting DNA, what would be the easiest way to determine the genotype of this dog?
5. Based on the information provided in question 1, can you tell which, if either, of the alleles is wild-type?
Q3.3
An important part of Mendel’s experiments was the use of homozygous lines as parents for his crosses. How did he know they were homozygous, and why was the use of the lines important?
Q3.4
Does equal segregation of alleles into daughter cells happen during mitosis, meiosis, or both?
Q3.5
If your blood type is B, what are the possible genotypes of your parents at the locus that controls the ABO blood types?
Q3.6
In the table below, match the mouse hair color phenotypes with the term from the list that best explains the observed phenotype, given the genotypes shown. In this case, the allele symbols do not imply anything about the dominance relationships between the alleles. List of terms: haplosufficiency, haploinsufficiency, pleiotropy, incomplete dominance, co-dominance, incomplete penetrance, broad (variable) expressivity.
Table for Question 3.6
A1A1 A1A2 A2A2
1 all hairs black on the same individual: 50% of hairs are all black and 50% of hairs are all white all hairs white
2 all hairs black all hairs are the same shade of grey all hairs white
3 all hairs black all hairs black 50% of individuals have all white hairs and 50% of individuals have all black hairs
4 all hairs black all hairs black mice have no hair
5 all hairs black all hairs white all hairs white
6 all hairs black all hairs black all hairs white
7 all hairs black all hairs black hairs are a wide range of shades of grey
Q3.7
A rare dominant mutation causes a neurological disease that appears late in life in all people that carry the mutation. If a father has this disease, what is the probability that his daughter will also have the disease?
Q3.8
Make Punnett Squares to accompany the crosses shown in Figure 3.10.
Q3.9
Another cat hair colour gene is called White Spotting. This gene is autosomal. Cats that have the dominant S allele have white spots. What are the possible genotypes of cats that are:
1. entirely black
2. entirely orange
3. black and white
4. orange and white
5. orange and black (tortoiseshell)
6. orange, black, and white (calico)
Q3.10
Draw reciprocal crosses that would demonstrate that the turkey E-gene is on the Z chromosome.
Q3.11
Mendel’s First Law (as stated in class) does not apply to alleles of most genes located on sex chromosomes. Does the law apply to the chromosomes themselves?
Q3.12
What is the relationship between the O0 and OB alleles of the Orange gene in cats?
Q3.13
Make a diagram similar to those in Figures 3.9, 3.11, and 3.13 that shows the relationship between genotype and phenotype for the F8 gene in humans.
Answers
3.1 There is a maximum of two alleles for a normal autosomal locus in a diploid species.
3.2 a) In the F1 generation, the genotype of all individuals will be Ww and all of the dogs will have wirey hair.
b) In the F2 generation, there would be an expected 3:1 ratio of wirey-haired to smooth-haired dogs.
c) Although it is expected that only one out of every four dogs in the F2 generation would have smooth hair, large deviations from this ratio are possible, especially with small sample sizes. These deviations are due to the random nature in which gametes combine to produce offspring. Another example of this would be the fairly common observation that in some human families, all of the offspring are either girls, or boys, even though the expected ratio of the sexes is essentially 1:1.
d) You could do a test cross, i.e. cross the wirey-haired dog to a homozygous recessive dog (ww). Based on the phenotypes among the offspring, you might be able to infer the genotype of the wirey-haired parent.
e) From the information provided, we cannot be certain which, if either, allele is wild-type. Generally, dominant alleles are wild-type, and abnormal or mutant alleles are recessive.
3.3 Even before the idea of a homozygous genotype had really been formulated, Mendel was still able to assume that he was working with parental lines that contained the genetic material for only one variant of a trait (e.g. EITHER green seeds of yellow seeds), because these lines were pure-breeding. Pure-breeding means that the phenotype doesn’t change over several generations of self-pollination. If the parental lines had not been pure-breeding, it would have been very hard to make certain key inferences, such as that the F1 generation could contain the genetic information for two variants of a trait, although only one variant was expressed. This inference led eventually to Mendel’s First Law.
3.4 Equal segregation of alleles occurs only in meiosis. Although mitosis does produce daughter cells that are genetically equal, there is no segregation (i.e. separation) of alleles during mitosis; each daughter cell contains both of the alleles that were originally present.
3.5 If your blood type is B, then your genotype is either IBi or IBIB. If your genotype is IBi, then your parents could be any combination of genotypes, as long as one parent had at least one i allele, and the other parent had at least one IB allele. If your genotype was IB IB, then both parents would have to have at least one IB allele.
3.6 case 1 co-dominance
case 2 incomplete-dominance
case 3 incomplete penetrance
case 4 pleiotropy
case 5 haplosufficiency
case 6 haploinsufficiency
case 7 broad (variable) expressivity
3.7 If the gene is autosomal, the probability is 50%. If it is sex-linked, 100%. In both situations the probability would decrease if the penetrance was less than 100%.
3.8
3.9 Note that a semicolon is used to separate genes on different chromosomes.
Phenotype Genotype(s)
a entirely black OB / OB ; s / s OB / Y ; s / s
b entirely orange O0 / O0 ; s / s O0 / Y ; s / s
c black and white OB / OB ; S / _ OB / Y ; S / _
d orange and white O0 / O0 ; S / _ O0 / Y ; S / _
e orange and black (tortoiseshell) O0 / OB ; s / s
f orange, black, and white (calico) O0 / OB ; S / _
3.10
3.11 Because each egg or sperm cell receives exactly one sex chromosome (even though this can be either an X or Y, in the case of sperm), it could be argued that the sex chromosomes themselves do obey the law of equal segregation, even though the alleles they carry may not always segregate equally. However, this answer depends on how broadly you are willing to stretch Mendel’s First Law.
3.12 Co-dominance
3.13 People with hemophilia A use injections of recombinant Factor VIII proteins on demand (to control bleeding) or regularily (to limit damage to joints).
3.S: Genetic Analysis of Single Genes (Summary)
• A diploid can have up to two different alleles at a single locus. The alleles segregate equally between gametes during meiosis.
• Phenotype depends on the alleles that are present, their dominance relationships, and sometimes also interactions with the environment and other factors.
• Classical geneticists make use of true breeding lines, monohybrid crosses, Punnett squares, test crosses, reciprocal crosses, and the chi-square test.
• Sex-linked genes are an exception to standard Mendelian inheritance. Their phenotypes are influenced by the type of sex chromosome system and the type of dosage compensation system found in the species.
• The male/female phenotype (sex) can be determined by chromosomes, genes, or the environment.
Key Terms
allele
Mendel’s First Law
Law of Equal Segregation
homozygous
heterozygous
hemizygous
wild-type
variant
locus
genotype
phenotype
dominant
recessive
complete dominance
incomplete (semi) dominance
co-dominance
ABO blood group
haplosufficiency
haploinsufficiency
classical genetics
molecular genetics
true breeding lines
monohybrid cross
Punnett Square
test cross
tester
sex-linked
dosage compensation
X-linked genes
autosomal genes
reciprocal cross
Z-linked genes
hermaphrodites
parthenogenesis
hormonal
cell-autonomous
sexual gynandro-morphs
sexually dimorphic
gynandromorphy
mosaic
chimera
G + E = P
penetrance
expressivity
sampling effects
chi-square χ2 test | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/03%3A_Genetic_Analysis_of_Single_Genes/3.06%3A__Phenotypes_May_Not_Be_As_Expected_from_the_Genotype.txt |
The techniques of genetic analysis discussed in the previous chapters depend on the availability of two or more alleles for a gene of interest. Where do these alleles come from? The short answer is mutation. Humans have an interesting relationship with mutation. From our perspective, mutations can be extraordinarily useful, since mutations are need for evolution to occur.Mutation is also essential for the domestication and improvement of almost all of our food. On the other hand, mutations are the cause of many cancers and other diseases, and can be devastating to individuals. Yet, the vast majority of mutations probably go undetected.In this section, we will examine some of the causes and effects of mutations.
Thumbnail: The difference in appearance between pigmented and white peacocks is due to mutation. (Flickr-ecstaticist-CCANS)
04: Mutation and Variation
We have previously noted that an important property of DNA is its fidelity: most of the time it accurately passes the same information from one generation to the next. However, DNA sequences can also change. Changes in DNA sequences are called mutations. If a mutation changes the phenotype of an individual, the individual is said to be a mutant. Naturally occurring, but rare, sequence variants that are clearly different from a normal, wild-type sequence are also called mutations.On the other hand, many naturally occurring variants exist for traits for which no clearly normal type can be defined; thus, we use the term polymorphism to refer to variants of DNA sequences (and other phenotypes) that co-exist in a population at relatively high frequencies (>1%).
Polymorphisms and mutations arise through similar biochemical processes, but the use of the word “polymorphism” avoids implying that any particular allele is more normal or abnormal. For example, a change in a person’s DNA sequence that leads to a disease such as cancer is appropriately called a mutation, but a difference in DNA sequence that explains whether a person has red hair rather than brown or black hair is an example of polymorphism. Molecular markers, which we will discuss in Chapter 9, are a particularly useful type of polymorphism for some areas of genetics research.
4.02: Origins of Mutations
Types of mutations
Mutations may involve the loss (deletion), gain (insertion) of one or more base pairs, or else the substitution of one or more base pairs with another DNA sequence of equal length. These changes in DNA sequence can arise in many ways, some of which are spontaneous and due to natural processes, while others are induced by humans intentionally (or unintentionally) using mutagens. There are many ways to classify mutagens, which are the agents or processes that cause mutation or increase the frequency of mutations. We will classify mutagens here as being (1) biological, (2) chemical, or (3) physical.
Mutations of biological origin
A major source of spontaneous mutation is errors that arise during DNA replication. DNA polymerases are usually very accurate in adding a base to the growing strand that is the exact complement of the base on the template strand. However, occasionally, an incorrect base is inserted. Usually, the machinery of DNA replication will recognize and repair mispaired bases, but nevertheless, some errors become permanently incorporated in a daughter strand, and so become mutations that will be inherited by the cell’s descendents (Figure \(1\)).
Another type of error introduced during replication is caused by a rare, temporary misalignment of a few bases between the template strand and daughter strand (Figure \(2\)). This strand-slippage causes one or more bases on either strand to be temporarily displaced in a loop that is not paired with the opposite strand. If this loop forms on the template strand, the bases in the loop may not be replicated, and a deletion will be introduced in the growing daughter strand. Conversely, if a region of the daughter strand that has just been replicated becomes displaced in a loop, this region may be replicated again, leading to an insertion of additional sequence in the daughter strand, as compared to the template strand.
Consequences: Regions of DNA that have several repeats of the same few nucleotides in a row are especially prone to this type of error during replication. Thus regions with short-sequence repeats (SSRs) are tend to be highly polymorphic, and are therefore particularly useful in genetics. They are called microsatellites.
Mutations can also be caused by the insertion of viruses, transposable elements (transposons), see below, and other types of DNA that are naturally added at more or less random positions in chromosomes. The insertion may disrupt the coding or regulatory sequence of a gene, including the fusion of part of one gene with another. These insertions can occur spontaneously, or they may also be intentionally stimulated in the laboratory as a method of mutagenesis called transposon-tagging. For example, a type of transposable element called a P element is widely used in Drosophila as a biological mutagen. T-DNA, which is an insertional element modified from a bacterial pathogen, is used as a mutagen in some plant species.
Mutations due to Transposable Elements
Transposable elements (TEs) are also known as mobile genetic elements, or more informally as jumping genes. They are present throughout the chromosomes of almost all organisms. These DNA sequences have a unique ability to be cut or copied from their original location and inserted into new locations in the genome. This is called transposition. These insert locations are not entirely random, but TEs can, in principle, be inserted into almost any region of the genome. TEs can therefore insert into genes, disrupting its function and causing a mutation. Researchers have developed methods of artificially increasing the rate of transposition, which makes some TEs a useful type of mutagen. However, the biological importance of TEs extends far beyond their use in mutant screening. TEs are also important causes of disease and phenotypic instability, and they are a major mutational force in evolution.
There are two major classes of TEs in eukaryotes (Figure \(3\)).
• Class I elements include retrotransposons; these transpose by means of an RNA intermediate. The TE transcript is reverse transcribed into DNA before being inserted elsewhere in the genome through the action of enzymes such as integrase.
• Class II elements are known also as transposons. They do not use reverse transcriptase or an RNA intermediate for transposition. Instead, they use an enzyme called transposase to cut DNA from the original location and then this excised dsDNA fragment is inserted into a new location. Note that the name transposon is sometimes used incorrectly to refer to any type of TEs, but in this book we use transposon to refer specifically to Class II elements.
TEs are relatively short DNA sequences (100-10,000 bp), and encode no more than a few proteins (if any). Normally, the protein-coding genes within a TE are all related to the TE’s own transposition functions. These proteins may include reverse transcriptase, transposase, and integrase. However, some TEs (of either Class I or II) do not encode any proteins at all. These non-autonomous TEs can only transpose if they are supplied with enzymes produced by other, autonomous TEs located elsewhere in the genome. In all cases, enzymes for transposition recognize conserved nucleotide sequences within the TE, which dictate where the enzymes begin cutting or copying.
The human genome consists of nearly 45% TEs, the vast majority of which are families of Class I elements called LINEs and SINEs. The short, Alu type of SINE occurs in more than one million copies in the human genome (compare this to the approximately 21,000, non-TE, protein-coding genes in humans). Indeed, TEs make up a significant portion of the genomes of almost all eukaryotes. Class I elements, which usually transpose via an RNA copy-and-paste mechanism, tend to be more abundant than Class II elements, which mostly use a cut-and-paste mechanism. But even the cut-paste mechanism can lead to an increase in TE copy number. For example, if the site vacated by an excised transposon is repaired with a DNA template from a homologous chromosome that itself contains a copy of a transposon, then the total number of transposons in the genome will increase.
Besides greatly expanding the overall DNA content of genomes, TEs contribute to genome evolution in many other ways. As already mentioned, they may disrupt gene function by insertion into a gene’s coding region or regulatory region. More interestingly adjacent regions of chromosomal DNA are sometimes mistakenly transposed along with the TE; this can lead to gene duplication. The duplicated genes are then free to evolve independently, leading in some cases to the development of new functions. The breakage of strands by TE excision and integration can disrupt genes, and can lead to chromosome rearrangement or deletion if errors are made during strand rejoining. Furthermore, having so many similar TE sequences distributed throughout a chromosome sometimes allows mispairing of regions of homologous chromosomes at meiosis, which can cause unequal crossing-over, resulting in deletion or duplication of large segments of chromosomes. Thus, TEs are a potentially important evolutionary force, and may not be included as merely “junk DNA”, as they once were. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/04%3A_Mutation_and_Variation/4.01%3A_Mutation_and_Polymorphism.txt |
One way to identify genes that affect a particular biological process is to induce random mutations in a large population, and then look for mutants with phenotypes that might be caused by a disruption of a particular biochemical pathway. This strategy of mutant screening has been used very effectively to identify and understand the molecular components of hundreds of different biological processes. For example, to find the basic biological processes of memory and learning, researchers have screened mutagenized populations of Drosophila to recover flies (or larvae) that lack the normal ability to learn. They lack the ability to associate a particular odor with an electric shock. Because of the similarity of biology among all organisms, some of the genes identified by this mutant screen of a model organism may be relevant to learning and memory in humans, including conditions such as Alzheimer’s disease.
– Genetic Screens
In a typical mutant screen, researchers treat a parental population with a mutagen. This may involve soaking seeds in EMS, or mixing a mutagen with the food fed to flies. Usually, no phenotypes are visible among the individuals that are directly exposed to the mutagen because in all the cells every strand of DNA will be affected independently. Thus the induced mutations will be heterozygous and limited to single cells. However, what is most important to geneticists are the mutations in the germline of the mutagenized individuals. The germline is defined as the gametes and any of their developmental precursors, and is therefore distinct from the somatic cells (i.e. non-reproductive cells) of the body. Because most induced mutations are recessive, the progeny of mutagenized individuals must be mated in a way that allows the new mutations to become homozygous (or hemizygous). Strategies for doing this vary between organisms. In any case, the generation in which induced mutations are expected to occur can be examined for the presence of novel phenotypes. Once a relevant mutant has been identified, geneticists can begin to make inferences about what the normal function of the mutated gene is, based on its mutant phenotype. This can then be investigated further with molecular genetic techniques.
Exposure of an organism to a mutagen causes mutations in essentially random positions along the chromosomes. Most of the mutant phenotypes recovered from a genetic screen are caused by loss-of-function mutations. These alleles are due to changes in the DNA sequence that cause it to no longer produce the same level of active protein as the wild-type allele. Loss-of-function alleles tend to be recessive because the wild-type allele is haplosufficient (see Chapter 3). A loss-of-function allele that produces no active protein is called an amorph, or null. On the other hand, alleles with only a partial loss-of-function are called hypomorphic. More rarely, a mutant allele may have a gain-of-function, producing either more of the active protein (hypermorph) or producing an active protein with a new function (neomorph). Finally, antimorph alleles have an activity that is dominant and opposite to the wild-type function; antimorphs are also known as dominant negative mutations.
4.04: Types of Mutations
Mutations (changes in a gene sequence) can result in mutant alleles that no longer produce the same level or type of active product as the wild-type allele. Any mutant allele can be classified into one of five types: (1) amorph, (2) hypomorph, (3) hypermorph, (4) neomorph, and (5) antimorph.
• Amorph alleles are complete loss-of-function. They make no active product – zero function. The absence of function can be due to a lack of transcription (gene regulation mutation) or due to the production of a malfunctioning (protein coding mutation) product. These are also sometimes referred to as a Null allele.
• Hypomorph alleles are only a partial loss-of-function. They make an incompletely functioning product. This could occur via reduced transcription or via the production of a product that lacks complete activity. These alleles are sometimes referred to as Leaky mutations, because they provide some function, but not complete function.
Both amorphs and hypomorphs tend to be recessive to wild type because the wild type allele is usually able to supply sufficient product to produce a wild type phenotype (called haplo-sufficient - see Chapter 6). If the mutant allele is not haplo-sufficient, then it will be dominant to the wild type.
While the first two classes involve a loss-of-function, the next two involve a gain-of-function – quantity or quality. Gain-of-function alleles are almost always dominant to the wild type allele.
• Hypermorph alleles produce more of the same, active product. This can occur via increased transcription or by changing the product to make it more efficient/effective at its function.
• Neomorph alleles produce an active product with a new, different function, something that the wild type allele doesn’t do. It can be either new expression (new tissue or time) or a mutation in the product to create a new function (additional substrate or new binding site), not present in the wild type product.
Antimorph alleles are relatively rare, and have an activity that is dominant and opposite to the wild-type function. These alleles usually have no normal function of their own and they interfere with the function from the wild type allele. Thus, when an antimorph allele is heterozygous with wild type, the wild type allele function is reduced. While at the molecular level there are many ways this can happen, the simplest model to explain antimorph effect is that the product acts as a dimer (or any multimer) and one mutant subunit poisons the whole complex. Antimorphs are also known as dominant negative mutations.
Identifying Muller’s Morphs - All mutations can be sorted into one of the five morphs base on how they behave when heterozygous with other alleles – deletion alleles (zero function), wild type alleles (normal function), and duplication alleles (double normal function). | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/04%3A_Mutation_and_Variation/4.03%3A_Genetic_screening_for_mutations-_Forward_genetics.txt |
Silent Changes
After mutagen treatment, the vast majority of base pair changes (especially substitutions) have no effect on the phenotype. Often, this is because the change occurs in the DNA sequence of a non-coding region of the DNA, such as in inter-genic regions (between genes) or within an intron region. Also, even if the change occurs in a base within a codon, it may not change the amino acid that it encodes (recall that the genetic code is degenerate; for example, GCT, GCC, GCA, and GCG all encode alanine) and is referred to as a silent mutation. Additionally, the base substitution may change an amino acid, but this doesn’t alter the function of the product, so no phenotypic change would occur.
Environment and Genetic Redundancy
There are also situations where a mutation can cause a complete loss-of-function of a gene, yet not produce a change in the phenotype, even when the mutant allele is homozygous. The lack of a phenotypic change can be due to environmental effects: the loss of that gene product may not be apparent in that environment, but might in another. Alternatively, the lack of a phenotype might be attributed to genetic redundancy, i.e. the encoding of similarly functioning genes at more than one locus in the genome. Thus the loss of one gene is compensated by another. This important limitation of mutational analysis should be remembered: genes with redundant functions cannot be easily identified by mutant screening.
Essential Genes and Lethal Alleles
Some phenotypes require individuals to reach a particular developmental stage before they can be scored. For example, flower color can only be scored in plants that are mature enough to make flowers, and eye color can only be scored in flies that have developed eyes. However, some alleles may not develop sufficiently to be included among the progeny that are scored for a particular phenotype. Mutations in essential genes create recessive lethal alleles that arrest the development of an individual at an embryonic stage. This type of mutation may therefore go unnoticed in a typical mutant screen because they are absent from the progeny being screened. Furthermore, the progeny of a monohybrid cross involving an embryonic lethal recessive allele may therefore all be of a single phenotypic class, giving a phenotypic ratio of 1:0 (which is the same as 3:0). In this case the mutation may not be detected.
Naming Genes
Many genes have been first identified in mutant screens, and so they tend to be named after their mutant phenotypes, not the normal function or phenotype. This can cause some confusion for students of genetics. For example, we have already encountered an X-linked gene named white in fruit flies. Null mutants of the white gene have white eyes, but the normal white+ allele has red eyes. This tells us that the wild type (normal) function of this gene is actually to help make red eyes. Its product is a protein that imports a pigment precursor into developing cells of the eye. Why don’t we call it the “red” gene, since that is what its product does? Because there are more than one-dozen genes that when mutant alter the eye colour; e.g. violet, cinnabar, brown, scarlet, etc. For all these genes their function is also needed to make the eye wild type red and not the mutant colour. If we used the name “red” for all these genes it would be confusing, so we use the distinctive mutant phenotype as the gene name. However, this can be problematic, as with the “lethal” mutations described above. This problem is usually handled by giving numbers or locations to the gene name, or making up names that describe how they die (e.g. even-skipped, hunchback, hairy, runt, etc.) . | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/04%3A_Mutation_and_Variation/4.05%3A_Some_mutations_may_not_have_detectable_phenotypes.txt |
– One, or more than one gene?
As explained earlier in this chapter, mutant screening is one of the beginning steps geneticists use to investigate biological processes. When geneticists obtain two independently derived mutants (either from natural populations or during a mutant screen) with similar phenotypes, an immediate question is whether or not the mutant phenotype is due to a loss of function in the same gene, or are they mutant in different genes that both affect the same phenotype (e.g., in the same pathway). That is, are they allelic mutations, or non-allelic mutations, respectively? This question can be resolved using complementation tests, which bring together, or combine, the two mutations under consideration into the same organism to assess the combined phenotype.
hypothetical example of purple flowers
The easiest way to understand a complementation test is by example (Fig.4.9). The pigment in a purple flower could depend on a biochemical pathway much like the biochemical pathways leading to the production of arginine in Neurospora (review in Chapter 1). A plant that lacks the function of gene A (genotype aa) would produce mutant, white flowers that looked just like the flowers of a plant that lacked the function of gene B (genotype bb). (The genetics of two loci are discussed more in the following chapters.) Both A and B are enzymes in the same pathway that leads from a colorless compound#1, thorough colorless compound#2, to the purple pigment. Blocks at either step will result in a mutant white, not wild type purple, flower.
Strains with mutations in gene A can be represented as the genotype aa, while strains with mutations in gene B can be represented as bb. Given that there are two genes here, A and B, then each of these mutant strains can be more completely represented as aaBB and AAbb . (LEARNING NOTE: Student often forget that genotypes usually only show mutant loci, however, one must remember all the other genes are assumed to be wild type.)
If these two strains are crossed together the resulting progeny will all be AaBb. They will have both a wild type, functional A gene and B gene and will thus have a pigmented, purple flower, a wild type phenotype. This is an example of complementation. Together, each strain provides what the other is lacking (AaBb). The mutations are in different genes and are thus called non-allelic mutations.
Now, if we are presented with a third pure-breeding, independently derived white-flower mutant strain, we won't initially know if it is mutant in gene A or gene B (or possibly some other gene altogether). We can use complementation testing to determine which gene is mutated. To perform a complementation test, two homozygous individuals with similar mutant phenotypes are crossed (Figure \(10\)).
If the F1 progeny all have the same mutant phenotype (Case 1 - Figure \(10\)A), then we infer that the same gene is mutated in each parent. These mutations would then be called allelic mutations - in the same gene locus. These mutations FAIL to COMPLEMENT one another (still mutant). These could be either the exact same mutant alleles, or different mutations in the same gene (allelic).
Conversely, if the F1 progeny all appear to be wild-type (Case 2 - Figure \(10\)B), then each of the parents most likely carries a mutation in a different gene. These mutations would then be called non-allelic mutations - in a different gene locus. These mutations do COMPLEMENT one another.
Note: For mutations to be used in complementation tests they are (1) usually true-breeding (homozygous at the mutant locus), and (2) must be recessive mutations. Dominant mutation CANNOT be used in complementation tests. Also, remember, some mutant strains may have more than one gene locus mutated and thus would fail to complement mutants from more than one other locus (or group).
A. B.
Figure \(10\)A – Observation: In a typical complementation test, the genotypes of two parents are unknown (although they must be pure breeding, homozygous mutants). If the F1 progeny all have a mutant phenotype (Case 1), there is no complementation. If the F1 progeny are all wild-type, the mutations have successfully complemented each other.
Figure \(10\)B – Interpretation: The pure breeding, homozygous mutant parents had unknown genotypes before the complementation test, but it could be assumed that they had either mutations in the same genes (Case 1) or in different genes (Case 2). In Case 1, all of the progeny would have the mutant phenotype, because they would all have the same, homozygous genotype as the parents. In Case 2, each parent has a mutation in a different gene, therefore none of the F1 progeny would be homozygous mutant at either locus. Note that the genotype in Case 1 could be written as either aa or aaBB. (Original-Deyholos-CC:AN) | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/04%3A_Mutation_and_Variation/4.06%3A_Complementation_tests_and_Allelism.txt |
– Cystic Fibrosis (CF) – autosmal recessive
Cystic fibrosis (CF) is one of many diseases that geneticists have shown to be caused by mutation of a single, well-characterized gene. Cystic fibrosis is the most common (1/2,500) life-limiting autosomal recessive disease among people of European heritage, with ~ 1 in 25 people being carriers. The frequency varies in different populations. Most of the deaths caused by CF are the result of lung disease, but many CF patients also suffer from other disorders including infertility and gastrointestinal disease. The disease is due to a mutation in the CFTR (Cystic Fibrosis Transmembrane Conductance Regulator) gene, which was identified by Lap-chee Tsui’s group at the University of Toronto.
Epithelial tissues in some organs rely on the CFTR protein to transport ions (especially Cl-) across their cell membranes. The passage of ions through a six-sided channel is gated by another part of the CFTR protein, which binds to ATP. If there is insufficient activity of CFTR, an imbalance in ion concentration results, which disrupts the properties of the liquid layer that normally forms at the epithelial surface. In the lungs, this causes mucus to accumulate and can lead to infection. Defects in CFTR also affect pancreas, liver, intestines, and sweat glands, all of which need this ion transport. CFTR is also expressed at high levels in the salivary gland and bladder, but defects in CFTR function do not cause problems in these organs, probably because other ion transporters are able to compensate.
Over one thousand different mutant alleles of CFTR have been described. Any mutation that prevents CFTR from sufficiently transporting ions can lead to cystic fibrosis (CF). Worldwide, the most common CFTR allele among CF patients is called ΔF508 (delta-F508; or PHE508DEL), which is a deletion of three nucleotides that eliminates a phenylalanine from position 508 of the 1480 aa wild-type protein. Mutation ΔF508 causes CFTR to be folded improperly in the endoplasmic reticulum (ER), which prevents CFTR from reaching the cell membrane. ΔF508 accounts for approximately 70% of CF cases in North America, with ~1/25 people of European descent being carriers. The high frequency of the ΔF508 allele has led to speculation that it may confer some selective advantage to heterozygotes, perhaps by reducing dehydration during cholera epidemics, or by reducing susceptibility to certain pathogens that bind to epithelial membranes.
CFTR is also notable because it is one of the well-characterized genetic diseases for which a drug has been developed that compensates for the effects of a specific mutation. The drug, Kalydeco, was approved by the FDA and Health Canada in 2012, decades after the CFTR gene was first mapped to DNA markers (in 1985) and cloned (in 1989). Kalydeco is effective on only some CFTR mutations, most notably G551D (i.e. where glycine is substituted by aspartic acid at position 551 of the protein; GLY551ASP). This mutation is found in less than 5% of CF patients. The G551D mutation affects the ability of ATP to bind to CFTR and open the channel it for transport. Kalydeco compensates for mutation by binding to CFTR and holding it in an open conformation. Kalydeco is expected to cost approximately \$250,000 per patient per year. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/04%3A_Mutation_and_Variation/4.07%3A_Example_of_Human_Mutations.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Q4.1
How are polymorphisms and mutations alike? How are they different?
Q4.2
What are some of the ways a substitution can occur in a DNA sequence?
Q4.3
What are some of the ways a deletion can occur in a DNA sequence?
Q4.4
What are all of the ways an insertion can occur in a DNA sequence?
Q4.5
In the context of this chapter, explain the health hazards of smoking tobacco.
Q4.6
You have a female fruit fly, whose father was exposed to a mutagen (she, herself, wasn’t). Mating this female fly with another non-mutagenized, wild type male produces offspring that all appear to be completely normal, except there are twice as many daughters as sons in the F1 progeny of this cross.
1. Propose a hypothesis to explain these observations.
2. How could you test your hypothesis?
Q4.7
You decide to use genetics to investigate how your favourite plant makes its flowers smell good.
1. What steps will you take to identify some genes that are required for production of the sweet floral scent? Assume that this plant is a self-pollinating diploid.
2. One of the recessive mutants you identified has fishy-smelling flowers, so you name the mutant (and the mutated gene) fishy. What do you hypothesize about the normal function of the wild-type fishy gene?
3. Another recessive mutant lacks floral scent altogether, so you call it nosmell. What could you hypothesize about the normal function of this gene?
Q4.8
Suppose you are only interested in finding dominant mutations that affect floral scent.
1. What do you expect to be the relative frequency of dominant mutations, as compared to recessive mutations, and why?
2. How will you design your screen differently than in the previous question, in order to detect dominant mutations specifically?
3. Which kind of mutagen is most likely to produce dominant mutations, a mutagen that produces point mutations, or a mutagen that produces large deletions?
Q4.9
hich types of transposable elements are transcribed?
Q4.10
You are interested in finding genes involved in synthesis of proline (Pro), an amino acid that is normally synthesizes by a particular model organism.
1. How would you design a mutant screen to identify genes required for Pro synthesis?
2. Imagine that your screen identified ten mutants (#1 through #10) that grew poorly unless supplemented with Pro. How could you determine the number of different genes represented by these mutants?
3. If each of the four mutants represents a different gene, what will be the phenotype of the F1 progeny if any pair of the four mutants are crossed?
4. If each of the four mutants represents the same gene, what will be the phenotype of the F1 progeny if any pair of the four mutants are crossed?
Chapter 4 - Answers
4.1 Polymorphisms and mutations are both variations in DNA sequence and can arise through the same mechanisms. We use the term polymorphism to refer to DNA variants that are relatively common in populations. Mutations affect the phenotype.
4.2 Misreading of bases during replication can lead to substitution and can be caused by things like tautomerism, DNA alkylating agents, and irradiation.
4.3 Looping out of DNA on the template strand during replication; strand breakage, due to radiation and other mutagens; and (discussed in earlier chapters) chromosomal aberrations such as deletions and translocations.
4.4 Looping out of DNA on the growing strand during replication; transposition; and (discussed in earlier chapters) chromosomal aberrations such as duplications, insertions, and translocation.
4.5 Benzopyrene is one of many hazardous compounds present in smoke. Benzopyrene is an intercalating agent, which slides between the bases of the DNA molecule, distorting the shape of the double helix, which disrupts transcription and replication and can lead to mutation.
4.6 a) One possible explanation is that original mutagenesis resulted in a loss-of-function mutation in a gene that is essential for early embryonic development, and that this mutation is X-linked recessive in the female. Because half of the sons will inherit the X chromosome that bears this mutation, half of the sons will fail to develop beyond very early development and will not be detected among the F1 progeny. The proportion of male flies that were affected depends on what fraction of the female parent’s gametes carried the mutation. In this case, it appears that half of the female’s gametes carried the mutation.
b) To test whether a gene is X-linked, you can usually do a reciprocal cross. However, in this case it would be impossible to obtain adult male flies that carry the mutation; they are dead. If the hypothesis proposed in a) above is correct, then half of the females, and none of the living males in the F1 should carry the mutant allele. You could therefore cross F1 females to wild type males, and see whether the expected ratios were observed among the offspring (e.g. half of the F1 females should have a fewer male offspring than expected, while the other half of the F1 females and all of the males should have a roughly equal numbers of male and female offspring).
4.7 a) Treat a population of seeds with a mutagen such as EMS. Allow these seeds to self-pollinate, and then allow the F1 generation to also self-pollinate. In the F2 generation, smell each flower to find individuals with abnormal scent.
b) The fishy gene appears to be required to make the normal floral scent. Because the flowers smell fishy in the absence of this gene, one possibility explanation of this is that fishy makes an enzyme that converts a fishy-smelling intermediate into a chemical that gives flowers their normal, sweet smell.
Note that although we show this biochemical pathway as leading from the fishy-smelling chemical to the sweet-smelling chemical in one step, it is likely that there are many other enzymes that act after the fishy enzyme to make the final, sweet-smelling product. In either case, blocking the pathway at the step catalyzed by the fishy enzyme would explain the fishy smell.
c) In nosmell plants, the normal sweet smell disappears. Unlike fishy, the sweet smell is not replaced by any intermediate chemical that we can easily detect. Thus, we cannot conclude where in the biochemical pathway the nosmell mutant is blocked; nosmell may normally therefore act either before or after fishy normally acts in the pathway:
Alternatively, nosmell may not be part of the biosynthetic pathway for the sweet smelling chemical at all. It is possible that the normal function of this gene is to transport the sweet-smelling chemical into the cells from which it is released into the air, or maybe it is required for the development of those cells in the first place. It could even be something as general as keeping the plants healthy enough that they have enough energy to do things like produce floral scent.
4.8 a) Dominant mutations are generally much rarer than recessive mutations. This is because mutation of a gene tends to cause a loss of the normal function of this gene. In most cases, having just one normal (wt) allele is sufficient for normal biological function, so the mutant allele is recessive to the wt allele. Very rarely, rather than destroying normal gene function, the random act of mutation will cause a gene to gain a new function (e.g. to catalyze a new enzymatic reaction), which can be dominant (since it performs this new function whether the wt allele is present or not). This type of gain-of-function dominant mutation is very rare because there are many more ways to randomly destroy something than by random action to give it a new function (think of the example given in class of stomping on an iPod).
b) Dominant mutations should be detectable in the F1 generation, so the F1 generation, rather than the F2 generation can be screened for the phenotype of interest.
c) Large deletions, such as those caused by some types of radiation, are generally less likely than point mutations to introduce a new function into a protein: it is hard for a protein to gain a new function if the entire gene has been removed from the genome by deletion.
4.9 Class I. see Figure 4.5 on Transposable Elements.
4.10 a) Mutagenize a wild type (auxotrophic) strain and screen for mutations that fail to grow on minimal media, but grow well on minimal media supplemented with proline.
b) Take mutants #1-#10) and characterize them, based on (1) genetic mapping of the mutants (different locations indicate different genes); (2) different response to proline precursors (a different response suggests different genes); (3) complementation tests among the mutations (if they complement then they are mutations in different genes).
c) If the mutations are in different genes then the F1 progeny would be wild type (able to grow on minimal medium without proline).
d) If the mutations are in the same gene then the F1 progeny would NOT be wild type (unable to grow on minimal medium without proline).
4.S: Mutation and Variation (Summary)
• When a variation in DNA sequence originated recently, and is rare in a population, we call that change a mutation.
• When variations in DNA sequence co-exist in a population, and neither one can be meaningfully defined as wild-type, we call the variations polymorphisms.
• Mutations may either occur spontaneously, or may be induced by exposure to mutagens.
• Mutations may result in either substitutions, deletions, or insertions.
• Mutation usually causes either a partial or complete loss of function, but sometimes results in a gain of function,including new functions.
• Spontaneous mutations arise from many sources including natural errors in DNA replication, usually associated with base mispairing, or else insertion deletion especially within repetitive sequences.
• Induced mutations result from mispairing, DNA damage, or sequence interruptions caused by chemical, biological, or physical mutagens.
• By randomly inducing mutations, then screening for a specific phenotype, it is possible to identify genes associated with specific biological pathways.
• Transposable elements are dynamic, abundant components of eukaryotic genomes and important forces in evolution.
• Transposable elements are dynamic, abundant components of eukaryotic genomes and important forces in evolution.
• The efficiency of mutant screening is limited by silent mutations, redundancy, and embyronic lethality.
• Mutation of different genes can produce a similar phenotype.
• Complementation testing determines whether two mutants are the result of mutation of the same gene (allelic mutations), or if each mutant is caused by mutation of a different gene (non-allelic mutations).
Key Terms
mutation
mutant
polymorphism
insertion
deletion
substitution
mutagen
DNA replication error
Strand slippage
biological mutagen
chemical mutagen
physical mutagen
mispairing
loop
SSR
insertional mutagen
Class I, Class II
transposon
retrotransposon
reverse transcriptase
transposase
non-autonomous
autonomous
SINE, LINE, Alu
P-element
T-DNA
copy-and-paste
cut-and-paste
alkylation agent
EMS
intercalating agent
benzopyrene
carcinogenic
ethidium bromide
thymine dimer
mutant screen
loss-of-function
gain-of-function
amorph
null
hypomorph
hypermorph
neopmorph
dominant negative
somatic cells
germline cells
silent mutation
inter-genic region
redundancy
essential gene
recessive lethal allele
complementation testing
allelic / non-allelic
cM0, M1, M2
redundancy
lethality
allelic
non-allelic
complementation group
CFTR
Cystic Fibrosis (CF)
DF508(PHE508DEL)
Kalydeco | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/04%3A_Mutation_and_Variation/4.E%3A_Mutation_and_Variation_%28Exercises%29.txt |
The basic concepts of genetics described in the preceding chapters can be applied to almost any eukaryotic organism. However, some techniques, such as test crosses, can only be performed with model organisms or other species that can be experimentally manipulated. To study the inheritance patterns of genes in humans and other species for which controlled matings are not possible, geneticists use the analysis of pedigrees and populations.
• 5.1: Prelude to Pedigrees and Populations
To study the inheritance patterns of genes in humans and other species for which controlled matings are not possible, geneticists use the analysis of pedigrees and populations.
• 5.2: Pedigree Analysis
Pedigree charts are diagrams that show the phenotypes and/or genotypes for a particular organism and its ancestors. While commonly used in human families to track genetic diseases, they can be used for any species and any inherited trait.
• 5.3: Inferring the Mode of Inheritance
Given a pedigree of an uncharacterized disease or trait, one of the first tasks is to determine which modes of inheritance are possible and then which mode of inheritance is most likely. This information is essential in calculating the probability that the trait will be inherited in any future offspring. We will mostly consider five major types of inheritance: autosomal dominant (AD), autosomal recessive (AR), X-linked dominant (XD), X-linked recessive (XR), and Y-linked (Y).
• 5.4: Sporadic and Non-Heritable Diseases
Many diseases that have a heritable component, have more complex inheritance patterns due to (1) the involvement of multiple genes, and/or (2) environmental factors.
• 5.5: Calculating Probabilities
Probabilities in pedigrees are calculated using knowledge of Mendelian inheritance and the same basic methods as are used in other fields. The first formula is the product rule: the joint probability of two independent events is the product of their individual probabilities; this is the probability of one event AND another event occurring.
• 5.6: Population Genetics
A population is a large group of individuals of the same species, who are capable of mating with each other. It is useful to know the frequency of particular alleles within a population, since this information can be used to calculate disease risks. Population genetics is also important in ecology and evolution, since changes in allele frequencies may be associated with migration or natural selection.
• 5.E: Pedigrees and Populations (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap.
• 5.S: Pedigrees and Populations (Summary)
05: Pedigrees and Populations
To study the inheritance patterns of genes in humans and other species for which controlled matings are not possible, geneticists use the analysis of pedigrees and populations.
5.02: Pedigree Analysis
Pedigree charts are diagrams that show the phenotypes and/or genotypes for a particular organism and its ancestors. While commonly used in human families to track genetic diseases, they can be used for any species and any inherited trait. Geneticists use a standardized set of symbols to represent an individual’s sex, family relationships and phenotype. These diagrams are used to determine the mode of inheritance of a particular disease or trait, and to predict the probability of its appearance among offspring. Pedigree analysis is therefore an important tool in both basic research and genetic counseling.
Each pedigree chart represents all of the available information about the inheritance of a single trait (most often a disease) within a family. The pedigree chart is therefore drawn using factual information, but there is always some possibility of errors in this information, especially when relying on family members’ recollections or even clinical diagnoses. In real pedigrees, further complications can arise due to incomplete penetrance (including age of onset) and variable expressivity of disease alleles, but for the examples presented in this book, we will presume complete accuracy of the pedigrees. A pedigree may be drawn when trying to determine the nature of a newly discovered disease, or when an individual with a family history of a disease wants to know the probability of passing the disease on to their children. In either case, a tree is drawn, as shown in Figure \(2\), with circles to represent females, and squares to represent males. Matings are drawn as a line joining a male and female, while a consanguineous mating (closely related is two lines.
The affected individual that brings the family to the attention of a geneticist is called the proband (or propositus). If an individual is known to have symptoms of the disease (affected), the symbol is filled in. Sometimes a half-filled in symbol is used to indicate a known carrier of a disease; this is someone who does not have any symptoms of the disease, but who passed the disease on to subsequent generations because they are a heterozygote. Note that when a pedigree is constructed, it is often unknown whether a particular individual is a carrier or not, so not all carriers are always explicitly indicated in a pedigree. For simplicity, in this chapter we will assume that the pedigrees presented are accurate, and represent fully penetrant traits. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/05%3A_Pedigrees_and_Populations/5.01%3A_Prelude_to_Pedigrees_and_Populations.txt |
Given a pedigree of an uncharacterized disease or trait, one of the first tasks is to determine which modes of inheritance are possible and then which mode of inheritance is most likely. This information is essential in calculating the probability that the trait will be inherited in any future offspring. We will mostly consider five major types of inheritance: autosomal dominant (AD), autosomal recessive (AR), X-linked dominant (XD), X-linked recessive (XR), and Y-linked (Y).
Autosomal Dominant (AD)
When a disease is caused by a dominant allele of a gene, every person with that allele will show symptoms of the disease (assuming complete penetrance), and only one disease allele needs to be inherited for an individual to be affected. Thus, every affected individual must have an affected parent. A pedigree with affected individuals in every generation is typical of AD diseases. However, beware that other modes of inheritance can also show the disease in every generation, as described below. It is also possible for an affected individual with an AD disease to have a family without any affected children, if the affected parent is a heterozygote. This is particularly true in small families, where the probability of every child inheriting the normal, rather than disease allele is not extremely small. Note that AD diseases are usually rare in populations, therefore affected individuals with AD diseases tend to be heterozygotes (otherwise, both parents would have had to been affected with the same rare disease). Achondroplastic dwarfism, and polydactyly are both examples of human conditions that may follow an AD mode of inheritance.
X-linked dominant (XD)
In X-linked dominant inheritance, the gene responsible for the disease is located on the X-chromosome, and the allele that causes the disease is dominant to the normal allele in females. Because females have twice as many X-chromosomes as males, females tend to be more frequently affected than males in the population. However, not all pedigrees provide sufficient information to distinguish XD and AD. One definitive indication that a trait is inherited as AD, and not XD, is that an affected father passes the disease to a son; this type of transmission is not possible with XD, since males inherit their X chromosome from their mothers.
Autosomal recessive (AR)
Diseases that are inherited in an autosomal recessive pattern require that both parents of an affected individual carry at least one copy of the disease allele. With AR traits, many individuals in a pedigree can be carriers, probably without knowing it. Compared to pedigrees of dominant traits, AR pedigrees tend to show fewer affected individuals and are more likely than AD or XD to “skip a generation”. Thus, the major feature that distinguishes AR from AD or XD is that unaffected individuals can have affected offspring.
X-linked recessive (XR)
Because males have only one X-chromosome, any male that inherits an X-linked recessive disease allele will be affected by it (assuming complete penetrance). Therefore, in XR modes of inheritance, males tend to be affected more frequently than females in a population. This is in contrast to AR and AD, where both sexes tend to be affected equally, and XD, in which females are affected more frequently. Note, however, in the small sample sizes typical of human families, it is usually not possible to accurately determine whether one sex is affected more frequently than others. On the other hand, one feature of a pedigree that can be used to definitively establish that an inheritance pattern is not XR is the presence of an affected daughter from unaffected parents; because she would have had to inherit one X-chromosome from her father, he would also have been affected in XR.
Y-linked and Mitochondrial Inheritance.
Two additional modes are Y-linked and Mitochondrial inheritance.
Only males are affected in human Y-linked inheritance (and other species with the X/Y sex determining system). There is only father to son transmission. This is the easiest mode of inheritance to identify, but it is one of the rarest because there are so few genes located on the Y-chromosome. An example of Y-linked inheritance is the hairy-ear-rim phenotype seen in some Indian families. As expected this trait is passed on from father to all sons and no daughters. Y-chromosome DNA polymorphisms can be used to follow the male lineage in large families or through ancient ancestral lineages. For example, the Y-chromosome of Mongolian ruler Genghis Khan (1162-1227 CE), and his male relatives, accounts for ~8% of the Y-chromosome lineage of men in Asia (0.5% world wide).
Mutations in Mitochondrial DNA are inherited through the maternal line (from your mother). There are some human diseases associated with mutations in mitochondria genes. These mutations can affect both males and females, but males cannot pass them on as the mitochondria are inherited via the egg, not the sperm. Mitochondrial DNA polymorphisms are also used to investigate evolutionary lineages, both ancient and recent. Because of the relative similarity of sequence mtDNA is also used in species identification in ecology studies. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/05%3A_Pedigrees_and_Populations/5.03%3A_Inferring_the_Mode_of_Inheritance.txt |
Not all of the characterized human traits and diseases are attributed to mutant alleles at a single gene locus. Many diseases that have a heritable component, have more complex inheritance patterns due to (1) the involvement of multiple genes, and/or (2) environmental factors. On the other hand, some non-genetic diseases may appear to be heritable because they affect multiple members of the same family, but this is due to the family members being exposed to the same toxins or other environmental factors (e.g. in their homes).
Finally, diseases with similar symptoms may have different causes, some of which may be genetic while others are not. One example of this is ALS (amyotrophic lateral sclerosis); approximately 5-10% of cases are inherited in an AD pattern, while the majority of the remaining cases appear to be sporadic, in other words, not caused by a mutation inherited from a parent. We now know that different genes or proteins are affected in the inherited and sporadic forms of ALS. The physicist Stephen Hawking (Figure \(10\)) and baseball player Lou Gehrig both suffered from sporadic ALS.
5.05: Calculating Probabilities
Once the mode of inheritance of a disease or trait is identified, some inferences about the genotype of individuals in a pedigree can be made, based on their phenotypes and where they appear in the family tree. Given these genotypes, it is possible to calculate the probability of a particular genotype being inherited in subsequent generations. This can be useful in genetic counseling, for example when prospective parents wish to know the likelihood of their offspring inheriting a disease for which they have a family history.
Probabilities in pedigrees are calculated using knowledge of Mendelian inheritance and the same basic methods as are used in other fields. The first formula is the product rule: the joint probability of two independent events is the product of their individual probabilities; this is the probability of one event AND another event occurring. For example, the probability of a rolling a “five” with a single throw of a single six-sided die is 1/6, and the probability of rolling “five” in each of three successive rolls is 1/6 x 1/6 x 1/6 = 1/216. The second useful formula is the sum rule, which states that the combined probability of two independent events is the sum of their individual probabilities. This is the probability of one event OR another event occurring. For example, the probability of rolling a five or six in a single throw of a dice is 1/6 + 1/6 = 1/3.
With these rules in mind, we can calculate the probability that two carriers (i.e. heterozygotes) of an AR disease will have a child affected with the disease as ½ x ½ = ¼, since for each parent, the probability of any gametes carrying the disease allele is ½. This is consistent with what we already know from calculating probabilities using a Punnett Square (e.g. in a monohybrid cross Aa x Aa, ¼ of the offspring are aa).
We can likewise calculate probabilities in the more complex pedigree shown in Figure \(11\).
Assuming the disease has an AR pattern of inheritance, what is the probability that individual 14 will be affected? We can assume that individuals #1, #2, #3 and #4 are heterozygotes (Aa), because they each had at least one affected (aa) child, but they are not affected themselves. This means that there is a 2/3 chance that individual #6 is also Aa. This is because according to Mendelian inheritance, when two heterozygotes mate, there is a 1:2:1 distribution of genotypes AA:Aa:aa. However, because #6 is unaffected, he can’t be aa, so he is either Aa or AA, but the probability of him being Aa is twice as likely as AA. By the same reasoning, there is likewise a 2/3 chance that #9 is a heterozygous carrier of the disease allele.
If individual 6 is a heterozygous for the disease allele, then there is a ½ chance that #12 will also be a heterozygote (i.e. if the mating of #6 and #7 is Aa × AA, half of the progeny will be Aa; we are also assuming that #7, who is unrelated, does not carry any disease alleles). Therefore, the combined probability that #12 is also a heterozygote is 2/3 x 1/2 = 1/3. This reasoning also applies to individual #13, i.e. there is a 1/3 probability that he is a heterozygote for the disease. Thus, the overall probability that both individual #12 and #13 are heterozygous, and that a particular offspring of theirs will be homozygous for the disease alleles is 1/3 x 1/3 x 1/4 = 1/36. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/05%3A_Pedigrees_and_Populations/5.04%3A_Sporadic_and_Non-Heritable_Diseases.txt |
A population is a large group of individuals of the same species, who are capable of mating with each other. It is useful to know the frequency of particular alleles within a population, since this information can be used to calculate disease risks. Population genetics is also important in ecology and evolution, since changes in allele frequencies may be associated with migration or natural selection.
Allele frequencies may also be studied at the population level
The frequency of different alleles in a population can be determined from the frequency of the various phenotypes in the population. In the simplest system, with two alleles of the same locus (e.g. A,a), we use the symbol p to represent the frequency of the dominant allele within the population, and q for the frequency of the recessive allele. Because there are only two possible alleles, we can say that the frequency of p and q together represent 100% of the alleles in the population (p+q=1).
We can calculate the values of p and q, in a representative sample of individuals from a population, by simply counting the alleles and dividing by the total number of alleles examined. For a given allele, homozygotes will count for twice as much as heterozygotes.
For example:
genotype number of individuals
AA
Aa
aa
320
160
20
A (p)
a (q)
A (p)
p2
pq
a (q)
pq
q2
\begin{align} \mathbf{p} &= \mathbf{\dfrac{2(AA) + Aa}{total\: alleles\: counted}}\ &= \dfrac{2(320) + 160}{2(320) + 2(160) + 2(20)} = 0.8 \end{align}
\begin{align} \mathbf{q} &= \mathbf{\dfrac{2(aa) + Aa}{total\: alleles\: counted}}\ &= \dfrac{2(20) + 160}{2(320) + 2(160) + 2(20)} = 0.2 \end{align}
Hardy-Weinberg formula
With the allele frequencies of a population we can use an extension of the Punnett Square, and the product rule, to calculate the expected frequency of each genotype following random matings within the entire population. This is the basis of the Hardy-Weinberg formula:
$p^2 + 2pq + q^2=1 \label{HW}$
Here $p^2$ is the frequency of homozygotes AA, $2pq$ is the frequency of the heterozygotes, and $q^2$ is the frequency of homozygotes aa. Notice that if we substitute the allele frequencies we calculated above (p=0.8, q=0.2) into the Equation \ref{HW}, we obtain expected probabilities for each of the genotypes that exactly match our original observations:
\begin{align} &\mathrm{p^2=0.8^2=0.64} &&0.64 \times 500 = 320 \ &\mathrm{2pq= 2(0.8)(0.2)=0.32} &&0.32 \times 500 = 160 \ &\mathrm{q^2=0.2^2=0.04} &&0.04 \times 500 = 20 \end{align}
This is a demonstration of the Hardy-Weinberg Equilibrium, where both the genotype frequencies and allele frequencies in a population remain unchanged following successive matings within a population, if certain conditions are met. These conditions are listed in Table $1$. Few natural populations actually satisfy all of these conditions. Nevertheless, large populations of many species, including humans, appear to approach Hardy-Weinberg equilibrium for many loci. In these situations, deviations of a particular gene from Hardy-Weinberg equilibirum can be an indication that one of the alleles affects the reproductive success of organism, for example through natural selection or assortative mating.
Conditions for the Hardy-Weinberg equilibrium
• Random mating: Individuals of all genotypes mate together with equal frequency. Assortative mating, in which certain genotypes preferentially mate together, is a type of non-random mating.
• No natural selection: All genotypes have equal fitness.
• No migration: Individuals do not leave or enter the population.
• No mutation: The allele frequencies do not change due to mutation.
• Large population: Random sampling effects in mating (i.e. genetic drift) are insignificant in large populations.
The Hardy-Weinberg formula can also be used to estimate allele frequencies, when only the frequency of one of the genotypic classes is known. For example, if 0.04% of the population is affected by a particular genetic condition, and all of the affected individuals have the genotype aa, then we assume that q2 = 0.0004 and we can calculate p, q, and 2pq as follows:
$\mathrm{q^2 = 0.04\% = 0.0004}$
$\mathrm{q = \sqrt{0.0004} = 0.02}$
$\mathrm{p= 1-q = 0.98}$
$\mathrm{2pq = 2(0.98)(0.02) = 0.04}$
Thus, approximately 4% of the population is expected to be heterozygous (i.e. a carrier) of this genetic condition. Note that while we recognize that the population is probably not exactly in Hardy-Weinberg equilibrium for this locus, application of the Hardy-Weinberg formula nevertheless can give a reasonable estimate of allele frequencies, in the absence of any other information. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/05%3A_Pedigrees_and_Populations/5.06%3A_Population_Genetics.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
1. What are some of the modes of inheritance that are consistent with this pedigree?
2. In this pedigree in question 1, the mode of inheritance cannot be determined unamibguously.What are some examples of data (e.g. from other generations) that, if added to the pedigree would help determine the mode of inheritance?
3. For each of the following pedigrees, name the most likely mode of inheritance (AR=autosomal recessive, AD=autosomal dominant, XR=X-linked recessive, XD=X-linked dominant).(These pedigrees were obtained from various external sources).
a)
b)
c)
d)
4. The following pedigree represents a rare, autosomal recessive disease. What are the genotypes of the individuals who are indicated by letters?
5. If individual #1 in the following pedigree is a heterozygote for a rare, AR disease, what is the probability that individual #7 will be affected by the disease? Assume that #2 and the spouses of #3 and #4 are not carriers.
6. You are studying a population in which the frequency of individuals with a recessive homozygous genotype is 1%. Assuming the population is in Hardy-Weinberg equilibrium, calculate:
a) The frequency of the recessive allele.
b) The frequency of dominant allele.
c) The frequency of the heterozygous phenotype.
d) The frequency of the homozygous dominant phenotype.
7. Determine whether the following population is in Hardy-Weinberg equilibrium.
genotype number of individuals
AA
Aa
aa
432
676
92
8. Out of 1200 individuals examined, 432 are homozygous dominant (AA)for a particular gene. What numbers of individuals of the other two genotypic classes (Aa, aa) would be expected if the population is in Hardy-Weinberg equilibrium?
9. Propose an explanation for the deviation between the genotypic frequencies calculated in question 8 and those observed in the table in question 7.
Chapter 5 - Answers
5.1 The pedigree could show an AD, AR or XR mode of inheritance. It is most likely AD. It could be AR if the mother was a carrier, and the father was a homozygote. It could be XR if the mother was a carrier, and the father was a hemizygote. It cannot be XD, since the daughter (#2) would have necessarily inherited the disease allele on the X chromosome she received from her father.
5.2 There are many possible answers. Here are some possibilities: if neither of the parents of the father were affected (i.e. the paternal grandparents of children 1, 2, 3), then the disease could not be dominant. If only the paternal grandfather was affected, then the disease could only be X-linked recessive if the paternal grandmother was a heterozygote (which would be unlikely given that this is a rare disease allele).
5.3 a) The mode of inheritance is most likely AD, since every affected individual has an affected parent, and the disease is inherited even in four different matings to unrelated, unaffected individuals. It is very unlikely that it is XD or XR, in part because affected father had an affected son.
b) The mode of inheritance cannot be AD or XD, because affected individuals must have an affected parent when a disease allele is dominant. Neither can it be XR, because there is an affected daughter of a normal father. Therefore, it must be AR, and this is consistent with the pedigree.
c) The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It is not XR, because there are unaffected sons of an affected mother. It is therefore likely AR, but note that the recessive alleles for this condition appear to be relatively common in the population (note that two of the marriages were to unrelated, affected individuals).
d) The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It could be either XR or AR, but because all of the affected individuals are male, and no affected males pass the disease to their sons, it is likely XR.
5.4 If a represents the disease allele, individuals a, d, f (who all married into this unusual family) are AA, while b, c, e, g, h, i, j are all Aa, and k is aa.
5.5 There is a ½ chance that an offspring of any mating Aa x AA will be a carrier (Aa). So, there is a ½ chance that #3 will be Aa, and likewise for #4. If #3 is a carrier, there is again a ½ chance that #5 will be a carrier, and likewise for #6. If #5 and #6 are both Aa, then there is a ¼ chance that this monohybrid cross will result in #7 having the genotype aa, and therefore being affected by the disease. Thus, the joint probability is 1/2 x 1/2 x 1/2 x 1/2 x 1/4 =1/64.
5.6 a) q = -0.01. = 0.1
b) 1-q = p; 1-0.1 = 0.9
c) 2pq = 2(0.1)(0.9) = 0.18
d) p2 = 0.81
5.7 First, calculate allele frequencies:
$\mathrm{p = \dfrac{2(AA) + (Aa)}{total\: number\: of\: alleles\: scored} = \dfrac{2(432) + 676}{2(432+676+92)} = 0.6417}$
$\mathrm{q = \dfrac{2(aa) + (Aa)}{total\: number\: of\: alleles\: scored} = \dfrac{2(92) + 676}{2(432+676+92)} = 0.3583}$
Next, given these observed allele frequencies, calculate the genotypic frequencies that would be expected if the population was in Hardy-Weinberg equilibrium.
$\mathrm{p^2 = 0.6417^2 = 0.4118}$
$\mathrm{2pq = 2(0.6417)(0.3583) = 0.4598}$
$\mathrm{q^2 = 0.3583^2 = 0.1284}$
Finally, given these expected frequencies of each class, calculate the expected numbers of each in your sample of 1200 individuals, and compare these to your actual observations.
expected observed (reported in the original question)
AA 0.4118 × 1200 = 494
Aa 0.4598 × 1200 = 552
aa 0.1284 × 1200 = 154
432
676
92
The population does not appear to be at Hardy-Weinberg equilibrium, since the observed genotypic frequencies do not match the expectations. Of course, you could do a chi-square test to determine how significant the discrepancy is between observed and expected.
5.8 If in this theoretical question, the frequency of genotype of AA is set at 432/1200 and we are asked what frequencies of the other classes would fit a Hardy-Weinberg equilibrium. So, given that p2 = 432/1200, then p=0.6, and q=0.4. Given these allele frequencies and a sample size of 1200 individuals, then there should be 576 Aa individuals (2pq × 1200 = 2(0.6)(0.4) × 1200=576) and 192 aa individuals (q2 × 1200 = 0.42 × 1200 = 192), if the population was at Hardy-Weinberg equilibrium with 432 AA individuals.
5.9 The actual population appear to have more heterozygotes and few recessive homozygotes than would be expected for Hardy-Weinberg equilibrium. There are many possible reasons that a population may not be in equilibrium (see Table 5.1). In this case, there is possibly some selection against homozygous recessive genotypes, in favour of heterozygotes in particular. Perhaps the heterozygotes have some selective advantage that increases their fitness.
It is also worth noting the discrepancies between the allele frequencies calculated in 5.8 and 5.7. In question 5.7, we calculated the frequencies directly from the genotypes (this is the most accurate method, and does not require the population to be in equilibrium). In 5.8, we essentially estimated the frequency base on one of the phenotypic classes. The discrepancy between these calculations shows the limitations of using phenotypes to estimate allele frequencies, when a population is not in equilibrium.
5.S: Pedigrees and Populations (Summary)
• Pedigree analysis can be used to determine the mode of inheritance of specific traits such as diseases. Loci can be X- or Y-linked or autosomal in location and alleles either dominant or recessive with respect to wild type.
• If the mode of inheritance is known, a pedigree can be used to calculate the probability of inheritance of a particular genotype by an individual.
