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4.3
(BPA) Answer the following questions with reference to the figure below.
Figure for 4.3 Reassociation of nucleic acids, sheared to 500-nucleotide fragments, from various sources [Derived fom R. J. Britten and D. Kohne, Science, 161,529 (1968).]
1. How many of these DNA preparations contain more than one frequency class of sequences? Explain your answer.
2. If the genome size of E. coliis taken to be 4.5 x 106 nucleotide pairs, what is the genome size of T4?
3. What is the complexity of mouse satellite DNA?
4. Mouse satellite DNA represents 10% of the mouse genome. What is the repetition number for mouse satellite sequences, given that the haploid genome size is 3.2 x 109 nucleotide pairs?
5. The calf genome is the same size as the mouse genome. What fraction of the calf genome is composed of unique sequences?
4.4
Let’s imagine that you obtained a DNA sample from an armadillo and measured the kinetics of renaturation of the genomic DNA. A standard of bacterial DNA (N= 3 x 106 bp) was also renatured under identical conditions. Three kinetic components were seen in the armadillo DNA C0t curve, renaturing fast, medium or slow. The fraction of the genome occupied by each component (f) and the C0tvalue for half-renaturation (Cot1/2(measured)) are as follows:
Component f Cot1/2(measured)
fast 0.2 \(10^{-4}\)
medium 0.4 \(10^{-1}\)
slow 0.4 \(10^{4}\)
1. Use the information provided to calculate the Cot1/2(pure), the complexity (N), and the repetition frequency (R) for each component. Assume that the slowly renaturing component is single copy.
2. Calculate the genome size (G) of the armadillo under the assumption that the slowly renaturing component is single copy.
3. Which of the following sequences could be a member of the fast renaturing component?
GACTCAGACTCAGACTCA
ATATATATATATATATAT
ACTGCCACGGGATACTGC
GCGCGC
4.S: Genomes and Chromosomes (Summary)
Additional Readings
1. Britten RJ, Kohne DE. (1968) Repeated sequences in DNA. Hundreds of thousands of copies of DNA sequences have been incorporated into the genomes of higher organisms. Science 161:529-540
2. Wetmur and Davidson (1968)The rate constant for renaturation is inversely proportional to sequence complexity. J. Molecular Biology 34:349-370.
3. Davidson EH, Hough BR, Amenson CS, Britten RJ. (1973) General interspersion of repetitive with non-repetitive sequence elements in the DNA of Xenopus. J. Molecular Biology 77:1-23.
4. Fleischmann RD, Adams MD, White O, Clayton RA, Kirkness EF, Kerlavage AR, Bult CJ, Tomb JF, Dougherty BA, Merrick JM, et al. (1995) Whole-genome random sequencing and assembly of Haemophilus influenzae Rd. Science. 269:496-512
5. Adams MD, Celniker SE, Holt RA, Evans CA, Gocayne JD, Amanatides PG, Scherer SE, Li PW, Hoskins RA, Galle RF, George RA, Lewis SE, Richards S, Ashburner M, Henderson SN, Sutton GG, Wortman JR, Yandell MD, Zhang Q, Chen LX, Brandon RC, Rogers YH, Blazej RG, Champe M, Pfeiffer BD, Wan KH, Doyle C, Baxter EG, Helt G, Nelson CR, Gabor GL, Abril JF, Agbayani A, An HJ, Andrews-Pfannkoch C, Baldwin D, Ballew RM, Basu A, Baxendale J, Bayraktaroglu L, Beasley EM, Beeson KY, Benos PV, Berman BP, Bhandari D, Bolshakov S, Borkova D, Botchan MR, Bouck J, Brokstein P, Brottier P, Burtis KC, Busam DA, Butler H, Cadieu E, Center A, Chandra I, Cherry JM, Cawley S, Dahlke C, Davenport LB, Davies P, de Pablos B, Delcher A, Deng Z, Mays AD, Dew I, Dietz SM, Dodson K, Doup LE, Downes M,
6. Dugan-Rocha S, Dunkov BC, Dunn P, Durbin KJ, Evangelista CC, Ferraz C, Ferriera S, Fleischmann W, Fosler C, Gabrielian AE, Garg NS, Gelbart WM, Glasser K, Glodek A, Gong F, Gorrell JH, Gu Z, Guan P, Harris M, Harris NL, Harvey D, Heiman TJ, Hernandez JR, Houck J, Hostin D, Houston KA, Howland TJ, Wei MH, Ibegwam C, Jalali M, Kalush F, Karpen GH, Ke Z, Kennison JA, Ketchum KA, Kimmel BE, Kodira CD, Kraft C, Kravitz S, Kulp D, Lai Z, Lasko P, Lei Y, Levitsky AA, Li J, Li Z, Liang Y, Lin X, Liu X, Mattei B, McIntosh TC, McLeod MP, McPherson D, Merkulov G, Milshina NV, Mobarry C, Morris J, Moshrefi A, Mount SM, Moy M,
7. Murphy B, Murphy L, Muzny DM, Nelson DL, Nelson DR, Nelson KA, Nixon K, Nusskern DR, Pacleb JM, Palazzolo M, Pittman GS, Pan S, Pollard J, Puri V, Reese MG, Reinert K, Remington K, Saunders RD, Scheeler F, Shen H, Shue BC, Siden-Kiamos I, Simpson M, Skupski MP, Smith T, Spier E, Spradling AC, Stapleton M, Strong R, Sun E, Svirskas R, Tector C, Turner R, Venter E, Wang AH, Wang X, Wang ZY, Wassarman DA, Weinstock GM, Weissenbach J, Williams SM, WoodageT, Worley KC, Wu D, Yang S, Yao QA, Ye J, Yeh RF, Zaveri JS, Zhan M, Zhang G, Zhao Q, Zheng L, Zheng XH, Zhong FN, Zhong W, Zhou X, Zhu S, Zhu X, Smith HO, Gibbs RA, Myers EW, Rubin GM, Venter JC. (2000) The genome sequence of Drosophila melanogaster. Science 287:2185-2195
8. International Human Genome Sequencing Consortium, I. H. G. S. (2001). Initial sequencing and analysis of the human genome. Nature 409: 860-921.
9. Rubin, G. M., Yandell, M. D., Wortman, J. R., Gabor Miklos, G. L., Nelson, C. R., Hariharan, I. K., Fortini, M. E., Li, P. W., Apweiler, R., Fleischmann, W., Cherry, J. M., Henikoff, S., Skupski, M. P., Misra, S., Ashburner, M., Birney, E., Boguski, M. S., Brody, T., Brokstein, P., Celniker, S. E., Chervitz, S. A., Coates, D., Cravchik, A., Gabrielian, A., Galle, R. F., Gelbart, W. M., George, R. A., Goldstein, L. S., Gong, F., Guan, P., Harris, N. L., Hay, B. A., Hoskins, R. A., Li, J., Li, Z., Hynes, R. O., Jones, S. J., Kuehl, P. M., Lemaitre, B., Littleton, J. T., Morrison, D. K., Mungall, C., O'Farrell, P. H., Pickeral, O. K., Shue, C., Vosshall, L. B., Zhang, J., Zhao, Q., Zheng, X. H., Zhong, F., Zhong, W., Gibbs, R., Venter, J. C., Adams, M. D. and Lewis, S. (2000). Comparative genomics of the eukaryotes. Science 287: 2204-15.
10. Venter, J. C., Adams, M. D., Myers, E. W., Li, P. W., Mural, R. J., Sutton, G. G., Smith, H. O., Yandell, M., Evans, C. A., Holt, R. A., Gocayne, J. D., Amanatides, P., Ballew, R. M., Huson, D. H., Wortman, J. R., Zhang, Q., Kodira, C. D., Zheng, X. H., Chen, L., Skupski, M., Subramanian, G., Thomas, P. D., Zhang, J., Gabor Miklos, G. L., Nelson, C., Broder, S., Clark, A. G., Nadeau, J., McKusick, V. A., Zinder, N., Levine, A. J., Roberts, R. J., Simon, M., Slayman, C., Hunkapiller, M., Bolanos, R., Delcher, A., Dew, I., Fasulo, D., Flanigan, M., Florea, L., Halpern, A., Hannenhalli, S., Kravitz, S., Levy, S., Mobarry, C., Reinert, K., Remington, K., Abu-Threideh, J., Beasley, E., Biddick, K., Bonazzi, V., Brandon, R., Cargill, M., Chandramouliswaran, I., Charlab, R., Chaturvedi, K., Deng, Z., Di Francesco, V., Dunn, P., Eilbeck, K., Evangelista, C., Gabrielian, A. E., Gan, W., Ge, W., Gong, F., Gu, Z., Guan, P., Heiman, T. J., Higgins, M. E., Ji, R. R., Ke, Z., Ketchum, K. A., Lai, Z., Lei, Y., Li, Z., Li, J., Liang, Y., Lin, X., Lu, F., Merkulov, G. V., Milshina, N., Moore, H. M., Naik, A. K., Narayan, V. A., Neelam, B., Nusskern, D., Rusch, D. B., Salzberg, S., Shao, W., Shue, B., Sun, J., Wang, Z., Wang, A., Wang, X., Wang, J., Wei, M., Wides, R., Xiao, C., Yan, C. (2001). The sequence of the human genome. Science 291: 1304-1351.
11. The Arabidopsis Genome Initiative (2000) Sequence of the Arabidopsis thaliana genome. Nature 408:796-815. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_I%3A_Genes_Nucleic_Acids_Genomes_and_Chromosomes/4%3A_Genomes_and_Chromosomes/4.E%3A_Genomes_and_Chromosomes_%28Exercises%29.txt |
A fundamental property of living organisms is their ability to reproduce. Bacteria and fungi can divide to produce daughter cells that are identical to the parental cells. Sexually reproducing organisms produce offspring that are similar to themselves. On a cellular level, this reproduction occurs by mitosis, the process by which a single parental cell divides to produce two identical daughter cells. In the germ line of sexually reproducing organisms, a parental cell with a diploid genome produces four germ cells with a haploid genome via a specialized process called meiosis. In both of these processes, the genetic material must be duplicated prior to cell division so that the daughter cells receive a full complement of the genetic information. Thus accurate and complete replication of the DNA is essential to the ability of a cell organism to reproduce.
In this chapter and the next, we will examine the process of replication. After describing the basic mechanism of DNA replication, we discuss the various techniques researchers have used to achieve a more complete understanding of replication. Indeed, a theme of this chapter is the combination of genetic and biochemical approaches that has allowed us to uncover the mechanism and physiology of DNA replication. In the remaining sections of the chapter, we focus on the enzymes that mediate DNA replication. In these descriptions, you will encounter several cases of structure suggesting a particular function. We will point out parallels and homologies between bacterial and eukaryotic replication components. This chapter covers the basic process and enzymology of DNA synthesis, and the next chapter will cover regulation of DNA replication.
Further readings
• A. Kornberg and T. Baker (1992) DNA Replication, 2nd Edition, W.H. Freeman and Company, New York.
• A. Kornberg, I. R. Lerman, M. J. Bessman, and E. S. Simms (1956) "Enzymic synthesis of deoxyribonucleic acid" Biochimica et Biophysica Acta 21:197-198.
• M. Meselson and F. W. Stahl (1958) "The replication of DNA in Escherichia coli." Proceedings of the National Academy of Sciences, USA 44:671-682.
• R. Okazaki, T. Okazaki, K. Sakabe, K. Sugimoto, and A. Sugino (1968) "Mechanism of DNA Chain Growth, I. Possible Discontinuity and Unusual Secondary Structure of Newly Synthesized Chains" Proceedings of the National Academy of Sciences, USA 59: 598-605.
• P. De Lucia and J. Cairns (1969) Isolation of an E. colistrain with a mutation affecting DNA polymerase. Nature 224:1164-1166.
• J. Gross and M. Gross (1969) Genetic analysis of an E. colistrain with a mutation affecting DNA polymerase. Nature 224:1166-1168
• R. Sousa (1996) Trends in Biochemical Sciences 21:186-190. Similarities in structure among DNA polymerases
• Herendeen and Kelly (1996) Cell 84:5-8. Subunits and mechanism of DNA polymerase III.
5. DNA replication I: Enzymes and mechanism
Question 5.8
Imagine you are investigating the replication of a bacterial species called B. mulligan. The bacteria is grown for several generations in medium containing a heavy density label, [15N] NH4Cl. The bacteria are then shifted to medium containing normal density [14N] NH4Cl. DNA is extracted after each generation and analyzed on a CsCl gradient. From the results shown below, what is the mode of replication in B. mulligan? Explain your conclusion.
Question 5.9
How many turns must be unwound during replication of the E. coli chromosome? The chromosome contains 4.64 x 106 base pairs.
Question 5.10
Which of the following comments about Okazaki fragments are true or false? Okazaki fragments:
1. are short segments of newly synthesized DNA.
2. are formed by synthesis on the leading strand of DNA.
3. have a short stretch of RNA, or a mixture of ribonucleotides and deoxyribonucleotides, at their 5' end.
4. account for overall synthesis of one DNA strand in a 3' to 5' direction.
Question 5.11.
The following experimental results are from A. Sugino and R. Okazaki (1972) "Mechanisms of DNA Chain Growth VII. Direction and rate of growth of T4 nascent short DNA chains" J. Mol. Biol. 64: 61-85.
a. E. colicells were infected with bacteriophage T4 and then chilled to 4oC to slow the rate of replication. Replicating DNA in the infected cells was pulse-labeled with [3H]-thymidine (a) or [3H]-thymine (b) for 5 sec (black-filled circles), 30 sec (open circles with vertical line), 60 sec (open circles with dot) or 300 sec (open circles). The pulse labeling was stopped with potassium cyanide and ice, and the DNA was extracted, denatured in NaOH, and separated on an alkaline sucrose gradient. Fractions from the gradient were collected and assayed for the amount of 3H in the DNA (as material that bound to a filter after washing in (a) and as acid-insoluble material in (b)). The sedimentation value in Svedbergs (S) is given along the x-axis; faster sedimenting material is toward the right. What do these data tell you about the sizes of nascent (newly synthesized) DNA at the various pulse labeling times?
(b) Sugino and Okazaki used a method to break the isolated short nascent chains (completed Okazaki fragments) randomly and recover only the oligonucleotides from the 5Ã ends. They found that at very short labeling times (e.g. 5 sec) the [3H] thymidine was not at the 5' ends of the DNA (hence it was internal and at the 3' ends). After longer labeling times, the [3H] thymidine was found in the oligonucleotides at the 5' end. What do you conclude is the direction of chain growth of the nascent chains? Explain your logic.
Question 5.12
We have covered two experiments from the Okazaki lab using pulse labeling for increasing times to follow the synthesis of new DNA. How would you design a pulse-chase experiment to monitor not only the initial production of Okazaki fragments, but also their incorporation into larger DNA molecules?
Question 5.13
Which enzymes, substrates, and cofactors are used in common and which ones are distinctive for synthesis of leading strands and lagging strands of DNA at the replication fork of E. coli?
Question 5.14
Which subunit or complex within E. coli DNA polymerase III holoenzyme has each the following functions?
1. Catalyzes 5'to 3' polymerization of new DNA.
2. Has the proofreading function (3' to 5' exonuclease).
3. Dimerizes the two catalytic cores.
4. Forms the clamp that is thought to account for its high processivity.
5. Loads and unloads the sliding clamp.
Question 5.15
What are the components of the multiprotein complex known as the primosome in E. coli? What does it do? In what direction does it travel?
Question 5.16
Which eukaryotic nuclear DNA polymerase(s) is (are) thought to account for leading strand and lagging strand synthesis? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/5._DNA_replication_I%3A_Enzymes_and_mechanism/5.E%3A_DNA_replication_I%3A_Enzymes_and_Mechanism_%.txt |
DNA Replication is Semiconservative
We begin our investigation by describing the basic model for how nucleotides are joined in a specific order during DNA replication. By the early 1950’s, it was clear that DNA was a linear string of deoxyribonucleotides. At that point, one could postulate three different ways to replicate the DNA of a cell. First, a cell might have a DNA-synthesizing "machine" which could be programmed to make a particular string of nucleotides for each chromosome. A second possibility is that the process of replication could break the parental DNA into pieces and use them to seed synthesis of new DNA.
A third model could be proposed from the DNA structure deduced by Watson and Crick. When they described the double-helical structure of DNA in a one-page article in Nature in 1953, they included this brief statement of a third model:
"It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material."
A subsequent paper elaborated on this mechanism. The complementarity between base pairs (A with T and G with C) not only holds the two strands of the double helix together, but the sequence of one strand is sufficient to determine the sequence of the other. Hence a third possibility for a mechanism of DNA replication was clear - one parental strand could serve as atemplate directing synthesis of a complementary strand in the daughter DNA molecules. This 1953 paper is of course most famous for its description of the double-helical structure of DNA held together by base complementarity, but it is also important because the proposed structure suggested a testable model for how a particular process occurs, in this case replication.
These three models make different predictions about the behavior of the two strands of the parental DNA during replication (Figure 5.1). In the first, programmed machine model, the two strands of the parental DNA can remain together, because they are not needed to determine the sequence of the daughter strands. This model of replication is called conservative: the parental DNA molecules are the same in the progeny as in the parent cell. In the second model, the each strand of the daughter DNA molecules would be a combination of old and new DNA. This type of replication is referred to as random (or dispersive). The third model, in which one strand of the parental DNA serves as a template directing the order of nucleotides on the new DNA strand, is a semiconservative mode of replication, because half of each parent duplex (i.e. one strand) remains intact in the daughter molecules.
When they were graduate students at the California Institute of Technology, Matthew Meselson and Franklin Stahl realized that they could test these three models for replication by distinguishing experimentally between old and new strands of DNA. They labeled the old or parental DNA with nucleotides composed of a heavy isotope of nitrogen (15N) by growing E. colicells for several generations in media containing [15N] NH4Cl. Ammonia is a precursor in the biosynthesis of the purine and pyrimidine bases, and hence this procedure labeled the nitrogen in the nucleotide bases in the DNA of the E. colicells with 15N. The cells were then shifted to grow in media containing the highly abundant, light isotope of nitrogen, 14N, in the NH4Cl, so that newly synthesized DNA would have a "light" density. The labeled, heavy (old) DNA could be separated from the unlabeled, light (new) DNA on a CsCl density gradient, in which the DNA bands at the position on the gradient where the concentration of CsCl has a density equal to that of the macromolecule. At progressive times after the shift to growth in [14N] NH4Cl, samples of the cells were collected, then DNA was isolated from the cells and separated on a CsCl gradient.
A. B.
Figure 5.2. Results of the Meselson and Stahl experiment demonstrating semiconservative replication of DNA. A. The left panel (a) shows ultraviolet absorption photographs of DNA after equilibrium sedimentation in a CsCl gradient, as a function of the number of generations from the shift from media that labeled DNA with a high density (15N-labeled) to a medium in which the DNA is normal, or light density (14N-DNA). The density of the CsCl gradient increases to the right. The panel on the right (b) shows a trace of the amount of DNA along the gradient. The number of generations since the shift to the media with 14N substrates is shown at the far right. Mixing experiments at the bottom show the positions of uniformly light and heavy DNA (generations 0 and 4.1 mixed) and the mixture of those plus hybrid light and heavy DNA (generations 0 and 1.9 mixed). Parental DNA forms a band at the heavy density (15N-labeled), whereas after one generation in light (14N) media, all the DNA forms a band at a hybrid density (between heavy and light). Continued growth in light media leads to the synthesis of DNA that is only light density. B. The interpretation of the experimental results as demonstrating a semiconservative model of replication. Part A of this figure is Figure 4 and Part B is Figure 6 from M. Meselson and F. Stahl (1958) “The Replication of DNA in Escherichia coli” Proceedings of the National Academy of Sciences, USA 44:671-682.
The results fit the pattern expected for semiconservative replication (Figure 5.2). To quote from Meselson and Stahl, “until one generation time has elapsed, half-labeled molecules accumulate, while fully labeled DNA is depleted. One generation time after the addition of 14N, these half-labeled or ‘hybrid’ molecules alone are observed. Subsequently, only half-labeled DNA and completely unlabeled DNA are found. When two generation times have elapsed after the addition of 14N, half-labeled and unlabeled DNA are present in equal amounts.” A conservative mode of replication is ruled out by the observation that all the DNA formed a band at a hybrid density after one generation in the [14N] NH4Cl-containing medium. However, it is consistent with either the semiconservative or random models. As expected for semiconservative replication, half of the DNA was at a hybrid density and half was at a light density after two generations in [14N] NH4Cl-containing medium. Further growth in the 14N medium resulted in an increase in the amount of DNA in the LL band.
Exercise
Question 5.1: What data from this experiment rule out a random mode of replication?
These experiments demonstrated that each parental DNA strand is used as a template directing synthesis of a new strand during DNA replication. The synthesis of new DNA is directed by base complementarity. The enzymes that carry out replication are not programmed “machines” with an inherent specificity to synthesize a given sequence, but rather the template strand of DNA determines the order of nucleotides along the newly synthesized DNA strand (Figure 5.3).
The association of a parental DNA strand with a newly synthesized DNA strand observed in this important experimental result is consistent with the use of each parental DNA strand as a template to direct the replication machinery to place nucleotides in a particular order. Watson and Crick proposed that base complementarity would guide the replication machinery to insert an A opposite a T, a T opposite an A, a G opposite a C, and C opposite a G (Figure 5.3). This was verified once the enzymes carrying out DNA synthesis were isolated, and the chemical composition of the products of replication was compared with that of the templates. These enzymes are discussed in detail later in the chapter, as will be the chemistry of the process of adding individual nucleotides to the growing DNA chain (a process called elongation). You may recall that these enzymes were also used to demonstrate the antiparallel arrangement of the DNA strands predicted by Watson and Crick (recall problem 2.5). With this understanding of how the sequence of nucleotides is specified, we can examine the types of DNA structures found during replication. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/5._DNA_replication_I%3A_Enzymes_and_mechanism/Basic_Mechanisms_of_Replication.txt |
Making primers for DNA synthesis
The enzyme primase catalyzes the synthesis of the primers from which DNA polymerases can begin synthesis (Figure 5.21). Primers are short oligonucleotides, ranging from 6 to 60 nucleotides long. They can be made of ribonucleotides or a mixture of deoxyribonucleotides and ribonucleotides. The principal primase in E. coli is the 60 kDa protein called DnaG protein, the product of the dnaG gene. The major primase in eukaryotic cells is DNA polymerase $\alpha$.
The primase of E. coli, DnaG protein, cannot synthesize primers by itself, but rather it is part of much larger complex called the primosome. The primosome acts repeatedly during lagging strand synthesis, finding a primer-binding site on the SSB-coated single-stranded template strand and synthesizing a primer. Identification of the components of the primosome was aided by the convenient model system of in vitrosynthesis of fX174 DNA. fX174 is a single-stranded bacteriophage; the DNA found in the virus is termed the plus strand. After infection of E. coli, this plus strand is converted to a double-stranded replicative form (Figure 5.24 A). The conversion of single‑stranded phage DNA to duplex DNA occurs by the synthesis of several Okazaki fragments, and hence it is a good model for discontinuous synthesis on the lagging strand. This reaction can be carried out in vitro, which allowed the biochemical dissection of the various steps in primosome assembly and movement.
Table 5.3. Components of the primosome in E. coli
protein gene activities and functions
PriA priA helicase, 3' to 5' movement, site recognition
PriB priB
PriC priC
DnaT dnaT needed to add DnaB‑DnaC complex to preprimosome
DnaC dnaC forms complex with DnaB
DnaB dnaB helicase, 5' to 3' movement. DNA dependent ATPase.
DnaG dnaG synthesize primer
Five different proteins are found in a prepriming complex, PriA, PriB, PriC, DnaT, and DnaB (Table 5.4). A sixth protein, DnaC, is needed for the assembly of this complex. In the case of fX174 viral DNA template coated with SSB, PriA (Figure 5.24 B) recognizes a primer assembly site. The proteins PriBand PriCare then added to form a complex. The hexameric protein DnaBis in a complex with six molecules of DnaCwhen it is not on the DNA. In an ATP-dependent process, and with help from DnaT, DnaB is transferred to the template and DnaC is released.
The prepriming complex is now ready for the primase, DnaG, to bind and make the active primosome. Although the role of each of the proteins in the primosome is not yet clear, information is available on some of the steps in primosome action. The preferred binding site on the template for primase is CTG, with the T being used as the template for the first nucleotide of the primers. A high affinity complex between DnaB and ATP forms or stabilizes a secondary structure in the single-stranded template DNA that is used by primase; this is thought to be how DnaB "activates" the primase to begin synthesis. After ATP hydrolysis by DnaB, the low affinity ADP-DnaB complex dissociates from the template. The primosome can now move to the next site for primer synthesis.
The primosome contains two helicases than can move along single-stranded DNA with opposite polarity. PriA moves in a 3' to 5' direction, whereas DnaB moves in a 5' to 3' direction. When tested in vitrowith a substrate similar to that shown in Figure 5.22, fragments from each end are displaced, indicating that the primosome moved in one direction on some molecules and in the other direction on others. Figure 5.24 B shows the migration as driven by the DnaB helicase, but movement can also occur in the other direction as well.
DNA topology during replication
Figure 5.22B shows the results of the tracking assay for a helicase called PriA. In what direction does it track along the single-stranded DNA? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/5._DNA_replication_I%3A_Enzymes_and_mechanism/DNA_Primers_for_Synthesis.txt |
Eukaryotic replication proteins have functions analogous to those found in bacteria.
The template-primer junctions are recognized by the multisubunit replication factor C, or RFC. Like the g complex in E. coli, this enzyme is an ATPase, and it helps to load on the processivity factor PCNA. Thus RFC is carrying out a similar function to the bacterial g-complex.
One of the first eukaryotic polymerases to be isolated was DNA polymerase a, which is now recognized as a catalyst of primer synthesis. This enzyme contains four polypeptide subunits, one with a polymerase activity (170 kDa), two that comprise a primase activity (50 and 60 kDa), and another subunit of (currently) undetermined function (70 kDa). DNA polymerase a has low processivity but high fidelity. This high fidelity is surprising because no 3' to 5' exonuclease is associated with the enzyme. Polymerase a, possibly with additional primases, catalyzes the synthesis of short segments of DNA and RNA that serve as primers for the replicative polymerases.
DNA polymerase e is related to polymerase d, and it may play a role in lagging strand synthesis. It is also dependent on PCNA, in vivo. However, no requirement has been identified for it in viral replication systems in vitro.
The compound aphidicolin will block the growth of mammalian cells. It does this by preventing DNA replication, and the targets of this drug are DNA polymerases a and d (as well as e). The fact that inhibition of these DNA polymerases with aphidicolin also stops DNA replication in mammalian cells argues that indeed, a and d are responsible for replication of nuclear DNA in eukaryotic cells. This conclusion is strongly supported by the phenotype of conditional loss-of-function mutations in the genes encoding the homologs to these polymerases in yeast. Such mutants do not grow at the restrictive temperature, indicating that d and a are the replicative polymerases. The biochemical evidence implicates polymerase a in primer formation, and d appears to be the major polymerases used to synthesize the new strands of DNA.
Table 5.4: Analogous components of the replication machinery in E. coliand eukaryotic cells.
Function
Bacterial (E. coli)
Number of subunits
Eukaryotic replication (SV40)
Number of subunits
Leading and lagging strand synthesis
asymmetric dimer, E. colipolymerase III
10 (3 in core)
polymerase d
2
Sliding clamp
b subunit
2
PCNA
3
Clamp loader
g-complex
6
RFC
multiple
Primase
DnaG
1
Polymerase a
4
Helicase
DnaB
6
T-antigen (SV40)
6
Bind single-stranded DNA
SSB
1
RFA
3
Swivel
Gyrase
4, A2B2
Topo I
or Topo II
1
2 (homodimer)
The parallels between bacterial and eukaryotic DNA replication are striking. The overall strategy of synthesis is similar, and analogous proteins carry out similar functions, as listed in Table 5.4. It is difficult to determine whether the proteins carrying out similar functions are actually homologous proteins, i.e. encoded by genes descended from the same gene in the last common ancestor. The protein sequence identities are marginal, and frequently the analogous proteins have different numbers of subunits. These differences complicate the analysis considerably, because different subunits in bacteria or mammals may have similar functions. However, the functional similarities are convincing.
Several other DNA polymerases have been isolated from eukaryotic cells. DNA polymerase b and e are involved in repair of nuclear DNA. DNA polymerase b is a single polypeptide of 36 kDa, and has no 3' to 5' exonuclease. DNA polymerase g replicates mitochondrial DNA.
Reverse transcriptaseis frequently referred to as an RNA-dependent DNA polymerase because it can use RNA as a template, but in fact it can use either RNA or DNA as a template. It is encoded by retroviruses, and hence it is present in cells infected with a retrovirus. This enzyme has widespread use in the laboratory for making complementary copies of RNA, called cDNA. Active copies of LINE1 repetitive elements (in mammals) or Ty1 repeats (in yeast), also encode reverse transcriptase. Thus in cells where these retrotransposable elements are being transcribed, active reverse transcriptase is also present. Reverse transcriptase also has an RNase H activity, which will digest away RNA from an RNA-DNA duplex.
In contrast to the other DNA polymerases discussed in this chapter, terminal deoxynucleotidyl transferasedoes not require a template. It adds dNTPs (as dNMP) to the 3' end of DNA, using that 3' hydroxyl as a primer. It is found in differentiating lymphocytes, and appears to be used physiologically to introduce somatic mutations into immunoglobulin genes. In the laboratory, it is used to add "homopolymer tails" to the ends of DNA molecules by incubating a linear DNA with one particular dNTP and terminal deoxynucleotidyl transferase.
As will be discussed in more detail in the next chapter, the ends of linear chromosomes (telomeres) must be expanded at each replication or they will eventually become shortened. The enzyme telomerasecatalyzes the addition of many tandem copies of a simple sequence to the ends of the chromosomes. The template for this reaction is an RNA that is a component of the enzyme. Thus telomerase is a reverse transcriptase that only makes copies of the template that it carries, using the 3' end of a chromosomal DNA strand as the primer. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/5._DNA_replication_I%3A_Enzymes_and_mechanism/Eukaryotic_Replication_Proteins.txt |
Of all the enzymatic functions needed for replication of DNA, the ability to catalyze the incorporation of deoxynucleotides into DNA is most central. Enzymes that catalyze this reaction, DNA polymerases, have been isolated from many species, and many species have multiple DNA polymerases. Our earliest and most complete understanding of the mechanism of these enzymes comes from studies of the first DNA polymerase isolated, called DNA polymerase I.
Mechanism of nucleotide addition by DNA polymerases
In 1956 Arthur Kornberg and his co-workers isolated a protein from E. colithat has many of the properties expected for a DNA polymerase used in replication. In particular, it catalyzes synthesis of DNA from deoxynucleotides, it requires a template and it synthesizes the complement of the template. It is a single polypeptide chain of 928 amino acids, and it is the product of the polAgene. We now understand that this an abundant polymerase, but rather than synthesizing new DNA at the replication fork, it is used during the process of joining Okazaki fragments after synthesis and in DNA repair. Detailed studies of DNA polymerase I have been invaluable to our understanding of the mechanisms of polymerization. Although DNA polymerase I is not the replicative polymerase in E. coli, homologous enzymes are used in replication in other species. Also, the story of how the replicative DNA polymerases were detected in E. coliis a classic illustration of the power of combining biochemistry and genetics to achieve a more complete understanding of an important cellular process.
DNA polymerase I catalyzes the polymerization of dNTPs into DNA. This occurs by the addition of a dNTP (as dNMP) to the 3' end of a DNA chain, hence chain growth occurs in a 5' to 3' direction (Figure 5.11). In this reaction, the 3' hydroxyl at the end of the growing chain is a nucleophile, attacking the phosphorus atom in the a-phosphate of the incoming dNTP. The reaction proceeds by forming a phosphoester between the 3' end of the growing chain and the 5' phosphate of the incoming nucleotide, forming a phosphodiester linkage with the new nucleotide and liberating pyrophosphate (abbreviated PPi). Thus in this reaction, a phosphoanhydride bond in the dNTP is broken, and a phosphodiester is formed. The free energy change for breaking and forming these covalent bonds is slightly unfavorable for the reaction as shown. However, additional noncovalent interactions, such as hydrogen bonding of the new nucleotide to its complementary nucleotide and base-stacking interactions with neighboring nucleotides, contribute to make a total free energy change that is favorable to the reaction in the synthetic direction. Nevertheless, at high concentrations of pyrophosphate, the reaction can be reversed. In the reaction in the reverse direction, nucleotides are progressively removed and released as dNTP in a pyrophosphorolysisreaction. This is unlikely to be of large physiological significance, because a ubiquitous pyrophosphatase catalyzes the hydrolysis of the pyrophosphate to molecules of phosphate. This latter reaction is strongly favored thermodynamically in the direction of hydrolysis. Thus the combined reactions of adding a new nucleotide to a growing DNA chain and pyrophosphate hydrolysis insure that the overall reactions favors DNA synthesis. The basic chemistry of addition of nucleotides to a growing polynucleotide chain outlined in Figure 5.11 is common to virtually all DNA and RNA polymerases.
The DNA synthesis reaction catalyzed by DNA polymerase I requires Mg2+, which is a cofactor for catalysis, and the four deoxynucleoside triphosphates (dNTPs), which are the monomeric building blocks for the growing polymer. The reaction also requires a template strand of DNA to direct synthesis of the new strand, as predicted by the double helical model for DNA and confirmed by the Meselson and Stahl experiment.
This reaction also requires a primer, which is a molecule (usually a chain of DNA or RNA) that provides the 3’ hydroxyl to which the incoming nucleotide is added. DNA polymerases cannot start synthesis on a template by simply joining two nucleotides. Instead, they catalyze the addition of a dNTP to a pre-existing chain of nucleotides; this previously synthesized chain is the primer. The primer is complementary to the template, and the 3’ end of the primer binds to the enzyme at the active site for polymerization (Figure 5.12). When a new DNA chain is being made, once a new nucleotide has been added to the growing chain, its 3' hydroxyl is now the end of the primer. The polymerase moves forward one nucleotide so that this new primer end is at the active site for polymerization. The alternative view, that the DNA primer-template moves while the DNA polymerase remains fixed, is also possible. In both cases the last nucleotide added is now the 3' end of the primer, and the next nucleotide on the template is ready to direct binding of another nucleoside triphosphate.
For the initial synthesis of the beginning of a new DNA chain, a primer has to be generated by a different enzyme; this will be discussed in more detail later in the chapter. For example, short oligoribonucleotides are the primers for the Okazaki fragments; these are found at the 5' ends of the Okazaki fragments and are made by an enzyme called primase. The RNA primers are removed and replaced with DNA (by DNA polymerase I) before ligation.
These requirements for Mg2+, deoxynucleotides and two types of DNA strands (template and primer) were discovered in studies of DNA polymerase I. We now realize that they are also required by all DNA polymerases.
The polymerization active site for DNA polymerase I has a specific dNTP-binding site (Figure 5.12), and the active site adjusts to the deoxynucleotide on the template strand to favor binding of the complementary deoxynucleotide at the active site. Thus the polymerase catalyzes addition to the growing chain of the deoxynucleotide complementary to the deoxynucleotide in the template strand.
In the reaction catalyzed by DNA polymerase I, and all other DNA polymerases studied, the incoming deoxynucleotide is activated. The phosphoanhydride bonds in the triphosphate form of the deoxynucleotide are high-energy bonds (i.e., they have a negative, or favored, free energy of hydrolysis), and the b- and g- phosphates make a good leaving group (as pyrophosphate) after the nucleophilic attack. In contrast, the end of the growing DNA chain is not activated; it is a simple 3'-hydroxyl on the last deoxynucleotide added. This addition of an activated monomer to an unactivated growing polymer is called a tail-growth mechanism. DNA polymerases using this mechanism can only synthesize in a 5' to 3' direction, and all known DNA and RNA polymerases do this. Some other macromolecules, such as proteins, are made by a head-growth mechanism. In this case, the nonactivated end of a monomer attacks the activated end of the polymer. The lengthened chain again contains an activated head (from the last monomer added).
Exercise
Question 5.4. Describe a hypothetical head-growth mechanism for DNA synthesis. In which direction does chain synthesis occur in this mechanism?
Proofreading the newly synthesized DNA by a 3’ to 5’ exonuclease that is part of the DNA polymerase
The protein DNA polymerase I has additional enzymatic activities related to DNA synthesis. One, a 3’ to 5’ exonuclease, is intimately involved in the accuracy of replication. Nucleases are enzymes that catalyze the breakdown of DNA or RNA into smaller fragments and/or nucleotides. An exonuclease catalyzes cleavage of nucleotides from the end of a DNA or RNA polymer. An endonuclease catalyzes cutting within a DNA or RNA polymer. These two activities can be distinguished by the ability of an endonuclease, but not an exonuclease, to cut a circular substrate. A 3’ to 5’ exonuclease removes nucleotides from the 3’ end of a DNA or RNA molecule.
DNA synthesis must be highly accurate to insure that the genetic information is passed on to progeny largely unaltered. Bacteria such as E. colican have a mutation rate, as low as one nucleotide substitution in about 109 to 1010 nucleotides. This low error frequency is accomplished by a strong preference of the polymerase for the nucleotide complementary to the template, which allows about one substitution every 104 to 105 nucleotides. The accuracy of DNA synthesis is enhanced by a proofreading function in the polymerase that removes incorrectly incorporated nucleotides at the end of the growing chain. With proofreading, the accuracy of DNA synthesis is improved by a factor of 102 to 103, so the combined effects of nucleotide discrimination at the polymerization active site plus proofreading allows only about one substitution in 106 to 108 nucleotides. Further reduction in the error rate is achieved by mismatch repair (Chapter 7).
The proofreading function of DNA polymerase I is carried out by a 3' to 5' exonuclease (Figure 5.13). It is located in a different region of the enzyme from the active site for polymerization. When an incorrect nucleotide is added to the 3' end of a growing chain, the rate of polymerization decreases greatly. The primer-template moves to a different active site on the enzyme, the one with the 3’ to 5’ exonucleolytic activity. The incorrect nucleotide is cleaved, and the primer-template moves back to the polymerization active site to resume synthesis.The enzyme distinguishes between correct and incorrect nucleotides at the 3’ end of the primer, such that the 3' to 5' exonuclease much more active when the terminus of the growing chain is not base paired correctly, but the polymerase activity exceeds that of the 3' to 5' exonuclease activity when the correct nucleotide is added.
The polymerizing activity and the proofreading 3' to 5' exonuclease found in DNA polymerase I are also found in most other DNA polymerases. These are central activities to DNA replication.
Tail growth mechanisms allow proofreading and subsequent elongation. If the end of the growing chain were activated (as in head growth), then proofreading would eliminate the activated end and elongation could not continue.
D.
A.
C.
B.
Exercise
Question 5.4 Removal of a nucleotide from the 3’ end of the growing chain by a 3’ to 5’ exonuclease is not the reverse of the polymerase reaction. Can you state what the difference is?
Removal of nucleotides by a 5' to 3' exonuclease that is part of DNA polymerase I
In addition to the polymerase and 3’ to 5’ exonuclease common to most DNA polymerases, DNA polymerase I has an unusual 5' to 3' exonucleolytic activity. This enzyme catalyzes the removal of nucleotides in base-paired regions and can excise either DNA or RNA. It is used by the cell to remove RNA primers from Okazaki fragments and in repair of damaged DNA.
This 5' to 3' exonuclease, in combination with the polymerase, has useful applications in the laboratory. One common use is to label DNA in vitro by nick translation (Figure 5.14). In this process, DNA polymerase I will remove the DNA from a nicked strand by the 5' to 3' exonuclease, and then use the exposed 3' hydroxyl at the nick as a primer for new DNA synthesis by the 5' to 3' polymerase, thereby replacing the old DNA. The result is also a movement, or translation, of the nick from one point on the DNA to another, hence the process is called nick translation. If the reaction is carried out in the presence of one or more radiolabeled deoxynucleoside triphosphates (e.g., [a32P] dNTPs), then the new DNA will be radioactively labeled.
A similar process can be used to repair DNA in a cell. As will be discussed in Chapter 7, specific enzymes recognize a damaged nucleotide and cleave upstream of the damage. One way to remove the damaged DNA and replace it with the correct sequence is with the 5' to 3' exonuclease of DNA polymerase I and accompanying DNA synthesis.
[a32P]
Figure 5.14.The 5' to 3' exonuclease of DNA polymerase I can be used in nick translation to label DNA in vitro.
Structural domains of DNA polymerase I
Further understanding of the mechanism of the three enzymatic functions of DNA polymerase can be obtained from a study of the three-dimensional (3-D) structure of the protein. Much of our knowledge of the structure of DNA polymerase I has come from biochemical characterization and more recently by determination of the 3-D structure using X-ray crystallography. These studies have shown that distinct structural domains of DNA polymerase I contain the different catalytic activities. Also, the 3-D structure provided the first look at what is now recognized as a common structure for many polymerases.
Mild treatment with the protease subtilisin cleaves DNA polymerase I into two fragments. The small fragment contains the 5' to 3' exonuclease, and the larger, or "Klenow," fragment (named for the biochemist who did the cleavage analysis) contains the polymerase and the proofreading 3' to 5' exonuclease (Figure 5.15). Thus the two activities common to most polymerases are together in the Klenow fragment, whereas the distinctive 5' to 3' exonuclease is in a separable domain. The fact that a mild treatment with a protease without a precise sequence specificity indicates that an exposed, readily cleaved domain connects the large and small fragments. Both these observations suggest that the 5' to 3' exonuclease was an active domain added to a polymerase plus proofreading domain during the evolution of E. coli. The Klenow polymerase is used in several applications in the laboratory, e.g., labeling the ends of restriction fragments by filling in the overhangs and sequencing by the dideoxynucleotide chain termination method.
The 3-D structure of the large fragment of DNA polymerase I, determined by crystallography, provides additional insight into the enzymatic functions of key structural components. The large fragment has a deep cleft, about 30 Å deep, into which the template strand and primer bind. This cleft resembles a "cupped right hand" as illustrated in Figure 5.16. The "palm" is formed by a series of b-sheets and the thumb and fingers are made by a-helices. The polymerase active site has been mapped within the deep cleft, with contributions from the b-sheets that form the palm and the a-helices forming the fingers. You can see more detailed views of the structure of the Klenow fragment at the Course/Book web site (currently www.bmb.psu.edu/courses/bmb400/default.htm. Click on the link to kinetic images, download the MAGE program and the kinemage file for DNA polymerase I, and view them on your own computer.)
The 3' to 5' proofreading exonuclease is located in another part of the structure of the Klenow fragment, about 25 Å from the polymerase active site. Thus the primer terminus has to move this distance in order for the enzyme to remove misincorporated nucleotides.
The large Klenow fragment of the E. coliDNA polymerase I lacks the 5’ to 3’ exonuclease, so the 3-D structure of the Klenow fragment gives no information about that exonuclease. However, the 5’ to 3’ exonuclease domain can be seen in the structure of DNA polymerase from the thermophilic bacterium Thermus aquaticus. This protein structure is very similar to that of DNA polymerase I of E. coliin the polymerase and 3’ to 5’ exonuclease domains, and it has an additional 5' to 3' exonuclease domain located about 70 Å from the polymerase active site. This is a large distance, but remember that this exonuclease is working on a different region of the DNA molecule than the polymerase. The 5’ to 3’ exonuclease uses one part of the DNA molecule as a substrate for excising primers or removing damaged DNA, whereas the polymerase uses a different part of the DNA molecule as a template to direct synthesis of a new strand.
Curiously, a region homologous to the proofreading 3' to 5' exonuclease domain of DNA polymerase I is present in the Thermus aquaticuspolymerase structure, but it is no longer functional. The absence of proofreading accounts for the elevated error rate in this polymerase used very commonly for amplification of DNA by PCR. Of course, this polymerase is used in PCR because it is stable at the high temperatures encountered during the cycles of PCR. Some other thermostable polymerases with a lower error rate have become available more recently for use in PCR.
Similar "cupped right hand" structures occur in the tertiary structure of T7 RNA polymerase and the HIV reverse transcriptase. Thus DNA polymerase I was the first member described in what we now realize is a large class of nucleic acid polymerases. This family includes single unit polymerases for both RNA and DNA synthesis. You can access a tutorial on the T7 DNA polymerase at www.clunet.edu/BioDev/omm/exh...s.htm#displays. This structure has some similarities to that of DNA polymerase I.
Physiological role of DNA polymerase I
Although studies of DNA polymerase I have provided much information about the mechanism of DNA synthesis, genetic analysis has shown that the polymerase function of this enzyme is not required for DNA replication. DNA polymerase I is encoded by the polAgene in E. coli.However, no mutant allele of polAwas isolated in screens for conditional mutants defective in DNA replication. The most compelling argument that this polymerase is not required for replication came from an examination of thousands of E. colimutants, assaying them for DNA polymerase I activity. A mutant polA strain was isolated (Figure 5.16). This mutant allele, called polA1, contained a nonsense codon, leading to premature termination of synthesis of the product polypeptide and hence a loss of polymerase function. However, the mutant strain grew at a normal rate, which shows that DNA polymerase I is notrequired for DNA synthesis. The most striking phenotype of the polA1mutant was its strongly reduced ability to repair DNA damage. Further investigation led to the isolation of conditional lethal alleles of the polAgene. The mutant DNA polymerase I proteins encoded by these conditional lethal alleles are defective in the 5' to 3' exonuclease activity, demonstrating that this activity is required for cell viability. The 5' to 3' exonuclease activity removes RNA primers during synthesis of the lagging strand at the replication fork, and it is used in DNA repair.
DNA polymerase III is a highly processive, replicative polymerase
The conclusion that DNA polymerase I is not the replicative polymerase for E. coliled to the obvious question of what enzyme is actually used during replication. Investigation of the genes isolated in screens for mutants that are conditionally deficient in replication led to the answer. The replicative polymerase in E. coliis DNA polymerase III.
DNA polymerase I is more abundant than other polymerases in E. coliand obscures their activity. Thus the depletion of DNA polymerase I activity in polA1mutant cells (Figure 5.17) provided the opportunity to observe the other DNA polymerases. DNA polymerases II and III were isolated from extracts of polA1cells, named in the order of their discovery.
DNA polymerase IIis a single polypeptide chain whose function is uncertain. Strains having a mutated gene for DNA polymerase II (polB1) show no defect in growth or replication. However, the activity of DNA polymerase II is increased during induced repair of DNA, and it may function to synthesize DNA opposite a deleted base on the template strand.
Genetic evidence clearly shows that DNA polymerase IIIis used to replicate the E. coli chromosome. This enzyme is composed of multiple polypeptide subunits. Several of the genes encoding these polypeptide subunits were identified in screens for conditional lethal mutants defective in DNA replication. Loss of function of these dnagenes blocks replication, showing that their products are required for replication.
Low abundance and high processivity of DNA polymerase III
DNA polymerase III has many of the properties expected for a replicative polymerase. One of the complications to studies of DNA polymerase III is that different forms were isolated by various procedures. We now realize that these forms differ in the number of subunits present in the isolated enzyme. For enzymes with multiple subunits, we refer to the complex with all the subunits needed for its major function as the holoenzyme or holocomplex. The DNA polymerase holoenzyme has ten subunits, which will be discussed in detail in the next section.
It is the DNA polymerase holoenzyme that has the properties expected for a replicative polymerase, whereas DNA polymerase I does not (see comparison in Table 5.1). It is less abundantthan DNA polymerase I, but large number of replicative DNA polymerases are not needed in the cell. Only one or two polymerases can be used at each replication fork, so the 10 molecules of the DNA polymerase III holoenzyme will suffice. DNA polymerase III catalyzes DNA synthesis at a considerably higher ratethan DNA polymerase I, by a factor of about 70. The elongation rate measured for the DNA polymerase III holoenzyme (42,000 nucleotides per min) is close to the rate of replication fork movement measured in vivoin E. coli(60,000 nucleotides per min).
A key property for a replicative DNA polymerase is high processivity, which is a striking characteristic of the DNA polymerase III holoenzyme. Processivity is the amount of polymerization catalyzed by an enzyme each time it binds to an appropriate template, or primer-template in the case of DNA polymerases. It is measured in nucleotides polymerized per binding event. In order to replicate the 4.5 megabase chromosome of E. coliin 30 to 40 min, DNA polymerase needs to synthesize DNA rapidly, and in a highly processive manner. DNA polymerase I synthesizes less than 200 nucleotides per binding event, but as the holoenzyme, DNA polymerase III is much more processive, exceeding the limits of the assay used to obtain the results summarized in Table 5.1. In contrast, the DNA polymerase III core, which has only three subunits (see next section), has very low processivity.
Table 5.1. Comparison of DNA polymerases I and III (Pol I and Pol III)
Property
Pol I
Pol III core
Pol III holoenzyme
molecules per cell
400
40
10
nucleotides polymerized min-1 (molecule enzyme)-1
600
9000
42,000
processivity [nucleotides polymerized per initiation]
3-188
10
>105
5' to 3' polymerase
+
+
+
3' to 5' exonuclease, proofreading
+
+
+
5' to 3' exonuclease
+
-
-
Note: + and – refer to the presence or absence of the stated activity in the enzyme.
Question 5.6. If the rate of replication fork movement measured in vivo in E. coliis 60,000 nucleotides per min, how many forks are needed to replicate the chromosome in 40 min? Recall that the size of the E. colichromosome is 4.64 ´ 106 bp.
Subunits and mechanism of DNA polymerase III
The DNA polymerase III enzyme has four distinct functional components, and several of these contain multiple subunits, as listed in Table 5.2 and illustrated in Figure 5.18. The a and e subunits contain the major polymerizing and proofreading activities, respectively. They combine with the q subunit to form the catalytic core of the polymerase. This core can be dimerized by the t2 linker protein to form a subassembly called DNA polymerase III'. Addition of the third functional component, the g complex, generates another subassembly denoted DNA polymerase III*. All of these subassemblies have been isolated from E. coliand have been characterized extensively. The final component is the b2 dimer, which when combined with DNA polymerase III* forms the holoenzyme.
The various activities of DNA polymerase III can be assigned to individual subunits (Table 5.2). For instance, the major polymerase is in the a subunit, which is encoded by the dnaEgene (also known as polC). The 3' to 5' exonuclease is in the e subunit, which is encoded by the dnaQgene (also known as the mutDgene). However, maximal activity is obtained with combinations of subunits. The DNA polymerase III core is a complex of the a, e and q subunits, and the activity of the core in both polymerase and 3' to 5' exonuclease assays is higher in than in the isolated subunits.
Table 5.2. Subassemblies of DNA polymerase III, major subunits, genes and functions
Functional component
Subunit
Mass (kDa)
Gene
Activity or function
Core polymerase
a
129.9
polC=dnaE
5' to 3' polymerase
e
27.5
dnaQ=mutD
3'-5' exonuclease
q
8.6
Stimulates e exonuclease
Linker protein
t
71.1
dnaX
Dimerizes cores
Clamp loader
g
47.5
dnaX
Binds ATP
(or g complex)
d
38.7
Binds to b
(ATPase)
d'
36.9
Binds to g and b
c
16.6
Binds to SSB
y
15.2
Binds to c and g
Sliding clamp
b
40.6
dnaN
Processivity factor
The activities of the subunits can be measured in vitroby appropriate biochemical assays. In addition, the phenotype of mutations in the gene encoding a given subunit can show that subunit is required for a particular process. Mutant a subunits are the product of conditional lethal alleles discovered in screens for dnagenes, but they also were discovered as the product of polymerase-defective alleles defining the polCgene. Thus the dnaEgene is the same as same as the polCgene, showing that this subunit with polymerase activity is needed in replication. Similarly, the phenotype of mutations in the gene encoding the e subunit shows that it is needed for proofreading. Mutant alleles of the dnaQgene were identified in a screen for mutator genes, which generate a high frequency of mutants in bacteria when defective. These alleles defined a gene mutD, which was subsequently shown to be the same as dnaQ. The mutator phenotype of mutant dnaQ/mutDstrains results from a lack of proofreading by the e subunit during replication, allowing more frequent incorporation of incorrect nucleotides into DNA.
The b2 dimer is the key protein that confers highprocessivityon DNA polymerase III. Association of the b2 dimer with DNA polymerase III increases the processivity from about 10 nucleotides polymerized per binding event to over 100,000 nucleotides polymerized per binding event (Table 5.1). This dimeric protein forms a ring through which the duplex DNA can pass; the ring will slide easily along DNA unless impeded, as, for example, by proteins bound to the template DNA. Thus the b2 dimer acts as a sliding clamp, holding the polymerase onto the DNA being copied. Once DNA polymerase III is associated with the clamp on DNA, it will polymerize until it reaches the next primer for an Okazaki fragment during lagging strand synthesis. For leading strand synthesis, the DNA polymerase presumably remains associated with the DNA via the b2 clamp until the chromosomal DNA is completely replicated. The 3-D structure of the b2 dimer, determined by X-ray crystallography, shows a macromolecular ring. This structure can be viewed at the web site for the course and at the Online Museum of Macromolecules (www.clunet.edu/BioDev/omm/exh...s.htm#displays).
The g-complex contains several subunits: two molecules of g subunits and one molecule each of d, d', c, and y. It loadsthe b2 dimer clamp onto a primer-template, in a process that requires ATP hydrolysis (Figure 5.19). The catalytic core of DNA polymerase III will then link to the template-bound clamp and will initiate highly processive replication. The g-complex also serves to unload the clamp once an Okazaki fragment is completed during lagging strand synthesis; hence it is both a clamp loader and unloader, allowing the polymerase and the clamp to cycle repeatedly from one Okazaki fragment to another.
The g-complex carries out these opposite activities on different structures, loading on the clamp at a template-primer and unloading the clamp at the end of a completed Okazaki fragment. For instance, encountering the 5' end of the previously synthesized Okazaki fragment may be the distinctive structure that shifts the g-complex into its unloading mode. It does not unload the clamp while DNA polymerase III is catalyzing polymerization.
Figure 5.19 illustrates the proposed steps in this process. The g-complex in the ATP-bound form binds the b2 clamp, whereas the g-complex in the ADP-bound form releases the b2 clamp. Thus loading and unloading depend on a round of ATP hydrolysis. When the g-complex in the ATP-bound form binds the b2 clamp, the DNA polymerase III holoenzyme is in a conformation that allows it to find a primer-template. The ring of the b2 clamp is held open by the g-complex-ATP, allowing it to bind around a primer-template. Hydrolysis of ATP by the g-complex leaves it in an ADP-bound form. In this new, ADP-bound conformation of the g-complex, it dissociates from the b2 clamp, thereby allowing the b2 clamp to bind to the catalytic core of the holoenzyme and also close around the primer-template. The holoenzyme is now ready to catalyze processive DNA synthesis. Elongation continues until the holoenzyme encounters a previously synthesized Okazaki fragment. Now the g-complex binds ATP (presumably by an ADP-ATP exchange reaction) and shifts into the conformation for binding to the b2 clamp and taking it off the DNA template. This half of the holoenzyme is now able to dissociate from the template and find the next primer-template junction to begin synthesis of another Okazaki fragment.
The clamp loading and unloading activities of the g-complex are a cycle of changes in protein associations. These changes occur because of the enzymatic activities of the complex, which in turn alter the conformations of the proteins and their preferred interactions. As shown in Figure 5.19, the g-complex is an ATPase, which is an enzyme that catalyzes the hydrolysis of ATP to ADP and phosphate. It is also an ATP-ADP exchange factor.
Changes in conformation and activity of proteins depending on whether they are bound to a nucleoside triphosphate (ATP or GTP) or a nucleoside diphosphate (ADP or GDP) is a common theme in biochemistry. The GTP-bound forms of proteins, which can be turned off by GTP-hydrolysis and reactivated by GDP-GTP exchange proteins, mediate critical cell signaling events. As will be seen in Chapter 14, GTP- and GDP-bound forms of translation factors carry out opposite functions. Proteins assume different conformations depending on the cofactor bound (in this case a nucleotide), and each conformation has a distinct activity. The ability to change the conformation by a hydrolytic activity (converting ATP to ADP and phosphate) allows the protein to shift activities readily.
The two catalytic cores of DNA polymerase III are joined together by the t subunits to make an asymmetric dimer (see Figure 5.18). The half of the holoenzyme without the g complex is proposed to synthesize the leading strand of new DNA, and the core with the g complex is proposed to synthesize the lagging strand. Both of the cores in the asymmetric dimer are associated with a b2 clamp at the replication fork. In this model, synthesis of boththe leading and lagging strands is catalyzed by the sameDNA polymerase III complex, thereby coordinating synthesis of both new strands strand. Note that if the template for lagging strand synthesis is looped around the enzyme, then leading and lagging strand synthesis would be occurring in the same direction as replication fork movement (Figure 5.20), despite the opposite polarities of the two template strands. Thus the asymmetric dimer model suggests a means to couple both leading strand and lagging strand synthesis.
Replication machinery
C.
B.
A. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/5._DNA_replication_I%3A_Enzymes_and_mechanism/Polymerases.txt |
Regulation is largely exerted at the initiation of replication, and methods for finding origins and termini of replication will be covered. The proteins involved in control of replication initiation in E. coli and yeast will be discussed. One solution to the problem of completing the synthesis of linear DNAs in eukaryotes will be described - that of making telomeres. Some of the factors controlling the rate of initiation of replication will be discussed briefly.
6. DNA replication II: Start stop and control
It is critical that all the DNA in a cell be replicated once, and only once, per cell cycle. Jacob, Brenner and Cuzin defined a replicon as the unit in which the cell controls individual acts of replication. The replicon initiates and completes synthesis once per cell cycle. Control is exerted primarily at initiation. They proposed that an initiator protein interacted with a DNA sequence, called a replicator, to start replication. The replicator can be identified genetically as a DNA sequence required for replication, whereas the origin is defined by physical or biochemical methods as the DNA sequence at which replication begins. For many replicons, such as the E. coli oriC and the autonomously replicating sequences (or ARS) in yeast, the replicator is also an origin. However, this need not be the case: the replicon for amplified chorion genes in silkmoths has an origin close to, but separable from, the replicator. Initiator proteins have now been identified for some replicons, such as the DnaA protein in E. coli and the Origin Recognition Complex in the yeast Saccharomyces cerevisiae. In both cases, they bind to the replicators, which are also origins in these two species.
The replicator is a sequence of DNA needed for synthesis of the rest of the DNA in a replicon. It is a control element that affects the chromosome on which it lies. We say that this element acts in cis, since the replicator and the replicon are on the same chromosome. In contrast, the initiator is a protein that can be encoded on any chromosome in a cell. Thus is acts in trans, since it does not have to be encoded on the same chromosome as the replicon that it controls. In general, a trans-acting factor is an entity, usually a protein, that can diffuse through the cell to act in regulation of a certain target, whereas a cis-acting DNA sequence is on the same chromosome as the target of control. This pattern of a trans-acting protein binding to a cis-acting site on the DNA is also seen in transcriptional control.
Exercise
Although E. coli has a single origin in a single replicon, eukaryotic chromosomes have multiple origins, and multiple replicons. Consider a line of mammalian cells growing in culture that has an S phase of 5 hr, i.e. all the genome is replicated in 5 hr. The haploid genome size is 3 x 109 bp. If the rate of replication fork movement in this cell lines is 2000 bp per min, how many replication origins are required to replicate the entire haploid genome during S phase? Assume that two replication forks emerge from each origin (this is bidirectional replication, see below).
Experimental approaches to map origins and termini of replication and to distinguish between uni‑ and bidirectional replication
Several experimental techniques have been established for finding where replication begins and ends on chromosomes, and for distinguishing between unidirectional and bidirectional replication. We will cover two major techniques.
6.2: Structural analysis of pulse-labeled DNA m
Table 6.1. Appearance of radiolabel into restriction fragments of completed SV40 DNA molecules. The relative amount of pulse label from each restriction fragment is given below (the relative amount of pulse label is the 3H/32P ratio of each fragment, corrected for thymidine content and normalized to 1 for fragment A).
Relative amount of pulse label
Fragment
5 min
10 min
15 min
A
1.0
1.0
1.0
B
3.9
3.0
2.3
C
0
0.75
0.75
D
0.92
0.86
1.1
E
1.8
2.0
1.7
F
4.0
3.1
2.4
G
5.4
4.2
2.6
H
1.7
2.5
2.0
I
2.7
3.0
2.2
J
4.9
3.7
2.6
K
2.4
2.9
1.9
Question 6.2: What would the pattern be for unidirectional replication?
Question 6.3: What would be the pattern if there were two origins, say in fragments E and H, with bidirectional replication from each?
6.3: Twodimensional gels to analyze the number
Question 6.4. A restriction map is shown for a portion of a chromosome below, along with the patterns on two-dimensional gels for the replication intermediates formed by each fragment. Where are the origins and termini? Can you deduce the direction of replication fork movement?
Question 6.5. How can you calculate the position of an origin within a DNA fragment from an asymmetric fork/bubble pattern on a 2-D gel of replicating DNA molecules? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/6._DNA_replication_II%3A_Start_stop_and_control/6.1%3A_The_Replicon.txt |
6.12
At what step is the rate of DNA replication in E. coli is regulated - initiation, elongation or termination?
6.13
The following problem further illustrates the analysis of replication by pulse-labeling, using a hypothetical virus and constructed data. Consider the replication of a circular viral DNA in infected cells. The infected cells were pulse labeled with [3H] thymidine for 1, 2, 3 and 4 min; it takes 4 min for the DNA molecules to be replicated in this system (from initiation to termination). Those DNA molecules that had completed synthesis at each time point were isolated, cut with a restriction endonuclease, and assayed for radioactivity in each fragment. This restriction endonuclease cleaves the circular DNA into 6 fragments, named A, B, C, D, E, and F in a clockwise orientation around the genome. The following results were obtained; a plus (+) means the fragment was radioactively labeled, and a minus (-) means it was not labeled.
Fragment Time of labeling (min)
1 2 3 4
A - - + +
B - - - +
C - - + +
D - + + +
E + + + +
F - + + +
1. What restriction fragment has the origin and which has the terminus of replication?
2. In which direction(s) does this viral DNA replicate?
6.14
The two-dimensional gels developed by Brewer and Fangman were used to examine the origin of replication of a DNA molecule. In this system, replicating molecules are cleaved with a restriction endonuclease and separated in two dimensions. The first dimension separates on the basis of size, and the second separates on the basis of shape (more pronounced deviations from linearity move slower in the second dimension). After blotting the DNA onto a membrane, it is probed with fragments from the replicon under study. Restriction fragment P gives the pattern shown on the left, and the adjacent fragment Q gives the pattern shown on the right. The dotted line denotes the diagonal expected if all molecules were linear. Assuming both P and Q are in the same replicon, what can you conclude about the positions of origins of replication?