• The frequencies of all alleles and genotypes remain unchanged through successive generations of a population in Hardy-Weinberg equilibrium.
• Populations in true Hardy-Weinberg equilibrium have random mating, and no genetic drift, no migration, no mutation, and no selection with respect to the gene of interest.
• The Hardy-Weinberg formula can be used to estimate allele and genotype frequencies given only limited information about a population.
Key Terms
Pedigree charts
mode of inheritance
genetic counseling
incomplete penetrance
variable expressivity
proband
affected
carrier
autosomal dominant
autosomal recessive
X-linked dominant
X-linked recessive
Y-linked inheritance
Y-linked
mitochondrial inheritance (mtDNA)
sporadic
product rule
sum rule
population
p+q=1
Hardy-Weinberg formula
p2 + 2pq + q2=1
Hardy-Weinberg equilibrium
assortative mating
random mating
migration
genetic drift | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/05%3A_Pedigrees_and_Populations/5.E%3A_Pedigrees_and_Populations_%28Exercises%29.txt |
The principles of genetic analysis that we have described for a single locus can be extended to the study of alleles at two loci simultaneously. Analysis of two loci in parallel is required for genetic mapping and can also reveal gene interactions. These techniques are very useful for both basic and applied research. Before discussing these techniques, we will first revisit Mendel’s classical experiments.
• 6.1: Dihybrid Crosses
Mendel's Second Law, also called the Law of Independent Assortment, argues that two loci assort independently of each other during gamete formation. The commonly observed 9:3:3:1 phenotypic ratio that is predicted from this law can also be obtained using Punnett Square
• 6.2: Epistasis and Other Gene Interactions
Epistasis (which means “standing upon”) occurs when the phenotype of one locus masks, or prevents, the phenotype of another locus. Thus, following a dihybrid cross fewer than the typical four phenotypic classes will be observed with epistasis. As we have already discussed, in the absence of epistasis, there are four phenotypic classes among the progeny of a dihybrid cross.
• 6.3: Example of Multiple Genes Affecting One Character
Most aspects of the fur phenotypes of common cats can be explained by the action of just a few genes. Other genes, not described here, may further modify these traits and account for the phenotypes seen in tabby cats and in more exotic breeds, such as Siamese.
• 6.E: Genetic Analysis of Multiple Genes (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
• 6.S: Genetic Analysis of Multiple Genes (Summary)
These are the summary for Chapter 6 of Nickle and Barrette-Ng's "Online Open Genetics" TextMap.
Thumbnail: Coat color in animals is an example of a trait that is controlled by more than one locus. (Flickr-Gossamer1013-CC:AND).
06: Genetic Analysis of Multiple Genes
Mendel’s Second Law
Before Mendel, it had not yet been established that heritable traits were controlled by discrete factors. Therefor an important question was therefore whether distinct traits were controlled by discrete factors that were inherited independently of each other? To answer this, Mendel took two apparently unrelated traits, such as seed shape and seed color, and studied their inheritance together in one individual. He studied two variants of each trait: seed color was either green or yellow, and seed shape was either round or wrinkled. (He studied seven traits in all.) When either of these traits was studied individually, the phenotypes segregated in the classical 3:1 ratio among the progeny of a monohybrid cross (Figure \(2\)), with ¾ of the seeds green and ¼ yellow in one cross, and ¾ round and ¼ wrinkled in the other cross. Would this be true when both were in the same individual?
To analyze the segregation of both traits at the same time in the same individual, he crossed a pure breeding line of green, wrinkled peas with a pure breeding line of yellow, round peas to produce F1 progeny that were all green and round, and which were also dihybrids; they carried two alleles at each of two loci (Figure \(3\)).
If the inheritance of seed color was truly independent of seed shape, then when the F1 dihybrids were crossed to each other, a 3:1 ratio of one trait should be observed within each phenotypic class of the other trait (Figure \(3\)). Using the product law, we would therefore predict that if ¾ of the progeny were green, and ¾ of the progeny were round, then ¾ × ¾ = 9/16 of the progeny would be both round and green (Table \(1\)). Likewise, ¾ × ¼ = 3/16 of the progeny would be both round and yellow, and so on. By applying the product rule to all of these combinations of phenotypes, we can predict a 9:3:3:1 phenotypic ratio among the progeny of a dihybrid cross, if certain conditions are met, including the independent segregation of the alleles at each locus. Indeed, 9:3:3:1 is very close to the ratio Mendel observed in his studies of dihybrid crosses, leading him to state his Second Law, the Law of Independent Assortment, which we now express as follows: two loci assort independently of each other during gamete formation.
Definition: Mendel's Second Law
Two loci assort independently of each other during gamete formation.
Table \(1\): Phenotypic classes expected in monohybrid and dihybrid crosses for two seed traits in pea.
Frequency of phenotypic crosses within separate monohybrid crosses:
seed shape: ¾ round ¼ wrinkled
seed color: ¾ yellow ¼ green
Frequency of phenotypic crosses within a dihybrid cross:
¾ round × ¾ yellow = 9/16 round & yellow
¾ round × ¼ green = 3/16 round & green
¼ wrinkled × ¾ yellow = 3/16 wrinkled & yellow
¼ wrinkled × ¼ green = 1/16 wrinkled & green
The 9:3:3:1 phenotypic ratio that we calculated using the product rule can also be obtained using Punnett Square (Figure \(4\)). First, we list the genotypes of the possible gametes along each axis of the Punnett Square. In a diploid with two heterozygous genes of interest, there are up to four combinations of alleles in the gametes of each parent. The gametes from the respective rows and column are then combined in the each cell of the array. When working with two loci, genotypes are written with the symbols for both alleles of one locus, followed by both alleles of the next locus (e.g. AaBb, not ABab). Note that the order in which the loci are written does not imply anything about the actual position of the loci on the chromosomes.
To calculate the expected phenotypic ratios, we assign a phenotype to each of the 16 genotypes in the Punnett Square, based on our knowledge of the alleles and their dominance relationships. In the case of Mendel’s seeds, any genotype with at least one R allele and one Y allele will be round and yellow; these genotypes are shown in the nine, green-shaded cells in Figure \(4\). We can represent all of four of the different genotypes shown in these cells with the notation (R_Y_), where the blank line (__), means “any allele”. The three offspring that have at least one R allele and are homozygous recessive for y (i.e. R_yy) will have a round, green phenotype. Conversely the three progeny that are homozygous recessive r, but have at least one Y allele (rrY_) will have wrinkled, yellow seeds. Finally, the rarest phenotypic class of wrinkled, yellow seeds is produced by the doubly homozygous recessive genotype, rryy, which is expected to occur in only one of the sixteen possible offspring represented in the square.
Assumptions of the 9:3:3:1 ratio
Both the product rule and the Punnett Square approaches showed that a 9:3:3:1 phenotypic ratio is expected among the progeny of a dihybrid cross such as Mendel’s RrYy × RrYy. In making these calculations, we assumed that:
1. both loci assort independently;
2. one allele at each locus is completely dominant; and
3. each of four possible phenotypes can be distinguished unambiguously, with no interactions between the two genes that would alter the phenotypes.
Deviations from the 9:3:3:1 phenotypic ratio may indicate that one or more of the above conditions has not been met. Modified ratios in the progeny of a dihybrid cross can therefore reveal useful information about the genes involved.
Linkage is one of the most important reasons for distortion of the ratios expected from independent assortment. Linked genes are located close together on the same chromosome. This close proximity alters the frequency of allele combinations in the gametes. We will return to the concept of linkage in Chapter 7. Deviations from 9:3:3:1 ratios can also be due to interactions between genes. These interactions will be discussed in the remainder of this chapter. For simplicity, we will focus on examples that involve easily scored phenotypes, such as pigmentation. Nevertheless, keep in mind that the analysis of segregation ratios of any markers can provide insight into a wide range of biological processes they represent. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/06%3A_Genetic_Analysis_of_Multiple_Genes/6.01%3A_Dihybrid_Crosses.txt |
Some dihybrid crosses produce a phenotypic ratio that differs from 9:3:3:1, such as 9:3:4, 12:3:1, 9:7, or 15:1. Note that each of these modified ratios can be obtained by summing one or more of the 9:3:3:1 classes expected from our original dihybrid cross. In the following sections, we will look at some modified phenotypic ratios obtained from dihybrid crosses and what they might tell us about interactions between genes.
Recessive epistasis
Epistasis (which means “standing upon”) occurs when the phenotype of one locus masks, or prevents, the phenotype of another locus. Thus, following a dihybrid cross fewer than the typical four phenotypic classes will be observed with epistasis. As we have already discussed, in the absence of epistasis, there are four phenotypic classes among the progeny of a dihybrid cross. The four phenotypic classes correspond to the genotypes: A_B_, A_bb, aaB_, and aabb. If either of the singly homozygous recessive genotypes (i.e. A_bb or aaB_) has the same phenotype as the double homozygous recessive (aabb), then a 9:3:4 phenotypic ratio will be obtained. For example, in the Labrador Retriever breed of dogs (Figure \(5\)), the B locus encodes a gene for an important step in the production of melanin. The dominant allele, B is more efficient at pigment production than the recessive b allele, thus B_ hair appears black, and bb hair appears brown. A second locus, which we will call E, controls the deposition of melanin in the hairs. At least one functional E allele is required to deposit any pigment, whether it is black or brown. Thus, all retrievers that are ee fail to deposit any melanin (and so appear pale yellow), regardless of the genotype at the B locus (Figure \(6\)).
The ee genotype is therefore said to be epistatic to both the B and b alleles, since the homozygous ee phenotype masks the phenotype of the B locus. The B/b locus is said to be hypostatic to the ee genotype. Because the masking allele is in this case is recessive, this is called recessive epistasis.
Dominant epistasis
In some cases, a dominant allele at one locus may mask the phenotype of a second locus. This is called dominant epistasis, which produces a segregation ratio such as 12:3:1, which can be viewed as a modification of the 9:3:3:1 ratio in which the A_B_ class is combined with one of the other genotypic classes that contains a dominant allele. One of the best known examples of a 12:3:1 segregation ratio is fruit color in some types of squash (Figure \(7\)). Alleles of a locus that we will call B produce either yellow (B_) or green (bb) fruit. However, in the presence of a dominant allele at a second locus that we call A, no pigment is produced at all, and fruit are white. The dominant A allele is therefore epistatic to both B and bb combinations (Figure \(8\)). One possible biological interpretation of this segregation pattern is that the function of the A allele somehow blocks an early stage of pigment synthesis, before neither yellow or green pigments are produced.
Figure \(8\): Genotypes and phenotypes among the progeny of a dihybrid cross of squash plants heterozygous for two loci affecting fruit color. (Original-Deyholos-CC:AN) | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/06%3A_Genetic_Analysis_of_Multiple_Genes/6.02%3A_Epistasis_and_Other_Gene_Interactions.txt |
Cat Fur Genetics
Most aspects of the fur phenotypes of common cats can be explained by the action of just a few genes (Table 6-2). Other genes, not described here, may further modify these traits and account for the phenotypes seen in tabby cats and in more exotic breeds, such as Siamese.
For example, the X-linked Orange gene has two allelic forms. The OO allele produces orange fur, while the OB alleles produce non-orange (often black) fur. Note however, that because of X-chromosome inactivation the result is mosaicism in expression. In OO / OB female heterozygotes patches of black and orange are seen, which produces the tortoise shell pattern (Figure 6-13 A,B). This is a rare example of co-dominance since the phenotype of both alleles can be seen. Note that the cat in part A has short fur compared to the cat in part B; recessive alleles at an independent locus (L/l) produce long (ll) rather than short (L_) fur.
Alleles of the dilute gene affect the intensity of pigmentation, regardless of whether that pigmentation is due to black or orange pigment. Part C shows a black cat with at least one dominant allele of dilute (D_), in contrast to the cat in D, which is grey rather than black, because it has the dd genotype.
Epistasis is demonstrated by an allele of only one of the genes in Table \(2\). One dominant allele of white masking (W) prevents normal development of melanocytes (pigment producing cells). Therefore, cats with genotype (W_) will have entirely white fur regardless of the genotype at the Orange or dilute loci (part E). Although this locus produces a white colour, W_ is not the same as albinism, which is a much rarer phenotype caused by mutations in other genes. Albino cats can be distinguished by having red eyes, while W_ cats have eyes that are not red.
Piebald spotting is the occurrence of patches of white fur. These patches vary in size due to many reasons, including genotype. Homozygous cats with genotype ss do not have any patches of white, while cats of genotype Ss and SS do have patches of white, and the homozygotes tend to have a larger proportion of white fur than heterozygotes (part F). The combination of piebald spotting and tortoise shell patterning produce a calico cat, which has separate patches of orange, black, and white fur.
Table \(2\): Summary of simplified cat fur phenotypes and genotypes.
Trait Phenotype Genotype Comments
fur length short LL or Ll L is completely dominant
long ll
all white fur (non-albino) 100% white fur WW or Ww If the cat has red eyes it is albino, not W_. W is epistatic to all other fur color genes; if cat is W_, can’t infer genotypes for any other fur color genes.
ww
piebald spotting > 50% white patches (but not 100%) SS S is incompletely dominant and shows variable expressivity
< 50% white patches Ss
no white patches ss
orange fur all orange fur XOXO or XOY O is X-linked
tortoise shell variegation XOXB
no orange fur (often black) XBXB or XBY
dilute pigmentation pigmentation is intense Dd or dd D is completely dominant
pigmentation is dilute (e.g. gray rather than black; cream rather than orange; light brown rather than brown) dd
tabby tabby pattern AA or Aa This is a simplification of the tabby phenotype, which involves multiple genes
solid coloration aa
sex female XX
male XY | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/06%3A_Genetic_Analysis_of_Multiple_Genes/6.03%3A_Example_of_Multiple_Genes_Affecting_One_Character.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
Answer questions 6.1 -6.3 using the following biochemical pathway for fruit color.Assume all mutations (lower case allele symbols) are recessive, and that either precursor 1 or precursor 2 can be used to produce precursor 3.If the alleles for a particular gene are not listed in a genotype, you can assume that they are wild-type.
6.1 If1 and 2 and 3 are all colorless, and 4 is red, what will be the phenotypes associated with the following genotypes?
1. aa
2. bb
3. dd
4. aabb
5. aadd
6. bbdd
7. aabbdd
8. What will be the phenotypic ratios among the offspring of a cross AaBb × AaBb?
9. What will be the phenotypic ratios among the offspring of a cross BbDd× BbDd?
10. What will be the phenotypic ratios among the offspring of a cross AaDd × AaDd?
6.2 If1 and 2 are both colorless, and 3 is blue and 4 is red, what will be the phenotypes associated with the following genotypes?
1. aa
2. bb
3. dd
4. aabb
5. aadd
6. bbdd
7. aabbdd
8. What will be the phenotypic ratios among the offspring of a cross AaBb × AaBb?
9. What will be the phenotypic ratios among the offspring of a cross BbDd× BbDd?
10. What will be the phenotypic ratios among the offspring of a cross AaDd × AaDd?
6.3 If1 is colorless, 2 is yellow and 3 is blue and 4 is red, what will be the phenotypes associated with the following genotypes?
1. aa
2. bb
3. dd
4. aabb
5. aadd
6. bbdd
7. aabbdd
8. What will be the phenotypic ratios among the offspring of a cross AaBb × AaBb?
9. What will be the phenotypic ratios among the offspring of a cross BbDd× BbDd?
10. What will be the phenotypic ratios among the offspring of a cross AaDd × AaDd?
6.4 Which of the situations in questions 6.1 – 6.3 demonstrate epistasis?
6.5 If the genotypes written within the Punnett Square are from the F2 generation, what would be the phenotypes and genotypes of the F1 and P generations for:
1. Figure 6.6
2. Figure 6.8
3. Figure 6.10
4. Figure 6.12
6.6 To better understand how genes control the development of three‐dimensional structures, you conducted a mutant screen in Arabidopsis and identified a recessive point mutation allele of a single gene (g) that causes leaves to develop as narrow tubes rather than the broad flat surfaces that develop in wild‐type (G). Allele g causes a complete loss of function. Now you want to identify more genes involved in the same process. Diagram a process you could use to identify other genes that interact with gene g. Show all of the possible genotypes that could arise in the Fgeneration.
6.7 With reference to question 6.6, if the recessive allele, g is mutated again to make allele g*, what are the possible phenotypes of a homozygous g* g* individual?
6.8 Again, in reference to question 6.7, what are the possible phenotypes of a homozygous aagg individual, where a is a recessive allele of a second gene? In each case, also specify the phenotypic ratios that would be observed among the F1 progeny of a cross of AaGg x AaGg
6.9 Calculate the phenotypic ratios from a dihybrid cross involving the two loci shown in Figure 6.13.There may be more than one possible set of ratios, depending on the assumptions you make about the phenotype of allele b.
6.10 Use the product rule to calculate the phenotypic ratios expected from a trihybrid cross.Assume independent assortment and no epistasis/gene interactions.
Chapter 6 - Answers
6.1 If 1 and 2 and 3 are all colorless, and 4 is red, what will be the phenotypes associated with the following genotypes? All of these mutations are recessive. As always, if the genotype for a particular gene is not listed, you can assume that alleles for that gene are wild-type.
1. red(because A and B are redundant, so products 3 and then 4 can be made)
2. red(because A and B are redundant, so products 3 and then 4 can be made)
3. white (because product3 will accumulate and it is colorless)
4. white (because only product 1 and 2 will be present and both are colorless)
5. white (because only product 1 and 3 will be present and both are colorless)
6. white (because only product 2 and 3 will be present and both are colorless)
7. white (because only product 1 and 2 will be present and both are colorless)
8. 15 red : 1 white
9. 12 red : 4 white
10. 12 red :4 white
6.2
1. red(because A and B are redundant, so products 3 and then 4 can be made)
2. red(because A and B are redundant, so products 3 and then 4 can be made)
3. blue (because product 3 will accumulate, and it is blue)
4. white (because only product 1 and 2 will be present and both are colorless)
5. blue (because only product 1 and 3 will be present and 1 is colorless and 3 is blue)
6. blue(because only product 2 and 3 will be present and 2 is colorless and 3 is blue)
7. white (because only product 1 and 2 will be present and both are colorless)
8. 15 red : 1 white
9. 12 red : 4 blue
10. 12 red : 4 blue
6.3
1. red(because A and B are redundant, so products 3 and then 4 can be made)
2. red(because A and B are redundant, so products 3 and then 4 can be made)
3. blue (because product 3 will accumulate, and it is blue)
4. yellow (because only product 1 and 2 will be present and 1 is colorless and 2 is yellow)
5. blue (because only product 1 and 3 will be present and 1 is colorless and 3 is blue)
6. green? (because only product 2 and 3 will be present and 2 is yellow and 3 is blue, so probably the fruit will be some combination of those two colors)
7. yellow (because only product 1 and 2 will be present and 1 is colorless and 2 is yellow)
8. 15 red : 1 yellow
9. 12 red : 3 blue:1 green
10. 12 red : 4 blue
6.4 Epistasis is demonstrated when the phenotype for a homozygous mutant in one gene is the same as the phenotype for a homozygous mutant in two genes. So, the following situations from questions 6.1-6.3 demonstrated epistasis:
6.1 No epistasis is evident from the phenotypes, even though we know from the pathway provided that D is downstream of A and B.
6.2 The phenotypes show that D is epistatic to A, because aadd looks like AAdd or Aadd. Also D is epistatic to B, because bbdd looks like BBdd or Bbdd.
6.3 The phenotypes show that D is epistatic to A, because aadd looks like AAdd or Aadd. The phenotypes do not provide evidence for epistasis between B and D.
6.5 The answer is the same for a) – d)
P could have been either: AABB xaabboraaBB x AAbb;
F1 was : AaBb x AaBb
6.6 Conduct an enhancer/suppressor screen (which can also result in the identification of revertants, as well)
allow the plants to self‐pollinate in order to make any new, recessive mutations homozygous
6.7 Depending which amino acids were altered, and how they were altered, a second mutation in g*g* could either have no effect (in which case the phenotype would be the same as gg), or it could possibly cause a reversion of the phenotype to wild‐type, so that g*g* and GG have the same phenotype.
6.8 Depending on the normal function of gene A, and which amino acids were altered in allele a, there are many potential phenotypes for aagg:
Case 1: If the normal function of gene A is in an unrelated process (e.g. A is required for
root development, but not the development of leaves), then the phenotype of aagg will
be: short roots and narrow leaves. The phenotypic ratios among the progeny of a
dihybrid cross will be:
9
3
3
1
A_G_
A_gg
aaG_
aagg
wild-type
tubular leaves
normal roots
short roots
normal leaves
tubular leaves
short roots
Case 2: If the normal function of gene A is in the same process as G, such that a is a recessive allele that increases the severity of the gg mutant (i.e. a is an enhancer of g) then the phenotype of aagg could be : no leaves. The phenotypic ratios among the progeny of a dihybrid cross depend on whether aa mutants have a phenotype independent of gg, in other words, do aaG_ plants have a phenotype that is different from wild‐type or from A_gg. There is no way to know this without doing the experiment, since it depends on the biology of the particular gene, mutation and pathway involved, so there are three possible outcomes:
Case 2a) If aa is an enhancer of gg, and aaG_ plants have a mutant phenotype that differs from wild‐type or (A_gg) then the phenotypic ratios among the progeny of a dihybrid cross will be:
9
3
3
1
A_G_
A_gg
aaG_
aagg
wild-type
tubular leaves
(some phenotype that differs from gg; maybe small twisted leaves)
abnormal leaves
no leaves
Case 2b) If aa is an enhancer of gg, and aaG_ plants have a mutant phenotype that is the same as A_gg , the phenotypic ratios among the progeny of a dihybrid cross will be:
9
6
1
A_G_
A_gg aaG_
aagg
wild-type
tubular leaves
no leaves
Case 2c) If aa is an enhancer of gg, and aaG_ do not have a phenotype that differs from wild‐type then the phenotypic ratios among the progeny of a dihybrid cross will be:
12
3
1
A_G_ aaG_
A_gg
aagg
wild-type
tubular leaves
no leaves
Case 3: If the normal function of gene A is in the same process as G, such that a is a recessive allele that decreases the severity of the gg mutant (i.e. a is an suppressor of g) then the phenotype of aagg could be : wild‐type. The phenotypic ratios among the progeny of a dihybrid cross depend on whether aa mutants have a phenotype independent of gg, in other words, do aaG_ plants have a phenotype that is different from wild‐type or from A_gg. There is no way to know this without doing the experiment, since it depends on the biology of the particular gene, mutation and pathway involved, so there are three possible outcomes:
Case 3a) If aa is a suppressor of gg, and aaG_ plants have a mutant phenotype that differs from wild‐type or (A_gg) then the phenotypic ratios among the progeny of a dihybrid cross will be:
10
3
3
A_G_ aagg
A_gg
aaG_
wild-type
tubular leaves (some phenotype that differs from gg)
no leaves
Case 3b) If aa is an suppressor of gg, and aaG_ plants have a mutant phenotype that is the same as A_gg the phenotypic ratios among the progeny of a dihybrid cross will be:
10
6
A_G_ aagg
A_gg aaG_
wild-type
tubular leaves
Case 3c) If aa is an suppressor of gg, and aaG_ plants do not have a phenotype that differs from wild‐type then the phenotypic ratios among the progeny of a dihybrid cross will be:
13
3
A_G_ aaG_ aagg
A_gg
wild-type
tubular leaves
Case 4: If the normal function of gene A is in the same process as G, such that a is a
recessive allele that with a phenotype that is epistatic to the gg mutant then the
phenotype of both aaG_ and aagg could be : no leaves. The phenotypic ratios among
the progeny of a dihybrid cross will be:
9
4
3
A_G_
aaG_ aagg
A_gg
wild-type
no leaves
tubular leaves
Case … ?: There are many more phenotypes and ratios that could be imagined (e.g.
different types of dominance relationships, different types of epistasis, lethality…etc).
Isn’t genetics wonderful? It is sometimes shocking that more people don’t want to
become geneticists.
The point of this exercise is to show that many different ratios can be generated,
depending on the biology of the genes involved. On an exam, you could be asked to
calculate the ratio, given particular biological parameters. So, this exercise is also meant
to demonstrate that it is better to learn how to calculate ratios than just trying to
memorize which ratios match which parameters. In a real genetic screen, you would
observe the ratios, and then try to deduce something about the biology from those
ratios.
6.9 Calculate the phenotypic ratios from a dihybrid cross involving the two loci shown in Figure 6.13. There may be more than one possible set of ratios, depending on the assumptions you make about the phenotype of allele b.
Assuming that bb has no phenotype on its own (i.e. A_bb looks like A_B_), then aaB_ will have the mutant phenotype, and A_bb, A_B_, and aabb will appear phenotypically wild-type. The phenotypic ratio will be 13 wild-type : 3 mutant.
6.10 For a dihybrid cross, there are 4 classes, 9:3:3:1. In a trihybrid cross without gene interactions, each of these 4 classes will be further split into a 3:1 ratio based on the phenotype at the third locus. For example, 9 x 3 =27 and 9 x 1 = 9. This explains the first two terms of the complete ratio: 27:9:9:9:3:3:3:1.
6.S: Genetic Analysis of Multiple Genes (Summary)
• The alleles of different loci are inherited independently of each other, unless they are genetically linked.
• The expected phenotypic ratio of a dihybrid cross is 9:3:3:1, except in cases of linkage or gene interactions that modify this ratio.
• Modified ratios from 9:3:3:1 are seen in the case of recessive and dominant epistasis, duplicate genes, and complementary gene action. This usually indicates that the two genes interact within the same biological pathway.
Key Terms
Mendel’s Second Law
Dihybrid
9:3:3:19:3:4
12:3:1
independent assortment
linkage
recessive epistasis
dominant epistasis
complementary action
redundancy
duplicate gene action
Orange, long, dilute, White masking, piebald spotting, calico | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/06%3A_Genetic_Analysis_of_Multiple_Genes/6.E%3A_Genetic_Analysis_of_Multiple_Genes_%28Exercises%29.txt |
As we learned in Chapter 6, Mendel reported that the pairs of loci he observed behaved independently of each other; for example, the segregation of seed color alleles was independent from the segregation of alleles for seed shape. This observation was the basis for his Second Law (Independent Assortment), and contributed greatly to our understanding of heredity. However, further research showed that Mendel’s Second Law did not apply to every pair of genes that could be studied. In fact, we now know that alleles of loci that are located close together on the same chromosome tend to be inherited together. This phenomenon is called linkage, and is a major exception to Mendel’s Second Law of Independent Assortment. Researchers use linkage to determine the location of genes along chromosomes in a process called genetic mapping. The concept of gene linkage is important to the natural processes of heredity and evolution.
Figure \(1\): Linkage affects the frequency at which some combinations of traits are observed. (Wikipedia-Abiyoyo-CC:AN)
• 7.1: Linkage
Mendel’s Second Law does not apply to every pair of genes that could be studied. In fact, we now know that alleles of loci that are located close together on the same chromosome tend to be inherited together. This phenomenon is called linkage, and is a major exception to Mendel’s Second Law of Independent Assortment. Researchers use linkage to determine the location of genes along chromosomes in a process called genetic mapping and is important to natural processes of heredity and evolution.
• 7.2: Recombination
In heredity, recombination is any process that results in gametes with combinations of alleles that were not present in the gametes of a previous generation. Interchromosomal recombination occurs either through independent assortment of alleles whose loci are on different chromosomes. Intrachromosomal recombination occurs through crossovers between loci on the same chromosomes . In both cases, recombination is a process that occurs during meiosis (mitotic recombination is relatively rare).
• 7.3: Linkage Reduces Recombination Frequency
• 7.4: Crossovers Allow Recombination of Linked Loci
• 7.5: Inferring Recombination From Genetic Data
• 7.6: Genetic Mapping
Because the frequency of recombination between two loci (up to 50%) is roughly proportional to the chromosomal distance between them, we can use recombination frequencies to produce genetic maps of all the loci along a chromosome and ultimately in the whole genome.
• 7.7: Mapping With Three-Point Crosses
A particularly efficient method of mapping three genes at once is the three-point cross, which allows the order and distance between three potentially linked genes to be determined in a single cross experiment. This is particularly useful when mapping a new mutation with an unknown location to two previously mapped loci.
• 7.E: Linkage and Mapping (Exercises)
• 7.S: Linkage and Mapping (Summary)
07: Linkage and Mapping
As we learned in Chapter 6, Mendel reported that the pairs of loci he observed behaved independently of each other; for example, the segregation of seed color alleles was independent from the segregation of alleles for seed shape. This observation was the basis for his Second Law (Independent Assortment), and contributed greatly to our understanding of heredity. However, further research showed that Mendel’s Second Law did not apply to every pair of genes that could be studied. In fact, we now know that alleles of loci that are located close together on the same chromosome tend to be inherited together. This phenomenon is called linkage, and is a major exception to Mendel’s Second Law of Independent Assortment. Researchers use linkage to determine the location of genes along chromosomes in a process called genetic mapping. The concept of gene linkage is important to the natural processes of heredity and evolution. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/07%3A_Linkage_and_Mapping/7.01%3A__Linkage.txt |
The term “recombination” is used in several different contexts in genetics. In reference to heredity, recombination is defined as any process that results in gametes with combinations of alleles that were not present in the gametes of a previous generation (see Figure \(2\)). Interchromosomal recombination occurs either through independent assortment of alleles whose loci are on different chromosomes (Chapter 6). Intrachromosomal recombination occurs through crossovers between loci on the same chromosomes (as described below). It is important to remember that in both of these cases, recombination is a process that occurs during meiosis (mitotic recombination may also occur in some species, but it is relatively rare). If meiosis results in recombination, the products are said to have a recombinant genotype. On the other hand, if no recombination occurs during meiosis, the products have their original combinations and are said to have a non-recombinant, or parental genotype. Recombination is important because it contributes to the genetic variation that may be observed between individuals within a population and acted upon by selection to produce evolution.
As an example of interchromosomal recombination, consider loci on two different chromosomes as shown in Figure \(2\). We know that if these loci are on different chromosomes, there are no physical connections between them, so they are unlinked and will segregate independently as did Mendel’s traits. The segregation depends on the relative orientation of each pair of chromosomes at metaphase. Since the orientation is random and independent of other chromosomes, each of the arrangements (and their meiotic products) is equally possible for two unlinked loci as shown in Figure \(2\). More precisely, there is a 50% probability for recombinant genotypes, and a 50% probability for parental genotypes within the gametes produced by a meiocyte with unlinked loci. Indeed, if we examined all of the gametes that could be produced by this individual (which are the products of multiple independent meioses), we would note that approximately 50% of the gametes would be recombinant, and 50% would be parental. Recombination frequency (RF) is simply the number of recombinant gametes, divided by the total number of gametes. A frequency of approximately 50% recombination is therefore a defining characteristic of unlinked loci. Thus the greatest recombinant frequency expected is ~50%.
7.03: Linkage Reduces Recombination Frequency
Having considered unlinked loci above, let us turn to the opposite situation, in which two loci are so close together on a chromosome that the parental combinations of alleles always segregate together (Figure \(3\)). This is because during meiosis they are so close that there are no crossover events between the two loci and the alleles at the two loci are physically attached on the same chromatid and so they always segregate together into the same gamete. In this case, no recombinants will be present following meiosis, and the recombination frequency will be 0%. This is complete (or absolute) linkage and is rare, as the loci must be so close together that crossovers are never detected between them.
7.04: Crossovers Allow Recombination of Linked Loci
Thus far, we have only considered situations with either no linkage (50% recombination) or complete linkage (0% recombination). It is also possible to obtain recombination frequencies between 0% and 50%, which is a situation we call incomplete (or partial) linkage. Incomplete linkage occurs when two loci are located on the same chromosome but the loci are far enough apart so that crossovers occur between them during some, but not all, meioses. Genes that are on the same chromosome are said to be syntenic regardless of whether they are completely or incompletely linked. All linked genes are syntenic, but not all syntenic genes are linked, as we will learn later.
Crossovers occur during prophase I of meiosis, when pairs of homologous chromosomes have aligned with each other in a process called synapsis. Crossing over begins with the breakage of DNA of a pair of non-sister chromatids. The breaks occur at corresponding positions on two non-sister chromatids, and then the ends of non-sister chromatids are connected to each other resulting in a reciprocal exchange of double-stranded DNA (Figure \(4\)). Generally every pair of chromosomes has at least one (and often more) crossovers during meioses (Figure \(5\)).
Figure \(5\): A crossover between two linked loci can generate recombinant genotypes (AB, ab), from the chromatids involved in the crossover. Remember that multiple, independent meioses occur in each organism, so this particular pattern of recombination will not be observed among all the meioses from this individual. (Original-Deyholos-CC:AN)
Because the location of crossovers is essentially random along the chromosome, the greater the distance between two loci, the more likely a crossover will occur between them. Furthermore, loci that are on the same chromosome, but are sufficiently far apart from each other, will on average have multiple crossovers between them and they will behave as though they are completely unlinked. A recombination frequency of 50% is therefore the maximum recombination frequency that can be observed, and is indicative of loci that are either on separate chromosomes, or are located very far apart on the same chromosome. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/07%3A_Linkage_and_Mapping/7.02%3A__Recombination.txt |
In the preceding examples, we had the advantage of knowing the approximate chromosomal positions of each allele involved, before we calculated the recombination frequencies. Knowing this information beforehand made it relatively easy to define the parental and recombinant genotypes, and to calculate recombination frequencies. However, in most experiments, we cannot directly examine the chromosomes, or even the gametes, so we must infer the arrangement of alleles from the phenotypes over two or more generations. Importantly, it is generally not sufficient to know the genotype of individuals in just one generation; for example, given an individual with the genotype AaBb, we do not know from the genotype alone whether the loci are located on the same chromosome, and if so, whether the arrangement of alleles on each chromosome is AB and ab or Ab and aB (Figure $6$). The top cell has the two dominant alleles together and the two recessive alleles together and is said to have the genes in the coupling (or cis) configuration. The alternative shown in the cell below is that the genes are in the repulsion (or trans) configuration.
Fortunately for geneticists, the arrangement of alleles can sometimes be inferred if the genotypes of a previous generation are known. For example, if the parents of AaBb had genotypes AABB and aabb respectively, then the parental gametes that fused to produce AaBb would have been genotype AB and genotype ab. Therefore, prior to meiosis in the dihybrid, the arrangement of alleles would likewise be AB and ab (Figure $7$). Conversely, if the parents of AaBb had genotypes aaBB and AAbb, then the arrangement of alleles on the chromosomes of the dihybrid would be aB and Ab. Thus, the genotype of the previous generation can determine which of an individual’s gametes are considered recombinant, and which are considered parental.
Let us now consider a complete experiment in which our objective is to measure recombination frequency (Figure $8$). We need at least two alleles for each of two genes, and we must know which combinations of alleles were present in the parental gametes. The simplest way to do this is to start with pure-breeding lines that have contrasting alleles at two loci. For example, we could cross short-tailed mice, brown mice (aaBB) with long-tailed, white mice (AAbb). Based on the genotypes of the parents, we know that the parental gametes will be aB or Ab (but not ab or AB), and all of the progeny will be dihybrids, AaBb. We do not know at this point whether the two loci are on different pairs of homologous chromosomes, or whether they are on the same chromosome, and if so, how close together they are.
The recombination events that may be detected will occur during meiosis in the dihybrid individual. If the loci are completely or partially linked, then prior to meiosis, alleles aB will be located on one chromosome, and alleles Ab will be on the other chromosome (based on our knowledge of the genotypes of the gametes that produced the dihybrid). Thus, recombinant gametes produced by the dihybrid will have the genotypes ab or AB, and non-recombinant (i.e. parental) gametes will have the genotypes aB or Ab.
How do we determine the genotype of the gametes produced by the dihybrid individual? The most practical method is to use a testcross (Figure $8$), in other words to mate AaBb to an individual that has only recessive alleles at both loci (aabb). This will give a different phenotype in the F2 generation for each of the four possible combinations of alleles in the gametes of the dihybrid. We can then infer unambiguously the genotype of the gametes produced by the dihybrid individual, and therefore calculate the recombination frequency between these two loci. For example, if only two phenotypic classes were observed in the F2 (i.e. short tails and brown fur (aaBb), and white fur with long tails (Aabb) we would know that the only gametes produced following meiosis of the dihybrid individual were of the parental type: aB and Ab, and the recombination frequency would therefore be 0%. Alternatively, we may observe multiple classes of phenotypes in the F2 in ratios such as shown in Table $1$:
Table $1$: An example of quantitative data that may be observed in a genetic mapping experiment involving two loci. The data correspond to the F2 generation in the cross shown in Figure $8$.
tail phenotype
fur phenotype
number of progeny
gamete from dihybrid
genotype of F2 from test cross
(P)arental or
(R)ecombinant
short
brown
48
aB
aaBb
P
long
white
42
Ab
Aabb
P
short
white
13
ab
aabb
R
long
brown
17
AB
AaBb
R
Given the data in Table $1$, the calculation of recombination frequency is straightforward:
\begin{align} \textrm{recombination frequency} &= \mathrm{\dfrac{number\: of\: recombinant\: gametes}{total\: number\: of\: gametes\: scored}}\ \textrm{R.F.} &= \dfrac{13+17}{48+42+13+17}\ &=25\% \end{align} | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/07%3A_Linkage_and_Mapping/7.05%3A__Inferring_Recombination_From_Genetic_Data.txt |
Because the frequency of recombination between two loci (up to 50%) is roughly proportional to the chromosomal distance between them, we can use recombination frequencies to produce genetic maps of all the loci along a chromosome and ultimately in the whole genome. The units of genetic distance are called map units (mu) or centiMorgans (cM), in honor of Thomas Hunt Morgan by his student, Alfred Sturtevant, who developed the concept. Geneticists routinely convert recombination frequencies into cM: the recombination frequency in percent is approximately the same as the map distance in cM. For example, if two loci have a recombination frequency of 25% they are said to be ~25cM apart on a chromosome (Figure \(9\)). Note: this approximation works well for small distances (RF<30%) but progressively fails at longer distances because the RF reaches a maximum at 50%. Some chromosomes are >100 cM long but loci at the tips only have an RF of 50%. The method for mapping of these long chromosomes is shown below.
Note that the map distance of two loci alone does not tell us anything about the orientation of these loci relative to other features, such as centromeres or telomeres, on the chromosome.
Map distances are always calculated for one pair of loci at a time. However, by combining the results of multiple pairwise calculations, a genetic map of many loci on a chromosome can be produced (Figure \(10\)). A genetic map shows the map distance, in cM, that separates any two loci, and the position of these loci relative to all other mapped loci. The genetic map distance is roughly proportional to the physical distance, i.e. the amount of DNA between two loci. For example, in Arabidopsis, 1.0 cM corresponds to approximately 150,000bp and contains approximately 50 genes. The exact number of DNA bases in a cM depends on the organism, and on the particular position in the chromosome; some parts of chromosomes (“crossover hot spots”) have higher rates of recombination than others, while other regions have reduced crossing over and often correspond to large regions of heterochromatin.
When a novel gene or locus is identified by mutation or polymorphism, its approximate position on a chromosome can be determined by crossing it with previously mapped genes, and then calculating the recombination frequency. If the novel gene and the previously mapped genes show complete or partial linkage, the recombination frequency will indicate the approximate position of the novel gene within the genetic map. This information is useful in isolating (i.e. cloning) the specific fragment of DNA that encodes the novel gene, through a process called map-based cloning.
Genetic maps are also useful to track genes/alleles in breeding crops and animals, in studying evolutionary relationships between species, and in determining the causes and individual susceptibility of some human diseases.
Genetic maps are useful for showing the order of loci along a chromosome, but the distances are only an approximation. The correlation between recombination frequency and actual chromosomal distance is more accurate for short distances (low RF values) than long distances. Observed recombination frequencies between two relatively distant markers tend to underestimate the actual number of crossovers that occurred. This is because as the distance between loci increases, so does the possibility of having a second (or more) crossovers occur between the loci. This is a problem for geneticists, because with respect to the loci being studied, these double-crossovers produce gametes with the same genotypes as if no recombination events had occurred (Figure \(11\)) – they have parental genotypes. Thus a double crossover will appear to be a parental type and not be counted as a recombinant, despite having two (or more) crossovers. Geneticists will sometimes use specific mathematical formulae to adjust large recombination frequencies to account for the possibility of multiple crossovers and thus get a better estimate of the actual distance between two loci. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/07%3A_Linkage_and_Mapping/7.06%3A__Genetic_Mapping.txt |
A particularly efficient method of mapping three genes at once is the three-point cross, which allows the order and distance between three potentially linked genes to be determined in a single cross experiment (Figure $12$). This is particularly useful when mapping a new mutation with an unknown location to two previously mapped loci. The basic strategy is the same as for the dihybrid mapping experiment; pure breeding lines with contrasting genotypes are crossed to produce an individual heterozygous at three loci (a trihybrid), which is then testcrossed to determine the recombination frequency between each pair of genes.
One useful feature of the three-point cross is that the order of the loci relative to each other can usually be determined by a simple visual inspection of the F2 segregation data. If the genes are linked, there will often be two phenotypic classes that are much more infrequent than any of the others. In these cases, the rare phenotypic classes are usually those that arose from two crossover events, in which the locus in the middle is flanked by a crossover on either side of it. Thus, among the two rarest recombinant phenotypic classes, the one allele that differs from the other two alleles relative to the parental genotypes likely represents the locus that is in the middle of the other two loci. For example, based on the phenotypes of the pure-breeding parents in Figure $12$, the parental genotypes are aBC and AbC (remember the order of the loci is unknown, and it is not necessarily the alphabetical order in which we wrote the genotypes). Because we can deduce from the outcome of the testcross (Table $2$) that the rarest genotypes were abC and ABc, we can conclude that locus A that is most likely located between the other two loci, since it would require a recombination event between both A and B and between A and C in order to generate these gametes. Thus, the order of loci is BAC (which is equivalent to CAB).
Table $2$: An example of data that might be obtained from the F2 generation of the three-point cross shown in Figure $12$. The rarest phenotypic classes correspond to double recombinant gametes ABc and abC. Each phenotypic class and the gamete from the trihybrid that produced it can also be classified as parental (P) or recombinant (R) with respect to each pair of loci (A,B), (A,C), (B,C) analyzed in the experiment.
tail phenotype
fur phenotype
whisker phenotype
number of progeny
gamete from trihybrid
genotype of F2 from test cross
loci A, B
loci A, C
loci B, C
short
brown
long
5
aBC
aaBbCc
P
R
R
long
white
long
38
AbC
AabbCc
P
P
P
short
white
long
1
abC
aabbCc
R
R
P
long
brown
long
16
ABC
AaBbCc
R
P
R
short
brown
short
42
aBc
aaBbcc
P
P
P
long
white
short
5
Abc
Aabbcc
P
R
R
short
white
short
12
abc
aabbcc
R
P
R
long
brown
short
1
ABc
AaBbcc
R
R
P
Recombination frequencies may be calculated for each pair of loci in the three-point cross as we did before for one pair of loci in our dihybrid (Figure 7. 8).
\begin{alignat}{2} \textrm{loci A,B R.F.} = &\dfrac{1+16+12+1}{120} &&= 25\%\ \textrm{loci A,C R.F.} = &\dfrac{1+5+1+5}{120} &&= 10\%\ \textrm{loci B,C R.F.} = &\dfrac{5+16+12+5}{120} &&= 32\%\ \textrm{(not corrected for double}\ \textrm{crossovers)}\hspace{40px} \end{alignat}
However, note that in the three point cross, the sum of the distances between A-B and A-C (35%) is less than the distance calculated for B-C (32%)(Figure $13$). this is because of double crossovers between B and C, which were undetected when we considered only pairwise data for B and C. We can easily account for some of these double crossovers, and include them in calculating the map distance between B and C, as follows. We already deduced that the map order must be BAC (or CAB), based on the genotypes of the two rarest phenotypic classes in Table $2$. However, these double recombinants, ABc and abC, were not included in our calculations of recombination frequency between loci B and C. If we included these double recombinant classes (multiplied by 2, since they each represent two recombination events), the calculation of recombination frequency between B and C is as follows, and the result is now more consistent with the sum of map distances between A-B and A-C.
\begin{align} \textrm{loci B,C R.F.} &= \dfrac{5+16+12+5+2(1)+2(1)}{120} = 35\%\ \textrm{(corrected for double}&\ \textrm{recombinants)}& \end{align}
Thus, the three point cross was useful for:
1. determining the order of three loci relative to each other,
2. calculating map distances between the loci, and
3. detecting some of the double crossover events that would otherwise lead to an underestimation of map distance.
However, it is possible that other, double crossovers events remain undetected, for example double crossovers between loci A,B or between loci A,C. Geneticists have developed a variety of mathematical procedures to try to correct for things like double crossovers during large-scale mapping experiments.
As more and more genes are mapped a better genetic map can be constructed. Then, when a new gene is discovered, it can be mapped relative to other genes of known location to determine its location. All that is needed to map a gene is two alleles, a wild type allele (e.g. A) and a mutant allele (e.g. 'a'). | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/07%3A_Linkage_and_Mapping/7.07%3A__Mapping_With_Three-Point_Crosses.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
7.1 Compare recombination and crossover. How are these similar? How are they different?
7.2 Explain why it usually necessary to start with pure-breeding lines when measuring genetic linkage by the methods presented in this chapter.
7.3 If you knew that a locus that affected earlobe shape was tightly linked to a locus that affected susceptibility to cardiovascular disease human, under what circumstances would this information be clinically useful?
7.4 In a previous chapter, we said a 9:3:3:1 phenotypic ratio was expected among the progeny of a dihybrid cross, in absence of gene interaction.
a) What does this ratio assume about the linkage between the two loci in the dihybrid cross?
b) What ratio would be expected if the loci were completely linked? Be sure to consider every possible configuration of alleles in the dihybrids.
7.5 Given a dihybrid with the genotype CcEe:
a) If the alleles are in coupling (cis) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?
b) If the alleles are in repulsion (trans) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?
7.6 Imagine the white flowers are recessive to purple flowers, and yellow seeds are recessive to green seeds. If a green-seeded, purple-flowered dihybrid is testcrossed, and half of the progeny have yellow seeds, what can you conclude about linkage between these loci? What do you need to know about the parents of the dihybrid in this case?
7.7 In corn (i.e. maize, a diploid species), imagine that alleles for resistance to a particular pathogen are recessive and are linked to a locus that affects tassel length (short tassels are recessive to long tassels). Design a series of crosses to determine the map distance between these two loci. You can start with any genotypes you want, but be sure to specify the phenotypes of individuals at each stage of the process. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.
7.8 In a mutant screen in Drosophila, you identified a gene related to memory, as evidenced by the inability of recessive homozygotes to learn to associate a particular scent with the availability of food. Given another line of flies with an autosomal mutation that produces orange eyes, design a series of crosses to determine the map distance between these two loci. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.
7.9 Image that methionine heterotrophy, chlorosis (loss of chlorophyll), and absence of leaf hairs (trichomes) are each caused by recessive mutations at three different loci in Arabidopsis. Given a triple mutant, and assuming the loci are on the same chromosome, explain how you would determine the order of the loci relative to each other.
7.10 If the progeny of the cross aaBB x AAbb is testcrossed, and the following genotypes are observed among the progeny of the testcross, what is the frequency of recombination between these loci?
AaBb 135
Aabb 430
aaBb 390
aabb 120
7.11 Three loci are linked in the order B-C-A. If the A-B map distance is 1cM, and the B-C map distance is 0.6cM, given the lines AaBbCc and aabbcc, what will be the frequency of Aabb genotypes among their progeny if one of the parents of the dihybrid had the genotypes AABBCC?
7.12 Genes for body color (B black dominant to b yellow) and wing shape (C straight dominant to c curved) are located on the same chromosome in flies. If single mutants for each of these traits are crossed (i.e. a yellow fly crossed to a curved-wing fly), and their progeny is testcrossed, the following phenotypic ratios are observed among their progeny.
black, straight
yellow, curved
black, curved
yellow, straight
17
12
337
364
a) Calculate the map distance between B and C.
b) Why are the frequencies of the two smallest classes not exactly the same?
7.13 Given the map distance you calculated between B-C in question 12, if you crossed a double mutant (i.e. yellow body and curved wing) with a wild-type fly, and testcrossed the progeny, what phenotypes in what proportions would you expect to observe among the F2 generation?
7.14 In a three-point cross, individuals AAbbcc and aaBBCC are crossed, and their F1 progeny is testcrossed. Answer the following questions based on these F2 frequency data.
aaBbCc
480
AaBbcc
15
AaBbCc
10
aaBbcc
1
aabbCc
13
Aabbcc
472
AabbCc
1
aabbcc
8
a) Without calculating recombination frequencies, determine the relative order of these genes.
b) Calculate pair-wise recombination frequencies (without considering double cross overs) and produce a genetic map.
c) Recalculate recombination frequencies accounting for double recombinants.
7.15 Wild-type mice have brown fur and short tails. Loss of function of a particular gene produces white fur, while loss of function of another gene produces long tails, and loss of function at a third locus produces agitated behaviour. Each of these loss of function alleles is recessive. If a wild-type mouse is crossed with a triple mutant, and their F1 progeny is test-crossed, the following recombination frequencies are observed among their progeny. Produce a genetic map for these loci.
Chapter 7 - Answers
7.1
Crossovers are defined cytologically; they are observed directly under the microscope.
Recombination is defined genetically; it is calculated from observed phenotypic proportions.
Some crossovers lead to recombination, but not all crossovers result in recombination.
Some recombinations involve crossovers, but not all recombinations result from crossovers.
Crossovers happen between sister and non-sister chromatids. If the chromatids involved the crossover have identical alleles, there will not be any recombination.
Crossovers can also happen without causing recombination when there are two crossovers between the loci being scored for recombination.
Recombination can occur without crossover when loci are on different chromosomes.
7.2
The use of pure breeding lines allows the researcher to be sure that he/she is working with homozygous genotypes. If a parent is known to be homozygous, then all of its gametes will have the same genotype. This simplifies the definition of parental genotypes and therefore the calculation of recombination frequencies.
7.3
This would suggest that individuals with a particular earlobe phenotype may also carry one or more alleles that increased their risk of cardiovascular disease. These individuals could therefore be informed of their increased risk and have an opportunity to seek increased monitoring and reduce other risk factors.
7.4 a)
It assumes that the loci are completely unlinked.
b)
If the parental gametes were AB and ab, then the gametes produced by the dihybrids would also be AB and ab, and the offspring of a cross between the two dihybrids would all be genotype AABB:AaBb:aabb,in a 1:2:1 ratio.
If the parental gametes were Ab and aB, then the gametes produced by the dihybrids would also be Ab and aB, and the offspring of a cross between the two dihybrids would all be genotype AAbb:AaBb:aaBB, in a 1:2:1 ratio.
fur
tail
behaviour
white
short
normal
16
brown
short
agitated
0
brown
short
normal
955
white
short
agitated
36
white
long
normal
0
brown
long
agitated
14
brown
long
normal
46
white
long
agitated
933
7.5 a) Parental: CcEe and ccee; Recombinant: Ccee and ccEe.
b) Parental: Ccee and ccEe; Recombinant: CcEe and ccee.
7.6 Let WwYy be the genotype of a purple-flowered (W), green seeded (Y) dihybrid . Half of the progeny of the cross WwYy × wwyy will have yellow seeds whether the loci are linked or not. The proportion of the seeds that are also either white or purple flowered would help you to know about the linkage between the two loci only if the genotypes of the parents of the dihybrid were also known.
7.7
Let tt be the genotype of a short tassels, and rr is the genotype of pathogen resistant plants. We need to start with homozygous lines with contrasting combinations of alleles, for example:
P: RRtt (pathogen sensitive, short tassels) × rrTT (pathogen resistant, long tassels)
F1: RrTt (sensitive, long) × rrtt (resistant, short)
F2: parental Rrtt (sensitive, short) , rrTt (resistant, long)
Recombinant rrtt (resistant, short) , RrTt (sensitive, long)
7.8 Let mm be the genotype of a mutants that fail to learn, and ee is the genotype of orange eyes. We need to start with homozygous lines with contrasting combinations of alleles, for example (wt means wild-type):
P: MMEE (wt eyes, wt learning) × mmee (orange eyes, failure to learn)
F1: MmEe (wt eyes, wt learning) × mmee (orange eyes, failure to learn)
F2: parental MmEe (wt eyes, wt learning) , mmee (orange eyes, failure to learn)
Recombinant Mmee (wt eyes, failure to learn) , mmEe (orange eyes, wt learning)
7.9 Given a triple mutant aabbcc , cross this to a homozygote with contrasting genotypes, i.e. AABBCC, then testcross the trihybrid progeny, i.e.
P: AABBCC × aabbcc
F1: AaBbCc × aabbcc
Then, in the F2 progeny, find the two rarest phenotypic classes; these should have reciprocal genotypes, e.g. aaBbCc and AAbbcc. Find out which of the three possible orders of loci (i.e. A-B-C, B-A-C, or B-C-A) would, following a double crossover that flanked the middle marker, produce gametes that correspond to the two rarest phenotypic classes. For example, if the rarest phenotypic classes were produced by genotypes aaBbCc and AAbbcc, then the dihybrid’s contribution to these genotypes was aBC and Abc. Since the parental gametes were ABC and abc the only gene order that is consistent with aBC and Abc being produced by a double crossover flanking a middle marker is B-A-C (which is equivalent to C-A-B).
7.10 If the progeny of the cross aaBB x AAbb is testcrossed, and the following genotypes are observed among the progeny of the testcross, what is the frequency of recombination between these loci?
AaBb 135
Aabb 430
aaBb 390
aabb 120
(135 + 120)/(135+120+390+430)= 24%
7.11
Based on the information given, the recombinant genotypes with respect to these loci will be Aabb and aaBb. The frequency of recombination between A-B is 1cM=1%, based on the information given in the question, so each of the two recombinant genotypes should be present at a frequency of about 0.5%. Thus, the answer is 0.5%.
7.12
a) 4cM
b) Random sampling effects; the same reason that many human families do not have an equal number of boys and girls.
7.13
There would be approximately 2% of each of the recombinants: (yellow, straight) and (black, curved), and approximately 48% of each of the parentals: (yellow, curved) and (black, straight).
7.14
a) Without calculating recombination frequencies, determine the relative order of these genes.
A-C-B
b)
A-B 4.6%
A-C 2%
B-C 3%
B C A
|--------------|---------|
3cM 2cM
A-B
A-C
B-C
aBC
0
0
0
ABc
15
0
15
ABC
10
10
0
aBc
0
1
1
abC
13
0
13
Abc
0
0
0
AbC
0
1
1
abc
8
8
0
TOTAL
46
20
30
%
4.6
2
3
c) Recalculate recombination frequencies accounting for double recombinants
A-B
A-C
B-C
aBC
0
0
0
ABc
15
0
15
ABC
10
10
0
aBc
1 x 2
1
1
abC
13
0
13
Abc
0
0
0
AbC
1 x 2
1
1
abc
8
8
0
TOTAL
50
20
30
%
5
2
3
7.15
A is fur color locus
B is tail length locus
C is behaviour locus
fur (A)
tail (B)
behaviour (C)
AB
AC
BC
white
short
normal
16
aBC
R
R
P
brown
short
agitated
0
ABc
P
R
R
brown
short
normal
955
ABC
P
P
P
white
short
agitated
36
aBc
R
P
R
white
long
normal
0
abC
P
R
R
brown
long
agitated
14
Abc
R
R
P
brown
long
normal
46
AbC
R
P
R
white
long
agitated
933
abc
P
P
P
B C A
|--------------|---------|
4.1cM 1.5cM
Pairwise recombination frequencies are as follows (calculations are shown below):
A-B 5.6%
A-C 1.5%
B-C 4.1%
AB
AC
BC
16
16
0
0
0
0
0
0
0
36
0
36
0
0
0
14
14
0
46
0
46
0
0
0
112
30
82
5.6%
1.5%
4.1%
7.S: Linkage and Mapping (Summary)
• Recombination is defined as any process that results in gametes with combinations of alleles that were not present in the gametes of a previous generation.
• The recombination frequency between any two loci depends on their relative chromosomal locations.
• Unlinked loci show a maximum 50% recombination frequency.
• Loci that are close together on a chromosome are linked and tend to segregate with the same combinations of alleles that were present in their parents.
• Crossovers are a normal part of most meioses, and allow for recombination between linked loci.
• Measuring recombination frequency is easiest when starting with pure-breeding lines with two alleles for each locus, and with suitable lines for test crossing.
• Because recombination frequency is usually proportional to the distance between loci, recombination frequencies can be used to create genetic maps.
• Recombination frequencies tend to underestimate map distances, especially over long distances, since double crossovers may be genetically indistinguishable from non-recombinants.
• Three-point crosses can be used to determine the order and map distance between of three loci, and can correct for some of the double crossovers between the two outer loci.
Key Terms
linkage
recombination
independent assortment
crossover
recombinant genotype
parental genotype
unlinked
recombination frequency (RF)
complete (absolute) linkage
incomplete (partial) linkage
syntenic
synapsis
coupling (cis) configuration
repulsion(trans) configuration
repulsion
map units (mu)
centiMorgans (cM)
genetic map
conserved synteny
double-crossover
three-point cross | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/07%3A_Linkage_and_Mapping/7.E%3A_Linkage_and_Mapping_%28Exercises%29.txt |
Genetics is the study of the inheritance and variation of biological traits. We have previously noted that it is possible to conduct genetic research without directly studying DNA. Indeed some of the greatest geneticists had no special knowledge of DNA at all, but relied instead on analysis of phenotypes, inheritance patterns, and their ratios in carefully designed crosses.
• 8.1: Prelude to Molecular Genetics
Today, classical genetics is often complemented by molecular biology, to give molecular genetics, which involves the study of DNA and other macromolecules that have been isolated from an organism. Usually, molecular genetics experiments involve some combination of techniques to isolate and analyze the DNA or RNA transcribed from a particular gene.
• 8.2: Isolating Genomic DNA
DNA purification strategies rely on the chemical properties of DNA that distinguish it from other molecules in the cell, namely that it is a very long, negatively charged molecule. To extract purified DNA from a tissue sample, cells are broken open by grinding or lysing in a solution that contains chemicals that protect the DNA while disrupting other components of the cell (Figure 8.2). These chemicals may include detergents, which dissolve lipid membranes and denature proteins.
• 8.3: Isolating or Detecting a Specific Sequence by PCR
The Polymerase Chain Reaction (PCR) is a method of DNA replication that is performed in a test tube (i.e. in vitro). Here “polymerase” refers to a DNA polymerase enzyme extracted and purified from bacteria, and “chain reaction” refers to the ability of this technique produce millions of copies of a DNA molecule, by using each newly replicated double helix as a template to synthesize two new DNA double helices. PCR is therefore a very efficient method of amplifying DNA.
• 8.4: Cutting and Pasting DNA- Restriction Digests and DNA Ligation
Many bacteria have enzymes that recognize specific DNA sequences and then cut the double stranded DNA helix at this sequence. These enzymes are called site-specific restriction endonucleases, or more simply “restriction enzymes”, and they naturally function as part of bacterial defenses against viruses and other sources of foreign DNA. To cut DNA at known locations, researchers use restriction enzymes from various bacterial species, and which can be purchased from various commercial sources.
• 8.5: Cloning DNA - Plasmid Vectors
Many bacteria contain extra-chromosomal DNA elements called plasmids. These are usually small (a few 1000 bp), circular, double stranded molecules that replicate independently of the chromosome and can be present in high copy numbers within a cell. In the wild, plasmids can be transferred between individuals during bacterial mating and are sometimes even transferred between different species. Plasmids often carry genes for pathogenicity and drug-resistance.
• 8.6: DNA Analysis - Gel Electrophoresis
A solution of DNA is colorless, and except for being viscous at high concentrations, is visually indistinguishable from water. Therefore, techniques such as gel electrophoresis have been developed to detect and analyze DNA.
• 8.7: DNA Analysis- Blotting and Hybridization
Bands of DNA in an electrophoretic gel form only if most of the DNA molecules are of the same size, such as following a PCR reaction, or restriction digestion of a plasmid. In other situations, such as after restriction digestion of chromosomal (genomic) DNA, there will be a large number of variable size fragments in the digest and it will appear as a continuous smear of DNA, rather than distinct bands.
• 8.8: Transgenic organisms
Transgenic organisms contain foreign DNA that has been introduced using biotechnology. Foreign DNA (the transgene) is defined here as DNA from another species, or else recombinant DNA from the same species that has been manipulated in the laboratory then reintroduced. The terms transgenic organism and genetically modified organism (GMO) are generally synonymous.
• 8.9: Techniques of Molecular Genetics (Exercises)
• 8.S: Techniques of Molecular Genetics (Summary)
08: Techniques of Molecular Genetics
Genetics is the study of the inheritance and variation of biological traits. We have previously noted that it is possible to conduct genetic research without directly studying DNA. Indeed some of the greatest geneticists had no special knowledge of DNA at all, but relied instead on analysis of phenotypes, inheritance patterns, and their ratios in carefully designed crosses. Today, classical genetics is often complemented by molecular biology, to give molecular genetics, which involves the study of DNA and other macromolecules that have been isolated from an organism. Usually, molecular genetics experiments involve some combination of techniques to isolate and analyze the DNA or RNA transcribed from a particular gene. In some cases, the DNA may be subsequently manipulated by mutation or by recombination with other DNA fragments. Techniques of molecular genetics have wide application in many fields of biology, as well as forensics, biotechnology, and medicine.