6.15
Dr. Howard Cedar and his colleagues at the Hebrew University in Jerusalem have developed a replication direction assay to map origins of replication on chromosomes (Kitsberg et al., Nature 366: 588-590, 1993). Growing cells are treated with the drug emetine to inhibit lagging strand synthesis. Leading strand synthesis continues, and this newly synthesized DNA is density labeled by incorporating 5-bromodeoxyuridylate (5-bromodeoxyuridine is added to the medium). The DNA is then sheared and denatured, and the newly synthesized leading strand DNA is separated from the rest of the DNA by sedimentation equilibrium on Cs2SO4 gradients. Samples of the heavy density DNA (containing 5-bromodeoxyuridylate) are spotted onto a membrane, and equal amounts are hybridized to labeled, separated strands of restriction fragments throughout a region.
Use of this approach to map replication origins in the human b-like globin gene cluster led to results like those below. The names of the genes are given above the line, and the names of the restriction fragments are given below the line. A + means that the leading strand (with 5-bromodeoxyuridylate incorporated) hybridized preferentially to a labeled probe corresponding to the designated strand, whereas a - means that the leading strand DNA did not hybridize to the designated probe. The genes are transcribed from left to right in this diagram, so the "top" strand reads the same as the mRNA in the coding regions (our convention is "nontemplate") and the "bottom" strand (abbreviated "bot") is complementary to the top strand ("template” or "antisense" strand).
e Gg Ag yh d b
|_|_-->_|____|_-->_|_-->_|____|__-->_|___|_-->_|___|__|_-->_|____|
A B C D E F G H I J K L M
+ + + + + + + + + + + - - top
- - - - - - - - - - - + + bot
1. Which restriction fragment(s) contain(s) the origin of replication?
2. Is replication from this origin uni- or bi-directional?
3. Explain how the data led you to your answers to a and b.
4. What direction is the replication fork moving for fragments A through K?
5. What direction is the replication fork moving for fragments L and M?
6. Name a possible target enzyme that could specifically block lagging strand synthesis when inhibited.
7. What cloning vector would be useful for generating the separated strands of the restriction fragments?
6.16
Let's imagine that you have isolated a new virus with a double-stranded, circular DNA that is 6000 bp long. The restriction endonuclease HhaI cleaves the DNA as shown below to generate 6 fragments.
You initially use a pulse-labeling procedure to map the origin and terminus of replication. Infected cells were first allowed to incorporate [32P] phosphate into the DNA for several hours to uniformly label the DNA, and then [3H] thymidine was added for short periods of time (pulse labels), i.e. 5, 10 and 15 min. Completed viral DNA molecules were isolated, cut with HhaI, and separated on polyacrylamide gels. The amount of [32P] and [3H] in each fragment was determined for each period of pulse label and is tabulated below. The data are corrected for thymidine content and normalized so that fragment A has a ratio of 1.
Relative amount of pulse label
Fragment 5 min 10 min 15 min
A 1.0 1.0 1.0
B 0.5 0.7 1.0
C 0 0.5 0.8
D 5.0 4.1 2.3
E 4.2 3.2 1.7
F 2.9 2.1 1.4
1. Which HhaI fragment(s) contain(s) the origin and terminus of replication?
2. What is the mode (uni- or bi-directional, or other) and direction(s) of replication (i.e. clockwise and/or counterclockwise)?
3. To confirm this result and map the origin and terminus more precisely, you analyzed the replicative intermediates on 2-dimensional gels. The DNA from infected cells, containing viral DNAs at all stages of synthesis, was digested with HhaI and then run initially on a gel that separates on the basis of size and then in a perpendicular direction in a gel that accentuates separations based on shape (Brewer and Fangman gels). The DNA in the gel was blotted onto a nylon membrane and hybridized with radiolabeled probes for the viral DNA fragments. The hybridization patterns obtained for HhaI fragments A, C and D are shown. The hypothetical line for linear intermediates of a fragment expanding from unit length to twice unit length is provided as a guide. How do you interpret these data, and what do you learn about the origin and terminus? Please indicate the significance of any transitions in the patterns.
1. (d) You also used a replication direction assay to examine the replication origin. Virally infected cells were treated with the drug emetine to inhibit lagging strand synthesis. Leading strand synthesis continued during the drug treatment, and this newly synthesized DNA was density labeled by incorporating 5-bromodeoxyuridylate (5-bromodeoxyuridine is added to the medium). The DNA was sheared and denatured, and the newly synthesized leading strand DNA was separated from the rest of the DNA by sedimentation equilibrium on Cs2SO4 gradients. Samples of the heavy density DNA (containing 5-bromodeoxyuridylate) were spotted onto a membrane, and equal amounts are hybridized to labeled, separated strands of restriction fragments throughout the virus. To keep track of strands and orientation in this problem, lets imagine the duplex circle to have an outerstrand oriented 5' to 3' in a clockwise direction and an innerstrand oriented 5' to 3' in a counterclockwise direction, as diagrammed below.
A grid of samples of heavy density DNA (containing 5-bromodeoxyuridylate, and enriched for leading strand DNA) immobilized on the filter is shown below, with each rectangle representing an equal loading of the heavy density DNA. What will be the pattern of hybridization to the indicated strands of each of the restriction fragments?
What does this experiment tell you about the origin and terminus of replication?
6.17
Are the following statements about the function of the DnaA protein true or false?
1. DnaA protein binds to 9-mer (nonamer) repeats at the origin for chromosomal replication.
2. DnaA protein catalyzes the formation of the primers for leading strand synthesis at the origin.
3. About 20 to 40 monomers of the DnaA protein form a large complex at the origin.
4. DnaA protein melts DNA at a series of 13-mer repeats at the origin.
6.18
Consider a bacterium with a circular chromosome with one replication origin. It takes 30 min for bi-directional replication to copy its chromosome (the elongation time or C period) and 10 min from the end of DNA synthesis until the cell divides (the D period). How many replication forks are needed per chromosome to allow a culture of this bacterium to double in cell number every 20 min? Follow the molecules through a complete cell division cycle.
6.19
In many eukaryotes, actively transcribed genes are replicated early in S phase and inactive genes are replicated late. One assay to determine replication timing is in situ hybridization of cells with a gene-specific, fluorescent probe, followed by examination of the number of signals per nucleus. In diploid cells, an unreplicated gene will be seen as 2 fluorescent dots per nucleus, whereas a replicated gene will be seen as 4 dots. They look like 2 doublets, indicating that the replicated chromatids are close in the nucleus.
The types of pattern one can see at various stages of the cell cycle are shown below. Each dark dot is a fluorescent signal, the larger circle is the cell, and the smaller circle is the nucleus.
The fraction of cells in an asynchronous population with 2 dots or 4 dots is then tabulated. In an asynchronous population, the number of cells in each phase of the cell cycle is directly proportional to the length of that phase. If GENEAwere replicated 1 hr after entry into S phase, and GENEBwere replicated 1 hr before the end of S phase, what fraction of cells would show 4 dots (two doublets) for each? The length of each phase of the cell cycle is given in the figure, and the vertical arrowhead shows the time of synthesis. The time from synthesis of each gene until the beginning of G2 is shown above a horizontal line. Consider cells in M to have 4 dots (i.e., assume that the transition from 4 dots to 2 occurs at the M to G1 boundary). | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/6._DNA_replication_II%3A_Start_stop_and_control/6.E%3A_DNA_replication_II%3A_Start_stop_and_contr.txt |
A new round of replication will initiate on the E. coli chromosome at oriC only when the growth conditions permit it. The dam methylase and features of its sites of action are used to prevent premature re-initiation. The dam methylase of E. coli recognizes the tetranucleotide GATC in DNA and transfers a methyl group (from S‑adenosyl methionine) to the amino group at position 6 of the adenine in that sequence. Note that GATC is a pseudopalindrome, so both strands read the same for these four nucletides in DNA.
Palindrome and Pseudopalindromes
A nucleotide sequence is said to be a palindrome if it has an even number of base pairs and is equal to the reverse of its complementary sequence. For example, in a single strand of DNA the sequence of bases CCATTAATGG is palindromic because the sequence of bases in the complementary strand is GGTAATTACC, its reverse.
A pseudopalindrome is a DNA sequence with an odd number of base pairs yielding a symmetrical complement except at the central base-pair. For example, the DNA sequence ACCTGGT is pseudopalindromic, because its complement on the other strand is TGGACCA, which is its reverse except for the central element.
Thus a GATC in duplex DNA can be unmethylated on either strand, methylated on only one strand (referred to as hemimethylated) or methylated on both strands (referred to as fully methylated), as shown in Figure \(1\).
The methylation status of the 11 GATC motifs at oriCregulate whether replication can intitiate. When the GATCs are fully methylated, oriCDNA serves as an origin (in the presence of Dna A and the other proteins discussed above). However, when the GATCs are hemimethylated, it is not active as an origin. The reason for this is not fully known. One hint comes from the behavior of unmethylated oriC(from dam- strains). This unmethylated oriC is active, showing that methylation of the GATC is not a requirement for initiation, and further suggesting that some inhibitor of initiation recognizes the hemimethylated form.
\(1\)
How do these results lead to this conclusion? Let’s explore this by posing the opposite hypotheses.
1. If methylation of the GATC motifs at oriC were needed for initiation, what would the result have been?
2. If some activator recognized the fully methylated form, what would the result have been.?
Re-methylation of oriCby the dam methylase is quite slow. Thus for some period the GATCs at oriCare hemimethylated, and the origin is inactive. This provides a means to delay the use of oriCto start another round of replication. Thus the methylation of the GATCs is part of a mechanism to regulate the timing of firing of oriC. In the next chapter, we will also see the use of methylation of GATCs in post-replicative repair.
Linear Templates
The problem of linear templates
The requirement of DNA polymerases to have a primer causes a problem at the ends of linear templates. As illustrated in Figure 6.13, leading strand synthesis can proceed to the end of its template strand, but lagging strand synthesis cannot. As lagging strand synthesis nears the end of its template, at some point no binding site will be available for primase, and part of the 3’ end of the template for lagging strand synthesis will not be copied. Hence a 3’ overhang is left after the replication fork has finished, and part of the chromosome is not copied into new DNA. If nothing else were done, the chromosome would become progressively shorter after each round of replication.
At least three different types of solution to this problem have been discovered in various organisms. One, utilized by bacteriophage such as l and T4, is to convert the linear template to a circle. For instance, the linear chromosome of bacteriophage l has cohesive ends (complementary single strands at each end, generated by a phage endonuclease) that can anneal upon infection, thereby forming a circular template for replication. Other viruses, such as adenovirus, attach a protein to the end of unreplicated DNA to serve as a primer. Such an attached protein obviates the requirement for using the unreplicated DNA as a template, and the entire viral chromosome can be replicated.
A third solution is to make the ends a series of simple repeats that are synthesized in a process distinct from DNA replication. Indeed, the ends of the linear chromosomes of most (perhaps all) eukaryotes, called telomeres, are composed of many copies of a simple repetitive sequence. This sequence is distinctive for different organisms, but in all cases one strand is rich in G and the other is rich in C. The repeating unit for human telomeres is 5' AGGGTT 3' running from the centromeric end of the repeats to the telomeric end), and the repeating units for the ciliate Tetrahymena is 5' GGGGTT 3’.
New copies of the telomeric repeats can be synthesized each time the chromosome replicates (Figure 6.14). This re-synthesis of the telomeric repeats counteracts the progressive shortening of the linear chromosomes that would occur if only the replication forks were used to synthesize new chromosomes.
In this figure, the string of "a" at the ends of the chromosome is the tandem repeat of simple sequence, in duplex form. For instance, for a human chromosome, "a" would be
CEN ... 5' AGGGTT 3' ... TEL
3' TCCCAA 5'
or for a Tetrahymena chromosome, "a" would be
CEN ... 5' GGGGTT 3' ... TEL
3' CCCCAA 5'
In each case, the "a" or monomer is repeated thousands of times in tandem.
Addition of new telomeric repeats is catalyzed by the enzyme telomerase. As illustrated in Figure 6.15, this enzyme catalyzes many successive rounds of synthesis, adding many copies of the simple repeat to the ends of the chromosomes. The enzyme is a ribonucleoprotein, i.e. it has both a polypeptide and an RNA component. The RNA serves as a template to direct addition of nucleotides to the 3' end of the G+T rich strand, and the polypeptide acts as a reverse transcriptase to make a DNA copy of a hexanuclotide segment of the RNA. For instance, the telomerase from Tetrahymena will copy the 3’CCCCAA in the RNA template into 5’GGGGTT telomeric repeat. Then the enzyme shifts over and synthesizes another hexanucleotide. The fact that the RNA serves as the template was demonstrated by exchanging the RNA component of isolated telomerase with the telomerase RNA from a second species. This exchange led to the addition of telomeres with sequences characteristic of that of the second species, showing that the telomerase RNA is the determinant of the sequence of the telomere. The protein component provides the reverse transcriptase activity.
Once many copies of the G+T-rich strand of telomeres have been synthesized by telomerase, the long single strand forms a specialized structure toward the 3’ end. Some evidence indicates that a “G-quartet” is formed, in which four guanine nucleotides form a hydrogen-bonded complex. Examination of the ends of replicating chromosomes in the electron microscope show a circular structure. Although details of the structure at the end of this strand are not fully established, it is likely that a primer to support synthesis of the C+A-rich strand is made effectively by turning the G+T-rich strand around. Conventional synthesis by DNA polymerases can then copy the G+T-rich strand to make the complementary strand. Some processing, e.g. nucleases acting at the end, can convert the specialized structure or hairpin into a linear duplex.
How processive is telomerase?
Not all replicating cells have telomerase activity. This activity is higher in some transformed cells than in nontransformed cells. Also, older cells tend to have shorter telomeres. Thus telomeres are being actively investigated as possibly playing roles in both aging and in tumorigenic transformation.
Telomeres are important for stabilizing chromosomes. Some chromosomal deletions remove the ends of the chromosome, including the telomere, and these shortened chromosomes are less stable than their wild-type counterparts. Directed mutations have been made in mice to eliminate telomerase activity. These mice are viable for several generations, but they show many broken and abnormal chromosomes, demonstrating the importance of this activity. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/6._DNA_replication_II%3A_Start_stop_and_control/Control_of_initiation_at_oriC_by_methylation.txt |
Cellular control of replication in bacteria
We have seen that the initiator protein DnaA and the replicator element oriCare needed for the initiation of replication, and that the slow rate of methylation at GATC motifs prevents re-initiation for some time. The bacterial cell can sense when the nutritional conditions, levels of nucleotide pools, and protein concentrations are adequate to support a round of replication. The details of this monitoring are beyond the scope of this presentation, and can be explored in references such as Niedhart et al. In general, initiation is triggered by the increase in cell mass. Initiation occurs at a constant ratio of cell mass to the number of origins. This suggests that a mechanism exists to titrate out some regulatory molecule as the cell mass increases, but the molecule and mechanism have not been elucidated.
The result of this monitoring and signalling is the formation of an active DnaA complex at oriC, followed by unwinding the DNA and the other events discussed above.
Depending on the growth conditions, bacteria can divide rapidly or slowly. In rich media, the cell number can double every 18 min, whereas when nutrients are scare, the doubling time can be long as 180 min. The bacterial cells accomplish this by varying the rate of re-initiation of replication. Re-initiation has to occur at the same frequency as the cell doubling time.
Although the frequency of re-initiation can be varied 10-fold, the time required for the replication cycle is constant. This cycle consists of two periods called C and D. The elongation time, or C period, is the time required to replicate the bacterial chromosome. From initiation to termination, this is about 40 min. The division time, or D period, is the time that elapses between completion of a round of DNA replication and completion of cell division. This is about 20 min. Hence the time for the replication cycle (C period plus D period) is essentially constant in bacterial cultures with doubling times shorter than 60 min.
In order to accommodate the variation in cell doubling time within the constraints of the constant time for replication (C+D), rapidly growing bacteria have chromosomes with multiple replication forks. The constant replication cycle time means that a round of replication must be initiated 60 min (i.e. C+D) before cell division. However, re-initiation can occur before 60 min has past. This is illustrated in Figure 6.16 for cells in a culture dividing every 30 min. When the cell doubling time is less than 60 min, a cycle of replication must initiate before the end of the preceding cycle. This results in chromosomes with more than one replication fork.
Exercise \(1\)
If the time required for two replication forks traveling in opposite directions to traverse the entire E. coli chromosome at 37oC is about 40 min, regardless of the culture conditions and the time required for cell division (D period) is 20 min, how many replication forks will be present on each DNA molecule in the culture?
Answer
TBA
Replication in Eukaryotes
Passage from one phase to the next is a highly regulated event. Critical control points, or checkpoints, are found at the G1 to S transition and at the G2 to M transition. The checkpoint in late G1 is the time for the cell to assess whether it has enough nucleotides, proteins and other materials to make two cells. The checkpoint prior to the G2/M transition allows any necessary repairs or corrections in the DNA to be made prior to mitosis. Loss of control of the G0 to G1 transition, or at the other checkpoints, generates cells that grow in an uncontrolled manner. This inappropriate expansion in the number of cells is fundamental to the progression of cancers, and hence the study of the molecular events at these checkpoints is an intensely active area of research in cell biology and biochemistry. A full treatment of this important topic is beyond the scope of this course. In general, cell cycle progression is regulated by environmental signals (such as extracellular growth factors) and intracellular monitors of metabolic state, intactness of DNA, and so forth. These disparate signals eventually impinge on highly regulated protein kinases. Activation of particular protein kinases is required for progression through each checkpoint. In general, two types of regulation have been seen.
1. Control of the amount of key proteins. The concentration of proteins called cyclins rise and fall through the cell cycle. Some of the cyclins are components of protein kinases whose activity regulates passage through the checkpoints. The cyclins must be present at a sufficiently high concentration for the kinase to be active.
2. Control of the state of phosphorylation. Proteins regulating the cell cycle (as is true of many regulatory proteins) can be covalently modified, e.g. by phosphorylation in a process catalyzed by protein kinases. The state of phosphorylation will determine the level of activity of the protein. So for instance a key protein kinase regulating passage through the G1 to S checkpoint must have its catalytic subunit in the correct state of phosphorylation, as well as having sufficient amounts of its cyclin subunit.
Many lines of investigation are being pursued to understand better the regulation of the cell cycle. One fundamental approach has been be isolation of scores of conditional yeast mutants that are defective in their progression through the cell cycle at the restrictive temperature. These mutants have particular phenotypes depending on which stage of the cell cycle they arrest in under nonpermissive conditions. The complementation groups defined by such mutants are called CDC, for cell division cycle phenotypes, followed by a number. For example, a protein kinase whose activity is needed for both the G1/S and the G2/M transition in S. cerevisiaeis the product of the CDC28 gene, and the polypeptide is called Cdc28p.
Once a cell has entered S phase, each origin of replication must fire once, but only once. As discussed above, the ORC is required for initiation of replication at an origin, but what determines when the origin fires? This is a matter of considerable current study, and many of the details are still unknown. In S. cerevisiae, the ORC binds to specific DNA sequences, the origins of replication, throughout the cell cycle, not just during S phase when the origins are active. During G1 phase, ORC recruits other proteins, such as Cdc6 and Mcm (minichromosome maintenace) proteins, to form a prereplication complex. At the G1/S transition, additional factors associate with this complex, and a cyclin-dependent kinase (CDK) activity stimulates intitiation of replication in S phase. After initiation, the Cdc6 and Mcm proteins are released from the prereplication complex, leaving the ORC still bound to the origin but unable to reinitiate replication until the next cell cycle. In mammals, an intact ORC is not stably bound to the origin, but rather one of the subunit, ORC1, is recruited to the origin at a defined time during G1. However, in both yeast and mammals, events in G1 involving the preinitiation complex mark an origin for firing in the next S phase.
As discussed above, many origins of replication, and hence many replicons, are used to replicate each chromosome. These origins do not all fire at the same time. In fact, replicons can initiate at different times during S phase. Replicons containing genes that are actively expressed in a given cells tend to replicate earlier in S phase than do replicons containing nonexpressed genes. This is an example of tissue-specific variation in replication timing. The time during S phase at which a particular origin will fire is determined early in G1, at the time that chromatin domains are repositioned in the nucleus following mitosis and before the preinitiation complex forms. Events important to the regulation of initiation at replication origins occur at various times during G1, but the full range of proteins and activities carrying out these events is still a matter of study.
Stages of DNA synthesis
The synthesis of any macromolecule proceeds in three stages: initiation, elongation and termination. This is true for DNA replication as well. During initiation, DNA synthesis begins at a specific site, called an origin of replication. The circular E. coli chromosome has a single origin, called oriC. Many bacteria have circular chromosomes with single origins of replication. However, other chromosomes, especially those in eukaryotes, can have multiple origins. During elongation, nucleotides are added to the growing DNA strand as the replication fork moves along the chromosome. Termination are the final steps that occur when all or an appropriate portion (replicon, see below) of the chromosome has been replicated.
The primary control of replication is exerted during initiation. This is economical, of course, since little benefit would come from initiating replication that will never be completed. As will be covered later in this chapter, an examination of the DNA structures, proteins and enzymes needed for initiation show that it is highly regulated. Initiation is an active process, requiring the accumulation of ATP-bound DNA binding proteins at a specific site prior to the start of replication. Both the activity of the initiator proteins and the state of covalent modification of the DNA at the origin are part of the control process. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/6._DNA_replication_II%3A_Start_stop_and_control/Replication_in_Bacteria.txt |
Most biological molecules have a limited lifetime. Many proteins, lipids and RNAs are degraded when they are no longer needed or damaged, and smaller molecules such as sugars are metabolized to compounds to make or store energy. In contrast, DNA is the most stable biological molecule known, befitting its role in storage of genetic information. The DNA is passed from one generation to another, and it is degraded only when cells die. However, it can change, i.e. it is mutable. Mutations, or changes in the nucleotide sequence, can result from errors during DNA replication, from covalent changes in structure because of reaction with chemical or physical agents in the environment, or from transposition. Most of the sequence alterations are repaired in cells. Some of the major avenues for changing DNA sequences and repairing those mutations will be discussed in this chapter.
7: Mutation and Repair of DNA
Sequence alteration in the genomic DNA is the fuel driving the course of evolution. Without such mutations, no changes would occur in populations of species to allow them to adapt to changes in the environment. Mutations in the DNA of germline cells fall into three categories with respect to their impact on evolution. Most have no effect on phenotype; these include sequence changes in the large portion of the genome that neither codes for protein, or is involved in gene regulation or any other process. Some of these neutral mutations will become prevalent in a population of organisms (or fixed) over long periods of time by stochastic processes. Other mutations do have a phenotype, one that is advantageous to the individuals carrying it. These mutations are fixed in populations rapidly (i.e. they are subject to positive selection). Other mutations have a detrimental phenotype, and these are cleared from the population quickly. They are subject to negative or purifying selection.
A red Darwin hybrid tulip "Apeldoorn" with a mutation that resulted in half of a petal being yellow. (CC BY-SA 3.0; LepoRello).
Whether a mutation is neutral, disadvantageous or useful is determined by where it is in the genome, what the type of change is, and the particulars of the environmental forces operating on the locus. For our purposes, it is important to realize that sequence changes are a natural part of DNA metabolism. However, the amount and types of mutations that accumulate in a genome are determined by the types and concentrations of mutagens to which a cell or organism is exposed, the efficiency of relevant repair processes, and the effect on phenotype in the organism.
7.1: Mutations and Mutagens
Mutations commonly are substitutions, in which a single nucleotide is changed into a different nucleotide. Other mutations result in the loss (deletion) or addition (insertion) of one or more nucleotides. These insertions or deletions can range from one to tens of thousands of nucleotides. Often an insertion or deletion is inferred from comparison of two homologous sequences, and it may be impossible to ascertain from the data given whether the presence of a segment in one sequence but not another resulted from an insertion of a deletion (in this case, it can be referred to as an "indel"). One mechanism for large insertions is the transposition of a sequence from one place in a genome to another (described in Chapter 9).
Types of muations
Nucleotide substitutions are one of two classes. In a transition, a purine nucleotide is replaced with a purine nucleotide, or a pyrimidine nucleotide is replaced with a pyrimidine nucleotide. In other words, the base in the new nucleotide is in the same chemical class as that of the original nucleotide. In a transversion, the chemical class of the base changes, i.e. a purine nucleotide is replaced with a pyrimidine nucleotide, or a pyrimidine nucleotide is replaced with a purine nucleotide.
Comparison of the sequences of homologous genes between species reveals a pronounced preference for transitions over transversions (about 10-fold), indicating that transitions occur much more frequently than transversions.
Errors in Replication
Despite effective proofreading functions in many DNA polymerases, occasionally the wrong nucleotide is incorporated. It is estimated that E. coliDNA polymerase III holoenzyme (with a fully functional proofreading activity) uses the wrong nucleotide during elongation about 1 in 108 times. It is more likely for an incorrect pyrimidine nucleotide to be incorporated opposite a purine nucleotide in the template strand, and for a purine nucleotide to be incorporated opposite a pyrimidine nucleotide. Thus these misincorporations resulting in a transition substitution are more common. However, incorporation of a pyrimidine nucleotide opposite another pyrimidine nucleotide, or a purine nucleotide opposite another purine nucleotide, can occur, albeit at progressively lower frequencies. These rarer misincorporations lead to transversions.
Exercise \(1\)
If a dCTP is incorporated into a growing DNA strand opposite an A in the template strand, what mutation will result? Is it a transition or a transversion?
Exercise \(2\)
If a dCTP is incorporated into a growing DNA strand opposite a T in the template strand, what mutation will result? Is it a transition or a transversion?
A change in the isomeric form of a purine or pyrimidine base in a nucleotide can result in a mutation. The base-pairing rules are based on the hydrogen-bonding capacity of nucleotides with their bases in the ketotautomer. A nucleotide whose base is in the enoltautomer can pair with the "wrong" base in another nucleotide. For example, a T in the rare enolisomer will pair with a keto G (Figure \(2\)), and an enolG will pair with a ketoT.
The enoltautomers of the normal deoxynucleotides guanidylate and thymidylate are rare, meaning that a single molecule is in the ketoform most of the time, or within a population of molecules, most of them are in the ketoform. However, certain nucleoside and base analogs adopt these alternative isomers more readily. For instance 5-bromo-deoxyuridine (or 5-BrdU) is an analog of deoxythymidine (dT) that is in the enoltautomer more frequently than dT is (although most of the time it is in the ketotautomer).
Thus the frequency of misincorporation can be increased by growth in the presence of base and nucleoside analogs. For example, growth in the presence of 5-BrdU results in an increase in the incorporation of G opposite a T in the DNA, as illustrated in Figure \(3\). After cells take up the nucleoside 5-BrdU, it is converted to 5-BrdUTP by nucleotide salvage enzymes that add phosphates to its 5’ end. During replication, 5-BrdUTP (in the ketotautomer) will incorporate opposite an A in DNA. The 5-BrdU can shift into the enolform while in DNA, so that when it serves as a template during the next round of replication (arrow 1 in the diagram below), it will direct incorporation of a G in the complementary strand. This G will in turn direct incorporation of a C in the top strand in the next round of replication (arrow 2). This leaves a C:G base pair where there was a T:A base pair in the parental DNA. Once the pyrimidine shifts back to the favored ketotautomer, it can direct incorporation of an A, to give the second product in the diagram below (with a BrU-A base pair).
Exercise \(3\)
Where are the hydrogen bonds in a base pair between enol –guanidine and keto-thymidine in DNA?
Likewise, misincorporation of A and C can occur when they are in the rare iminotautomers rather than the favored aminotautomers. In particular, iminoC will pair with aminoA, and iminoA will pair with aminoC (Figure \(4\)).
Misincorporation during replication is the major pathway for introducing transversions into DNA. Normally, DNA is a series of purine:pyrimidine base pairs, but in order to have a transversion, a pyrimidine has to be paired with another pyrimidine, or a purine with a purine. The DNA has to undergo local structural changes to accommodate these unusual base pairs. One way this can happen for a purine-purine base pair is for one of the purine nucleotides to shift from the preferred anticonformation to the synconformation. Atoms on the "back side" of the purine nucleotide in the syn-isomer can form hydrogen bonds with atoms in the rare tautomer of the purine nucleotide, still in the preferred anticonformation. For example, an A nucleotide in the syn-, amino- isomer can pair with an A nucleotide in the anti-, imino- form (Figure \(5\)). Thus the transversion required a shift in the tautomeric form of the base in one nucleotide as well as a change in the base-sugar conformation (antito syn) of the other nucleotide.
Exercise \(4\)
Why does the shift of a purine nucleotide from anti to synhelp allow a purine:purine base pair? Is this needed for a pyrimidine:pyrimidine base pair?
Errors in replication are not limited to substitutions. Slippage errors during replication will add or delete nucleotides. A DNA polymerase can insert additional nucleotides, more commonly when tandem short repeats are the template (e.g. repeating CA dinucleotides). Sometimes the template strand can loop out and form a secondary structure that the DNA polymerase does not read. In this case, a deletion in the nascent strand will result. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/7%3A_Mutation_and_Repair_of_DNA/7.0%3A_Prelude_to_Mutations.txt |
Many mutations do not result from errors in replication. Chemical reagents can oxidize and alkylate the bases in DNA, sometimes changing their base-pairing properties. Radiation can also damage DNA. Examples of these mutagenic reactions will be discussed in this section.
Chemical Modification by Oxidation
When the amino bases, adenine and cytosine, are oxidized, they also lose an amino group. Thus the amine is replaced by a keto group in the product of this oxidative deamination reaction. For instance, oxidation of cytosine produces uracil, which base pairs with adenine (shown for deoxycytidine in Figure 7.6). Likewise, oxidation of adenine yields hypoxanthine, which base pairs with cytosine (Figure 7.7.A). Thus the products of these chemical reactions will be mutations in the DNA, if not repaired. Oxidation of guanine yields xanthine (Figure 7.7.B). In DNA, xanthine will pair with cytosine, as does the original guanine, so this particular alteration is not mutagenic.
Exercise
Both hypoxanthine and xanthine can base pair with cytosine in DNA. Why is this?
Oxidation of C to U occurs spontaneously at a high rate. The frequency is such that 1 in 1000 Cs in the human genome would become Us during a lifetime, if they were not repaired. As will be discussed later, repair mechanisms have evolved to replace a U in DNA with a T.
Methylation of C prior to its oxidative deamination will effectively mask it from the repair processes to remove U’s from DNA. This has a substantial impact on the genomes of organisms that methylate C. In many eukaryotes, including vertebrates and plants (but not yeast or Drosophila), the principal DNA methyl transferase recognizes the dinucleotide CpG in DNA as the substrate, forming 5-methyl-CpG (Figure 7.8). When 5-methyl cytosine undergoes oxidative deamination, the result is 5-methyl uracil, which is the same as thymine. The surveillance system that recognizes U’s in DNA does nothing to the T, since it is a normal component of DNA. Hence the oxidation of 5-methyl CpG to TpG, followed by a round of replication, results in a C:G to T:A transition at former CpG sites (Figure 7.8). This spontaneous deamination is quite frequent; indeed, C to T transitions at CpG dinucleotides are the most common mutations in humans. Since this transition is not repaired, over time the number of CpG dinucleotides is greatly diminished in the genomes of vertebrates and plants.