Figure \(1\): Disposable tips for a pipette are used to distribute microliter volumes of liquid in molecular biology. (Flickr-estherase-CC:ANS)
8.02: Isolating Genomic DNA
DNA purification strategies rely on the chemical properties of DNA that distinguish it from other molecules in the cell, namely that it is a very long, negatively charged molecule. To extract purified DNA from a tissue sample, cells are broken open by grinding or lysing in a solution that contains chemicals that protect the DNA while disrupting other components of the cell (Figure \(2\)). These chemicals may include detergents, which dissolve lipid membranes and denature proteins. A cation such as Na+ helps to stabilize the negatively charged DNA and separate it from proteins such as histones. A chelating agent, such as EDTA, is added to protect DNA by sequestering Mg2+ ions, which can otherwise serve as a necessary co-factor for nucleases (enzymes that digest DNA). As a result, free, double-stranded DNA molecules are released from the chromatin into the extraction buffer, which also contains proteins and all other cellular components. (The basics of this procedure can be done with household chemicals and are presented on YouTube.)
The free DNA molecules are subsequently isolated by one of several methods. Commonly, proteins are removed by adjusting the salt concentration so they precipitate. The supernatant, which contains DNA and other, smaller metabolites, is then mixed with ethanol, which causes the DNA to precipitate. A small pellet of DNA can be collected by centrifugation, and after removal of the ethanol, the DNA pellet can be dissolved in water (usually with a small amount of EDTA and a pH buffer) for the use in other reactions. Note that this process has purified all of the DNA from a tissue sample; if we want to further isolate a specific gene or DNA fragment, we must use additional techniques, such as PCR. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/08%3A_Techniques_of_Molecular_Genetics/8.01%3A_Prelude_to_Molecular_Genetics.txt |
Components of the PCR Reaction
The Polymerase Chain Reaction (PCR) is a method of DNA replication that is performed in a test tube (i.e. in vitro). Here “polymerase” refers to a DNA polymerase enzyme extracted and purified from bacteria, and “chain reaction” refers to the ability of this technique produce millions of copies of a DNA molecule, by using each newly replicated double helix as a template to synthesize two new DNA double helices. PCR is therefore a very efficient method of amplifying DNA.
Besides its ability to make large amounts of DNA, there is a second characteristic of PCR that makes it extremely useful. Recall that most DNA polymerases can only add nucleotides to the end of an existing strand of DNA, and therefore require a primer to initiate the process of replication. For PCR, chemically synthesized primers of about 20 nucleotides are used. In an ideal PCR, primers only hybridize to their exact complementary sequence on the template strand (Figure \(3\)).
The experimenter can therefore control exactly what region of a DNA template is amplified by controlling the sequence of the primers used in the reaction.
To conduct a PCR amplification, an experimenter combines in a small, thin-walled tube (Figure \(4\)), all of the necessary components for DNA replication, including DNA polymerase and solutions containing nucleotides (dATP, dCTP, dGTP, dTTP), a DNA template, DNA primers, a pH buffer, and ions (e.g. Mg2+) required by the polymerase. Successful PCR reactions have been conducted using only a single DNA molecule as a template, but in practice, most PCR reactions contain many thousands of template molecules. The template DNA (e.g. total genomic DNA) has usually already been purified from cells or tissues using the techniques described above. However, in some situations it is possible to put whole cells directly in a PCR reaction for use as a template.
An essential aspect of PCR is thermal-cycling, meaning the exposure of the reaction to a series of precisely defined temperatures (Figure \(5\)). The reaction mixture is first heated to 95°C. This causes the hydrogen bonds between the strands of the template DNA molecules to melt, or denature. This produces two single-stranded DNA molecules from each double helix (Figure \(6\)). In the next step (annealing), the mixture is cooled to 45-65°C. The exact temperature depends on the primer sequence used and the objectives of the experiment. This allows the formation of double stranded helices between complementary DNA molecules, including the annealing of primers to the template. In the final step (extension) the mixture is heated to 72°C. This is the temperature at which the particular DNA polymerase used in PCR is most active. During extension, the new DNA strand is synthesized, starting from the 3' end of the primer, along the length of the template strand. The entire PCR process is very quick, with each temperature phase usually lasting 30 seconds or less. Each cycle of three temperatures (denaturation, annealing, extension) is usually repeated about 30 times, amplifying the target region approximately 230-fold. Notice from the figure that most of the newly synthesized strands in PCR begin and end with sequences either identical to or complementary to the primer sequences; although a few strands are longer than this, they are in such a small minority that they can almost always be ignored.
Figure \(6\): PCR with the three phases of the thermalcycle numbered. The template strand (blue) is replicated from primers (red), with newly synthesized strands in green. The green strands flanked by two primer binding sites will increase in abundance exponentially through successive PCR cycles. (Wikipedia-madprime-GFDL)
The earliest PCR reactions used a polymerase from E. coli. Because the high temperature of the denaturation step destroyed the enzyme, new polymerase had to be added after each cycle. To overcome this, researchers identified thermostable DNA polymerases such as Taq DNA pol, from Thermus acquaticus, a thermophilic bacterium that lives in hot springs. Taq, and similar thermostable polymerases from other hot environments, are able to remain functional in the repeated cycles of amplification. Taq polymerase cannot usually amplify fragments longer than about 3kbp, but under some specialized conditions, PCR can amplify fragments up to approximately 10kbp. Other polymerases, either by themselves or in combination with Taq, are used to increase the length of amplified fragments or to increase the fidelity of the replication.
After completion of the thermalcycling (amplification), an aliquot from the PCR reaction is usually loaded onto an electrophoretic gel (described below) to determine whether a DNA fragment of the expected length was successfully amplified or not. Usually, the original template DNA will be so dilute that it will not be visible on the gel, only the amplified PCR product. The presence of a sharp band of the expected length indicates that PCR was able to amplify its target. If the purpose of the PCR was to test for the presence of a particular template sequence, this is the end of the experiment. Otherwise, the remaining PCR product can be used as starting material for a variety of other techniques such as sequencing or cloning.
An Application of PCR: the StarLink Affair
PCR is very sensitive (meaning it can amplify very small starting amounts of DNA), and specific (meaning it can amplify only the target sequence from a mixture of many DNA sequences). This made PCR the perfect tool to test whether genetically modified corn was present in consumer products on supermarket shelves. Although currently (2013) 85% of corn in the United States is genetically modified, and contains genes that government regulators have approved for human consumption, back in 2000, environmental groups showed that a strain of genetically modified corn, which had only been approved for use as animal feed, had been mixed in with corn used in producing human food, like taco shells.
To do this, the groups purchased taco shells from stores in the Washington DC area, extracted DNA from the taco shells and used it as a template in a PCR reaction with primers specific to the unauthorized gene (Cry9C). Their suspicions were confirmed when they ran this PCR product on an agarose gel and saw a band of expected size. The PCR test was able to detect one transgenic kernel in a whole bushel of corn (1 per 100,000). The company (Aventis) that sold the transgenic seed to farmers had to pay for the destruction of large amounts of corn, and was the target of a class action law-suit by angry consumers who claimed they had been made sick by the taco shells. While no legitimate cases of harm were ever proven, and the plaintiffs were awarded \$9 million, of which \$3 million went to the legal fees, and the remainder of the judgment went to the consumers in the form of coupons for taco shells. The affair damaged the company, and exposed a weakness in the way the genetically modified crops were handled in the United States at the time. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/08%3A_Techniques_of_Molecular_Genetics/8.03%3A__Isolating_or_Detecting_a_Specific_Sequence_by_PCR.txt |
Restriction Enzymes
Many bacteria have enzymes that recognize specific DNA sequences (usually 4 or 6 nucleotides) and then cut the double stranded DNA helix at this sequence (Figure \(7\)). These enzymes are called site-specific restriction endonucleases, or more simply “restriction enzymes”, and they naturally function as part of bacterial defenses against viruses and other sources of foreign DNA. To cut DNA at known locations, researchers use restriction enzymes that have been purified from various bacterial species, and which can be purchased from various commercial sources. These enzymes are usually named after the bacterium from which they were first isolated. For example, EcoRI and EcoRV are both enzymes from E. coli. EcoRI cuts double stranded DNA at the sequence GAATTC, but note that this enzyme, like many others, does not cut in exactly the middle of the restriction sequence (Figure \(8\)). The ends of a molecule cut by EcoRI have an overhanging region of single stranded DNA, and so are sometimes called sticky-ends. On the other hand, EcoRV is an example of an enzyme that cuts both strands in exactly the middle of its recognition sequence, producing what are called blunt-ends, which lack overhangs.
DNA Ligation
The process of DNA ligation occurs when DNA strands are covalently joined, end-to-end through the action of an enzyme called DNA ligase. Sticky-ended molecules with complementary overhanging sequences are said to have compatible ends, which facilitate their joining to form recombinant DNA. Likewise, two blunt-ended sequences are also considered compatible to join together, although they do not ligate together as efficiently as sticky-ends. Note: sticky-ended molecules with non-complementary sequences will not ligate together with DNA ligase. Ligation is therefore central to the production of recombinant DNA, including the insertion of a double stranded DNA fragment into a plasmid vector.
8.05: Cloning DNA - Plasmid Vectors
Plasmids are Naturally Present in Some Bacteria
Many bacteria contain extra-chromosomal DNA elements called plasmids. These are usually small (a few 1000 bp), circular, double stranded molecules that replicate independently of the chromosome and can be present in high copy numbers within a cell. In the wild, plasmids can be transferred between individuals during bacterial mating and are sometimes even transferred between different species. Plasmids are particularly important in medicine because they often carry genes for pathogenicity and drug-resistance. In the lab, plasmids can be inserted into bacteria in a process called transformation.
Using Plasmids as Cloning Vectors
To insert a DNA fragment into a plasmid, both the fragment and the circular plasmid are cut using a restriction enzyme that produces compatible ends (Figure \(1\)). Given the large number of restriction enzymes that are currently available, it is usually not too difficult to find an enzyme for which corresponding recognition sequences are present in both the plasmid and the DNA fragment, particularly because most plasmid vectors used in molecular biology have been engineered to contain recognition sites for a large number of restriction endonucleases.
After restriction digestion, the desired fragments may be further purified or selected before they are mixed together with ligase to join them together. Following a short incubation, the newly ligated plasmids, containing the gene of interest are transformed into E. coli. Transformation is accomplished by mixing the ligated DNA with E. coli cells that have been specially prepared (i.e. made competent) to uptake DNA. Competent cells can be made by exposure to compounds such as CaCl2 or to electrical fields (electroporation). Because only a small fraction of cells that are mixed with DNA will actually be transformed, a selectable marker, such as a gene for antibiotic resistance, is usually also present on the plasmid. After transformation (combining DNA with competent cells), bacteria are spread on a bacterial agar plate containing an appropriate antibiotic so that only those cells that have actually incorporated the plasmid will be able to grow and form colonies. This can then be picked and used for further study.
Molecular biologists use plasmids as vectors to contain, amplify, transfer, and sometimes express genes of interest that are present in the cloned DNA. Often, the first step in a molecular biology experiment is to clone (i.e. copy) a gene into a plasmid, then transform this recombinant plasmid back into bacteria so that essentially unlimited copies of the gene (and the plasmid that carries it) can be made as the bacteria reproduce. This is a practical necessity for further manipulations of the DNA, since most techniques of molecular biology are not sensitive enough to work with just a single molecule at a time. Many molecular cloning and recombination experiments are therefore iterative processes in which:
1. a DNA fragment (usually isolated by PCR and/or restriction digestion) is cloned into a plasmid cut with a compatible restriction enzyme
2. the recombinant plasmid is transformed into bacteria
3. the bacteria are allowed to multiply, usually in liquid culture
4. a large quantity of the recombinant plasmid DNA is isolated from the bacterial culture
5. further manipulations (such as site directed mutagenesis or the introduction of another piece of DNA) are conducted on the recombinant plasmid
6. the modified plasmid is again transformed into bacteria, prior to further manipulations, or for expression
An Application of Molecular Cloning: Production of Recombinant Insulin
Purified insulin protein is critical to the treatment of diabetes. Prior to ~1980, insulin for clinical use was isolated from human cadavers or from slaughtered animals such as pigs. Human-derived insulin generally had better pharmacological properties, but was in limited supply and carried risks of disease transmission. By cloning the human insulin gene and expressing it in E. coli, large quantities of insulin identical to the human hormone could be produced in fermenters, safely and efficiently. Production of recombinant insulin also allows specialized variants of the protein to be produced: for example, by changing a few amino acids, longer-acting forms of the hormone can be made. The active insulin hormone contains two peptide fragments of 21 and 30 amino acids, respectively. Today, essentially all insulin is produced from recombinant sources (Figure \(2\)), i.e. human genes and their derivatives expressed in bacteria or yeast. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/08%3A_Techniques_of_Molecular_Genetics/8.04%3A_Cutting_and_Pasting_DNA-_Restriction_Digests_and_DNA_Ligation.txt |
A solution of DNA is colorless, and except for being viscous at high concentrations, is visually indistinguishable from water. Therefore, techniques such as gel electrophoresis have been developed to detect and analyze DNA (Figure \(11\)).
This analysis starts when a solution of DNA is deposited at one end of a gel slab. This gel is made from polymers such as agarose, which is a polysaccharide isolated from seaweed. The DNA is then forced through the gel by an electrical current, with DNA molecules moving toward the positive electrode (Figure \(12\)).
As it migrates, each piece of DNA threads its way through the pores, which form between the polymers in the gel. Because shorter pieces can move through these pores faster than longer pieces, gel electrophoresis separates molecules based on their size (length), with smaller DNA pieces moving faster than long ones. DNA molecules of a similar size migrate to a similar location in each gel, called a band. This feature makes it easy to see DNA after staining the DNA with a fluorescent dye such as ethidium bromide (Figure \(13\)). By separating a mixture of DNA molecules of known size (size markers) in adjacent lanes on the same gel, the length of an uncharacterized DNA fragment can be estimated. Gel segments containing the DNA bands can also be cut out of the gel, and the size-selected DNA extracted and used in other types of reactions, such as sequencing and cloning.
8.07: DNA Analysis- Blotting and Hybridization
Bands of DNA in an electrophoretic gel form only if most of the DNA molecules are of the same size, such as following a PCR reaction, or restriction digestion of a plasmid. In other situations, such as after restriction digestion of chromosomal (genomic) DNA, there will be a large number of variable size fragments in the digest and it will appear as a continuous smear of DNA, rather than distinct bands. In this case, it is necessary to use additional techniques to detect the presence of a specific DNA sequence within the smear of DNA separated on an electrophoretic gel. This can be done using a “Southern Blot”.
Southern Blots
A Southern blot (also called a Southern Transfer) is named after Ed Southern, its inventor. In the first step, DNA is digested with restriction enzymes and separated by gel electrophoresis (as discussed above). Then a sheet or membrane of nylon or similar material is laid under the gel and the DNA, in its separated position (bands or smear), is transferred to the membrane by drawing the liquid out of the gel, in a process called blotting (Figure \(1\)). The blotted DNA is usually covalently attached to the nylon membrane by briefly exposing the blot to UV light. Transferring the DNA to the sturdy membrane is necessary because the fragile gel would fall apart during the next two steps in the process. Next, the membrane is bathed in a solution to denature (double stranded made single stranded) the attached DNA. Then a hybridization solution containing a small amount of single-stranded probe DNA that is complementary in sequence to a target molecule on the membrane. This probe DNA is labeled using fluorescent or radioactive molecules, and if the hybridization is performed properly, the probe DNA will form a stable duplex only with those DNA molecules on the membrane that are exactly complementary to it. Then, the unhybridized probe is washed off and remaining radioactive or fluorescent signal will appear in a distinct band when appropriately detected. The band represents the presence of a particular DNA sequence within the mixture of DNA fragments.
The probe is sequence specific (requires complementarity). However, variation in hybridization temperature and washing solutions can alter the stringency of the probe. At maximum stringency (higher temperature) hybridization conditions, probes will only hybridize with the exact target sequences that are perfectly complementary (maximum number of hydrogen bonds). At lower temperatures, probes will be able to hybridize to targets to which they do not match exactly, but only are roughly complementary for part of the sequence.
Southern blotting is useful not only for detecting the presence of a DNA sequence within a mixture of DNA molecules, but also for determining the size of a restriction fragment in a DNA sample. Southern blots are useful for detecting fragments larger than those normally amplified by PCR, and when trying to detect fragments that may be only distantly related to a known sequence. Applications of Southern blotting will be discussed further in the context of molecular markers in a subsequent chapter. Southern blotting was invented before PCR, but PCR has replaced blotting in many applications because of its simplicity, speed, and convenience. Following the development of the Southern blot, other types of blotting techniques were invented.
Northern Blots
The Northern blot involves the size separation of RNA in gels like that of DNA. Because we wish to determine the native size of the RNA transcript (and because RNA is single stranded) no restriction enzymes are ever used. Because most RNA is single stranded and can fold into various conformations thorough intra-molecular base pairing, the electrophoresis separation is more haphazard and the bands are often less sharp, compared to that of double stranded DNA.
Western Blots
In a Western blot, protein is size separated on a gel (usually an acrylamide gel) before transferring to a membrane, which is then probed with an antibody that specifically binds to an antigenic site on the target protein. This antibody is then detected by other antibodies with some fluorescent or color production marker system. It will also give bands proportional to the amount and size of the target protein (Figure \(2\)).
A comparison of all three blotting methods is shown in Figure \(3\). | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/08%3A_Techniques_of_Molecular_Genetics/8.06%3A__DNA_Analysis_-_Gel_Electrophoresis.txt |
General principles of transgenesis
Transgenic organisms contain foreign DNA that has been introduced using biotechnology. Foreign DNA (the transgene) is defined here as DNA from another species, or else recombinant DNA from the same species that has been manipulated in the laboratory then reintroduced. The terms transgenic organism and genetically modified organism (GMO) are generally synonymous. The process of creating transgenic organisms or cells to be come whole organisms with a permanent change to their germline has been called either transformation or transfection. (Unfortunately, both words have alternate meanings. Transformation also refers to the process of mammalian cell becoming cancerous, while transfection also refers to the process of introducing DNA into cells in culture, either bacterial or eukaryote, for a temporary use, not germ line changes.) Transgenic organisms are important research tools, and are often used when exploring a gene’s function. Transgenesis is also related to the medical practice of gene therapy, in which DNA is transferred into a patient’s cells to treat disease. Transgenic organisms are widespread in agriculture. Approximately 90% of canola, cotton, corn, soybean, and sugar beets grown in North America are transgenic. No other transgenic livestock or crops (except some squash, papaya, and alfalfa) are currently produced in North America.
To make a transgenic cell, DNA must first be transferred across the cell membrane, (and, if present, across the cell wall), without destroying the cell. In some cases, naked DNA (meaning plasmid or linear DNA that is not bound to any type of carrier) may be transferred into the cell by adding DNA to the medium and temporarily increasing the porosity of the membrane, for example by electroporation. When working with larger cells, naked DNA can also be microinjected into a cell using a specialized needle. Other methods use vectors to transport DNA across the membrane. Note that the word “vector” as used here refers to any type of carrier, and not just plasmid vectors. Vectors for transformation/transfection include vesicles made of lipids or other polymers that surround DNA; various types of particles that carry DNA on their surface; and infectious viruses and bacteria that naturally transfer their own DNA into a host cell, but which have been engineered to transfer any DNA molecule of interest. Usually the foreign DNA is a complete expression unit that includes its own cis-regulators (e.g. promoter) as well as the gene that is to be transcribed.
When the objective of an experiment is to produce a stable (i.e. heritable) transgenic eukaryote, the foreign DNA must be incorporated into the host’s chromosomes. For this to occur, the foreign DNA must enter the host’s nucleus, and recombine with one of the host’s chromatids. In some species, the foreign DNA is inserted at a random location in a chromatid, probably wherever strand breakage and non-homologous end joining happen to occur. In other species, the foreign DNA can be targeted to a particular locus, by flanking the foreign DNA with DNA that is homologous to the host’s DNA at that locus. The foreign DNA is then incorporated into the host’s chromosomes through homologous recombination.
Furthermore, to produce multicellular organisms in which all cells are transgenic and the transgene is stably inherited, the cell that was originally transformed must be either a gamete or must develop into tissues that produce gametes. Transgenic gametes can eventually be mated to produce homozygous, transgenic offspring. The presence of the transgene in the offspring is typically confirmed using PCR or Southern blotting, and the expression of the transgene can be measured using reverse-transcription PCR (RT-PCR), RNA blotting, and Western (protein blotting).
The rate of transcription of a transgene is highly dependent on the state of the chromatin into which it is inserted (i.e. position effects), as well as other factors. Therefore, researchers often generate several independently transformed/transfected lines with the same transgene, and then screen for the lines with the highest expression. It is also good practice to clone and sequence the transgenic locus from a newly generated transgenic organism, since errors (truncations, rearrangements, and other mutations) can be introduced during transformation/transfection.
Producing a transgenic plant
The most common method for producing transgenic plants is Agrobacterium-mediated transformation (Figure \(1\)). Agrobacterium tumifaciens is a soil bacterium that, as part of its natural pathogenesis, injects its own tumor-inducing (Ti) plasmid into cells of a host plant. The natural Ti plasmid encodes growth-promoting genes that cause a gall (i.e. tumor) to form on the plant, which also provides an environment for the pathogen to proliferate. Molecular biologists have engineered the Ti plasmid by removing the tumor-inducing genes and adding restriction sites that make it convenient to insert any DNA of interest. This engineered version is called a T-DNA (transfer-DNA) plasmid; the bacterium transfers a linear fragment of this plasmid that includes the conserved “left-border (LB)”, and right-border (RB)” DNA sequences, and anything in between them (up to about 10 kb). The linear T-DNA fragment is transported into the nucleus, where it recombines with the host-DNA, probably wherever random breakages occur in the host’s chromosomes.
In Arabidopsis and a few other species, flowers can simply be dipped in a suspension of Agrobacterium, and ~1% of the resulting seeds will be transformed. In most other plant species, cells are induced by hormones to form a mass of undifferentiated tissues called a callus. The Agrobacterium is applied to a callus and a few cells are transformed, which can then be induced by other hormones to regenerate whole plants (Figure \(2\)). Some plant species are resistant (i.e. “recalcitrant”) to transformation by Agrobacterium. In these situations, other techniques must be used such as particle bombardment, whereby DNA is non-covalently attached to small metallic particles, which are accelerated by compressed air into callus tissue, from which complete transgenic plants can sometimes be regenerated. In all transformation methods, the presence of a selectable marker (e.g. a gene that confers antibiotic resistance or herbicide resistance) is useful for distinguishing transgenic cells from non-transgenic cells at an early stage of the transformation process.
Producing a transgenic mouse
In a commonly used method for producing a transgenic mouse, stem cells are removed from a mouse embryo, and a transgenic DNA construct is transferred into the stem cells using electroporation, and some of this transgenic DNA enters the nucleus, where it may undergo homologous recombination (Figure \(3\)). The transgenic DNA construct contains DNA homologous to either side of a locus that is to be targeted for replacement. If the objective of the experiment is simply to delete (“knock-out”) the targeted locus, the host’s DNA can simply be replaced by selectable marker, as shown. It is also possible to replace the host’s DNA at this locus with a different version of the same gene, or a completely different gene, depending on how the transgenic construct is made. Cells that have been transfected and express the selectable marker (i.e. resistance to the antibiotic neomycin resistance, neoR, in this example) are distinguished from unsuccessfully transfected cells by their ability to survive in the presence of the selective agent (e.g. an antibiotic). Transfected cells are then injected into early stage embryos, and then are transferred to a foster mother. The resulting pups are chimeras, meaning that only some of their cells are transgenic. Some of the chimeras will produce gametes that are transgenic, which when mated with a wild-type gamete, will produce mice that are hemizygous for the transgene. Unlike the chimeras, these hemizygotes carry the transgene in all of their cells. Through further breeding, mice that are homozygous for the transgene can be obtained.
Human gene therapy
Many different strategies for human gene therapy are under development. In theory, either the germline or somatic cells may be targeted for transfection, but most research has focused on somatic cell transfection, because of risks and ethical issues associated with germline transformation. Gene therapy approaches may be further classified as either ex vivo or in vivo, with the former meaning that cells (e.g. stem cells) are transfected in isolation before being introduced to the body, where they replace defective cells. Ex vivo gene therapies for several blood disorders (e.g. immunodeficiencies, thalassemias) are undergoing clinical trials. For in vivo therapies, the transfection occurs within the patient. The objective may be either stable integration, or non-integrative transfection. As described above, stable transfection involves integration into the host genome. In the clinical context, stable integration may not be necessary, and carries with it higher risk of inducing mutations in either the transgene or host genome). In contrast, transient transfection does not involve integration into the host genome and the transgene may therefore be delivered to the cell as either RNA or DNA. Advantages of RNA delivery include that no promoter is needed to drive expression of the transgene. Besides mRNA transgenes, which could provide a functional version of a mutant protein, there is great interest in delivery of siRNA (small-inhibitory RNAs), which can be used to silence specific genes in the host cell’s genome.
Vectors for in vivo gene therapy must be capable of delivering DNA or RNA to a large proportion of the targeted cells, without inducing a significant immune response, or having any toxic effects. Ideally, the vectors should also have high specificity for the targeted cell type. Vectors based on viruses (e.g. lentiviruses) are being developed for in both in vivo and ex vivo gene therapies. Other, non-viral vectors (e.g. vesicles and nanoparticles) are also being developed for gene therapy as well. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/08%3A_Techniques_of_Molecular_Genetics/8.08%3A__Transgenic_organisms.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
8.1 What information, and what reagents would you need to use PCR to detect HIV in a blood sample?
8.2 A 6.0 kbp PCR fragment flanked by recognition sites for the HindIII restriction enzyme is cut with HindIII then ligated with a 3kb plasmid vector that has also been cut with HindIII. This recombinant plasmid is transformed into E. coli. From one colony a plasmid is prepared and digested with HindIII.
a) When the product of the HindIII digestion is analyzed by gel electrophoresis, what will be the size of the bands observed?
b) What bands would be observed if the recombinant plasmid was cut with EcoRI, which has only one site, directly in the middle of the PCR fragment?
c) What bands would be observed if the recombinant plasmid was cut with both EcoRI and HindIII at the same time?
8.3 If you started with 10 molecules of double stranded DNA template, what is the maximum number of molecules you would you have after 10 PCR cycles?
8.4 What is present in a PCR tube at the end of a successful amplification reaction? With this in mind, why do you usually only see a single, sharp band on a gel when it is analyzed by electrophoresis?
8.5 A coat protein from a particular virus can be used to immunize children against further infection. However, inoculation of children with proteins extracted from natural viruses sometimes causes fatal disease, due to contamination with live viruses. How could you use molecular biology to produce an optimal vaccine?
8.6 How would cloning be different if there were no selectable markers?
8.7 Research shows that a particular form of cancer is caused by a 200bp deletion in a particular human gene that is normally 2kb long. Only one mutant copy is needed to cause the disease.
a) Explain how you would use Southern blotting to diagnose the disease.
b) How would any of the blots appear if you hybridized and washed at very low temperature?
8.8 Refer to question 8.7.
a) Explain how you would detect the presence of the same deletion using PCR, rather than a Southern blot.
b) How would PCR products appear if you annealed at very low temperature?
8.9 You have a PCR fragment for a human olfactory receptor gene (perception of smells). You want to know what genes a dog might have that are related to this human gene. How can you use your PCR fragment and genomic DNA from a dog to find this out? Do you think dogs have more or less of these genes?
8.10 You add ligase to a reaction containing a sticky-ended plasmid and sticky-ended insert fragment, which both have compatible ends. Unbeknownst to you, someone in the lab left the stock of ligase enzyme out of the freezer overnight and it degraded (no longer works). Explain in detail what will happen in your ligation experiment in this situation should you try and transform with it.
8.S: Techniques of Molecular Genetics (Summary)
• Molecular biology involves the isolation and analysis of DNA and other macromolecules
• Isolation of total genomic DNA involves separating DNA from protein and other cellular components, for example by ethanol precipitation of DNA.
• PCR can be used as part of a sensitive method to detect the presence of a particular DNA sequence
• PCR can also be used as part of a method to isolate and prepare large quantities of a particular DNA sequence
• Restriction enzymes are natural endonucleases used in molecular biology to cut DNA sequences at specific sites.
• DNA fragments with compatible ends can be joined together through ligation.If the ligation produces a sequence not found in nature, the molecule is said to be recombinant.
• Transformation is the introduction of DNA (usually recombinant plasmids) into bacteria.
• Cloning of genes in E. coli is a common technique in molecular biology, since it allows large quantities of a DNA for gene to made, which allows further analysis or manipulation
• Cloning can also be used to produce useful proteins, such as insulin, in microbes.
• Southern blotting can be used to detect the presence of any sequence that matches a probe, within a mixture of DNA (such as total genomic DNA).
• The stringency of hybridization in blotting and in PCR is dependent on physical parameters such as temperature and washing solution content.
Key Terms:
macromolecules
lysis
detergent
chelating agent
EDTA
nuclease
pellet
PCR
primer
thermalcycle
denature
anneal
extend
Taq DNApol
electrophoretic gel
restriction endonuclease
EcoRI
sticky end
blunt end
compatible end
ligation
ligase
plasmid
transformation
competent
electroporation
selectable marker
agarose
vector
band
ethidium bromide
Southern blot
membrane
denaturation
hybridization
washing
probe
stringency
northern blot
western blot
transgene
GMO
transformation
transfection
naked DNA
vector
electroporation
microinjection
vesicles
callus
knock-out
recalcitrant
Agrobacterium
particle bombardment
T-DNA
Ti plasmid
position effects
stem cells | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/08%3A_Techniques_of_Molecular_Genetics/8.09%3A_Techniques_of_Molecular_Genetics_%28Exercises%29.txt |
Previous chapters described chromosomes as simple linear DNA molecules on which genes are located. For example, your largest chromosome, chromosome 1, has about 3536 genes. To ensure that each of your cells possesses these genes the chromosome has features that allow it to be passed on during cell division. Origins of replication found along its length provide places for DNA replication to start, telomeres protect each end of the chromosome, and a single centromere near the middle provides a place for microtubules to attach and move the chromosome during mitosis and meiosis. This chapter examines: (1) changes in the number of whole chromosomes and how they affect the phenotype of an organism and (2) changes in the structure of individual chromosomes and how they affect meiotic pairing. Human examples will be used to show the phenotypic consequences and methods for detection.
• 9.1: Changes in Chromosome Number
If something goes wrong during cell division, an entire chromosome may be lost and the cell will lack all of these genes. The causes behind these chromosome abnormalites and the consequences they have for the cell and the organism is the subject of this section.
• 9.2: Changes in Chromosome Structure
If the chromosome is altered, but still retains the three critical features of a chromosome (centromeres, telomeres, and origin of replication), it will continue to be inherited during subsequent cell divisions, however the daughter cell may not retain all the genes. For example, if a segment of the chromosome has been lost, the cell may be missing some genes. The causes of chromosome structural abnormalites, which involves breaks in the DNA that makes up the chromosome.
• 9.3: Chromosome Abnormalities in Humans
To better understand the consequences let's consider those that affect people. As you will recall humans are 2n=46. The convention when describing a person's karyotype (chromosome composition) is to list the total number of chromosomes, then the sex chromosomes, and then anything out of the ordinary. Most of us are 46,XX or 46,XY. What follows are some examples of chromosome number and chromosome structure abnormalities.
• 9.4: Diagnosing Human Chromosome Abnormalities
How can we confirm that a person has a specific chromosomal abnormality? The first method was simply to obtain a sample of their cells, stain the chromosomes with Giemsa dye, and examine the results with a light microscope. Each chromosome can be recognized by its length, the location of its centromere, and the characteristic pattern of purple bands produced by the Giemsa.
• 9.E: Changes in Chromosome Number and Structure (Exercises)
• 9.S: Changes in Chromosome Number and Structure (Summary)
Thumbnail: (Wikipedia-Pmx-CC:AS)
09: Changes in Chromosome Number and Structure
If something goes wrong during cell division, an entire chromosome may be lost and the cell will lack all of these genes. The causes behind these chromosome abnormalites and the consequences they have for the cell and the organism is the subject of this section.
Cause: Nondisjunction During Mitosis or Meiosis
Segregation occurs in anaphase. In mitosis and meiosis II, sister chromatids (of replicated chromosomes) are normally pulled to opposite ends of the cell (see Figure \(\PageIndex{1a}\)). In Meiosis I, it is homologous chromosomes, which are synapsed at that time, that segregate and move apart.
In rare cases, the sister chromatids (or paired chromosomes in Meiosis I) fail to separate, or dis-join. This failure to segregate properly is called nondisjunction and it can happen during mitosis, meiosis I, or meiosis II. This nondisjunction results in both chromatids (or chromosomes) moving to one pole and none at the other. One cell will have an extra copy and the other will lack a copy. Thus failure to segregate properly leads to unbalanced products.
Consequence: Decreased Viability
The result of a non-disjunction event is daughter cells that have an abnormal number of chromosomes. Cells, such as the parent cell in Figure \(\PageIndex{1a}\), which have the proper number of chromosomes, are said to be euploid. The daughter cells have one too many or one too few chromosomes and are thus aneuploid. Even though both product cells have at least one copy of all genes, both cells will probably die. The reason is due to the loss or gain of a large number of genes. Genes produce an standard amount of product - either functional RNAs or proteins. The parent cell shown has a balanced genotype because it has two copies of all of its genes. Each of its genes produces half of the products needed by the cell. But if one of these cells suddenly had only one copy (or three copies) of an important gene, the amount of product would be either 50% (or 150%) of what was required. Such a change for a single gene could probably be tolerated by the cell and it would probably survive. But the sudden change to one copy (or three copies) of the hundreds or thousands of genes on an entire chromosome the results would be more than tolerable and be catastrophic for the daughter cells. They have what’s called an unbalanced genotype, which usually decreases their viability.
If a first division or second division nondisjunction event occurs during meiosis the result is an unbalanced gamete (Figure \(\PageIndex{2b}\) and \(\PageIndex{2c}\)). The gamete will often be functional, but after fertilization the embryo will be genetically unbalanced. This usually leads to the death of the embryo. There are some exceptions to this in humans and these will be presented later in this chapter. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/09%3A__Changes_in_Chromosome_Number_and_Structure/9.01%3A__Changes_in_Chromosome_Number.txt |
If the chromosome is altered, but still retains the three critical features of a chromosome (centromeres, telomeres, and origin of replication), it will continue to be inherited during subsequent cell divisions, however the daughter cell may not retain all the genes. For example, if a segment of the chromosome has been lost, the cell may be missing some genes. The causes of chromosome structural abnormalites and the consequences they have for the cell and the organism is shown below. They all involve breaks in the DNA that makes up the chromosome.
Cause #1: Incorrect Repair of Double Strand DNA Breaks During Interphase
A chromosome is a very long but very thin molecule. In the phopho-diester backbone there are only two covalent bonds holding each base pair to the next. If one of these covalent bonds is broken the chromosome will still remain intact, although a DNA Ligase will be needed to repair the nick (Figure \(\PageIndex{1a}\)). Problems arise when both strands are broken at or near the same location. This double strand break will cleave the chromosome into two independent pieces (Figure \(\PageIndex{1b}\)). Because these events do occur in cells there is a repair system called the non-homologous end joining (NHEJ) system to fix them. Proteins bind to each broken end of the DNA and reattach them with new covalent bonds. This system is not perfect and sometimes leads to chromosome rearrangements (see next section).
The NHEJ system proteins only function if required. If the chromosomes within an interphase nucleus are all intact the system is not active. The telomeres at the natural ends of chromosomes prevent the NHEJ system from attempting to join the normal ends of chromosomes together. If there is one double strand break the two broken ends can be recognized and joined. But if there are two double strand breaks at the same time there will be four broken ends in total. The NHEJ system proteins may join the ends together correctly but if they do not the result is a chromosome rearrangement (Figure \(2\)).
The Four Types of Chromosome rearrangements
Errors during the repair of multiple double strand breaks can cause four types of chromosome rearrangements. The type of chromosome rearrangement is dependent upon where the two breaks were originally and how they are rejoined. Figure \(5\) shows some possibilities but more are shown below. In these there is a double strand DNA break between the B and C genes (shown here as a red X). A second DNA break occurs and the NHEJ proteins mend the damage incorrectly by joining the ends shown with the blue arrows. The chromosomes are drawn unreplicated as they are in G1 phase but these events can happen anytime during interphase.
There are four major types of rearrangements:
a) Deletions arise when both breaks are on one chromosome. If the ends are joined in this way the piece of DNA with the B gene on it does not have a centromere and will be lost during the next cell division.
b) Inversions also occur when both breaks are on one chromosome. If the ends are joined in this way, part of the chromosome is inverted. This example shows a paracentric inversion, named because the inverted section does not include the centromere (para = beside). If the breaks occur on different chromosome arms the inverted section includes the centromere and the result is a pericentric inversion (peri = around).
c) Duplications can occur from two DNA breaks at different places in sister chromatids (in a replicated chromosome). The ends are joined together incorrectly to give a chromosome with a duplication (two “B” regions as shown above). Note: the reciprocal product has a deletion.
d) Translocations result from two breaks on different chromosomes (not homologs) and incorrect rejoining. This example shows a reciprocal translocation - two chromosomes have 'swapped' arms, the E gene is now part of the white chromosome and the C gene is now part of the shaded chromosome. Robertsonian translocations are those rare situations in which all of the genes end up together on one chromosome and the other chromosome is so small that it is typically lost.
Cause #2: Incorrect Crossovers During Meiosis
Meiotic crossovers occur at the beginning of meiosis for two reasons. They help hold the homologous chromosomes together until separation occurs during anaphase I (see Chapter 2). They also allow recombination to occur between linked genes (see Chapter 7). The event itself takes place during prophase I when a double strand break on one piece of DNA is joined with a double strand break on another piece of DNA and the ends are put together (Figure \(\PageIndex{3a}\)). Most of the time the breaks are on non-sister chromatids and most of the time the breaks are at the same relative locations.
Problems occur when the wrong pieces of DNA are matched up along the chromosomes during crossover events. This can happen if the same or similar DNA sequence is found at multiple sites on the chromosomes (Figure \(\PageIndex{3b}\)). For example, if there are two Alu transposable elements on a chromosome. When the homologous chromosomes pair during prophase I the wrong Alu sequences might line up. A crossover may occur in this region. If so, when the chromosomes separate during anaphase I one of the chromatids will have a duplication and one will have a deletion. Ultimately, of the four cells produced by this meiosis, two will be normal, one will have a chromosome with extra genes, and one will have a chromosome missing some genes. Errors of this type can also cause inversions and translocations.
Consequence #1 - Rearrangements Show Abnormal pairing at Meiosis
Homologous regions of chromosomes pair at meiosis I (prophase I). With rearranged chromosomes this can lead to visible abnormalities and segregation abnormalities.
Deletion chromosomes will pair up with a normal homolog along the shared regions and at the missing segment, the normal homolog will loop out (nothing to pair with) to form a deletion loop. This can be used to locate the deletion cytologically. The deleted region is also pseudo-dominant, in that it permits the mutant expression of recessive alleles on the normal homolog. Deletion mutations don’t revert - nothing to replace the missing DNA.
When an inversion chromosome is paired up in meiosis there is an inversion loop formed. If there is a crossover within the loop then abnormal products will result and abnormal, unbalanced gametes will be produced. For example, a crossover event within the loop of a paracentric inversion will lead to a di-centric product that will break into deletion products and produce unbalanced gametes (Figure \(4\)). Similarly, with a pericentric inversion, a crossover event leads to duplicate/deletion products that are unbalanced (Figure \(5\)).
If joined with a normal gamete, they will result in an unbalanced zygote, which are usually lethal. The consequence for this is that crossover products (recombinants) are lost and thus inversions appear to suppress crossovers within the inverted region.
With both types of inversions, crossovers outside the loop are possible and fully viable as they don’t alter the gene balance.
Duplications also produce a cytologically visible loop at meiotic pairing. Duplications can revert at a relatively high frequency by unequal crossing over. Duplicated genes offer new possibilities for mutational divergence followed by natural selection in the course of evolution.
For translocations, a consequence for the two chromosomes involved is that when they pair at meiosis both replicated chromosome pairs will be together, which can be seen cytologically as a tetrad. This tetrad can segregate in three ways. Two of which are shown below. This set of paired, replicated chromosomes can segregate as Alternate (balanced) where both normal and both translocated chromosomes go to the same polls. Or the chromosomes can segregate as Adjacent-1 (unbalanced) where the normal and translocation chromosomes segregate as shown below. Each of these possibilities is approximately equally frequent and thus only about half the time do the gametes end up unbalanced (Figure \(6\)).
Consequence #2: Decreased Viability
All of the chromosome rearrangements shown above produce functional chromosomes. Each has one centromere, two telomeres, and thousands of origins of replication. Because inversions and translocations do not change the number of genes in a cell or organism they are said to be balanced rearrangements. Unless one of the breakpoints occurred in the middle of a gene the cells will not be affected. On the other hand, deletions and duplications are unbalanced rearrangements. The larger they are (more genes involved) the more disruption they cause to the proper functioning of the cell or organism. As explained in Section 9.1.2 above having too much or too little gene action for a large number of genes can disrupt the cellular metabolism to generate a phenotype or reduce viability.
Consequence #3: Decreased Fertility
Recall that during meiosis I homologous chromosomes pair up. If a cell has a chromosome with a rearrangement this chromosome will have to pair with its normal homolog.
Cells heterozygous for balanced rearrangements actually have more difficulties in prophase I. Consider the chromosomes shown in Figure \(7\). There are different ways they might pair during prophase I - one is shown in Figure \(8\). But if a crossover occurs in the inverted region the result will be unbalanced gametes. Embryos made with unbalanced gametes rarely survive. The consequence is that the heterozygous organism will have reduced fertility.
Note that an organism homozygous for this inversion chromosome will not be affected in this way because no loops are formed. The chromosomes can pair along their entire length and crossovers will not produce any unbalanced gametes. This is a general property of inversions and translocations. In heterozygotes there are problems during meiosis resulting in a lot of the gametes being unbalanced and an overall reduction in fertility. In homozygotes the rearranged chromosomes pair with one another just fine and there is no effect on fertility.
Consequence #4: Cancer
Some chromosome rearrangements have breakpoints within genes leading to the creation of hybrid genes – the first part of one gene with the last part of another. If the hybrid gene inappropriately promotes cell replication, the cell can become cancerous. An example of this is shown in Figure \(1\) where the chromosomes are from a patient with leukemia caused by a translocation between chromosomes 9 and 22 (the red and green spots side-by-side).
Consequence #5: Evolution
Those chromosome changes that duplicate genes are important for evolution. If an organism has an extra copy of important genes, one gene can be retained for their original function while others can mutate and potentially acquire new functions (Figure \(9\)). An example of this is the multiple copies of the globin genes found in mammals (see Figure \(13\)).
Chromosome rearrangements that decrease fertility are also important for the origin of new species. If a rearrangement, such as the inversion shown in Figure \(9\), becomes common in a small isolated population, that population has 100% fertility if they mate within their group, but a reduced fertility if they mate with members of the larger population. As rearrangements accumulate the small population will become more and more reproductively isolated. When members are incapable of forming viable, fertile offspring with the original population the group will have become a new species. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/09%3A__Changes_in_Chromosome_Number_and_Structure/9.02%3A__Changes_in_Chromosome_Structure.txt |
The problems described above can affect all eukaryotes, unicellular and multicellular. To better understand the consequences let us consider those that affect people. As you will recall from Figure \(12\), humans are 2n=46. The convention when describing a person's karyotype (chromosome composition) is to list the total number of chromosomes, then the sex chromosomes, and then anything out of the ordinary. Most of us are 46,XX or 46,XY. What follows are some examples of chromosome number and chromosome structure abnormalities.
Down Syndrome
The most common chromosome number abnormality is trisomy-21 or, as it is more commonly known, Down syndrome. Having an extra copy of the smallest human chromosome, chromosome 21, causes substantial health problems.It is present in about 1 in 800 births. Infants with this condition have three copies of chromosome 21 rather than the normal two. Don't confuse trisomy - having three copies of one chromosome (i.e. 2n+1) with triploidy - having three entire chromosome sets (3x; see Section 2.6.) Females with trisomy-21 are 47,XX,+21 while males are 47,XY,+21. In general, people with Down syndrome are 47,sex,+21 where the word 'sex' signifies that the sex chromosomes may be XX or XY.
Trisomy-21 may arise from a nondisjunction event during meioisis in either parent or during mitosis very early during embryogenesis. However, most cases are due to a first division non-disjunction event occurring in the female parent (Figure \(1\)).
Having an extra copy of the smallest human chromosome, chromosome 21, causes substantial health problems. People with Down syndrome have various degrees of intellectual disability and often have other health problems such as heart defects. The disease was first described by John Down in 1866 but it was not until 1959 when its chromosomal basis was discovered. Current research suggests that at least some of the mental problems are due to having three copies of the DYRK gene on chromosome 21. This gene is active in the brain and there is evidence from humans and from mice that neurons are damaged if there is too much DYRK protein synthesized.
XYY and XXX
While fetuses trisomic for one of the other larger autosomes seldom survive to term, the situation is quite different for the sex chromosomes. Approximately 1 in 1000 males have an extra Y chromosome and yet most are unaware of it! There is little harm in having two Y chromosomes because they have relatively few genes. Similarly, 1 in 1000 females have an extra X chromosome. This situation is also relatively harmless although for a different reason. Normally in female mammals one of the two X chromosomes is inactivated in each cell so that there can be genetic balance (Figure \(2\) and see Section 3.5.2). In 46,XX females one of the X chromosomes is inactivated while in 47,XXX females two are inactivated.
Turner Syndrome
Monosomy (2n-1) for autosomal chromosomes does occur at conception, but these embryos almost never survive to term. Similarly, embryos that are 45,Y are also non-viable because they lack many essential genes found on the X chromosome. The only viable monosomy in humans is 45,X, also known as Turner syndrome. These people are phenotypically female because they do not have a Y chromosome (Section 2.5.2). They are viable because in females only one X is active in most cells anyway. People with this condition do have health problems though: they are shorter than average, they have an elevated risk of heart defects, and they are infertile.
The reason for the health problems is that there are a few genes that are found on both the X and the Y chromosome. Because these genes are found in two copies in both XY males and XX females they are in what is called the pseudoautosomal region. This region escapes X chromosome inactivation. One of the genes in this region is called SHOX. It makes a protein that promotes bone growth. 46,XX and 46,XY people have two functioning copies and have average height. People with 47,XYY and 47,XXX genomes have three copies and are taller than average. And people with 45,X have one copy and are short. It is the single copy of SHOX and a few of the other genes in the pseudoautosomal region that causes health problems for women with Turner syndrome.
The reason for the infertility is that the X chromosome inactivation system only acts in somatic cells - it is not needed in the germline cells. Ovaries naturally have two functional X chromosomes. Women with Turner syndrome can not perform oogenesis because this process only works if there are two active X chromosomes. Recently, it has become possible for these women to become pregnant with donated eggs and in vitro fertilization.
Klinefelter Syndrome
There are four common sex-chromosome aneuploidies: 47,XYY, 47,XXX, 45,X, and 47,XXY. This last situation is known as Klinefelter syndrome. These people are male (because they have a Y chromosome) and tall (because they have three SHOX genes). They do not have health problems because the X chromosome inactivation system is independent of sex. In the embryonic nuclei the X chromosome are counted and all but one are shut down. It does not matter whether the embryo is male or female. Men with Klinefelter syndrome have a Barr bodies in their nuclei, the same as 46,XX females. They do have fertility problems because there are two active X chromosome in their testes and this interferes with spermatogenesis. They make enough sperm to conceive children using intracytoplasmic sperm injection though.
Cri-du-Chat Syndrome
Cri-du-chat syndrome occurs when a child inherits a defective chromosome 5 from one parent (Figure \(16\)n). This condition is rare - it is present in only 1 in 20,000 to 1 in 50,000 births but it does account for 1% of cases of profound intellectual disability. The specific defect is a deletion that removes 2 Mb or more from the tip of the short arm of the chromosome. In most cases the deletion is the result of a chromosomal rearrangement in one of the parent's germ line cells. People with cri-du-chat have a karyotype of 46,sex,deletion(5).
As with Down syndrome this condition is associated with intellectual disability and other health problems. These problems include an improperly formed larynx which leads to infants making high pitched cat-like crying sounds (hence the name "cry of the cat"). It is suspected that at least some of the intellectual disability phenotype is due to having only a single copy of the CTNND2 gene. This gene is active during embryogenesis and makes a protein essential for neuron migration. Down syndrome and cri-du-chat syndrome are two examples of the need for genomes to contain the proper number of genes. Having too many copies of key genes (Down syndrome) or too few (cri-du-chat syndrome) can lead to substantial developmental problems.
Inversion(9)
The most common chromosome rearrangements in humans are inversions of chromosome 9. About 2% of the world's population is heterozygous or homozygous for inversion(9). This rearrangement does not affect a person's health because the genes on the chromosome are all present - all that has changed is their relative locations. Inversion(9) is different from deletion(5) in two main respects. As mentioned above because it is a balanced rearrangement it does not cause harm. And because of this nearly everyone with an inversion(9) chromosome has inherited it from a parent who had inherited it from one of their parents and so on. In contrast, most cases of deletion(5) are due to new mutations occurring in a parent. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/09%3A__Changes_in_Chromosome_Number_and_Structure/9.03%3A__Chromosome_Abnormalities_in_Humans.txt |
Bright Field Microscopy
How can we confirm that a person has a specific chromosomal abnormality? The first method was simply to obtain a sample of their cells, stain the chromosomes with Giemsa dye, and examine the results with a light microscope (Figure \(1\)). Each chromosome can be recognized by its length, the location of its centromere, and the characteristic pattern of purple bands produced by the Giemsa. For example, if mitotic cells from a person consistently contained forty seven chromosomes in total with three chromosome 21s this would be indicative of Down syndrome. Bright field microscopy does has its limitations though - it only works with mitotic chromosomes and many chromosome rearrangements are either too subtle or too complex for even a skilled cytogeneticist to discern.
Fluorescence In Situ Hybridization
The solution to these problems was fluorescence in situ hybridization (FISH). The technique is similar to a Southern blot in that a single stranded DNA probe is allowed to hybridize to denatured target DNA (see Section 8.6). However, instead of the probe being radioactive it is fluorescent and instead of the target DNA being restriction fragments on a nylon membrane it is denatured chromosomes on a glass slide. Because there are several fluorescent colours available it is common to use more than one probe at the same time. Typically the chromosomes are also labeled with a fluorescent stain called DAPI which gives them a uniform blue colour. If the chromosomes have come from a mitotic cell it is possible to see all forty six of them spread out in a small area. Alternatively, if the chromosomes are within the nucleus of an interphase cell they appear together within a large blue circle.
Using FISH to Diagnose Down Syndrome
Most pregnancies result in healthy children. However in some cases there is an elevated chance that the fetus has trisomy-21. Older women are at a higher risk because the non-disjunction events that lead to trisomy become more frequent with age. The second consideration is what the fetus looks like during an ultrasound examination. Fetuses with trisomy-21 and some other chromosome abnormalities have a swelling in the back of the neck called a nuchal translucency. If either or both factors is present the woman may choose amniocentesis. In this test some amniotic fluid is withdrawn so that the fetal cells within it can be examined. Figure \(2\) shows a positive result for trisomy-21. Based upon this image the fetus has two X chromosomes and three chromosome 21s and therefore has a karyotype of 47,XX,+21.
Using FISH to Diagnose Cri-du-Chat Syndrome
A physician may suspect that a patient has a specific genetic condition based upon the patient's physical appearance, mental abilities, health problems, and other factors. FISH can be used to confirm the diagnosis. For example, Figure \(3\) shows a positive result for cri-du-chat syndrome. The probes are binding to two long arms of chromosome 5 but only one short arm. One of the chromosome 5s must therefore be missing part of its short arm.
Newer Techniques
FISH is an elegant technique that produces dramatic images of our chromosomes. Unfortunately, FISH is also expensive, time consuming, and requires a high degree of skill. For these reasons, FISH is slowly being replaced with PCR and DNA chip based methods. Versions of these techniques have been developed that can accurately quantify a person's DNA. For example a sample of DNA from a person with Down syndrome will contain 150% more DNA from chromosome 21 than the other chromosomes. Likewise DNA from a person with cri-du-chat syndrome will contain 50% less DNA from the end of chromosome 5. These techniques are very useful if the suspected abnormality is a deletion, a duplication, or a change in chromosome number. They are less useful for diagnosing chromosome inversions and translocations because these rearrangements often involve no net loss or gain of genes.
In the future all of these techniques will likely be replaced with DNA sequencing. Each new generation of genome sequencing machines can sequence more DNA in less time. Eventually it will be cheaper just to sequence a patient's entire genome than to use FISH or PCR to test for specific chromosome defects.
9.E: Changes in Chromosome Number and Structure (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
9.1
Make diagrams showing how an improper crossover event during meiosis can lead to: (a) an inversion or (b) a translocation.
9.2
Make a diagram showing how a nondisjunction event can lead to a child with a 47,XYY karyotype.
9.3
How many Barr bodies would you expect to see in cells from people who are: (a) 46, XY, (b) 46,XX, (c) 47, XYY, (d) 47,XXX, (e) 45,X, and (f) 47,XXY ?
9.4
Why can people survive with trisomy-21 (47,sex,+21) but not monosomy-21 (45,sex,-21)?
9.5
If Drosophila geneticists want to generate mutant strains with deletions they expose flies to gamma rays. What does this imply about gamma rays?
9.6
What would happen if there was a nondisjunction event involving chromosome 21 in a 46,XY zygote?
9.7
Design a FISH based experiment to find out if your lab partner is a 47,XXX female or a 47,XYY male.
9.8
What would Figure 9.18 look like if it also showed metaphase chromosomes from another cell?
9.S: Changes in Chromosome Number and Structure (Summary)
• Errors during anaphase in mitosis or meiosis can lead to trisomy and other forms of aneuploidy.
• Errors during the repair of DNA breaks or during meiotic crossing over can lead to chromosome rearrangements.
• Five common forms of aneuploidy in humans are 47,sex,+21 (Down syndrome), 47,XYY, 47,XXX, 45,X (Turner syndrome) and 47,XXY (Klinefelter syndrome).
• Deletion(5) causes a serious condition (cri-du-chat syndrome) because deletions are unbalanced chromosome rearrangements.
• Inversion(9) causes few health consequences because inversions are balanced chromosome rearrangements.
• Bright field microscopy can be used to detect chromosome number abnormalities and some chromosome rearrangements.
• Fluorescence in situ hybridization can be used to detect all types of chromosome abnormalities.
• PCR and DNA chip based techniques can be used to detect chromosome number abnormalities, deletions, and duplications.
Key Terms:
origin of replication
telomere
centromere
non-disjunction
euploid
aneuploid
balanced
unbalanced
first division nondisjunction
second division nondisjunction
double strand break
nonhomologous end joining
DNA repair system
chromosome rearrangement
deletion
inversion
paracentric inversion
pericentric inversion
tandem duplication
translocation
reciprocal translocation
Robertsonian translocation
meiotic crossover
deletion loop
karyotype
46,XX
46,XY
47,sex,+21 (Down syndrome)
trisomy
47,XYY
47,XXX
monosomy
45,X (Turner syndrome)
pseudoautosomal region
47,XXY (Klinefelter syndrome)
46,sex,deletion(5) (cri-du-chat syndrome)
46,sex,inversion(9)
bright field microscopy
Giemsa stain
fluorescence in situ hybridization
fluorescent probe
DAPI stain
amniocentesis | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/09%3A__Changes_in_Chromosome_Number_and_Structure/9.04%3A__Diagnosing_Human_Chromosome_Abnormalities.txt |
Figure \(1\): Many traits, such as body mass, show continuous variation, rather than discrete variation. Although environment obviously also affects this trait, some of the variation observed between individuals is heritable, and is dependent on the interactions of multiple alleles at multiple loci. The study of quantitative traits is one of many applications of molecular markers. (Flickr-Jamie Golombek-CC:AND)
• 10.1: Some Variations in the Genome Affect Complex Traits
The genomic sequences of almost any two unrelated people differ at millions of nucleotide positions. Some of these differences would be found in the regions of genes that code for proteins. Others might affect the amount of transcript that is made for a particular gene. A person’s health, appearance, behavior, and other characteristics depend in part on these polymorphisms.
• 10.2: Origins of Molecular Polymorphisms
Some of mutations occur during DNA replication processes, resulting in an insertion, deletion, or substitution of one or a few nucleotides. Larger mutations can be caused by mobile genetic elements such as transposons, which are inserted more or less randomly into chromosomal DNA, sometimes occurring in clusters.
• 10.3: Classification and Detection of Molecular Markers
Mutations that do not affect the function of protein sequences or gene expression are likely to persist in a population as polymorphisms, since there will be no selection either in favor or against them (i.e. they are neutral). Note that the although the rate of spontaneous mutation in natural populations is sufficiently high so as to generate millions of polymorphisms that accumulate over thousands of generations, the rate of mutation is slow.
• 10.4: Applications of Molecular Markers
Several characteristics of molecular markers make them useful to geneticists. DNA polymorphisms are a natural part of most genomes. Geneticists discover these polymorphisms in various ways, including comparison of random DNA sequence fragments from several individuals in a population. Once molecular markers have been identified, they can be used in many ways.
• 10.5: Quantitative Trait Locus (QTL) Analysis
We can use molecular markers to identify at least some of the genes (those with a major influence) that affect a given quantitative trait. This is essentially an extension of the mapping techniques we have already considered for discrete traits.
• 10.E: Molecular Markers and Quantitative Traits (Exercises)
• 10.S: Molecular Markers and Quantitative Traits (Summary)
10: Molecular Markers and Quantitative Traits
Imagine that you could compare the complete genomic DNA sequence of any two people you meet today. Although their sequences would be very similar on the whole, they would certainly not be identical at each of the 3 billion base pair positions you examined (unless, perhaps, your subjects were identical twins – but even they may have some somatic differences). In fact, the genomic sequences of almost any two unrelated people differ at millions of nucleotide positions. Some of these differences would be found in the regions of genes that code for proteins. Others might affect the amount of transcript that is made for a particular gene. A person’s health, appearance, behavior, and other characteristics depend in part on these polymorphisms.
Most difference, however, have no effect at all. They have no effect on gene sequences or expression, because they occur within regions of DNA that neither encode proteins, nor regulate the expression of genes. These polymorphisms are nevertheless very useful because they can be used as molecular markers in medicine, forensics, ecology, agriculture, and many other fields. In most situations, molecular markers obey the same rules of inheritance that we have already described for other types of loci, and so can be used to create genetic maps and to identify linked genes.
10.02: Origins of Molecular Polymorphisms
Mutations of DNA sequences can arise in many ways (Chapter 4). Some of these changes occur during DNA replication processes, resulting in an insertion, deletion, or substitution of one or a few nucleotides. Larger mutations can be caused by mobile genetic elements such as transposons, which are inserted more or less randomly into chromosomal DNA, sometimes occurring in clusters. In these and other types of repetitive DNA sequences, the number of repeated units is highly prone to change through unequal crossovers and other replication events.
10.03: Classification and Detection of Molecular Markers
Regardless of their origins, molecular markers can be classified as polymorphisms that either vary in the length of a DNA sequence, or vary only in the identity of nucleotides at a particular position on a chromosome (Figure \(1\)). In both cases, because two or more alternative versions of the DNA sequence exist, we can treat each variant as a different allele of a single locus. Each allele gives a different molecular phenotype. For example, polymorphisms of SSRs (short sequence repeats) can be distinguished based on the length of PCR products: one allele of a particular SSR locus might produce a 100bp band, while the same primers used with a different allele as a template might produce a 120bp band (Figure \(2\)). A different type of marker, called a SNP (single nucleotide polymorphism), is an example of polymorphism that varies in nucleotide identity, but not length. SNPs are the most common of any molecular markers, and the genotypes of thousands of SNP loci can be determined in parallel, using new, hybridization based instruments. Note that the alleles of most molecular markers are co-dominant, since it is possible to distinguish the molecular phenotype of a heterozygote from either homozygote.
Mutations that do not affect the function of protein sequences or gene expression are likely to persist in a population as polymorphisms, since there will be no selection either in favor or against them (i.e. they are neutral). Note that the although the rate of spontaneous mutation in natural populations is sufficiently high so as to generate millions of polymorphisms that accumulate over thousands of generations, the rate of mutation is on the other hand sufficiently low that existing polymorphisms are stable throughout the few generations we study in a typical genetic experiment. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/10%3A__Molecular_Markers_and_Quantitative_Traits/10.01%3A_Some_Variations_in_the_Genome_Affect_Complex_Traits.txt |
Several characteristics of molecular markers make them useful to geneticists. First, because of the way DNA polymorphisms arise and are retained, they are frequent throughout the genome. Second, because they are phenotypically neutral, it is relatively easy to find markers that differ between two individuals. Third, their neutrality also makes it possible to study hundreds of loci without worrying about gene interactions or other influences that make it difficult to infer genotype from phenotype. Lastly, unlike visible traits such as eye color or petal color, the phenotype of a molecular marker can be detected in any tissue or developmental stage, and the same type of assay can be used to score molecular phenotypes at millions of different loci. Thus, the neutrality, high density, high degree of polymorphism, co-dominance, and ease of detection of molecular markers has lead to their wide adoption in many areas of research.
It is worth emphasizing again that DNA polymorphisms are a natural part of most genomes. Geneticists discover these polymorphisms in various ways, including comparison of random DNA sequence fragments from several individuals in a population. Once molecular markers have been identified, they can be used in many ways, including:
DNA fingerprinting
By comparing the allelic genotypes at multiple molecular marker loci, it is possible to determine the likelihood of similarity between two DNA samples. If markers differ, then clearly the DNA is from different sources. If they don’t differ, then one can estimate the unlikelihood of them coming from different sources – eg they are from the same source. For example, a forensic scientist can demonstrate that a blood sample found on a weapon came from a particular suspect. Similarly, that leaves in the back of a suspect's pick-up truck came from a particular tree at a crime scene. DNA fingerprinting is also useful in paternity testing (Figure \(1\)) and in commercial applications such as verification of species of origin of certain foods and herbal products.
Construction of genetic linkage maps
By calculating the recombination frequency between pairs of molecular markers, a map of each chromosome can be generated for almost any organism (Figure \(2\)). These maps are calculated using the same mapping techniques described for genes in Chapter 7, however, the high density and ease with which molecular markers can be genotyped makes them more useful than other phenotypes for constructing genetic maps. These maps are useful in further studies, including map-based cloning of protein coding genes that were identified by mutation.
Population studies
As described in Chapter 5, the observed frequency of alleles, including alleles of molecular markers, can be compared to frequencies expected for populations in Hardy-Weinberg equilibrium to determine whether the population is in equilibrium. By monitoring molecular markers, ecologists and wildlife biologists can make inferences about migration, selection, diversity, and other population-level parameters.
Molecular markers can also be used by anthropologists to study migration events in human ancestry. There is a large commercial business available that will genotype people and determine their deep genetic heritage for ~\$100. This can be examined through the maternal line via sequencing their mitochondrial genome and through the paternal line via genotyping their Y-chromosome.
For example, about 8% of the men in parts of Asia (about 0.5% of the men in the world) have a Y-chromosomal lineage belonging to Genghis Khan (the haplogroup C3) and his decendents.
Identification of linked traits
It is often possible to correlate, or link, an allele of a molecular marker with a particular disease or other trait of interest. One way to make this correlation is to obtain genomic DNA samples from hundreds of individuals with a particular disease, as well as samples from a control population of healthy individuals. The genotype of each individual is scored at hundreds or thousands of molecular marker loci (e.g. SNPs), to find alleles that are usually present in persons with the disease, but not in healthy subjects. The molecular marker is presumed to be tightly linked to the gene that causes the disease, although this protein-coding gene may itself be as yet unknown. The presence of a particular molecular polymorphism may therefore be used to diagnose a disease, or to advise an individual of susceptibility to a disease.
Molecular markers may also be used in a similar way in agriculture to track desired traits. For example, markers can be identified by screening both the traits and molecular marker genotypes of hundreds of individuals. Markers that are linked to desirable traits can then be used during breeding to select varieties with economically useful combinations of traits, even when the genes underlying the traits are not known.
Quantitative trait locus (QTL) mapping
Molecular markers can be used to identify multiple different regions of chromosomes that contain genes that act together to produce complex traits. This process involves finding combinations of alleles of molecular markers that are correlated with a quantitative phenotype such as body mass, height, or intelligence. QTL mapping is described in more detail in the following section. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/10%3A__Molecular_Markers_and_Quantitative_Traits/10.04%3A_Applications_of_Molecular_Markers.txt |
Most of the phenotypic traits commonly used in introductory genetics are qualitative, meaning that the phenotype exists in only two (or possibly a few more) discrete, alternative forms, such as either purple or white flowers, or red or white eyes. These qualitative traits are therefore said to exhibit discrete variation. On the other hand, many interesting and important traits exhibit continuous variation; these exhibit a continuous range of phenotypes that are usually measured quantitatively, such as intelligence, body mass, blood pressure in animals (including humans), and yield, water use, or vitamin content in crops. Traits with continuous variation are often complex, and do not show the simple Mendelian segregation ratios (e.g. 3:1) observed with some qualitative traits. Many complex traits are also influenced heavily by the environment. Nevertheless, complex traits can often be shown to have a component that is heritable, and which must therefore involve one or more genes.
How can genes, which are inherited (in the case of a diploid) as at most two variants each, explain the wide range of continuous variation observed for many traits? The lack of an immediately obvious explanation to this question was one of the early objections to Mendel's explanation of the mechanisms of heredity. However, upon further consideration, it becomes clear that the more loci that contribute to trait, the more phenotypic classes may be observed for that trait (Figure \(1\)).
If the number of phenotypic classes is sufficiently large (as with three or more loci), individual classes may become indistinguishable from each other (particularly when environmental effects are included), and the phenotype appears as a continuous variation (Figure \(2\)). Thus, quantitative traits are sometimes called polygenic traits, because it is assumed that their phenotypes are controlled by the combined activity of many genes. Note that this does not imply that each of the individual genes has an equal influence on a polygenic trait – some may have major effect, while others only minor. Furthermore, any single gene may influence more than one trait, whether these traits are quantitative or qualitative traits.
We can use molecular markers to identify at least some of the genes (those with a major influence) that affect a given quantitative trait. This is essentially an extension of the mapping techniques we have already considered for discrete traits. A QTL mapping experiment will ideally start with two pure-breeding lines that differ greatly from each other in respect to one or more quantitative traits (Figure \(3\)). The parents and all of their progeny should be raised under as close to the same environmental conditions as possible, to ensure that observed variation is due to genetic rather than external environmental factors. These parental lines must also be polymorphic for a large number of molecular loci, meaning that they must have different alleles from each other at hundreds of loci. The parental lines are crossed, and then this F1 individual, in which recombination between parental chromosomes has occurred is self-fertilized (or back-crossed). Because of recombination (both crossing over and independent assortment), each of the F2 individuals will contain a different combination of molecular markers, and also a different combination of alleles for the genes that control the quantitative trait of interest (Table \(1\)).
Table \(1\) Genotypes and quantitative data for some individuals from the crosses shown in Figure \(8\)
By comparing the molecular marker genotypes of several hundred F2 individuals with their quantitative phenotypes, a researcher can identify molecular markers for which the presence of particular alleles is always associated with extreme values of the trait. In this way, regions of chromosomes that contain genes that contribute to quantitative traits can be identified. (Figure \(4\)) It then takes much more work (further mapping and other experimentation) to identify the individual genes in each of the regions that control the quantitative trait. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/10%3A__Molecular_Markers_and_Quantitative_Traits/10.05%3A_Quantitative_Trait_Locus_%28QTL%29_Analysis.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
10.1 Three different polymorphisms have been identified at a particular molecular marker locus. A single pair of PCR primers will amplify either a 50bp fragment (B2), a 60bp fragment (B3), or a 100bp fragment (B4).
Draw the PCR bands that would be expected if these primers were used to amplify DNA from individuals with each of the following genotypes:
a) B2B2
b) B4B4
c) B2B3
d) B2B4
10.2 In addition to the primers used to genotype locus B (described above), a separate pair of primers can amplify another polymorphic SSR locus E, with either a 60bp product (E1) or a 90bp (E2) product. DNA was extracted from six individuals (#1- #6), and DNA from each individual was used as a template in separate PCR reactions with primers for either locus B or primers for locus E, and the PCR products were visualized on electrophoretic gels as shown below.
Based on the following PCR banding patterns, what is the full genotype of each of the six individuals?
10.3 Based on the genotypes you recorded in Question 10.2, can you determine which of the individuals could be a parent of individual #1?
10.4 Here is part of the DNA sequence of a chromosome:
TAAAGGAATCAATTACTTCTGTGTGTGTGTGTGTGTGTGTGTGTTCTTAGTTGTTTAAGTTTTAAGTTGTGA
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
ATTTCCTTAGTTAATGAAGACACACACACACACACACACACACAAGAATCAACAAATTCAAAATTCAACACT
Identify the following features on the sequence:
a) the region of the fragment that is most likely to be polymorphic
b) any simple sequence repeats
c) the best target sites for PCR primers that could be used to detect polymorphisms in the length of the simple sequence repeat region in different individuals
10.5 In a particular diploid plant, seed color is a polygenic trait. If true-breeding plants that produce red seeds are crossed with true breeding plants that produce white seeds, the F1 produces seeds that are intermediate in color (i.e. pink). When an F1 plant self-fertilizes, white seeds are observed in the next generation. How many genes are involved in seed color for each of the following frequencies of white seeds in the F2 generation?
a) 1/4 white seeds
b) 1/16 white seeds
c) 1/64 white seeds
d) 1/256 white seeds
10.6 If height in humans is a polygenic trait, explain why it occasionally happens that two tall parents have a child who grows up to be much shorter than either of them.
10.7 In quantitative trait (QTL) mapping, researchers cross two parents that differ in expression of some quantitative trait, then allow chromosomes from these parents to recombine randomly, and after several generations of inbreeding, produce a large number of offspring (“recombinant inbred lines”). Because the position of crossovers is random, each of the offspring contain a different combination of chromosomal regions from each of the two parents. The researchers then use molecular markers to determine which chromosomal regions have the greatest influence on the quantitative trait, e.g. in tall offspring, which chromosomal regions always come from the tall parent?
Imagine that two mice strains have been identified that differ in the time required to complete a maze, which may be an indication of intelligence. The time for maze completion is heritable and these parental strains “breed true” for the same completion time in each generation. Imagine also that their chromosomes are different colors and we can track the inheritance of chromosomal regions from each parent based on this color.
Based on the following diagrams of one chromosome from each individual in a pedigree, identify a chromosomal region that may contain a gene that affects time to complete a maze. The time for each individual is shown below each chromosome. Assume that all individuals are homozygous for all loci.
Parents:
Selected individuals from among F8 progeny of the above parents:
10.8 In a more realistic situation (as compared to question 7), where you could not distinguish the parental origin of different chromosomal regions just by appearance of chromosomes, explain how you could identify which parent was the source of a particular region of a chromosome in recombinant offspring.
10.S: Molecular Markers and Quantitative Traits (Summary)
• Natural variations in the length or identity of DNA sequences occur at millions of locations throughout most genomes.
• DNA polymorphisms are often neutral, but because of linkage may be used as molecular markers to identify regions of genomes that contain genes of interest.
• Molecular markers are useful because of their neutrality, co-dominance, density, allele frequencies, ease of detection, and expression in all tissues.
• Molecular markers can be used for any application in which the identity of two DNA samples is to be compared, or when a particular region of a chromosome is to be correlated with inheritance of a trait.
• Many important traits show continuous, rather than discrete variation.These are also called quantitative traits.
• Many quantitative traits are influenced by a combination of environment and genetics.
• The heritable component of quantitative traits can best be studied under controlled conditions, with pure-breeding parents that are polymorphic for both a quantitative trait and a large number of molecular markers.
• Molecular markers can be identified for which specific alleles are tightly correlated with the quantitative value of a particular phenotype.The genes that are linked to these markers can be identified through subsequent research. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/10%3A__Molecular_Markers_and_Quantitative_Traits/10.E%3A_Molecular_Markers_and_Quantitative_Traits_%28Exercises%29.txt |
Imagine that you could identify and quantify every molecule within a cell (Figure \(1\)) in a single assay. You could use this ability to better understand almost any aspect of biology. For example, by comparing the molecular profiles of plants that differed in their resistance to drought, you might discover which combination of genes or proteins makes a crop drought tolerant. Although it is not currently possible to study literally every molecule in a cell in a single experiment, recent advances in molecular biology have made it possible to study many genes (or their products) in parallel.
Figure \(1\): An artist’s depiction of part of an E.coli cell, showing many different types of molecules in their typical abundance. mRNA appears as white lines associated with purple ribosomes, while DNA and proteins such as histones are yellow. (Goodsell, Scripps-EDU)
11: Genomics and Systems Biology
The complete set of DNA within an organism is called its genome. Genomics is therefore the large-scale description, using techniques of molecular biology of many genes or even whole genomes at once. This type of research is facilitated by technologies that increase throughput (i.e. rate of analysis), and decrease cost. The –omics suffix has been used to indicate high-throughput analysis of many types of molecules, including transcripts (transcriptomics), proteins (proteomics), and the products of enzymatic reactions, or metabolites (metabolomics; Figure \(1\)). Interpretation of the large data sets generated by –omics research depends on a combination of computational, biological, and statistical knowledge provided by experts in bioinformatics. Attempts to combine information from different types of –‘omics studies is sometimes called systems biology.
11.02: DNA Sequencing
DNA sequencing determines the order of nucleotide bases within a given fragment of DNA. This information can be used to infer the RNA or protein sequence encoded by the gene, from which further inferences may be made about the gene’s function and its relationship to other genes and gene products. DNA sequence information is also useful in studying the regulation of gene expression. If DNA sequencing is applied to the study of many genes, or even a whole genome, it is considered an example of genomics.
Dideoxy sequencing
Recall that DNA polymerases incorporate nucleotides (dNTPs) into a growing strand of DNA, based on the sequence of a template strand. DNA polymerases add a new base only to the 3’-OH group of an existing strand of DNA; this is why primers are required in natural DNA synthesis and in techniques such as PCR. Most of the currently used DNA sequencing techniques rely on the random incorporation of modified nucleotides called terminators. Examples of terminators are the dideoxy nucleotides (ddNTPs), which lack a 3’-OH group and therefore cannot serve as an attachment site for the addition of new bases to a growing strand of DNA (Figure \(1\)). After a ddNTP is incorporated into a strand of DNA, no further elongation can occur. Terminators are labeled with one of four fluorescent dyes, each specific for one the four nucleotide bases.
To sequence a DNA fragment, you need many copies of that fragment (Figure \(2\)). Unlike PCR, DNA sequencing does not amplify the target sequence and only one primer is used. This primer is hybridized to the denatured template DNA, and determines where on the template strand the sequencing reaction will begin. A mixture of dNTPs, fluorescently labeled terminators, and DNA polymerase is added to a tube containing the primer-template hybrid. The DNA polymerase will then synthesize a new strand of DNA until a fluorescently labeled nucleotide is incorporated, at which point extension is terminated. Because the reaction contains millions of template molecules, a corresponding number of shorter molecules is synthesized, each ending in a fluorescent label that corresponds to the last base incorporated.
The newly synthesized strands can be denatured from the template, and then separated electrophoretically based on their length (Figure \(3\)). Since each band differs in length by one nucleotide, and the identity of that nucleotide is known from its fluorescence, the DNA sequence can be read simply from the order of the colors in successive bands. In practice, the maximum length of sequence that can be read from a single sequencing reaction is about 700 bp.
A particularly sensitive electrophoresis method used in the analysis of DNA sequencing reactions is called capillary electrophoresis (Figure \(6\)). In this method, a current pulls the sequencing products through a gel-like matrix that is encased in a fine tube of clear plastic. As in conventional electrophoresis, the smallest fragments move through the capillary the fastest. As they pass through a point near the end of the capillary, the fluorescent intensity of each dye is read. This produces a graph called a chromatogram. The sequence is determined by identifying the highest peak (i.e. the dye with the most intense fluorescent signal) at each position.
Next-generation sequencing
Advances in technology over the past two decades have increased the speed and quality of sequencing, while decreasing the cost. This has become especially true with the most recently developed methods called next-generation sequencing. Not all of these new methods rely on terminators, but one that does is a method used in instruments sold by a company called Illumina. Illumina sequencers use a special variant of PCR called bridge PCR to make many thousands of copies of a short (45bp) template fragment. Each of these short template fragments is attached in a cluster in a small spot on a reaction surface. Millions of other clusters, each made by different template fragment, are located at other positions on the reaction surface. DNA synthesis at each template strand then proceeds using dye-labeled terminators that are used are reversible. Synthesis is therefore terminated (temporarily) after the incorporation of each nucleotide. Thus, after the first nucleotide is incorporated in each strand, a camera records the color of fluorescence emitted from each cluster. The terminators are then modified, and a second nucleotide is incorporated in each strand, and again the reaction surface is photographed. This cycle is repeated a total of 45 times. Because millions of 45 bp templates are sequenced in parallel in a single process, Illumina sequencing is very efficient compared to other sequencing techniques. However, the short length of the templates currently limits the application of this technology. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/11%3A_Genomics_and_Systems_Biology/11.01%3A__Omics__Technologies.txt |
The need for assembly
Given that the length of a single, individual sequencing read is somewhere between 45bp and 700bp, we are faced with a problem determining the sequence of longer fragments, such as the chromosomes in an entire genome of humans (3 x109 bp). Obviously, we need to break the genome into smaller fragments. There are two different strategies for doing this:
1. clone-by-clone sequencing, which relies on the creation of a physical map first then sequencing, and
2. whole genome shotgun sequencing, which sequences first and does not require a physical map.
Physical mapping
A physical map is a representation of a genome, comprised of cloned fragments of DNA. The map is therefore made from physical entities (pieces of DNA) rather than abstract concepts such as the linkage frequencies and genes that make up a genetic map (Figure \(1\)). It is usually possible to correlate genetic and physical maps, for example by identifying the clone that contains a particular molecular marker. The connection between physical and genetic maps allows the genes underlying particular mutations to be identified through a process call map-based cloning.
To create a physical map, large fragments of the genome are cloned into plasmid vectors, or into larger vectors called bacterial artificial chromosomes (BACs). BACs can contain approximately 100kb fragments. The set of BACs produced in a cloning reaction will be redundant, meaning that different clones will contain DNA from the same part of the genome. Because of this redundancy, it is useful to select the minimum set of clones that represent the entire genome, and to order these clones respective to the sequence of the original chromosome. Note that this is all to be done without knowing the complete sequence of each BAC. Making a physical map may therefore rely on techniques related to Southern blotting: DNA from the ends of one BAC is used as a probe to find clones that contain the same sequence. These clones are then assumed to overlap each other. A set of overlapping clones is called a contig.
Clone-by-clone sequencing
Physical mapping of cloned sequences was once considered a pre-requisite for genome sequencing. The process would begin by breaking the genome into BAC-sized pieces, arranging these BACs into a map, then breaking each BAC up into a series of smaller clones, which were usually then also mapped. Eventually, a minimum set of smaller clones would be identified, each of which was small enough to be sequenced (Figure \(8\)). Because the order of clones relative to the complete chromosome was known prior to sequencing, the resulting sequence information could be easily assembled into one complete chromosome at the end of the project. Clone-by-clone sequencing therefore minimizes the number of sequencing reactions that must be performed, and makes sequence assembly straightforward and reliable. However, a drawback of this strategy is the tedious process of building physical map prior to any sequencing.
Whole genome shotgun sequencing
This strategy breaks the genome into fragments that are small enough to be sequenced, then reassembles them simply by looking for overlaps in the sequence of each fragment. It avoids the laborious process of making a physical map (Figure \(2\)). However, it requires many more sequencing reactions than the clone-by-clone method, because, in the shotgun approach, there is no way to avoid sequencing redundant fragments. There is also a question of the feasibility of assembling complete chromosomes based simply on the sequence overlaps of many small fragments. This is particularly a problem when the size of the fragments is smaller than the length of a repetitive region of DNA. Nevertheless, this method has now been successfully demonstrated in the nearly complete sequencing of many large genomes (rice, human, and many others). It is the current standard methodology.
However, shotgun assemblies are rarely able to complete entire genomes. The human genome, for example, relied on a combination of shotgun sequence and physical mapping to produce contiguous sequence for the length of each arm of each chromosome. Note that because of the highly repetitive nature of centromeric and telomeric DNA, sequencing projects rarely include these heterochromatic, gene poor regions.
Genome analysis
An assembled genome is a string of millions of A’s,C’s,G’s,T’s. Which of these represent nucleotides that encode proteins, and which of these represent other features of genes and their regulatory elements? The process of genome annotation relies on computers to define features such a start and stop codons, introns, exons, and splice sites. However, few of the predictions made by these programs is entirely accurate, and most must be verified experimentally for any gene of particular importance or interest. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/11%3A_Genomics_and_Systems_Biology/11.03%3A_Whole_Genome_Sequencing.txt |
Having identified putative genes within a genome sequence, how do we determine their function? Techniques of functional genomics are an experimental approach to address this question. One widely used technique in functional genomics is called microarray analysis (Figure \(1\)). Microarrays can measure the abundance of mRNA for hundreds or thousands of genes at once. The abundance of mRNA of a particular gene is usually correlated with the activity of that gene. For example, genes that are involved in neuronal development likely produce more mRNA in brain tissue than in heart tissue. We can therefore learn about the relationship between particular genes and particular processes by comparing transcript abundance under different conditions. This can identify tissue specific expression (e.g. the nerve/heart example above), as well as differences in temporal expression (development), or exposure to external agents (eg, disease, hormones, drugs, etc.).
11.E: Genomics and Systems Biology (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
11.1 What are the advantages of high-throughput –omics techniques compared to studying a single gene or protein at a time? What are the disadvantages
11.2 What would the chromatogram from a capillary sequencer look like if you accidentally added only template, primers, polymerase, and fluorescent terminators to the sequencing reaction?
11.3 What are the advantages and disadvantages of clone-by-clone vs. whole genome shotgun sequencing?
11.4 How could you use DNA sequencing to identify new species of marine microorganisms?
11.5 Explain how you could use a microarray to identify wheat genes that have altered expression during drought?
11.6 A microarray identified 100 genes whose transcripts are abundant in tumors, but absent in normal tissues. Do any or all of these transcripts cause cancer? Explain your answer.
11.7 How could you ensure that each spot printed on a microarray contains DNA for only one gene?
11.8 What would the spots look like on a microarray after hybridization, if each spot contained a random mixture of genes?
11.9 What would the spots look like if the hybridization of green and red labeled DNA was done at low stringency?
11.S: Genomics and Systems Biology (Summary)
• Genomics and related technologies differ from other techniques of molecular biology largely because of their scale; they allow many different genes (or gene products) to be studied in parallel.
• DNA sequencing can be applied to either a single gene, or in the case of genomics, to a large number of genes.
• Most DNA sequencing relies on the incorporation of dye-labeled terminator molecules, which create products that differ in length and end in a known nucleotide. The products can then be separated based on length, and the identity of the last based in each fragment can be determined based on fluorescence.
• Next-generation sequencing technologies have further reduced costs of sequencing, through miniaturization and parallelization.
• Physical maps are ordered sets of clones containing overlapping pieces of DNA, which together represent large pieces of chromosomes.
• Whole genomes may be sequenced using either a clone-by-clone approach, which required a physical map, or by a shotgun approach, in which small fragments are randomly sequenced.
• Genome analysis does not end after sequence acquisition; various features of the genome including genes (and their introns, exons, etc.) must be identified through a process called annotation.
• Functional genomics techniques including microarray analysis correlate transcript abundance with particular tissue samples. Genes whose transcripts are highly abundant under certain biological conditions may cause or respond to that condition.
Key Terms
genome
genomics
proteomics
transcriptomics
ddNTP
terminator nucleotide
capillary electrophoresis
chromatogram
next-generation sequencing
Ilumina
physical map
BAC
clone-by-clone sequencing
whole genome shotgun
genome annotation
functional genomics
microarray | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/11%3A_Genomics_and_Systems_Biology/11.04%3A_Functional_Genomics__Determining_Function%28s%29.txt |
Within most multicellular organisms, every cell contains essentially the same genomic sequence. How then do cells develop and function differently from each other? The answer lies in the regulation of gene expression. Only a subset of all the genes is expressed (i.e. are functionally active) in any given cell participating in a particular biological process. Gene expression is regulated at many different steps along the process that converts DNA information into active proteins. In the first stage, transcript abundance can be controlled by regulating the rate of transcription initiation and processing, as well as the degradation of transcripts. In many cases, higher abundance of a gene’s transcripts is correlated with its increased expression. In this chapter, we will focus on transcriptional regulation. Be aware, however, that cells also regulate the overall activity of genes in other ways. For example, by controlling the rate of mRNA translation, processing, and degradation, as well as the post-translational modification of proteins and protein complexes.
• 12.1: The lac Operon
Early insights into mechanisms of transcriptional regulation came from studies of E. coli by researchers Francois Jacob & Jacques Monod. In E. coli, and many other bacteria, genes encoding several different proteins may be located on a single transcription unit called an operon. The genes in an operon share the same transcriptional regulation, but are translated individually. Eukaryotes generally do not group genes together as operons (exception is C. elegans and a few other species).
• 12.2: The Use of Mutants to Study the lac Operon
The lac operon and its regulators were first characterized by studying mutants of E. coli that exhibited various abnormalities in lactose metabolism.
• 12.3: Eukaryotic Gene Regulation
• 12.4: Regulatory Elements in Evolution
• 12.5: Additional Levels of Regulating Transcription
Eukaryotes regulate transcription via promoter sequences close to the transcription unit (as in prokaryotes) and also use more distant enhancer sequences to provide more variation in the timing, level, and location of transcription, however, there are still additional levels of genetic control. This consists of two major mechanism: (1) large-scale changes in chromatin structure, and (2) modification of bases in the DNA sequence. These two are often inter-connected.
• 12.6: Epigenetics
The word “epigenetics” has become popular in the last decade and its meaning has become confused. The term epigenetics describes any heritable change in phenotype that is not associated with a change the chromosomal DNA sequence.
• 12.7: Regulation of Gene Expression (Exercises)
• 12.S: Regulation of Gene Expression (Summary)
Thumbnail: The stickleback is an example of an organism in which mutations cause changes in the regulation of gene expression. These mutations confer a selective advantage in some environments. Natural selection acts on mutations altering gene expression as well as those changing the coding regions of genes. (Flickr-frequency-CC:AND)
12: Regulation of Gene Expression
Early insights into mechanisms of transcriptional regulation came from studies of E. coli by researchers Francois Jacob & Jacques Monod. In E. coli, and many other bacteria, genes encoding several different proteins may be located on a single transcription unit called an operon. The genes in an operon share the same transcriptional regulation, but are translated individually. Eukaryotes generally do not group genes together as operons (exception is C. elegans and a few other species).
Basic lac Operon structure
E. coli encounters many different sugars in its environment. These sugars, such as lactose and glucose, require different enzymes for their metabolism. Three of the enzymes for lactose metabolism are grouped in the lac operon: lacZ, lacY, and lacA (Figure \(1\)). LacZ encodes an enzyme called β-galactosidase, which digests lactose into its two constituent sugars: glucose and galactose. lacY is a permease that helps to transfer lactose into the cell. Finally, lacA is a trans-acetylase; the relevance of which in lactose metabolism is not entirely clear. Transcription of the lac operon normally occurs only when lactose is available for it to digest. Presumably, this avoids wasting energy in the synthesis of enzymes for which no substrate is present. A single mRNA transcript includes all three enzyme-coding sequences and is called polycistronic. A cistron is equivalent to a gene.
cis- and transRegulators
In addition to the three protein-coding genes, the lac operon contains short DNA sequences that do not encode proteins, but are instead binding sites for proteins involved in transcriptional regulation of the operon. In the lac operon, these sequences are called P (promoter), O (operator), and CBS (CAP-binding site). Collectively, sequence elements such as these are called cis-elements because they must be located on the same piece of DNA as the genes they regulate. On the other hand, the proteins that bind to these cis-elements are called trans-regulators because (as diffusible molecules) they do not necessarily need to be encoded on the same piece of DNA as the genes they regulate.
lacI is an allosterically regulated repressor
One of the major trans-regulators of the lac operon is encoded by lacI. Four identical molecules of lacI proteins assemble together to form a homotetramer called a repressor (Figure \(2\)). This repressor binds to two operator sequences adjacent to the promoter of the lac operon. Binding of the repressor prevents RNA polymerase from binding to the promoter (Figure \(3\)). Therefore, the operon will not be transcribed when the operator is occupied by a repressor.
Besides its ability to bind to specific DNA sequences at the operator, another important property of the lacI protein is its ability to bind to lactose. When lactose is bound to lacI, the shape of the protein changes in a way that prevents it from binding to the operator. Therefore, in the presence of lactose, RNA polymerase is able to bind to the promoter and transcribe the lac operon, leading to a moderate level of expression of the lacZ, lacY, and lacA genes. Proteins such as lacI that change their shape and functional properties after binding to a ligand are said to be regulated through an allosteric mechanism. The role of lacI in regulating the lac operon is summarized in Figure \(4\).
CAP is an allosteric activator of the lac operon
A second aspect of lac operon regulation is conferred by a trans-factor called cAMP binding protein (CAP, Figure \(4\)). CAP is another example of an allosterically regulated trans-factor. Only when the CAP protein is bound to cAMP can another part of the protein bind to a specific cis-element within the lac promoter called the CAP binding sequence (CBS). CBS is located very close to the promoter (P). When CAP is bound to at CBS, RNA polymerase is better able to bind to the promoter and initiate transcription. Thus, the presence of cAMP ultimately leads to a further increase in lac operon transcription.
The physiological significance of regulation by cAMP becomes more obvious in the context of the following information. The concentration of cAMP is inversely proportional to the abundance of glucose: when glucose concentrations are low, an enzyme called adenylate cyclase is able to produce cAMP from ATP. Evidently, E. coli prefers glucose over lactose, and so expresses the lac operon at high levels only when glucose is absent and lactose is present. This provides another layer of logical control of lac operon expression: only in the presence of lactose, and in the absence of glucose is the operon expressed at its highest levels. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/12%3A_Regulation_of_Gene_Expression/12.01%3A_The_lac_Operon.txt |
Single mutants of the lac operon
The lac operon and its regulators were first characterized by studying mutants of E. coli that exhibited various abnormalities in lactose metabolism. Some mutants expressed the lac operon genes constitutively, meaning the operon was expressed whether or not lactose was present in the medium. Such mutant are called constitutive mutants.
The operator locus (lacO) - One example is Oc, in which a mutation in an operator sequence and reduces or precludes the repressor (the lacI gene product) from recognizing and binding to the operator sequence. Thus, in Oc mutants, lacZ, lacY, and lacA are expressed whether or not lactose is present.
The lacI locus – One type of mutant allele of lacI (callled I-) prevents either the production of a repressor polypeptide or produces a polypeptide that cannot bind to the operator sequence. This is also a constitutive expresser of the lac operon because absence of repressor binding permits transcription.
Another type of mutant of lacI called Is prevents the repressor polypeptide from binding lactose, and thus will bind to the operator and be non-inducible.. This mutant constitutively represses the lac operon whether lactose is present or not. The lac operon is not expressed and this mutant is called a “super-suppressor”.
The F-factor and two lac operons in a single cell – partial diploid in E.coli
More can be learned about the regulation of the lac operon when two different copies are present in one cell. This can be accomplished by using the F-factor to carry one copy, while the other is on the genomic E. coli chromosome. This results in a partial diploid in E. coli.
The F-factor is an episome that is capable of being either a free plasmid or integrated into the host bacterial chromosome. This switching is accomplished by IS elements where unequal crossing over can recombine the F-factor and adjacent DNA sequences (genes) in and out of the host chromosome. Researchers have used this genetic tool to create partial diploids (merozygotes) that allow them to test the regulation with different combinations of different mutations in one cell. For example, the F-factor copy may have a IS mutation while the genomic copy might have an OC mutation. How would this cell respond to the presence/absence of lactose (or glucose)? This partial diploid can be used to determine that IS is dominant to I+, which in turn is dominant to I-. It can also be used to show the OC mutation only acts in cis- while the lacI mutation can act in trans- .
12.03: Eukaryotic Gene Regulation
Like prokaryotes, transcriptional regulation in eukaryotes involves both cis-elements and trans-factors, only there are more of them and they interact in a more complex way. A diagram of a typical eukaryotic gene, including several types of cis-elements, is shown in Figure \(7\).
Proximal Regulatory Sequences
As in prokaryotes the RNA polymerase binds to the gene at its promoter to begin transcription. In eukaryotes, however, RNApol is part of a large protein complex that includes additional proteins that bind to one or more specific cis-elements in the promoter region, including GC-rich boxes, CAAT boxes, and TATA boxes. High levels of transcription require both the presence of this protein complex at the promoter, as well as their interaction with other trans-factors described below. The approximate position of these elements relative to the transcription start site (+1) is shown in Figure \(7\), but it should be emphasized that the distance between any of these elements and the transcription start site can vary, but are typically within ~200 base pairs of the start of transcription. This contrasts the next set of elements.
Distal Regulatory elements
Even more variation is observed in the position and orientation of the second major type of cis-regulatory element in eukaryotes, which are called enhancer elements. Regulatory trans-factor proteins called transcription factors bind to enhancer sequences, then, while still bound to DNA, these proteins interact with RNApol and other proteins at the promoter to enhance the rate of transcription. There are a wide variety of different transcription factors and each recognizes a specific DNA sequence (enhancer element) to promote gene expression in the adjacent gene under specific circumstances. Because DNA is a flexible molecule, enhancers can be located near (~100s of bp) or far (~10K of bp), and either upstream or downstream, from the promoter (Figures \(7\) and \(8\)).
Example 1: Drosophila yellow gene
The yellow gene of Drosophila provides an example of the modular nature of enhancers. This gene encodes an enzyme in the pathway that produces a dark pigment in the insect’s exoskeleton. Mutants have a yellow cuticle rather than the wild type darker pigmented cuticle. (Why call the gene “yellow”: recall that genes are often named after their mutant phenotype.) Figure \(9\) shows three enhancer elements (left side - wing, body, mouth), each binds a different tissue specific transcription factor to enhancer transcription of yellow+ in that tissue and makes the pigment. So, the wing cells will have a transcription factor that binds to the wing enhancer to drive expression; likewise in the body and mouth cells. Thus, specific combinations of cis-elements and trans-factors control the differential, tissue-specific expression of genes. This type of combinatorial action of enhancers is typical of the transcriptional activation of most eukaryotic genes: specific transcription factors activate the transcription of target genes under specific conditions.
While enhancer sequences promote expression, there is an oppositely acting type of element, called silencers. These elements function in much the same manner, with transcription factors that bind to DNA sequences, but they act to silence or reduce transcription from the adjacent gene.
Again, a gene’s expression profile (transcription level, tissue specific, temporal specific) is a combination of various enhancer and silencer elements.
Example 2: Gal4-UAS system from yeast – a genetic tool
Genetic researchers have taken advantage of a yeast distal enhancer sequence to make the GAL4-UAS system, a powerful technique for studying the expression of genes in other eukaryotes. It relies on two parts: a “driver” and a “responder” (Figure \(10\)). The driver part is a gene encoding a yeast transcriptional activator protein called Gal4. It is separate from the responder part, which contains the enhancer sequence, or upstream activation sequence (UAS, as it is called in yeast) to which the Gal4 protein specifically binds. This UAS is placed upstream (using genetic engineering) from a promoter transcribing a reporter gene, or other gene of interest, such as GFP (green fluorescent protein).
Both parts must be present in the same cell for the system to express the responder gene. If the driver is absent, the responder product will not be expressed. However, both are in the same cell (or organism) the pattern of expression of the driver part will induce the responder part’s expression in the same pattern.
This system works is a variety of eukaryotes, including humans. It has been especially well exploited in Drosophila where 100’s (1,000’s ? ) of differently expressing driver lines are available. These lines permit the tissue specific expression of any responder gene to examine its effect on development. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/12%3A_Regulation_of_Gene_Expression/12.02%3A_The_Use_of_Mutants_to_Study_the_lac_Operon.txt |
Mutations can occur in both cis-elements and trans-factors; both can result in altered patterns of gene expression. If an altered pattern of gene expression results in a selective advantage (or at least do not produce a major disadvantage), they may be selected and maintained in future populations. They may even contribute to the evolution of new species. An example of a sequence change in an enhancer is found in the Pitx gene.
Example: Pitx expression in Stickleback
The three-spined stickleback (Figure $11$) provides an example of natural selection of a mutation in a cis-regulatory element. This fish occurs in two forms: (1) populations that inhabit deep, open water and have a spiny pelvic fin that deters larger predator fish from feeding on them; (2) populations from shallow water environments and lack this spiny pelvic fin. In shallow water, it appears that a long, spiny pelvic fin would be a disadvantage because it frequently contacts the sediment at the bottom of the pond and allows parasitic insects in the sediment to invade the stickleback. Researchers compared gene sequences of individuals from both deep and shallow water environments as shown in Figure $11$. They observed that in embryos from the deep-water population, a gene called Pitx was expressed in several groups of cells, including those that developed into the pelvic fin. Embryos from the shallow-water population expressed Pitx in the same groups of cells as the other population, with an important exception: Pitx was not expressed in the pelvic fin primordium in the shallow-water population. Further genetic analysis showed that the absence of Pitx gene expression from the developing pelvic fin of shallow-water stickleback was due to the absence (mutation) of a particular enhancer element upstream of Pitx.
Example: Hemoglobin expression in placental mammals.
Hemoglobin is the oxygen-carrying component of red blood cells (erythrocytes). Hemoglobin usually exists as tetramers of four non-covalently bound hemoglobin molecules (Fig 12.12). Each hemoglobin molecule consists of a globin polypeptide with a covalently attached heme molecule. Heme is made through a specialized metabolic pathway and is then bound to globin polypeptide through post-translational modification.
The composition of the tetramers changes during development (Figure $13$). From early childhood onward, most tetramers are of the type $\mathbf{\alpha}$2$\mathbf{\beta}$2, which means they contain of two copies of each of two slightly different globin proteins named $\alpha$ and $\beta$. A small amount of adult hemoglobin is $\alpha$2$\delta$2, which has $\delta$ globin instead of the more common $\beta$ globin. Other tetrameric combinations predominate before birth: $\zeta$2$\varepsilon$2 is most abundant in embryos, and $\alpha$2$\gamma$2 is most abundant in fetuses. Although the six globin proteins ($\alpha$ = alpha, $\beta$ = beta , $\gamma$ = gamma, $\delta$ =delta, $\varepsilon$ =epsilon , $\zeta$ = zeta) are very similar to each other, they do have slightly different functional properties. For example, fetal hemoglobin has a higher oxygen affinity than adult hemoglobin, allowing the fetus to more effectively extract oxygen from maternal blood. The specialized $\gamma$ globin genes that are characteristic of fetal hemoglobin are found only in placental mammals.
Each of these globin polypeptides is encoded by a different gene. In humans, globin genes are located in clusters on two chromosomes (Figure $14$). We can infer that these clusters arose through a series of duplications of an ancestral globin gene. Gene duplication events can occur through rare errors in processes such as DNA replication, meiosis, or transposition. The duplicated genes can accumulate mutations independently of each other. Mutations can occur in either the regulatory regions (e.g. promoter regions), or in the coding regions, or both. In this way, the promoters of globin genes have evolved to be expressed at different phases of development, and to produce proteins optimized for the prenatal environment.
Of course, not all mutations are beneficial: some mutations can lead to inactivation of one or more of the products of a gene duplication. This can produce what is called a pseudogene. Examples of pseudogenes ($\psi$) are also found in the globin clusters. Pseudogenes have mutations that prevent them from being expressed at all. The globin genes provide an example of how gene duplication and mutation, followed by selection, allows genes to evolve specialized expression patterns and functions. Many genes have evolved as gene families in this way, although they are not always clustered together as are the globins.
12.05: Additional Levels of Regulating Transcription
Eukaryotes regulate transcription via promoter sequences close to the transcription unit (as in prokaryotes) and also use more distant enhancer sequences to provide more variation in the timing, level, and location of transcription, however, there are still additional levels of genetic control. This consists of two major mechanism: (1) large-scale changes in chromatin structure, and (2) modification of bases in the DNA sequence. These two are often inter-connected.
Chromatin Dynamics
Despite the simplified way in which we often represent DNA in figures such as those in this chapter, DNA is almost always associated with various chromatin proteins. For example, histones remain associated with the DNA even during transcription. Thus the rate of transcription is also controlled by the accessibility of DNA to RNApol and regulatory proteins. So, in regions were the chromatin is highly compacted, it is unlikely that any gene will be transcribed, even if all the necessary cis- and trans- factors are present in the nucleus. The extent of chromatin compaction in various regions is regulated through the action of chromatin remodeling proteins. These protein complexes include enzymes that add or remove chemical tags, such as methyl or acetyl groups, to various DNA bound proteins. These modifications alter the local chromatin density and thus the availability for transcription. Acetylated histones, for example, tend to be associated with actively transcribed genes, whereas deacetylated histone are associated with genes that are silenced (Figure \(15\)).
Likewise, methylation of DNA itself is also associated with transcription regulation. Cytosine bases, particularly when followed by a guanine (CpG sites) are important targets for DNA methylation (Figure \(16\)). Methylated cytosine within clusters of CpG sites is often associated with transcriptionally inactive DNA.
The modification of DNA and its associated proteins is enzymatically reversible (acetylation/deacetylation; methylation/demethylation) and thus a cyclical activity. Regulation of this provides another layer through which eukaryotic cells control the transcription of specific genes. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/12%3A_Regulation_of_Gene_Expression/12.04%3A_Regulatory_Elements_in_Evolution.txt |
The word “epigenetics” has become popular in the last decade and its meaning has become confused. The term epigenetics describes any heritable change in phenotype that is not associated with a change the chromosomal DNA sequence.
Originally it meant the processes through which the genes were expressed to give the phenotype; that is, the changes in gene expression that occur during normal development of multicellular organisms. This includes the change in transcriptional state of a DNA sequence (gene) via DNA or chromatin protein reversible modifications. Thus, DNA methylation and chromatin protein methylation, phosphorylation, and acetylation have been targeted as mechanisms for “heritable” changes in cells as they grow from a single cell (zygote) and differentiate to a multicellular organism. Here, dividing cells commit to differentiate into different tissues such as muscle, neuron, and fibroblast due to the genes that they express or silence. Some genes are irreversibly silenced, through epigenetic mechanisms, in some cell types, but not in others. This doesn’t involve any change in DNA sequence.
Remember, these epigenetic effects are not permanent changes and thus cannot be selectable in an evolutionary context. However, mutations in the genes that regulate the epigenetic effect can be selected.
Definition: Epigenetics
Epigenetics describes any heritable change in phenotype that is not associated with a change the chromosomal DNA sequence.
Some heritable information can be passed on independent of the DNA sequence
More recently however, researchers have found many cases of environmentally induced changes in gene expression that can be passed on to subsequent generations – a multi-generational effect. These cases have also been called “epigenetics”, and probably involve similar reversible changes to the DNA and chromatin proteins.These altered expression patterns represent the diversity of expression for a genome. This “extended” phenotype, the ability to influence traits in the next generation, is a topic of current research and only some examples will be discussed here.
One example comes from the grandchildren of famine victims are known to have lower birth weight than children without a family history of famine. This heritability of altered state of gene expression is surprising, since it appears not to involve typical changes in the sequence of DNA.The term epigenetics is applied here since the apparently heritable change in phenotype is associated with something other than chromosomal DNA sequence.
This change is inherited from one generation to the next and is thus transgenerational. In develpmental epigenetics, the expression state (developmentally differentiated state) is conserved only from one mitosis to the next, but is erased or rest at meiosis (the boundary of one generation to the next). The basis of at least some types of epigenetic inheritance appears to be replication of patterns of histone and DNA methylation that occurs in parallel with the replication of the primary DNA sequence. It is becoming clear that epigenetics is an important part of biology, and can serve as a type of cellular memory, sometimes within an individual, or sometimes across a few generations, at least.
The permanence of this “change” is not the same as changes in the DNA sequence itself. What is clear is that epigenetics is an important part of regulating gene expression, and can serve as a type of cellular memory, certainly within an individual, or across a few generations in some cases.
Imprinting and Parent-of-Origin Effects
For some genes, the allele inherited from the female parent is expressed differently than the allele that is inherited from the male parent. This is distinct from sex-linkage and is true even if both alleles are wild-type and autosomal. During gamete development (gametogenesis), each parent imprints epigenetic information on some genes that will affect the activity of the gene in the offspring. Imprinting does not change the DNA sequence, but does involve methylation of DNA and histones, and generally silences the expression of one of the parent’s alleles. In humans, some genes are expressed only from the paternal allele, and other genes are expressed only from the maternal allele. The imprinting marks are reprogrammed before the next generation of gametes are formed. Thus, although a male inherits epigenetic information from both his mother and father, this information is erased before sperm development, and he passes only one pattern of imprinting to both his sons and daughters. Most examples of imprinting come from placental mammals, and many imprinted genes control growth rate, such as IGF2 (insulin-like growth factor 2).
Imprinting appears to explain many different parent-of-origin effects. For example, Prader-Willi Syndrome (PWS) and Angelman Syndrome (AS) are two phenotypically different conditions in humans that result from deletion of a specific region of chromosome 15, which contains several genes. Whether the deletion results in PWS or in AS depends on the parent-of-origin. If the deletion is inherited from the father, PWS results.Conversely, if the deletion is inherited from the mother, AS is the result. The gene(s) involved in PWS is maternally silenced by imprinting, therefore the deletion of its paternally-inherited allele results in a complete deficiency of a required protein.On the other hand, the paternal allele of the gene involved in AS is silenced by imprinting, so deletion of the maternal allele results in deficiency of the protein encoded by that gene.
Transgenerational inheritance of nutritional influences
Nutrition is one aspect of the environment that has been particularly well-studied from an epigenetic perspective in both mice and humans. People alive today who experienced the Dutch famine of 1944-1945 as fetuses have IGF2 genes that are less-methylated than their siblings. Methylation of IGF2 (and birth rate) is also lower in children of mothers who do not take folic acid supplements as compared those who do.Furthermore, an individual’s phenotype can be influenced by the nutrition of parents or even grandparents.This transgenerational inheritance of nutritional effects appears to involve epigenetic mechanisms.
The mouse agouti gene produces a signaling molecule that regulates pigment-producing cells and brain cells that affect feeding and body weight. Normally, agouti is silenced by methylation, and these mice are brown and have a normal weight. When agouti is demethylated by feeding certain chemicals or by mutating a gene that controls methylation, some mice become yellow and overweight, although their DNA sequence remains unchanged.Methylation of agouti and normal weight and pigmentation of offspring can be restored if their mothers are fed folic acid and other vitamins during pregnancy.
A study of an isolated Swedish village called Överkalix provides an example of transgenerational inheritance of nutritional factors. Detailed historical records allowed researchers to infer the nutritional status of villagers going back to 1890.The researchers then studied the health of two generations of these villagers’ offspring, using medical records. A significant correlation was found between the mortality risk of grandsons and the food availability of their paternal grandfathers.This effect was not seen in the granddaughters. Furthermore, the nutrition of paternal grandmothers, or either of the maternal grandparents did not affect the health of the grandsons. It was therefore proposed that epigenetic information affecting health (specifically diabetes and heart disease) was passed from the grandfathers, to the grandsons, through the male line.
Vernalization as an example of epigenetics
Many plant species in temperate regions are winter annuals, meaning that their seeds germinate in the late summer, and grow vegetatively through early fall before entering a dormant phase during the winter, often under a cover of snow. In the spring, the plant resumes growth and is able to produce seeds before other species that germinated in the spring. In order for this life strategy to work, the winter annual must not resume growth or start flower production until winter has ended. Vernalization is the name given to the requirement to experience a long period of cold temperatures prior to flowering.
How does a plant sense that winter has passed? The signal for resuming growth cannot simply be warm air temperature, since occasional warm days, followed by long periods of freezing, are common in temperate climates. Researchers have discovered that winter annuals use epigenetic mechanisms to sense and “remember” that winter has occurred
Fortunately for the researchers who were interested in vernalization, some varieties of Arabidopsis are winter annuals. Through mutational analysis of Arabidopsis, researchers found that a gene called FLC (FLOWERING LOCUS C) encodes a transcription repressor acting on several of the genes involved in early stages of flowering (Figure \(16\)).In the fall and under other warm conditions, the histones associated with FLC are acetylated and so FLC is transcribed at high levels; expression of flowering genes is therefore entirely repressed. However, in response to cold temperatures, enzymes gradually deacetylate the histones associated with FLC. The longer the cold temperatures persist, the more acetyl groups are removed from the FLC-associated histones, until finally the FLC locus is no longer transcribed and the flowering genes are free to respond to other environmental and hormonal signals that induce flowering later in the spring. Because the deacetylated state of FLC is inherited as cells divide and the plant grows in the early spring, this is an example of a type of cellular memory mediated by an epigenetic mechanism. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/12%3A_Regulation_of_Gene_Expression/12.06%3A_Epigenetics.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
12.1 List all the mechanisms that can be used to regulate gene expression in eukaryotes.
12.2 With respect to the expression of β-galactosidase, what would be the phenotype of each of the following strains of E. coli?
a) I+, O+, Z+, Y+ (no glucose, no lactose)
b) I+, O+, Z+, Y+ (no glucose, high lactose)
c) I+, O+, Z+, Y+ (high glucose, no lactose)
d) I+, O+, Z+, Y+ (high glucose, high lactose)
e) I+, O+, Z-, Y+ (no glucose, no lactose)
f) I+, O+, Z-, Y+ (high glucose, high lactose)
g) I+, O+, Z+, Y- (high glucose, high lactose)
h) I+, Oc, Z+, Y+ (no glucose, no lactose)
i) I+, Oc,Z+, Y+ (no glucose, high lactose)
j) I+, Oc, Z+, Y+ (high glucose, no lactose)
k) I+, Oc, Z+, Y+ (high glucose, high lactose)
l) I-, O+, Z+, Y+ (no glucose, no lactose)
m) I-, O+, Z+, Y+ (no glucose, high lactose)
n) I-, O+, Z+, Y+ (high glucose, no lactose)
o) I-, O+, Z+, Y+ (high glucose, high lactose)
p) Is, O+, Z+, Y+ (no glucose, no lactose)
q) Is, O+, Z+, Y+ (no glucose, high lactose)
r) Is, O+, Z+, Y+ (high glucose, no lactose)
s) Is, O+, Z+, Y+ (high glucose, high lactose)
12.3 In the E. coli strains listed below, some genes are present on both the chromosome, and the extrachromosomal F-factor episome. The genotypes of the chromosome and episome are separated by a slash. What will be the β-galactosidase phenotype of these strains? All of the strains are grown in media that lacks glucose.
a) I+, O+, Z+, Y+ / O-, Z-, Y- (high lactose)
b) I+, O+, Z+, Y+ / O-, Z-, Y- (no lactose)
c) I+, O+, Z-, Y+ / O-, Z+, Y+ (high lactose)
d) I+, O+, Z-, Y+ / O-, Z+, Y+ (no lactose)
e) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (high lactose)
f) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (no lactose)
g) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
h) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
i) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
j) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
k) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (high lactose)
l) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (no lactose)
m) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (high lactose)
n) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (no lactose)
o) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
p) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
12.1 Transcriptional: initiation, processing & splicing, degradation
Translational: initiation, processing, degradation
Post-translational: modifications (e.g. phosphorylation), localization
Others: histone modification, other chromatin remodeling, DNA methylation
12.2 Legend:
+++ Lots of β-galactosidase activity
+ Moderate β-galactosidase activity
-- No β-galactosidase activity
-- a) I+, O+, Z+, Y+ (no glucose, no lactose)
+++ b) I+, O+, Z+, Y+ (no glucose, high lactose)
-- c) I+, O+, Z+, Y+ (high glucose, no lactose)
+ d) I+, O+, Z+, Y+ (high glucose, high lactose)
-- e) I+, O+, Z-, Y+ (no glucose, no lactose)
-- f) I+, O+, Z-, Y+ (high glucose, high lactose)
+ g) I+, O+, Z+, Y- (high glucose, high lactose)
+++ h) I+, Oc, Z+, Y+ (no glucose, no lactose)
+++ i) I+, Oc,Z+, Y+ (no glucose, high lactose)
+ j) I+, Oc, Z+, Y+ (high glucose, no lactose)
+ k) I+, Oc, Z+, Y+ (high glucose, high lactose)
+++ l) I-, O+, Z+, Y+ (no glucose, no lactose)
+++ m) I-, O+, Z+, Y+ (no glucose, high lactose)
+ n) I-, O+, Z+, Y+ (high glucose, no lactose)
+ o) I-, O+, Z+, Y+ (high glucose, high lactose)
-- p) Is, O+, Z+, Y+ (no glucose, no lactose)
-- q) Is, O+, Z+, Y+ (no glucose, high lactose)
-- r) Is, O+, Z+, Y+ (high glucose, no lactose)
-- s) Is, O+, Z+, Y+ (high glucose, high lactose)
12.3 Legend:
+++ Lots of β-galactosidase activity
+ Moderate β-galactosidase activity
-- No β-galactosidase activity
+++ a) I+, O+, Z+, Y+ / O-, Z-, Y- (high lactose)
-- b) I+, O+, Z+, Y+ / O-, Z-, Y- (no lactose)
+++ c) I+, O+, Z-, Y+ / O-, Z+, Y+ (high lactose)
+ d) I+, O+, Z-, Y+ / O-, Z+, Y+ (no lactose)
+++ e) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (high lactose)
-- f) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (no lactose)
+++ g) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
-- h) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
+++ i) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
+++ j) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
+++ k) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (high lactose)
+++ l) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (no lactose)
-- m) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (high lactose)
-- n) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (no lactose)
-- o) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
-- p) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
12.4 You could demonstrate this with just I+OcZ-/I+O+Z+. The fact that this does not have constitutive lactose expression shows that the operator only acts on the same piece of DNA on which it is located. There are also other possible answers.
12.5 You could also demonstrate this with just I+O+Z-/I-O+Z+. The fact that this has the same lactose-inducible phenotype as wild-type hows that a functional lacI gene can act on operators on both the same piece of DNA from which it is transcribed, or on a different piece of DNA. There are also other possible answers.
12.6 For all of these, the answer is the same: The lacoperon would be inducible by lactose, but only moderate expression of the lac operon would be possible, even in the absence of glucose
a) loss-of-function of adenylate cyclase
b) loss of DNA binding ability of CAP
c) loss of cAMP binding ability of CAP
d) mutation of CAP binding site (CBS) cis-element so that CAP could not bind
12.7 Both involve trans-factors binding to corresponding cis-elements to regulate the initiation of transcription by recruiting or stabilizing the binding of RNApol and related transcriptional proteins at the promoter. In prokaryotes, genes may be regulated as a single operon. In eukaryotes, enhancers may be located much further from the promoter than in prokaryotes.
12.8 These fish would all have spiny tales like the deep-water population.
12.9 These could have arisen from loss-of-function mutation in FLC, or in the cis-element to which FLC normally binds.
12.10 If there was no deacetylation of FLC by HDAC, transcription of FLC might continue constantly, leading to constant suppression of flowering, even after winter.
12.S: Regulation of Gene Expression (Summary)
• Regulation of gene expression is essential to the normal development and efficient functioning of cells
• Gene expression may be regulated by many mechanisms, including those affecting transcript abundance, protein abundance, and post-translational modifications
• Regulation of transcript abundance may involve controlling the rate of initiation and elongation of transcription, as well as transcript splicing, stability, and turnover
• The rate of initiation of transcription is related to the presence of RNA polymerase and associated proteins at the promoter.
• RNApol may be blocked from the promoter by repressors, or may be recruited or stabilized at the promoter by other proteins including transcription factors
• The lac operon is a classic, fundamental paradigm demonstrating both positive and negative regulation through allosteric effects on trans-factors.
• In eukaryotes, cis-elements that are usually called enhancers bind to specific trans-factors to regulate transcriptional initiation.
• Enhancers may be modular, with each enhancer and its transcription factor regulating a distinct component of a gene’s expression pattern, as in the yellow gene.
• Sticklebacks provide examples of recent evolutionary events in which mutation of an enhancer produced a change in morphology and a selective advantage.
• Chromatin structure, including reversible modifications such as acetylation of histones, and methylation DNA CpG sites also regulates the initiation of transcription.
• Chromatin modifications or DNA methylation of some genes are heritable over many mitotic, and sometimes even meiotic divisions.
• Heritable changes in phenotype that do not result from a change in DNA sequence are called epigenetic. Many epigenetic phenomena involve regulation of gene expression by chromatin modification and/or DNA methylation.
Key Terms:
gene expression
transcriptional regulation
operon
lactose
glucose
lac operon
lacZ
lacY
lacA
galactosidase
permease
trans-acetylase
P / promoter
O / operator
CBS
CAP-binding site
cis-elements
trans-regulators
lacI
homotetramer
repressor
allosteric
cAMP binding protein
CAP
CAP binding sequence
CBS
adenylate cyclase
constitutive
Oc / I- / Is
F-factor / episome
GC boxes
CAAT boxes
TATA boxes
GAL4-UAS
Driver/responder
transcription start site
enhancers/silencers
transcription factors
hemoglobin/heme/globin
pseudogene
gene families
stickleback
primordium
chromatin remodeling
acetylation/deacetylation
methylation/demethylation
CpG sites
epigenetics
winter annual
vernalization
FLC | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/12%3A_Regulation_of_Gene_Expression/12.07%3A_Regulation_of_Gene_Expression_%28Exercises%29.txt |
Cancers can be classified based on the tissues in which they originate. Sarcomas are cancers that originate in mesoderm tissues, such as bone or muscle, and cancers arising in glandular tissues (e.g. breast, prostate) are classified as adenocarcinomas. Carcinomas originate in epithelial cells (both inside the body and on its surface) and are the most common types of cancer (~85%). Each of these classifications may be further sub-‐divided. For example, squamous cell carcinoma (SCC), basal cell carcinoma (BCC), and melanoma are all types of skin cancers originating respectively in the squamous cells, basal cells, or melanocytes of the skin.
13.02: Cancer Cell Biology
Cancer is a progressive disease that usually begins with increased frequency of cell division (Figure \(2\)). Under the microscope, this may be detectable as increased cellular and nuclear size, and an increased proportion of cells undergoing mitosis. As the disease progresses, cells typically lose their normal shape and tissue organization. Tissues with increased cell division and abnormal tissue organization exhibit dysplasia. Eventually a tumor develops, which can grow rapidly and expand into adjacent tissues.
As cellular damage accumulates and additional control mechanisms are lost, some cells may break free of the primary tumor, pass into the blood or lymph system, and be transported to another organ, where they develop into new tumors (Figure \(3\)). The early detection of tumors is important so that they can be treated or removed before the onset of metastasis, but note that not all tumors will lead to cancer. Tumors that do not metastasize are classified as benign, and are not usually considered life threatening. In contrast, malignant tumors become invasive, and ultimately result in cancer.
13.03: Hallmarks of Cancer
Researchers have identified six molecular and cellular traits that characterize most cancers. These six hallmarks of cancer are summarized in Table \(1\). In this chapter, we will focus on the first two hallmarks, namely growth signal autonomy and insensitivity to anti-‐growth signals.
Table \(1\) Ten Hallmarks of Cancer (Hanahan and Weinberg, 2000; Hanahan 2011)
1. Growth signal autonomy
Cancer cells can divide without the external signals normally required to stimulate division.
2. Insensitivity to growth inhibitory signals
Cancer cells are unaffected by external signals that inhibit division of normal cells.
3. Evasion of apoptosis
When excessive DNA damage and other abnormalities are detected, apoptosis (a type of programmed cell death) is induced in normal cells, but not in cancer cells.
4. Reproductive potential not limited by telomeres
Each division of a normal cell reduces the length of its telomeres. Normal cells arrest further division once telomeres reach a certain length. Cancer cells avoid this arrest and/or maintain the length of their telomeres.
5. Sustained angiogenesis
Most cancers require the growth of new blood vessels into the tumor. Normal angiogenesis is regulated by both inhibitory and stimulatory signals not required in cancer cells.
6. Tissue invasion and metastasis
Normal cells generally do not migrate (except in embryo development). Cancer cells invade other tissues including vital organs.
7. Deregulated metabolic pathways
Cancer cells use an abnormal metabolism to satisfy a high demand for energy and nutrients.
8. Evasion of the immune system
Cancer cells are able to evade the immune system.
9. Chromosomal instability
Severe chromosomal abnormalities are found in most cancers.
10. Inflammation
Local chronic inflammation is associated with many types of cancer. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/13%3A_Cancer_Genetics/13.01%3A__Classification_of_Cancers.txt |
A carcinogen is any agent that directly increases the incidence of cancer. Most, but not all carcinogens are mutagens. Carcinogens that do not directly damage DNA include substances that accelerate cell division, thereby leaving less opportunity for cell to repair induced mutations, or errors in replication. Carcinogens that act as mutagens may be biological, physical, or chemical in nature, although the term is most often used in relation to chemical substances.
Human Papilloma Virus (HPV, Figure \(4\)) is an example of a biological carcinogen. Almost all cervical cancers begin with infection by HPV, which contains genes that disrupt the normal pattern of cell division within the host cell. Any gene that leads to an uncontrolled increase in cell division is called an oncogene. The HPV E6 and E7 genes are considered oncogenes because they inhibit the host cell’s natural tumor suppressing proteins (include p53, described below). The product of the E5 gene mimics the host’s own signals for cell division, and these and other viral gene products may contribute to dysplasia, which is detected during a Pap smear (Figure \(5\)). Detection of abnormal cell morphology in a Pap smear is not necessarily evidence of cancer. It must be emphasized again that cells have many regulatory mechanisms to limit division and growth, and for cancer to occur, each of these mechanisms must be disrupted. This is one reason why only a minority of individuals with HPV infections ultimately develop cancer. Although most HPV-related cancers are cervical, HPV infection can also lead to cancer in other tissues, in both women and men.