Me
--CG-- --CG-- [O] --TG-- + NH3 --TG--
|||||| ® |||||| ® ||o||| ® |||||| +wt
--GC-- --GC-- --GC-- --AC--
Methyl- Replicate
transferase mutation
Figure 7.8. Methylation of CpG dinucleotides followed by oxidative deamination results in TpG dinucleotides.
Some regions of plant and vertebrate genomes do not show the usual depletion of CpG dinucleotides. Instead, the frequency of CpG approaches that of GpC or the frequency expected from the individual frequency of G and C in the genome. One working definition of these CpG islandsis that they are segments of genomic DNA at least 100 bp long with a CpG to GpC ratio of at least 0.6. These islands can be even longer and have a CpG/GpC > 0.75. They are distinctive regions of these genomes and are often found in promoters and other regulatory regions of genes. Examination of several of these CpG islands has shown that they are not methylated in any tissue, unlike most of the other CpGs in the genome. Current areas of research include investigating how the CpG islands escape methylation and their role in regulation of gene expression.
Exercise
If a CpG island were to be methylated in the germ line, what would be consequences be over many generations?
The rate of oxidation of bases in DNA can be increased by treating with appropriate reagents, such as nitrous acid (HNO2). Thus treatment with nitrous acid will increase the oxidation of C to U, and hence lead to C:G to T:A transitions in DNA. It will also increase the oxidation of adenine to hypoxanthine, leading to A:T to G:C transitions in DNA.
Chemical Modification by Alkylation
Many mutagens are alkylating agents. This means that they will add an alkyl group, such as methyl or ethyl, to a base in DNA. Examples of commonly used alkylating agents in laboratory work are N-methyl-nitrosoguanidine and N-methyl-N'-nitro-nitrosoguanidine (MNNG, Figure 7.9.A.). The chemical warfare agents sulfur mustard and nitrogen mustard are also alkylating agents.
N-methyl-nitrosoguanidine and MNNG transfer a methyl group to guanine (e.g. to the O6 position) and other bases (e.g. forming 3-methyladenine from adenine). The additional methyl (or other alkyl group) causes a distortion in the helix. The distorted helix can alter the base pairing properties. For instance, O6-methylguanine will sometimes base pair with thymine (Figure 7.9.B.).
A. N-methyl-N'-nitro-N-nitrosoguanidine (MNNG)
B. 6-O-methyl-G will pair with T
The order of reactivity of nucleophilic centers in purines follows roughly this series:
N7-G >> N3-A > N1-A @ N3-G @ O6-G.
A common laboratory reagent for purines in DNA is dimethylsulfate, or DMS. The products of this reaction are primarily N7-guanine, but N3-adenine is also detectable. This reaction is used to identify protein-binding sites in DNA, since interaction with a protein can cause decreased reactivity to DMS of guanines within the binding site but enhanced reactivity adjacent to the site. Methylation to form N7-methyl-guanine does not cause miscoding in the DNA, since this modified purine still pairs with C.
Chemicals that Cause Deletions
Some compounds cause a loss of nucleotides from DNA. If these deletions occur in a protein-coding region of the genomic DNA, and are not an integral multiple of 3, they result in a frameshift mutation. These are commonly more severe loss-of-function mutations than are simple substitutions. Frameshift mutagens such as proflavin or ethidium bromide have flat, polycyclic ring structures (Figure 7.10.A.). They may bind to and intercalate within the DNA, i.e. they can insert between stacked base pairs. If a segment of the template DNA is the looped out, DNA polymerase can replicate past it, thereby generating a deletion. Intercalating agents can stabilize secondary structures in the loop, thereby increasing the chance that this segment stays in the loop and is not copied during replication (Figure 7.10.B.) Thus growth of cells in the presence of such intercalating agents increase the probability of generating a deletion.
A.
B.
7.2: Reaction with Mutagens
Nitrogen mustards were produced in the 1920s and 1930s as potential chemical warfare weapons and are similar to sulfur mustard.
Skeletal formula of tris(2-chloroethyl)amine. Image used wtih permission (Public Domain; Benjah-bmm27).
Sulfur Mustard
Sulfur mustard is a type of chemical warfare agent. These kinds of agents cause blistering of the skin and mucous membranes on contact. They are called vesicants or blistering agents. Sulfur mustard is also known as “mustard gas or mustard agent,” or by the military designations H, HD, and HT. Sulfur mustard sometimes smells like garlic, onions, or mustard and sometimes has no odor. It can be a vapor (the gaseous form of a liquid), an oily-textured liquid, or a solid.
Ball-and-stick model of the sulfur mustard molecule. (Public Domain; Ben Mills).
Reactions
Sulfur mustard is the organic compound with formula \(\ce{(Cl-CH2CH2)2S}\). In the Depretz method, sulfur mustard is synthesized by treating sulfur dichloride with ethylene:
\[\ce{SCl2 + 2 C2H4 → (Cl-CH2CH2)2S }\]
In the Levinstein process, sulfur monochloride is used instead:
\[\ce{8 S2Cl2 + 16 C2H4 → 8 (Cl-CH2CH2)2S + S8}\]
In the Meyer method, thiodiglycol is produced from chloroethanol and potassium sulfide and chlorinated with phosphorus trichloride:
\[\ce{3 (HO-CH2CH2)2S + 2 PCl3 → 3 (Cl-CH2CH2)2S + 2 P(OH)3}\]
In the Meyer-Clarke method, concentrated hydrochloric acid (HCl) instead of PCl3 is used as the chlorinating agent:
\[\ce{(HO-CH2CH2)2S + 2 HCl → (Cl-CH2CH2)2S + 2 H2O}\]
Thionyl chloride and phosgene have also been used as chlorinating agents. It is a viscous liquid at normal temperatures. The pure compound has a melting point of 14 °C (57 °F) and decomposes before boiling at 218 °C (424.4 °F).
7.3: Ionizing Radiation
High energy radiation, such as X-rays, $\gamma$-rays, and $\beta$ particles (or electrons) are powerful mutagens. Since they can change the number of electrons on an atom, converting a compound to an ionized form, they are referred to as ionizing radiation. They can cause a number of chemical changes in DNA, including directly break phosphodiester backbone of DNA, leading to deletions. Ionizing radiation can also break open the imidazole ring of purines. Subsequent removal of the damaged purine from DNA by a glycosylase generates an apurinic site.
Ultraviolet Radiation
Ultraviolet radiation with a wavelength of 260 nm will form pyrimidine dimers between adjacent pyrimidines in the DNA. The dimers can be one of two types (Figure 7.11). The major product is a cytobutane-containing thymine dimer (between C5 and C6 of adjacent T's). The other product has a covalent bond between position 6 on one pyrimidine and position 4 on the adjacent pyrimidine, hence it is called the "6-4" photoproduct.
The pyrimidine dimers cause a distortion in the DNA double helix. This distortion blocks replication and transcription.
Exercise $1$
What is the physical basis for this distortion in the DNA double helix? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/7%3A_Mutation_and_Repair_of_DNA/7.2%3A_Reaction_with_Mutagens/Nitrogen_Mustard.txt |
The second part of this chapter examines the major classes of DNA repair processes. These are:
• reversal of damage,
• nucleotide excision repair,
• base excision repair,
• mismatch repair,
• recombinational repair, and
• error-prone repair.
Many of these processes were first studies in bacteria such as E. coli, however only a few are limited to this species. For instance, nucleotide excision repair and base excision repair are found in virtually all organisms, and they have been well characterized in bacteria, yeast, and mammals. Like DNA replication itself, repair of damage and misincorporation is a very old process.
Reversal of damage
Some kinds of covalent alteration to bases in DNA can be directly reversed. This occurs by specific enzyme systems recognizing the altered base and breaking bonds to remove the adduct or change the base back to its normal structure.
Photoreactivation is a light-dependent process used by bacteria to reverse pyrimidine dimers formed by UV radiation. The enzyme photolyase binds to a pyrimidine dimer and catalyzes a second photochemical reaction (this time using visible light) that breaks the cyclobutane ring and reforms the two adjacent thymidylates in DNA. Note that this is not formally the reverse of the reaction that formed the pyrimidine dimers, since energy from visible light is used to break the bonds between the pyrimidines, and no UV radiation is released. However, the result is that the DNA structure has been returned to its state prior to damage by UV. The photolyase enzyme has two subunits, which are encoded by the phrA and phrBgenes in E. coli.
A second example of the reversal of damage is the removal of methyl groups. For instance, the enzyme O6‑methylguanine methyltransferase, encoded by the adagene in E. coli, recognizes O6‑methylguanine in duplex DNA. It then removes the methyl group, transferring it to an amino acid of the enzyme. The methylated enzyme is no longer active, hence this has been referred to as a suicide mechanism for the enzyme.
Excision Repair
The most common means of repairing damage or a mismatch is to cut it out of the duplex DNA and recopy the remaining complementary strand of DNA, as outlined in Figure 7.12. Three different types of excision repair have been characterized: nucleotide excision repair, base excision repair, and mismatch repair. All utilize a cut, copy, and paste mechanism. In the cuttingstage, an enzyme or complex removes a damaged base or a string of nucleotides from the DNA. For the copying, a DNA polymerase (DNA polymerase I in E. coli) will copy the template to replace the excised, damaged strand. The DNA polymerase can initiate synthesis from 3' OH at the single-strand break (nick) or gap in the DNA remaining at the site of damage after excision. Finally, in the pastingstage, DNA ligase seals the remaining nick to give an intact, repaired DNA.
Nucleotide Excision Repair (NER)
In nucleotide excision repair, damaged bases are cut out within a string of nucleotides, and replaced with DNA as directed by the undamaged template strand. This repair system is used to remove pyrimidine dimers formed by UV radiation as well as nucleotides modified by bulky chemical adducts. The common feature of damage that is repaired by nucleotide excision is that the modified nucleotides cause a significant distortion in the DNA helix. NER occurs in almost all organisms examined.
Some of the best-characterized enzymes catalyzing this process are the UvrABC excinuclease and the UvrD helicase in E. coli. The genes encoding this repair function were discovered as mutants that are highly sensitive to UV damage, indicating that the mutants are defective in UV repair. As illustrated in Figure 7.13, wild type E. coli cells are killed only at higher doses of UV radiation. Mutant strains can be identified that are substantially more sensitive to UV radiation; these are defective in the functions needed for UV-resistance, abbreviated uvr. By collecting large numbers of such mutants and testing them for their ability to restore resistance to UV radiation in combination, complementation groups were identified. Four of the complementation groups, or genes, encode proteins that play major rules in NER; they are uvrA, uvrB, uvrCand uvrD.
The enzymes encoded by the uvrgenes have been studied in detail. The polypeptide products of the uvrA, uvrB, and uvrCgenes are subunits of a multisubunit enzyme called the UvrABC excinuclease. UvrA is the protein encoded by uvrA, UvrB is encoded by uvrB, and so on. The UvrABC complex recognizes damage-induced structural distortions in the DNA, such as pyrimidine dimers. It then cleaves on both sides of the damage. Then UvrD (also called helicase II), the product of the uvrDgene, unwinds the DNA, releasing the damaged segment. Thus for this system, the UvrABC and UvrD proteins carry out a series of steps in the cutting phase of excision repair. This leaves a gapped substrate for copying by DNA polymerase and pasting by DNA ligase.
The UvrABC proteins form a dynamic complex that recognizes damage and makes endonucleolytic cuts on both sides. The two cuts around the damage allow the single-stranded segment containing the damage to be excised by the helicase activity of UvrD. Thus the UvrABC dynamic complex and the UvrBC complex can be called excinucleases. After the damaged segment has been excised, a gap of 12 to 13 nucleotides remains in the DNA. This can be filled in by DNA polymerase and the remaining nick sealed by DNA ligase. Since the undamaged template directs the synthesis by DNA polymerase, the resulting duplex DNA is no longer damaged.
In more detail, the process goes as follows (Figure 7.14). UvrA2 (a dimer) and Uvr B recognize the damaged site as a (UvrA)2UvrB complex. UvrA2 then dissociates, in a step that requires ATP hydrolysis. This is an autocatalytic reaction, since it is catalyzed by UvrA, which is itself an ATPase. After UvrA has dissociated, UvrB (at the damaged site) forms a complex with UvrC. The UvrBC complex is the active nuclease. It makes the incisions on each side of the damage, in another step that requires ATP. The phosphodiester backbone is cleaved 8 nucleotides to the 5' side of the damage and 4-5 nucleotides on the 3' side. Finally, the UvrD helicase then unwinds DNA so the damaged segment is removed. The damaged DNA segment dissociates attached to the UvrBC complex. Like all helicase reactions, the unwinding requires ATP hydrolysis to disrupt the base pairs. Thus ATP hydrolysis is required at three steps of this series of reactions.
Exercise 7.9
How does an excinuclease differ from an exonuclease and an endonuclease?
Nucleotide excision repair is very active in mammalian cells, as well as cells from may other organisms. The DNA of a normal skin cell exposed to sunlight would accumulate thousands of dimers per day if this repair process did not remove them! One human genetic disease, called xeroderma pigmentosum (XP), is a skin disease caused by defect in enzymes that remove UV lesions. Fibroblasts isolated from individual XP patients are markedly sensitive to UV radiation when grown in culture, similar to the phenotype shown by E. coliuvrmutants. These XP cell lines can be fused in culture and tested for the ability to restore resistance to UV damage. XP cells lines that do so fall into different complementation groups. Several complementation groups, or genes, have been defined in this way. Considerable progress has been made recently in identifying the proteins encoded by each XP gene (Table 7.2). Note the tight analogy to bacterial functions needed for NER. Similar functions are also found in yeast (Table 7.2). Additional proteins utilized in eukaryotic NER include hHR23B (which forms a complex with the DNA-damage sensor XPC), ERCCI (which forms a complex with the XPF to catalyze incision 5’ to the site of damage), the several other subunits of TFIIH (see Chapter 10) and the single-strand binding protein RPA.
Table 7.2: Genes affected in XP patients, and encoded proteins
Human Gene Protein Function Homologous to S. cerevisiae Analogous to E. coli
XPA Binds damaged DNA Rad14 UvrA/UvrB
XPB 3’ to 5’ helicase, component of TFIIH Rad25 UvrD
XPC DNA-damage sensor (in complex with hHR23B) Rad4
XPD 5’ to 3’ helicase, component of TFIIH Rad3 UvrD
XPE Binds damaged DNA UvrA/UvrB
XPF Works with ERRC1 to cut DNA on 5’ side of damage Rad1 UvrB/UvrC
XPG Cuts DNA on 3’ side of damage Rad2 UvrB/UvrC
NER occurs in two modes in many organisms, including bacteria, yeast and mammals. One is the global repair that acts throughout the genome, and the second is a specialized activity is that is coupled to transcription. Most of the XP gene products listed in Table 2 function in both modes of NER in mammalian cells. However, XPC (acting in a complex with another protein called hHR23B) is a DNA-damage sensor that is specific for global genome NER. In transcription coupled NER, the elongating RNA polymerase stalls at a lesion on the template strand; perhaps this is the damage recognition activity for this mode of NER. One of the basal transcription factors that associates with RNA polymerase II, TFIIH (see Chapter 10), also plays a role in both types of NER. A rare genetic disorder in humans, Cockayne syndrome (CS), is associated with a defect specific to transcription coupled repair. Two complementation groups have been identified, CSAand CSB. Determination of the nature and activity of the proteins encoded by them will provide additional insight into the efficient repair of transcribed DNA strands. The phenotype of CS patients is pleiotropic, showing both photosensitivity and severe neurological and other developmental disorders, including premature aging. These symptoms are more severe than those seen for XP patients with no detectable NER, indicating that transcription-coupled repair or the CS proteins have functions in addition to those for NER.
Other genetic diseases also result from a deficiency in a DNA repair function, such as Bloom's syndrome and Fanconi's anemia. These are intensive areas of current research. A good resource for updated information on these and other inherited diseases, as well as human genes in general, is the Online Mendelian Inheritance in Man, or OMIM, accessible at http://www.ncbi.nlm.nih.gov.
Ataxia telangiectasia, or AT, illustrates the effect of alterations in a protein not directly involved in repair, but perhaps signaling that is necessary for proper repair of DNA. AT is a recessive, rare genetic disease marked by uneven gait (ataxia), dilation of blood vessels (telangiectasia) in the eyes and face, cerebellar degeneration, progressive mental retardation, immune deficiencies, premature aging and about a 100-fold increase in susceptibility to cancers. That latter phenotype is driving much of the interest in this locus, since heterozygotes, which comprise about 1% of the population, also have an increased risk of cancer, and may account for as much as 9% of breast cancers in the United States. The gene that is mutated in AT (hence called "ATM") was isolated in 1995 and localized to chromosome 11q22-23.
The ATM gene does not appear to encode a protein that participates directly in DNA repair (unlike the genes that cause XP upon mutation). Rather, AT is caused by a defect in a cellular signaling pathway. Based on homologies to other proteins, the ATM gene product may be involved in the regulation of telomere length and cell cycle progression. The C-terminal domain is homologous to phosphatidylinositol-3-kinase (which is also a Ser/Thr protein kinase) - hence the connection to signaling pathways. The ATM protein also has regions of homology to DNA-dependent protein kinases, which require breaks, nicks or gaps to bind DNA (via subunit Ku); binding to DNA is required for the protein kinase activity. This suggests that ATM protein could be involved in targeting the repair machinery to such damage.
Base Excision Repair
Base excision repair differs from nucleotide excision repair in the types substrates recognized and in the initial cleavage event. Unlike NER, the base excision machinery recognizes damaged bases that do not cause a significant distortion to the DNA helix, such as the products of oxidizing agents. For example, base excision can remove uridines from DNA, even though a G:U base pair does not distort the DNA. Base excision repair is versatile, and this process also can remove some damaged bases that do distort the DNA, such as methylated purines. In general, the initial recognition is a specific damaged base, not a helical distortion in the DNA. A second major difference is that the initial cleavage is directed at the glycosidic bond connecting the purine or pyrimidine base to a deoxyribose in DNA. This contrasts with the initial cleavage of a phosphodiester bond in NER.
Cells contain a large number of specific glycosylases that recognize damaged or inappropriate bases, such as uracil, from the DNA. The glycosylase removes the damaged or inappropriate base by catalyzing cleavage of the N-glycosidic bond that attaches the base to the sugar-phosphate backbone. For instance, uracil-N-glycosylase, the product of the ung gene, recognizes uracil in DNA and cuts the N-glycosidic bond between the base and deoxyribose (Figure 7.15). Other glycosylases recognize and cleave damaged bases. For instance methylpurine glycosylase removes methylated G and A from DNA. The result of the activity of these glycosylases is an apurinic/apyrimidinic site, or AP site (Figure 7.15). At an AP site, the DNA is still an intact duplex, i.e. there are no breaks in the phosphodiester backbone, but one base is gone.
Next, an AP endonuclease nicks the DNA just 5’ to the AP site, thereby providing a primer for DNA polymerase. In E. coli, the 5' to 3' exonuclease function of DNA polymerase I removes the damaged region, and fills in with correct DNA (using the 5' to 3' polymerase, directed by the sequence of the undamaged complementary strand).
Additional mechanisms have evolved for keeping U’s out of DNA. E. colialso has a dUTPase, encoded by the dutgene, which catalyzes the hydrolysis of dUTP to dUMP. The product dUMP is the substrate for thymidylate synthetase, which catalyzes conversion of dUMP to dTMP. This keeps the concentration of dUTP in the cell low, reducing the chance that it will be used in DNA synthesis. Thus the combined action of the products of the dut+ unggenes helps prevent the accumulation of U's in DNA.
Exercise
In base excision repair, which enzymes are specific for a particular kind of damage and which are used for all repair by this pathway?
Mismatch Repair
The third type of excision repair we will consider is mismatch repair, which is used to repair errors that occur during DNA synthesis. Proofreading during replication is good but not perfect. Even with a functional e subunit, DNA polymerase III allows the wrong nucleotide to be incorporated about once in every 108 bp synthesized in E. coli. However, the measured mutation rate in bacteria is as low as one mistake per 1010 or 1011 bp. The enzymes that catalyze mismatch repairare responsible for this final degree of accuracy. They recognize misincorporated nucleotides, excise them and replace them with the correct nucleotides. In contrast to nucleotide excision repair, mismatch repair does not operate on bulky adducts or major distortions to the DNA helix. Most of the mismatches are substitutes within a chemical class, e.g. a C incorporated instead of a T. This causes only a subtle helical distortions in the DNA, and the misincorporated nucleotide is a normal component of DNA. The ability of a cell to recognize a mismatch reflects the exquisite specificity of MutS, which can distinguish normal base pairs from those resulting from misincorporation. Of course, the repair machinery needs to know which of the nucleotides at a mismatch pair is the correct one and which was misincorporated. It does this by determining which strand was more recently synthesized, and repairing the mismatch on the nascent strand.
In E. coli, the methylation of A in a GATC motif provides a covalent marker for the parental strand, thus methylation of DNA is used to discriminate parental from progeny strands. Recall that the dam methylase catalyzes the transfer of a methyl group to the A of the pseudopalindromic sequence GATC in duplex DNA. Methylation is delayed for several minutes after replication. IN this interval before methylation of the new DNA strand, the mismatch repair system can find mismatches and direct its repair activity to nucleotides on the unmethylated, newly replicated strand. Thus replication errors are removed preferentially.
The enzyme complex MutH-MutL-MutS , or MutHLS, catalyzes mismatch repair in E. coli. The genes that encode these enzymes, mutH, mutLand mutS, were discovered because strains carrying mutations in them have a high frequency of new mutations. This is called a mutator phenotype, and hence the name mutwas given to these genes. Not all mutator genes are involved in mismatch repair; e.g., mutations in the gene encoding the proofreading enzyme of DNA polymerase III also have a mutator phenotype. This gene was independently discovered in screens for defects in DNA replication (dnaQ ) and mutator genes (mutD). Three complementation groups within the set of mutator alleles have been implicated primarily in mismatch repair; these are mutH, mutLand mutS.
MutS will recognize seven of the eight possible mismatched base pairs (except for C:C) and bind at that site in the duplex DNA (Figure 7.16). MutHand MutL (with ATP bound) then join the complex, which then moves along the DNA in either direction until it finds a hemimethylated GATC motif, which can be as far a few thousand base pairs away. Until this point, the nuclease function of MutH has been dormant, but it is activated in the presence of ATP at a hemimethylated GATC. It cleaves the unmethylated DNA strand, leaving a nick 5' to the G on the strand containing the unmethylated GATC (i.e. the new DNA strand). The same strand is nicked on the other side of the mismatch. Enzymes involved in other processes of repair and replication catalyze the remaining steps. The segment of single-stranded DNA containing the incorrect nucleotide is to be excised by UvrD, also known as helicase II and MutU. SSB and exonuclease I are also involved in the excision. As the excision process forms the gap, it is filled in by the concerted action of DNA polymerase III (Figure 7.16.).
Mismatch repair is highly conserved, and investigation of this process in mice and humans is providing new clues about mutations that cause cancer.Homologs to the E. coli genes mutLand mutShave been identified in many other species, including mammals. The key breakthrough came from analysis of mutations that cause one of the most common hereditary cancers, hereditary nonpolyposis colon cancer(HNPCC). Some of the genes that, when mutated, cause this disease encode proteins whose amino acid sequences are significantly similar to those of two of the E. colimismatch repair enzymes. The human genes are called hMLH1(for human mutLhomolog 1), hMSH1, and hMSH2(for human mutS homolog 1 and 2, respectively). Subsequent work has shown that these enzymes in humans are involved in mismatch repair. Presumably the increased frequency of mutation in cells deficient in mismatch repair leads to the accumulation of mutations in proto-oncogenes, resulting in dysregulation of the cell cycle and loss of normal control over the rate of cell division.
Exercise
The human homologs to bacterial enzymes involved in mismatch repair are also implicated in homologous functions. Given the human homologs discussed above, which enzymatic functions found in bacterial mismatch repair are also found in humans? What functions are missing, and hence are likely carried out by an enzyme not homologous to those used in bacterial mismatch repair?
Recombination Repair (Retrieval system)
In the three types of excision repair, the damaged or misincorporated nucleotides are cut out of DNA, and the remaining strand of DNA is used for synthesis of the correct DNA sequence. However, this complementary strand is not always available. Sometimes DNA polymerase has to synthesize past a lesion, such as a pyrimidine dimer or an AP site. One way it can do this is to stop on one side of the lesion and then resume synthesis about 1000 nucleotides further down. This leaves a gap in the strand opposite the lesion (Figure 7.17).
The information needed at the gap is retrieved from the normal daughter molecule by bringing in a single strand of DNA, using RecA-mediated recombination (see Chapter VIII). This fills the gap opposite the dimer, and the dimer can now be replaced by excision repair (Figure 7.17). The resulting gap in the (previously) normal daughter can be filled in by DNA polymerase, using the good template.
Translesion Synthesis
As just described, DNA polymerase can skip past a lesion on the template strand, leaving behind a gap. It has another option when such a lesion is encountered, which is to synthesis DNA in a non-template directed manner. This is called translesion synthesis, bypass synthesis, or error-prone repair. This is the last resort for DNA repair, e.g. when repair has not occurred prior to replication. In translesion replication, the DNA polymerase shifts from template directed synthesis to catalyzing the incorporation of random nucleotides. These random nucleotides are usually mutations (i.e. in three out of four times), hence this process is also designated error-prone repair.
Translesion synthesis uses the products of the umuCand umuDgenes. These genes are named for the UV nonmutable phenotype of mutants defective in these genes
Exercise
Question 7.11. Why do mutations in genes required for translesion synthesis (error prone repair) lead to a nonmutable phenotype?
UmuD forms a homodimer that also complexes with UmuC. When the concentration of single-stranded DNA and RecA are increased (by DNA damage, see next section), RecA stimulates an autoprotease activity in UmuD2 to form UmuD’2. This cleaved form is now active in translesional synthesis. UmuC itself is a DNA polymerase. A multisubunit complex containing UmuC, the activated UmuD’2 and the a subunit of DNA polymerase III catalyze translesional synthesis. Homologs of the UmuC polymerase are found in yeast (RAD30) and humans (XP-V).
The SOSresponse
A coordinated battery of responses to DNA damage in E. coliis referred to as the SOS response. This name is derived from the maritime distress call, “SOS” for "Save Our Ship". Accumulating damage to DNA, e.g. from high doses of radiation that break the DNA backbone, will generate single-stranded regions in DNA. The increasing amounts of single-stranded DNA induce SOS functions, which stimulate both the recombination repair and the translesional synthesis just discussed.
Key proteins in the SOS response are RecA and LexA. RecA binds to single stranded regions in DNA, which activates new functions in the protein. One of these is a capacity to further activate a latent proteolytic activity found in several proteins, including the LexA repressor, the UmuDprotein and the repressor encoded by bacteriophage lambda (Figure 7.18). RecA activated by binding to single-stranded DNA is not itself a protease, but rather it serves as a co-protease, activating the latent proteolytic function in LexA, UmuD and some other proteins.
In the absence of appreciable DNA damage, the LexA protein represses many operons, including several genes needed for DNA repair: recA, lexA, uvrA, uvrB, and umuC.When the activated RecA stimulates its proteolytic activity, it cleaves itself (and other proteins), leading to coordinate induction of the SOS regulated operons (Figure 7.18).
Restriction/Modification systems
The DNA repair systems discussed above operate by surveillance of the genome for damage or misincorporation and then bring in enzymatic machines to repair the defects. Other systems of surveillance in bacterial genomes are restriction/modification systems. These look for foreign DNA that has invaded the cell, and then destroy it. In effect, this is another means of protecting the genome from the damage that could result from the integration of foreign DNA.
These systems for safeguarding the bacterial cell from invasion by foreign DNA use a combination of covalent modification and restriction by an endonuclease. Each species of bacteria modifies its DNA by methylation at specific sites (Figure 7.19). This protects the DNA from cleavage by the corresponding restriction endonuclease. However, any foreign DNA (e.g. from an infecting bacteriophage or from a different species of bacteria) will not be methylated at that site, and the restriction endonuclease will cleave there. The result is that invading DNA will be cut up and inactivated, while not damaging the host DNA.
Any DNA that escapes the restriction endonuclease will be a substrate for the methylase. Once methylated, the bacterium now treats it like its own DNA, i.e. does not cleave it. This process can be controlled genetically and biochemically to aid in recombinant DNA work. Generally, the restriction endonuclease is encoded at the rlocus and the methyl transferase is encoded at the m locus. Thus passing a plasmid DNA through an r‑m+strain (defective in restriction but competent for modification) will make it resistant to restriction by strains with a wildtype r+gene. For some restriction/modification systems, both the endonuclease and the methyl transferase are available commercially. In these cases, one can modify the foreign DNA (e.g. from humans) prior to ligating into cloning vectors to protect it from cleavage by the restriction endonucleases it may encounter after transformation into bacteria.
For the type II restriction/modification systems, the methylation and restriction occurs at the same, pseudopalindromic site. These are the most common systems, with a different sequence specificity for each bacterial species. This has provided the large variety of restriction endonucleases that are so commonly used in molecular biology. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/7%3A_Mutation_and_Repair_of_DNA/7.4%3A_Repair_Mechanisms.txt |
Q7.12
If the top strand of the segment of DNA GGTCGTT were targeted for reaction with nitrous acid, and then it underwent two rounds of replication, what are the likely products?
Q7.13
Are the following statements about nucleotide excision repair in E. coli true or false?
1. UvrA and UvrB recognize structural distortions resulting from damage in the DNA helix.
2. In a complex with UvrB, UvrC cleaves the damaged strand on each side of the lesion.
3. The helicase UvrD unwinds the DNA, thereby dissociating the damaged patch.
Q7.14
Are the following statements about mismatch repair in E. coli true or false?
1. MutS will recognize a mismatch.
2. MutL, in a complex with ATP, will bind to the MutS (bound to the mismatched region) and activate MutH.
3. MutH will cleave 5' to the G of the nearest methylated GATC motif (GmeATC).
4. The mismatch repair system can discriminate between old versus newly synthesized strands of DNA.
For the next 6 problems, consider the following DNA sequence, from the first exon of the HRAS gene. A transversion of G to T at position 24 confers anchorage independence and tumorigenicity to NIH 3T3 cells (fibroblasts). This mutation is one step in tumorigenic transformation of bladder cells, and it likely plays a role in other cancers.
10 20 30
5' TAAGCTGGTG GTGGTGGGCG CCGGCGGTGT
3' ATTCGACCAC CACCACCCGC GGCCGCCACA
Q7.15
hat would the sequence be if the G at position 14 (top strand) were alkylated at the O6 position by MNNG and then went through 2 rounds of replication?
Q7.16
hat would the sequence be if the C at position 24 (bottom strand) were oxidized by HNO2 and then went through 2 rounds of replication?
Q7.17
What would happen if this sequence were irradiated with UV at a wavelength of 260 nm?
Q7.18
If you were in charge of maintaining this DNA sequence, and you had the enzymatic tools known in E. coli, how would you repair the damage from question 7.15? Consider what would happen if the damage were corrected before or after replication.
Q7.19
How could
1. the oxidative damage in problem 7.16 or
2. the UV products in problem 7.17 be repaired?
7.S: Mutation and Repair of DNA (Summary)
Summary: Causes of Transitions and Transversions
Table 7.1 lists several causes of mutations in DNA, including mutagens as well as mutator strains in bacteria. Note that some of these mutations lead to mispairing (substitutions), others lead to distortions of the helix, and some lead to both. Transitions can be generated both by damage to the DNA and by misincorporation during replication. Transversions occur primarily by misincorporation during replication. The frequency of such errors is greatly increased in mutator strains, e.g. lacking a proofreading function in the replicative DNA polymerase. Also, after a bacterial cell has sustained sufficient damage to induce the SOS response, the DNA polymerase shifts into a an error-prone mode of replication. This can also be a source of mutant alleles.
Table. 7.1. Summary of effects of various agents that alter DNA sequences (mutagens and mutator genes)
Agent (mutagen, etc.) Example Result
Nucleotide analogs BrdUTP transitions, e.g. A:T to G:C
Oxidizing agents nitrous acid transitions, e.g. C:G to T:A
Alkylating agents nitrosoguanidine transitions, e.g. G:C to A:T
Frameshift mutagens Benz(a)pyrene deletions (short)
Ionizing radiation X-rays, g-rays breaks and deletions (large)
UV UV, 260 nm Y-dimers, block replication
Misincorporation:
Altered DNA Pol III mutD=dnaQ; e subunit of DNA PolIII transitions, transversions and frameshifts in mutant strains
Error-prone repair Need UmuC, UmuD, DNA PolIII transitions and transversions in wild-type during SOS
Other mutator genes mutM, mutT, mutY transversions in the mutant strains
Additional Readings
• Friedberg, E. C., Walker, G. C., and Siede, W. (1995) DNA repair and mutagenesis, ASM Press, Washington, D.C.
• Kornberg, A. and Baker, T. (1992) DNA Replication, 2nd Edition, W. H. Freeman and Company, New York.
• Zakian, V. (1995) ATM-related genes: What do they tell us about functions of the human gene? Cell 82: 685-687.
• Kolodner, R. (1996) Biochemistry and genetics of eukaryotic mismatch repair. Genes & Development10:1433-1442.
• Sutton MD, Smith BT, Godoy VG, Walker GC. (2000) The SOS response: recent insights into umuDC-dependent mutagenesis and DNA damage tolerance.Annu Rev Genet34:479-497.
• De Laat, W. L., Jaspers, N. C. J. and Hoeijmakers, J. H. J. (1999) Molecular mechanism of nucleotide excision repair. Genes & Development13: 768-785. This review focuses on nucleotide excision repair in mammals. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/7%3A_Mutation_and_Repair_of_DNA/7.E_%3A_Mutation_and_Repair_of_DNA_%28Exercises%29.txt |
The chapter on mutation and repair of DNA dealt mainly with small changes in DNA sequence, usually single base pairs, resulting from errors in replication or damage to DNA. The DNA sequence of a chromosome can change in large segments as well, by the processes of recombination and transposition. Recombination is the production of new DNA molecule(s) from two parental DNA molecules or different segments of the same DNA molecule; this will be the topic of this chapter. Transposition is a highly specialized form of recombination in which a segment of DNA moves from one location to another, either on the same chromosome or a different chromosome; this will be discussed in the next chapter.
• 8.1: Types and Examples of Recombination
At least four types of naturally occurring recombination have been identified in living organisms: (1) General or homologous recombination, (2) Illegitimate or nonhomologous recombination, (3) Site-specific recombination, and (4) replicative recombination.
• 8.2: Detecting Recombination
Mendel’s Second Law described the random assortment of alleles of pairs of genes. However, certain pairs of genes show deviations from this random assortment, leading to the conclusion that those genes are linked on a chromosome. The linkage is not always complete, meaning that nonparental genotypes are seen in a proportion of the progeny. This is explained by crossing over between the gene pairs during meiosis in the parents.
• 8.3: Meiotic Recombination
The ability of homologous chromosomes to be paired during the first phase of meiosis is fundamental to the success of this process, which maintains a correct haploid set of chromosomes in the germ cell. Recombination is an integral part of the pairing of homologous chromosomes. It occurs between non-sister chromatids during the pachytene stage of meiosis I (the first stage of meiosis) and possibly before, when the homologous chromosomes are aligned in zygotene.
• 8.4: Advantages of Genetic Recombination
Not only is recombination needed for homologous pairing during meiosis, but recombination has at least two additional benefits for sexual species. It makes new combinations of alleles along chromosomes, and it restricts the effects of mutations largely to the region around a gene, not the whole chromosome. Since each chromosome undergoes at least one recombination event during meiosis, new combinations of alleles are generated.
• 8.5: Evidence for Heteroduplexes from Recombination in Fungi
The mechanism by which recombination occurs has been studied primarily in fungi, such as the budding yeast Saccharomyces cerevisiae and the filamentous fungus Ascomycetes, and in bacteria. The fungi undergo meiosis, and hence some aspects of their recombination systems may be more similar to that of plants and animals than is that of bacteria. However, the enzymatic functions discovered by genetic and biochemical studies of recombination in bacteria have counterparts in eukaryotic organisms too.
• 8.6: Holliday Model for General Recombination - Single Strand Invasion
In 1964, Robin Holliday proposed a model that accounted for heteroduplex formation and gene conversion during recombination. Although it has been supplanted by the double-strand break model (at least for recombination in yeast and higher organisms), it is a useful place to start. It illustrates the critical steps of pairing of homologous duplexes, formation of a heteroduplex, formation of the recombination joint, branch migration and resolution.
• 8.7: Double-strand-break model for Recombination
Several lines of evidence, primarily from studies of recombination in yeast, required changes to the reciprocal exchange of DNA chains initiated at single-strand nicks. As just mentioned, one DNA duplex tended to be the donor of information and the other the recipient, in contrast to the equal exchange predicted by the original Holliday model. Also, in yeast, recombination could be initiated by double-strand breaks.
• 8.8: Enzymes required for recombination in E. coli
The initial steps in finding enzymes that carry out recombination were genetic screens for mutants of E. coli that are defective in recombination. Assays were developed to test for recombination, and mutants that showed a decrease in recombination frequency were isolated. These were assigned to complementation groups called recA, recB, recC, recD, and so forth. Roughly 20 different genes (different rec complementation groups) have been identified in E. coli.
• 8.9: Generation of Single Strands
One of the major pathways for generating 3’ single-stranded termini uses the RecBCD enzyme, also known as exonuclease V. The three subunits of this enzyme are encoded by the genes recB, recC, and recD. Each model for recombination requires a single-strand with with a free end for strand invasion, and this enzyme does so, but with several unexpected features.
• 8.10: Synapsis and Invasion of Single Strands
The pairing of the two recombining DNA molecules (synapsis) and invasion of a single strand from the initiating duplex into the other duplex are both catalyzed by the multi-functional protein RecA. This invasion of the duplex DNA by a single stranded DNA results in the replacement of one of the strands of the original duplex with the invading strand, and the replaced strand is displaced from the duplex. Hence this reaction can also be called strand assimilation or strand exchange.
• 8.11: Branch Migration
The movement of a Holliday junction to generate additional heteroduplex requires two proteins. One is the RuvA tetramer, which recognizes the structure of the Holliday junction. The other is RuvB, which is an ATPase. It forms hexameric rings that provide the motor for branch migration.
• 8.12: Resolution
RuvC is the endonuclease that cleaves the Holliday junctions. It forms dimers that bind to the Holliday junction; recent data indicate an interaction among RuvA, RuvB and RuvC as a complex at the Holliday junction. The structure of the RuvA-Holliday junction complex suggests that the open structure of the junction stabilized by the binding of RuvA may expose a surface that is recognized by Ruv C for cleavage.
• 8.E: Recombination of DNA (Exercises)
Suggested readings
• Holliday, R. (1964) A mechanism for gene conversion in fungi. Genetics Research 5: 282-304.
• Orr-Weaver, T. L., Szostak, J. W. and Rothstein, R. J. (1981) Yeast transformation: a model system for the study of recombination. Proc. Natl. Acad. Sci. USA 78: 6354-6358.
• Szostak, J. W., Orr-Weaver, T. L., Rothstein, R. J. and Stahl, F. W. (1983) The double-strand-break repair model for recombination. Cell 33: 25-35.
• Stahl, F. W. (1994) The Holliday junction on its thirtieth anniversary. Genetics 138: 241-246.
• Kowalczykowski, S.C., Dixon, D. A., Eggleston, A. K., Lauder, S. D. and Rehrauer, W. M. (1994) Microbiological Reviews 58:401-465.
• Eggleston, A. K. and West, S. C. (1996) Exchanging partners: recombination in E. coli. Treand in Genetics 12: 20-25.
• Edelmann, W. and Kucherlapati, R. (1996) Role of recombination enzymes in mammalian cell survival. Proc. Natl. Acad. Sci. USA 93: 6225-6227.
8: Recombination of DNA
At least four types of naturally occurring recombination have been identified in living organisms (Figure 8.1).
1. General or homologous recombination occurs between DNA molecules of very similar sequence, such as homologous chromosomes in diploid organisms. General recombination can occur throughout the genome of diploid organisms, using one or a small number of common enzymatic pathways. This chapter will be concerned almost entirely with general recombination.
2. Illegitimate or nonhomologous recombination occurs in regions where no large-scale sequence similarity is apparent, e.g. translocations between different chromosomes or deletions that remove several genes along a chromosome. However, when the DNA sequence at the breakpoints for these events is analyzed, short regions of sequence similarity are found in some cases. For instance, recombination between two similar genes that are several million bp apart can lead to deletion of the intervening genes in somatic cells.
3. Site-specific recombination occurs between particular short sequences (about 12 to 24 bp) present on otherwise dissimilar parental molecules. Site-specific recombination requires a special enzymatic machinery, basically one enzyme or enzyme system for each particular site. Good examples are the systems for integration of some bacteriophage, such as l, into a bacterial chromosome and the rearrangement of immunoglobulin genes in vertebrate animals.
4. The third type is replicative recombination, which generates a new copy of a segment of DNA. Many transposable elements use a process of replicative recombination to generate a new copy of the transposable element at a new location.
Recombinant DNA technology uses two other types of recombination. The directed cutting and rejoining of different DNA molecules in vitro using restriction endonucleases and DNA ligases is well-known, as covered in Chapter 2. Once made, these recombinant DNA molecules are then introduced into a host organism, often a bacterium. If the recombinant DNA is a plasmid, phage or other molecule capable of replicating in the host, it will stay extrachromosomal. However, one can introduce the recombinant DNA molecule into a host in which it cannot replicate, such as a plant, an animal cell in culture, or a fertilized mouse egg. In order for the host to be stably transformed, the introduced DNA has to be taken up into a host chromosome. In bacteria and yeast, this can occur by homologous recombination at a reasonably high frequency. However, this does not occur in plant or animal cells. In contrast, at a low frequency, some of these introduced DNA molecules are incorporated into random locations in the chromosomes of the host cell. Thus random recombination into chromosomes can make stably transfected cells and transgenic plants and animals. The mechanism of this recombination during transformation or transfection is not well understood, although it is commonly used in the laboratory.
General recombination is an integral part of the complex process of meiosis in sexually reproducing organisms. It results in a crossing over between pairs of genes along a chromosome, which are revealed in appropriate matings (Chapter 1). The chiasmata that link homologous chromosomes during meiosis are the likely sites of the crossovers that result in recombination. General recombination also occurs in nonsexual organisms when two copies of a chromosome or chromosomal segment are present. We have encountered this as recombination during F-factor mediated conjugal transfer of parts of chromosomes in E. coli (Chapter 1). Recombination between two phage during a mixed infection of bacteria is another example. Also, the retrieval system for post-replicative repair (Chapter 7) involves general recombination.
The mechanism of recombination has been intensively studied in bacteria and fungi, and some of the enzymes involved have been well characterized. However, a full picture of the mechanism, or mechanisms, of recombination has yet to be achieved. We will discuss the general properties of recombination, cover two models of recombination, and discuss some of the properties of key enzymes in the pathways of recombination.
Reciprocal and Nonreciprocal Recombination
General recombination can appear to result in either an equal or an unequal exchange of genetic information. Equal exchange is referred to as reciprocal recombination, as illustrated in Figure 8.1. In this example, two homologous chromosomes are distinguished by having wild type alleles on one chromosome (A+, B+ and C+) and mutant alleles on the other (A-, B- and C-). Homologous recombination between genes A and B exchanges the segment of one chromosome containing the wild type alleles of genes B and C (B+ and C+) for the segment containing the mutant alleles (B- and C-) on the homologous chromosome. This could be explained by breaking and rejoining of the two homologous chromosomes during meiosis; however, we will see later that the enzymatic mechanism is more complex than simple cutting and ligation. The DNA that is removed from the top (thin dark blue) chromosome is joined with the bottom (thick light blue) chromosome, and the DNA removed from the bottom chromosome is added to the top chromosome. This process resulting in new DNA molecules that carry genetic information derived from both parental DNA molecules is called reciprocal recombination. The number of alleles for each gene remains the same in the products of this recombination, only their arrangement has changed.
General recombination can also result in a one-way transfer of genetic information, resulting in an allele of a gene on one chromosome being changed to the allele on the homologous chromosome. This is called gene conversion. As illustrated in Figure 8.2, recombination between two homologous chromosomes A+B+C+ and A-B-C- can result in a new arrangement, A-B+C-, without a change in the parental A+B+C+. In this case, the allele of gene B on the bottom chromosome has changed from B- to B+ without a reciprocal change on the other chromosome. Thus, in contrast to reciprocal recombination, the number of types of alleles for gene B has changed in the products of this recombination; now there is only one (B+). This is an example of interchromosomal gene conversion, i.e. between homologous chromosomes. Similar copies of genes can be on the same chromosome, and these can undergo gene conversion as well. Cases of intrachromosomal gene conversion have been documented for the gamma-globin genes of humans. The occurrence of gene conversion during general recombination is one indication that the enzymatic mechanism is not a simple cutting and pasting.
Exercise 8.1
Why would you not interpret the A-B+C- chromosome as resulting from two reciprocal crossovers, one on each side of gene B? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/8%3A_Recombination_of_DNA/8.01%3A_Types_and_Examples_of_Recombination.txt |
Mendel’s Second Law described the random assortment of alleles of pairs of genes. However, certain pairs of genes show deviations from this random assortment, leading to the conclusion that those genes are linked on a chromosome. The linkage is not always complete, meaning that nonparental genotypes are seen in a proportion of the progeny. This is explained by crossing over between the gene pairs during meiosis in the parents.
Let’s think about the general recombination shown in Figure 8.1 in this context. The two chromosomes outlined in the figure are in a heterozygous parent, with the wild type alleles for genes A and B (A+ and B+) are on one chromosome and the mutant alleles (A- and B-) are on the homologous chromosome (we can ignore gene C for this discussion.) Homologous recombination during meiosis can generate the new chromosomes shown, now with A+ and B- on one chromosome and A- and B+ on the other. However, this crossover will not occur between genes A and B on all chromosomes undergoing meiosis in this parent. Although recombination is an essential part of meiosis (see next section), the sites of recombination on a particular chromosome varies from cell to cell. In fact, the probability that a crossover will occur between two genes is a measure of the genetic distance between them (reviewed in Chapter 1). The recombinant chromosomes resulting from a crossover are revealed in a mating between the heterozygous parent (A+B+/A-B-) and a homozygous recessive individual (A-B-/A-B-). Most of the germ cells contributed by the heterozygous parent will have one of the parental chromosomes A+B+ or A-B-, but those germ cells resulting from the crossover between genes A and B will have the recombinant chromosomes (either A+B- or A-B+). The homozygous recessive parent will contribute only A-B- chromosomes. Thus in the progeny, one sees mainly offspring whose phenotype is determined by one of the chromosomes in the heterozygous parent, either wild type A and B (genotype of A+B+/A-B-) or mutant A and B (genotype A-B-/A-B-). However, some of the progeny will show a wild type A and a mutant B phenotype, or vice versa. These carry the chromosomes resulting from the crossover (genotype of A+B-/A-B- or A-B+/A-B-). The frequency with which one sees progeny with nonparental phenotypes is related to their distance apart on the chromosome; this measure is referred to as a genetic distance or a recombination distance.
8.03: Meiotic Recombination
A diploid organism has two copies of each chromosome. If it has four chromosomes, there are two pairs, A and A’ and B and B’, not four different chromosomes A, B, C and D. One copy of each chromosome came from its father (e.g. A and B) and one copy of each came from its mother (e.g. A’ and B’). Meiosis is the process of reductive division whereby a diploid organism generates haploid germ cells (in this case, with two chromosomes), and each germ cell has a single copy of each chromosome. In this example, meiosis does not generate germ cells with A and A’ or B and B’, rather it produces cells with A and B, or A and B’, or A’ and B, or A’ and B’. The homologous chromosomes, each consisting of two sister chromatids, are paired during the first phase of meiosis, e.g., A with A’ and B with B’ (Figure 8.3; see also Figs. 1.3 and 1.4). Then the homologous chromosomes are moved to separate cells at the end of the first phase, insuring that the two homologs do not stay together during reductive division in the second phase of meiosis. Thus each germ cell receives the haploid complement of the genetic material, i.e. one copy of each chromosome. The combination of two haploid sets of chromosomes during fertilization restores the diploid state, and the cycle can resume. Failure to distribute one copy of each chromosome to each germ cell has severe consequences. Absence of one copy of a chromosome in an otherwise diploid zygote is likely fatal. Having an extra copy of a chromosome (trisomy) also causes problems. In humans, trisomy for chromosomes 15 or 18 results in perinatal death and trisomy 21 leads to developmental defects known as Down’s syndrome.
Exercise 8.2
If this diploid organism with chromosomes A, A’, B and B’ underwent meiosis without homologous pairing and separation of the homologs to different cells, what fraction of the resulting haploid cells would have an A-type chromosome (A or A’) and a B-type chromosome (B or B’)?
The ability of homologous chromosomes to be paired during the first phase of meiosis is fundamental to the success of this process, which maintains a correct haploid set of chromosomes in the germ cell. Recombination is an integral part of the pairing of homologous chromosomes. It occurs between non-sister chromatids during the pachytene stage of meiosis I (the first stage of meiosis) and possibly before, when the homologous chromosomes are aligned in zygotene (Figure 8.3). The crossovers of recombination are visible in the diplotene phase. During this phase, the homologous chromosomes partially separate, but they are still held together at joints called chiasmata; these are likely the actual crossovers between chromatids of homologous chromosomes. The chiasmata are progressively broken as meiosis I is completed, corresponding to resolution of the recombination intermediates. During anaphase and telophase of meiosis I, each homologous chromosome moves to a different cell, i.e. A and A’ in different cells, B and B’ in different cells in our example. Thus recombinations occur in every meiosis, resulting in at least one exchange between pairs of homologous chromosomes per meiosis.
Recent genetic evidence demonstrates that recombination is required for homologous pairing of chromosomes during meiosis. Genetic screens have revealed mutants of yeast and Drosophila that block pairing of homologous chromosomes. These are also defective in recombination. Likewise, mutants defective in some aspects of recombination are also defective in pairing. Indeed, the process of synapsis (or pairing) between homologous chromosomes in zygotene, crossing over between homologs in pachytene, and resolution of the crossovers in the latter phases of meiosis I (diakinesis, metaphase I, and anaphase I) correspond to the synapsis, formation of a recombinant joint and resolution that mark the progression of recombination, as will be explained below.
8.04: Advantages of Genetic Recombination
Not only is recombination needed for homologous pairing during meiosis, but recombination has at least two additional benefits for sexual species. It makes new combinations of alleles along chromosomes, and it restricts the effects of mutations largely to the region around a gene, not the whole chromosome.
Since each chromosome undergoes at least one recombination event during meiosis, new combinations of alleles are generated. The arrangement of alleles inherited from each parent are not preserved, but rather the new germ cells carry chromosomes with new combinations of alleles of the genes (Figure 8.4). This remixing of combinations of alleles is a rich source of diversity in a population.
Over time, recombination will separate alleles at one locus from alleles at a linked locus. A chromosome through generations is not fixed, but rather it is "fluid," having many different combinations of alleles. This allows nonfunctional (less functional) alleles to be cleared from a population. If recombination did not occur, then one deleterious mutant allele would cause an entire chromosome to be eliminated from the population. However, with recombination, the mutant allele can be separated from the other genes on that chromosome. Then negative selection can remove defective alleles of a gene from a population while affecting the frequency of alleles only of genes in tight linkage to the mutant gene. Conversely, the rare beneficial alleles of genes can be tested in a population without being irreversibly linked to any potentially deleterious mutant alleles of nearby genes. This keeps the effective target size for mutation close to that of a gene, not the whole chromosome. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/8%3A_Recombination_of_DNA/8.02%3A_Detecting_Recombination.txt |
The mechanism by which recombination occurs has been studied primarily in fungi, such as the budding yeast Saccharomyces cerevisiae and the filamentous fungus Ascomycetes, and in bacteria. The fungi undergo meiosis, and hence some aspects of their recombination systems may be more similar to that of plants and animals than is that of bacteria. However, the enzymatic functions discovered by genetic and biochemical studies of recombination in bacteria are also proving to have counterparts in eukaryotic organisms as well. We will refer to studies mainly in fungi for the models of recombination, and to studies mainly in bacteria for the enzymatic pathways.
Many important insights into the mechanism of recombination have come from studies in fungi. One fundamental observation is that recombination proceeds by the formation of a region of heteroduplex, i.e. the recombination products have a region with one strand from one chromosome and the complementary strand from the other chromosome. Thus recombination is not a simple cut and paste operation, unlike the joining of two different molecules by recombinant DNA technology. The two recombining molecules are joined and form a hybrid, or heteroduplex, over part of their lengths.
The anatomy and physiology of the filamentous fungus Ascomycetes allows one to observe this heteroduplex formed during recombination. A cell undergoing meiosis starts with a 4n complement of chromosomes (i.e. twice the diploid number) and undergoes two rounds of cell division to form four haploid cells. In fungi these haploid germ cells are spores, and they are found together in an ascus. They can be separated by dissection and plated individually to examine the phenotype of the four products of meiosis. This is called tetrad analysis.
The fungus Ascomycetes goes one step further. After meiosis is completed, the germ cells undergo one further round of replication and mitosis. This separates each individual polynucleotide chain (or “strand” in the sense used in nucleic acid biochemistry) of each DNA duplex in the meiotic products into a separate spore. The eight spores in the ascus reflect the genetic composition of each of the eight polynucleotide chains in the four homologous chromosomes. (The two sister chromatids in each homologous chromosome become two chromosomes after meiosis, and each chromosome is a duplex of two polynucleotide chains.)
The order of the eight spores in the ascus of Ascomycetes reflects the descent of the spores from the homologous chromosomes. As shown in Figure 8.5, a heterozygote with a “blue” allele on one homologous chromosome and a “red” allele on the other will normally produce four “blue” spores and four “red” spores. The four spores with the same phenotype were derived from one homologous chromosome and are adjacent to each other in the ascus. This is called a 4:4 parental ratio, i.e. with respect to the phenotypes of the parent of the heterozygote.
The evidence for heteroduplex formation comes from deviations from the normal 4:4 ratio. Sometimes a 3:5 parental ratio is seen for a particular genetic marker. This shows that one polynucleotide chain of one allele has been lost (giving 4-1=3 spores with the corresponding phenotype in the ascus) and replaced by the polynucleotide chain of the other allele (giving 4+1=5 spores with the corresponding phenotype). As illustrated in Figure 8.5, this is 3 blue spores and 5 red spores. The segment of the chromosome containing this gene was a heteroduplex with one chain from each of two alleles. The round of replication and mitosis that follows meiosis in this fungus allows the two chains to be separated into two alleles that generated a different phenotype in a plating assay. Thus this 3:5 ratio results from post-meiotic segregation of the two chains of the different alleles. In this fungus, a region of heteroduplex can be directly observed by a plating assay.
The region of heteroduplex is associated with a recombination between the chromosomes. Other genes flank the region of heteroduplex shown in Figure 8.5. In many cases, the arrangement of alleles of these flanking genes has changed from that on the parental chromosomes, reflecting a recombination. For instance, let the region of heteroduplex be in a gene B, flanked by gene A in the left and gene C on the right. Each gene has a blue allele and a red allele, making the parental chromosomes AbBbCb and ArBrCr. If one monitored the phenotypes of determined by genes A and C (in addition to B) in the third and fourth spores (derived from the chromosome with the heteroduplex), they would see the phenotypes for the nonparental chromosomes AbBbCr and AbBrCr. This change in the flanking markers (genes A and C) reflects a recombination. Thus the heteroduplex can be found between markers that have undergone recombination.
Other markers can show a 2:6 parental ratio. This means that one of the alleles (formerly blue in fig. 8.5) has been changed to the other allele (now red), in a process called gene conversion. This can occur between flanking markers that have been switched because of recombination. Thus like the heteroduplex, the region of gene conversion is associated with recombination. Models for recombination need to incorporate both phenomenon into their proposed mechanism.
Exercise 8.3.
Imagine that you are studying a fungus that generates an ascus with 8 spores like Ascomycetes, in which the products of meiosis complete an additional round of replication and mitosis. You generate a heterozyous strain by mating a parent that was homozyous for the markers leu+, SmR, ade8+ and another that was leu-, SmS, ade8-. Previous studies had shown that all three markers are linked in the order given. Each of these pairs of alleles can be distinguished in a plating assay. The allele leu+ confers leucine auxotrophy whereas leu- confers leucine prototrophy. The allele SmR confers resistance to spectinomycin whereas SmS is sensitive to this antibiotic. Colonies of fungi with the ade8+ allele give a red color in under appropriate conditions in a plate, but those with the ade8- are white. Analysis of the individual spores from an ascus gave the following phenotypes results. The spores are numbered in the order they were in the ascus. What are the corresponding genotypes of the chromosome in each spore? How do you interpret these results with respect to recombination?
Spore leucine Spectinomycin Color in ade test
1 prototroph resistant red
2 prototroph resistant red
3 prototroph resistant white
4 prototroph sensitive white
5 auxotroph sensitive red
6 auxotroph sensitive red
7 auxotroph sensitive white
8 auxotroph sensitive white | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/8%3A_Recombination_of_DNA/8.05%3A_Evidence_for_Heteroduplexes_from_Recombination_in_Fungi.txt |
In 1964, Robin Holliday proposed a model that accounted for heteroduplex formation and gene conversion during recombination. Although it has been supplanted by the double-strand break model (at least for recombination in yeast and higher organisms), it is a useful place to start. It illustrates the critical steps of pairing of homologous duplexes, formation of a heteroduplex, formation of the recombination joint, branch migration and resolution.
The steps in the Holliday Model are illustrated in Figure 8.6.
1. Two homologous chromosomes, each composed of duplex DNA, are paired with similar sequences adjacent to each other.
2. An endonuclease nicks at corresponding regions of homologous strands of the paired duplexes. This is shown for the strands with the arrow to the right in the figure.
3. The nicked ends dissociate from their complementary strands and each single strand invades the other duplex. This occurs in a reciprocal manner to produce a heteroduplex regionderived from one strand from each parental duplex.
4. DNA ligase seals the nicks. The result is a stable joint molecule, in which one strand of each parental duplex crosses over into the other duplex.This X-shaped joint is called a Holliday intermediate or Chi structure.
5. Branch migration then expands the region of heteroduplex. The stable joint can move along the paired duplexes, feeding in more of each invading strand and extending the region of heteroduplex.
6. The recombination intermediate is then resolved by nicking a strand in each duplex and ligation.
Resolution can occur in either of two ways, only one of which results in an exchange of flanking markers after recombination. The two modes of resolution can be visualized by rotating the duplexes so that no strands cross over each other in the illustration (Figure 8.6). In the “horizontal” mode of resolution, the nicks are made in the same DNA strands that were originally nicked in the parental duplexes. After ligation of the two ends, this produces two duplex molecules with a patch of heteroduplex, but no recombination of flanking regions. In contrast, for the “vertical” mode of resolution, the nicks are made in the other strands, i.e. those not nicked in the original parental duplexes. Ligation of these two ends also leaves a patch of heteroduplex, but additionally causes recombination of flanking regions. Note that “horizontal” and “vertical” are just convenient designations for the two modes based on the two-dimensional drawings that we can make. The important distinction in terms of genetic outcome is whether the resolution steps target the strands initially cleaved or the other strand.
The steps in this model of general recombination can be viewed in a dynamic form by visiting a web site maintained by geneticists at the University of Wisconsin (URL is www.wisc.edu/genetics/Holliday/index.html). This shows the steps in the Holliday model in a movie, illustrating the actions much more vividly than static diagrams.
The recombinant joint proposed by Holliday has been visualized in electron micrographs of recombining DNA duplexes (Figure 8.7A). It has the proposed X shape. Although this joint is drawn with some distance between the duplexes in illustrations, in fact the two duplexes are juxtaposed, and only a very few base pairs are broken in the Holliday intermediate (Figure 8.7B). The structure is symmetrical , and it is likely that the choice between “horizontal” and “vertical” resolution is a random event by the resolving nuclease. It chooses two strands, but it cannot tell which were initially cleaved and which were not.
Studies of recombination between chromosomes with limited homology have shown that the minimum length of the region required to establish the connection between the recombining duplexes is about 75 bp. If the homology region is shorter than this, the rate of recombination is substantially reduced.
The patch of heteroduplex can be replicated (Figure 8.8) or repaired to generate a gene conversion event. As shown in Figure 8.8, replication through the products of horizontal resolution (from Figure 8.6) will generate a duplex from each strand of the heteroduplex. If we consider the parental chromosomes to be A+C+B+ and A-C-B-, and the heteroduplex to be in gene C, the products of replication can have a the parental C+ converted to a C- but still flanked by A+ and B+ or C- converted to C+ but still flanked by A- and B-. In either case, gene C has changed to a new allele without affecting the flanking markers.
Although the original Holliday model accounted for many important aspects of recombination (all that were known at the time), some additional information requires changes to the model. For instance, the Holliday model treats both duplexes equally; both are the invader and the target of the strand invasion. Also, no new DNA synthesis is required in the Holliday model. However, subsequent work showed that one of the duplex molecules is the used preferentially as the donor of genetic information. Hence additional models, such as one from Meselson and Radding, incorporated new DNA synthesis at the site of the nick to make and degradation of a strand of the other duplex to generate asymmetry into the two duplexes, with one the donor the other the recipient of DNA. These ideas and others have been incorporated into a new model of recombination involving double strand breaks in the DNAs. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/8%3A_Recombination_of_DNA/8.06%3A_Holliday_Model_for_General_Recombination_-_Single_Strand_Invasi.txt |
Several lines of evidence, primarily from studies of recombination in yeast, required changes to the reciprocal exchange of DNA chains initiated at single-strand nicks. As just mentioned, one DNA duplex tended to be the donor of information and the other the recipient, in contrast to the equal exchange predicted by the original Holliday model. Also, in yeast, recombination could be initiated by double-strand breaks. For instance, both DNA strands are cleaved (by the HO endonuclease) to initiate recombination between the MATand HML(R) loci in mating type switching in yeast. Using plasmids transformed into yeast, it was shown that a double-strand gap in the “aggressor” duplex could be used to initiate recombination, and the gap was repaired during the recombination (this experiment is explored in problem 8.___). In this case, the gap in one duplex was filled by DNA donated from the other substrate. All this evidence was incorporated into a major new model for recombination from Jack Szostak and colleagues in 1983. It is called the double-strand-break model. New features in this model (contrasting with the Holliday model) are initiation at double-strand breaks, nuclease digestion of the aggressor duplex, new synthesis and gap repair. However, the fundamental Holliday junction, branch migration and resolution are retained, albeit with somewhat greater complexity because of the additional numbers of Holliday junctions. Although many aspects of the recombination mechanism differ
The steps in the double-strand-break model up to the formation of the joint molecules are diagrammed in Figure 8.9.