Figure \(4\): Electron micrograph of HPV. (Wikipedia-Unknown-PD)
Figure \(5\): Dysplastic (left) and normal (right) cells from a Pap smear. (Flickr-Ed Uthman-CC:AS)
Radiation is a well-known physical carcinogen, because of its potential to induce DNA damage within the body. The most damaging type of radiation is ionizing, meaning waves or particles with sufficient energy to strip electrons from the molecules they encounter, including DNA or molecules that can subsequently react with DNA. Ionizing radiation, which includes x-rays, gamma rays, and some wavelengths of ultraviolet rays, is distinct from the non-ionizing radiation of microwave ovens, cell phones, and radios. As with other carcinogens, mutation of multiple, independent genes that normally regulate cell division is required before cancer develops.
Chemical carcinogens (Table \(2\)) can be either natural or synthetic compounds that, based on animal feeding trials or epidemiological (i.e. human population) studies, increase the incidence of cancer. The definition of a chemical as a carcinogen is problematic for several reasons. Some chemicals become carcinogenic only after they are metabolized into another compound in the body; not all species or individuals may metabolize chemicals in the same way. Also, the carcinogenic properties of a compound are usually dependent on its dose. It can be difficult to define a relevant dose for both lab animals and humans. Nevertheless, when a correlation between cancer incidence and chemical exposure is observed, it is usually possible to find ways to reduce exposure to that chemical.
Table \(2\): Some classes of chemical carcinogens (Pecorino 2008)
1. PAHs (polycyclic aromatic hydrocarbons)
e.g. benzo[a]pyrene and several other components of the smoke of cigarettes, wood, and fossil fuels
2. Aromatic amines
e.g. formed in food when meat (including fish, poultry) are cooked at high temperature
3. Nitrosamines and nitrosamides
e.g. found in tobacco and in some smoked meat and fish
4. Azo dyes
e.g. various dyes and pigments used in textiles, leather, paints.
5. Carbamates
e.g. ethyl carbamate (urethane) found in some distilled beverages and fermented foods
6. Halogenated compounds
e.g. pentachlorophenol used in some wood preservatives and pesticides.
7. Inorganic compounds
e.g. asbestos; may induce chronic inflammation and reactive oxygen species
8. Miscellaneous compounds
e.g. alkylating agents, phenolics | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/13%3A_Cancer_Genetics/13.04%3A_Mutagens_and_Carcinogens.txt |
The control of cell division involves many different genes. Some of these genes act as signaling molecules to activate normal progression through the cell cycle. One of the pre-requisites for cancer occurs when one or more of these activators of cell division become mutated.
The mutation may involve a change in the coding sequence of the protein, so that it is more active than normal, or a change in the regulation of its expression, so that it is produced at higher levels than normal, or persists in the cell longer than normal. Genes that are a part of the normal regulation of cell division, but which after mutation contribute to cancer, are called proto-oncogenes. Once a proto-oncogene has been abnormally activated by mutation, it is called an oncogene. More than 100 genes have been defined as proto-oncogenes. These include genes at almost every step of the signaling pathways that normally induce cell to divide, including growth factors, receptors, signal transducers, and transcription factors.
ras is an example of a proto-oncogene. ras acts as a switch within signal transduction pathways, including the regulation of cell division. When a receptor protein receives a signal for cell division, the receptor activates ras, which in turn activates other signaling components, ultimately leading to activation of genes involved in cell division. Certain mutations of the ras sequence causes it to be in a permanently active form, which can lead to constitutive activation of the cell cycle. This mutation is dominant as are most oncogenes. An example of the role of ras in relaying a signal for cell division in the EGF pathway is shown in Figure \(7\).
13.06: Tumor Suppressor Genes
More than 30 genes are classified as tumor suppressors. The normal functions of these genes include repair of DNA, induction of programmed cell death (apoptosis) and prevention of abnormal cell division. In contrast to proto-oncogenes, in tumor suppressors it is loss-of-function mutations that contribute to the progression of cancer. This means that tumor suppressor mutations tend to be recessive, and thus both alleles must be mutated in order to allow abnormal growth to proceed. It is perhaps not surprising that mutations in tumor suppressor genes, are more likely than oncogenes to be inherited. An example is the tumor suppressor gene, BRCA1, which is involved in DNA-repair. Inherited mutations in BRCA1 increase a woman’s lifetime risk of breast cancer by up to seven times, although these heritable mutations account for only about 10% of breast cancer. Thus, sporadic rather than inherited mutations are the most common sources of both oncogenes and disabled tumor suppressor genes.
An important tumor suppressor gene is a transcription factor named p53. Other proteins in the cell sense DNA damage, or abnormalities in the cell cycle and activate p53 through several mechanisms including phosphorylation (attachment of phosphate to specific site on the protein) and transport into the nucleus. In its active form, p53 induces the transcription of genes with several different types of tumor suppressing functions, including DNA repair, cell cycle arrest, and apoptosis. Over 50% of human tumors contain mutations in p53. People who inherit only one function copy of p53 have a greatly increased incidence of early onset cancer. However, as with the other cancer related genes we have discussed, most mutations in p53 are sporadic, rather than inherited. Mutation of p53, through formation of pyrimidine dimers in the genes following exposure to UV light, has been causally linked to squamous cell and basal cell carcinomas (but not melanomas, highlighting the variety and complexities of mechanisms that can cause cancer). | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/13%3A_Cancer_Genetics/13.05%3A_Oncogenes.txt |
Chronic myelogenous leukemia (CML)
Chronic myelogenous leukemia (CML) is a type of cancer of white blood cells, myeloid cells, that are mutated and proliferate uncontrollably through three stages (chronic, accelerated, and blast crisis) and lead eventually to death. Cytogenetics showed the myeloid cells of CML patients usually also have a consistent chromosome translocation (the mutant event) between the long arms of chromosomes 9 and 22, t(9:22)(q34;q11). It is also known as the Philadelphia chromosome (Ph+). This translocation involves breaks in two genes, c-abl and bcr, on chromosomes 9 and 22, respectively. The fusion of the translocation breaks result in a chimeric gene, called bcr-abl, that contains exons 1 and/or 2 from bcr (this varies from patient to patient) and 2-11 from abl and it produces a chimeric protein (BCR-ABL or p185bcr-abl) that is transcribed like bcr and contains abl enzyme sequences. This chimeric protein has a tyrosine-kinase from the abl gene sequences that is unique to the CML mutant cell. The consistent, unregulated expression of this gene and its kinase product causes activation of a variety of intracellular signaling pathways, promoting the uncontrolled proliferative and survival properties of CML cells (the cancer). Thus the BCR-ABL tyrosine kinase enzyme exists only in cancer cells (and not in healthy cells) and a drug that inhibits this activity could be used to target and prevent the uncontrolled growth of the cancerous CML cells.
Inhibiting the Bcr-Abl tyrosine kinase activity
Knowing that the kinase activity was the key to treatment, pharmaceutical companies screened chemical libraries of potential kinase inhibitory compounds. After initially finding low potency inhibitors, a relationship between structure and activity suggested other compounds that were optimized to inhibit the BCR-ABL tyrosine kinase activity. The lead compound was STI571, now called Gleevec™ or imatinib (Figure \(9\)). This drug was shown to inhibit the BCR-ABL tyrosine kinase activity and to inhibit CML cell proliferation in vitro and in vivo. Gleevec™ works via targeted therapy—only the kinase activity in cancer cells was targeted and thereby killed through the drug's action. In this regard, Gleevec™ was one of the first cancer therapies to show the potential for this type of targeted action. It was dependent upon the genetic identification of the cause and protein target and is often cited as a paradigm for genetic research in cancer therapeutics.
Figure \(9\): Biochemical structure of Gleevec™ or Imatinib. . (Wikipedia-Fuse809-CC:AN)
Caution
This is a simplified presentation of the CML/cancer targeting by the drug Gleevec™. There are many more details than could be presented here. It is represents as a model of finding a drug for each type of cancer, rather than the one, single “magic bullet” that kills all cancers. Remember, there are always complexities in this type of research to treatment process, such as patient genetic and environmental variation that leads to differences in drug metabolism, uptake, and binding. Also, changes in drug dose, mutation of the bcr-abl gene, and other events can affect the effectiveness of the treatment and the relapse rate. Biological systems are extremely complex and difficult to modulate in the specific, targeted manner necessary to treat cancer ideally.
Remember, the drug, Gleevec™, is not a cure, but only a treatment. It prevents the uncontrolled proliferation of the CML cells, but doesn’t kill them directly. The arrested cells will die eventually, but there is always a small pool of CML cells that will proliferate if the drug is discontinued. While sustained use of this expensive drug is beneficial to the pharmaceutical companies, it is certainly not the ideal situation for the patient.
13.E: Cancer Genetics (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
13.1 Why do oncogenes tend to be dominant, but mutations in tumor suppressors tend to be recessive?
13.2 What tumor suppressing functions are controlled by p53? How can a single gene affect so many different biological pathways?
13.3 Are all carcinogens mutagens? Are all mutagens carcinogens? Explain why or why not.
13.4 Imagine that a laboratory reports that feeding a chocolate to laboratory rats increases the incidence of cancer. What other details would you want to know before you stopped eating chocolate?
13.5 Do all women with HPV get cancer? Why or why not? Do all women with mutations in BRCA1 get cancer? Why or why not?
13.S: Cancer Genetics (Summary)
• Cancer is the name given to a class of different diseases that share common properties.
• Most cancers require accumulation of mutations in several different genes.
• Most cancer causing mutations are sporadic, rather than inherited, and most are caused by environmental carcinogens, including virus, radiation, and certain chemicals.
• Oncogenes are hyperactivated regulators of cell division, and are often derived from gain-of-function mutations in proto-oncogenes.
• Tumor suppressor genes normal help to repair DNA damage, arrest cell division, or to kill over proliferating cells. Loss-of-function of these genes contributes to the progression of cancer.
• Genetic research into cancer can provide enzyme targets for drug investigation and potential treatment. E.g. Gleevec™
Key Terms:
metastasis
dysplasia
carcinogen
HPV
oncogene
ionizing
epidemiology
proto-oncogene
receptor
signal transduction
ras
apoptosis
BRC1A
p53
tumor suppressor
phosphorylation
CML
Gleevec
bcr-abl | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/13%3A_Cancer_Genetics/13.07%3A__The_Poster_Boy_of_Genetic_Research_Leading_to_a_Cancer_Treatment__Gleevec_%28Imatinib%29.txt |
You have probably encountered symbols in high school genetics and perhaps in previous chapters assigned to you in this course. Chapter 1 points out that a position on a chromosome is a gene locus, which simply means the location at which a gene is found. At that position, the gene could be one of several variants that we call alleles. The set of alleles comprise an organism’s genotype, which might affect the organism’s phenotype through the action of the genotype.
When you are solving genetics problems you have to keep track of different alleles of one or more genes. This can become confusing, so understanding how a gene works, how alleles vary, how they are expressed, and the influence they have on phenotype is important. A consistent nomenclature system – a set of rules dictating how you name alleles – becomes immensely helpful.
14: Appendices
Word and practical problems in biology can get confusing in a hurry, particularly if you’re distracted by something like exam stress! When you are manipulating several ideas, it is good practice to be thoughtful and follow rules that keep you consistent in your interpretation. As an instructor, I have seen work in which the student clearly got flustered and forgot that the mutation he or she was working on was dominant. This often leads to an answer that is inconsistent with the data.
There are a few simple rules we can use for nomenclature. This first appendix uses a simplified system to communicate allele characteristics. The next appendix will show a more complicated system that carries even more information in the gene symbols.
14.02: Biochemistry of GeneBehavior
The Mechanics of Gene Expression
We often think of genes as “made of DNA”; they reside in the nucleus and endosymbiotic organelles of eukaryotes or the nucleoid region of prokaryotes. As described in Chapter 13, they are transcribed into an RNA message by RNA polymerase then interpreted by ribosomes that assemble particular amino acids into a polypeptide strand (also known as a protein) based on the sequence of nucleotides. In a cell, proteins can act as enzymes, structural features, pigments, and a host of other functions, including regulating the expression of other genes. This expression of genes leads to how an organism looks – its phenotype.
Dominant and Recessive; Homozygous, Heterozygous, and Hemizygous
Alleles themselves do not directly exert an effect on the phenotype of an organism. You will recall from Chapter 3 that genes are instructions, often for proteins. It is the effect of the protein that causes an organism to take on particular traits. In this chapter we will look at how to design symbols appropriate to communicate the characteristics of alleles you want to investigate, but keep in mind that the allele instructs what kind of protein to make.
Chapter 1 points out that organisms usually fall into the classes of being diploid or haploid. Humans and eukaryotic genetic systems usually assume the organism is diploid, which means that most chromosomes are represented as pairs. Each pair has homologous loci: the term homologous means “the same information” and refers to the gene at each locus. Note that although the loci, or genes, are the same, the alleles that comprise them may be different!
If an organism is diploid and homozygous for an allele (Figure A1.2, left and right), the gene at the same position of the homologous chromosomes is the same allele. Only one type of protein is made. If an organism is diploid and heterozygous (Figure A1.2, middle), and the protein from one allele influences the phenotype more than the protein from the other allele, we use the terms dominant and recessive, respectively.
Figure A1.2: Relationship between genotype and phenotype for an allele that is completely dominant to another allele. (Original-M. Deyholos -CC:AN – from chapter 13)
If one copy of an allele makes enough protein to compensate for the absence of protein from the other allele in a heterozygote, it will influence the phenotype. If this phenotype looks identical to that of an organism homozygous for the “functional” allele, we consider the “functional” allele to be dominant to the “null”. We could also say the “null” is recessive to the “functional” allele. Keep in mind that the alleles themselves aren’t doing anything, but it’s common practice to label the alleles as dominant or recessive, although in reality we are talking about the expression of those alleles.
Why this is important is how the proteins from two alleles interact. If both proteins are identical (from a homozygous genotype) the phenotype that results will be that of the action of one “type” of protein, even if though there are two copies of the gene – they are both the same allele. There’s no real interaction because the proteins do the same thing. It’s in heterozygotes that we can see whether a particular allele is dominant over another or otherwise influences the phenotype in an interesting way.
Here’s another point about the “normal” allele: it was honed by natural selection over a long period of time. Genes are instructions for the protein tools of an organism’s cells. For this reason we often call the “normal” allele of a gene the “wild type” allele. This would be the allele most common in the wild, presumably because it provides a benefit to the organism. Thus, most mutations are likely to reduce the effectiveness of the wild type allele, although the process of evolution allows (and, in fact, requires) an occasional beneficial allele to permeate a population if it provides a selective advantage.
Note that we do not use the terms homozygous or heterozygous in haploid organisms. If they are haploid, their phenotype will reflect the genotype of the only allele present. The proper term for their genotype is hemizygous to reflect only one copy of each gene.
There are Many Kinds of Alleles for a Gene
A mutation is a change in nucleotide sequence. Chapter 11 goes into more detail of what this means . What’s important now is that you understand that the amino acids of a protein can be different if we compare different alleles of a gene and they may behave differently – often one protein will “work better” than the other. If the promoter of a “functional” allele of a gene is damaged, the allele that is created might not even create an mRNA so no protein will be encoded by that allele. This is called an amorphic or “null” mutant (See Chapter 1).
Keep also in mind that a gene can be mutated in different parts of its sequence to create different alleles. A diploid organism can have a maximum of two alleles (aside from gene duplication or abnormal chromosome structure, but ignore that for now). In a population, though, there can be many, many different alleles. Perhaps allele one decreases protein function and allele 2 is even less effective. This describes an allelic series as follows: wild-‐type > allele 1 > allele 2 > any null allele. Wild-‐type alleles in this case encode the most effective protein. Null alleles represent catastrophic mutations that eliminate transcription or produces proteins that can’t function at all. See section 3.2 for another example of an allelic series.
Gender and Gene Interactions
Finally, here are a couple points about other interactions before we move on to gene symbols. Some organisms have a pair of sex chromosomes that dictate gender (see Chapter 4). We’ll ignore incorporating ways to communicate if a chromosome is a sex chromosome in this chapter; Appendix 2 will show how we can indicate those using a symbol. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/14%3A_Appendices/14.01%3A_Why_Use_Nomenclature.txt |
Single Letter System
Sometimes what you want to do is a little rough work for investigating your genetic model. A genetic model is a diagram of the logic that you propose for a particular type of inheritance. For instance, if you cross a true‐breeding purple plant with a true‐breeding white plant (Figure A1.2), you will get a heterozygote (the middle plant). If we name the gene after the mutation (a is the first letter in “albino”), we know that the heterozygote will have one capital letter “A” and a lower case “a”. The heterozygote is the F1 generation (“first filial”, which means it’s the first child from parents that are crossed). The F1 is purple, which means the “a” allele is recessive; only one copy of the “A” allele is needed for enough purple pigment to make it identical to one true-breeding parent. This is complete dominance.
We cannot know from the information given which allele is wild type or mutant. One hypothesis, though, is that purple pigments are required to attract pollinators and therefore would help the plant in the wild. Albino plants could be a mutant and might not generate as many seeds for lack of pollinators. If the context of your genetics problem doesn’t indicate which allele is wild‐type, it’s good practice to name your allele based on the recessive trait. Often the recessive allele is the mutation. We might consider that the “a” allele is null and makes no pigment at all. One or two “A” alleles make enough protein to cause the plant to be purple.
We can’t know from the information given which allele is wild type or mutant. One hypothesis, though, is that purple pigments are required to attract pollinators and therefore would help the plant in the wild. Albino plants could be a mutant and might not generate as many seeds for lack of pollinators. If the context of your genetics problem doesn’t indicate which allele is wild‐type, it’s good practice to name your allele based on the recessive trait. Often the recessive allele is the mutation. We might consider that the “a” allele is null and makes no pigment at all. One or two “A” alleles make enough protein to cause the plant to be purple.
Figure A1.2 already assumes that the capital letter (A) stands for an allele that encodes a protein for purple pigment and the recessive allele (a) doesn’t make pigment. Thus the Aa heterozygote is sufficient evidence to adopt upper- and lower-case letter “A”s to communicate the characteristics of purple and white alleles.
A note of caution. When you’re writing down gene symbols for homework or on an exam, be sure to make the characters distinct. A typewritten “y” is easy to distinguish from the upper case “Y” but not as easy when writing it down. Instructors who ask you to show your work need to be able to follow your logic train. More important than that is YOU have to be able to follow your own reasoning. Students often switch symbols and come up with an answer that is inconsistent with the data given because of this. Consider underlining your capitals or putting a line through one of them to make it distinct (e.g. Y for the dominant allele; y for the recessive).
Name the gene after the mutant phenotype
Some instructors would accept “P” for “purple” for the previous cross. However, the better answer is to follow an established system. During “exam fog”, it’s easy to get lost if you are inconsistent with how you develop your symbols. During your study period and when you’re practicing genetics problems, be thoughtful about the gene names you choose.
Let’s always choose a letter based on the mutant phenotype for our gene symbol. If we are presented with a ladybug mutant that is small, we might choose “d” for “dwarf”. Geneticists sometimes set up a research program based on unusual phenotypes of the organism they are studying. The fact that a mutant phenotype that is heritable exists tells us that there is a genetic control for the trait and that it might be isolated in the lab. When you look at your classmates, you don’t necessarily note that none of them has an arm growing from the tops of their heads. If one student had this trait, however, you couldn’t help but notice it. If you found out it showed up in that student’s ancestry in a predictable fashion, you might reasonably suggest that there is a genetic basis for that. If it happened to be controlled by a single gene, you might call the gene “extra arm” or “arm head”. If it happened to be a dominant trait, you might use the letter “X” or “A” for the mutant allele. The wild type allele would be “x” or “a”, respectively.
you don’t know which is mutant, use the recessive trait for the gene name
What if you don’t know which allele is mutant? What if you’re presented with two true-‐ breeding frogs: one that is gold and one that is yellow. If you don’t know what the predominant colour in nature is you can’t know which one is mutant. If you crossed them and all the progeny are gold, then you know the dominant allele encodes a protein to make it gold. The recessive, therefore, is “yellow” and you should name the gene “y” after the recessive phenotype. This means the dominant allele would be “Y”. Your offspring would therefore be Yy and the gold parent would be YY. The yellow parent would be yy.
Apply these ideas at the Online Open Genetics exercises.
1.3 One Letter System
Superscripts
Sometimes a letter is used as the name of a gene, and superscripts can modify it to indicate the different alleles. One common single letter code for an allelic series is “I". Red blood cells can have their cell membranes modified by sugar tags that give rise to our blood type. One allele of I gives rise to blood type A and is therefore called IA. An enzyme encoded by IB modifies sugars to create blood type B. A heterozygote IAIB demonstrates both sugar tags because those alleles are expressed – they are codominant. People with blood type O only possess alleles for the I gene that don’t work and are therefore recessive – they don’t modify the extracellular sugar tags. Because it is recessive, individuals are homozygous for i: they are ii. Chapter 13, Section 7 has more detail on this allelic series.
Sometimes a superscript “plus sign” is used to denote the wild type allele. One might use the symbol W+ to indicate a wild-‐type allele that promotes wing growth. Note that the generic “wing” gene name isn’t a best practice – name the gene after the mutant phenotype! A wingless mutant would be W . You should never use a “+” and shift the case of the letter unless you are dealing with a special case such as the codominance in the blood type example above. The capital “I” letter indicates it is dominant to “i". The superscript A and B for the codominant alleles indicate the dominant alleles are different from each other.
Superscripts can be symbols, a single letter, or many letters. They modify the gene name only in the superscripted symbols: the regular-‐sized letters are identical between them (see Table 1). This means that Abc+ and abc would be different genes (i.e. not allelic); Abc+ and Abc are alleles, as are abc+ and abc. Note that a superscript is not mandatory for all alleles of that gene, depending on the convention.
Alleles of bacterial genes are typically indicated with a superscript + or -. For example, a bacterial allele that creates an enzyme that makes methionine would be met+, and a defective allele of that gene is met .
Table 1: Examples of genes using a superscript modifier.
whiteapr or whiteapricot An allele of the white gene which has an “apricot” phenotype
Abc+ Abc Two alleles for the Abc gene (wild-type and mutant, respectively). Note the mutant allele is dominant.
w+; wa; w Three alleles in a series for the w gene. The first is wild type; the second two are different mutant alleles.
bio+; bio- A wild-type allele of a biotin gene and its recessive counterpart. This is likely a bacterial gene due to the convention. | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/14%3A_Appendices/14.03%3A_Basic_Nomenclature.txt |
It’s perfectly acceptable to use a single letter or even two letters. Sometimes, though, multiple traits spelled with the same first letter can get confusing. Using three letters for a gene symbol can make it easier to remember what the letters stand for. In fact, for some model systems those who study them adopt a defined nomenclature system. A plant often used for genetic studies, called Arabidopsis thaliana, has a three‐letter code (https://www.arabidopsis.org/portals/nomenclature/namerule.jsp).
Just as we saw for the one‐letter symbols, the dominant allele has the first letter capitalized and the last two letters are lower-case. Recessive alleles are all lower-case. With three letters, you can make gene names that are easier to keep track of. For example, you might see a fly with an extra set of wings. Instead of calling it “w” for “wings” (which is a poor choice because it doesn’t represent the mutant phenotype), you can instead call it “exw” for “extra wings”. Then, when you see it, you can sound out the abbreviation and remember that it stands for the mutation. If the wild-type allele for this is dominant, then you would write that one “Exw”. The first letter indicates that it is the dominant allele. What you can’t tell just from these examples is that wild-type allele is dominant! Appendix 2 will go into an extension of this system so you can add that information to the gene symbol. Hint: remember in Section 3.2 that the “+” superscript indicates the wild‐type allele.
1.4 Three Letter System
14.05: Linked Genes
Mendel was lucky. He studied a variety of traits in pea plants and his data were consistent with his idea of traits being encoded by pairs of discrete heritable units. He did not call these “genes” and had no idea about their chromosomal origins or chemical makeup. It turns out that the genes he studied were either on different chromosomes and so assorted independently (Chapter 9), or so far apart on the same chromosome that linkage could not be detected (Chapter 10).
Symbols for a gene can be drawn on a page to communicate their position on a chromosome. To do this, we use a forward slash (/) to demonstrate what is on each chromosome. Figure A1.4 shows how we might conceptualize the position of a gene on two chromosomes by collapsing the chromosomes into a single line.
There’s no question about where the gene is located when only one trait is under investigation: it will be at the same position on each homolog. Two genes, however, can be one of three possibilities. Each possibility has implications for gene mapping and predicting ratios from a dihybrid cross. Figure A1.5 shows the positions of genes for an unlinked situation as well as linked genes in coupling and repulsion configurations. If genes are unlinked, put the allele symbols for one gene on either side of one slash followed by a semicolon (indicating that it’s unlinked) and the other gene with the alleles separated by a second slash (A/a; B/b). When genes are linked, only one slash is used: remember, the slash stands for a pair of homologous chromosomes. Genes in coupling would have the dominant genes together on one side of the slash and recessives on the other side (AB/ab). Repulsion would represent the other arrangement (Ab/aB).
Practice your skills with identifying linked and unlinked genes online in module 1.5. Some examples of different forms of gene symbols are shown in Table 2. Keep in mind that sometimes you have flexibility in which system of nomenclature you use, but sometimes it is dictated to you, for example in publications or other formal submissions. You are discouraged from inventing your own system or mixing up different systems because it will confuse your readers (or graders!).
Table 2: Examples of symbols used to represent genes and alleles.
Examples Interpretation
A and a Uppercase letters represent dominant alleles and lowercase letters indicate recessive alleles. Mendel invented this system but it is not commonly used in publications because not all alleles show complete dominance and many genes have more than two alleles. It’s quick and easy for you to use when working out genetics problems when you are sure each gene involves only two alleles.
a+ and a Superscripts are used to indicate alleles. For wild type alleles the symbol is a superscript +. The mutant allele of gene a would be recessive.
met+ and met- This is typical of a prokaryote gene symbol. It could be referring to wild-type (functional) and mutant (nonfunctional) alleles of a gene that makes a protein in the methionine synthesis pathway.
AA or A/A Sometimes a forward slash is used to indicate that the two symbols are alleles of the same gene, but on homologous chromosomes. Both representations in this row are identical: it represents a homozygous dominant.
Aa/Aa or Aa/aa Note that this example shows two alleles of the gene Aa. We know that the gene symbol is two letters because the slash separates the allele found on each of the homologous chromosomes. We cannot tell if the mutant phenotype is recessive because there’s no indication which is wild type.
Grn+ shr/Grn shr The three-letter system is used here. “Grn” might mean that the phenotype is “green”, but we can’t be sure. What we do know is that the mutant allele codes for a protein leading to a dominant phenotype. The wild-type allele must be recessive to the mutant allele. Maybe “shr” means “shrunken” or “short”, but we know that the mutant phenotype can only be seen in the homozygous recessive configuration. The phenotype for this organism is mutant for both Grn and shr traits. Final note: the genes are on the same chromosome based on the position of the slash.
bob+/bob; mia/mia This also uses the three-letter system. The organism is heterozygous for bob but shows the wild-type trait in its phenotype. It is homozygous recessive for mia and therefore shows that mutant phenotype. The genes are unlinked.
A more advanced system of nomenclature is outlined in Appendix 2. New rules are introduced to help you identify sex linked genes and predict phenotypes from gene symbols alone.
14.06: Summary
• Symbols are used to denote the alleles, or genotype, of a locus.
• Phenotype depends on the alleles that are present, their dominance relationships, and sometimes also interactions with the environment and other factors.
• In a diploid organism, the alleles can be homozygous, heterozygous or hemizygous.
• Allelic interactions at a locus can be described as dominant vs. recessive, incomplete dominance, or co-dominance.
• Two different genes can be on the same chromosome (linked) or on different ones (unlinked)
• Steps:
• Can you identify the mutant? Name the gene after its phenotype
• If you cannot tell which allele is mutant, name the gene after the recessive allele
• Capitalize the first letter for the dominant allele; use lower-case for the recessive allele
• If genes are linked, write the gene symbols together on each side of the slash. If genes are unlinked, they appear on different sides of a semicolon | textbooks/bio/Genetics/Online_Open_Genetics_(Nickle_and_Barrette-Ng)/14%3A_Appendices/14.04%3A_Three-Letter_Symbols.txt |
Gene mapping by conjugal transfer
Conjugal transfer can also be used for genetic mapping. By using many different hfr strains, each with the F factor integrated at a different part of the E. colichromosome, the positions of many genes were mapped. These studies showed that the genetic map of the E. coli chromosome is circular. During conjugal transfer, genes closer to the site of F integration are transferred first. By disrupting the mating at different times, one can determine which genes are closer to the integration site. Thus on the E. colichromosome, genes are mapped in terms of minutes (i.e., the time it takes to transfer to recipient).
For example, for an hfr strain with the F factor integrated at 0 min on the E. colimap, conjugal transfer to a female recipient would transfer
• leuACBD at 1.7 min
• pyrH at 4.6 min
• proAB at 5.9 min
• bioABFCD at 17.5 min.
Bacteriophage
Bacteriophageare viruses that infect bacteria. Because of their very large number of progeny and ability to recombine in mixed infections (more than one strain of bacteria in an infection), they have been used extensively in high-resolution definition of genes. Much of what we know about genetic fine structure, prior to the advent of techniques for isolating and sequencing genes, derive form studies in bacteriophage.
Bacteriophage have been a powerful model genetic system, because they have small genomes, have a short life cycle, and produce many progeny from an infected cell. They provide a very efficient means for transfer of DNA into or between cells. The large number of progeny makes it possible to measure very rare recombination events.
Lytic bacteriophage form plaqueson lawns of bacteria; these are regions of clearing where infected bacteria have lysed. Early work focused on mutants with different plaque morphology, e.g. T2 r, which shows rapid lysis and generates larger plaques, or on mutants with different host range, e.g. T2 h, which will kill both host strains B and B/2.
A cis-trans complementation test defines a cistron, which is a gene
Seymour Benzer used the rIIlocus of phage T4 to define genes by virtue of their behavior in a complementation test, and also to provide fundamental insight into the structure of genes (in particular, the arrangement of mutable sites - see the next section). The difference in plaque morphology between rand r+phage is easy to see (large versus small, respectively), and Benzer isolated many r mutants of phage T4. The wild type, but not any rIImutants, will grow on E. colistrain K12(l), whereas both wild type and mutant phage grow equally well on E. colistrain B. Thus the wild phenotype is readily detected by its ability to grow in strain K12 (l).
If E. colistrain K12 (l) is co-infected with 2 phage carrying mutations at different positions in rIIA, you get no multiplication of the phage (except the extremely rare wild type recombinants, which occur at about 1 in 106 progeny). In the diagram below, each line represents the chromosome from one of the parental phage.
rIIA rIIB
phage 1 _|__x______|________|_
phage 2 _|_______x_|________|_
Likewise, if the two phage in the co-infection carry mutations at different positions in rIIB, you get no multiplication of the phage (except the extremely rare wild type recombinants, about 1 in 106).
rIIA rIIB
phage 3 _|_________|_x______|_
phage 4 _|_________|______x_|_
However, if one of the co-infecting phage carries a mutation in rIIAand the other a mutation in rIIB, then you see multiplication of the phage, forming a very large number of plaques on E. colistrain K12 (l).
rIIA rIIB
phage 1 _|__x______|________|_ Provides wt rIIB protein
phage 4 _|_________|______x_|_ Provides wt rIIA protein
Together these two phage provide all the phage functions - they complementeach other. This is a positive complementation test. The first two examples show no complementation, and we place them in the same complementation group. Mutants that do not complement are placed in the same complementation group; they are different mutant alleles of the same gene. Benzer showed that there were two complementation groups (and therefore two genes) at the r II locus, which he called A and B.
In the mixed infection with phage 1 and phage 4, you also obtain the rare wild type recombinants, but there are more recombinants than are seen in the co-infections with different mutant alleles. Why?
Consider the results of infection of a bacterial culture with two mutant alleles of gene rIIA.
T4rIIA6 _|_______________________x______|_
and T4rIIA27 _|_______x______________________|_
(x marks the position of the mutation in each allele).
Progeny phage from this infection include those with a parental genotype (in the great majority), and at a much lower frequency, two types of recombinants:
wild type T4 r+ _|______________________________|_
double mutant T4rIIA6 rIIA27 _|_______x_______________x______|_
The wild type is easily scored because it, and not any rIImutants, will grow on E. coli strain K12(l), whereas both wild type and mutant phage grow equally well on E. coli strain B. Thus you can selectfor the wild type (and you will see only the desired recombinant). Finding the double mutants is more laborious, because they are obtained only by screening through the progeny, testing for phage that when backcrossed with the parental phage result in no wild type recombinant progeny.
Equal numbers of wild type and double mutant recombinants were obtained, showing that recombination can occur within a gene, and that this occurs by reciprocal crossing over. If recombination were only between genes, then no wild type phage would result. A large spectrum of recombination values was obtained in crosses for different alleles, just like you obtain for crosses between mutants in separate genes.
Several major conclusions could be made as a result of these experiments on recombination within the rIIgenes.
1. A large number of mutable sites occur within a gene, exceeding some 500 for the rIIA and rIIBgenes. We now realize that these correspond to the individual base pairs within the gene.
2. The genetic maps are clearly linear, indicating that the gene is linear. Now we know a gene is a linear polymer of nucleotides.
3. Most mutations are changes at one mutable site (point mutations). Many genes can be restored to wild type by undergoing a reverse mutation at the same site (reversion).
4. Other mutations cause the deletionof one or more mutable sites, reflecting a physical loss of part of the rII gene. Deletions of one or more mutable site (base pair) are extremely unlikely to revert back to the original wild type.
One gene encodes one polypeptide
One of the fundamental insights into how genes function is that one gene encodes one enzyme (or more precisely, one polypeptide). Beadle and Tatum reached this conclusion based on their complementation analysis of the genes required for arginine biosynthesis in fungi. They showed that a mutation in each gene led to a loss of activity of one enzyme in the multistep pathway of arginine biosynthesis. As discussed above in the section on genetic dissection, a large number of Arg auxotrophs (requiring Arg for growth) were isolated, and then organized into a set of complementation groups, where each complementation groups represents a gene.
The classic work of Beadle and Tatum demonstrated a direct relationship between the genes defined by the auxotrophic mutants and the enzymes required for Arg biosynthesis. They showed that a mutation in one gene resulted in the loss of one particular enzymatic activity, e.g. in the generalized scheme below, a mutation in gene 2 led to a loss of activity of enzyme 2. This led to an accumulation of the substrate for that reaction (intermediate N in the diagram below). If there were 4 complementation groups for the Arg auxotrophs, i.e. 4 genes, then 4 enzymes were found in the pathway for Arg biosynthesis. Each enzyme was affected by mutations in one of the complementation groups.
Intermediates:
M ® N ® O ® P ® Arg
enzyme 1 enzyme 2 enzyme 3 enzyme 4
gene 1 gene 2 gene 3 gene 4
Figure 1.15. A general scheme showing the relationships among metabolic intermediates (M, N, O, P), and end product (Arg), enzymes and the genes that encode them.
In general, each step in a metabolic pathway is catalyzed by an enzyme (identified biochemically) that is the product of a particular gene (identified by mutants unable to synthesize the end product, or unable to break down the starting compound, of a pathway). The number of genes that can generate auxotrophic mutants is (usually) the same as the number of enzymatic steps in the pathway. Auxotrophic mutants in a given gene are missing the corresponding enzyme. Thus Beadle and Tatum concluded that one gene encodes one enzyme. Sometimes more than one gene is required to encode an enzyme because the enzyme has multiple, different polypeptide subunits. Thus each polypeptide is encoded by a gene.
The metabolic intermediates that accumulate in each mutant can be used to place the enzymes in their order of actionin a pathway. In the diagram in Figure 1.15, mutants in gene 3 accumulated substance O. Feeding substance O to mutants in gene 1 or in gene 2 allows growth in the absence of Arg. We conclude that the defects in enzyme 1 or enzyme 2, respectively, are upstream of enzyme 3. In contrast, feeding substance O to mutants in gene 4 will not allow growth in the absence of Arg. Even though this mutant can convert substance O to substance P, it does not have an active enzyme 4 to convert P to Arg. The inability of mutants in gene 4 to grow on substance O shows that enzyme 4 is downstream of enzyme 3.
Imagine that you are studying serine biosynthesis in a fungus. You isolate serine auxotrophs, do all the pairwise crosses of the mutants and discover that the auxotrophs can be grouped into three complementation groups, called A, B and C. You also discover that a different metabolic intermediate accumulates in members of each complementation group - substance A in auxotrophs in the A complementation group, substance B in the B complementation group and substance C in the C complementation group. Each of the intermediates is fed to auxotrophs from each of the three complementation groups as tabulated below. A + means that the auxotroph was able to grow in media in the absence of serine when fed the indicated substance; a - denotes no growth in the absence of serine.
Fed:
mutant in complementation group A
mutant in complementation group B
mutant in complementation group C
substance A
-
+
+
substance B
-
-
-
substance C
-
+
-
Eukaryotic mRNAs have covalent attachment of nucleotides at the 5' and 3' ends, and in some cases nucleotides are added internally (a process called RNA editing). Recent work shows that additional nucleotides are added post‑transcriptionally to some bacterial mRNAs as well.
Regulatory signals can be considered parts of genes
In order to express a gene at the correct time, the DNA also carries signals to start transcription (e.g. promoters), signals for regulating the efficiency of starting transcription (e.g. operators, enhancers or silencers), and signals to stop transcription (e.g. terminators). Minimally, a gene includes the transcription unit, which is the segment of DNA that is copied into RNA in the primary transcript. The signals directing RNA polymerase to start at the correct site, and other DNA segments that influence the efficiency of this process are regulatory elements for the gene. One can also consider them to be part of the gene, along with the transcription unit.
A contemporary problem - finding the function of genes
Genes were originally detected by the heritable phenotype generated by their mutant alleles, such as the white eyes in the normally red-eyed Drosophilaor the sickle cell form of hemoglobin (HbS) in humans. Now that we have the ability to isolate virtually any, and perhaps all, segments of DNA from the genome of an organism, the issue arises as to which of those segments are genes, and what is the function of those genes. (The genomeis all the DNA in the chromosomes of an organism.) Earlier geneticists knew what the function of the genes were that they were studying (at least in terms of some macroscopic phenotype), even when they had no idea what the nature of the genetic material was. Now molecular biologists are confronted with the opposite problem - we can find and study lots of DNA, but which regions are functions? Many computational approaches are being developed to guide in this analysis, but eventually we come back to that classical definition, i.e. that appropriate mutations in any functional gene should generate a detectable phenotype. The approach of biochemically making mutations in DNA in the laboratory and then testing for the effects in living cells or whole organisms is called "reverse genetics."
Additional Readings
• Griffiths, A. J. F., Miller, J. H., Suzuki, D. T., Lewontin, R. C. and Gelbart, W. M. (1993) An Introduction to Genetic Analysis, Fifth Edition (W. H. Freeman and Company, New York).
• Cairns, J., Stent, G. S. and Watson, J. D., editors (1992) Phage and the Origins of Molecular Biology, Expanded Edition (Cold Spring Harbor Laboratory Press, Plainview, NY).
• Brock, T. D. (1990) The Emergence of Bacterial Genetics (Cold Spring Harbor Laboratory Press, Plainview, NY).
• Benzer, S. (1955) Fine structure of a genetic region in bacteriophage. Proceedings of the National Academy of Sciences, USA 47: 344-354.
• Yanofsky, C. (1963) Amino acid replacements associated with mutation and recombination in the A gene and their relationship to in vitro coding data. Cold Spring Harbor Symposia on Quantitative Biology 18: 133-134.
• Crick, F. (1970) Central dogma of molecular biology. Nature 227:561-563
Questions
The actual results from Luria and Delbrück are summarized in the following table. They examined 87 cultures, each with 0.2 ml of bacteria, for phage resistant colonies.
Number of resistant bacteria
Number of cultures
0
29
1
17
2
4
3
3
4
3
5
2
6-10
5
11-20
6
21-50
7
51-100
5
101-200
2
201-500
4
501-1000
0
Interested students may wish to read about the re-examination of the origin of mutations by Cairns, Overbaugh and Miller (1988, The origin of mutants. Nature 335:142-145). Using a non-lethal selective agent (lactose), they obtained results indicating both pre-adaptive (spontaneous) mutations as well as some apparently induced by the selective agent.
1: Fundamental Properties of Genes
Introduction
Species share many traits in common from generation to generation. The bluebird nestlings in the box in my yard will look much like their parents when they are full-grown. The tomato plants that we set out will produce fruits that look, and hopefully taste, like those of their parents. Observable features of organisms, like color, size, and shape, comprise their phenotype. Adult male bluebirds share the phenotype of blue wings and a red breast.
A phenotype can be determined by inherited factors, by the environment, and often by both. For example, you are similar to your parents in many aspects of your appearance, your intelligence, and your susceptibility to some diseases, but you are not identical to them in all aspects of these traits. These three traits are clearly the product of both inherited and environmental factors. Considering appearance, I have crooked lower teeth and thinning gray hair, just like my father, but unlike me, neither of my parents has a scar on their knee from a childhood cut. The hair phenotype is inherited, whereas scars are from environmental influences. Quantitative studies show that intellectual capacity is about equally influenced by genetic and environmental factors. Susceptibility to diabetes is partially inherited, but a viral infection may trigger the autoimmune response at its core.
The genetic determinants of the inherited component of a phenotype are called genes. The set of genes that make up an organism is its genotype. In practice, we will consider only a small subset of the genes in an organism, which comprise a partial genotype. Likewise, an organism’s phenotype is all the traits it possesses, but we will only consider partial phenotypes, such as the blue wings of a bluebird or the color of the eyes of a fly.
This chapter will explore some of the basic characteristics of genes, and the experimental evidence for them. Some of the major points include the following.
• Genes are the units of heredity
• They are arranged in a linear fashion along chromosomes.
• Recombination can occur both between and within genes.
• Mutations in different genes required for a phenotype will complement each other in a diploid. This is the basis for genetic dissection of a pathway.
• A gene is composed of a series of mutable sites that are also sites for recombination (now recognized as nucleotides).
• One gene encodes one polypeptide.
• The gene and the polypeptide are colinear.
• Single amino acids are specified by a set of three adjacent mutable sites; this set is called a codon.
In considering experimental evidence for these points, some general genetic techniques as well as genetic techniques for bacteria and phage will be discussed. The first experiments that eventually hypothesized the existence of genes were conducted by Gregor Mendel (see Mendel's Laws).
Genes are mutable
We know that genes are mutable because they appear in different forms, called alleles. An allele that encodes a normal, functional product (found in nature or a standard laboratory stock) is called the wild type allele. Other alleles are altered in a way such that the encoded product differs in function from the wild type. This type of allele is mutated or mutant (adjective). The alteration in the gene is a mutation, and an organism showing the altered phenotype is a mutant (noun). Many mutated alleles encode a product that is nonfunctional or less functional than is the wild type, or normal, product; it is easier to break something than to improve it. A loss-of-function allele usually shows a recessive phenotype, which means that when it is present in the same cell as an allele that produces a different phenotype, the phenotype of the other allele is obtained. If no functional product is made, this loss-of-function allele is a null mutation; this can result from no expression or expression of a completely nonfunctional product. Other loss-of-function mutants make less than the normal amount of product, these are called hypomorphs. Another class of mutated allele encodes a product that provides an altered or new function. These gain-of-function mutations usually show a dominant phenotype; e.g. when the gain-of-function allele is in the same cell as a loss-of-function allele, the phenotype of the gain-of-function allele is observed. Another class of gain-of-function mutants makes more than the normal amount of product; these are called hypermorphs.
Within a population, the number of alleles at a given locus can vary considerably. Mutant alleles that cause a loss or detrimental change in the function of a gene are selected against, and they are rare in a wild population. In the laboratory, one can utilize growth conditions that select for certain mutants or that maintain mutants, so mutant organisms that would be rare or non-existent in the wild are encountered quite frequently in the laboratory. In many cases, however, alternate forms of genes, i.e. different alleles, have no particular effect on gene function. These variants can be found quite frequently in a population. One common examples of such genetically determined, apparently neutral variation is the ability of some persons to "roll" their tongue. In general, these common alleles are roughly equivalent in function to the wild type allele. Thus they are not providing a strong selective advantage or disadvantage. All the common alleles can be considered the wild type allele. Variant alleles that occur in greater than 5% of population are called polymorphisms. The term variant includes all alternative forms of a gene, whether they have an effect on function or not. The term mutant allele sometimes implies an altered function for the gene.
As will become clearer when we study the fine structure of genes, it is possible to change the structure of the gene (the nucleotide sequence in DNA) without changing the structure of the encoded polypeptide (the amino acid sequence). These silent substitutions also generate different alleles, but they can only be detected by examining the structure of the gene; the phenotypes of alleles that differ by silent substitutions are usually identical.
Another possibility is that a mutant allele not only causes a loss of function of the encoded protein, but this altered protein interferes with the activity of other proteins. One way this can happen is by the polypeptide product forming a complex with other polypeptides (e.g. in a heteromultimeric enzyme complex). Sometimes the mutant polypeptide will prevent formation of an active complex with the partner, even in the presence of wild-type polypeptide, thereby leading to a dominant negative phenotype. These are of considerable utility now in designing mutant genes and proteins to try to disrupt some cellular function. They are most commonly made in a vector that will drive a high level of expression of the mutant gene, and usually over-expression is needed to generate the dominant negative phenotype. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/1%3A_Fundamental_Properties_of_Genes/1.1%3A_Introduction_to_Genes.txt |
Mendel's First Law: Alleles segregate equally
The original experiments by Gregor Mendel involved phenotypic traits (physical, observable characteristics) controlled by single genes. The first one we'll consider is seed color, which can be yellow or green. The dominant allele, denoted Y, generates yellow peas in either the homozygous (YY) or heterozygous (Yy) state, whereas the recessive allele, denoted y, generates green peas only in the homozygous state (yy). (In plants and flies, the dominant allele is denoted by a capitalized abbreviation and the recessive allele is denoted by a lower case abbreviation.) In a cross between two parents, one homozygous for the dominant allele (YY) and the other homozygous for the recessive allele (yy), Mendel showed that the F1 progeny were all yellow, i.e. they had had the same phenotype as the parent with the dominant allele. The recessive allele was not contributing to the phenotype.
Had it been lost during the cross? No, when the F1 is crossed with itself, both parental phenotypes were seen in the F2 progeny. The effect of the recessive allele reappeared in the second cross, showing that it was still present in the F1 hybrids, but was having no effect. In the F2 progeny, the dominant phenotype (yellow) was observed in 75% of the progeny and the recessive (green) appeared in only 25% of the progeny.
Note that discrete phenotypes were obtained (yellow or green), not a continuum of phenotypes. The genes are behaving as units, not as some continuous function.
The results can be explained by hypothesizing that each parent has two copies of the gene (i.e., two alleles) that segregate equally, one per gamete. Since they are homozygous, each parent can form only type of gamete (Y or y,respectively). When the gametes join in the zygotes of the F1 generation, each individual receives one dominant allele and one recessive allele (Yy), and thus all of the F1 generation shows the dominant phenotype (e.g. yellow peas). This is the uniform phenotype observed for the F1 generation.
The two alleles did not alter one another when present together in the F1 generation, because when F1 is crossed with F1, the two parental phenotypes are obtained in the F2 generation.
The ratio of 3:1 dominant: recessive observed in the F2 is expected for the equal segregation of the alleles from the F1(Y and y) and their random rejoining in the zygotes of the F2, producing the genotypes 1 YY, 2 Yy, and 1 yy. Again the genes are behaving as discrete units. These precise mathematical ratios (3:1 for phenotypes in this cross, or 1:2:1 for the genotype) provide the evidence that genes, units of heredity, are determining the phenotypes observed.
Not all loci show the property of complete dominance illustrated by the Y locus in peas. Sometimes partial dominance is observed, in which an intermediate phenotype seen in a heterozygote. An example is the pink color of snapdragons obtained when white and red are crossed. However, the parental phenotypes reappear in the F2 generation, showing that the alleles were not altered in the heterozygote. In this case, gene dosage is important in determining the phenotype; two wild-type alleles produce a red flower, but only one wild-type allele produces a pink flower. Sometimes co‑dominance is observed, in which both alleles contribute equally to the phenotype. An example is the ABO blood group locus. Heterozygotes have both the A and B form of the glycoprotein that is encoded by the different alleles of the gene.
Mendel's Second Law: Different Genes Assort Independently
Now that we have some understanding of the behavior of the different alleles of a single gene, let's consider how two different genes behave during a cross. Do they tend to stay together, or do they assort independently?
Mendel examined two different traits, seed color (as described in the previous section) and seed shape. Two alleles at the locus controlling seed shape were studied, the dominant round (R) and recessive wrinkled (r) alleles. Mendel crossed one parent that was homozygous for the dominant alleles of these two different genes (round yellow RRYY) with another parent that was homozygous for the recessive alleles of those two genes (wrinkled green rryy) (see Figure 1.2).
Re-stating the basic question, do the alleles at each locus always stay together (i.e. round with yellow, wrinkled with green) or do they appear in new combinations in the progeny? As expected from the 1st law, the F1 generation shows a uniform round yellow phenotype, since one dominant and one recessive allele was inherited from the parents. When the F2 progeny are obtained by crossing the F1 generation, the parental phenotypes reappear (as expected from the first law), but two nonparental phenotypes also appear that differ from the parents: wrinkled yellow and round green!
The results can be explained by the alleles of each different gene assorting into gametes independently. For example, in the gametes from the F1 generation, R can assort with Yor y, and r can assort with Y or y, so that four types of gametes form: RY, Ry, rY, and ry. These can rejoin randomly with other gametes from the F1 generation, producing the results in the grid shown in Figure 1.2. The alternative, that R always assorted with Y, etc. was not observed.
Again, the genes are behaving as units, and the gene for one trait (e.g. color) does not affect a gene for another trait (e.g. shape). Further breeding shows that many nonparental genotypes are present, some of which give a parental phenotype (e.g. RrYy). These results are obtained for genes that are not linked on chromosomes. Linkage can lead to deviations from these expected ratios in a mating, and this can be used to map the locations of genes on chromosomes, as discussed in the next section. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/1%3A_Fundamental_Properties_of_Genes/1.2%3A_Genes_are_the_Units_of_Heredity%3A_Mendel%27s_Laws.txt |
Genes are on Chromosomes
In 1902, Sutton and Boveri independently realized that the behavior of genes in Mendelian crosses mimics the movement of chromosomes during meiosis and fertilization. They surmised that the two alleles of each gene correlated with the homologous pair of chromosomes. The equal segregation of alleles could be explained by the separation of homologous chromosomes at anaphase I of meiosis. As diagrammed in Figure 1.3, the chromosome with the R allele would go to a different cell than its homolog with the r allele at the end of meiosis I, and likewise for the Yand y alleles. The rejoining of alleles corresponded to the joining of chromosomes, one from each parent, at fertilization. The independent assortment of different genes mimics the independent separation of homologs of different chromosomes in meiosis. For instance, the paternal copy of chromosome 1 may assort with the maternal copy of chromosome 21 in formation of a gamete. Figure 1.3 shows the dark blue chromosome with the R allele assorting with the light red chromosome with the yallele, but it is equally likely that it will assort with the dark red chromosome with the Yallele. As shown in Figure 1.4, the completion of meiosis results in 4 germ cells for each cell that entered meiosis. All the combinations of alleles of different genes diagrammed in Figure 1.2 can be formed in this process.
This correlation of the behavior of alleles in matings and the movement of chromosomes during meiosis and fertilization produced the chromosomal theory of inheritance. One could think of the alleles discerned in genetic crosses as being located at the same locus on the different homologs of a chromosome.
Linked genes lie along chromosomes in a linear array
The proponents of the chromosome theory of heredity realized that that the number of genes would probably greatly exceed the number of chromosomes. However, many early genetic studies showed independent assortment between genes with no evidence of linkage. This led to a proposal that a chromosome broke down during meiosis into smaller parts consisting only of individual genes, but such disassembly of chromosomes during meiosis was never observed. Evidence for linkage did eventually come from a demonstration of the absence of independent assortment between different genes. In complementary work, McClintock and Creighton demonstrated an association between different genes and a particular chromosome in 1931.
The behavior of two genes carried on the same chromosome may deviate from the predictions of Mendel's 2nd law. The proportion of parental genotypes in the F2 may be greater than expected because of a reduction in nonparental genotypes. This propensity of some characters to remain associated instead of assorting independently is called linkage. When deduced from studies of a population, it is called linkage disequilibrium. Figure 1.5. illustrates a cross that shows linkage.
An F1 heterozygote (AaBb) is made by crossing a homozygous dominant parent (AABB) with a homozygous recessive parent (aabb). A backcross is then made between the F1 heterozygote (AaBb) and a recessive homozygote (aabb), so that the alleles of the recessive parent make no contribution to the phenotype of the progeny. (This is a fairly common cross in genetics, since the genotype of an individual can be ascertained by crossing with such an individual, homozygous recessive at both loci.)
1. As shown in part A of Figure 1.5, if there is no linkage, one expects 50% parental phenotypes (from genotypes AaBb and aabb) and 50% nonparental phenotypes (from genotypes Aabband aaBb). This fits with the expectations of Mendel's law of independent assortment of different genes for this backcross. (Sometimes the nonparental phenotypes are called "recombinant" but that confuses this reassortment with events that involve crossovers in the DNA.)
2. If the two genes are linked and there is no recombination between them, then all progeny will have a parental phenotype. In particular, if genes A and B are linked, then the backcross AB/ab x ab/ab yields AB/ab progeny 50% of the time and ab/ab progeny 50% of the time, in the absence of recombination. [In this notation, the alleles to the left of the slash (/) are linked on one chromosome and the alleles to the right of the slash are linked on the homologous chromosome.] Thus only the parental phenotypes are found in the progeny of this cross (i.e. the progeny will show either the dominant characters at each locus or the recessive characters at each locus). Another way of looking at this is that, in the absence of recombination between the homologous chromosomes, all the progeny of this cross will be one of the first two types shown in panel B of Figure 1.5.
3. Note that the dominant alleles can be in the opposite phase, with the dominant Aallele linked to the recessive b allele. For instance, the F1 heterozygote could be formed by a cross between the parents Ab/Aband aB/aBto generate Ab/aB. In this case, the backcross Ab/aBx ab/ab will still generate only progeny with parental phenotypes but a new, nonparental genotype (i.e. Ab/aband aB/ab; these look like the parents Ab/Ab and aB/aB). The phase with both dominant alleles on the same chromosome is called the "coupling conformation”, whereas the opposite phase is called the "repulsing conformation."
4. But in most cases, recombination can occur between linked genes. In part B of Figure 1.5, there is an increase in parental types (from the 50% expected for unlinked genes to the observed 70%) and a decreasein nonparental types (30%), showing that allele Atends to stay with allele B, in contrast to the prediction of the 2nd law. Thus these genes are not assorting independently, and one concludes there is linkage between genes Aand B.
The frequency of parental types is not as high as expected for linkage without recombination (which would have been 100%, as discussed above). Indeed, the nonparental types in this experiment result from a physical crossover (breaking and rejoining) between the two homologous chromosomes during meiosis in the AB/ab parent. This is a recombination event in the DNA.
(5) We conclude that genes A and Bare linked, and have a recombination frequency of 30%.
map distance = x 100
1 map unit = 1 centiMorgan = 1% recombination
1 centiMorgan = 1 cM = about 1 Mb for human chromosomes
Exercise \(1\)
In their genetic studies of the fruitfly Drosophila melanogaster, Thomas Hunt Morgan and his co-workers found many examples of genes that associated together in groups. One example is the gene for purple eye color (the mutant allele is abbreviated pr) that is recessive to the allele for normal red eyes (pr+) and the gene for vestigial, or shortened, wings (the mutant allele is abbreviated vg) that is recessive to the normal allele for long wings (vg+). When a homozygous purple vestigialfly is crossed to a homozygous red-eyed long-winged fly, the heterozygous F1 generation shows a normal phenotype. When male heterozygotes are backcrossed to females that are homozygous purple vestigial (i.e. homozygous recessive at both loci), only two phenotypes appear in the progeny: the homozygous recessive purple vestigialflies and the normal flies.
1. What are the predictions of the backcross if the two genes are not linked?
2. What do the results of the backcross tell you?
3. If the heterozygotes F1 in the backcross are female, then purple long-winged and red-eyed vestigialflies appear in the progeny. The combined frequency of these recombinant types is 15.2 %. What does this tell you about the arrangement of the genes?
Answer
TBA
Question 1.5 provides some practice in calculating recombination frequencies.
Individual map distances are (roughly) additive
A‑‑10‑‑B‑5‑C
‑‑‑-‑‑‑-‑15‑‑‑‑
The recombination distances are not strictly additive if multiple crossovers can occur (see questions 1.6 and 1.7.) Recombination between linked genes occurs by the process of crossing over between chromosomes, at chiasma during meiosis. The mechanism of recombination is considered in Chapter 8. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/1%3A_Fundamental_Properties_of_Genes/1.3%3A_Properties_of_Genes.txt |
What is Complementation?
A general definition of complementation is the ability of two mutants in combination to restore a normal phenotype. Dominance observed in heterozygotes reflects the ability of wild-type alleles to complement loss-of-function alleles. You know that a dominant allele will determine the phenotype of a heterozygote composed of a dominant and a recessive allele. Often, recessive alleles are loss-of-function mutations, whereas the dominant allele is the wild type, encoding a functional enzyme. Using the example that led to Mendel's First Law, a cross between YY (yellow) peas and yy (green) peas yielded yellow peas in the F1 heterozygote (Yy). In this case the chromosome carrying the Y allele encodes the enzymatic function missing in the product of the recessive y allele, and the pathway for pigment biosynthesis continues on to make a yellow product. Thus you could say that the dominant Y allele complements the recessive y allele - it provides the missing function.
We can continue the analogy to the classic cross for Mendel's Second Law. Let's look at the same genes, but a different arrangement of alleles. Consider a cross between round green (RRyy) and wrinkled yellow (rrYY) peas; in this case each parent is providing a dominant allele of one gene and a recessive allele of the other. The F1 heterozygote is round yellow (RrYy), i.e., the phenotypes of the dominant alleles are seen. But you could also describe this situation as the chromosomes from rrYYpeas complementing the deficiency in the RRyy chromosomes, and vice versa. In particular, the Yallele from the rrYYparent provides the function missing in the y allele from the RRyy parent, and the R allele from the RRyy parent provides the function missing in the rallele from the rrYY parent. If the phenotype you are looking for is a round yellow pea, you could conclude that mutants in the R-gene complement mutants in the Y-gene. Since in a heterozygote, the functional allele will provide the activity missing in the mutant allele (if the mutation is a loss-of-function), one could say that dominant alleles complement recessive alleles. Thus dominant alleles determine the phenotype in a heterozygote with both dominant and recessive alleles.
Complementation distinguishes between mutations in the same gene or in different genes
The ability of complementation analysis to determine whether mutations are in the same or different genes is the basis for genetic dissection. In this process, one finds the genes whose products are required in a pathway. In the examples from peas, the metabolic pathway to yellow pigments is distinctly different from the pathway to round peas, which is the starch biosynthesis pathway. Complementation analysis is useful in dissecting the steps in a pathway, starting with many mutants that generate the same phenotype. This is a more conventional example of complementation.
Many fungi can propagate as haploids but can also mate to form diploids prior to sporulation. Thus one can screen for mutants in haploids and obtain recessive mutants, and then test their behavior in combination with other mutants in the diploid state. Let's say that a haploid strain of a fungus was mutagenized and screened for arginine auxotrophs, i.e. mutants that require arginine to grow. Six of the mutants were mated to form all the possible diploid combinations, and tested for the ability of the diploids to grow in the absence of arginine (prototrophy). The results are tabulated below, with a + designating growth in the absence of arginine, and a - designating no growth.
Table 1.1. Growth of the diploids in the absence of arginine
Mutant number
1
2
3
4
5
6
1
-
+
+
-
+
+
2
-
-
+
+
+
3
-
+
+
+
4
-
+
+
5
-
+
6
-
As you would expect, when mutant 1 is mated with itself, the resulting diploid is still an auxotroph; this is the same as being homozygous for the defective allele of a gene. But when mutant 1 is mated with mutant 2 (so their chromosomes are combined), the resulting diploid has prototrophy restored, i.e. it can make its own arginine. This is true for allthe progeny. We conclude that mutant 1 will complement mutant 2. If we say that mutant 1 has a mutation in gene 1 of the pathway for arginine biosynthesis, and mutant 2 has a mutation in gene 2 of this pathway, then the diagram in Figure 1.6 describes the situation in the haploids and the diploid. (Note that if the organism has more than one chromosome, then genes 1 and 2 need not be on the same chromosome.) Since the enzymes encoded by genes 1 and 2 are needed for arginine biosynthesis, neither mutant in the haploid state can make arginine. But when these chromosomes are combined in the diploid state, the chromosome from mutant 1 will provide a normal product of gene 2, and the chromosome from mutant 2 will provide a normal product of gene 1. Since each provides what is missing in the other, they complement. Mutant 1 will also complement mutant 3, and one concludes that these strains are carrying mutations in different genes required for arginine biosynthesis.
In contrast, the diploid resulting from mating mutant 1 with mutant 4 is still an auxotroph; it will not grow in the absence of arginine. Assuming that both these mutants are recessive (i.e. contain loss-of-function alleles), then we conclude that the mutations are in the same gene (gene 1 in the above diagram). We place these mutants in the same complementation group. Likewise, mutants 2 and 3 fail to complement, and they are in the same complementation group. Thus mutant 2 and mutant 3 are carrying different mutant alleles of the same gene (gene 2).
Mutant 5 will complement all the other mutants, so it is in a different gene, and the same is true for mutant 6. Thus this mutation and complementation analysis shows that this fungus has at least 4 genes involved in arginine biosynthesis: gene 1 (defined by mutants alleles in strains 1 and 4), gene 2 (defined by mutants alleles in strains 2 and 3), and two other genes, one mutated in strain 5 and the other mutated in strain 6.
Genetic dissection by complementation is very powerful. An investigator can start with a large number of mutants, all of which have the same phenotype, and then group them into sets of mutant alleles of different genes. Groups of mutations that do not complement each other constitute a complementation group, which is equivalent to a gene. Each mutation in a given complementation group is a mutant allele of the gene. The product of each gene, whether a polypeptide or RNA, is needed for the cellular function that, when altered, generates the phenotype that was the basis for the initial screen. The number of different complementation groups, or genes, gives an approximation of the number of polypeptides or RNA molecules utilized in generating the cellular function.
Question 1.2. Consider the following complementation analysis. Five mutations in a biosynthetic pathway (producing auxotrophs in a haploid state) were placed pairwise in a cell in trans(diploid analysis). The diploid cells were then assayed for reconstitution of the biosynthetic pathway; complementing mutations were able to grow in the absence of the end product of the pathway (i.e. they now had a functional biosynthetic pathway). A + indicates a complementing pair of mutations; a - means that the two mutations did not complement.
Mutation number
1 2 3 4 5
1 - + - + -
2 - + + +
3 - + -
4 - +
5 -
a) Which mutations are in the same complementation group (representing mutant alleles of the same gene)?
b) What is the minimal number of enzymatic steps in the biosynthetic pathway?
Recombination
Note that all the diploid progeny fungi from the mating of mutant strains 1 and 2 have the ability to grow on arginine, and this complementation does not require any change in the two chromosomes (Figure 1.6.). The only thing that is happening is that the functional alleles of each gene are providing active enzymes. If genes 1 and 2 are on the same chromosome, at a low frequency, recombinations between the two chromosomes in the diploid can lead to crossovers, resulting in one chromosome with wild-type alleles of each gene and another chromosome with the mutant alleles of each gene (Figure 1.7). This can be observed in fungi by inducing sporulation of the diploid. Each spore is haploid, and the vast majority will carry one of the two parental chromosomes, and hence be defective in either gene 1 or gene 2. But wild type recombinants can be observed at a low frequency; these will be prototrophs. The double-mutant recombinants will be auxotrophs, of course, but these can be distinguished from the parental single mutants by the inability of the double mutants to complement either mutant strain 1 or strain 2.
Note that this recombination is a physical alteration in the chromosomes. The frequency of its occurrence is directly proportional to the distance the genes are apart, which is the basis for mapping genes by their recombination distances. Recombination occurs in a small fraction of the progeny, whereas all the progeny of a complementing diploid have the previously lost function restored. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/1%3A_Fundamental_Properties_of_Genes/1.4%3A_Complementation_and_Recombination.txt |
Question 1.5. Calculating recombination frequencies
Corn kernels can be colored or white, determined by the alleles C(colored, which is dominant) or c(white, which is recessive) of the coloredgene. Likewise, alleles of the shrunkengene determine whether the kernels are nonshrunken (Sh, dominant) or shrunken (sh, recessive). The geneticist Hutchison crossed a homozygous colored shrunken strain (CC shsh) to a homozygous white nonshrunken strain (cc ShSh) and obtained the heterozygous colored nonshrunken F1. The F1 was backcrossed to a homozygous recessive white shrunken strain (cc shsh). Four phenotypes were observed in the F2 progeny, in the numbers shown below.
Phenotype Number of plants
colored shrunken 21,379
white nonshrunken 21,096
colored nonshrunken 638
white shrunken 672
a) What are the predicted frequencies of these phenotypes if the coloredand shrunkengenes are not linked?
b) Are these genes linked, and if so, what is the recombination frequency between them?
Question 1.6. Constructing a linkage map:
Consider three genes, A, B and C, that are located on the same chromosome. The arrangement of the three genes can be determined by a series of three crosses, each following two of the genes (referred to as two-factor crosses). In each cross, a parental strain that is homozygous for the dominant alleles of the two genes (e.g. AB/AB) is crossed with a strain that is homozygous for the recessive alleles of the two genes (e.g. ab/ab), to yield an F1 that is heterozygous for both of the genes (e.g. AB/ab). In this notation, the slash (/) separates the alleles of genes on one chromosome from those on the homologous chromosome. The F1 (AB/ab) contains one chromosome from each parent. It is then backcrossed to a strain that is homozygous for the recessive alleles (ab/ab) so that the fates of the parental chromosomes can be easily followed. Let's say the resulting progeny in the F2 (second) generation showed the parental phenotypes (AB and ab) 70% of the time. That is, 70% of the progeny showed only the dominant characters (AB) or only the recessive characters (ab), which reflect the haploid genotypes AB/aband ab/ab, respectively, in the F2 progeny. The remaining 30% of the progeny showed recombinant phenotypes (Aband aB) reflecting the genotypes Ab/aband aB/abin the F2 progeny. Similar crosses using F1's from parental AC/ACand ac/acbackcrossed to a homozygous recessive strain (ac/ac) generated recombinant phenotypes Acand aCin 10% of the progeny. And finally, crosses using F1's from parental BC/BCand bc/bcbackcrossed to a homozygous recessive strain (bc/bc) generated recombinant phenotypes Bcand bCin 25% of the progeny.
a. What accounts for the appearance of the recombinant phenotypes in the F2 progeny?
b. Which genes are closer to each other and which ones are further away?
c. What is a linkage map that is consistent with the data given?
Question 1.7
Why are the distances in the previous problem not exactly additive, e.g. why is the distance between the outside markers (A and B) not 35 map units (or 35% recombination)? There are several possible explanations, and this problem explores the effects of multiple crossovers. The basic idea is that the further apart two genes are, the more likely that recombination can occur multiple times between them. Of course, two (or any even number of) crossover events between two genes will restore the parental arrangement, whereas three (or any odd number of) crossover events will give a recombinant arrangement, thereby effectively decreasing the observed number of recombinants in the progeny of a cross.
For the case examined in the previous problem, with genes in the order A___C_______B, let the term abrefer to the frequency of recombination between genes Aand B, and likewise let acrefer to the frequency of recombination between genes Aand C, and cbrefer to the frequency of recombination between genes Cand B.
a) What is the probability that when recombination occurs in the interval between Aand C, an independent recombination event also occurs in the interval between Cand B?
b) What is the probability that when recombination occurs in the interval between Cand B, an independent recombination event also occurs in the interval between Aand C?
c) The two probabilities, or frequencies, in a and b above will effectively lower the actual recombination between the outside markers Aand Bto that observed in the experiment. What is an equation that expresses this relationship, and does it fit the data in problem 3?
d. What is the better estimate for the distance between genes Aand Bin the previous problem?
Question 1.8 Complementation and recombination in microbes.
The State College Bar Association has commissioned you to study an organism, Alcophila latrobus, which thrives on Rolling Rock beer and is ruining the local shipments. You find three mutants that have lost the ability to grow on Rolling Rock (RR).
a) Recombination between the mutants can restore the ability to grow on RR. From the following recombination frequencies, construct a linkage map for mutations 1, 2, and 3.
Recombination between Frequency
1- and 2- 0.100
1- and 3- 0.099
2- and 3- 0.001
b) The following diploid constructions were tested for their ability to grow on RR. What do these data tell you about mutations 1, 2, and 3?
Grow on RR?
1) 1- 2+ / 1+ 2- yes
2) 1- 3+ / 1+ 3- yes
3) 2- 3+ / 2+ 3- no
Question 1.9 Using recombination frequencies and complementation to deduce maps and pathways in phage.
A set of four mutant phage that were unable to grow in a particular bacterial host (lets call it restrictive) were isolated; however, both mutant and wild type phage will grow in another, permissive host. To get information about the genes required for growth on the restrictive host, this host was co-infected with pairs of mutant phage, and the number of phage obtained after infection was measured. The top number for each co-infection gives the total number of phage released (grown on the permissive host) and the bottom number gives the number of wild-type recombinant phage (grown on the restrictive host). The wild-type parental phage gives 1010 phage after infecting either host. The limit of detection is 102 phage.
Phenotypes of phage, problem 1.9:
Assays after co-infection with mutant phage:
Results of assays, problem 1.9:
Number of phage
mutant 1 mutant 2 mutant 3 mutant 4
mutant 1 total <102
recombinants <102
mutant 2 total 1010 <102
recombinants 5x106 <102
mutant 3 total 1010 1010 <102
recombinants 107 5x106 <102
mutant 4 total 105 1010 1010 <102
recombinants 105 5x106 107 <102
a) Which mutants are in the same complementation group? What is the minimum number of genes in the pathway for growth on the restrictive host?
b) Which mutations have the shortest distance between them?
c) Which mutations have the greatest distance between them?
d) Draw a map of the genes in the pathway required for growth on the restrictive host. Show the positions of the genes, the positions of the mutations and the relative distances between them.
Question 1.10
One of the classic experiments in bacterial genetics is the fluctuation analysisof Luria and Delbrück (1943, Mutations of bacteria from virus sensitivity to virus resistance, Genetics 28: 491-511). These authors wanted to determine whether mutations arose spontaneouslywhile bacteria grew in culture, or if the mutations were inducedby the conditions used to select for them. They knew that bacteria resistant to phage infection could be isolated from infected cultures. When a bacterial culture is infected with a lytic phage, initially it “clears” because virtually all the cells are lysed, but after several hours phage-resistant bacteria will start to grow.
Luria and Delbrück realized that the two hypothesis for the source of the mutations could be distinguished by a quantitative analysis of the number of the phage-resistant bacteria found in many infected cultures. The experimental approach is outlined in the figure below. Many cultures of bacteria are grown, then infected with a dose of phage T1 that is sufficient to kill all the cells, except those that have acquired resistance. These resistant bacteria grow into colonies on plates and can be counted.
a. What are the predictions for the distribution of the number of resistant bacteria in the two models? Assume that on average, about 1 in 107 bacteria are resistant to infection by phage T1.
b. What do results like those in the figure and table tell you about which model is correct?
Figure for question 1.10.
The actual results from Luria and Delbrück are summarized in the following table. They examined 87 cultures, each with 0.2 ml of bacteria, for phage resistant colonies.
Number of resistant bacteria
Number of cultures
0
29
1
17
2
4
3
3
4
3
5
2
6-10
5
11-20
6
21-50
7
51-100
5
101-200
2
201-500
4
501-1000
0 | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/1%3A_Fundamental_Properties_of_Genes/1.E%3A_Fundamental_Properties_of_Genes_%28Exercises%29.txt |
Additional Readings
• Griffiths, A. J. F., Miller, J. H., Suzuki, D. T., Lewontin, R. C. and Gelbart, W. M. (1993) An Introduction to Genetic Analysis, Fifth Edition (W. H. Freeman and Company, New York).
• Cairns, J., Stent, G. S. and Watson, J. D., editors (1992) Phage and the Origins of Molecular Biology, Expanded Edition (Cold Spring Harbor Laboratory Press, Plainview, NY).
• Brock, T. D. (1990) The Emergence of Bacterial Genetics (Cold Spring Harbor Laboratory Press, Plainview, NY).
• Benzer, S. (1955) Fine structure of a genetic region in bacteriophage. Proceedings of the National Academy of Sciences, USA 47: 344-354.
• Yanofsky, C. (1963) Amino acid replacements associated with mutation and recombination in the A gene and their relationship to in vitro coding data. Cold Spring Harbor Symposia on Quantitative Biology 18: 133-134.
• Crick, F. (1970) Central dogma of molecular biology. Nature 227:561-563
Transcription and mRNA structure
A few years after he and James Watson had proposed the double helical structure for DNA, Francis Crick (with other collaborators) proposed that a less stable nucleic acid, RNA, served as a messenger RNA that provided a transient copy of the genetic material that could be translated into the protein product encoded by the gene. Such mRNAs were indeed found. These and other studies led Francis Crick to formulate this “central dogma” of molecular biology (Figure 1.21).
This model states that DNA serves as the repository of genetic information. It can be replicated accurately and indefinitely. The genetic information is expressed by the DNA first serving as a template for the synthesis of (messenger) RNA; this occurs in a process called transcription. The mRNA then serves as a template, which is read by ribosomes and translatedinto protein. The protein products can be enzymes that catalyze the many metabolic transformations in the cell, or they can be structural proteins.
Although there have been some additional steps added since its formulation, the central dogma has stood the test of time and myriad experiments. It provides a strong unifying theme to molecular genetics and information flow in cell biology and biochemistry.
Although in many cases a gene encodes one polypeptide, other genes encode a functional RNA. Some genes encode tRNAs and rRNAs needed for translation, others encode other structural and catalytic RNAs. Genes encode some product that is used in the cell, i.e. that when altered generates an identifiable phenotype. More generally, genes encode RNAs, some of which are functional as transcribed (or with minor alterations via processing) such as tRNAs and rRNAs, and others are messengers that are then translated into proteins. These proteins can provide structural, catalytic and regulatory roles in the cell.
Note the static role of DNA in this process. Implicit in this model is the idea that DNA does not provide an active cellular function, but rather it encodes macromolecules that are functional. However, the expression of virtually all genes is highly regulated. The sites on the DNA where this control is exerted are indeed functional entities, such as promoters and enhancers. In this case, the DNA is directly functional (cis‑regulatory sites), but the genes being regulated by these sites still encode some functional product (RNA or protein).
Studies of retroviruses lead Dulbecco to argue that the flow of information is not unidirectional, but in fact RNA can be converted into DNA (some viral RNA genomes are converted into DNA proviruses integrated into the genome). Subsequently Temin and Baltimore discovered the enzyme that can make a DNA copy of RNA, i.e. reverse transcriptase.
Central Dogma: DNA to RNA to protein
Several aspects of the structure of genes can be illustrated by examining the general features of a bacterial gene as now understood.
A gene is a string of nucleotides in the duplex DNA that encodes a mRNA, which itself codes for protein. Only one strand of the duplex DNA is copied into mRNA (Figure 1.22). Sometimes genes overlap, and in some of those cases each strand of DNA is copied, but each for a different mRNA. The strand of DNA that reads the same as the sequence of mRNA is the nontemplate strand. The strand that reads as the reverse complement of the mRNA is the template strand.
Note
The term "sense strand" has two opposite uses (unfortunately). Sidney Brenner first used it to designate the strand that served as the template to make RNA (bottom strand above), and this is still used in many genetics texts. However, now many authors use the term to refer to the strand that reads the same as the mRNA (top strand above). The same confusion applies to the term "coding strand" which can refer to the strand encoding mRNA (bottom strand) or the strand "encoding" the protein (top strand). Interestingly, "antisense" is used exclusively to refer to the strand that is the reverse complement of the mRNA (bottom strand).
Figure 1.22 helps illustrate the origin of terms used in gene expression. Copying the information of DNA into RNA stays in the same "language" in that both of these polymers are nucleic acids, hence the process is called transcription. An analogy would be writing exercises where you had to copy, e.g. a poem, from a book onto your paper - you transcribed the poem, but it is still in English. Converting the information from RNA into DNA is equivalent to converting from one "language" to another, in this case from one type of polymer (the nucleic acid RNA) to a different one (a polypeptide or protein). Hence the process is called translation. This is analogous to translating a poem written in French into English.
Figure 1.23 illustrates the point that a gene may be longer than the region coding for the protein because of 5' and/or 3' untranslated regions.
Eukaryotic mRNAs have covalent attachment of nucleotides at the 5' and 3' ends, and in some cases nucleotides are added internally (a process called RNA editing). Recent work shows that additional nucleotides are added post‑transcriptionally to some bacterial mRNAs as well.
Regulatory signals can be considered parts of genes
In order to express a gene at the correct time, the DNA also carries signals to start transcription (e.g. promoters), signals for regulating the efficiency of starting transcription (e.g. operators, enhancers or silencers), and signals to stop transcription (e.g. terminators). Minimally, a gene includes the transcription unit, which is the segment of DNA that is copied into RNA in the primary transcript. The signals directing RNA polymerase to start at the correct site, and other DNA segments that influence the efficiency of this process are regulatory elements for the gene. One can also consider them to be part of the gene, along with the transcription unit. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/1%3A_Fundamental_Properties_of_Genes/1.S%3A_Additional_Readings_%28Summary%29.txt |
The genetic systems found in bacteria and fungi are particularly powerful. The small size of the genome (all the genetic material in an organism), the ability to examine both haploid and diploid forms, and the ease of large-scale screens have made them the method of choice for many investigations.
Genetic methods in microorganisms
Bacteriophage are viruses that infect bacteria. Because of their very large number of progeny and ability to recombine in mixed infections (more than one strain of bacteria in an infection), they have been used extensively in high-resolution definition of genes. Much of what we know about genetic fine structure, prior to the advent of techniques for isolating and sequencing genes, derive form studies in bacteriophage.
Bacteriophage have been a powerful model genetic system, because they have small genomes, have a short life cycle, and produce many progeny from an infected cell. They provide a very efficient means for transfer of DNA into or between cells. The large number of progeny makes it possible to measure very rare recombination events.
Lytic bacteriophage form plaques on lawns of bacteria; these are regions of clearing where infected bacteria have lysed. Early work focused on mutants with different plaque morphology, e.g. T2 r, which shows rapid lysis and generates larger plaques, or on mutants with different host range, e.g. T2 h, which will kill both host strains B and B/2.
A cis-trans complementation test defines a cistron, which is a gene
Seymour Benzer used the rIIlocus of phage T4 to define genes by virtue of their behavior in a complementation test, and also to provide fundamental insight into the structure of genes (in particular, the arrangement of mutable sites - see the next section). The difference in plaque morphology between rand r+phage is easy to see (large versus small, respectively), and Benzer isolated many r mutants of phage T4. The wild type, but not any rIImutants, will grow on E. colistrain K12(l), whereas both wild type and mutant phage grow equally well on E. colistrain B. Thus the wild phenotype is readily detected by its ability to grow in strain K12 (l).
If E. colistrain K12 (l) is co-infected with 2 phage carrying mutations at different positions in rIIA, you get no multiplication of the phage (except the extremely rare wild type recombinants, which occur at about 1 in 106 progeny). In the diagram below, each line represents the chromosome from one of the parental phage.
rIIA rIIB
phage 1 _|__x______|________|_
phage 2 _|_______x_|________|_
Likewise, if the two phage in the co-infection carry mutations at different positions in rIIB, you get no multiplication of the phage (except the extremely rare wild type recombinants, about 1 in 106).
rIIA rIIB
phage 3 _|_________|_x______|_
phage 4 _|_________|______x_|_
However, if one of the co-infecting phage carries a mutation in rIIAand the other a mutation in rIIB, then you see multiplication of the phage, forming a very large number of plaques on E. colistrain K12 (l).
rIIA rIIB
phage 1 _|__x______|________|_ Provides wt rIIB protein
phage 4 _|_________|______x_|_ Provides wt rIIA protein
Together these two phage provide all the phage functions - they complement each other. This is a positive complementation test. The first two examples show no complementation, and we place them in the same complementation group. Mutants that do not complement are placed in the same complementation group; they are different mutant alleles of the same gene. Benzer showed that there were two complementation groups (and therefore two genes) at the r II locus, which he called A and B.
Exercise 1.3
In the mixed infection with phage 1 and phage 4, you also obtain the rare wild type recombinants, but there are more recombinants than are seen in the co-infections with different mutant alleles. Why?
Benzer’s experiments analyzing the rIIlocus of bacteriophage T4 formalized the idea of a cis‑transcomplementation test to define a cistron, which is an operational definition of a gene. First, let’s define cis and transwhen used to refer to genes. In the cis configuration, both mutations are on the same chromosome. In the trans configuration, each mutation is on a different chromosome
Mutations in the same gene will not complement in trans, whereas mutations in different genes will complement in trans (Figure 1.12). In the cis configuration, the other chromosome is wild type, and wild‑type will complement any recessive mutation. The complementation group corresponds to a genetic entity we call a cistron, it is equivalent to a gene.
This test requires a diploid situation. This can be a natural diploid (2 copies of each chromosome) or a partial, or merodiploid, e.g. by conjugating with a cell carrying an F' factor. Some bacteriophage carry pieces of the host chromosome; these are called transducing phage. Infection of E. coli with a transducing phage carrying a mutation in a host gene is another way to create a merodiploid in the laboratory for complementation analysis.
Recombination within genes allows construction of a linear map of mutable sites that constitute a gene
Once the recombination analysis made it clear that chromosomes were linear arrays of genes, these were thought of as "string of pearls" with the genes, or "pearls," separated by some non‑genetic material (Figure 1.13). This putative non-genetic material was thought to be the site of recombination, whereas the genes, the units of inheritance, were thought to be resistant to recombination. However, by examining the large number of progeny of bacteriophage infections, one can demonstrate that recombination can occur within a gene. This supports the second model shown in Figure 1.13. Because of the tight packing of coding regions in phage genomes, recombination almost always occurs within genes in bacteriophage, but in genomes with considerable non-coding regions between genes, recombination can occur between genes as well.
The tests between these two models required screening for genetic markers (mutations) that are very close to each other. When two markers are very close to each other, the recombination frequency is extremely low, so enough progeny have to be examined to resolve map distances of, say 0.02 centiMorgans = 0.02 map units = 0.02 % recombinants. This means that 2 out of 10,000 progeny will show recombination between two markers that are 0.02 map units apart, and obviously one has to examine at least 10,000 progeny to reliably score this recombination. That's the power of microbial genetics ‑ you actually can select or screen through this many progeny, sometimes quite easily.
An example of recombination in phage is shown in Figure 1.14. Wild type T2 phage forms small plaques and kills only E. colistrain B. Thus different alleles of hcan be distinguished by plating on a mixture of E. colistrains B and B/2. The phage carrying mutant hallele will generate clear plaques, since they kill both strains. Phage with the wild type h+ give turbid plaques, since the B/2 cells are not lysed but B cells are. When a mixture of E. colistrains B and B/2 are co-infected with both T2 hrand T2 h+r+, four types of plaques are obtained. Most have the parental phenotypes, clear and large or turbid and small. These plaques contain progeny phage that retain the parental genotypes T2 hrand T2 h+r+, respectively. The other two phenotypes are nonparental, i.e. clear and small or turbid and large. These are from progeny with recombinant genotypes, i.e. T2 hr+and T2 h+r. In this mixed infection, recombination occurred between two phage genomes in the same cell.
The first demonstration of recombination within a gene came from work on the rIIAand rIIBgenes of phage T4. These experiments from Seymour Benzer, published in 1955, used techniques like that diagrammed in Figure 1.14. Remember that mutations in the rgene cause rapid lysis of infected cells, i.e. the length of the lytic cycle is shorter. The difference in plaque morphology between rand r+phage is easy to see (large versus small, respectively). These two genes are very close together, and many mutations were independently isolated in each. This was summarized in the discussion on complementation above.
Consider the results of infection of a bacterial culture with two mutant alleles of gene rIIA.
T4rIIA6 _|_______________________x______|_
and T4rIIA27 _|_______x______________________|_
(x marks the position of the mutation in each allele).
Progeny phage from this infection include those with a parental genotype (in the great majority), and at a much lower frequency, two types of recombinants:
wild type T4 r+ _|______________________________|_
double mutant T4rIIA6 rIIA27 _|_______x_______________x______|_
The wild type is easily scored because it, and not any rIImutants, will grow on E. coli strain K12(l), whereas both wild type and mutant phage grow equally well on E. coli strain B. Thus you can selectfor the wild type (and you will see only the desired recombinant). Finding the double mutants is more laborious, because they are obtained only by screening through the progeny, testing for phage that when backcrossed with the parental phage result in no wild type recombinant progeny.
Equal numbers of wild type and double mutant recombinants were obtained, showing that recombination can occur within a gene, and that this occurs by reciprocal crossing over. If recombination were only between genes, then no wild type phage would result. A large spectrum of recombination values was obtained in crosses for different alleles, just like you obtain for crosses between mutants in separate genes.
Several major conclusions could be made as a result of these experiments on recombination within the rIIgenes.
1. A large number of mutable sites occur within a gene, exceeding some 500 for the rIIA and rIIBgenes. We now realize that these correspond to the individual base pairs within the gene.
2. The genetic maps are clearly linear, indicating that the gene is linear. Now we know a gene is a linear polymer of nucleotides.
3. Most mutations are changes at one mutable site (point mutations). Many genes can be restored to wild type by undergoing a reverse mutation at the same site (reversion).
4. Other mutations cause the deletionof one or more mutable sites, reflecting a physical loss of part of the rII gene. Deletions of one or more mutable site (base pair) are extremely unlikely to revert back to the original wild type.
One gene encodes one polypeptide
One of the fundamental insights into how genes function is that one gene encodes one enzyme (or more precisely, one polypeptide). Beadle and Tatum reached this conclusion based on their complementation analysis of the genes required for arginine biosynthesis in fungi. They showed that a mutation in each gene led to a loss of activity of one enzyme in the multistep pathway of arginine biosynthesis. As discussed above in the section on genetic dissection, a large number of Arg auxotrophs (requiring Arg for growth) were isolated, and then organized into a set of complementation groups, where each complementation groups represents a gene.
The classic work of Beadle and Tatum demonstrated a direct relationship between the genes defined by the auxotrophic mutants and the enzymes required for Arg biosynthesis. They showed that a mutation in one gene resulted in the loss of one particular enzymatic activity, e.g. in the generalized scheme below, a mutation in gene 2 led to a loss of activity of enzyme 2. This led to an accumulation of the substrate for that reaction (intermediate N in the diagram below). If there were 4 complementation groups for the Arg auxotrophs, i.e. 4 genes, then 4 enzymes were found in the pathway for Arg biosynthesis. Each enzyme was affected by mutations in one of the complementation groups.
Intermediates:
M ® N ® O ® P ® Arg
enzyme 1 enzyme 2 enzyme 3 enzyme 4
gene 1 gene 2 gene 3 gene 4
Figure 1.15. A general scheme showing the relationships among metabolic intermediates (M, N, O, P), and end product (Arg), enzymes and the genes that encode them.
In general, each step in a metabolic pathway is catalyzed by an enzyme (identified biochemically) that is the product of a particular gene (identified by mutants unable to synthesize the end product, or unable to break down the starting compound, of a pathway). The number of genes that can generate auxotrophic mutants is (usually) the same as the number of enzymatic steps in the pathway. Auxotrophic mutants in a given gene are missing the corresponding enzyme. Thus Beadle and Tatum concluded that one gene encodes one enzyme. Sometimes more than one gene is required to encode an enzyme because the enzyme has multiple, different polypeptide subunits. Thus each polypeptide is encoded by a gene.
The metabolic intermediates that accumulate in each mutant can be used to place the enzymes in their order of actionin a pathway. In the diagram in Figure 1.15, mutants in gene 3 accumulated substance O. Feeding substance O to mutants in gene 1 or in gene 2 allows growth in the absence of Arg. We conclude that the defects in enzyme 1 or enzyme 2, respectively, are upstream of enzyme 3. In contrast, feeding substance O to mutants in gene 4 will not allow growth in the absence of Arg. Even though this mutant can convert substance O to substance P, it does not have an active enzyme 4 to convert P to Arg. The inability of mutants in gene 4 to grow on substance O shows that enzyme 4 is downstream of enzyme 3.
Exercise 1.4
Imagine that you are studying serine biosynthesis in a fungus. You isolate serine auxotrophs, do all the pairwise crosses of the mutants and discover that the auxotrophs can be grouped into three complementation groups, called A, B and C. You also discover that a different metabolic intermediate accumulates in members of each complementation group - substance A in auxotrophs in the A complementation group, substance B in the B complementation group and substance C in the C complementation group. Each of the intermediates is fed to auxotrophs from each of the three complementation groups as tabulated below. A + means that the auxotroph was able to grow in media in the absence of serine when fed the indicated substance; a - denotes no growth in the absence of serine.
Fed:
mutant in complementation group A
mutant in complementation group B
mutant in complementation group C
substance A
-
+
+
substance B
-
-
-
substance C
-
+
-
In the biosynthetic pathway to serine in this fungus, what is the order of the enzymes encoded in the three complementation groups? Enzyme A is encoded by the gene that when altered generates mutants that fall into complementation group A, etc.
The gene and its polypeptide product are colinear
Once it was determined that a gene was a linear array of mutable sites, that genes are composed of a string of nucleotides called DNA (see Chapter 2), and that each gene encoded a polypeptide, the issue remained to be determined how exactly that string of nucleotides coded for a particular amino acid sequence. This problem was studied along several avenues, culminating in a major achievement of the last half of the 20th century – the deciphering of the genetic code. The detailed assignment of particular codons (triplets of adjacent nucleotides) will be discussed in Chapter 13. In the next few sections of this chapter, we will examine how some of the basic features of the genetic code were deciphered.
A priori, the coding units within a gene couldencode both the composition and the address for each amino acid, as illustrated in Model 1 of Figure 1.17. In this model, the coding units could be scrambled and still specify the same protein. In such a situation, the polypeptide would not be colinear with the gene.
In an alternative model (Model 2 in Figure 1.16), the coding units only specify the composition, but not the position, of an amino acid. The "address" of the amino acid is derived from the position of the coding unit within the gene. This model would predict that the gene and its polypeptide product would be colinear - e.g. mutation in the 5th coding unit would affect the 5th amino acid of the protein, etc.
Charles Yanofsky and his co-workers (1964) tested these two models and determined that the gene and the polypeptide product are indeed colinear. They used recombination frequencies to map the positions of different mutant alleles in the gene that encodes a particular subunit of the enzyme tryptophan synthase. They then determined the amino acid sequence of the wild type and mutant polypeptides. As illustrated in Figure 1.17, the position of a mutant allele on the recombination map of the gene corresponds with the position of the amino acid altered in the mutant polypeptide product. For instance, allele A101 maps to one end of the gene, and the corresponding Glu ® Val replacement is close to the N terminus of the polypeptide. Allele A64maps close to the other end of the gene, and the corresponding Ser ® Leu replacement is close to the C terminus of the polypeptide. This correspondence between the positions of the mutations in each allele and the positions of the consequent changes in the polypeptide show that Model 1 can be eliminated and Model 2 is supported.
Mutable sites are base pairs along the double helix
The large number of mutable sites found in each gene, and between which recombination can occur, leads one to conclude that the mutable sites are base pairs along the DNA. Sequence determination of the wild type and mutant genes confirms this conclusion.
Single amino acids are specified by three adjacent nucleotides, which are a codons
This conclusion requires three pieces of information.
First of all, adjacent mutable sites specify amino acids. Reaching this conclusion required investigation of the fine structure of a gene, including rare recombination between very closely linked mutations within a gene. Yanofsky and his colleagues, working with mutations the trpA gene of E. coli, encoding tryptophan synthase, showed that different alleles mutated in the same codon could recombine (albeit at very low frequency). (This is the same laboratory and same system that was used to show that a gene and its polypeptide product are colinear.) Thus recombination between two different alleles can occur within a codon, which means that a codon must have more than one mutable site. We now recognize that a mutable site is a nucleotide in the DNA. Thus adjacent mutable sites (nucleotides) encode a single amino acid.
Let’s look at this in more detail (Figure 1.18). Yanofsky and colleagues examined two different mutant alleles of trpA, each of which caused alteration in amino acid 211 of tryptophan synthase. In the mutant allele A23, wild type Gly is converted to mutant Arg. In the mutant allele A46, wild type Gly is converted to mutant Glu.
GGA (Gly 211) --> AGA (Arg 211) mutant allele A23
GGA (Gly 211) --> GAA (Glu 211) mutant allele A46
A23 ´ A46 AGA ´ GAA ® GGA (wild type Gly 211 in 2 out of 100,000 progeny)
Figure 1.18.Recombination can occur between two mutant alleles affecting the same codon.
Alleles A23and A46are not alternative forms of the same mutable site, because recombination to yield wild type occurs, albeit at a very low frequency (0.002%; the sites are very close together, in fact in the same codon!). If they involved the same mutable site, one would never see the wild-type recombinant.
The second observation is that the genetic code is non-overlapping. This was shown by demonstrating that a mutation at a single site alters only one amino acid. This conflicts with the predictions of an overlapping code (see Figure 1.19), and thus the code must be non-overlapping.
The third observation is that the genetic code is read in triplets from a fixed starting point. This was shown by examining the effect of frameshift mutations. As shown in Figure 1.19, a code lacking punctuation has a certain reading frame. Insertions or deletions of nucleotides are predicted to have a drastic effect on the encoded protein because they will change that reading frame. The fact that this was observed was one of the major reasons to conclude that the mRNA molecules encoded by genes are read in successive blocks of three nucleotides in a particular reading frame.
For the sequence shown in Figure 1.20, insertion of an A shifts the reading frame, so all amino acids after the insertion differ from the wild type sequence. (The 4th amino acid is still a Gly because of degeneracy in the code: both GGC and GGG code for Gly.) Similarly, deletion of a U alters the entire sequence after the deletion.
These observations show that the nucleotide sequence is read, or translated, from a fixed starting point without punctuation. An alternative model is that the group of nucleotides encoding an amino acid (the codon) could also include a signal for the end of the codon (Model 2 in Figure 1.19). This could be considered a "comma" at the end of each codon. If that were the case, insertions or deletions would only affect the codon in which they occur. However, the data show that all codons, including and after the one containing the insertion or deletion, are altered. Thus the genetic code is not punctuated, but is read in a particular frame that is defined by a fixed starting point (Model 3 in Figure 1.19). That starting point is a particular AUG, encoding methionine. (More about this will be covered in Chapter 13).
The results of frame-shift mutations are so drastic that the proteins are usually not functional. Hence a screen or selection for loss-of-function mutants frequently reveals these frameshift mutants. Simple nucleotide substitutions that lead to amino acid replacements often have very little effect on the protein, and hence have little, or subtle, phenotypes.
A double mutant generated by crossing over between the insertion (+) and deletion (‑) results in an (almost) normal phenotype, i.e. reversion of insertion or deletion.
A gene containing three closely spaced insertions (or deletions) of single nucleotides will produce a functional product. However, four or five insertions or deletions do not give a functional product (Crick, Barnett, Brenner and Watts‑Tobin, 1961). This provided the best evidence that the genetic code is read in groups of three nucleotides(not two or four). Over the next 5 years the code was worked out (by 1966) and this inference was confirmed definitively. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/1%3A_Fundamental_Properties_of_Genes/Genetic_methods_in_microorganisms/Bacteriophage.txt |
DNA and RNA are both nucleic acids, which are the polymeric acids isolated from the nucleus of cells. DNA and RNA can be represented as simple strings of letters, where each letter corresponds to a particular nucleotide, the monomeric component of the nucleic acid polymers. Although this conveys almost all the information content of the nucleic acids, it does not tell you anything about the underlying chemical structures. This chapter will be review the evidence that nucleic acids are the genetic material, and then exploring the chemical structure of nucleic acids.
2: Structures of Nucleic Acids
Three major forms of DNA are double stranded and connected by interactions between complementary base pairs. These are terms A-form, B-form,and Z-form DNA.
B-form DNA
The information from the base composition of DNA, the knowledge of dinucleotide structure, and the insight that the X‑ray crystallography suggested a helical periodicity were combined by Watson and Crick in 1953 in their proposed model for a double helical structure for DNA. They proposed two strands of DNA -- each in a right‑hand helix -- wound around the same axis. The two strands are held together by H‑bonding between the bases (in anti conformation) as shown in Figure \(1\).
Major groove Major groove
Minor groove Minor groove
Figure \(1\): (left) An A:T base pair and (right) a G:C base pair
Bases fit in the double helical model if pyrimidine on one strand is always paired with purine on the other. From Chargaff's rules, the two strands will pair A with T and G with C. This pairs a keto base with an amino base, a purine with a pyrimidine. Two H‑bonds can form between A and T, and three can form between G and C. This third H-bond in the G:C base pair is between the additional exocyclic amino group on G and the C2 keto group on C. The pyrimidine C2 keto group is not involved in hydrogen bonding in the A:T base pair.
These are the complementary base pairs. The base‑pairing scheme immediately suggests a way to replicate and copy the the genetic information.
The two strands are not in a simple side‑by‑side arrangement, which would be called a paranemic joint (Figure \(3\)). (This will be encountered during recombination in Chapter 8.) Rather the two strands are coiled around the same helical axis and are intertwined with themselves (which is referred to as a plectonemic coil). One consequence of this intertwining is that the two strands cannot be separated without the DNA rotating, one turn of the DNA for every "untwisting" of the two strands.
Dimensions of B-form (the most common) of DNA
• 0.34 nm between bp, 3.4 nm per turn, about 10 bp per turn
• 1.9 nm (about 2.0 nm or 20 Angstroms) in diameter
Major and minor groove
The major groove is wider than the minor groove in DNA (Figure \(\PageIndex{2d}\)), and many sequence specific proteins interact in the major groove. The N7 and C6 groups of purines and the C4 and C5 groups of pyrimidines face into the major groove, thus they can make specific contacts with amino acids in DNA-binding proteins. Thus specific amino acids serve as H‑bond donors and acceptors to form H-bonds with specific nucleotides in the DNA. H‑bond donors and acceptors are also in the minor groove, and indeed some proteins bind specifically in the minor groove. Base pairs stack, with some rotation between them.
A‑form nucleic acids and Z‑DNA
Three different forms of duplex nucleic acid have been described. The most common form, present in most DNA at neutral pH and physiological salt concentrations, is B-form. That is the classic, right-handed double helical structure we have been discussing. A thicker right-handed duplex with a shorter distance between the base pairs has been described for RNA-DNA duplexes and RNA-RNA duplexes. This is called A-form nucleic acid.
A third form of duplex DNA has a strikingly different, left-handed helical structure. This Z DNA is formed by stretches of alternating purines and pyrimidines, e.g. GCGCGC, especially in negatively supercoiled DNA. A small amount of the DNA in a cell exists in the Z form. It has been tantalizing to propose that this different structure is involved in some way in regulation of some cellular function, such as transcription or regulation, but conclusive evidence for or against this proposal is not available yet.
Differences between A-form and B-form nucleic acid
The major difference between A-form and B-form nucleic acid is in the conformation of the deoxyribose sugar ring. It is in the C2' endoconformation for B-form, whereas it is in the C3' endoconformation in A-form. As shown in Figure \(4\), if you consider the plane defined by the C4'-O-C1' atoms of the deoxyribose, in the C2' endoconformation, the C2' atom is above the plane, whereas the C3' atom is above the plane in the C3' endoconformation. The latter conformation brings the 5' and 3' hydroxyls (both esterified to the phosphates linking to the next nucleotides) closer together than is seen in the C2' endoconfromation (Figure 2.16). Thus the distance between adjacent nucleotides is reduced by about 1 Angstrom in A-form relative to B-form nucleic acid (Figure \(4\)).
A second major difference between A-form and B-form nucleic acid is the placement of base-pairs within the duplex. In B-form, the base-pairs are almost centered over the helical axis (Figure \(4\)), but in A-form, they are displaced away from the central axis and closer to the major groove. The result is a ribbon-like helix with a more open cylindrical core in A-form.
Z-form DNA
Z-DNA is a radically different duplex structure, with the two strands coiling in left-handed helices and a pronounced zig-zag (hence the name) pattern in the phosphodiester backbone. As previously mentioned, Z-DNA can form when the DNA is in an alternating purine-pyrimidine sequence such as GCGCGC, and indeed the G and C nucleotides are in different conformations, leading to the zig-zag pattern. The big difference is at the G nucleotide. It has the sugar in the C3' endoconformation (like A-form nucleic acid, and in contrast to B-form DNA) and the guanine base is in the synconformation. This places the guanine back over the sugar ring, in contrast to the usual anticonformation seen in A- and B-form nucleic acid. Note that having the base in the anticonformation places it in the position where it can readily form H-bonds with the complementary base on the opposite strand. The duplex in Z-DNA has to accomodate the distortion of this G nucleotide in the synconformation. The cytosine in the adjacent nucleotide of Z-DNA is in the "normal" C2' endo, anticonformation.
Even classic B-DNA is not completely uniform in its structure. X-ray diffraction analysis of crystals of duplex oligonucleotides shows that a given sequence will adopt a distinctive structure. These variations in B-DNA may differ in the propeller twist (between bases within a pair) to optimize base stacking, or in the 3 ways that 2 successive base pairs can move relative to each other: twist, roll, or slide.
Table \(1\) Comparisons of B-form, A-form and Z‑DNA
B-Form A-Form Z-Form
helix sense Right Handed Right Handed Left Handed
base pairs per turn 10 11 12
vertical rise per bp 3.4 Å 2.56 Å 19 Å
rotation per bp +36° +33° -30°
helical diameter 19 Å 19 Å 19 Å | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/2%3A_Structures_of_Nucleic_Acids/2.5%3A_B-Form_A-Form_and_Z-Form_of_DNA.txt |
DNA and RNA are both nucleic acids, which are the polymeric acids isolated from the nucleus of cells. DNA and RNA can be represented as simple strings of letters, where each letter corresponds to a particular nucleotide, the monomeric component of the nucleic acid polymers. Although this conveys almost all the information content of the nucleic acids, it does not tell you anything about the underlying chemical structures. This chapter will be review the evidence that nucleic acids are the genetic material, and then exploring the chemical structure of nucleic acids.
Southern blot-hybridizations
After separation by electrophoresis, DNA fragments are transferred to a membrane (nylon or nitrocellulose) and immobilized; this replica of the DNA pattern in the gel is called a "blot." A specific labeled probe is hybridized to the blot to detect related sequences. After nonspecifically bound probe is washed away, the specific hybrids are detected by autoradiography of the blot.
Restriction sites can be used as genetic markers. One can identify restriction fragment length polymorphisms (RFLPs) that are linked to a particular locus. This can be be used to
1. Develop a diagnostic test for a disease locus (e.g. sickle cell disease)
2. Help isolate the gene.
3. DNA fingerprinting for highly variable loci.
Sizes of DNAs and chromosomes, and methods to resolve them
The next figure presents views of chromosomes and DNA segments on four different, expanding scales. The top level compares the sizes of intact chromosomes from four of the organisms we will be discussing in this course. The scale on yeast chromosome III is then expanded so that it can be compared to some of the viral and plasmid genomes that are in common use. Next, a higher resolution view of the plasmid pBR322 is given, and finally the highest resolution that we are usually concerned with, i.e. the nucleotide sequence.
Determining the sequence of DNA and RNA
The basic approach is to generate a nested set of DNA fragmentsthat start a common site and end in either A, G ,C or T. These sets of (labeled) DNA fragments are separated on a denaturing polyacrylamide gel that has a resolution of 1 bp. The resulting pattern allows the sequence to be read. Base-specific chemical modification and degradation, developed by Maxam and Gilbert, was a widely used approach. Nucleotide-specific cleavage of RNA by a set of Rnases can be used to sequence RNA. We will focus on the most common method of sequencing DNA, that of nucleotide-specific chain termination.
The dideoxynucleotide chain termination methodwas developed in the laboratory of Fred Sanger at Cambridge. A 2’, 3’ dideoxynucleotide can be incorporated into DNA, as directed by the template strand. However, the missing 3’-OH precludes further polymerization. Hence the newly synthesized chain of nucleotides ends at base-specific, chain terminating dideoxynucleotide. Reactions are run such that all the products end in a G, a C, an A, or aT, but they all begin at the same place. This generates a nested set of products whose length is a measure of the position of all G’s in a target sequence, or all C’s, etc. Thus one can deduce that the target sequence is complementary to, e.g. G at position 1, T at positon 2, C at positions 3 and 4, etc. for hundreds of nucleotides per run.
In more detail, a specific primer is annealed to the template, upstream from the region to be sequenced. DNA polymerase will catalyze the synthesis of new DNA from the 3' end of that primer (elongation). The primer therefore generates a common end to all the product fragments. (This is the basis for the nested set in this approach).
The synthesized DNA is labeled with either a radioactive nucleotide, such as [a35S]deoxy‑thio‑ATP, or a fluorescent dye, often attached to the primer.
A base-specific chain‑terminator is included in each of four reactions:
• 2',3' dideoxyGTP in the "G" reaction.
• 2',3' dideoxyATP in the "A" reaction.
• 2',3' dideoxyTTP in the "T" reaction.
• 2',3' dideoxyCTP in the "C" reaction.
The DNA polymerase will elongate from each annealed primer until it incorporates a 2', 3' dideoxynucleotide. No additional nucleotides can be added to this product, since it has no 3' OH, thus it is a chain-terminator. This termination occurs only at G residues (complementary to C's in the template) in the "G" reaction, only at A residues in the "A" reaction, etc. Thus the products of each reaction comprise a nested set of fragments, with the specific primer at the 5' end and the base-specific chain terminator at the 3' end. The products are resolved on a sequencing gel, exposed to X-ray film and the sequence read, as in Figure 2.30.
The dideoxynucleotide chain-termination approach is the method used in automated sequenators. Different color fluorescent dyes (usually attached to the primer) are included in each base-specific reaction. Therefore the products of all four can be run in 1 lane of the resolving gel, allowing >20 sequencing sets to be analyzed at one time. A laser scans continuously along one zone of the gel, and records when a (e.g.) red, green, blue or yellow fluoresence is detected in each lane, meaning that the primer extended to a (e.g.) A, G, C or T is passing through the detection zone. These data are automatically processed, and a readout is generated with the peaks for each fluorescent dye as function of time of the gel running and the deduced sequence. An example of the output is shown below in black-and-white; the original output is in color (a different color for each nucleotide). Manual editing of the deduced sequence can be done based on the raw data, but in large scale sequencing projects, each region is determined about 8 different times and other software is used to determine the most frequently ocurring nucleotide at each position.
The capacity of automated sequencing machines is extraordinary. New machines using capillary gel electrophoresis are used to generated millions of nucleotides per day in the major sequencing centers. This technology allows large, complex genomes to be sequenced rapidly, as discussed in Chapter 4.
Supercoiling of Topologically Constrained DNA
Topologically closed DNA can be circular (covalently closed circles) or loops that are constrained at the base. The coiling (or wrapping) of duplex DNA around its own axis is called supercoiling (Figure 2.32 middle).
• Negative supercoils twist the DNA about its axis in the opposite direction from the clockwise turns of the right-handed (R-H) double helix.
• Negatively supercoiled DNA is underwound (and thus favors unwinding of duplex).
• Negatively supercoiled DNA has R-H supercoil turns (Figure 2.32).
• Positive supercoils twist the DNA in the same direction as the turns of the R-H double helix.
• Positively supercoiled DNA is overwound (helix is wound more tightly).
• Positively supercoiled DNA has L-H supercoil turns.
The clockwise turns of R-H double helix (A or B form) generate a positive Twist (T); see Figure 2.32 left. The couterclockwise (ccw) turns of L-H helix (Z ) generate a negative T.
T= Twisting number
• For B form DNA, it is + (# bp/10 bp per twist)
• For A form DNA, it is + (# bp/11 bp per twist)
• For Z DNA, it is - (# bp/12 bp per twist)
W= Writhing Numberis the turning of the axis of the DNA duplex in space
• Relaxed molecule W=0
• Negative supercoils, W is negative
• Positive supercoils, W is positive
L= Linking number= total number of times one strand of the double helix (of a closed molecule) encircles (or links) the other.
$L = W + T$
• L cannot change unless one or both strands are broken and reformed.
• A change in the linking number, DL, is partitioned between T and W (Figure 2.32 right). Thus:
$DL=DW+DT$
if $DL = 0$, $DW=-DT$
Ethidium Bromide intercalates in DNA, and untwists (or unwinds) the duplex by -27° per molecule of ethidium bromide intercalated. Thus intercalation of 14 molecules of ethidium bromide will untwist the duplex by 378o, i.e. slightly more than one full twist (which would be 360°). For this process of intercalation, DL=0, since no covalent bonds in the DNA are broken or reformed. The change in twist, DT, is negative, and thus DW is positive. Thus intercalation of ethidium bromide can relax a negatively supercoiled circle, and further intercalation will make the DNA positively supercoiled (Figure 2.33).
It is useful to have an expression for supercoiling that is independent of length. The superhelical density is simply the number of superhelical (S.H.) turns per turn (or twist) of double helix.
$\text{Superhelical density} = s= \dfrac{W}{T}$
This is -0.05 for natural bacterial DNA. i.e., in bacterial DNA, there is 1 negative S.H. turn per 200 bp (calculated from 1 negative S.H. turn per 20 twists = 1 negative S.H. turn per 200 bp)
Negative supercoiled DNA has energy stored that favors unwinding, or a transition from B-form to Z DNA.
For s = -0.05, $\Delta G=-9 Kcal/mole$, which favoring unwinding
Thus negative supercoiling could favor initiation of transcription and initiation of replication.
Topoisomerases
Topoisomerases catalyze a change in the linking number of DNA.
• Topo I = nicking-closing enzyme, can relax positive or negative supercoiled DNA, makes a transient break in 1 strand. E. coli Topo I specifically relaxes negatively supercoiled DNA. Calf thymus Topo I works on both negatively and positively supercoiled DNA.
• Topo II = gyrase: uses the energy of ATP hydrolysis to introduce negative supercoils. Its mechanism of action is to make a transient double strand break, pass a duplex DNA through the break, and then re-seal the break.
Measuring a change in linking number
One can measure a change in linking number (DL) by sedimentation, electrophoresis, or electron microscopy, as illustrated in Figure 2.34. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/2%3A_Structures_of_Nucleic_Acids/2.8%3A_Intro.txt |
The first two chapters covered many important aspects of genes, such as how they function in inheritance, how they code for protein (in general terms) and their chemical nature. All this was learned without having a single gene purified. A full understanding of a gene, or the entire set of genes in a genome, requires that they be isolated and then studied intensively. Once a gene is “in hand”, in principal one can determine both its biochemical structures and its function(s) in an organism. One of the goals of biochemistry and molecular genetics is to assign particular functions to individual or composite structures. This chapter covers some of the techniques commonly used to isolate genes and illustrates some of the analyses that can be done on isolated genes.
• 3.1: Recombinant DNA, Polymerase Chain Reaction and Applications to Eukaryotic Gene Structure and Function
• 3.2: Overview of Recombinant DNA Technology
• 3.3: Introduction of recombinant DNA into cell and replication: Vectors
Vectors used to move DNA between species, or from the lab bench into a living cell, must meet three requirements: (1) They must be autonomously replicating DNA molecules in the host cell. (2) They must contain a selectable marker so cells containing the recombinant DNA can be distinguished from those that do not. (3) They must have an insertion site to accommodate foreign DNA. Usually a unique restriction cleavage site in a nonessential region of the vector DNA.
• 3.4: Introducting Recombinant DNA into Host Cells
• 3.5: Polymerase Chain Reaction (PCR)
The polymerase chain reaction (PCR) is now one of the most commonly used assays for obtaining a particular segment of DNA or RNA. It is rapid and extremely sensitive. By amplifying a designated segment of DNA, it provides a means to isolate that particular DNA segment or gene. This method requires knowledge of the nucleotide sequence at the ends of the region that you wish to amplify.
• 3.6: cDNA
Construction of cDNA clones involves the synthesis of complementary DNA from mRNA and then inserting a duplex copy of that into a cloning vector, followed by transformation of bacteria.
• 3.7: Genomic DNA clones
Clones of genomic DNA, containing individual fragments of chromosomal DNA, are needed for many purposes
• 3.8: Eukaryotic Gene Structure
Much can be learned about any gene after it has been isolated by recombinant DNA techniques. The structure of coding and noncoding regions, the DNA sequence, and more can be deduced. This is true for bacterial and viral genes, as well as eukaryotic cellular genes. The next sections of this chapter will focus on analysis of eukaryotic genes, showing the power of examining purified copies of genes.
• 3.9: Introns and Exons
Far more exons and introns have been discovered (or more accurately, predicted) throught the analysis of genomic DNA sequences than could ever be discovered by direct experimentation. The different types of exons, the enormous length of introns, and other factors have complicated the task of finding reliable diagnostic signatures for exons in genomic sequences. However, considerable progress has been made and continues in current research.
• 3.10: Functional analysis of isolated genes
• 3.E: Isolating and Analyzing Genes (Exercises)
3: Isolating and Analyzing Genes
Methods to purify some abundant proteins were developed early in the 20th century, and some of the experiments on the fine structure of the gene (colinearity of gene and protein for trpA and tryptophan synthase) used microbial genetics and proteins sequencing. However, methods to isolate genes were not developed until the 1960’s, and the were applicable to only a few genes.
All this changed in the late 1970’s with the development of recombinant DNA technology, or molecular cloning. This technique enabled researchers to isolate any gene from any organism from which one could isolate intact DNA (or RNA). The full potential to provide access to all genes of organisms is now being realized as full genomes are sequenced. One of the by-products of the intense investigation of individual DNA molecules after the advent of recombinant DNA was a procedure to isolate any DNA for which one knows the sequence. This technique, called the polymerase chain reaction (PCR), is far easier than traditional molecular cloning methods, and it has become a staple of many laboratories in the life sciences. After covering the basic techniques in recombinant DNA technology and PCR, their application to studies of eukaryotic gene structure and function will be discussed.
Like many advances in molecular genetics, recombinant DNA technology has its roots in bacterial genetics.
Transducing Phage
The first genes isolated were bacterial genes that could be picked up by bacteriophage. By isolating these hybrid bacteriophage, the DNA for the bacterial gene could be recovered in a highly enriched form. This is the basic principal behind recombinant DNA technology.
Some bacteriophage will integrate into a bacterial chromosome and reside in a dormant state (Figure 3.1). The integrated phage DNA is called a prophage, and the bacterium is now a lysogen. Phage that do this are lysogenic. Induction of the lysogen will result in excision of the prophage and multiplication to produce many progeny, i.e. it enters a lytic phasein which the bacteria are broken open and destroyed. The nomenclature is descriptive. The bacteria carrying the prophage show no obvious signs of the phage (except immunity to superinfection with the same phage, covered later in Part Four), but when induced (e.g. by stress or UV radiation) they will generate a lytic state, hence they are called lysogens. Induced lysogens make phage from the prophage that was integrated. Phage that always multiply when they infect a cell are called lytic.
Excision of a prophage from a lysogen is notalways precise. Usually only the phage DNA is cut out of the bacterial chromosome, but occassionally some adjacent host DNA is included with the excised phage DNA and encapsidated in the progeny. These transducing phage are usually biologically inactive because the piece of the bacterial chromosome replaces part of the phage chromosome; these can be propagated in the presence of helper phage that provide the missing genes when co-infected into the same bacteria. When DNA from the transducing phage is inserted into the newly infected cell, the bacterial genes can recombine into the host chromosome, thereby bringing in new alleles or even new genes and genetically altering the infected cell. This process is called transduction.
Note that the transducing phage are carrying one or a small number of bacterial genes. This is a way of isolating the genes. The bacterial gene in the transducing phage has been separated from the other 4000 bacterial genes (in E. coli). By isolating large numbers of the transducing phage, the phage DNA, including the bacterial genes, can be obtained in large quantitiesfor biochemical investigation. One can isolate mg or mg quantities of a single DNA molecule, which allows for precise structural determination and detailed investigation.
A generalized transducing phagecan integrate at many different locations on the bacterial chromosome. Imprecise excision from any of those locations generates a particular transducing phage, carrying a short sections of the bacterial genome adjacent to the integration site. Thus a generalized transducing phage such as P1 can pick up many different parts of the E. coli genome.
A specialized transducing phageintegrates into only one or very few sites in the host genome. Hence it can carryonly a few specific bacterial genes, e.g., l lac(Figure 3.2).
This process of isolating a particular bacterial gene on a transducing phage is mimicked in recombinant DNA technology, in which a gene or genome fragment from any organism is isolated on a recombinant phage or plasmid. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.01%3A_Recombinant_DNA_Polymerase_Chain_Reaction_and_Applications_to_Eukaryotic_G.txt |
Recombinant DNA technology utilizes the power of microbiological selection and screening procedures to allow investigators to isolate a gene that represents as little as 1 part in a million of the genetic material in an organism. The DNA from the organism of interest is divided into small pieces that are then placed into individual cells (usually bacterial). These can then be separated as individual colonies on plates, and they can be screened through rapidly to find the gene of interest. This process is called molecular cloning.
Joining DNA in vitro to form recombinant molecules
Restriction endonucleasescut at defined sequences of (usually) 4 or 6 bp. This allows the DNA of interest to be cut at specific locations. The physiological function of restriction endonucleases is to serve as part of system to protect bacteria from invasion by viruses or other organisms. (See Chapter 7)
Table 3.1. List of restriction endonucleases and their cleavage sites. A ' means that the nuclease cuts between these 2 nucleotides to generate a 3' hydroxyl and a 5' phosphate.
Enzyme Site Enzyme Site
AluI AG'CT NotI GC'GGCCGC
BamHI G'GATCC PstI CTGCA'G
BglII A'GATCT PvuII CAG'CTG
EcoRI G'AATTC SalI G'TCGAC
HaeIII GG'CC Sau3AI 'GATC
HhaI GCG'C SmaI CCC'GGG
HincII GTY'RAC SpeI A'CTAGT
HindIII A'AGCTT TaqI T'CGA
HinfI G'ANTC XbaI T'CTAGA
HpaII C'CGG XhoI C'TCGAG
KpnI GGTAC'C XmaI C'CCGGG
MboI 'GATC
N = A,G,C or T
R = A or G
Y = C or T
S = G or C
W = A or T
a. Sticky ends
(1) Since the recognition sequences for restriction endonucleases are pseudopalindromes, an off-center cleavage in the recognition site will generate either a 5' overhang or a 3' overhang with self-complementary (or "sticky") ends.
e.g. 5' overhang EcoRI G'AATTC
BamHI G'GATCC
3' overhangPstI CTGCA'G
(2) When the ends of the restriction fragments are complementary,
e.g. for EcoRI 5'‑‑‑G AATTC‑‑‑3'
3'‑‑‑CTTAA G‑‑‑5'
the ends can anneal to each other. Any two fragments, regardless of their origin (animal, plant, fungal, bacterial) can be joined in vitro to form recombinant molecules (Figure 3.3).
Figure 3.3.
b. Blunt ends
(1) The restriction endonuclease cleaves in the center of the pseudopalindromic recognition site to generate blunt (or flush) ends.
(2) E.g. HaeIII GG'CC
HincII GTY'RAC
T4 DNA ligase is used to tie together fragments of DNA (Figure 3.4). Note that the annealed "sticky" ends of restriction fragments have nicks(usually 4 bp apart). Nicks are breaks in the phosphodiester backbone, but all nucleotides are present. Gapsin one strand are missing a string of nucleotides.
T4 DNA ligase uses ATP as source of adenylyl group attached to 5' end of the nick, which is a good leaving group after attack by the 3' OH. (See Chapter 5 on Replication).
At high concentration of DNA ends and of ligase, the enzyme can also ligate together blunt‑ended DNA fragments. Thus any two blunt‑ended fragments can be ligated together. Note: Any fragment with a 5' overhang can be readily converted to a blunt‑ended molecule by fill‑in synthesis catalyzed by a DNA polymerase (often the Klenow fragment of DNA polymerase I). Then it can be ligated to another blunt‑ended fragment.
Linkers are short duplex oligonucleotides that contain a restriction endonuclease cleavage site. They can be ligated onto any blunt‑ended molecule, thereby generating a new restriction cleavage site on the ends of the molecule. Ligation of a linker on a restriction fragment followed by cleavage with the restriction endonuclease is one of several ways to generate an end that is easy to ligate to another DNA fragment.
Annealing of homopolymer tails are another way to joint two different DNA molecules.
The enzyme terminal deoxynucleotidyl transferasewill catalyze the addition of a string of nucleotides to the 3' end of a DNA fragment. Thus by incubating each DNA fragment with the appropriate dNTP and terminal deoxynucleotidyl transferase, one can add complementary homopolymers to the ends of the DNAs that one wants to combine. E.g., one can add a string of G's to the 3' ends of one fragment and a string of C's to the 3' ends of the other fragment. Now the two fragments will join together via the homopolymer tails. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.02%3A_Overview_of_Recombinant_DNA_Technology.txt |
Vectors used to move DNA between species, or from the lab bench into a living cell, must meet three requirements (Figure \(1\)):
1. They must be autonomously replicating DNA molecules in the host cell. The most common vectors are designed for replicating in bacteria or yeast, but there are vectors for plants, animals and other species.