1. An endonuclease cleaves both strands of one of the homologous DNA duplexes, shown as thin blue lines in Figure 8.9. This is the aggressor duplex, since it initiates the recombination. It is also the recipientof genetic information, as will be apparent as we go through the model.
2. The cut is enlarged by an exonuclease to generate a gap with 3' single-stranded termini on the strands.
3. One of the free 3' ends invades a homologous region on the other duplex (shown as thick red lines), called the donor duplex. The formation of heteroduplex also generates a D-loop(a displacement loop), in which one strand of the donor duplex is displaced.
4. The D-loop is extended as a result of repair synthesis primed by the invading 3' end. The D-loop eventually gets large enough to cover the entire gap on the aggressor duplex, i.e. the one initially cleaved by the endonuclease. The newly synthesized DNA uses the DNA from the invaded DNA duplex (thick red line) as the template, so the new DNA has the sequence specified by the invaded DNA.
5. When the displaced strand from the donor (red) extends as far as the other side of the gap on the recipient (thin blue), it will anneal with the other 3' single stranded end at that end of the gap. The displaced strand has now filled the gap on the aggressor duplex, donating its sequence to the duplex that was initially cleaved. Repair synthesiscatalyzed by DNA polymerase converts the donor D-loop to duplex DNA. During steps 4 and 5, the duplex that was initially invaded serves as the donorduplex; i.e. it provides genetic information during this phase of repair synthesis. Conversely, the aggressor duplex is the recipient of genetic information. Note that the single strand invasion models predict the opposite, where the initial invading strand is the donor of the genetic information.
6. DNA ligase will seal the nicks, one on the left side of the diagram in Figure 8.9 and the other on the right side. Although the latter is between a strand on the bottom duplex and a strand on the top duplex, it is equivalent to the ligation in the first nick (the apparent physical separation is an artifact of the drawing). In both cases, sealing the nick forms a Holliday junction.
At this point, the recombination intermediate has two recombinant joints (Holliday junctions). The original gap in the aggressor duplex has been filled with DNA donated by the invaded duplex. The filled gap is now flanked by heteroduplex. The heteroduplexes are arranged asymmetrically, with one to the left of the filled gap on the aggressor duplex and one to the right of the filled gap on the donor duplex. Branch migration can extend the regions of heteroduplex from each Holliday junction.
The recombination intermediate can now be resolved. The presence of two recombination joints adds some complexity, but the process is essentially the same as discussed for the Holliday model. Each joint can be resolved horizontally or vertically. The key factor is whether the joints are resolved in the same mode or sense (both horizontally or both vertically) or in different modes.
If both joints are resolved the same sense (Figure 8.10), the original duplexes will be released, each with a region of altered genetic information that is a "footprint" of the exchange event. That region of altered information is the original gap, plus or minus the regions covered by branch migration. For instance, if both joints are resolved by cutting the originally cleaved strands ("horizontally" in our diagram of the Holliday model), then you have no crossover at either joint. If both joints are resolved by cleaving the strands not cut originally ("vertically" in our diagram of the Holliday model), then you have a crossover at both joints. This closely spaced double crossover will produce no recombination of flanking markers.
In contrast, if each joint is resolved in opposite directions (Figure 8.11), then there will be recombination between flanking markers. That is, one joint will not give a crossover and the other one will.
Several features distinguish the double-strand-break model from the single-strand nick model initially proposed by Holliday. In the double-strand-break model, the region corresponding to the original gap now has the sequence of the donor duplex in both molecules. This is flanked by heteroduplexes at each end, one on each duplex. Hence the arrangement of heteroduplex is asymmetric; i.e. there is a different heteroduplex in each duplex molecule. Part of one duplex molecule has been converted to the sequence of the other (the recipient, initiating duplex has been converted to the sequence of the donor). In the single strand invasion model, each DNA duplex has heteroduplex material covering the region from the initial site of exchange to the migrating branch, i.e. the heteroduplexes are symmetric. In variations of the model (Meselson-Radding) in which some DNA is degraded and re‑synthesized, the initiating chromosome is the donor of the genetic information.
These models also have many important features in common. Steps that are common to all the models include the generation of a single strand of DNA at an end, a search for homology, strand invasion or strand exchange to form a joint molecule, branch migration, and resolution. Enzymes catalyzing each of these steps have been isolated and characterized. This is the topic of the rest of this chapter. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/8%3A_Recombination_of_DNA/8.07%3A_Double-strand-break_model_for_Recombination.txt |
The initial steps in finding enzymes that carry out recombination were genetic screens for mutants of E. coli that are defective in recombination. Assays were developed to test for recombination, and mutants that showed a decrease in recombination frequency were isolated. These were assigned to complementation groups called recA, recB, recC, recD, and so forth. Roughly 20 different genes (different rec complementation groups) have been identified in E. coli. Each gene encodes an enzyme or enzyme subunit required for recombination.
Many of these genes have been cloned and their encoded products characterized in terms of a variety of enzymatic functions. However, we still do not have a clear picture of how all these enzymes work together to carry out recombination, nor has recombination has been reconstituted in vitro from purified components. Further complicating matters is the presence of multiple pathways for recombination. Much work remains to be done to completely understand recombination at a biochemical level. Despite this, the array of recombination enzymes gives us at least a partial view of the mechanisms of recombination. Also, the enzymes characterized in E. coli have homologs and counterparts in other species. Some aspects of the recombination machinery appear to be conserved across a wide phylogenetic range.
The major enzymatic steps are outlined in Figure 8.12. Three different pathways have been characterized that differ in the steps used to generate the invading single strand of DNA. All three pathways use RecA for homologous pairing and strand exchange, RuvA and RuvB for branch migration, and RuvC and DNA ligase for resolution. These steps and enzymes will be considered individually in the following sections.
8.09: Generation of Single Strands
One of the major pathways for generating 3’ single-stranded termini uses the RecBCD enzyme, also known as exonuclease V (Figure 8.13). The three subunits of this enzyme are encoded by the genes recB, recC, and recD. Each model for recombination requires a single-strand with with a free end for strand invasion, and this enzyme does so, but with several unexpected features.
RecBCD has multiple functions, and it can switch activities. It is a helicase (in the presence of SSB), an ATPase and a nuclease. The nuclease can be a 3’ to 5’ exonuclease, and endonuclease or a 5’ to 3’ exonuclease, at different steps of the process.
The helicase activity of the RecBCD enzyme initiates unwinding only on DNA containing a free duplex end. It binds to the duplex end, using the energy of ATP hydrolysis to travel along the duplex, unwinding the DNA. The enzyme complex tracks along the top strand faster than it does on the bottom strand, so single‑stranded loops emerge, getting progressively larger as it moves down the duplex. These loops can be visualized in electron micrographs. RecBCD is also a 3' to 5' exonuclease during this phase, removing the end of one of the unwound strands (Figure 8.13).
The activities of the RecBCD enzyme change at particular sequences in the DNA called chi sites(for the Greek letter c). The sequence of a chi site is 5' GCTGGTGG; this occurs about once every 4 kb on the E. coligenome. Genetic experiments show that RecBCD promotes recombination most frequently at chi sites. These sites were first discovered as mutations in bacteriophage l that led to increased recombination at those sites. These mutations altered the l sequence at the site of the mutation to become a chi site (GCTGGTGG).
When the RecBCD enzyme encounters a chi site, it will leave an extruded single strand close to this site (4 to 6 nucleotides 3' to it). A chi site serves as a signal to RecBCD to shift the polarity of its exonuclease function. Before reaching the chi site, RecBCD acts primarily as a 3’ to 5’ exonuclease, e.g. working on the top strand in Figure 8.13. At the chi site, the 3’ to 5’ exonuclease function is suppressed, and afterthe chi site, RecBCD converts to a 5’ to 3’ exonuclease, now working on the other strand (e.g. the bottom strand in Figure 8.13). Presumably, the strand that will be the substrate for the 5’ to 3’ exonuclease is nicked in concert with this conversion in polarity of the exonuclease. This process leaves the chi site at the 3’ end of a single stranded DNA. This is the substrate to which RecA can bind to initiate strand exchange (see below).
Some tests of the models for recombination have examined whether chi sites serve preferentially as either donors or recipients of the DNA during recombination. However, both results have been obtained, which makes it difficult to tie this activity precisely into either model for recombination. The genetic evidence is clear, however, that it is needed for one major pathway of recombination.
Exercise 8.4
What are the predictions of the Holliday model and the double-strand-break model for whether chi sites would be used as donors or recipients of genetic information during recombination?
An alternative pathway for generating single-strand ends for recombination uses the enzyme RecE, also known as exonuclease VIII. This pathway is revealed in recBCD- mutants. RecE is a 5’ to 3’ exonuclease that digests double‑stranded linear DNA, thereby generating single‑stranded 3' tails. RecE is encoded on a cryptic plasmid in E. coli. It is similar to the redexonuclease encoded by bacteriophage l.
A third pathway used the RecQhelicase, which is also a DNA-dependent ATPase. This pathway is revealed in recBCD- recE-mutants. The result of its helicase activity, in the presence of SSB, is the formation of a DNA molecule with single-stranded 3' tails, which can be used for strand invasion. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/8%3A_Recombination_of_DNA/8.08%3A_Enzymes_required_for_recombination_in_E._coli.txt |
The pairing of the two recombining DNA molecules (synapsis) and invasion of a single strand from the initiating duplex into the other duplex are both catalyzed by the multi-functional protein RecA. This invasion of the duplex DNA by a single stranded DNA results in the replacement of one of the strands of the original duplex with the invading strand, and the replaced strand is displaced from the duplex. Hence this reaction can also be called strand assimilation or strand exchange. RecA has many activities, including stimulating the protease function of LexA and UmuD (see Chapter 7), binding to and coating single-stranded DNA, stimulating homologous pairing between single-stranded and duplex DNA, assimilating single-stranded DNA into a duplex, and catalyzing the hydrolysis of ATP in the presence of DNA (i.e. it is a DNA-dependent ATPase). It is required in all 3 pathways for recombination. For instance, the DNA molecule with a single-stranded 3’ end generated by the RecBCD enzyme can be assimilated into a homologous region of another duplex, catalyzed by RecA and requiring the hydrolysis of ATP (Figure 8.14).
The process of single-strand assimilation occurs in three steps, as illustrated in Figure 8.15. First, RecA polymerizes onto single-stranded DNA in the presence of ATP to form the presynaptic filament. The single strand of DNA lies within a deep groove of the RecA protein, and many RecA-ATP molecules coat the single-stranded DNA. One molecule of the RecA protein covers 3 to 5 nucleotides of single-stranded DNA. The nucleotides are extended axially so they are about 5 Angstroms apart in the single-stranded DNA, about 1.5 times longer than in the absence of RecA-ATP.
Next, the presynaptic filament aligns with homologous regions in the duplex DNA. A substantial length of the three strands are held together by a polumer of RecA-ATP molecules. The aligned duplex and single strand forms a paranemic joint, meaning that the single strand is not intertwined with the double strand at this point. The duplex DNA, like the single-stranded DNA, is extended to about 1.5 times longer than in normal B form DNA (18.6 bp per turn). This extension is thought to be important in homologous pairing.
Finally, the strands are exchanged from to form a plectonemic joint. In this stage, the invading single strand is now intertwined with the complementary strand in the duplex, and one strand of the invaded duplex is now displaced. In E. coli, exchange occurs in a 5' to 3' direction relative to the single strand and requires ATP hydrolysis. In contrast, the yeast homolog, Rad51, causes the single-strand to invade with the opposite polarity, i.e. 3' to 5'. Thus the direction of this polarity is not a universally conserved feature of recombination mechanisms.
The product of strand assimilation is a heteroduplex in which one strand of the duplex was the original single-stranded DNA. The other strand of the original duplex is displaced.
Many details of the activity of RecA have been revealed by in vitroassays for single strand assimilation, or strand exchange. The DNA substrates for strand exchange catalyzed by RecA must meet three requirements. There must be a region of single stranded DNA on which RecA can bind and polymerize, the two molecules undergoing strand exchange must have a region of homology, and there must be a afree end within the region of homology. The latter requirement can be overcome by providing a topoisomerase.
One such assay is the conversion of a single-stranded circular DNA to a duplex circle (Figure 8.16). The substrates for this reaction are a circular single-stranded DNA and a homologous linear duplex. These are mixed together in the presence of RecA and ATP. Many RecA-ATP molecules coat the single-stranded circle to form the nucleoprotein presynaptic filament, as discussed above. During synapsis, annealing is initiated with the 3' end of the strand complementary to the single-stranded circle. Thus the single strand invades with 5' to 3' polarity (with reference to its own polarity). Strand displacement, driven by ATP hydrolysis to dissociate the RecA, results in the formation of a nicked circle (one strand of which was the original single-stranded circle) and a linear single strand of DNA.
Exercise
Try to relate this in vitro assay to the steps in the double-strand-break model for recombination. What step(s) in the model does this mimic? What else is needed for to get to the recombinant joints (Holliday junctions)?
The structure of E. coliRecA bound by ADP, both monomer and polymer, have been solved by X-ray crystallography. As shown in Figure 8.17, the central domain has the binding site for ATP and ADP, and is presumably the site of binding of the single-stranded and double-stranded DNA. The domains extending away from the central region are involved in polymerization of RecA proteins and in interactions between the presynaptic fibers.
Proteins homologous to the E. coli RecA are found in yeast (Rad51 and Dmc1) and in mice (Rad51). Given the universality of recombination, it is likely that homologs will be found in virtually all species. Mutations in the E. coli recA gene reduce conjugational recombination by as much as 10,000 fold, so it is clear that RecA plays a central role in recombination. However, null mutations in recA are not lethal, nor are null mutations in the yeast homologs RAD51 and DMC1. In contrast, mice homozygous for a knockout mutation in the Rad51 gene die very early in development, at the 4-cell stage. This indicates that in mice, this RecA homolog is playing a novel role in replication or repair, presumably in addition to its role in recombination. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/8%3A_Recombination_of_DNA/8.10%3A_Synapsis_and_Invasion_of_Single_Strands.txt |
The movement of a Holliday junction to generate additional heteroduplex requires two proteins. One is the RuvA tetramer, which recognizes the structure of the Holliday junction. A rendering of the structure derived from X-ray crystallographic analysis of the RuvA-Holliday junction crystals is shown in Figure 8.18.
RuvB is an ATPase. It forms hexameric rings that provide the motor for branch migration. As illustrated in Figure 8.19, RuvA tetramers recognize the Holliday junction, and RuvB uses the energy of ATP hydrolysis to unwind the parental duplexes and form heteroduplexes between them.
8.12: Resolution
RuvC is the endonuclease that cleaves the Holliday junctions (Figure 8.20). It forms dimers that bind to the Holliday junction; recent data indicate an interaction among RuvA, RuvB and RuvC as a complex at the Holliday junction. The structure of the RuvA-Holliday junction complex (Figure 8.18) suggests that the open structure of the junction stabilized by the binding of RuvA may expose a surface that is recognized by Ruv C for cleavage. RuvC cleaves symmetrically, in two strands with the same nearly identical sequences, thereby producing ligatable products.
The preferred site of cleavage by RuvC is 5’ WTT’S, where W = A or T and S = G or C, and ‘ is the site of cleavage. RuvC can cut strands for either horizontal or vertical resolution. Strand choice is influenced by the sequence preference and also by the presence of RecA protein, which favors vertical cleavage (i.e. to cause recombination of flanking markers).
8.E: Recombination of DNA (Exercises)
Question 8.6. According to the Holliday model for genetic recombination, what factor determines the length of the heteroduplex in the recombination intermediate?
Question 8.7. Holliday junctions can be resolved in two different ways. What are the consequences of the strand choice used in resolution?
Question 8.8. Why do models for recombination include the generation of heteroduplexes in the products?
Question 8.9. Consider two DNA duplexes that undergo recombination by the double-strand break mechanism. The parental duplex indicated by thin lines has dominant alleles for genes M, N, O, P, and Q, and the parental duplex shown in thick lines has recessive alleles, indicated by the lower case letters. The recombination intermediate with two Holliday structures is also shown.
• a) What duplexes result from resolution of the left Holliday junction vertically and the right junction horizontally?
• b) After the vertical-horizontal resolution, what will the genotype be of the recombination products with respect to the flanking markers M and Q? In answering, use a slash to separate the designation for the 2 chromosomes, each of which is indicated by a line (i.e. the parental arrangement is M___Q / m___q).
• c) If the products of the vertical-horizontal resolution were separated by meiosis, and then replicated by mitosis to generate 8 spores in an ordered array (as in the Ascomycetefungi), what would be the phenotype of the spores with respect to alleles of gene O? Assume that the sister chromatids of these chromosomes did not undergo recombination in this region (i.e. one parental duplex from each homologous chromosome remains from the 4n stage).
For the next 3 problems, consider two DNA duplexes that undergo recombination by the double-strand break mechanism. The parental duplex denoted by thin black lines has dominant alleles (capital letters) for genes (or loci) K, L, and M, and the parental duplex denoted by thick gray lines has recessive alleles, indicated by k, l, m. The genes are shown as boxes with gray outlines. In the diagram on the right, the double strand break has been made in the L gene in the black duplex and expanded by the action of exonucleases.
Question 8.10. When recombination proceeds by the double-strand break mechanism, what is the structure of the intermediate with Holliday junctions, prior to branch migration? Please draw the structure, and distinguish between the DNA chains from the parental duplexes.
Question 8.11. If the recombination intermediates are resolved to generate a chromosome with the dominant K allele of the K gene and the recessive m allele of the M gene on the same chromosome (K___m), which allele (dominant L or recessive l) will be be at the L, or middle, gene?
Question 8.12. If the left Holliday junction slid leftward by branch migration all the way through the K gene (K allele on the black duplex, k allele on the gray duplex), what will the structure of the product be, prior to resolution?
Question 8.13. According to the original Holliday model and the double-strand break model for recombination, what are the predicted outcomes of recombination between a linear duplex chromosome and a (formerly) circular duplex carrying a gap in the region of homology? The homology is denoted by the boxes labeled ABC on the linear duplex and ac on the gapped circle. The regions flanking the homology (P and Q versus X and Y) are not homologous.
The results of an experiment like this are reported in Orr-Weaver, T. L., Szostak, J. W. and Rothstein, R. J. (1981) Yeast transformation: a model system for the study of recombination. Proc. Natl. Acad. Sci. USA 78: 6354-6358. These data were instrumental in formulating the double-strand-break model for recombination.
Question 8.14.A variety of in vitro assays have been developed for strand exchange catalyzed by RecA. For each of the substrates shown below, what are the expected products when incubated with RecA and ATP (and SSB to facilitate removal of secondary structures from single-stranded DNA)? In practice, the reactions proceed in stages and one can see intermediates, but answer in terms of the final products after the reaction has gone to completion.
In each case, the molecule with at least partical single stranded region is shown with thick blue strands, and the duplex that will be invaded is shown with thin red lines. The DNA substrates are as follows.
• A. Single-stranded circle and duplex linear. The two substrates are the same length and are homologous throughout.
• B. Single-stranded short linear fragments and duplex circle. The short fragments are homologous to the circle.
• C. Single-stranded linear and duplex linear. The two substrates are the same length and are homologous throughout.
• D. Gapped circle and duplex linear. The intact strand of the circle is the same length as the linear and is homologous throughout. The gapped strand of the circle is complementary to the intact strand, of course, but is just shorter. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/8%3A_Recombination_of_DNA/8.11%3A_Branch_Migration.txt |
The final method of changing the DNA in a genome that we will consider is transposition, which is the movement of DNA from one location to another. Segments of DNA with this ability to move are called transposable elements. Transposable elements were formerly thought to be found only in a few species, but now they are recognized as components of the genomes of virtually all species.
• 9.1: Transposable Elements (Transposons)
Transposable elements (both active and inactive) occupy approximately half the human genome and a substantially greater fraction of some plant genomes! These movable elements are ubiquitous in the biosphere, and are highly successful in propagating themselves. We now realize that some transposable elements are also viruses, for instance, some retroviruses can integrate into a host genome to form endogenous retroviruses.
• 9.2: Are Transposons Parasites or Symbionts?
Do the transposable elements confer some selective advantage on the "host"? Or are they merely parasitic or "selfish," existing only to increase the number of copies of the element? This critical issue is a continuing controversy.
• 9.5: Transposition occurs by Insertion into Staggered Breaks
A common property of virtually all transposable elements is that they move by inserting into a staggered break in a chromosome, i.e. one strand is slightly longer than the other at the break. The first indication of this was the observation that the same short DNA sequence is found on each side of a transposable element. The sequence within these flanking direct repeats is distinctive for each copy of the transposable element, but the size is characteristic of a particular family of elements.
• 9.6: Classes of Transposable Elements
The two major classes of transposable elements are defined by the intermediates in the transposition process. One class moves by DNA intermediates, using transposases and DNA polymerases to catalyze transposition. The other class moves by RNA intermediates, using RNA polymerase, endonucleases and reverse transcriptase to catalyze the process. Both classes are abundant in many species, but some groups of organisms have a preponderance of one or the other.
• 9.E: Transposition of DNA (Exercises)
• Additional consequences of transposition
• Dissociation Elements
• Mechanism of DNA-mediated transposition
• Mechanism of Retrotransposition
Although the mechanism of retrotransposition is not completely understood, it is clear that at least two enzymatic activities are utilized. One is an integrase, which is an endonuclease that cleaves at the site of integration to generate a staggered break. The other is RNA-dependent DNA polymerase, also called reverse transcriptase. These activities are encoded in some autonomous retrotransposons, including both LTR-retrotransposons and non-LTR-retrotransposons.
• Unstable Alleles
The insertion of a controlling element can generate an unstable allele of a locus, designated mutable. This instability can be seen both in somatic and in germline tissues. The instability can result from reversion of a mutation, due to the excision and transposition of the controlling element. After excision and re-integration, the transposable element can alter the expression of a gene at the new location. This new phenotype indicated that the element was mobile.
9. Transposition of DNA
Transposable elements (both active and inactive) occupy approximately half the human genome and a substantially greater fraction of some plant genomes! These movable elements are ubiquitous in the biosphere, and are highly successful in propagating themselves. We now realize that some transposable elements are also viruses, for instance, some retroviruses can integrate into a host genome to form endogenous retroviruses. Indeed, some viruses may be derived from natural transposable elements and vice versa. Since viruses move between individuals, at least some transposable elements can move between genomes (between individuals) as well as within an individual’s genome. Given their prevalence in genomes, the function (if any) of transposable elements has been much discussed but is little understood. It is not even clear whether transposable elements should be considered an integral part of a species’ genome, or if they are successful parasites. They do have important effects on genes and their phenotypes, and they are the subject of intense investigation.
Transposition is related to replication, recombination and repair. The process of moving from one place to another involves a type of recombination, insertions of transposable elements can cause mutations, and some transpositions are replicative, generating a new copy while leaving the old copy intact. However, this ability to move is a unique property of transposable elements, and warrants treatment by itself.
Properties and effects of transposable elements
The defining property of transposable elements is their mobility; i.e. they are genetic elements that can move from one position to another in the genome. Beyond the common property of mobility, transposable elements show considerable diversity. Some move by DNA intermediates, and others move by RNA intermediates. Much of the mechanism of transposition is distinctive for these two classes, but all transposable elements effectively insert at staggered breaks in chromosomes. Some transposable elements move in a replicative manner, whereas others are nonreplicative, i.e. they move without making a copy of themselves.
Transposable elements are major forces in the evolution and rearrangement of genomes (Figure 9.1). Some transposition events inactivate genes, since the coding potential or expression of a gene is disrupted by insertion of the transposable element. A classic example is the r allele (rugosus) of the gene encoding a starch branching enzyme in peas is nonfunctional due to the insertion of a transposable element. This allele causes the wrinkled pea phenotype in homozygotes originally studied by Mendel. In other cases, transposition can activate nearby genes by bringing an enhancer of transcription (within the transposable element) close enough to a gene to stimulate its expression. If the target gene is not usually expressed in a certain cell type, this activation can lead to pathology, such as activation of a proto-oncogene causing a cell to become cancerous. In other cases, no obvious phenotype results from the transposition. A particular type of transposable element can activate, inactivate or have no effect on nearby genes, depending on exactly where it inserts, it’s orientation and other factors.
Transposable elements can cause deletions or inversions of DNA. When transposition generates two copies of the same sequence in the same orientation, recombination can delete the DNA between them. If the two copies are in the opposite orientations, recombination will invert the DNA between them.
As part of the mechanism of transposition, additional DNA sequences can be mobilized. DNA located between two copies of a transposable element can be moved together with them when they move. In this manner, transposition can move DNA sequences that are not normally part of a transposable element to new locations. Indeed, "host" sequences can be acquired by viruses and propagated by infection of other individuals. This may be a natural means for evolving new strains of viruses. One of the most striking examples is the acquisition and modification of a proto-oncogene, such as cellular c-src, by a retrovirus to generate a modified, transforming form of the gene, called v-src. These and related observations provided insights into the progression of events that turn a normal cell into a cancerous one. They also point to the continual acquisition (and possibly deletion) of information from host genomes as a natural part of the evolution of viruses. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/9._Transposition_of_DNA/9.1%3A_Transposable_Elements_%28Transposons%29.txt |
Do the transposable elements confer some selective advantage on the "host"? Or are they merely parasitic or "selfish," existing only to increase the number of copies of the element? This critical issue is a continuing controversy. As just mentioned, certain results of transposition can be detrimental, leading to a loss of function or changes in regulation of the genes at the site of integration after movement. Also, we are starting to appreciate the intimate connection between viruses and transposable elements. Thus one can view many transposable elements as parasites on the genome. The number of transposable elements can expand rapidly in a genome. For instance, it appears that transposable elements making up a majority of the genome of maize are not abundant in the wild parent, teosinte. Thus this massive expansion has occurred since the domestication of corn, roughly within the past 10,000 years.
However, other studies indicate that the presence of transposable elements is beneficial to an organism. Two strains of bacteria, one with a normal number of transposable elements and the other with many fewer, can be grown in competitive conditions. The strain with the higher number of transposable elements has a growth advantage under these conditions. Various proposals have been made as to the nature of that advantage. One intriguing possibility is that the mechanism of transposition affords an opportunity to seal chromosome breaks. Other possible benefits have not been excluded. Thus the relationship between transposable elements and their hosts may be as much symbiotic as parasitic. Resolving these issues is an interesting challenge for future research.
9.5: Transposition occurs by Insertion into Staggered Breaks
Staggered Break in a chromosome
A common property of virtually all transposable elements is that they move by inserting into a staggered break in a chromosome, i.e. one strand is slightly longer than the other at the break (Figure 9.9). The first indication of this was the observation that the same short DNA sequence is found on each side of a transposable element. The sequence within these flanking direct repeats (FDRs) is distinctive for each copy of the transposable element, but the size of the FDR is characteristic of a particular family of transposable elements. Some families of transposable elements have FDRs as short as 4 bp and other families have FDRs as long as 12 bp. However, within a particular family, the sequence of the FDR will differ between individual copies. These FDRs are hallmarks of transposable elements.
Since the FDRs are distinctive for each copy, they are not part of the transposable element themselves. Some families of transposable elements do have repeated sequences at their flanks that are identical for all members of the family, but these are integral parts of the transposable element. The variation in sequence of the FDRs indicates that they are generated from the target sites for the transposition events. If the transposable element inserted into a break in the chromosome that left a short overhang (one strand longer than the other), and this overhang were filled in by DNA polymerase as part of the transposition, then the sequence of that overhang would be duplicated on each side of the new copy. Such a break with an overhang is called a staggered break. The size of the staggered break would determine the size of the FDR.
Mechanistic studies of the enzymes used for transposition have shown that such staggered breaks are made at the target site prior to integration and are repaired as part of the process of transposition (see below). The staggered breaks are used in transposition both by DNA intermediates and by RNA intermediates. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/9._Transposition_of_DNA/9.2%3A_Are_Transposons_Parasites_or_Symbionts.txt |
The two major classes of transposable elements are defined by the intermediates in the transposition process. One class moves by DNA intermediates, using transposases and DNA polymerases to catalyze transposition. The other class moves by RNA intermediates, using RNA polymerase, endonucleases and reverse transcriptase to catalyze the process. Both classes are abundant in many species, but some groups of organisms have a preponderance of one or the other. For instance, bacteria have mainly the DNA intermediate class of transposable elements, whereas the predominant transposable elements in mammalian genomes move by RNA intermediates.
• Transposable elements that move via DNA intermediates
• Transposable elements that move via RNA intermediates
Transposable elements that move via DNA intermediates
Among the most thoroughly characterized transposable elements are those that move by DNA intermediates. In bacteria, these are either short insertion sequences or longer transposons.
An insertion sequences, or IS, is a short DNA sequence that moves from one location to another. They were first recognized by the mutations they cause by inserting into bacterial genes. Different insertion sequences range in size from about 800 bp to 2000 bp. The DNA sequence of an IS has inverted repeats (about 10 to 40 bp) at its termini (Figure 9.10A.). Note that this is different from the FDRs, which are duplications of the target site. The inverted repeats are part of the IS element itself. The sequences of the inverted repeats at each end of the IS are very similar but not necessarily identical. Each family of insertion sequence in a species is named IS followed by a number, e.g. IS1, IS10, etc.
An insertion sequence encodes a transposase enzyme that catalyzes the transposition. The amount of transposase is well regulated and is the primary determinant of the rate of transposition. Transposons are larger transposable elements, ranging in size from 2500 to 21,000 bp. They usually encode a drug resistance gene or other marker besides the functions required for transposition (Figure 9.10.B.). One type of transposon, called a composite transposon, has an IS element at each end (Figure 9.10.C.). One or both IS elements may be functional; these encode the transposition function for this class of transposons. The IS elements flank the drug resistance gene (or other selectable marker). It is likely that the composite transposon evolved when two IS elements inserted on both sides of a gene. The IS elements at the end could either move by themselves or they can recognize the ends of the closely spaced IS elements and move them together with the DNA between them. If the DNA between the IS elements confers a selective advantage when transposed, then it will become fixed in a population.
Exercise 9.3
What are the predictions of this model for formation of a composite transposon for the situation in which a transposon in a small circular replicon, such as a plasmid?
The TnA family of transposons has been intensively studied for the mechanism of transposition. Members of the TnA family have terminal inverted repeats, but lack terminal IS elements (Figure 9.10). The tnpAgene of the TnA transposon encodes a transposase, and the tnpRgene encodes a resolvase. TnA also has a selectable marker, ApR, which encodes a beta-lactamase and makes the bacteria resistance to ampicillin.