2. They must contain a selectable marker so cells containing the recombinant DNA can be distinguished from those that do not. An example is drug resistance in bacteria.
3. They must have an insertion site to accommodate foreign DNA. Usually a unique restriction cleavage site in a nonessential region of the vector DNA. Later generation vectors have a set of about 15 or more unique restriction cleavage sites.
Plasmid Vectors
Plasmids are autonomously replicating circular DNA molecules found in bacteria. They have their own origin of replication, and they replicate independently of the origins on the "host" chromosome. Replication is usually dependent on host functions, such as DNA polymerases, but regulation of plasmid replication is distinct from that of the host chromosome. Plamsids, such as the sex-factor F, can be very large (94 kb), but others can be small (2‑4 kb). Plasmids do not encode an essential function to the bacterium, which distinguishes them from chromosomes. Plasmids can be present in a single copy, such as F, or in multiple copies, like those used as most cloning vectors, such as pBR322, pUC, and pBluescript.
In nature, plasmids provide carry some useful function, such as transfer (F), or antibiotic resistance. This is what keeps the plasmids in a population. In the absence of selection, plasmids are lost from bacteria. The antibiotic resistance genes on plasmids are often carried within, or are derived from, transposons, a types of transposable element. These are DNA segments that are capable of "jumping" or moving to new locations (Chapter 9).
A plasmid that was widely used in many recombinant DNA projects is pBR322 (Figure \(2\)). It replicates from an origin derived from a colicin-resistance plasmid (ColE1). This origin allows a fairly high copy number, about 100 copies of the plasmid per cell. Plasmid pBR322 carries two antibiotic resistance genes, each derived from different transposons. These transposons were initially found in R-factors, which are larger plasmids that confer antibiotic resistance.
Use of the TcR and ApR genes allows for easy screening for recombinants carrying inserts of foreign DNA. For instance, insertion of a restriction fragment in the BamHI site of the TcR gene inactivates that gene. One can still select for ApR colonies, and then screen to see which ones have lost TcR .
Exercise \(1\)
What effects on drug resistance are seen when you use the EcoRI or PstI sites in pBR322 for inserting foreign DNA?
A generation of vectors developed after pBR322 are designed for even more efficient screening for recombinant plasmids, i.e. those that have foreign DNA inserted. The pUC plasmids (named for plasmid universal cloning) and plasmids derived from them use a rapid screen for inactivation of the b-galactosidase gene to identify recombinants (Figure \(3\)).
One can screen for production of functional b‑galactosidase in a cell by using the chromogenic substrate X‑gal (a halogenated indoyl b‑galactoside). When cleaved by b‑galactosidase, the halogenated indoyl compound is liberated and forms a blue precipitate. The pUC vector has the b‑galactosidase gene {actually only part of it, but enough to form a functional enzyme with the rest of the gene that is encoded either on the E. coli chromosome or an F' factor}. When introduced into E. coli, the colonies are blue on plates containing X‑gal.
The multiple cloning sites (unique restriction sites) are in the b‑galactosidase gene (lacZ). When a restriction fragment is introduced into one or more of these sites, the b‑galactosidase activity is lost by this insertional mutation. Thus cells containing recombinant plasmids form white(not blue) colonies on plates containing X‑gal.
The replication origin is a modified ColE1 origin of replication that has been mutated to eliminate a negative control region. Hence the copy number is very high(several hundred or more plasmid molecules per cell), and one obtains an very high yield of plasmid DNA from cultures of transformed bacteria. The plasmid has ApR as a selectable marker. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.03%3A_Introduction_of_recombinant_DNA_into_cell_and_replication%3A_Vectors.txt |
Transformation in E. coli
E. coli does not have a natural system for taking up DNA, but when treated with \(\ce{CaCl2}\), the cells will take up the added DNA (Figure \(1\)). The recombinant vectors will give a new phenotype to the cells (usually drug resistance), so this process can be considered DNA-mediated transformation. An average efficiency is about 106 transformants per mg of DNA, although some more elaborate transformation cocktails procedures can give up to about 108 transformants per mg of DNA.
Usually one will transform with a mixture of recombinant vector molecules, most of which carry a different restriction fragment. Each transformed E. coli cell will pick up only one plasmid molecule, so the complex mixture of plasmids in the ligation mix has been separated into a population of transformed bacteria (Figre \(1\)). The bacterial cells are then plated at a sufficiently low density that individual colonies can be identified. Each colony (or transformant) carries a single plasmid, so as one screens the colonies, one is actually screening through individual DNA molecules. A colony is a visible group of bacterial cells on a plate, all of which are derived from a single bacterial cell. A group of identical cells derived from a single cell is called a clone. Since each clone carries a single type of recombinant DNA molecule, the process is called molecular cloning.
Phage Vectors
Phase vectors are a more efficient introduction of DNA into bacteria. Phage vectors such as those derived from bacteriophage l can carry larger inserts and can be introduced into bacteria more efficiently. l phage has a duplex DNA genome of about 50 kb. The internal 20 kb can be replaced with foreign DNA and still retain the lytic functions. Hence restriction fragments up to 20 kb can replace the l sequences, allowing larger genomic DNA fragments to be cloned (Figure \(2\)).
Recombinant bacteriophage can be introduction into E. coli by infection. DNA that has the cohesive ends of l can be packaged in vitro into infective phage particles. Being in a viral particle brings the efficiency of infection reliably over 108 plaque forming units per mg of recombinant DNA.
Some other bacteriphage vectors for cloning are derived from the virus M13. One can obtain single stranded DNA from M13 vectors and recombinants. M13 is a virus with a genome of single stranded DNA. It has a nonessential region into which foreign genes can be inserted. It has been modified to carry a gene for b‑galactosidase as a way to screen for recombinants. Introduction of recombinant M13 DNA into E. coli will lead to an infection of the host, and the progeny viral particles will contain single‑stranded DNA. The replicative form is duplex, allowing one to cleave with restriction enzymes and insert foreign DNA.
Some vectors are hybrids between plasmids and single‑strand phage; these are called phagemids. One example is pBluescript. Phagemids are plasmids (with the modified, high-copy number ColE1 origin) that also have an M13 origin of replication. Infection of transformed bacteria (containing the phagemid) with a helper virus (e.g. derived from M13) will cause the M13 origin to be activated, and progeny viruses carrying single‑stranded copies of the phagemid can be obtained. Hence one can easily obtain either double‑ or single‑stranded forms of thes plasmids. {The "blue" comes from the blue‑white screening for recombinants that can be done when the multiple cloning sites are in the b‑galactosidase gene. The "script" refers to the ability to make RNA copies of either strand in vitro with phage RNA polymerases.}
Vectors Designed to Carry Larger Inserts
Fragments even larger than those carried in l vectors are useful for studies of longer segments of chromosomes or whole genomes. Several vectors have been designed for cloning these very large fragments, 50 to 400 kb.
• Cosmids are plasmids that have the cohesive ends of l phage. They can be packaged in vitro into infective phage particles to give a more efficient delivery of the DNA into the cells. They can carry about 35 to 45 kb inserts.
• Yeast artificial chromosomes (YACs) are yeast vectors with centromeres and telomeres. They can carry about 200 kb or larger fragments (in principle up to 1000 kb = 1 Mb). Thus very large fragments of DNA can be cloned in yeast (Figure \(3\)). In practice, chimeric clones with fragments from different regions of the genome are obtained fairly often, and some of the inserts are unstable.
Vectors derived from bacteriophage P1 can carry fragments of about 100 kb. Fragments in a similar size range are also cloned into bacterial artificial chromosomes (BACs), which are derived from the F-factor (Figure \(4\)). These have a lower copy number (like F) but they are stable and relatively easy to work with in the laboratory. BACs have become one of the most frequently used vectors for large inserts in genome projects.
Shuttle vectors for testing functions of isolated genes
Shuttle vectors can replicate in two different organisms, e.g. bacteria and yeast, or mammalian cells and bacteria. They have the appropriate origins of replication. Hence one can, e.g. clone a gene in bacteria, maybe modify it or mutate it in bacteria, and test its function by introducing it into yeast or animal cells. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.04%3A_Introducting_Recombinant_DNA_into_Host_Cells.txt |
The polymerase chain reaction (PCR) is now one of the most commonly used assays for obtaining a particular segment of DNA or RNA. It is rapid and extremely sensitive. By amplifying a designated segment of DNA, it provides a means to isolate that particular DNA segment or gene. This method requires knowledge of the nucleotide sequence at the ends of the region that you wish to amplify. Once that is known, one can make large quantities of that region starting with miniscule amounts of material, such as the DNA within a single human hair. With the availability of almost complete or complete sequences of genomes from many species, the range of genes to which it can be applied is enormous. The applications of PCR are numerous, from diagnostics to forensics to isolation of genes to studies of their expression.
The power of PCR lies in the exponential increase in amount of DNA that results from repeated cycles of DNA synthesis from primers that flank a given region, one primer designed to direct synthesis complementary to the top strand, the other designed to direct synthesis complementary to the bottom strand (Figure \(1\)). When this is done repeatedly, there is roughly a 2-fold increase in the amount of synthesized DNA in each cycle. Thus it is possible to generate a million-fold increase in the amount of DNA from the amplified region with a sufficient number of cycles. This exponential increase in abundance is similar to a chemical chain reaction, hence it is called the polymerase chain reaction.
The events in the polymerase chain reaction are examined in more detail in Figure \(2\). The several panels show what happens in each cycle. Each cycle consists of a denaturation step at a temperature higher than the melting temperature of the duplex DNA (e.g. 95 oC ), then an annealing step at a temperature below the melting temperature for the primer-template (e.g. 55 oC), followed by extension of the primer by DNA polymerase using dNTPs provided in the reaction. This is done at the temperature optimum for the DNA polymerase (e.g. 70 oC for a thermostable polymerase). Thermocylers are commercially available for carrying out many cycles quickly and reliably (Figure \(3\)).
The template supplied for the reaction is the only one available in the first cycle, and it is still a major template in the second cycle. At the end of the second cycle, a product is made whose ends are defined by primers. This is the desired product, and it serves as the major template for the remaining cycles. The initial template is still present and can be used, but it does not undergo the exponential expansion observed for the desire product.
If nis the number of cycles, the amount of desired product is approximately 2n-1 –2 times the amount of input DNA (between the primers). Thus in 21 cycles, one can achieve a million-fold increase in the amount of that DNA (assuming all cycles are completely efficient). A sample with 0.1 pg of the segment of DNA between the primers can be amplified to 0.1 mg in 21 cycles, in theory. In practice, roughly 25-35 cycles are done in many PCR assays.
The ease if doing PCR was greatly increased by the discovery of DNA polymerases that were stable at high temperatures. These have been isolated from bacteria that grow in hot springs, such as those found in Yellowstone National Park, such as Thermus aquaticus. The Taq polymerase from this bacterium will retain activity even at the high temperatures needed for melting the templates, and it is active at a temperature between the melting and annealing temperature. This particular polymerase is rather error-prone, and other thermostable polymerases have been discovered that are more accurate. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.05%3A_Polymerase_Chain_Reaction_%28PCR%29.txt |
cDNA clones are copies of mRNAs
Construction of cDNA clones involves the synthesis of complementary DNA from mRNA and then inserting a duplex copy of that into a cloning vector, followed by transformation of bacteria (Figure \(1\)).
a. First strand synthesis: First, one anneals an oligo dT primer onto the 3' polyA tail of a population of mRNAs. Then reverse transcriptase will begin DNA synthesis at the primer, using dNTPs supplied in the reaction, and copy the mRNA into complementary DNA, abbreviated cDNA. The mRNA is degraded by the RNase H activity associated with reverse transcriptase and by subsequent treatment with alkali.
b. Second strand synthesis: For the primer to make the second strand of DNA (equivalent in sequence to the original mRNA), one can utilize a transient hairpin at the end of the cDNA. (The basis for its formation is not certain.) In other schemes, one generates a primer binding site and uses a primer directed to that site; one way to do this is by homopolymer tailing of the cDNA followed by use of a complementary primer. Random primers can also be used for second strand synthesis; although this precludes the generation of a full-length cDNA (i.e. a copy of the entire mRNA). However, it is rare to generate duplex copies of the entire mRNA by any means.
DNA polymerase (e.g. Klenow polymerase) is used to synthesize the second strand, complementary to the cDNA. The product is duplex cDNA.
If the hairpin was used to prime second strand synthesis, it must be opened by a single‑strand specific nuclease such as S1.
c. Insertion of the duplex cDNA into a cloning vector:
One method is to use terminal deoxynucleotidyl transferase to add a homopolymer such as poly-dC to the ends of the duplex cDNA and a complementary homopolymer such as poly-dG to the vector.
An alternative approach is to use linkers; these can be employed such that a linker carrying a cleavage site for one restriction endonuclease is on the 5' end of the duplex cDNA and a linker carrying a cleavage site for a different restriction endonuclease is on the 3' end. (In this context, 5’ and 3’ refer to the nontemplate, or "top" strand.) This allows "forced" cloning into the vector, and one has initial information about orientation, based on proximity to one cleavage site or the other.
The cDNA and vector are joined at the ends, using DNA ligase, to form recombinant cDNA plasmids (or phage).
d. The ligated cDNA plasmids are then transformed into E. coli. The resulting set of transformants is a library of cDNA clones.
Screening methods for cDNA clones
a. Brute force examination of individual cDNA plasmids.
If the mRNA is highly abundant in a given tissue, then many of the cDNA clones will be copies of that mRNA. One can examine DNA from individual clones and test for characteristic restriction cleavage patterns or a particular sequence. This was a common approach for screening cDNAs in the early days of recombinant DNA technology.
Starting in the mid-1990’s, cooperative efforts from corporations (such as Merck) and publicly funded genome centers (such as at Washington University) have generated the sequence of individual clones from large cDNA libraries from many tissues from human, mouse, and rat. Other consortia have sequenced cDNA libraries from other species. Each sequence is called an “expressed sequence tag” or EST. These are now a major source of partially or fully characterized cDNA clones. Hundreds of thousands of ESTs are available, and contain at part of the DNA sequence from many, if not most, human genes. The web site for NCBI (http://www.ncbi.nlm.nih.gov) is an excellent resource for examining the ESTs.
b. Hybridization with a gene‑specific probe.
If the sequence of the desired cDNA is known, or if the sequence from homologs from related species is known, one can use synthetic oligonucleotides (or other source of the diagnostic sequence) as a radiolabeled hybridization probe to identify the cDNA of interest.
If the amino acid sequence has been determined for all or even just parts of the protein product of the gene of interest, then one can chemically synthesize oligonucleotides based on the genetic code for those amino acids. The oligonucleotides need to be at least 18 nucleotides or longer (so that they will anneal to specific sites in the genome), and because the genetic code is degenerate (more than one codon per amino acid; discussed in Part Two), they have to be degenerate as well. The oligonucleotides can be used directly as hybridization probes, although it is becoming more common to amplify the region between two oligonucleotides using the polymerase chain reaction, and to use that amplification product as a labeled probe.
The process of hybridization screening is illustrated schematically in Figure 3.16. The colonies of bacteria, each with a single cDNA plasmid, are transferred to a solid substrate (such as a nylon or nitrocellulose membrane), lysed. and the released DNA immobilized onto the membrane. Hybridization of this membrane (with the DNA attached) to a specfic probe allows one to screen through thousands of colonies in a single experiment.
c. Express the cDNA, i.e. make the protein product encoded by the mRNA, and screen for that protein product (Figure \(3\)). This is often in bacteria by constructing the clones in a vector that has an active E. coli promoter (for transcription) and efficient translation signals upstream from the site at which the cDNAs were inserted. The transformed bacterial cells will express the encoded protein, and one tries to identify it. One can also screen for expression in yeast, plant or mammalian cells. The expression vector has to contain gene-regulatory signals (such as promoters and enhancers, see Part Three) that allow expression of the desired gene in the appropriate cell.
1. One can use specific antiserato detect the desired colony expressing the gene of interest.
2. One can use a labeled ligand that will bind to the expressed cDNA on the cell surface. For example, cDNAs for receptors can be expressed in an appropriate cell (usualy mammalian cells in culture) and identified by newly-acquired ability to bind a labeled hormone (such as growth hormone or erythropoietin)
3. by complementationof a known mutation in the host. E.g. a cDNA for the human homolog to yeast p34cdc2 was isolated by its ability to complement a yeast mutant that had lost the function of this key regulator of progress through the cell cycle.
4. Expression cloning can be done in mammalian cells, as long as one can screen or select for a new function generated by the expression. Use of this method to isolate the receptor for the glycoprotein hormone erythropoietin is illustrated in Figure \(4\).
d. Differential analysis:
Often one is interested in finding all the genes (or their mRNAs) that are expressed uniquely in some differentiated or induced state of cells. Two classic examples are (i) identifying the genes whose products regulate the determination process that causes a multipotential mouse cell line (like 10T1/2 cells) to differentiate into muscle cells, and (ii) ,using the fact that the T-cell receptor is expressed only in T-lymphocytes, but not in their sister lineage B-lymphocytes, to help isolate cDNA clones for that mRNA. Both of these projects used subtractive hybridization to highly enrich for the cDNA clones of interest.
In this technique, the cDNA from the differentiating or induced cell of interest is hybridized to mRNA from a related cell line, but which has not undergone the key differentiation step. This allows one to remove mRNA-cDNA duplexes that contain the cDNAs for all the genes expressed in common between the two types of cells. The resulting single-stranded are enriched for the cDNAs that are involved in the process under study.
The subtractive hybridization scheme used in isolation of the muscle determination gene MyoDis illustrated in Figure \(5\).
A conceptually equivalent strategy, using PCR (see next section) rather than cDNA cloning, is differential display of PCR products from cells that differ by some process (e.g. differentiation, induction, growth arrest versus stimulation, etc.). In this technique, one uses several sets of PCR primers annealed to cDNA to mRNA from the two types of cells that are being compared. The sets of primers are empirically designed to allow many regions of cDNA to be amplified. The amplification products are resolved (or displayed) on polyacrylamide gels, and the products specific to the cell type of interest are isolated and used to screen through cDNA libraries. This technique is also called representational difference analysis.
The advent of sequencing all or a very large number of genes from various organisms (e.g. E. coli, yeast, Drosophila, humans) has allowed the development of high-density microchip arrays of DNA from each gene. One can hybridize RNA from cells or tissues of interest, isolated under various metabolic conditions, to identify all (known) genes expressed. Even more useful are assays for genes whose expression changes during a shift in cell metabolism (cell cycle, heat shock, hormonal induction, etc.) or as a result of mutation of some other gene (e.g. a gene encoding a transcription factor of interest). This powerful new technology is being used more and more to examine global effects on gene expression.
For a description (and movie) of the Affymetrix GeneChip, go to http://www.affymetrix.com/technology/index.html | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.06%3A_cDNA.txt |
Clones of genomic DNA, containing individual fragments of chromosomal DNA, are needed for many purposes. Some examples include:
• to obtain detailed structures of genes,
• to identify regulatory regions, i.e. DNA sequences needed for correct expression of the gene,
• to map and analyze alterations to the genome, e.g. the isolate genes that when mutated cause a hereditary disease,
• to direct alterations in the genome, e.g. by homologous recombination to replace a wild-type allele with a mutant one (to test function of the gene in mouse) or vice versa(to cure a hereditary disease, perhaps eventually in humans).
Construction of libraries of genomic DNA fragments in cloning vectors
Genomic DNA is digested with restriction enzymes (Figure \(1\)). The more frequently an enzyme cuts (the shorter the recognition sequence), the smaller the average size of DNA fragments. Some enzymes cut very infrequently, such as NotI (8 bp recognition sequence) and can be used to generate very large fragments. Alternatively, one can do a partial digest (not all sites are cleaved) with a particular enzyme and isolate the products that are in the desired size range (e.g. 20 kb). A particularly clever way to do this is to digest partially with Sau3AI or MboI (both cut at 'GATC) and ligate these fragments into vector cut with BamHI (cuts at G'GATCC) ‑ i.e. they have the same sequence in the overhang (or sticky end). In this process one uses vectors that can accomodate large DNA fragments, such as l phage vectors, cosmids, YACs or P1 vectors.
Screening methods for genomic DNA clones
One method is to use complementation of a mutation in the host to select or screen for the desired gene. This works just like the situation for cDNA clones described above, and it requires that the cloned fragments be expressed in the host cell. Far more common is to screen by hybridization with gene‑specific probes (Figure \(2\)). Frequently the cDNA clone is found first, and the genomic clone then isolated by hybridization screening (using the cDNA clone as a probe) against a library of genomic DNA fragments.
3.08: Eukaryotic gene structure
Much can be learned about any gene after it has been isolated by recombinant DNA techniques. The structure of coding and noncoding regions, the DNA sequence, and more can be deduced. This is true for bacterial and viral genes, as well as eukaryotic cellular genes. The next sections of this chapter will focus on analysis of eukaryotic genes, showing the power of examining purified copies of genes.
Split genes and introns
Precursors to mRNA longer than mRNA
Initial indications of a complex structure to eukaryotic genes came from analysis of nuclear RNAs during the 1970’s. The precursors to messenger RNA, or pre-mRNAs, were found to be surprisingly long, considerably larger than the average mRNA size (Figure \(1\)).
Denaturing sucrose gradients (with high concentration of formamide, e.g. >50%) separate RNAs on the basis of size. Analysis of nuclear RNA showed that the average size was much larger than the average size of cytoplasmic RNA. Labeled RNA could be "chased" from the nucleus to the cytoplasm ‑ i.e. nuclear RNA was a precursor to mRNA and other cytoplasmic RNAs. Was the extra RNA at the ends? or in the middle of the pre‑mRNA? More precisely, one could examine specific RNAs by hybridizing fractions from the denaturing sucrose gradients to labeled copies of, e.g. globin mRNA. The hybridizing RNA from the nucleus was about 11S (as well as mature 8S message), whereas cytoplasmic RNA of about 8S hybridized. Thus the nuclear RNA encoding globin is larger than the cytoplasmic mRNA.
Visualization of mRNA-DNA heteroduplexes revealed extra sequences internal to the mRNA-coding segments
R-loops are hybrids between RNA and DNA that can be visualized in the EM, under conditions where DNA‑RNA duplexes are favored over DNA‑DNA duplexes (Figure \(2\)). For a simple gene structure, one sees a continuous RNA‑DNA duplex (smooth, slowly curving) and a displaced single strand of DNA (thinner, many more turns and curves – single stranded DNA is not a rigid as double stranded nucleic acid, either duplex DNA or RNA-DNA).
EM pictures of duplexes between purified adenovirus mRNAs and the genomic DNA showed extensions at both the 3' (poly A) and 5' ends, which are encoded elsewhere on the genome. All late mRNAs have the same sequence at the 5' end; this is dervied from from the tripartite leader. R‑loops between late mRNAs and adenovirus DNA fragments including the major late promoter showed duplexes with the leader segments, separated by loops of duplex DNA (Figure 3.23, bottom panel). The RNA-DNA hybrids identify regions of DNA that encode RNA. The surprising result is that RNA-coding portions of a gene are separated by loops of duplex DNA in the R-loop analysis. Examples of R-loops in genes with introns are shown in Figure \(3\).
These data showed that the adenovirus RNAs are encoded in different segments of the viral genome; i.e. the genes are split. The portion of a gene that encodes mRNA was termed an exon. The part of gene does not code for sequences in the mature mRNA is called an intron. These observations led to the Nobel Prize for Phil Sharp and Rich Roberts. Louise Chow and Sue Berget were also key players in the discovery of introns.
Interruptions in cellular genes were discovered subsequently, in the late 1970's, in globin genes, immunoglobulin genes and others. We now realize that mostgenes in complex eukaryotes are split by multiple introns.
Exons are more conserved than introns (in most cases), since alterations in protein-coding regions that alter or decrease function are selected against, whereas many sequences in introns can be altered without affecting the function of the gene product. Important sequences in introns (such as splice junctions, the branch point, and occassionally enhancers) are covered in some detail in Part Three.
Differences in restiction maps between cDNA and genomic clones reveal introns
Restriction maps based on copies of the mRNA (cDNA) were different from those in genomic DNA ‑ the genes were cleaved by some restriction endonucleases that the cDNAs were not, and some restriction sites were further apart in the genomic DNA. These observations were explained by the presence of intervening sequences or introns (Figure \(4\)).
The experimental procedures to do this involve making a restriction map of the clones of genomic DNA, and then identifying the regions that encode mRNA by hybridization of labeled cDNA probes to the restriction digests. Cloned genomic DNA digested with appropriate restriction endonucleases, separated by size on an agarose gel, and then transferred onto a nylon or nitrocellulose solid support. This Southern blot is then hybridized with a labeled probe specific to the cDNA (composed only of exons). The pattern of labeled fragments on the resulting autoradiogram shows the fragments that contain exons. Alignment of these with the restriction map of the gene gives an approximation of the position of the exons.
The blot-hybridization approach can be combined with a PCR (polymerase chain reaction) analysis for higher resolution. Primers are synthesized that will anneal to adjacent exons. The difference in size of the PCR amplification product between genomic DNA and cDNA is the size of the intron. The PCR product can be cloned and sequenced for more detailed information, e.g. to precisely define the exon/intron junctions.
Subsequently, the nucleotide sequence of exonic regions and preferably the entire gene is determined. The presence of introns were confirmed and their locations defined precisely in DNA sequences of isolated clones of the genes.
Types of Exons
Eukaryotic genes are a combination of introns and exons. However, not all exons do the same thing (Figure \(5\)). In particular, the protein-coding regions or genes are a subset of the sequences in exons. Exons include both the untranslated regions and the protein-coding, translated regions. Introns are the segments of genes that are present in the primary transcript (or precursor RNA) but are removed by splicing in the production of mature RNA. Methods used to detect coding regions will not find all exons. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.07%3A_Genomic_DNA_clones.txt |
Multiple, large introns can make some eukaryotic genes very large. Eukaryotic genes can be split into many (>60), sometimes very small exons (e.g. <60 bp, coding for <20 amino acids), separated by very large introns (as large as >100kb), resulting in some enormous genes (>500 kb). E.g. the DMDgene (which when mutated can cause Duchenne's muscular dystrophy) is almost 1 Mb, about 1/4 the size of the E. coli chromosome! The average size of genes from more complex organisms is considerably larger than those of simpler ones, but the avg. size of mRNA is about the same, reflecting the presence of more and larger introns in the more complex organisms. tRNA and rRNA genes also contain introns
Finding exons in long genomic sequences using computer programs
Far more exons and introns have been discovered (or more accurately, predicted) throught the analysis of genomic DNA sequences than could ever be discovered by direct experimentation. The different types of exons, the enormous length of introns, and other factors have complicated the task of finding reliable diagnostic signatures for exons in genomic sequences. However, considerable progress has been made and continues in current research. Some of the commonly used approaches are summarized in Figure 3.27.
Introns are removed by splicing RNA precursors
Alternative splicing generates more than one polypeptide from the same gene
Some segments of RNA may be included in the mature mRNA (exons) but not included on other spliced products. The alternative products may be made in different tissues or at different developmental stages ‑ i.e. alternative splicing can be regulated.
Split genes may enhance the rate of evolution
Many exons encode a unit very close to a protein domain, e.g. the exons of leghemoglobin, or the variable and constant regions of immunoglobulins, or domains (e.g. "kringle") in EGF precursor that are also found in part of the LDL receptor. The exon organization tends to be well conserved in highly divergent species. Introns tend to occur between those portions of genes that encode structural domains of proteins.
Duplication of the exons encoding structural domains and subsequent recombination can lead to more rapid evolution of a new protein, essentially using the parts from earlier evolved genes. Analogous to building a house from prefabricated parts, as opposed to one nail and one board at a time ‑ start with preassembled walls, roof joists etc.
However, the relationship between exons and structural domains of proteins is not exact, and some exon‑intron boundaries vary (a little) in genes for different species. A different model holds that the introns are transposable elements (some certainly are ‑ see later). They can insert anywhere in a gene, but they are least disruptive at domain boundaries, and these latter insertions are more likely to be fixed in a population than insertions into the middle of a region encoding a domain. So the results after long years of evolution is that the introns tend to be between region coding domains, but the gene was originally intact, not assembled from discrete exons.
Multigene families and gene clusters
Many eukaryotic genes are found in multiple copies. Some of them are developmentally regulated, such as HOXgene clusters and globin gene clusters .
A multigene family contains multiple genes of similar sequence encoding similar proteins; e.g. globin genes (Figure 3.30). Globin genes are expressed at different times of development. The order of developmental expression is the same as their order along the chromosome, e.g. the e-globin gene is expressed in early embryonic red cells, the g-globin gene is expressed at a high level in fetal red cells, and the b-globin gene is expressed in red cells after birth. As we will see later, this correlates with their distance from a dominant control element at the 5' end of the cluster, the Locus Control Region.
The order of HOXgenes is also aligned with their spatial expression in the embryo. This is another example of alignment between chromosomal position and regulation of expression.
Other multi‑gene families include those encoding histones, immunoglobulins, actins, cyclins, cyclin‑dependent protein kinases, and rRNAs. Some of these families are linked in gene clusters, but others are dispersed around the genome. Having multiple copies of genes may be more the rule than the exception in eukaryotic genomes.
Experimental techniques that reveal multigene families include the following.
Purification and analysis of a particular kind of protein, e.g. hemoglobins, immunoglobulins, and many enzymes, may reveal heterogeneity. Further purification (via chromatography and electrophoresis) and sequencing can show that the observed heterogeneity is a result of related but not identical proteins, and one deduces that these similar proteins are encoded by multiple genes with similar sequences, i.e. a multigene family.
Analysis of the clones obtained by screening a library of cloned genomic DNA may reveal multiple related sequences, each with a distinctive restriction map. In many cases these are clones of different, related genes that comprise a multigene family (Figure 3.31).
Southern blot‑hybridization of restriction‑cleaved genomic DNA can reveal multiple copies of genes, simply as multiple bands on the hybridized blot. Although the number of fragments generated from total genomic DNA is too many to resolve on a gel, after transfer to a membrane, particular fragments can be visualized by hybridization with a specific probe. The number of hybridizing fragments is roughly correlated with the number of copies of related genes. Some genes are cleaved by the restriction enzyme, producing multiple bands, but some fragments can have multiple genes. A true measure of the number of related genes comes from more detailed restriction mapping or sequencing.
Keeping multigene families homogeneous
Sometimes multiple copies of genes are maintained as virtually identical over the course of evolution: e.g. rRNA genes, histone genes, a‑globin genes (in primates). In these cases, the multiple copies are coevolving(concerted evolution).
sequence differences
Human: A | A | A | among human genes: 1%
between human & chimp5%
Chimp: A | A | A | among chimp genes: 1%
between chimp & monkey 10%
Monkey: A | A | A | among monkey genes: 1%
Since all three primates have 3 A genes, we infer that the common ancestor had 3 genes (the duplications preceded the speciation events). If in the time since human and chimp diverged, the A genes have diverged 5%, why haven't the A genes in human (e.g.) also diverged 5% from each other? They have been apart even longer than the human and chimp chromosomes carrying them! The A genes within a species are "talking to each other", or co‑evolving or evolving in concert.
Sequence homogeneity in a multigene family can arise because of recent gene amplification (Figure 3.32 part1). In this case the genes have not been separate from each other long enough to accumulate variation in their sequences. Other multigene families have existed for a long time, but maintain sequence homogeneity despite ample opportunity for divergence. Two mechanisms have been seen that maintain similarity. The first is multiple rounds of unequal crossing over. As illustrated in Figure 3.32, part 2, the expansions and contractions of repeated genes can result in a new variant predominanting in the gene cluster. The other method for maintaining homogeneity is gene conversion between homologs. When a new mutation arises, it can be removed by conversion with the unmutated allele, or the mutation can be passed on the the other allele. Either way, the sequences of the two alleles becomes the same.
Sometimes the products of the gene duplications, or duplicative transpositions, accumulate mutations so they are no longer functional. These remnants of once‑active genes are called pseudogenes. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.09%3A_Introns_and_Exons.txt |
Gene expression
"Northern blots" or RNA blot‑hybridization
In the reverse of Southern blot‑hybridizations, one can separate RNAs by size on a denaturing agarose gel, and transfer them to nylon or other appropriate solid support. Labeled DNA can then be used to visualize the corresponding mRNA (Figure 3.33). Ed Southern initially used labeled rRNA to find the complementary regions in immobilized, digested DNA, so this "reverse" of Southern blot-hybridizations, i.e. using a labeled DNA probe to hybridize to immobilized RNA, is often referred to as "Northern" blot‑hybridizations.
One can hybridize a labeled DNA clone to a panel of RNA samples from a wide variety of tissues to determine in what tissues a particular cloned gene is expressed (top panel of Figure 3.33. More precisely, this technique reveals the tissues in which the genes is transcribed into stable RNA. The results allow one to determine the tissue specificity of expression, e.g. a gene may only be expressed in liver, or only in erythroid cells (e.g. the b-globin gene). This helps give some general idea of the possible function of the gene, since it should reflect the function of that tissue. Other genes are expressed in almost all cells or tissue types (such as GAPDH); these are referred to as housekeeping genes. They are involved in functions common to all cells, such as basic energy metabolism, cell structure, etc. The relative amounts of RNA in the different lanes can be directly compared to see, e.g., which tissues express the gene most abundantly.
One can hybridize a labeled DNA clone to a panel of RNA samples from a progressive stages of development to determine the developmental stagewhen during development a particular cloned gene is expressed as RNA (bottom panel of Figure 3.33). For instance, a gene product may be required for determination decisions early in development, and only be expressed in early embryos.
Once the DNA sequence of the gene of interest is known, and its intron-exon structure determined, highly sensitive RT-PCR assayscan be designed (Figure 3.34). The RNA from the cell or tissue of interest is copied into cDNA using reverse transcriptase and dNTPs, and then primers are annealed for PCR. Ideally, the primers are in different exons so that the product of amplifying the cDNA will be smaller than the product of amplifying the genomic DNA.
In situ hybridizations / immunochemistry
In complementary approaches, the labeled DNA can be hybridized in situto thin sections of a tissue or embryo or other specimen, and the resulting pattern of grains visualized along the specimen in the microscope (Figure 3.35). Also, antibody probes against the protein product can be used to localize it in the specimen. This gives a more detailed picture of the pattern of expression, with resolution to the particular cells that are expressing the gene. The RNA blot-hybridization techniques described in a. above look at the RNA in all the cells from a tissue, and do not provide the level of resolution to single cells.
Microarrays
As large numbers of sequenced mRNAs and genes become available, technology has been developed to look at expression of very large numbers of genes simulatneously. DNA sequences specific for each gene in a bacterium or yeast can be spotted in a high-density array with 400 r more spots. Some technologies use many more spots, with mutliple sequences per gene. Microarrays, or “gene chips” are available for many species, some with tens of thousands of different sequences or “probes.” RNA from different tissues can be converted to cDNA with a distinctive fluorescent label, and then hybridized to the gene chip. Differences in level of expression can be measured. Thus global changes in gene expression can now be measured.
Database searches
An increasingly powerful approach is to determine candidates for the the function of your gene by searching the databaseswith the sequence, looking for matches to known proteins and genes. These matches provide clues as to protein function.
The power of this approach increases as the amount of sequences deposited in databases expand. Sequences of many genes are already known. The sequenced genes from more complex organisms, such as plants and animals, tend to be the ones more easily isolated using the techniques discussed in recombinant DNA technology. However, the sequences of genes expressed at a low level are starting to accumulate in the databases.
One remarkable advance in the past few years is the increasing number of organisms whose entire genome has been sequenced. About 10 bacterial genomes have been sequenced, and the number increases every few months. Genomics sequences for two eukaryotes are now available. That of the yeast Saccharomyces cerevisiaehas been known for a few years, and the genome of the nematode Caenorhabditis eleganswas completed in 1998. These sequences are being analyzed intensively, and a very high fraction of all the genes in each genome can be reliably detected using computational tools (one part of bioinformatics). It has become clear that many of the enzymes used in basic metabolism, regulation of the cell cycle, cellular signaling cascades, etc. are highly conserved across a broad phylogenetic spectrum. Thus it is common to find significant sequence matches in the genomes of model organisms when they are queried by the sequence of a previously unknown gene, e.g. from humans or mouse. The function already established for that gene in worms or yeast is a highly reliable guide to the function of the homologous gene in humans. The worm C. elegansis multicellular, and fate of each of its cells during development has been mapped. Thus it is possible that many functions involved in cellular interactions and cell-cell signaling will be conserved in this species, thus expanding the list of potential targets for a search in the databases.
This potential is being realized as working draft sequences of the human and mouse genomes are being analyzed. Within these data is a good approximation of sequences from virtually all human and mouse genes. Random clones have been partially sequenced from libraries of cDNAs from various human tissues, normalized to remove much of the products of abundant mRNAs and thus increasing the frequency of products of rare mRNAs. These sequences from the ends of the cDNA clones are called expressed sequence tags, or ESTs. The name is derived from the fact that since they are in cDNA libraries, they are obviously expressed at the level of mRNA, and some are used as tags in generating high-resolution maps of human chromosome. Hundreds of thousands of these have now been sequenced in collaborative efforts between pharmaceutical companies, other companies and universities. The database dbEST records all those in the public domain, and it is a strong complement to the databases recording all known sequences of genes. Many different parts of the same, or highly related, cDNAs, are recorded as separate entries in dbEST. Projects are underway to group all the sequences from the same (or highly related) gene into a a unified sequence. One example is the Unigene project at NCBI. The number of entries grows continually, but in the summer of 1998 there are about 50,000 entries, each representing about one gene. The number is higher now. Current estimates of the number of human genes are around 30,000, so it is possible that some UniGene clusters represent only parts of genes, and some genes match more than one cluster.
Very efficient search engines have been designed for handling queries to these databases, and several are freely available over the World Wide Web. One of the most popular and useful sites for this and related activities is maintained by the National Center for Biotechnology Information (http://www.ncbi.nlm.nih.gov/). Their Entrez browser provides integrated access to sequence, mapping and some functional information, PubMed provides access to abstracts of papers in journals in the National Library of Medicine, and the BLAST server allows rapid searches through various sequence databases. dbEST and the Unigene collection are maintained here, many genome maps are available, and three-dimensional structures of proteins and nucleic acids are available.
Make the protein product and analyze it
It is often possible to express the gene and make the encoded protein in large amounts. The protein can be purified and assayed for various enzymatic or other activities. Hypotheses for such activities may come from database searches.
Directed mutation
The previously describe approaches give some idea about gene function, but they do not firmly establish those functions. Indeed, this is a modern problem of trying to assign a function to an isolated gene. Several “reverse genetic” approaches can now be taken to tackle this problem. The most powerful approach to determining the physiological role(s) of a gene product is to mutate the gene in an appropriate organism and search for an altered phenotype.
The easiest experiment to do, but sometimes most difficult to interpret, is a gain of function assay. In this case, one forces expression of the gene in a transgenic organism, which often already has a wild type copy of the gene. One can look for a phenotype resulting from over-expression in tissues where it is normally expressed, or ectopic expression in tissues where it is normally silent.
In some organisms, it is possible to engineer a loss of functionof the gene. The most effective method is to use homologous recombination to replace the wild type gene with one engineered to have no function. This knock-outmutation will prevent expression of the endogenous gene and one can see the effects on the whole organism. Unfortunately, the efficiency of homologous recombination is low in many organisms and cell lines, so this is not always feasible. Other methods for knocking out expression are being developed, although the mechanism for their effect (when successful) is still being studied. In some cases, one can block expression of the endogenous gene by forcing production of antisenseRNA. Another method that is effective in some, but currently not all organisms, is the use of double-stranded, interfering RNA (RNAi). Duplex RNAs less than 30 nucleotide pairs long from the gene of interest can prevent expression of genes in worms, flies, and plants. Some success in mammals was recently reported.
Another way to generate a loss-of-function phenotype is to express dominant negative alleles of the gene. These mutant alleles encode stable proteins that form an aberrant structure that prevents functioning of the endogenous protein. This usually requires some protein-protein interaction (e.g. homodimers or heterodimers).
Localization on a genetic map
Sometimes the gene you have isolated maps to a region on a chromosome with a known function. Of course, many genes are probably located in that region, so it is critical to show that a candidate gene really is the one that when mutated causes an altered phenotype. This can be done by showing that a wild type copy of the candidate gene will restore a normal phenotype to the mutant. If a marker is known to be very tightly linked to the candidate gene, one can test whether this marker is always in linkage disequilibrium with the determinant of the mutant phenotype, i.e. in a large number of crosses, the marker for the candidate gene and the mutant phenotype never separated by recombination.
The mapping is often done with gene‑specific probes for in situ hybridizations to mitotic chromosomes. One then aligns the hybridization pattern with the chromosome banding patterns to map the isolated gene. Another method is to hybridize to a panel of DNAs from hybrid cells that contain only part of the chromosomal complement of the genome of interest. This is particularly powerful with radiation hybrid panels. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.10%3A_Functional_analysis_of_isolated_genes.txt |
3.2 Altering the ends of DNA fragments for ligation into vectors.
(Adapted from POB)
a) Draw the structure of the end of a linear DNA fragment that was generated by digesting with the restriction endonuclease EcoRI. Include those sequences remaining from the EcoRI recognition sequence.
b) Draw the structure resulting from the reaction of this end sequence with DNA polymerase I and the four deoxynucleoside triphosphates.
c) Draw the sequence produced at the junction if two ends with the structure derived in (b) are ligated.
d) Design two different short synthetic DNA fragments that would permit ligation of structure (a) with a DNA fragment produced by a PstI restriction digest. In one of these synthetic fragments, design the sequence so that the final junction contains the recognition sequences for both EcoRI and PstI. Design the sequence of the other fragment so that neither the EcoRI nor the PstI sequence appears in the junction.
3.3. What properties are required of vectors used in molecular cloning of DNA?
3.4. A student ligated a BamHI fragment containing a gene of interest to a pUC vector digested with BamHI, transformed E. coliwith the mixture of ligation products and plated the cells on plates containing the antibiotic ampicillin and the chromogenic substrate X‑gal. Which colonies should the student pick to find the ones containing the recombinant plasmid (with the gene of interest in pUC)?
3.5. Starting with an isolated mRNA, one wishes to make a double stranded copy of the mRNA and insert it at the PstI site of pBR322 via G-C homopolymer tailing. One then transforms E. coliwith this recombinant plasmid, selecting for tetracycline resistance. What are the four enzymatic steps used in preparing the cDNA insert? Name the enzymes and describe the intermediates.
3.6 A researcher needs to isolate a cDNA clone of giraffe actin mRNA, and she knows the size (Mr = 42,000) and partial amino acid sequence of giraffe actin protein and has specific antibodies against giraffe actin. After constructing a bank of cDNA plasmids from total mRNA of giraffe fibroblasts (dG-dC tailed into the PstI site of pBR322), what methods of screening the bank could be used to identify the actin cDNA clone?
3.7 The restriction map of pBR322 is
The distance in base pairs between restriction sites is as follows:
PstI to EcoRI 750 bp
EcoRI to HindIII 50 bp
HindIII to BamHI 260 bp
BamHI to PstI 3300 bp
A recombinant cDNA plasmid, pAlc-1, has double-stranded cDNA inserted at the PstI site of pBR322, using a technique that retains this cleavage site at both ends of the insert. Digestion of pBR322 and pAlc-1 with restriction endonucleases gives the following pattern after gel electrophoresis (left). The sizes of the fragments are given in base pairs. The DNA fragments were transferred out of the gel onto nitrocellulose and hybridized with radiolabeled cDNA from wild-type A. latrobus(a Southern blot-hybridizaton). Hybridizing fragments are shown in the autoradiogam diagram on the right.
a) What is the size of the cDNA insert?
b) What two restriction endonucleases cleave within the cDNA insert?
c) For those two restriction endonucleases, each DNA fragment in the single digest is cut by PstI into two DNA fragments in the double digest (i.e. the restriction endonuclease plus PstI). Determine which fragments each single digest fragment is cut into, and use this information to construct a map.
d) Draw a restriction map for pAlc-1, showing sites for PstI, EcoRI, BamHI and HindIII. Indicate the distance between sites and show the cDNA insert clearly.
3.8. You isolate and clone a KpnI fragment from A. latrobusgenomic DNA that encodes the mRNA cloned in pAlc-1 (as analyzed in question 3.7). The restriction map of the genomic fragment is
Each fragment that hybridizes to pAlc-1 is indicated by an asterisk. What does this map, especially when compared to that in problem 3.7, tell you about the structure of the gene? Be as quantitative as possible.
3.9. Some particular enzyme is composed of a polypeptide chain of 192 amino acids. The gene that encodes it has 1,440 nucleotide pairs. Explain the relationship between the number of amino acids in this polypeptide and the number of nucleotide pairs in its gene.
3.10. When viewed in the electron microscope, a hybrid between a cloned giraffe actin gene (genomic DNA) and mature actin mRNA looks like this:
What can you conclude about actin gene structure in the giraffe?
3.11. DNA complementary to pepper mRNA was synthesized using oligo (dT) as a primer for first strand synthesis. The second strand (synonymous with the mRNA) was then synthesized, and the population of double stranded cDNAs were ligated into a plasmid vector using a procedure that leaves PstI sites flanking the cDNA insert (i.e. the terminal PstI sites for each clone are not part of the cDNA). This cDNA library was screened for clones made from the mRNA from the pepper yellow gene. One clone was isolated, and subsequent analysis of the pattern of restriction endonuclease cleavage patterns showed it had the following structure:
The map shows the positions of restriction endonuclease cleavage sites and the distance between them in kilobases (kb). The map of the cDNA insert is shown with solid lines, and plasmid vector DNA flanking the cDNA is shown as dotted lines. The top strand is oriented 5' to 3' from left to right, and the bottom strand is oriented 5' to 3' from right to left. The positions and orientations of two oligonucleotides to prime synthesis for sequence determination are shown, and are placed adjacent to the strand that will be synthesized in the sequencing reaction.
a) Oligonucleotides that anneal to the plasmid vector sequences that flank the duplex cDNA insert were used to prime synthesis of DNA for sequencing by the Sanger dideoxynucleotide procedure. A primer that annealed to the vector sequences to the left of the map shown above generated the sequencing gel pattern shown below on the left. A primer that annealed to the vector sequences to the right of the map shown above generated the sequencing gel pattern shown below on the right. The gels were run from the negative electrode at the top to the positive electrode at the bottom, and the segment presented is past the PstI site (i.e. do not look for a PstI recognition site).
Left primer Right primer
G
A
T
C
G
A
T
C
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a) What is the DNA sequence of the left and right ends of the insert in the cDNA clone? Be sure to specify the 5' to 3' orientation, and the strand (top or bottom) whose sequence is reported. The terms left, right, top and bottom all refer to the map shown above for the cDNA clone.
b) Which end of the cDNA clone (left or right in the map above) is most likely to include the sequence synonymous with the 3' end of the mRNA?
c) What restriction endonuclease cleavage sites do you see in the sequencing data given?
3.12. Genomic DNA from the pepper plant was ligated into EcoRI sites in a l phage vector to construct a genomic DNA library. This library was screened by hybridization to the yellowcDNA clone. The pattern of EcoRI cleavage sites for one clone that hybridized to the yellowcDNA clone was analyzed in two experiments.
In the first experiment, the genomic DNA clone was digested to completion with EcoRI, the fragments separated on an agarose gel, transferred to a nylon filter, and hybridized with the radioactive yellowcDNA clone. The digest pattern (observed on the agarose gel) is shown in lane 1, and the pattern of hybridizing fragments (observed on an autoradiogram after hybridization) is shown in lane 2. Sizes of the EcoRI fragments are indicated in kb. The right arm of this l vector is 6 kb long, and the left arm is 30 kb.
In the second experiment, the genomic DNA clone was digested with a range of concentrations of EcoRI, so that the products ranged from a partial digest to a complete digest. The cleavage products were annealed to a radioactive oligonucleotide that hybridized only to the right cohesive end (cossite) of the l vector DNA. This simply places a radioactive tag at the right end of all the products of the reaction that extend to the right end of the l clone (partial or complete); digestion products that do not include the right end of the l clone will not be seen. The results of the digestion are shown above, on the right. Lane 1 is the clone of genomic DNA in l that has not been digested, lane 5 is the complete digest with EcoRI, and lanes 2, 3 and 4 are partial digests using increasing amounts of EcoRI. The sizes of the radioactive DNA fragments (in kb) are given, and the density of the fill in the boxes is proportional to the intensity of the signal on the autoradiogram.
a) What is the map of the EcoRI fragments in the genomic DNA clone, and which fragments encode mRNA for the yellowgene? You may wish to fill in the figure below; the left and right arms of the l vector are given. Show positions of the EcoRI cleavage sites, distances between them (in kb) and indicate the fragments that hybridize to the cDNA clone.
EcoRI EcoRI
Left arm ___| |_ Right arm
(30 kb) (6 kb)
In a third experiment, the pepper DNA from the genomic DNA clone was excised, hybridized with yellowmRNA under conditions that favor RNA-DNA duplexes and examined in the electron microscope to visualize R-loops. A pattern like the following was observed. The lines in the figure can be duplex DNA, RNA-DNA duplexes and single-stranded DNA.
b) What do the R-loop data indicate? Please draw an interpretation of the R-loops, showing clearly the two DNA strands and the mRNA and distinguishing between the template (bottom, or message complementary) and nontemplate (top, or message synonymous) strands.
The EcoRI fragments that hybridize to the yellowcDNA clone were isolated and digested with SalI (S in the figure below), HindIII (H), and the combination of SalI plus HindIII (S+H). The resulting patterns of DNA fragments are shown below; all will hybridize to the yellowcDNA clone. Cleavage of the 5 kb EcoRI fragment with SalI generates two fragments of 2.5 kb.
c) What are the maps of the SalI and HindIII site(s) in each of the EcoRI fragments? Show positions of the cleavage sites and distances between them on the diagram below.
5 kb EcoRI fragment: 4 kb EcoRI fragment:
EcoRI EcoRI EcoRI EcoRI
|___________________________| |___________________________|
d) Compare these restriction maps with that of the cDNA clone (problem 1.38) and the R-loops shown above. Assuming that the SalI and HindIII sites in the genomic DNA correspond to those in the cDNA clone, what can you deduce about the intron/exon structure of the yellowgene(s) contained within the 5 kb and 4 kb EcoRI fragments? Please diagram the exon-intron structure in as much detail as the data permit (i.e. show the size of the intron(s) and positions of intron/exon junctions as precisely as possible).
5 kb EcoRI fragment: 4 kb EcoRI fragment:
EcoRI EcoRI EcoRI EcoRI
|___________________________| |___________________________|
e) Considering all the data (maps of cDNA and genomic clones and R-loop analysis), what can you conclude about the number and location(s) of yellowgene(s) in this genomic clone?
3.13 You have isolated an 1100 base pair (bp) cDNA clone for a gene called azure that when mutated causes blue eyes in frogs. You also isolate a 3000 bp SalI genomic DNA fragment that hybridizes to the azurecDNA. The map of the azure cDNA is as follows, with sizes of fragments given in bp.
Digestion of the 3000 bp SalI fragment of genomic DNA with the indicated restriction endonucleases yields the following pattern of fragments, all of which hybridize to the azurecDNA. Remember that the starting fragment has SalI sites at each end. Sizes of fragments are in bp.
Restriction enzymes
BamHI Bam+Pst PstI Pst+Eco EcoRI Bam+Eco
2700
2300
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1100 1100
800
700 700 700
300 300 300
The SalI to SalI (3000 bp) genomic fragment was hybridized to the 1100 bp cDNA fragment, and the heteroduplexes were examined in the electron microscope. Measurements on a large number of molecules resulted in the determination of the sizes indicated in the structure on the left, i.e. duplex regions of 400 and 600 bp are interrupted by a single stranded loop of 1500 nucleotides and are flanked by single stranded regions of 500 and 100 nucleotides. When the same experiment is carried out with the 2700 bp SalI to EcoRI genomic DNA fragment hybridized to the cDNA fragment, the structure on the right is observed.
a) What is the restriction map of the 3000 bp SalI to SalI genomic DNA fragment from the azuregene? Specify distances between sites in base pairs.
b) How many introns are present in the azuregenomic DNA fragment?
c) Where are the exons in the azuregenomic DNA fragment? Draw the exons as boxes on the restriction map of the 3000 bp SalI to SalI genomic DNA fragment? Specify (in base pairs) the distances between restriction sites and the intron/exon boundaries.
3.14 The T-cell receptor is present only on T-lymphocytes, not on B-lymphocytes or other cells. Describe a strategy to isolate the T-cell receptor by subtractive hybridization, using RNA from T-lymphocytes and from B-lymphocytes.
3.15.How many exons are in the human insulin (INS) gene, how big are they, and how large are the introns that separate them? Use three different bioinformatic approaches to answer this.
a. Align the available genomic sequence containing INS(encoding insulin) with the sequence of the mRNA to find exons and introns in the INSgene. The sequence files are:
INSmRNA: accession number NM_000207
INS gene (includes part of THand IGF2in addition to INS): accession number L15440
Files can be obtained from NCBI (http://www.ncbi.nlm.nih.gov), or from the course web site (www.bmb.psu.edu/Courses/bmb400/default.htm)
Align the mRNA (cDNA) and genomic sequence using the BLAST2sequences server at
http://www.ncbi.nlm.nih.gov/blast/
and the sim4server at
pbil.univ-lyon1.fr/sim4.html
Sim4is designed to take into account terminal redundancy at the exon/intron junctions, whereas BLAST2does not. Do you see this effect in the output?
b. Use the ab initioexon finding program Genscan, available at
genes.mit.edu/GENSCAN.html
to predict exons in the INSgenomic sequence (L15440).
How does this compare with the results of analyzing with the program genscan?
c. What do you see for INSat the Human Genome Browser and Ensembl? They are accessed at:
http://genome.ucsc.edu/goldenPath/hgTracks.html
http://www.ensembl.org/ | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/3%3A_Isolating_and_Analyzing_Genes/3.E%3A_Isolating_and_Analyzing_Genes_%28Exercises%29.txt |
Chapter 4 has two parts: genomes and chromosomes. Initial studies on genome structure used the kinetics of hybridization of nucleic acids to determine the bulk features of genomes, e.g. how big is a particular genome, how much is single-copy and how much is repeated, and how much of that genome is transcribed into nuclear or mRNA in a particular tissue. More detailed whole-genome mapping and sequencing projects are now revolutionizing biology. Some of the information on whole-genome sequences of bacteria, the yeast Saccharomyces cerevisiae, worms, flies and mammals (humans and mice) will be reviewed. All this genomic DNA is packaged into chromosomes, and Chapter 4 will also review some of their cytological features, and discuss their packaging into nucleosomes and higher order structure. Transitions between types of chromatin structure are fundamental to issues of gene regulation in eukaryotes; this will be explored in more biochemical detail in Part Four of the text.
• 4.1: Reassociation kinetics measure sequence complexity
The components of complex genomes differ not only in repetition frequency (highly repetitive, moderately repetitive, single copy) but also in sequence complexity. Complexity (denoted by N) is the number of base pairs of unique or nonrepeating DNA in a given segment of DNA, or component of the genome. This is different from the length (L) of the sequence if some of the DNA is repeated.
• 4.2: Analysis of Renaturation curves with Multiple Components
this section, the analysis in Section 4.1 is applied quantitatively in an example of renaturation of genomic DNA. If an unknown DNA has a single kinetic component, meaning that the fraction renatured increases from 0.1 to 0.9 as the value of C0t increases 100-fold, then one can calculate its complexity easily.
• 4.3: RNA Abundance
The availability of cloned DNA probes for many genes has greatly facilitated the analysis of amounts of RNAs in different cells or under different conditions. For instance, it is very common to label a DNA probe that will hybridize to mRNA; the DNA comes from either a cDNA clone or a genomic clone containing an exon. The labeled probe is then hybridized to total or polyA-containing RNA (the latter is called polyA+ RNA, and is roughly equivalent to mRNA) from a cell.
• 4.4: Genome Analysis by Large Scale Sequencing
Whole genomes can be sequenced both by random shot-gun sequencing and by a directed approach using mapped clones.
• 4.5: Sizes of genomes - The C‑value paradox
The C-value paradox is basically this: how can we account for the amount of DNA in terms of known function? The C-value is the amount of DNA in the haploid genome of an organism. It varies over a very wide range, with a general increase in C-value with complexity of organism from prokaryotes to invertebrates, vertebrates, plants.
• 4.6: Large Scale Genome Organization
• 4.7: Comparative Genome Analysis
• 4.E: Genomes and Chromosomes (Exercises)
• 4.S: Genomes and Chromosomes (Summary)
4: Genomes and Chromosomes
Low complexity DNA Sequences Reanneal Faster than Do High Complexity Sequences
The components of complex genomes differ not only in repetition frequency (highly repetitive, moderately repetitive, single copy) but also in sequence complexity. Complexity (denoted by N) is the number of base pairs of unique or nonrepeating DNA in a given segment of DNA, or component of the genome. This is different from the length (L) of the sequence if some of the DNA is repeated, as illustrated in this example.
E.g. consider 1000 bp DNA.
• 500 bp is sequence a, present in a single copy.
• 500 bp is sequence b (100 bp) repeated 5 times:
a b b b b b
|___________|__|__|__|__|__| L = length = 1000 bp = a + 5b
N = complexity = 600 bp = a + b
Some viral and bacteriophage genomes have almost no repeated DNA, and L is approximately equal to N. But for many genomes, repeated DNA occupies 0.1 to 0.5 of the genome, as in this simple example. The key result for genome analysis is that less complex DNA sequences renature faster than do more complex sequences. Thus determining the rate of renaturation of genomic DNA allows one to determine how many kinetic components (sequences of different complexity) are in the genome, what fraction of the genome each occupies, and the repetition frequency of each component.
Before investigating this in detail, let's look at an example to illustrate this basic principle, i.e. the inverse relationship between reassociation kinetics and sequence complexity.
Inverse Relationship between Reassociation Kinetics and Sequence Complexity
Let a, b, ... z represent a string of base pairs in DNA that can hybridize (see Figure 4.2.). For simplicity in arithmetic, we will use 10 bp per letter.
• DNA 1 = ab (This is very low sequence complexity, 2 letters or 20 bp)
• DNA 2 = cdefghijklmnopqrstuv. (This is 10 times more complex (20 letters or 200 bp)).
• DNA 3 =izyajczkblqfreighttrainrunninsofastelizabethcottonqwftzxvbifyoudontbelieveimleavingyoujustcountthe
daysimgonerxcvwpowentdowntothecrossroadstriedtocatchariderobertjohnsonpzvmwcomeonhomeintomykitchentrad.
(This is 100 times more complex (200 letters or 2000 bp).
A solution of 1 mg DNA/ml is 0.0015 M (in terms of moles of bp per L) or 0.003 M (in terms of nucleotides per L). We'll use 0.003 M = 3 mM, i.e. 3 mmoles nts per L. (nts = nucleotides).
Consider a 1 mg/ml solution of each of the three DNAs. For DNA 1, this means that the sequence ab (20 nts) is present at 0.15 mM or 150 mM (calculated from 3 mM / 20 nt in the sequence). Likewise, DNA 2 (200 nts) is present at 15 mM, and DNA 3 is present at 1.5 mM. Melt the DNA (i.e. dissociate into separate strands) and then allow the solution to reanneal, i.e. let the complementary strand reassociate.
Since the rate of reassociation is determined by the rate of the initial encounter between complementary strands, the higher the concentration of those complementary strands, the faster the DNA will reassociate. So for a given overall DNA concentration, the simple sequence (ab) in low complexity DNA 1 will reassociate 100 times faster than the more complex sequence (izyajcsk ....trad) in the higher complexity DNA 3. Fast reassociating DNA is low complexity.
Kinetics of renaturation
In this section, we will develop the relationships among rates of renaturation, complexity, and repetition frequency more formally.
The time required for half renaturation is inversely proportional to the rate constant. Let C = concentration of single-stranded DNA at time $t$ (expressed as moles of nucleotides per liter). The rate of loss of single-stranded (ss) DNA during renaturation is given by the following expression for a second-order rate process:
$\dfrac{-dC}{dt}= kC^2$
or
$\dfrac{dC}{C^2}=-kdt$
Integration and some algebraic substitution shows that
$\dfrac{C}{C_o}=\dfrac{1}{1+kC_ot} \;\;\; \label{1}$
Thus, at half renaturation, when
$\dfrac{C}{C_o}=0.5 \; \text{at} \; t=t_{1/2}$
one obtains:
$C_ot_{1/2}=\dfrac{1}{k} \; \;\; \label{2}$
where $k$ is the rate constant in in liters (mole nt)-1 sec-1
The rate constant for renaturation is inversely proportional to sequence complexity.
The rate constant, k, shows the following proportionality:
$k \propto \dfrac{\sqrt{L}}{N} \;\;\; \label{3}$
where
• L = length and
• N = complexity.
Empirically, the rate constant k has been measured as
$k = 3x10^5 \dfrac{\sqrt{L}}{N}$
in 1.0 M Na+ at $T = T_m - 25^oC$
The time required for half renaturation (and thus Cot1/2) is directly proportional to sequence complexity.
From Equations \ref{2} and \ref{3},
$C _0 t_{1/2} \propto \dfrac{N}{\sqrt{L}} \;\;\;\; \label{4}$
For a renaturation measurement, one usually shears DNA to a constant fragment length L (e.g. 400 bp). Then L is no longer a variable, and
$C_o t_{1/2}\propto N \;\;\;\; \label{5}.$
The data for renaturation of genomic DNA are plotted as $C_0 t$ curves:
Renaturation of a single component is complete (0.1 to 0.9) over 2 logs of $C_0t$ (e.g., 1 to 100 for E. coli DNA), as predicted by Equation \ref{1}.
Sequence complexity is usually measured by a proportionality to a known standard
If you have a standard of known genome size, you can calculate $N$ from $C_0t_{1/2}$:
$\dfrac{N^{unknown}}{N^{known}} = \dfrac{C_0t_{1/2}^{unknown}}{C_0t_{1/2}^{known}} \;\;\;\; (6)$
A known standard could be
• E. coli with N = 4.639 x 106 bp
• pBR322 with N = 4362 bp
More complex DNA sequences renature more slowly than do less complex sequences. By measuring the rate of renaturation for each component of a genome, along with the rate for a known standard, one can measure the complexity of each component. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/4%3A_Genomes_and_Chromosomes/4.1%3A_Reassociation_kinetics_measure_sequence_complexity.txt |
In this section, the analysis in Section 4.2 is applied quantitatively in an example of renaturation of genomic DNA. If an unknown DNA has a single kinetic component, meaning that the fraction renatured increases from 0.1 to 0.9 as the value of C0t increases 100-fold, then one can calculate its complexity easily. Using equation (6), all one needs to know is its C0t1/2, plus the \(C_0t_{1/2}\) and complexity of a standard renatured under identical conditions (initial concentration of DNA, salt concentration, temperature, etc.).
The same logic applies to the analysis of a genome with multiple kinetic components. Some genomes reanneal over a range of C0tvalues covering many orders of magnitude, e.g. from 10-3 to 104. Some of the DNA renatures very fast; it has low complexity, and as we shall see, high repetition frequency. Other components in the DNA renature slowly; these have higher complexity and lower repetition frequency. The only new wrinkle to the analysis, however, is to treat each kinetic component independently. This is a reasonable approach, since the DNA is sheared to short fragments, e.g. 400 bp, and it is unlikely that a fast-renaturing DNA will be part of the same fragment as a slow-renaturing DNA.
Some terms and abbreviations need to be defined here.
• f = fraction of genome occupied by a component
• \(C_0t_{1/2}\) for pure component = (f) (\(C_0t_{1/2}\) measured in the mixture of components)
• R = repetition frequency
• G = genome size. G can be measured chemically (e.g. amount of DNA per nucleus of a cell) or kinetically (see below).
One can read and interpret the \(C_0t\) curve as follows. One has to estimate the number of components in the mixture that makes up the genome. In the hypothetical example in Figure 4.5, three components can be seen, and another is inferred because 10% of the genome has renatured as quickly as the first assay can be done. The three observable components are the three segments of the curve, each with an inflection point at the center of a part of the curve that covers a 100-fold increase in \(C_0t\) (sometimes called 2 logs of \(C_0t). The fraction of the genome occupied by a component, f, is measured as the fraction of the genome annealing in that component. The measured\(C_0t_{1/2}\) is the value of \(C_0t\) at which half the component has renatured. In Figure 4.5, component 2 renatures between \(C_0t\) values of 10-3 and 10-1, and the fraction of the genome renatured increased from 0.1 to 0.3 over this range. Thus f is 0.3-0.1=0.2. The C0t value at half-renaturation for this component is the value seen when the fraction renatured reached 0.2 (i.e. half-way between 0.1 and 0.3; this C0t value is 10-2, and it is referred to as the C0t1/2for component 2 (measured in the mixture of components). Values for the other components are tabulated in Figure 4.5.
All the components of the genome are present in the genomic DNA initially denatured. Thus the value for C0 is for all the genomic DNA, not for the individual components. But once one knows the fraction of the genome occupied by a component, one can calculate the C0 for each individual component, simply as C0 ´ f. Thus the \(C_0t_{1/2}\) for the individual component is the \(C_0t_{1/2}\) (measured in the mixture of components) ´ f. For example the \(C_0t_{1/2}\) for individual (pure) component 2 is 10-2 ´ 0.2 = 2 ´ 10-3 .
Knowing the measured \(C_0t_{1/2}\)for a DNA standard, one can calculate the complexity of each component.
\[ Nn= C_0t_{1/2}_{pure}, n \] ´
• where n refers to the particular component, i.e. (1, 2, 3, or 4)
The repetition frequency of a given component is the total number of base pairs in that componentdivided by the complexity of the component. The total number of base pairs in that component is given by fn ´ G.
Rn =
For the data in Figure 4.5, one can calculate the following values:
Component f \(C_0t_{1/2}\), mix \(C_0t_{1/2}\), pure N (bp) RR
1 foldback 0.1 < 10-4 < 10-4
2 fast 0.2 10-2 2 x 10-3 600 105
3 intermediate 0.1 1 0.1 3 x 104 103
4 slow (single copy) 0.6 103 600 1.8 x 108 1
std bacterial DNA 10 3 x 106 1
The genome size, G, can be calculated from the ratio of the complexity and the repetition frequency.
G=
E.g. If G = 3 x 108 bp, and component 2 occupies 0.2 of it, then component 2 contains 6 x 107 bp. But the complexity of component 2 is only 600 bp. Therefore it would take 105 copies of that 600 bp sequence to comprise 6 x 107 bp, and we surmise that R = 105.
Exercise 4.1
If one substitutes the equation for Nn and for G into the equation for Rn, a simple relationship for R can be derived in terms of \(C_0t_{1/2}\) values measured for the mixture of components . What is it?
Types of DNA in each kinetic component for complex genomes
Eukaryotic genomes usually have multiple components, which generates complex C0t curves. Figure 4.6 shows a schematic C0t curve that illustrates the different kinetic components of human DNA, and the following table gives some examples of members of the different components.
Table 4.2. Four principle kinetic components of complex genomes
Renaturation kinetics C0t descriptor Repetition frequency Examples
too rapid to measure "foldback" not applicable inverted repeats
fast renaturing low C0t highly repeated, > 105 copies per cell interspersed short repeats (e.g. human Alu repeats); tandem repeats of short sequences (centromeres)
intermediate renaturing mid C0t moderately repeated, 10-104 copies per cell families of interspersed repeats (e.g. human L1 long repeats); rRNA, 5S RNA, histone genes
slow renaturing high C0t low, 1-2 copies per cell, "single copy" most structural genes (with their introns); much of the intergenic DNA
N, R for repeated DNAs are averages for many families of repeats. Individual members of families of repeats are similar but not identical to each other.
The emerging picture of the human genome reveals approximately 30,000 genes encoding proteins and structural or functional RNAs. These are spread out over 22 autosomes and 2 sex chromosomes. Almost all have introns, some with a few short introns and others with very many long introns. Almost always a substantial amount of intergenic DNA separates the genes.
Several different families of repetitive DNA are interspersed throughout the the intergenic and intronic sequences. Almost all of these are repeats are vestiges of transposition events, and in some cases the source genes for these transposons have been found. Some of the most abundant families of repeats transposed via an RNA intermediate, and can be called retrotransposons. The most abundant repetitive family in humans are Alu repeats, named for a common restriction endonuclease site within them. They are about 300 bp long, and about 1 million copies are in the genome. They are probably derived from a modified gene for a small RNA called 7SL RNA. (This RNA is involved in translation of secreted and membrane bound proteins). Genomes of species from other mammalian orders (and indeed all vertebrates examined) have roughly comparable numbers of short interspersed repeats independently derived from genes encoding other short RNAs, such as transfer RNAs.
Another prominent class of repetitive retrotransposons are the longL1 repeats. Full-length copies of L1 repeats are about 7000 bp long, although many copies are truncated from the 5' end. About 50,000 copies are in the human genome. Full-length copies of recently transposed L1s and their sources genes have two open reading frames (i.e. can encode two proteins). One is a multifunctional protein similar to the pol gene of retroviruses. It encodes a functional reverse transcriptase. This enzyme may play a key role in the transposition of all retrotransposons. Repeats similar to L1s are found in all mammals and in other species, although the L1s within each mammalian order have features distinctive to that order. Thus both short interspersed repeats (or SINEs) and the L1 long interspersed repeats (or LINEs) have expanded and propogated independently in different mammalian orders.
Both types of retrotransposons are currently active, generating de novomutations in humans. A small subset of SINEs have been implicated as functional elements of the genome, providing post-transcriptional processing signals as well as protein-coding exons for a small number of genes.
Other classes of repeats, such as L2s (long repeats) and MIRS (short repeats named mammalian interspersed repeats), appear to predate the mammalian radiation, i.e. they appear to have been in the ancestral eutherian mammal. Other classes of repeats are transposable elements that move by a DNA intermediate.
Other common interspersed repeated sequences in humans | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/4%3A_Genomes_and_Chromosomes/4.2%3A_Analysis_of_Renaturation_curves_with_Multiple_Components.txt |
The availability of cloned DNA probes for many genes has greatly facilitated the analysis of amounts of RNAs in different cells or under different conditions. For instance, it is very common to label a DNA probe that will hybridize to mRNA; the DNA comes from either a cDNA clone or a genomic clone containing an exon. The labeled probe is then hybridized to total or polyA-containing RNA (the latter is called polyA+ RNA, and is roughly equivalent to mRNA) from a cell. The concentration of the probe is much greater than the concentration of the target mRNA for the specific gene, thus the probe is in vast excess and all mRNA from the gene of interest should be driven into a duplex with the probe. The amount of probe protected from digestion by a single-strand specific nuclease such as nuclease S1 gives a measure of the amount of the specific mRNA that is in the cell. (This situation differs in some important aspects from the materialon estimating numbers of genes expressed and abundance from the kinetics of RNA-driven reactions. In that material, one was looking at entire populations of mRNAs, whereas in this situation, one is looking at only one mRNA - the one complementary to the labeled probe.)
[Two technical notes: The diagnostic assay here measures the amount of labeled DNA in duplex and the unhybridized DNA is digested. If the DNA probe is originally double-stranded, it is initially denatured prior to hybridization, but now how do you distinguish between nuclease protection arising from DNA-mRNA duplexes versus those that arise from the two strands of DNA reannealing? The cleanest approach is to just synthesize and label the strand of DNA complementary to the mRNA; this can be done by appropriate choices of primers for synthesis of DNA from plasmids carrying the DNA used as a probe. Alternatively, a labeled duplex DNA probe can be prepared that extends past the mRNA coding portion of a gene, so that the DNA-DNA duplex resulting from reannealing is larger than the DNA-RNA duplex resulting from hybridization to mRNA. Also, hybridization conditions with high concentrations of salt and formamide are used that favor DNA-RNA duplexes over DNA-DNA duplexes. (2) An equivalent approach is to synthesize an RNA probe derived from the cloned DNA; this "complementary RNA" forms a stronger duplex with the mRNA than does cDNA; RNA-RNA duplexes are stronger than RNA-DNA duplexes under conditions of high salt and formamide concentrations. The fragments protected from digestion by RNases are then detected.]
a) Murine erythroleukemia (MEL) cells are equivalent to proerythroblasts, immortalized by the Friend virus complex so that they can grow continuously in culture. Treatment with small organic compounds like dimethylsulfoxide (DMSO) will induce them to mature on to erythroblasts, with a substantial increase in the expression of erythroid specific genes (the mechanism for this induction is still unknown). Let's say that you isolated total RNA from both uninduced (untreated) cells and an equal number of DMSO-induced cells. The RNA samples were hybridized to an excess of a radiolabeled DNA probe from a mouse b-globin gene, and the amount of probe hybridized to the mRNA was determined by treatment of the samples with nuclease S1, electrophoresis on a denaturing polyacrylamide gel, and measuring the amount of radioactivity in the fragment resulting from the mRNA-DNA duplex. An illustration of the heteroduplex, the nuclease S1 treatment, and the resultant autoradiograph of the gel are shown below. The protected fragment from uninduced cells had 10,000 cpm, and the protected fragment from induced cells had 500,000 cpm. A negative control with RNA from a T-lymphocytic cell line, which produces no globin mRNA, gave no protection, i.e. 0 cpm for the diagnostic fragment. The expression of this b-globin gene is induced how much in MEL cells treated with DMSO?
b) The previous assay gives the relative amounts of the mRNA under the two conditions, and this is an extremely powerful and widely used assay. But what does this mean in terms of mRNA molecules per cell, i.e. how does the abundance change upon induction? One can alter this assay somewhat to get a measure of abundance, similar in principle to the calculations in Section VIIF. First, one needs a measure of the number of mRNA molecules per cell. Let's say that you harvested 107 MEL cells and isolated 3 mg of polyA+ RNA (essentially mRNA). What is the total number of mRNA molecules per MEL cell, assuming an average length of mRNA of 2000 nucleotides?
c) If one labels the RNA in the MEL cells, e.g. by growing the cells in the presence of [3H] uridine, which is incorporated only into RNA, then the isolated, labeled polyA+ RNA can be hybridized to an excess of the (now unlabeled) DNA complementary to the mRNA of interest. RNA in duplex with DNA can be detected by its protection from digestion by nucleases such as RNase A and RNase T1; the resulting autoradiograph would look something like that shown below, with bands containing more radioactivity represented as a darker fill. Since the DNA is still in excess, all the mRNA complementary to the probe should be driven into duplex, and one can readily measure the fraction of polyA+ RNA complementary to each probe. The following table provides some representative, idealized data for polyA+ RNA from uninduced and induced MEL cells, including the total input RNA (not treated with nucleases) and the amount protected from nuclease digestion by hybridization with an excess of b-globin gene DNA, DNA encoding the erythroid transcription factor GATA1, and DNA encoding ovalbumin (which is not expressed in MEL cells, i.e. it is a negative control). What fraction of the mRNA (or polyA+ RNA) is composed of mRNA from these three genes, and what is their abundance in uninduced and induced cells?
DNA probe cpm protected uninduced MEL cells cpm protected induced MEL cells
[input labeled RNA] [1,000,000] [1,000,000]
b-globin 5,000 250,000
GATA1 25 25
ovalbumin 0 0
d) In general, what is the distribution of mRNAs in a particular type of differentiated cell, i.e. how abundant are the different complexity classes of mRNA?
Use of databases of sequences, mutations, and functional data
4.6 We used arginine biosynthesis to illustrate complementation analysis and construction of a pathway. The steps involved in arginine synthesis are also part of the urea cycle. One of the enzymes catalyzes the formation of citrulline from carbamoyl phosphate and ornithine. Let's find out more about this enzyme, called ornithine transcarbamoylase, or OTC.
Use your favorite Web browser to go to the URL for NCBI (National Center for Biotechnology Information).
• http://www.ncbi.nlm.nih.gov/
• Click on the Entrez button. Entrez provides a portal to many types of information at this server. Let's start with DNA and protein sequences.
• Click on the Nucleotides button.
• Enter "X00210" and press the Search button. Do not enter the quotation marks, and those are zeros and a one, not O or l.
• You should get a report on the gene for OTC in E. coli, called argI.
1. How large is the protein-coding region, from translation initiation codon to the termination codon? How big is the encoded protein?
2. Where is the argI gene on the E. colichromosome? Go back to the Entrez server (where you clicked on Nucleotides before). Click on Genomes, and then select Escherichia coli. Enter "argI" in the Search window (don't enter the quotes, and that is the letter I "eye" not a "one").
4.7 Is the E. coli OTC protein related to any other proteins in the sequence databases? You need to get the protein sequence, which you can do by clicking on argIwhile you are at the genome map, or you can go back to the entry for the gene (accession number X00210). If you are at the GenBank Report for entry X00210, you need to click on the Protein button at the top of the page, and then select FastA Report from the next page. (If you take the default path the GenPept Report, that is OK, you can get the FastA Report from there as well.) Make a copy of this OTC sequence in FastA format (you may want to save it in another program, e.g. your favorite word processor, for convenience).
Now click on the Blast button at the top of the page, and at the next page select Basic Blast search. At the Blast server, select blastp from the pull-down menu next to Program (this aligns protein sequences; the default blastn aligns nucleotide sequences), and paste the E. coliOTCsequence in FastA format into the input window. Note that the pull-down menu gives you the option of entering the accession number (40962) instead of the sequence. The default sequence databases are nr, the non-redundant compilation of databases from the US, Europe and Japan. We'll use that, but note that a pull-down menu allows you to select other databases.
a) Click on the Submit Query button. When the job finally runs (this can take a minute or more when the Server is busy) what do you see?
b) Is the E. coli OTC protein related to any human protein? Scroll down the table of hits, past many bacterial OTCs (Neisseria, Pyrococcus...) until you run into some mammalian hits. With a score of 172, you should find a hyperlink to sp|P00480|OTC_HUMAN ORNITHINE CARBAMOYLTRANSFERASE PRECURSOR. Click on this hyperlink.
4.8 The entry for human OTC (P00480, which is the same as 400687) is quite long.
a) What occupies much of the feature table? What does this tell you about the OTCgene in humans?
b) Using either the features table for the GenBank entry 400687 (or P00480) or better yet, go back to the home page for NCBI and click on the OMIM button to go to the On-Line Medelian Inheritance In Man (from Victor McKusick, M.D.). Where is the gene? What happens in OTC deficiency?
4.9 What do the aligned amino acid sequences of the bacterial and human proteins tell you? Do conserved regions correlate with functional regions? For instance, does mutation of any amino acids in the conserved regions lead to a phenotype in humans?
Since the Blast search generated so many hits with higher scores than the E. coli- human pair, we will have to use a different tool to see the alignment. At the Blast server top page (where you selected Basic Blast search before), select Blast 2 sequences. This utility allows you to enter any two sequences and generate a pairwise alignment by the program Blast2. You should use the human and E. coli OTC protein sequences or their accession numbers, and be sure to choose blastp as the program. When doing this in July of 1998, I ran into a problem with the utility making a duplicate of each sequence I entered (I don't know if that was a problem at my end or theirs); this is likely a temporary condition. If you encounter a problem, try a different Server, such as the Sequence Analysis Server at genome.cs.mtu.edu/sas.html. Choose Pairwise Sequence Alignment, enter your sequences and run GAP or SIM on protein sequences.
Chromatin
4.10 One of the important early pieces of evidence that helped define the structure of the nucleosome was the pattern of nuclease cleavage in chromatin. In this experiment, chromatin was treated briefly with an enzyme, micrococcal nuclease, that degrades DNA, then all protein was removed and the and the purified DNA resolved by electrophoresis. A regular pattern of broad bands was seen; the average sizes of the DNA fragments were multiples of 200 bp, i.e. 200, 400, 600, 800 bp, etc. What does this result tell you about chromatin structure? The bands of DNA bands were thick and spread out rather than sharp; what does this tell you about the positions of cleavage by micrococcal nuclease?
4.11 Which histones are in the core of the nucleosome? What are the protein-protein interactions in the core? What protein domains mediate these interactions?
4.12 The mammalian virus SV40 has minichromosomes in which the circular duplex DNA is packaged into nucleosomes. When histones are removed from the minichromosomes, the resulting DNA is found to be negatively supercoiled. What does this tell you about the state of the DNA in the minichrosomes and the path of the DNA around the nucleosome?
4.13 Are the following statements true or false?
1. The DNA coils around the histones about 1.65 turns per nucleosomal core.
2. The DNA in chromatin containing actively transcribed genes is usually more sensitive to DNases than is the DNA in nontranscribed chromatin.
4.14 The packing ratio of a nucleic acid-protein complex is the ratio between the length of the naked DNA in normal B form to the length of the protein-DNA structure. For instance, if a set of proteins folded a DNA molecule of 100 Å into a structure that is 25 Å long, this structure has a packing ratio of 4.
a) Given the dimensions of the nucleosome structure, what is the packing ratio for the DNA in the nucleosome core? Note that the pitch is the distance between the midpoints of the DNA duplex as it turns around the histones in the core.
b) If the nucleosomes are tight-packed into a solenoid with 6 nucleosomes per turn, what is the packing ratio now? Assume that each turn of the solenoid translates 110 Å, i.e. the distance between the midpoints of nucleosomes in successive turns of the solenoid is 110 Å.
4.15 How close are the edges of the DNA as it curves around the surface of the nucleosomal core? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/4%3A_Genomes_and_Chromosomes/4.3%3A_RNA_Abundance.txt |
Whole genomes can be sequenced both by random shot-gun sequencing and by a directed approach using mapped clones.
The results of the Celera and public collaboration on the fly sequence was published in early 2000, and descriptions of the human genome sequence were published separately by Celera and IHGSC in 2001. Neither genome is completely sequenced (as of 2001), but both are highly sequenced and are stimulating a major revolution in the life sciences.
The wisdom of which approach to take is still a matter of debate, and depends to some extent on how thoroughly one needs to sequence a complex genome. For instance, a publicly accessible sequence of the mouse genome at 3X coverage was recently generated by the shotgun approach. Other genomes will likely be “lightly sequenced” at a similar coverage. But a full, high quality sequence of mouse will likely use aspects of the more directed approach. Also, the Celera assembly (primarily shotgun sequence) used the public data on the human genome sequence as well. Thus current efforts use both the rapid sequencing by shotgun methods and as well as sequencing mapped clones.
Survey of sequenced genomes
The genome sequences are available for many species now, covering an impressive phylogenetic range. This includes more than 28 eubacteria, at least 6 archaea, a fungus (the yeast Saccharomyces cerevisiae), a protozoan (Plasmodium falciparum), a worm (the nematode Caenorhabditis elegans), an insect (the fruitfly Drosophila melanogaster), two plants (Arabadopsisand rice (soon)), and two mammals (human Homo sapiensand mouse Mus domesticus). Some information about these is listed in Table 4.4.
Table 4.4.Sequenced genomes. This table is derived from the listing of “Complete Genomes Mapped on the KEGG Pathways (Kyoto Encyclopedia of Genes and Genomes)” at
www.genome.ad.jp/kegg/java/org_list.html
Additional genomes have been added, but only samples of the bacterial sequences are listed.
Genes encoding
Species
Genome Size
(bp)
Protein
RNA
Total
Enzymes
Category
Eubacteria
Escherichia coli
4,639,221
4,289
108
1,254
gram negative
Haemophilus influenzae
1,830,135
1,717
74
571
gram negative
Helicobacter pylori
1,667,867
1,566
43
394
gram negative
Bacillus subtilis
4,214,814
4,100
121
819
gram positive
Mycoplasma genitalium
580,073
467
36
202
gram positive
Mycoplasma pneumoniae
816,394
677
33
226
gram positive
Mycobacterium tuberculosis
4,411,529
3,918
48
-
gram positive
Aquifex aeolicus
1,551,335
1,522
50
-
hyperthermophilic bacterium
Borrelia burgdorferi
1,230,663
1,256
23
176
lyme disease Spirochete
Synechocystis sp.
3,573,470
3,166
49
702
cyanobacterium
Archaebacteria
Archaeoglobus fulgidus
2,178,400
2,407
49
439
S-metabolizing archaea
Methanococcus jannaschii
1,739,934
1,735
43
441
archaea
Methanobacterium thermoautotrophicum
1,751,377
1,871
47
558
archaea
Eukaryotes
Saccharomyces cerevisiae
12,069,313
6,064
262
861
fungi
Caenorhabditis elegans
97,000,000
18,424
-
nematode
Drosophila melanogaster
180,000,000
13,601
insect, fly, 120 Mb sequenced
Arabidopsis thaliana
115,500,000
25,706
plant, complete
Homo sapiens
3,200,000,000
30,000-40,000
human, draft + finished
Mus domesticus
3,000,000,000
mouse, draft
Genome size
Bacterial genomes range in size from 0.58 to almost 5 million bp (Mb). E. coli and B. subtilis, two of the most intensively studied bacteria, have the largest genomes and largest numbers of genes. The genome of the yeast Saccharomyces cerevisiae is only 2.6 times as large as that of E. coli. The genome of humans is almost 700 times larger than that of E. coli. However, genome size is not a direct measure of genetic content over long phylogenetic distances. One needs to examine the fraction of the genome that codes for protein or contains other important information. Let’s look at sizes and numbers of genes in different genomes.
Gene size and number
The average gene size is similar among bacteria, averaging around 1100 bp. Very little DNA separates most bacterial genes; in E. colithere is an average of only 118 bp between genes. Since the gene size varies little, then the number of genes varies over as wide a range as the genome size, from 467 genes in M. genitaliumto 4289 in E. coli. Thus within bacteria, which have little noncoding DNA, the number of genes is proportional to the genome size.
Saccharomyces cerevisiaehas one gene every 1900 bp on average, which could reflect both an increase in size of gene as well as somewhat greater distance between genes. Both bacteria and yeast show a much denser packing of genes than is seen in more complex genomes.
Data on a large sample of human genes shows that they are much larger than bacterial genes, with the median being about 14 times larger than the 1 kb bacterial genes. This is not because most human proteins are substantially larger; both bacterial proteins average about 350 amino acids in length, which is similar to the median size of human proteins. The major difference is the large amount of intronic sequence in human genes.
Table 4.5.Average size of human genes and parts of genes. This is based on information in the IHGSC paper in Nature, and derived from analysis of 1804 human genes.
Median
Mean
Internal exon
122 bp
145 bp
Number of exons
7
8.8
Length of each intron
1023 bp
3365 bp
3’ UTR
400 bp
770 bp
5’ UTR
240 bp
300 bp
Coding sequence
1100 bp
1340 bp
Length of protein encoded
367 amino acids
447 amino acids
Genomic extent
14,000 bp
27,000 bp
Alternative splicing is common in human genes
A previous lower estimate is that alternative splicing occurs in 35% of human genes. However, recent data show this fraction is larger.
For Chromosome 22:
• 642 transcripts cover 245 genes, 2.6 txpts/gene
• 2 or more transcripts for 145 (59%) of genes
For Chromosome 19:
• 1859 transcripts cover 544 genes, 3.2 txpts/gene
This contrasts with the situation in worm, in which alternative splicing occurs in 22% of genes. The increased genetic diversity from alternative splicing may contribute considerably to the greater complexity of humans, not just the increase in the number of genes.
Estimates of number of human genes
The estimated number of human genes has varied greatly over recent years. Some of these numbers have been widely quoted, and it may be useful to list some of the sources of these estimates.
• mRNA complexity (association kinetics): 40,000 genes
• Avg size of gene 30,000 bp: 100,000 genes
• Number of CpG islands: 70,000 to 80,000
• Unigene clusters of ESTs: 35,000 to 125,000
• More rigorous EST clustering: 35,000 genes
• Comparison to pufferfish: 30,000 genes
• Extrapolate from gene counts on chromosomes 21 and 22 (which are finished): 30,000 to 35,500 genes
Using the draft human sequence from Juy 2000, the IHGSC constructed an Initial Gene Index for human. They use the Ensembl system at the Sanger Centre. They started with ab initio predictions by Genscan, then confirmed by similarity to proteins, mRNAs, ESTs, and protein motifs (Pfam database) from any organism. This led to an initial set of 35,500 genes and 44,860 transcripts in the Ensemble database. After reducing fragmentation, merging with known genes, and removing contaminating bacterial sequences, they were left with 31,778 genes. After taking into account residual fragmentation, and the rate at which true genes are found by a similar analysis, the estimate remains about 32,000 genes. However, it is an estimate and is subject to change as more annotation is completed..
Starting with this estimate that the human genome contains about 32,000 genes, one can calculate how much of the genome is coding and how much is transcribed. If the average coding length is 1400 bp, then 1.5%of human genome consists of coding sequence. If the average genomic extent per gene is 30 kb, then 33% of human genome is “transcribed”.
Summary of number of genes in eukaryotic species:
• Human: 32,000 “still uncertain”
• Fly: 13, 338
• Worm: 18,266
• Yeast: 6,144
• Mustard weed: 25,706
• Human: 2x number of genes in fly and worm
• Human: more alternative splicing, perhaps 5x number of proteins as in fly or worm
Assignment of functions to genes
Genes encoding proteins and RNAs can be detected with considerable accuracy using compuational tools. Note that even for an extensively studies organism like E. coli, the number of genes found by sequence analysis (4289 encoding proteins) is far greater than the number that can be assigned as encoding a particular enzyme (1254). The discrepancy between genes found in the sequence versus those with known function (i.e. assigned as encoding an enzyme) is greater for some poorly characterized organisms such as the lyme-disease causing Spirochete Borrelia burgdorferi.
The many genes with unassigned function present an exciting challenge both in bioinformatics and in biochemistry/cell biology/genetics. Large collaborations have been initiated for a comprehensive genetic and expression analysis of some organisms. For instance, projects are underway to make mutations in all detected genes in Saccharomyces cerevisiae and to quantify the level of stable RNA from each gene in a variety of growth conditions, through the cell cycle and in other conditions. Databases are already established that record the changes in RNA levels for all yeast genes when the organism is shifted from glucose to galactose as a carbon source. These large scale expression analysis use high density microchip arrays that contain characteristic sequences for all 6064 yeast genes. These gene arrays are then hybridized with fluorescently labeled RNA or cDNA from cells grown under the two different conditions. The hybridization signals are quantitated and compared automatically, analyzed. The plan is to store the results in public databases. Useful websites include:
• SGD
• MIPS: a database for genomes and protein sequences | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/4%3A_Genomes_and_Chromosomes/4.4%3A_Genome_Analysis_by_Large_Scale_Sequencing.txt |
The C-value is the amount of DNA in the haploid genome of an organism. It varies over a very wide range, with a general increase in C-value with complexity of organism from prokaryotes to invertebrates, vertebrates, plants. The C-value paradox is basically this: how can we account for the amount of DNA in terms of known function? Very similar organisms can show a large difference in C-values (e.g. amphibians). The amount of genomic DNA in complex eukaryotes is much greater than the amount needed to encode proteins. For example: Mammals have 30,000 to 50,000 genes, but their genome size (or C-value) is 3 x 109 bp.
$\dfrac{3 \times 10^9\, \text{base pairs}}{3000\, \text{base paires (average gene size)}} = 1 \times 10^6 \,\text{(“gene capacity”).}$
Drosophila melanogaster has about 5000 mutable loci (~genes). If the average size of an insect gene is 2000 bp, then
$\dfrac{1 \times 10^8 \text{base pairs}}{2 \times 10^3\, \text{base pairs}} = > 50,000 \text{ “gene capacity”}.$
Our current understanding of complex genomes reveals several factors that help explain the classic C-value paradox:
• Introns in genes
• Regulatory elements of genes
• Pseudogenes
• Multiple copies of genes
• Intergenic sequences
• Repetitive DNA
The facts that some of the genomic DNA from complex organisms is highly repetitive, and that some proteins are encoded by families of genes whereas others are encoded by single genes, mean that the genome can be considered to have several distinctive components. Analysis of the kinetics of DNA reassociation, largely in the 1970's, showed that such genomes have components that can be distinguished by their repetition frequency. The experimental basis for this will be reviewed in the first several sections of this chapter, along with application of hybridization kinetics to measurement of complexity and abundance of mRNAs. Advances in genomic sequencing have provided more detailed views of genome structure, and some of this information will be reviewed in the latter sections of this chapter.
Table $1$: Distinct components in complex genomes
R= repetition frequency
Highly repeated DNA R (repetition frequency) > 100,000 Almost no information, low complexity
Moderately repeated DNA 10<<em>R<10,000> Little information, moderate complexity
“Single copy” DNA R=1 or 2 Much information, high complexity
LTR-containing retrotransposons
• MaLR: mammalian, LTR retrotransposons
• Endogenous retroviruses
• MER4 (MEdium Reiterated repeat, family 4)
MER1 and MER2
Mariner repeats
Some of the repeats are clustered into tandem arrays and make up distinctive features of chromosomes (Figure $1$). In addtion to the interspersed repeats discussed above, another contributor to the moderately repetitive DNA fraction are the thousands of copies of rRNA genes. These are in extensive tandem arrays on a few chromosomes, and are condensed into heterochromatin. Other chromosomal structures with extensive arrays of tandem repeats are centromeres and telomeres.
The common way of finding repeats now is by sequence comparison to a database of repetitive DNA sequences, RepBase (from J. Jurka). One of the best tools for finding matches to these repaats is RepeatMasker (from Arian Smit and P. Green, U. Wash.). A web server for RepeatMasker can be accessed at: ftp.genome.washington.edu/cgi-bin/RepeatMasker
Exercise $2$
Try Repeat Masker on INSgene sequence. You can get the INSsequence either from NCBI (GenBank accession gi|307071|gb|L15440.1 or one can use LocusLink, query on ) or from the course website.
Very little of the nonrepetive DNA component is expressed as mRNA
Hybridization kinetic studies of RNA revealed several important insights. First, saturation experiments, in which an excess of unlabeled RNA was used to drive labeled, nonrepetitive DNA (tracer) into hybrid, showed that only a small fraction of the nonrepetitive DNA was present in mRNA. Classic experiments from Eric Davidson’s lab showed that only 2.70% of total nonrepetitive DNA correspondss to mRNA isolated from sea urchin gastrula (this is corrected for the fact that only one strand of DNA is copied into RNA; the actual amount driven into hybrid is half this, or 1.35%; Figure 4.8). The complexity of this nonrepetitive fraction is (Nsc ) is 6.1 x 108 bp, so only 1.64 x 107 bp of this DNA is present as mRNA in the cell. If an "average" mRNA is 2000 bases long, there are ~8200 mRNAs present in gastrula.
In contrast, if the nonrepetitive DNA is hybridized to nuclear RNA from the same tissue, 28% of the nonrepetitive fraction corresponds to RNA (Figure 4.8). The nuclear RNA is heterogeneous in size, and is sometimes referred to as heterogeneous nuclear RNA, or hnRNA. Some of it is quite large, much more so than most of the mRNA associated with ribosomes in the cytoplasm. The latter is called polysomal mRNA.
These data show that a substantial fraction of the genome (over one-fourth of the nonrepetitive fraction) is transcribed in nuclei at the gastrula stage, but much of this RNA never gets out of nucleus (or more formally, many more sequences from the DNA are represented in nuclear RNA than in cytoplasmic RNA). Thus much of the complexity in nuclear RNA stays in the nucleus; it is not processed into mRNA and is never translated into proteins.
Factors contributing to an explanation include
1. Genes may be transcribed but the RNA is not stable. (Even the cytoplasmic mRNA from different genes can show different stabilities; this is one level of regulation of expression. But there could also be genes whose transcripts are so unstable in some tissues that they are never processed into cytoplasmic mRNA, and thus never translated. In this latter case, the gene is transcribed but not expressed into protein.)
2. Intronic RNA is transcribed and turns over rapidly after splicing.
3. Genes are transcribed well past the poly A addition site. These transcripts through the 3' flanking, intergenic regions are usually very unstable.
4. Not all of this "extra" RNA in the nucleus is unstable. For instance, some RNAs are used in the nucleus, e.g.:
5. U2-Un RNAs in splicing (small nuclear RNAs, or snRNAs).
RNA may be a structural component of nuclear scaffold (S. Penman)
Thus, although 10 times as much RNA complexity is present in the nucleus compared to the cytoplasm, this does not mean that 10 times as many genes are being transcribed as are being translated. Some fraction (unknown presently) of this "excess" nuclear RNA may represent genes that are being transcribed but not expressed, but many other factors also contribute to this phenomenon.
mRNA populations in different tissues show considerable overlap:
• Housekeeping genes encode metabolic functions found in almost all cells.
• Specialized genes, or tissue-specific genes, are expressed in only 1 (or a small number of) tissues. These tissue-specific genes are sometimes expressed in large amounts.
Estimating numbers of genes expressed and mRNA abundance from the kinetics of RNA-driven reactions
Using principles similar to those for analysis of repetition classes in genomic DNA, one can determine from the kinetics of hybridization between a preparation of RNA and single copy DNA both the average number of genes represented in the RNA, as well as the abundance of the mRNAs. The details of the kinetic analysis will not be presented, but they are similar to those already discussed. Highly abundant RNAs (like high copy number DNA) will hybridize to genomic DNA faster than will low abundance RNA (like low copy number DNA). Only a few mRNAs are highly abundant, and they constitute a low complexity fraction. The bulk of the genes are represented by lower abundance mRNA, and these many mRNAs constitute a high complexity, slowly hybridizing fraction.
An example is summarized in Table $2$. an excess of mRNA from chick oviduct wash ybridized to a tracer of labeled cDNA (prepared from oviduct mRNA). Three principle components were found, ranging from the highly abundant ovalbumin mRNA to much rarer mRNAs from many genes.
Table $2$
Component
Kinetics of hybridization
N (nt)
# mRNAs
Abundance
Example
1
fast
2,000
1
120,000
Ovalbumin
2
medium
15,000
7-8
4,800
Ovomucoid, others
3
slow
2.6 x 107
13,000
6-7
Everything else
Preparation of normalized cDNA libraries for ESTs
Just like the mRNA populations used as the templates for reverse transcriptase, the cDNAs from a particular tissue or cell type will be composed of many copies of a very few, abundant mRNAs, a fairly large number of copies of the moderately abundant mRNAs, and a small number of copies of the rare mRNAs. Since most genes produce low abundance mRNA, a corresponding small number of cDNAs will be made from most genes. In an effort to obtain cDNAs from most genes, investigators have normalized the cDNA libraries to remove the most abundant mRNAs.
The cDNAs are hybridized to the template mRNA to a sufficiently high Rot (concentration of RNA ´ time) so that the moderately abundant mRNAs and cDNAs are in duplex, whereas the rare cDNAs are still single-stranded. The duplex mRNA-cDNA will stick to a hydroxyapatite column, and the desired single-stranded, low abundance cDNA will elute. This procedure can be repeated a few times to improve the separation. The low abundance, high complexity cDNA is then ligated into a cloning vector to construct the cDNA library.
This normalization is key to the success of a random sequencing approach. Random cDNA clones, hundreds of thousands of them, have been picked and sequenced. A single-pass sequence from one of these cDNA clones is called an expressed sequence tag, or EST (Figure 4.9). It is called a “tag” because it is a sequence of only part of the cDNA, and since it is in cDNA, which is derived from mRNA, it is from an expressed gene. If the cDNA libraries reflected the normal abundance of the mRNAs, then this approach would result in re-sequencing the abundant cDNAs over and over, and most of the rare cDNAs would never be sequenced. However, the normalization has been successful, and many genes, even with rare mRNAs, are represented in the EST database.
As of May, 2001, over 2,700,000 ESTs individual sequences of human cDNA clones have been deposited in dbEST. They are grouped into nonredundant sets (called Unigene clusters). Over 95,000 Unigene clusters have been assembled, and almost 20,000 of them contain known human genes. The estimated number of human genes is less than the number of Unigene clusters, presumably because some large genes are still represented in more than one Unigene cluster. It is likely that most human genes are represented in the EST databases. Exceptions include genes expressed only in tissues which have not been sampled in the cDNA libraries. For more information, see www.ncbi.nlm.nih.gov/UniGene/index.html
Databases for genomic analysis
• Nucleic acid sequences
• genomic and mRNA, including ESTs
• Protein sequences
• Protein structures
• Genetic and physical maps
Organism-specific databases
• MedLine (PubMed)
• Online Mendelian Inheritance in Man (OMIM)
Sequences and annotation of the human genome
Ensemble (European Bioinformatics Institute (EMBL) and Sanger Centre)
http://www.ensembl.org/
A.
Programs for sequence analysis
• BLAST to search rapidly through sequence databases
• PipMaker (to align 2 genomic DNA sequences)
• Gene finding by ab initio methods (GenScan, GRAIL, etc.)
• RepeatMasker | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/4%3A_Genomes_and_Chromosomes/4.5%3A_Sizes_of_genomes_-_The_Cvalue_paradox.txt |
How to get by with the smallest possible genome
The Mycoplasma species have the smallest genomes of any free-living species. They are most related to the Bacillaceae family, but have lost their cell walls and many other functions in a process of reductive evolution. They are obligate parasites, e.g. living in the lungs of humans. Their genomes encode many transport proteins, so that amino acids, sugars, etc. can be taken up from their hosts. They have very little metabolic capacity, utilizing only glycolysis in the case of M. genitalium. There is very little biosynthetic capacity, depending largely on uptake from the host for these nutrients.
Mycoplasma Haemofelis, Wright-Geiemsa Staining 100X. (CC BY-SA 3.0; Nr387241).
One might have thought that the Mycoplasmal species would retain only the most highly conserved genes in bacteria, under the premise that these are the most critical genes. However, they have retained a proportion of conserved and variable genes that is quite similar to the proportion seen in E. coli. This indicates that these bacteria are maintaining a balance between conserved and variable genes that perhaps reflects an equilibrium between the stability of major physiological processes and the need for environmental adaptability.
More information from E. coli
The complete sequence of the E. coli genome provides an overview of genome structure within a well-understood context. For more information, see Blattner et al. (1997) Science, vol. 277, pp. 1453- 1462.
(1) Organization with respect to direction of replication
Since replication proceeds bidirectionally from the origin (oriC) and ends at the terminus, one can divide the genome into two "replicores." The replication fork proceeds clockwise in Replicore 1 and counter-clockwise in Replicore 2 (Figure 4.19).
Several features of the genome are oriented with respect to replication. All the rRNA genes, 53 of 86 tRNA genes, and 55% of the protein coding genes are transcribed in the same direction as the replication fork moves. In other species, such as the Mycoplasma, the transcriptional polarity is even more pronounced, and it also corresponds to the direction of replication.
These replicores show a pronounced skew in base composition, such that an excess of G over C is seen on the top strand (i.e. the one presented in the sequence file) in Replicore 1 and the opposite in Replicore 2. This nucleotide bias is striking and unexpected. As will be appreciated more after we study DNA synthesis in Part Two, this means that the leading strand for both replication forks is richer in G than C. Such an nucleotide bias may reflect differential mutation in the leading and lagging strands as a result of the asymmetry inherent in the DNA replication mechanism.
The recombination hotspot chi (GCTGGTGG) also shows a prominent strand preference, being more abundant on the leading strand of each replicore. The role of chi sites in recombination is covered in Chapter 8.
(2) Repeats, prophage and transposable elements
The E. coli chromosome contains several prophages and remnants of prophage, including lambda and three lambdoid prophages. The genome is peppered with at least 18 families of repeated DNA. The longest are the 5 Rhselements, which are 5.7 to 9.6 kb in length. Others are as short as the 581 copies of the 40 bp palindromic REP repeat. Several families of insertion sequences, which are transposable elements, are found. Note that repetitive elements are common in bacteria as well as in eukaryotes.
(3) General categories of genes
Many of the genes are similar to other genes in E. coli. Homologous genes that have diverged because of gene duplications are paralogous. The genes that encode proteins of similar but not necessarily identical function are referred to as a paralogous family. About 1/3 of the E. coli genes (1345) have at least one paralogous sequence in the genome. Some paralogous groups are quite large, the largest being the ABC transporters with 80 members. The larger number of genes in E. coli could reflect some redundancy in function as well as greater diversification of function compared to other bacteria with fewer genes.
Based on current understanding of the function of the gene products, about 1/4 are involved in small-molecule metabolism, about 1/8 are used in large-molecule metabolism, and at least 1/5 are associated with cell structure and processes. A specific function has not been assigned to the products of about 40% of the E. coli genes. Segmental duplications are common, as illustrated in Figure 4.21 for chromosomes 22. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/4%3A_Genomes_and_Chromosomes/4.6%3A_Large_Scale_Genome_Organization.txt |
Paralogous Genes
• Genes that are similar because of descent from a common ancestor are homologous.
• Homologous genes that have diverged after speciation are orthologous.
• Homologous genes that have diverged after duplication are paralogous.
One can identify paralogous groups of genes encoding proteins of similar but not identical function in a species e.g., ABC transporters: 80 members in E. coli
Core proteomes vary little in size
Proteome: all the proteins encoded in a genome
To calculate the Core proteome:
Count each group of paralogous proteins only once
Number of distinct protein families in each organism
Species Number of genes Core proteome
Haemophilus 1709 1425
Yeast 6241 4383
Worm 18424 9453
Fly 13601 8065
Core proteomes are conserved
• Many of the proteins in the core proteomes are shared among eukaryotes
• 30% of fly genes have orthologs in worm
• 20% of fly genes have orthologs in both worm and yeast
• 50% of fly genes have likely orthologs in mammals
Function of proteins in flies (and worms and yeast) provides strong indicators of function in humans. Flies have orthologs to 177 of the 289 human disease genes
CHROMOSOMES and CHROMATIN
Chromosomes are the cytological package for genes. Genomes are much longer than the cellular compartment they occupy compartment dimensions length of DNA
• Phage T4: $0.065 \times 0.10 \,mm 55\, mm = 170\, kb$
• E. coli: $1.7 \times 0.65\, mm \,1.3\, mm = 4.6 \times 10^3\, kb$
• Nucleus (human): $6 mm \,diam. 1.8\, m = 6 \times 10^6\, kb$
Definition: Packing ratio
$\text{Packing ratio} = \dfrac{\text{length of DNA}}{\text{length of the unit that contains it}}.$
The smallest human chromosome contains about
$46 \times 10^6\, bp = 14,000\, mm = 1.4\,cm \,DNA.$
When condensed for mitosis, this chromosome is about. 2 mm long. The packing ratio is therefore about 7000!
Loops, matrix and the chromosome scaffold
When DNA is released from mitoticchromosomes by removing most of the proteins, long loops of DNA are seen, emanating from a central scaffold that resembles the remnants of the chromosome.
Biochemical preparations using salt and detergent to remove proteins and nuclease to remove most of the DNA leaves a "matrix" or "scaffold" preparation. Similar DNA sequences are found in these preparations; these sequences are called matrix attachment regions = MARs (or scaffold attachment regions = SARs). They tend to be A+T rich and have sites for cleavage by topoisomerase II. Topoisomerase II is one of the major components of the matrix preparation; but the composition of the matrix is still in need of further study.
Since it is attached at the base to the matrix, each loop is a separate topological domain and can accumulate supercoils of DNA.
From the measured sizes of loops, and calculations based on the amount of nicking required to relax DNA within the loops, we estimate that the average size of these loops is about 100 kb (85 kb based on nicking frequency for relaxation).
Some evidence suggests that replication and possibly some transcriptional control may be exerted at the bases of the loops.
Interphase chromatin and mitotic chromosomes
During interphase, i.e. between mitotic divisions, the highly condensed mitotic chromosomesspread out through the nucleus to form chromatin. Interphase chromatin is not very densely packed in most of the nucleus (euchromatin). In some regions it is very densely packed, comparable to a mitotic chromosome (heterochromatin).
Both interphase chromatin and mitotic chromosomes are made of a 30 nm fiber. The mitotic chromosome is much more coiled than interphase chromosomes.
Most transcription occurs in euchromatin.
• Constitutive heterochromatin = nonexpressed regions that are condensed (compact) in all cells (e.g. centromeric simple repeats)
• Facultative heterochromatin = inactive in only some cell lineages, active in others.
One example of heterochromatin is the inactive X chromosome in female mammals. The choice of which X chrosomosome to inactivate is random in various cell lineages, leading to a mosaic phenotypes for some X-linked traits. For instance, one genetic determinant of coat color in cats is X-linked, and the patchy coloration on calico cats results from this random inactivation of one of the X chromosomes, leading to the lack of expression of this determinant in some but not all hair cells.
Cytologically visible bands in chromosomes
G bands and R bands in mammalian mitotic chromosomes (Figure 4.27)
Giemsa‑dark (G) bands tend to be A+T rich, with a large number of L1 repeats.
Giemsa‑light bands tend to be more G+C rich, with very few L1 repeats and many Alu repeats.
(R bands are about the same as Giemsa-light bands. They are visualized by a different preparative procedure so that the "reverse" of the Giemsa-stained images are seen.)
T bands are adjacent to telomeres, do not stain with Giemsa, and are extremely G+C rich, with lots of genes and myriad Alu repeats.
The functional significance of these bands is still under active investigation.
One can localizea gene to a particular region of a chromosome by in situhybridization with a radioactive or, now more commonly, fluorescent probe for the gene. The region of hybridization is determined by simultaneously viewing the stained banding pattern and the hybridization pattern. Many spreads of mitotic chromosomes are viewed and scored, and the gene is localized to the chromosomal region with a significantly greater incidence of hybridization signal than that seen to the rest of the chromosomes.
Another common method of mapping the location of genes is by hybridization to DNA isolated from a panel of somatic cell hybrids, each hybrid cell carrying a small subset of, e.g., human chromosomes on a hamster background. Some hybrid cells carry broken human chromosomes, which allows even more precise localization (see Figure 1.8.2, "J-1 series").
Polytene chromosomes are visible in several Drosophilatissues
These contain many copies of the chromosomes, side by side in register. Thus most chromosomal regions are highly amplified in these tissues. Chromosomal stains reveal characteristic banding pattern, which is the basis for the cytological map. The cytological map (of polytene bands) combined with the genetic map gives a cytogenetic map, which is a wonderful guide to the Drosophila genome. One can localize a gene to a particular region by in situ hybridization (in fact the technique was invented using Drosophilapolytene chromoomes.
Multiple genes per band on mammalian chromosomes
Figure 4.27 gives a view of human chromosome 11 at several different levels of resolution. The region 11p15 has many genes of interest, including genes whose products regulate cell growh (HRAS), determination and differentiation of muscle cells (MYOD), carbohydrate metabolism (INS), and mineral metabolism (PTH). The b-globin gene (HBB) and its closely linked relatives are also in this region. A higher resolution view of 11p15, based on a compilation of genetic and physical mapping (Cytogenetics and Cell Genetics, 1995) is shown next to the classic ideogram (banding pattern). This is in a scale of millions of base pairs, and one can start to get a feel for gene density in this region. Interestingly, it varies quite a lot, with the gene-dense sub-bands near the telomeres; these may correspond to the T-bands discussed above. Other genes appear to be more widely separated. For instance, each of the b-like globin genes is separated by about 5 to 8 kb from each other (see the map of the YAC, or yeast artificial chromosome, carrying the b-like globin genes), and this gene cluster is about 1000 kb (i.e. 1 Mb) from the nearest genes on the map. However, further mapping will likely find many other genes in this region. Now even more information is available at the web sites mentioned earlier.
The relationship between recombination distances and physical distances varies substantially among organisms. In human, one centiMorgan (or cM) corresponds to roughly 1 Mb, whereas in yeast 1 cM corresponds to about 2 kb, and this value varies at least 10-fold along the different yeast chromosomes. This is a result of the different frequencies of recombination along the chromosomes.
Specialized regions of chromosomes
Centromere: region responsible for segregation of chromosomes at mitosis and meiosis. The centromere is a constricted region (usually) toward the center of the chromosome (although it can be located at the end, as with mouse chromosomes.) It contains a kinetochore, a fibrous region to which microtubules attach as they pull the chromosome to one pole of the dividing cell. DNA sequences in this region are highly repeated simple sequences (in Drosophila, the unit of the repeat is about 25 bp long, repeated hundreds of times). Specific proteins are at the centromere, and are now intensely investigated.
Telomere: forms the ends of the linear DNA molecule that makes up the chromosome. The telomeres are composed of thousands of repeats of CCCTAA in human. Variants of this sequence are found in the telomeres in other species. Telomeres are formed by telomerase; this enzyme catalyzed the synthesis of more ends at each round of replication to stabilize linear molecules.
The Principal Proteins in Chromatin are Histones
Composition of chromatin: Various biochemical methods are avialable to isolated chromatin from nuclei. Chemical analysis of chromatin reveals proteins and DNA, with the most abundant proteins being the histones. A complex set of less abundant histones are referred to as the nonhistone chromosomal proteins.
The histones and DNA present in equal masses.
Mass Ratio DNA: histones: nonhistone proteins: RNA = 1: 1: 1: 0.1
Histones are small, basic (positively charged), highly conserved proteins. They bind to each other to form specific complexes, around which DNA wraps to form nucleosomes. The nucleosomes are the fundamental repeating unit of chromatin.
There are 5 histones, 4 in the core of the nucleosome and one outside the core.
H3, H4: Arg rich, most conserved sequence ü
ý CORE Histones
H2A, H2B: Slightly Lys rich, fairly conservedþ
H1: very Lys rich, most variable in sequence between species.
X-ray diffraction studies of histone complexes and the nucleosome core have provided detailed insight into how histones interact with each other and with DNA in this fundamental entity of chromatin structure.
Key reference: "Crystal structure of the nucleosome core particle at 2.8 Å resolution" by Luger, K. Mader, A., Richmond, R.K., Sargent, D.F. & Richmond, T.J. in Nature 389: 251-260 (1997)
Histone Interactions via the Histone fold
The core histones have a highly positively charged amino-terminal tail, and most of the rest of the protein forms an a-helical domain. Each core histone has at least 3 a-helices.
The a-helical domain forms a characteristic histone fold, in which shorter a1 and a3 helices are perpendicular to the longer a2 helix. The a-helices are separated by two loops, L1 and L2. The histone fold is the dimerization domain between pairs of histones, mediating the formation of crescent-shaped heterodimers H3-H4 and H2A-H2B. The histone-fold motifs of the partners in a pair are antiparallel, so that the L1 loop of one is adjacent to the L2 loop of the other.
A structure very similar to the histone fold has now been seen in other nuclear proteins, such as some subunits of TFIID, a key component in the general transcription machinery of eukaryotes. It also serves as a dimerization domain for these proteins.
Two H3-H4 heterodimers bind together to form a tetramer.
Nucleosomes are the Subunits of the Chromatin Fiber
The most extended chromatin fiber is about 10 nm in diameter. It is composed of a series of histone-DNA complexes called nucleosomes.
Principal lines of evidence for this conclusion are:
1. Observations of this 10 nm fiber in the electron microscope showed a series of bodies that looked like beads on a string. We now recognize the beads as the nucleosomal cores and the string as the linker between them.
2. Digestion of DNA in chromatin or nuclei with micrococcal nuclease releases a series of products that contain DNA of discrete lengths. When the DNA from the products of micrococcal nuclease digestion was run on an agarose gel, the it was found to be a series of fragments of 200 bp, 400 bp, 600 bp, 800 bp, etc. , i.e. integral multiples of 200 bp. This showed that cleavage by this nuclease, which has very little sequence specificity, was restricted to discrete regions in chromatin. Those regions of cleavage are the linkers.
3. Physical studies, including both both neutron diffraction and electron diffraction data on fibers and most recently X-ray diffraction of crystals, have provided more detailed structural information.
2. The nucleosomal core is composed of an octamer of histones with 146 bp of duplex DNA wrapped around it in 1.65 very tight turns. The octamer of histones is actually a tetramer H32H42 at the central axis, flanked by two H2A-H2B dimers (one at each end of the core.
The 10 nm fiber is composed of a string of nucleosomal cores joined by linker DNA. The length of the linker DNA varies among tissues within an organism and between species, but a common value is about 60 bp. The nucleosome is the core plus the linker, and thus contains about 200 bp of DNA.
Detailed structure of the nucleosomal core.
Path of the DNA and tight packing
The 146 bp of DNA is wrapped around the histone octamer in 1.65 turns of a flat, left-handed torroidal superhelix. Thus 14 turns or "twists" of the DNA are in the 1.65 superhelical turns, presenting 14 major and 14 minor grooves to the histone octamer. Pancreatic DNase I will cleave DNA on the surface of the core about every 10 bp, when each twist of the DNA is exposed on the surface.
The DNA superhelix has an average radius of 41.8 Å and a pitch of 23.9 Å. This is a very tight wrapping of the DNA around the histones in the core - note that the duplex DNA on one turn is only a few Å from the DNA on the next turn! The DNA is not uniformly bent in this superhelix. As the DNA wraps around the histones, the major and then minor grooves are compressed, but not in a uniform manner for all twists of the DNA. G+C rich DNA favors the major groove compression, whereas A+T rich DNA favors the minor groove compression. This is an important feature in translational positioning of nucleosomes and could also affect the affinity of different DNAs for histones in nucleosomes.
The DNA phosphates have high mobility when not contacting histones; the DNA phosphates facing the solvent are much more mobile than is seen with other protein-DNA complexes.
The left-handed torroidal supercoils of DNA in nucleosomal cores is the equivalent of a right-handed, hence negative, supercoil. Thus the DNA in nucleosomes is effectively underwound.
Histones in the nucleosome core particle
The protein octamer is composed of four dimers (2 H2A-H2B pairs and 2 H3-H4 pairs) that interact through the "histone fold". The two H3-H4 pairs interact through a 4-helix bundle formed between the two H3 proteins to make the H32H42 tetramer. Each H2A-H2B pair interacts with the H32H42 tetramer through a second 4-helix bundle between H2B and H4 histone folds.
The histone-fold regions of the H32H42 tetramer bind to the center of of the DNA covering a total of about 6 twists of the DNA, or 3 twists of DNA per H3-H4 dimer. Those of the H2A-H2B dimers cover a comparable amount of DNA, 3 twists per dimer. Additional helical regions extend from the histone fold regions and are an integral part of the the core protein within the confines of the DNA superhelix.
Histone-DNA interactions in the core particle.
The histone-fold domain of the heterodimers (H3-H4 and H2A-H2B) bind 2.5 turns of DNA double helix, generating a 140˚ bend. The interaction with DNA occurs at two types of sites:
1. The L1 plus L2 loops at the narrowly tapered ends of each heterodimer form a similar DNA binding site for each histone pair. The L1-L2 loops interact with DNA at each end of the 2.5 turns of DNA.
2. The a1 helices of each partner in a pair form the convex surface in the center of the DNA binding site. The principal interactions are H-bonds between amino acids and the phosphate backbone of the DNA (there is little sequence specificity to histone-DNA binding). However, there are some exceptions, such a hydrophobic contact between H3Leu65 and the 5-methyl in thymine. An Arg side chain from a histone fold enters the minor groove at 10 of the 14 times it faces the histone octamer. The other 4 occurrences have Arg side chains from tail regions penetrating the minor groove.
Histone Tails
The histone N- and C-termial tails make up about 28% of the mass of the core histone proteins, and are seen over about 1/3 of their total length in the electron density map - i.e. that much of their length is relatively immobile in the structure.
The tails of H3 and H2B pass through channels in the DNA superhelix created by 2 juxtaposed minor grooves. One H4 tail segment makes a strong interparticle connection, perhaps relevant to the higher-order structure of nucleosomes.
The most N-terminal regions of the histone tails are not highly ordered in the X-ray crystal structure. These regions extend out from the nucleosome core and hence could be involved in interparticle interactions. The sites for acetylation and de-acetylation of specific lysines are in these segments of the tails that protrude from the core. Post-translational modifications such as acetylation have been implicated in "chromatin remodeling" to allow or aid transcription factor binding. It seems likely that these modifications are affecting interactions between nucleosomal cores, but not changing the structure of the core particle.
Outside Links
• Some excellent resources are available on the World Wide Webfor visualizing and further investigating chromatin structure and its involvment in nuclear processes.
• Dmitry Pruss maintains a site with many good images, including dynamic, step-by-step view of the nuclesomal core beginning with the histone fold domains and ending with a complete core, with DNA. www.average.org/~pruss/nucleosome.html
• Another good site is from J.R. Bone: rampages.onramp.net/~jrbone/chrom.html
Higher order chromatin structure
1. The 10 nm fiber composed of nucleosomal cores and spacers is folded into higher order structures for much of the DNA in chromatin. In fact, the 10 nm fiber with the beads-on-a-string appearance in the electron microscope was prepared at very low salt concentrations and is free of histone H1.
2. In the presence of H1 and at more physiological salt concentrations, chromatin forms a 30 nm fiber. The exact structure of this fiber remains a point of considerable debate, and one cannot rule the possibility of multiple structure in this fiber.
3. One reaonable model is that the 10 nm fiber coils around itself to generate a solenoid that is 30 nm in diameter, with 6 nucleosomes per turn of of the solenoid.
Histone H1 binds to the outer surface of the nucleosomal core, interacting at the points of DNA entry and exit. H1 molecules can be cross-linked to each other with chemical reagents, indicating that the H1 proteins also interact with each other. Interactions between H1 proteins, each bound to a nucleosomal core, may be one of the forces driving the formation of the 30 nm fiber.
Figure 4.34. Model for one turn of the solenoid in the 30 nm fiber.
4. Each level of chromatin structure produces a more compact arrangment of the DNA. This can be described in terms of a packing ratio, which is the length of the DNA in an extended state divided by the length of the DNA in the more compact state.
For the 10 nm fiber, the packing ratio is about 7, i.e. there are 7mm of DNA per mm of chromatin fiber. The packing ratio in the core is higher (see problems), but this does not include the additional, less compacted DNA in the spacer. In the 30 nm fiber, the packing ratio is about 40, i.e. there 40mm DNA per mm of chromatin fiber.
5. The 30 nm fiber is probably the basic constituent of both interphase chromatin and mitotic chromosomes. It can be compacted further by additional coils and loops. One of the key issues in gene regulation is the nature of the chromating fiber in transcriptionally acative euchromatin. Is it the 10 nm fiber? the 30 nm fiber? some modification of the latter? or even some higher order structure? These are topics for current research. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/4%3A_Genomes_and_Chromosomes/4.7%3A_Comparative_Genome_Analysis.txt |
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