Transposable elements that move via DNA intermediates are not limited to bacteria, but rather they are found in many species. The P elements and copiafamily of repeats are examples of such transposable elements in Drosophila, as are marinerelements in mammals and the controlling elements in plants. Indeed, the general structure of controlling elements in maize is similar to that of bacterial transposons. In particular, they end in inverted repeats and encode a transposase. As illustrated in Figure 9.11, the DNA sequences at the ends of an Ac element are very similar to those of a Dselement. However, internal regions, which normally encode the transposase, have been deleted. This is why Dselements cannot transpose by themselves, but rather they require the presence of the intact transposon, Ac, in the cell to provide the transposase. Since transposase works in trans, the Acelement can be anywhere in the genome, but it can act on Dselements at a variety of sites. Note that Ac is an autonomous transposon because it provides its own transposase and it has the inverted repeats needed to act as the substrate for transposase.
Mechanism of DNA-mediated transposition
Some families of transposable elements that move via a DNA intermediate do so in a replicative manner. In this case, transposition generates a new copy of the transposable element at the target site, while leaving a copy behind at the original site. A cointegrate structure is formed by fusion of the donor and recipient replicons, which is then resolved (Figure 9.12). Other families use a nonreplicative mechanism. In this case, the original copy excises from the original site and move to a new target site, leaving the original site vacant.
Studies of bacterial transposons have shown that replicative transposition and some types of nonreplicative transposition proceed through a strand-transfer intermediate (also known as a crossover structure), in which both the donor and recipient replicons are attached to the transposable element (Figure 9.13). For replicative transposition, DNA synthesis through the strand-transfer intermediate produces a transposable element at both the donor and target sites, forming the cointegrate intermediate. This is subsequently resolved to separate the replicons. DNA synthesis does not occur at the crossover structure in nonreplicative transposition, thus leaving a copy only at the new target site. In an alternative pathway for nonreplicative transposition, the transposon is excised by two double strand breaks, and is joined to the recipient at a staggered break (illustrated at the bottom of Figure 9.12).
In more detail, there are two steps in common for replicative and nonreplicative transposition, generating the strand-transfer intermediate (Figure 9.13).
1. The transposase encoded by a transposable element makes four nicks initially. Two nicks are made at the target site, one in each strand, to generate a staggered break with 5' extensions (3' recessed). The other two nicks flank the transposon; one nick is made in one DNA strand at one end of the transposon, and the other nick is made in the other DNA strand at the other end. Since the transposon has inverted repeats at each end, these two nicks that flank the transposon are cleavages in the same sequence. Thus the transposase has a sequence-specific nicking activity. For instance, the transposase from TnA binds to a sequence of about 25 bp located within the 38 bp of inverted terminal repeat (Figure 9.10). It nicks a single strand at each end of the transposon, as well as the target site (Figure 9.13). Note that although the target and transposon are shown apart in the two-dimensional drawing in Figure 9.13, they are juxtaposed during transposition.
2. At each end of the transposon, the 3' end of one strand of the transposon is joined to the 5' extension of one strand at the target site. This ligation is also catalyzed by transposase. ATP stimulates the reaction but it can occur in the absence of ATP if the substrate is supercoiled. Ligation of the ends of the transposon to the target site generates a strand-transfer intermediate, in which the donor and recipient replicons are now joined by the transposon.
After formation of the strand-transfer intermediate, two different pathways can be followed. For replicative transposition, the 3' ends of each strand of the staggered break (originally at the target site) serve as primers for repair synthesis (Figure 9.13). Replication followed by ligation leads to the formation of the cointegrate structure, which can then be resolved into the separate replicons, each with a copy of the transposon. The resolvase encoded by transposon TnA catalyzes the resolution of the cointegrate structure. The site for resolution (res) is located between the divergently transcribed genes for tnpA and tnpR (Figure 9.10). TnA resolvase also negatively regulates expression of both tnpA and tnpR (itself).
For nonreplicative transposition, the strand-transfer intermediate is released by nicking at the ends of the transposon not initially nicked. Repair synthesis is limited to the gap at the flanking direct repeats, and hence only one copy of the transposon is left. This copy is ligated to the new target site, leaving a vacant site in the donor molecule.
The enzyme transposase can recognize specific DNA sequences, cleave two duplex DNA molecules in four places, and ligate strands from the donor to the recipient. This enzyme has a remarkable ability to generate and manipulate the ends of DNA. A three-dimensional structure for the Tn5 transposase in complex with the ends of the Tn5 DNA has been solved by Rayment and colleagues. One static view of this protein DNA complex is in Figure 9.14.A. The transposase is a dimer, and each double-stranded DNA molecule (donor and target) is bound by both protein subunits. This orients the transposon ends into the active sites, as shown in the figure. Also, an image with just the DNA (Figure 9.14.B.) shows considerable distortion of the DNA helix at the ends. This recently determined structure is a good starting point to better understand the mechanism for strand cleavage and transfer.
Transposable elements that move via RNA intermediates
Transposable DNA sequences that move by an RNA intermediate are called retrotransposons. They are very common in eukaryotic organisms, but some examples have also been found in bacteria. Some retrotransposons have long terminal repeats (LTRs) that regulate expression (Figure 9.15). The LTRs were initially discovered in retroviruses. They have now been seen in some but not all retrotransposons. They have a strong promoter and enhancer, as well as signals for forming the 3’ end of mRNAs after transcription. The presence of the LTR is distinctive for this family, and members are referred to as LTR-containing retrotransposons. Examples include the yeast Ty-1family and retroviral proviruses in vertebrates. Retroviral proviruses encode a reverse transcriptase and an endonuclease, as well as other proteins, some of which are needed for viral assembly and structure.
Others retrotransposons are in the large and diverse class of non-LTR retrotransposons (Figure 9.15). One of the most prevalent examples is the family of long interspersed repetitive elements, or LINEs. It was initially found in mammals but has now been found in a broad range of phyla, including fungi. The first and most common LINE family in mammals is the LINE1 family, also called L1. An older family, but discovered later, is called LINE2. Full-length LINEs are about 7000 bp long, and there are about 10,000 copies in humans. Many other copies are truncated from the 5’ ends. Like retroviral proviruses, the full-length L1 encodes a reverse transcriptase and an endonuclease, as well as other proteins. However, the promoter is not an LTR. Other abundant non-LTR retrotransposons, initially discovered in mammals, are short interspersed repetitive elements, or SINEs. These are about 300 bp long. Alurepeats, with over a million copies, comprise the predominant class of SINEs in humans. Non-LTR retrotransposons besides LINEs are found in many other species, such as jockey repeats in Drosophila.
Extensive studies in of genomic DNA sequences have allowed the reconstruction of the history of transposable elements in humans and other mammals. The major approach has been to classify the various types of repeats (themselves transposable elements), align the sequences and determine how different the members of a family are from each other. Since the vast majority of the repeats are no longer active in transposition, and have no other obvious function, they will accumulate mutations rapidly, at the neutral rate. Thus the sequence of more recently transposing members are more similar to the source sequence than are the members that transposed earlier. The results of this analysis show that the different families of repeats have propagated in distinct waves through evolution (Figure 9.16). The LINE2 elements were abundant prior to the mammalian divergence, roughly 100 million years ago. Both LINE1 and Alu repeats have propagated more recently in humans. It is likely that the LINE1 elements, which encode a nuclease and a reverse transcriptase, provide functions needed for the transposition and expansion of Alu repeats. LINE1 elements have expanded in all orders of mammals, but each order has a distinctive SINE, all of which are derived from a gene transcribed by RNA polymerase III. This has led to the idea that LINE1 elements provide functions that other different transcription units use for transposition.
Mechanism of retrotransposition
Although the mechanism of retrotransposition is not completely understood, it is clear that at least two enzymatic activities are utilized. One is an integrase, which is an endonuclease that cleaves at the site of integration to generate a staggered break (Figure 9.17). The other is RNA-dependent DNA polymerase, also called reverse transcriptase. These activities are encoded in some autonomous retrotransposons, including both LTR-retrotransposons such as retroviral proviruses and non-LTR-retrotransposons such as LINE1 elements.
The RNA transcript of the transposable element interacts with the site of cleavage at the DNA target site. One strand of DNA at the cleaved integration site serves as the primer for reverse transcriptase. This DNA polymerase then copies the RNA into DNA. That cDNA copy of the retrotransposon must be converted to a double stranded product and inserted at a staggered break at the target site. The enzymes required for joining the reverse transcript (first strand of the new copy) to the other end of the staggered break and for second strand synthesis have not yet been established. Perhaps some cellular DNA repair functions are used.
The model shown in Figure 9.17 is consistent with any RNA serving as the template for synthesis of the cDNA from the staggered break. However, LINE1 mRNA is clearly used much more often than other RNAs. The basis for the preference of the retrotransposition machinery for LINE1 mRNA is still being studied. Perhaps the endonuclease and reverse transcriptase stay associated with the mRNA that encodes them after translation has been completed, so that they act in cis with respect to the LINE1 mRNA. Other repeats that have expanded recently, such as Alu repeats in humans, may share sequence determinants with LINE1 mRNA for this cis preference.
Clear evidence that retrotransposons can move via an RNA intermediate came from studies of the yeast Ty-1 elements by Gerald Fink and his colleagues. They placed a particular Ty-1 element, called TyH3 under control of a GAL promoter, so that its transcription (and transposition) could be induced by adding galactose to the media. They also marked TyH3 with an intron. After inducing transcription of TyH3, additional copies were found at new locations in the yeast strain. When these were examined structurally, it was discovered that the intron had been removed. If the RNA transcript is the intermediate in moving the Ty-1 element, it is subject to splicing and the intron can be removed. Hence, these results fit the prediction of an RNA-mediated transposition. They demonstrate that during transposition, the flow of Ty-1 sequence information is from DNA to RNA to DNA.
Exercise 9.4
If yeast Ty-1 moved by the mechanism illustrated for DNA-mediated replicative transposition in Figure 9.13, what would be predicted in the experiment just outlined? Also, would you expect an increase in transposition when transcription is induced?
Additional Consequences of Transposition
Not only can transposable elements interrupt genes or disrupt their regulation, but they can cause additional rearrangements in the genome. Homologous recombination can occur between any two nearly identical sequences. Thus when transposition makes a new copy of a transposable element, the two copies are now potential substrates for recombination. The outcome of recombination depends on the orientation of the two transposable elements relative to each other. Recombination between two transposable elements in the same orientation on the same chromosome leads to a deletion, whereas it results in an inversion if they are in opposite orientations (Figure 9.18).
The preference of the retrotransposition machinery for LINE1 mRNA does not appear to be absolute. Many processed genes have been found in eukaryotic genomes; these are genes that have no introns. In many cases, a homologous gene with introns is seen in the genome, so it appears that these processed genes have lost their introns. It is likely that these were formed when processed mRNA derived from the homologous gene with introns was copied into cDNA and reinserted into the genome. Many, but not all, of these processed genes are pseudogenes, i.e. they have been mutated such that they no longer encode proteins. Other examples of active processed genes have inserted next to promoters and encode functional proteins. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/9._Transposition_of_DNA/9.6%3A_Classes_of_Transposable_Elements.txt |
Question 9.5. Suppose you are studying a gene that is contained within a 5 kb EcoRI fragment for the wild type allele. When analyzing mutations in that gene, you found one that converted the 5 kb fragment to an 8 kb EcoRI fragment. Further analysis showed that the additional 3 kb of DNA was flanked by direct repeats of 6 bp, that the terminal 30 bp of the additional DNA was identical at each end but in an inverted orientation. Recombinant plasmids carrying the 8 kb EcoRI fragment conferred resistance to the antibiotic kanamycin in the host bacteria, whereas neither the parental cloning vector nor a recombinant plasmid carrying the 5 kb EcoRI fragment did. What do you conclude is the basis for this mutation? What other enzyme activities might you expect to be encoded in the additional DNA?
Use the following diagram to answer the next two questions. Transposase encoded by a transposable element (TE) has nicked on each side of the TE in the donor (black) replicon and made a staggered break in the recipient (gray) replicon, and the ends of the TE have been joined to the target (T) site in the recipient replicon. The strands of the replicons have been designated top (t) or bottom (b). The open triangles with 1 or 2 in them just refer to locations in the figure; they are not part of the structure.
Question 9.6. The action of DNA polymerase plus dNTPs, primed at positions 1, followed by ligase (with ATP or NAD) leads to what product or result? (In this scenario, nothing occurs at positions 2).
Question 9.7. The action of an endonuclease at the positions labeled 2 followed by DNA polymerase and dNTPs to fill in the gaps (from positions 1 to the next 5' ends of DNA fragments), and finally DNA ligase (with ATP or NAD) leads to what product or result?
Question 9.8. Refer to the model for a crossover intermediate in replicative transposition in Fig. 9.13. If the transposon moved to a second site on the same DNA molecule by replicative transposition (not to a different molecule as shown in the Figure), what are the consequences for the DNA between the donor and recipient sites?
Question 9.9.The technique of transposon tagging uses the integration of transposons to mutate a large numbers of genes while leaving a "tag" in the mutated gene to allow subsequent isolation of the gene using molecular probes (such as hybridization probes for the transposon). What is a good candidate for transposon tagging in mammalian cells?
Mechanism of DNA-mediated transposition
1. The transposase encoded by a transposable element makes four nicks initially. Two nicks are made at the target site, one in each strand, to generate a staggered break with 5' extensions (3' recessed). The other two nicks flank the transposon; one nick is made in one DNA strand at one end of the transposon, and the other nick is made in the other DNA strand at the other end. Since the transposon has inverted repeats at each end, these two nicks that flank the transposon are cleavages in the same sequence. Thus the transposase has a sequence-specific nicking activity. For instance, the transposase from TnA binds to a sequence of about 25 bp located within the 38 bp of inverted terminal repeat (Figure 9.10). It nicks a single strand at each end of the transposon, as well as the target site (Figure 9.13). Note that although the target and transposon are shown apart in the two-dimensional drawing in Figure 9.13, they are juxtaposed during transposition.
2. At each end of the transposon, the 3' end of one strand of the transposon is joined to the 5' extension of one strand at the target site. This ligation is also catalyzed by transposase. ATP stimulates the reaction but it can occur in the absence of ATP if the substrate is supercoiled. Ligation of the ends of the transposon to the target site generates a strand-transfer intermediate, in which the donor and recipient replicons are now joined by the transposon.
Mechanism of Retrotransposition
Although the mechanism of retrotransposition is not completely understood, it is clear that at least two enzymatic activities are utilized. One is an integrase, which is an endonuclease that cleaves at the site of integration to generate a staggered break (Figure 9.17). The other is RNA-dependent DNA polymerase, also called reverse transcriptase. These activities are encoded in some autonomous retrotransposons, including both LTR-retrotransposons such as retroviral proviruses and non-LTR-retrotransposons such as LINE1 elements.
The RNA transcript of the transposable element interacts with the site of cleavage at the DNA target site. One strand of DNA at the cleaved integration site serves as the primer for reverse transcriptase. This DNA polymerase then copies the RNA into DNA. That cDNA copy of the retrotransposon must be converted to a double stranded product and inserted at a staggered break at the target site. The enzymes required for joining the reverse transcript (first strand of the new copy) to the other end of the staggered break and for second strand synthesis have not yet been established. Perhaps some cellular DNA repair functions are used.
The model shown in Figure 9.17 is consistent with any RNA serving as the template for synthesis of the cDNA from the staggered break. However, LINE1 mRNA is clearly used much more often than other RNAs. The basis for the preference of the retrotransposition machinery for LINE1 mRNA is still being studied. Perhaps the endonuclease and reverse transcriptase stay associated with the mRNA that encodes them after translation has been completed, so that they act in ciswith respect to the LINE1 mRNA. Other repeats that have expanded recently, such as Alurepeats in humans, may share sequence determinants with LINE1 mRNA for this cispreference.
Clear evidence that retrotransposons can move via an RNA intermediate came from studies of the yeast Ty-1elements by Gerald Fink and his colleagues. They placed a particular Ty-1element, called TyH3under control of a GALpromoter, so that its transcription (and transposition) could be induced by adding galactose to the media. They also marked TyH3with an intron. After inducing transcription of TyH3, additional copies were found at new locations in the yeast strain. When these were examined structurally, it was discovered that the intron had been removed. If the RNA transcript is the intermediate in moving the Ty-1element, it is subject to splicing and the intron can be removed. Hence, these results fit the prediction of an RNA-mediated transposition. They demonstrate that during transposition, the flow of Ty-1sequence information is from DNA to RNA to DNA.
Exercise
If yeast Ty-1 moved by the mechanism illustrated for DNA-mediated replicative transposition in Figure 9.13, what would be predicted in the experiment just outlined? Also, would you expect an increase in transposition when transcription is induced? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/9._Transposition_of_DNA/9.E%3A_Transposition_of_DNA_%28Exercises%29.txt |
Insertion of a controlling element can generate an unstable allele of a locus
The insertion of a controlling element can generate an unstable allele of a locus, designated mutable. This instability can be seen both in somatic and in germline tissues. The instability can result from reversion of a mutation, due to the excision and transposition of the controlling element. After excision and re-integration, the transposable element can alter the expression of a gene at the new location. This new phenotype indicated that the element was mobile.
An example of the effects of integration and excision of a transposable element can be seen at the bronze locus in maize (Figure 9.8). The aleurone is the surface layer of endosperm in a kernel of maize. The wild type has a deep bluish-purple color. This is determined by the bronzelocus. The Bz allele is dominant and confers the bluish-purple color to the aleurone. The bzallele is recessive, and gives a bronze color to the aleurone when homozygous. In Bzkernels, anthocyanin is produced. Bzencodes UDPglucose:flavanoid 3-O-glucosyltransferase (UFGT), an enzyme needed for anthocyanin production. The loss-of-function bz alleles have no UFGT activity, and the bluish-purple anthocyanins are not produced. Some alleles of bronze show an unstable, or mutable, phenotype. In the bz-malleles, clones of cells regain the bluish-purple color characteristic of Bz cells. This produces patches of bluish-purple color in the aleurone of kernels (Figure 9.8).
This mutation in the bz-mallele is the insertion of the Ds(dissociation) transposable element. Dsdisrupts the function of the UFGT gene to give a bronze color to the seed kernel. In the presence of the Ac(activator) element, the Ds can excise from the locus, restoring a functional UFGT gene (and a bluish-purple color). This occurs in some but not all cells in the developing seed and is clonally inherited, resulting in the patches of blue on a bronze background for each kernel.
Current methods for observing transposition and transposable elements
Movement of DNA segments can be observed by a variety of modern techniques. In organisms with a short generation time, such as bacteria and yeast, one can simply monitor many generations for the number and positions of a family of repeated DNA elements by blot-hybridization analysis of genomic DNA. Using a Ty-1DNA fragment as a probe, about 20 hybridizing bands could be seen at the start of an experiment, meaning that about 20 copies were present in the yeast genome. The size of the restriction fragment containing each element was distinctive, as determined by restriction endonuclease cleavage sites that flanked the different locations of each element. After growing for many generations, some new bands were observed, showing that new Ty-1elements had been generated and moved to new locations. These observations led to this family of repeats being christened Ty-1, for transposable element, yeast, number 1.
Evidence for transposition in many organisms comes from analysis of new mutations. Transposable elements appear to be the major source of new mutation in Drosophila, and they have been shown to cause mutations in bacteria, fungi, plants and animals. One example from humans is a new mutation causing hemophilia. A patient from a family with no prior history was diagnosed with hemophilia, resulting from an absence of factor VIII. By molecular cloning techniques, Kazazian and his colleagues showed that the mutant factor VIII gene had a copy of a LINE1, or L1, repeat inserted. In contrast to most L1 repeats in the human genome, whose sequences have diverged from a predicted source gene, the sequence of this L1 was very close to that predicted for an active L1. Tests showed that the patient’s parents did not carry this mutation in their factor VIII genes. Screening a genomic library for L1s that were almost identical to the mutagenic L1 revealed a full-length, active L1 that was the source, on a different chromosome. The appearance of a new L1 in the factor VIII gene, making an allele that was not present in the parents, is a strong argument for transposition. The further studies identifying a source gene and showing that the source gene is active in transposition make the evidence unequivocal.
Now that it is recognized that most repetitive elements in many species result from transposition events, it is easy to find transposable elements or their progeny. A comprehensive database of repetitive elements in many species is maintained as RepBase (J. Jurka) and the program RepeatMasker (Green and Smit) will widely used to find matches to these repeats. Virtually all these repeats are made by transposition. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_II%3A_Replication_Maintenance_and_Alteration_of_the_Genetic_Material/9._Transposition_of_DNA/Unstable_Alleles.txt |
Recall the Central Dogma of molecular biology: DNA is transcribed into RNA, which is translated into protein. We will cover the material in that order, since that is the direction that information flows.
Introduction
The liberated pyrophosphate is cleaved in the cell to 2 Pi, an energetically favorable reaction that drives the reaction in the direction of synthesis. In the presence of excess PPi, the reverse reaction of pyrophosphorolysis can occur. Synthesis always proceeds in a 5' to 3' direction (with respect to the growing RNA chain). The template is read in a 3' to 5' direction.
B. E. coli RNA polymerase structure
1. This one RNA polymerase synthesizes all classes of RNA
mRNA, rRNA, tRNA
2. It is composed of four subunits.
a. Core and holoenzyme
a2bb's a2bb' + s
Holoenzyme = a2bb's = core + s = can initiatetranscription accurately as the proper site, as determined by the promoter
Core = a2bb' = can elongate a growing RNA chain
A promoter can be defined in two ways.
1. The sequence of DNA required for accurate, specific intiation of transcription
2. The sequence of DNA to which RNA polymerase binds to accurately initiate transcription.
b. Subunits
Subunit Size Gene Function
b' 160 kDa rpoC b' + b form the catalytic center.
b 155 kDa rpoB b' + b form the catalytic center.
a 40 kDa rpoA enzyme assembly; also binds UP sequence in the promoter
s 70 kDa (general) rpoD confers specificity for promoter; binds to -10 and -35 sites in the promoter
C. E. coli RNA polymerase mechanism
Mode of action of sfactors
The presence of the s factor causes the RNA polymerase holoenzyme to be selective in choosing the site of initiation. This is accomplished primarily through effects on the dissociation rate of RNA polymerase from DNA.
• a. Core has strong affinity for general DNA sequences. The t1/2 for dissociation of the complex of core‑DNA is about 60 min. This is useful during the elongation phase, but not during initiation.
• b. Holoenzyme has a reduced affinity for general DNA; it is decreased about 104 fold. The t1/2 for dissociation of holoenzyme from general DNA is reduced to about 1 sec.
• c. Holoenzyme has a greatly increased affinity for promoter sequences.The t1/2 for dissociation of holoenzyme from promoter sequences is of the order of hours.
Events at initiation of transcription
• a. RNA polymerase holoenzyme binds to the promoter to form a closed complex; at this stage there is no unwinding of DNA.
• b. The polymerase‑promoter complex undergoes the closed to open transition, which is a melting or unwinding of about 12 bp.
• c. The initiating nucleotides can bind to the enzyme, as directed by their complementary nucleotides in the DNA template strand, and the enzyme will catalyze formation of a phosphodiester bond between them. This polymerase‑DNA‑RNA complex is referred to as the ternary complex.
• d. During abortive initiation, the polymerase catalyzes synthesis of short transcripts about 6 or so nucleotides long and then releases them.
• e. This phase ends when the nascent RNA of ~6 nucleotides binds to a second RNA binding site on the enzyme; this second site is distinct from the catalytic center. This binding is associated with "resetting" the catalytic center so that the enzyme will now catalyze the synthesis of oligonucleotides 7-12 long.
• f. The enzyme now translocates to an new position on the template. During this process sigma leaves the complex. A conformational change in the enzyme associated with sigma leaving the complex lets the "thumb" wrap around the DNA template, locking in processivity. Thus the core enzyme catalyzes RNA synthesis during elongation, which continues until "signals" are encountered which indicate termination.
Figure 3.1.8. Events at initiation
3. Transcription cycle
d The incoming nucleotide (NTP) that will be added to the growing RNA chain binds adjacent to the 3' end of the growing RNA chain, as directed by the template, at the active site for polymerization.
e. The incoming nucleotide is linked to the growing RNA chain by nucleophilic attack of the 3' OH on the a phosphoryl of the NTP, with liberation of pyrophosphate.
f. The reaction progresses (the enzyme moves) about 50 nts per sec. This is much slower than the rate of replication (about 1000 nts per sec).
g. If the template is topologically constrained, the DNA ahead of the RNA polymerase becomes overwound (positive superhelical turns) and the DNA behind the RNA polymerase becomes underwound (negative superhelical turns).
The effect of the unwinding of the DNA template by RNA polymerase is to decrease T by 1 for every 10 bp unwound. Thus DT = -1, and since DL = 0, then DW = +1 for every 10 bp unwound. This effect of the increase in W will be exerted in the DNA ahead of the polymerase.
The effect of rewinding the DNA template by RNA polymerase is just the opposite, of course. T will increase by 1 for every 10 bp rewound. Thus DT= +1, and since DL = 0, then DW = -1 for every 10 bp rewound. This effect of the decrease in W will be exerted in the DNA behind the polymerase, since that is where the rewinding is occurring.
5. Inhibitors: useful reagents and clues to function
a. Rifamycins, e.g. rifampicin: bind the b subunit to block initiation. The drug prevents addition of the 3rd or 4th nucleotide, hence the initiation process cannot be completed.
How do we know the site of rifampicin action is the b subunit? Mutations that confer resistance to rifampicin map to the rpoBgene.
b. Streptolydigins: bind to the b subunit to inhibit chain elongation.
These effects of rifamycins and streptolydigins, and the fact that they act on the b subunit, argue that the b subunit is required for nucleotide addition to the growing chain.
c. Heparin, a polyanion, binds to the b' subunit to prevent binding to DNA in vitro
D. Eukaryotic RNA polymerases
1. Eukaryotes have 3 different RNA polymerases in their nuclei.
1. a. Each nuclear RNA polymerase is a large protein with about 8 to 14 subunits. MW is approximately 500,000 for each.
2. b. Each polymerase has a different function:
RNA polymerase
localization
synthesizes
effect of a‑amanitin
RNA polymerase I
nucleolus
pre‑rRNA
none
RNA polymerase II
nucleoplasm
pre‑mRNA
some snRNAs
inhibited by low concentrations (0.03 mg/ml)
RNA polymerase III
nucleoplasm
pre‑tRNA, other small RNAs
some snRNAs
inhibited by high concentrations (100 mg/ml)
2. Subunit structures
a. The genes and encoded proteins for the subunits of the yeast RNA polymerases have been isolated and the sequences determined, and some functional analysis has been done.
b. Some of the subunits are homologous to bacterial RNA polymerases: The largest two subunits are homologs of b and b'. The roughly 40 kDa subunit is the homolog of a.
c. Some subunits are common to all three RNA polymerases.
d. Example of yeast RNA polymerase II:
Approximate size (kDa)
subunits per polymerase
role / comment
220
1
related to b'
catalytic?
130
1
related to b
catalytic?
40
2
related to a
assembly?
35
< 1
30
2
common to all 3
27
1
common to all 3
24
< 1
20
1
common to all 3
14
2
10
1
e. The largest subunit has a carboxy‑terminal domain (CTD)with an unusual structure: tandem repeats of the sequence Tyr‑Ser‑Pro‑Thr‑Ser‑Pro‑Thr. The yeast enzyme has 26 tandem repeats and the mammalian enzyme has about 50. These can be phophorylated on Ser and Thr to give a highly charged CTD.
• RNA Pol IIa is not phosphorylated in the CTD.
• RNA Pol IIo is phosphorylated in the CTD.
4. RNA polymerases in chloroplasts (plastids) and mitochondria
• a. The RNA polymerase found in plastids is encoded on the plastid chromosome. In some species the mitochondrial RNA polymerase is encoded by the mitochondrial DNA.
• b. These organellar RNA polymerases are much more related to the bacterial RNA polymerases than to the nuclear RNA polymerases. This is a strong argument in favor of the origins of these organelles being bacterial, supporting the endo-symbiotnt model for acquisition of these organelles in eukaryotes.
• c. These RNA polymerases catalyze specific transcription of organellar genes.
E. General transcription factors for eukaryotic RNA polymerase II
It is not known if the same set of TAFs are in the TFIID for all promoters transcribed by RNA polymerase II, or if some are used only for certain types of promoters. TFIID is the only sequence‑specific general transcription factor so far characterized, and it binds in the minor groove of the DNA. It is also used at TATA‑less promoters, so the role of the sequence ‑specific binding is still under investigation.
3. Summary of general transcription factors for RNA polymerase II.
Factors for RNA polymerase II (human cells)
Factor
No. of
subunits
Molecular
mass (kDa)
Functions
Functions to
Recruit:
TFIID: TBP
1
38
Recognize core promoter (TATA)
TFIIB
TFIID: TAFs
12
15-250
Recognize core promoter (non-TATA); Positive and negative regulation
RNA Pol II?
TFIIA
2
12, 19, 35
Stabilize TBP-DNA binding; Anti-repression
TFIIB
1
35
Select start site for RNA Pol II
RNA PolII-TFIIF
RNA Pol II
12
10-220
Catalyze RNA synthesis
TFIIE
TFIIF
2
30, 74
Target RNA PolII to promoter; destabilize non-specific interactions between PolII and DNA
TFIIE
2
34, 57
Modulate TFIIH helicase, ATPase and kinase activities; Directly enhance promoter melting?
TFIIH
TFIIH
9
35-89
Helicase to melt promoter; CTD kinase; promoter clearance?
Roeder, R.G. (1996) TIBS 21: 327-335.
4. TFIIH is a multisubunit transcription factor also involved in DNA repair.
Subunits of the human factor
Gene
Molec. mass
of protein (kDa)
Function/ Structure
Proposed Role
XPB
89
helicase, tracks 3' to 5'
Unwind duplex for transcription/ Repair
XPD
80
helicase, tracks 5' to 3'
Unwind duplex, Repair
P62
62
unknown
P52
52
unknown
P44
44
Zn-finger
Binds DNA
P34
34
Zn-finger
MAT1
32
CDK assembly factor
Cyclin H
38
Cyclin partner for CDK7/MO15
CDK7/MO15
32
Protein kinase
Kinase for CTD
Table 3.1.6. RNA polymerase II holoenzyme and mediator
These studies show that RNA polymerase II can exist in several different states or complexes. One is in a very large holocomplex containing the mediator. In this state, it will accurately initiate transcription when directed by TFIID, and respond to activators (Table 3.1.6). The mediator subcomplex appears to be able to dissociate and reassociate with RNA polymerase II and GTFs. Indeed, this reassociation could be the step that was assayed in the identification of mediator. Without mediator, RNA polymerase II plus GTFs can initiate transcription at the correct place (as directed by TFIID), but they do not respond to activators. In the absence of GTFs, RNA polymerase II is capable of transcribing DNA templates, but it will not begin transcription at the correct site. Hence it is competent for elongation but not initiation.
Table 3.1.7. Expanding the functions of RNA polymerase II
Fig.3.1.17
If the holoenzyme is the primary enzyme involved in transcription initiation in eukaryotic cells, then the progressive assembly pathway observed in vitro(see section d above) may be of little relevance in vivo. Perhaps the holoenzyme will bind to promoters simply marked by binding of TBP (or TFIID) to the TATA box, in contrast to the progressive assembly model that has a more extensive, ordered assembly mechanism. In both models, TBP or TFIID binding is the initial step in assembly of the preinitiation complex. However, at this point one cannot rule out the possibility that the holoenzyme is used at some promoters, and progressive assembly occurs at others.
7. Targets for the activator proteins
The targets for transcriptional activator proteins may be some component of the initiation complex. One line of investigation is pointing to the TAFs in TFIID as well as TFIIB as targets for the activators. Thus the activators may facilitate the ordered assembly of the intiation complex by recruiting GTFs. However, the holoenzyme contains the "mediator" or SRB complex that can mediate response to activators. Thus the activators may serve to recruit the holoenzyme to the promoter. Further studies are required to establish whether one or the other is correct, or if these are separate paths to activation.
F. General transcription factors for eukaryotic RNA polymerases I and III
1. General transcription factors for RNA polymerase I
a. Core promoter covers the start site of transcription, plus an upstream control element located about 70 bp further 5'.
b. The factor UBF1 binds to a G+C rich sequence in both the upstream control element and in the core promoter.
c. A multisubunit complex called SL1 binds to the UBF1‑DNA complex, again at both the upstream and core elements.
d. One of the subuntis of SL1 is TBP ‑ the TATA‑binding protein from TFIID!
e. RNA polymerase I then binds to this complex of DNA+UBF1+SL1 to initiate transcription at the correct nucleotide and the elongate to make pre‑rRNA.
2. General transcription factors for RNA Pol III
• a. Internal control sequences are characteristic of genes transcribed by RNA Pol III (see below).
• b. TFIIIA: binds to the internal control region of genes that encode 5S RNA (type 1 internal promoter)
• c. TFIIIC: binds to internal control regions of genes for 5S RNA (alongside TFIIIA) and for tRNAs (type 2 internal promoters)
• d. TFIIIB: The binding of TFIIIC directs TFIIIB to bind to sequences (-40 to +11) that overlap the start site for transcription. One subunit of TFIIIB is TBP, even though no TATA box is required for transcription. TFIIIA and TFIIIC can now be removed without affecting the ability of RNA polymerase III to initiate transcription. Thus TFIIIA and TFIIIC are assembly factors, and TFIIIB is the initiation factor.
Figure 3.1.18.
e. RNA polymerase III binds to the complex of TFIIIB+DNA to accurately and efficiently initiated transcription.
3 Transcription factor used by all 3 RNA Pol'ases: TBP
TBP seems to play a common role in directing RNA polymerase (I, II and III) to initiate at the correct place. The multisubunit factors that contain TBP (TFIID, SL1 and TFIIIB) may serve as positioning factors for their respective polymerases.
10: Transcription: RNA polymerases
10.1 What is the role of the sigma factor in transcription, and how does it accomplish this?
10.2 Specific binding of E. coliRNA polymerase to a promoter. Which of the following statements are correct?
1. Completely envelopes the DNA duplex (both sides).
2. requires sigma factor to be part of the holoenzyme.
3. is enhanced by methylation of purine bases.
4. results in a temperature-dependent unwinding of about 10 base pairs.
10.3 (POB) RNA polymerase. How long would it take for the E. coliRNA polymerase to synthesize the primary transcript for E. colirRNAs (6500 bases), given that the rate of RNA chain growth is 50 nucleotides per second?
10.4 What is the maximum rate of initiation at a promoter, assuming that the diameter of RNA polymerase is about 204 Angstroms and the rate of RNA chain growth is 50 nucleotides per second?
10.5 Although three different eukaryotic RNA polymerases are used to transcribe nuclear genes, the enzymes and their promoters show several features in common. Are the following statements about common features of the polymerases and their mechanisms of initiation true or false?
1. All three purified polymerases need additional transcription factors for accurate initiation at promoter sequences.
2. All three polymerases catalyze the addition of a nucleotide "cap" to the 5' end of the RNA.
3. For all three polymerases, the TATA‑binding protein is a subunit of a transcription factor required for initiation (not necessarily the same factor for each polymerase).
4. All three polymerases are composed of multiple subunits.
10.6 What is common and what is distinctive to the reactions catalyzed by DNA polymerase, RNA polymerase, reverse transcriptase, and telomerase? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/10%3A_Transcription%3A_RNA_polymerases/10.E%3A_Transcription%3A_RNA_polymerases_%28Exercises%29.txt |
3. How do you label DNA at the ends?
• 5' end label: T4 polynucleotide kinase and [g 32P] ATP. The reaction is most efficient if the 5' phosphate is removed (by alkaline phosphatase) prior to the kinase treatment.
• 3' end label: Klenow DNA polymerase plus [a 32P] dNTP. The labeled dNTP is chosen to be complementary to the first position past the primer. A restriction fragment with a 5' overhang is ideal for this "fill‑in" labeling.
• Digestion with a second restriction endonuclease will frequently work to remove the label at the "other" end. One can also use electrophoretic gels that separate strands.
B. General methods for identifying the site for sequence‑specific binding proteins
1. Does a protein bind to a particular region?
a. Electrophoretic mobility shift assay (EMSA), or gel retardation assay: This assay will test for the ability of a particular sequence to form a complex with a protein. Many protein‑DNA complexes are sufficiently stable that they will remain together during electrophoresis through a (nondenaturing) polyacrylamide gel. A selected restriction fragment or synthetic duplex oligonucleotide is labeled (to make a probe) and mixed with a protein (or crude mixture of proteins). If the DNA fragment binds to the protein, the complex will migrate much slower in the gel than does the free probe; it moves with roughly the mobility of the bound protein. The presence of a slowly moving signal is indicative of a complex between the DNA probe and some protein(s). By incubating the probe and proteins in the presence of increasing amounts of competitor DNA fragments, one can test for specificity and even glean some information about the identity of the binding protein.
Figure 3.2.4. Diagram of results from an electrophoretic mobility shift assay
In this example, two proteins recognize sequences in the labeled probe, forming complexes A and B (lane 2). The proteins in complexes A and B recognize specificDNA sequences in the probe. This is shown by the competition assays in lanes 3-8. An excess of unlabeled oligonucleotide with the same sequence as the labeled probe (“self”) prevents formation of the complexes with labeled probe, whereas “nonspecific DNA” in the form of E. coli DNA does not compete effectively (compare lanes 6-7 with lanes 3-5).
This experiment also provides some information about the identity of the protein forming complex A. It recognizes an Sp1-binding site, as shown by the ability of an oligonucloetide with an Sp1-binding to compete for complex A, but not complex B (lanes 9-11). Hence the protein could be Sp1 or a relative of it.. The proteins forming complexes A and B do not recognize an Oct1-binding site (lanes 12-14).
b. Nitrocellulose binding: Free duplex DNA will not stick to a nitrocellulose membrane, but a protein‑DNA complex will bind.
2. To what sequence in the probe DNA is the protein binding?
The presence of a protein will either protect a segment of DNA from attack by a nuclease or other degradative reagent, or in some cases will enhance cleavage (e.g. to an adjacent sequence that is distorted from normal B‑form). An end‑labeled DNA fragment in complex with protein is treated with a nuclease (or other cleaving reagent), and the protected fragments are resolved on a denaturing polyacrylamide gel, and their sizes measured.
• a. Exonuclease protection assay: The protein will block the progress of an exonuclease, so the protected fragment extends from the labeled site to the edge of the protein furtherest from the labeled site. One can use a combination of a 3' to 5' exonuclease (ExoIII) and a 5' to 3' exonuclease (l exonuclease) to map both edges.
• b. DNase footprint analysis: DNase I will cut at many (but not all) phosphodiester bonds in the free DNA. The protein‑DNA complex is treated lightly with DNase I, so that on average each DNA molecule is cleaved once. The presence of a bound protein will block access of the DNase, and the bound region will be visible as a region of the gel that has no bands, i.e. that was not cleaved by the reagent.
4. DNA sequence‑affinity chromatography to purify DNA binding proteins
The specific binding sites (often 6 to 8 bp) can serve as an affinity ligand for chromatography. Multimers of the binding site are made by ligating together duplex oligonucleotides that contain the specific site. After a few crude initial steps (e.g. isolating all DNA‑binding proteins on DNA‑sepharose) the extract is applied to the affinity column. Most of the proteins do not bind, and subsequently the specifically bound proteins are eluted.
C. Promoters and the Initiation of Transcription: General Properties
1. A promoter is the DNA sequence required for correct initiation of transcription
2. Phenotype of promoter mutants
a. cis‑acting: A cis-acting regulatory element functions as a segment of DNA to affect the expression of genes on the same chromosome that it is located on. Cis-acting elements do not encode a diffusible product. The promoter is a cis-acting regulatory element.
Compare the phenotypes of mutations in the gene encoding b‑galactosidase (lacZ) versus mutations in its promoter (p).
Consider a heterozygote that is p+ lacZ‑ /p+ lacZ+ .
The phenotype is Lac+. lacZ+ complements lacZ‑ in trans. In this case, lacZ+ is dominant to lacZ-.
Consider a heterozygote that is p+ lacZ‑ /placZ+ .
The phenotype is Lac‑. p+ does not complement p‑ in trans.
p‑ operates in cis to prevent expression of lacZ+ on this chromosome. The mutant promoter is dominant over the wild-type when the mutant promoter is in cisto the wt lacZ.
Consider a heterozygote that isp+ lacZ+ /placZ‑ .
The phenotype is Lac+. lacZ+ now complements lacZ‑ in trans because it is driven by a functional promoter in cis, p+
b. Dominance in cis: the promoter “allele” that is in cisto the wild-type structural gene (lacZ) is dominant over the other promoter allele.
c. Promoter mutations affect the amount of product from the gene but do not affect the structure of the gene product.
D. Bacterial promoters
1. Bacterial promoters occur just 5' to and overlap the start site for transcription(usually)
2. Bacterial promoters are the binding site for E. coliRNA polymerase holoenzyme. The promoter covers about 70 bp from about ‑50 to about +20.
3. Consensus sequences in the E. colipromoter
a. ‑35 and ‑10 sequences
‑35 16‑19 bp ‑10+1
‑‑‑‑‑‑‑‑TTGACA‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑--TATAAT‑‑‑CAT
Recognition by Allows binary complex to convert
RNA polymerase from closed to open
holoenzyme
b. The sequences are conserved in all E. coli genes transcribed by holoenzyme with s70
4. Promoter mutants
• a. Tend to fall into or close to one of these hexanucleotides
• b. Affect the level of gene expression, not the structure of the gene product
• c. Down promoter mutations: decrease the level of transcription. Tend to make the promoter sequence less like the consensus.
• d. Up promoter mutations: increase the level of transcription. Tend to make the promoter sequence more like the consensus.
• e. Down promoter mutations in the ‑35 sequence: decrease the rate of formation of the closed complex, indicating this is the sequence needed for intial recognition by the polymerase holoenzyme.
• f. Down promoter mutations in the ‑10 sequence: decrease the rate of conversion from the closed to the open complex, again supporting the proposed role for this A+T rich hexanucleotide.
• g. The critical contact points between RNA polymerase and the promoter tend to be in or immediately upstream from the consensus ‑35 and ‑10 boxes. (See Figure 3.2.7). Thus the biochemical and genetic data all support the importance of these conserved sequences.
Figure 3.2.7. Correlation of conserved sequences, location of promoter mutants, and regions of contact with polymerase at bacterial promoters
5. Alternate s factors can control the expression of sets of genes
• a. Alternative s factors make complexes with the core polymerase to direct the new holoenzyme to a particular set of promoters that differ in sequence from the general E. coli promoter sequence. Thus the polymerase can be directed to trancribe a new set of genes. This is one way to control gene expression.
• b. Examples include s factors for heat‑shock response (s32), transcription of genes involved in chemotaxis and flagellar formation (s28), and nitrogen starvation (s54). The s factors are named by their size in kDa.
• c. Three of the E. colis factors have regions of sequence similarity (s70,s32, and s28 ) whereas s54 is a distinctly different molecule that works rather differently.
Factor
Gene
Use
‑35
Separation
‑10
s70
rpoD
General
TTGACA
16-19 bp
TATAAT
s32
rpoH
Heat shock
CCCTTGAA
13-15 bp
CCCGATNT
s28
fliA
Flagella
CTAAA
15 bp
GCCGATAA
s54
rpoN
Nitrogen starvation
CTGGNA
6 bp
TTGCA
E. Promoters for eukaryotic RNA polymerases
Promoters contain binding sites for nuclear proteins, but which of these binding sites have a function in gene expression? This requires a genetic approach for an answer.
1. Use of "surrogate genetics" to define the promoter
a. In vitro mutagenesis (deletions or point mutations)
1. Mutations of the binding sites for activator proteins lead to a decrease in the level of transcription of the gene. [Loss of function].
2. Addition of a DNA fragment containing these binding sites will activate (some) heterologous promoters. [Gain of function].
3. Sequences of the binding sites are frequently well conserved in promoters for homologous genes from related species.
4. A potential regulatory region is initially examined by constructing progressive deletions from the 5' end (with respect to the direction of transcription) and also from the 3' end. Subsequently one can make clusters of point mutations (e.g. by linker scanning mutagenesis) or individual point mutations.
b. Test in an expression assay
(1)The mutagenized promoter is linked to a reporter gene so that RNA or protein from that gene can be measured quantitatively
• (a) Gene itself ‑ measure RNA production by S1 protection, primer extension, or other assay that is specific for a particular RNA
• (b) Heterologous reporter gene: encodes an enzyme whose activity is easy to measure quantitatively. Note that these measures of expression require both transcription and translation, in contrast to measurement of RNA directly. E.g., the genes encoding:
1. b‑galactosidase: colorimetric assay, monitor the cleavage of o‑nitrophenyl‑b‑galactoside
2. chloramphenicol (Cm) acetyl transferase (CAT): measure the acetylation of Cm, ususally use [14C] Cm; this is the enzyme that confers resistance to Cm in bacteria
3. luciferase: monitor the emmission of photons resulting from the ATP‑dependent oxidation of luciferin; this is the enzyme that catalyzes light production in firefly tails
(2) The promoter‑reporter DNA constructs are introduced into an assay system that will allow the reporter to be expressed.
(a) Whole cells
microinjection into Xenopus oocytes
transfection of cell lines: introduce the DNA via electroporation or by getting the cells to take up a precipitate of DNA and Ca phosphate by pinocytosis
(b) Whole animals = transgenic animals
Introduce the DNA into the germ line of an animal, in mammals by microinjecting into a fertilized egg and placing that into a pseudopregnant female. This technology allows one to examine the effects of the mutation throughout the development of the animal.
(c) Cell‑free systems
Extracts of nuclei, or purified systems (i.e. with all the necessary components purified)
2. Promoter for RNA Pol II
a. The minimal promoter is needed for basal activity and accurate initiation.
1. Needed for assembly of the initiation complex at the correct site
2. DNA sequences
(a) TATA box
1. Initially identified as a well conserved sequence motif about 25 bp 5' to the cap site (The cap site is the usual start site for transcription)
2. The transcription factor TFIID binds to the TATA box
3. Mutations at the TATA box generates heterogeneous 5' ends of the mRNAs ‑ indicative of a loss of start site specificity
(b) Initiator
1. Sequences at the start site for transcription have consensus YANWYY (Y = C or t, W = T or A)
2. Mode of action is still under investigation. Recent data indicate that TFIID also binds to the initiator; binds to one of the TAFs (see below).
3. TATA plus initiator is the simplest minimal promoter.
b. The amount of expression is regulated via upstream elements.
1. Proteins bind to specific sequences (usually) 5' to the TATA box to regulate the efficiency of utilization of the promoter.
2. These are frequently activators, but proteins that exert negative control are also being characterized.
3. Examples of activator proteins
Sp1: binds GGGGCGGGG = GC box
Octn: binds ATTTGCAT = octamer motif
Oct1 is a general factor (ubiquitous)
Oct2 is specific for lymphoid cells
CP1, CTF = NF1, C/EBP bind to CCAAT = CCAAT box (pronounced "cat" box)
These are different families of proteins, CP1 and CTF are found in many cell types, C/EBP is found in liver and adipose tissue.
(4) These upstream control elements may be inducible (e.g. by hormones), may be cell‑type specific, or they may be present and active in virtually all cell types (i.e. ubiquitous and constitutive).
Figure 3.2.10.
3. Promoter for RNA Pol I
• a. The core promoter covers the start site of transcription, from about ‑40 to about +30. The promoter also contains an upstream control element located about 70 bp further 5', extending from ‑170 to ‑110.
• b. The factor UBF1 binds to a G+C rich sequence in both the upstream control element and in the core promoter. A multisubunit complex called SL1 binds to the UBF1‑DNA complex, again at both the upstream and core elements. One of the subuntis of SL1 is TBP.
• c. RNA polymerase I then binds to this complex of DNA+UBF1+SL1 to initiate transcription at the correct nucleotide and the elongate to make pre‑rRNA.
F. Enhancers
1. Enhancers are DNA sequences that cause an increase in the level of expression of a gene with an intact promoter. They may act to increasethe efficiency of utilization of a promoter, or they may increase the probability that a promoter is in a transcriptionally competent chromatin conformation. This will be explored further in Part Four.
2. They are operationally defined by their ability to act in either orientation and at a variety of positions and distances from a gene, i.e. act independently of orientation and position. This contrasts with promoters, that act (usually) in only one orientation and (usually) are at or close to the 5' end of the gene.
3. They consist of binding sites for specific activator proteins. Always have multiple binding sites, often for several different activator proteins.
4. Particular sets of genes can be regulated by their need for defined sets of activator proteins at their enhancers.
G. Elongation of transcription
1. RNA polymerase must be released from the initiation complex to transcribe the rest of the gene. Elongation must be highly processive, i.e. once the polymerase begins elongation, it must transcribe that template all the way to the end of the gene.
2. The factors required for initiation are not needed (and may inhibit) elongation, and they dissociate.
3. There is some indication that factors that increase the processivity of the transcription complexbind to the elongating polymerase. Examples include the following.
• NusA in bacteria
• GreA and GreB in bacteria
• TFIIS in eukaryotes, possibly many others.
a. r‑independent sites [Note: r = rho]
1. Identified in vitro
2. G+C rich hairpin followed by about 6 U's
3. Hairpin is thought to be a site at which RNA polymerase pauses, and the weak rU‑dA base pairs in the RNA‑DNA heteroduplex allow melting of the duplex and termination.
4. Some of the best examples of r-independent terminators are integral parts of the mechanism of regulation. Examples include the attenuators in the trpoperon and other amino acid biosynthetic operons. The r-independent terminators may be a specialized adaptation for regulation.
b. r‑dependent sites
1. C‑rich, G‑poor stretch
2. Requires the action of the protein r both in vitro and in vivo
3. The r-dependent terminators are used at the 3' ends of many eubacterial genes.
2. r factor
• a. Hexamer, each subunit 46 kDa
• b. RNA‑dependent ATPase
• c. Gene for r is essential for E. coli
3. Model for action of r factor
• a. r binds to protein‑free RNA and moves along it
• b. When it reaches a paused polymerase, it causes the polymerase to dissociate and unwinds the RNA‑DNA duplex, thereby terminating transcription. This last step utilizes the energy of ATP hydrolysis. The protein r serves as the ATPase.
Figure 3.2.18.
I. Termination of transcription in eukaryotes
1. Termination by RNA Pol II
• a. No clear evidence for a discrete terminator for RNA polymerase II
• b. 3' end of mRNA is generated by cleavage and polyadenylation
• c. Signal for cleavage and polyadenylation:
(1) AAUAAA, about 20 nt before the 3' end of the mRNA
(2) Other sequences 3' to cleavage site
• d. Cleavage enzyme not well characterized at this point; the U4 snRNP may play a role in cleavage. A polyA polymerase has been identified.
• e. Polyadenylation is required for termination by RNA Pol II; possibly also pausing by the RNA polymerase
2. Model for r action can explain why stopping translation can also lead to a cessation of transcription.
• a. Suppose a r‑dependent terminator of transcription is present in the first gene of an operon. Normally it does not cause transcription to stop because it is covered by ribosomes translating the mRNA, and the subsequent genes in the operon are transcribed. Recall that r requires protein‑free RNA to bind to and to move along.
• b. A nonsense mutation before the cryptic r‑dependent terminator would cause the ribosomes to dissociate, now exposing the cryptic terminator in a protein‑free stretch of RNA. The hexamer r can bind and move along the RNA, and when it encounters an RNA polymerase stalled, or paused, at the terminator, it will cause the RNA polymerase to dissociate and the RNA to be released, hence preventing transcription of the subsequent genes in the operon.
This general structure is true for almost all eukaryotic mRNAs. The cap structure is almost ubiquitous. A few examples of mRNAs without poly A at the 3' end have been found. Some of the most abundant mRNAs without poly A encode the histones. However, most mRNAs do have the 3' poly A tail.
The poly A tail at the 3' end can be used to purify mRNAs from other RNAs. Total RNA from a cell (which is about 90% rRNA and less than 10% mRNA) can be passed over an oligo(dT)-cellulose column. The poly A-containing mRNAs will bind, whereas other RNAs will elute.
11: Transcription: Promoters terminators and mRNA
Q11.1 Determining the sequences that encode the ends of mRNAs
A gene that determines eye color in salamanders, called almond, is contained within a 2000 bp KpnI fragment. After cloning the KpnI fragment in a plasmid, it was discovered that it has a BglII site 500 bp from the left KpnI site and an EcoRI site 300 bp from the right KpnI site, as shown in the map below.
bp 0 500 1000 1500 2000
| | | | |
| | | |
KpnI BglII EcoRI KpnI
In order to determine the positions that correspond to the 5' and 3' ends of the almondRNA, the EcoRI and BglII sites were labeled at the 5' or 3' end. The KpnI to BglII fragments (500 and 1500 bp) and the KpnI to EcoRI fragments (1700 and 300 bp) were isolated, hybridized to almondRNA and treated with the single‑strand specific nuclease S1. The sizes of the probe fragments protected from digestion in the RNA‑DNA duplex are shown below (in nucleotides); a 0 means that the probe was not protected by RNA.
5' end‑labeled probe 3' end‑labeled probe
protected protected
probe fragment probe fragment
KpnI‑BglII* 5000 KpnI‑BglII* 500 100
*BglII‑KpnI 1500 1300 *BglII‑KpnI 1500 0
KpnI‑EcoRI* 17000 KpnI‑EcoRI* 1700 1300
*EcoRI‑KpnI 300 100 *EcoRI‑KpnI 300 0
The asterisk denotes the end that was labeled.
1. What is the direction of transcription of the almondgene, relative to the map above?
2. What position on the map corresponds to the 5' end of the mRNA?
3. What position on the map corresponds to the 3' end of the mRNA?
Q11.2: Determining the sequences that encode the ends of mRNAs
The gene for histone H2A from armadillo can be isolated as a 1400 bp PstI fragment. The map is shown below; the armadillo PstI fragment is shown by the double dashed line, and the vector DNA is denoted by the single dashed lines. Sizes are in base pairs. The H2A gene clone was cleaved with HindIII, treated with alkaline phosphatase, and incubated with polynucleotide kinase and [32P] ATP in an appropriate buffer to introduce a radiolabel at the 5’ ends of the DNA fragments. The DNA was then extracted with phenol to remove the kinase, and then cut again with PstI. The labeled 600 bp and 800 bp PstI-HindIII fragments were separated by gel electrophoresis and isolated. The isolated fragments were denatured, hybridized to histone mRNA, and treated with nuclease S1. The S1-resistant labeled DNA fragments were identified by gel electrophoresis followed by radioautography. A 200 nucleotide protected fragment was observed when the 600 bp fragment was used in the S1 protection assay, but no protected fragment was observed when the 800 bp fragment was used.
|--------------600----------------|-------------800----------------|
PstI HindIII PstI
---------|====================|===================|-----------
| | | |
0 500 1000 1400
1. What is the direction of transcription of the histone H2A gene (relative to the restriction map above)?
2. With reference to the numbers below the restriction map, what is the position of the 5' end of the histone H2A mRNA?
3. What is the position of the 3' end of the mRNA?
Q11.3
A 400 bp DNA fragment containing the start site for transcription of the almond gene was investigated to find transcriptional control signals. The start site (+1 in the coordinate system) is 100 bp from the right end. The 400 bp fragment is sufficient to drive transcription of a reporter gene (for luciferase) in an appropriate cell line. Two series of 5' and 3' deletions were made in the 400 bp fragment and tested for their ability to drive transcription of the luciferase reporter gene. Each fragment in the 5' deletion series has a different 5' end, but all are fused to the luciferase gene at +100 (see diagram below). Each fragment in the 3' deletion series has a common 5' end at ‑300, but each is fused to the luciferase gene at the designated 3' position. The amount of luciferase (a measure of the level of transcription) for each construct is shown in the first two pairs of columns in the table. The intact reporter construct, with almondDNA (the horizontal line) fused to the luciferase gene, is diagrammed immediately below.
‑300‑250 ‑200‑150 ‑100‑50 +1+50 +100
| | | | | | | | |
________________________________________________|Luciferase>
|
start>
To further investigate the function of different regions, sub‑fragments of the almondDNA fragment were added to a construct in which the reporter gene was driven by a different promoter, as diagrammed below. The effects of the almond DNA fragments on this heterologous promoter are shown in the third pair of columns in the table.
Test fragment from almondDNA heterologous promoter Luciferase gene>
___________________________|‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑‑----‑|**************>
____________________ _____________________ _______________________
5' deletion endpoints Amount of expression 3' deletion endpoints Amount of expression Test fragment of almond Amount of expression
‑300 100 ‑200 0 ‑300 to ‑250 100
‑250 100 ‑150 0 ‑250 to ‑200 500
‑200 50 ‑100 0 ‑200 to ‑150 100
‑150 50 ‑50 0 ‑150 to ‑100 300
‑100 25 +1 100 ‑100 to ‑50 300
‑50 10 +50 100 ‑50 to ‑1 100
‑1 0 +100 100 none 100
1. What do you conclude is the role of the ‑250 to ‑200 fragment?
2. What do you conclude is the role of the ‑200 to ‑150 fragment?
3. What do you conclude is the role of the ‑150 to ‑100 fragment?
4. What is the role of the ‑50 to ‑1 fragment of the almondgene?
Q11.4
An electrophoretic mobility shift assay was used to test for the ability of a short restriction fragment to bind to proteins from the nuclei of kidney cells. The restriction fragment was labeled at one end, mixed with an extract containing the nuclear proteins, and run on a non-denaturing polyacrylamide gel. Lane 1 (below) shows the free probe and lane 2 shows the the probe plus extract; electrophoresis is from the top to the bottom. Complexes between proteins and the labeled DNA probe move more slowly on the gel than does the free probe. Further tests of specificity are shown in the competition lanes, in which the labeled probe was mixed with an increasing excess of other DNA before mixing with the nuclear proteins to test for binding. Competitor DNAs included the unlabeled probe (self competition, lanes 3-5; the triangle above the lanes indicates that an increasing amount of competitor is used in successive lanes), a completely different DNA (sheared E. coli DNA) as a nonspecific competitor (lanes 6-8), and two different duplex oligonucleotides, one containing the binding site for Sp1 (lanes 9-11) and the other containing the binding site for Oct1 (lanes 12-14). Thinner, less densely filled boxes denote bands of less intensity than the darker, thicker bands.
1. How many protein-DNA complexes are formed between the labeled DNA probe and the nuclear extract?
2. What do lanes 3-8 tell you about the protein-DNA complexes?
3. What do lanes 9-14 tell you about the protein-DNA complexes?
Q11.5
In order to determine the contact points between a regulatory protein and its binding site on the DNA, a small fragment of duplex DNA was end‑labeled (at the 5' terminus of the left end as written below) and treated with dimethyl sulfate so that each molecule on average has one G nucleotide methylated. The regulatory protein was mixed with the preparation of partially methylated DNA, and protein‑bound DNA was separated from unbound DNA. After cleaving the DNA at the methylated sites, the resultant fragments were resolved on a "sequencing gel". An autoradiogram of the results showed bands corresponding to all the G's in the labeled fragment for the unbound DNA, but the protein‑bound DNA did not have bands corresponding to the G's at positions 14 and 16 below. When the left end of the fragment was labeled at the 3' terminus, no band corresponding to the G (bottom strand) at position 18 (same numbering system as for top strand) was seen in the preparation of protein‑bound DNA.
5 10 15 20 25 30
| | | | | |
5' GATCCGCATGGATGAGTCACGTAACGTGTA
3' GCGTACCTACTCAGTGCATTGCACAT
What is the binding site for the regulatory protein?
Q11.6
Are the following statements about r and polar effects of some mutations in operons in E. colitrue or false?
1. Nonsense mutations (terminating translation) in the first gene of an operon can have no effect on the transcription of subsequent gene in the operon.
2. Mutations in the gene for r (rhogene) can suppress polarity.
3. The hexameric protein r binds to protein‑free RNA and moves along the RNA; when it encounters a stalled RNA polymerase it promoters termination of transcription.
4. The protein r is an RNA‑dependent ATPase. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/11%3A_Transcription%3A_Promoters_terminators_and_mRNA/11.E%3A_Transcription%3A_Promoters_terminators_and_mRNA_%28Exercises%29.txt |
RNA processing refers to any covalent modification to the RNA that occurs after transcription. This includes specific cleavage, addition of nucleotides, methylation or other modification of the nucleotides, and removal of introns by splicing.
12: RNA processing
RNA processing refers to any covalent modification to the RNA that occurs after transcription. This includes specific cleavage, addition of nucleotides, methylation or other modification of the nucleotides, and removal of introns by splicing.
Overview
RNA Precursor Modification Addition Cleavage Splicing
mRNA pre‑mRNA (hnRNA) methylation on 2'‑OH of ribose 5' cap 3' poly A cut at site for poly A; excise viral mRNA remove introns
rRNA pre‑rRNA methylation on 2'OH of ribose no excise products fr. precursor remove introns
tRNA pre‑tRNA extensive and varied CCA to 3' end yes remove introns
snRNAs ? ? 5' cap ? ?
12.1: Cutting and Trimming RNA
pre‑rRNA
In E. coli, the rrn operon is transcribed into a 30S precursor RNA, containing 3 rRNAs and 2 tRNAs.
The segment containing 16S rRNA (small ribosomal subunit) and the one containing 23S rRNA (large ribosomal subunit) are flanked by inverted repeats that form stem structure in the RNA. The stems are cleaved by RNase III. There is no apparent single sequence at which RNase III cleaves ‑ perhaps it recognizes a particular stem structure. This plus subsequent cleavage events (by an activity called M16) generates the mature 16S and 23S rRNAs. The rRNAs are also methylated.
tRNA is liberated by RNases P and F and 5S rRNA is liberated by RNases E and M5
In eukaryotes
The initial precursor is 47S and contains ETS1, 18S rRNA, ITS1, 5.8S rRNA, ITS2, and 28S rRNA, where ETS = extragenic transcribed spacer and ITS = intragenic transcribed spacer. Specific cleavage events followed by methylations generate the mature products. Also, some rRNA genes in some species have introns that must be spliced out.
pre-tRNA in E. coli
Sequence specific cleavage by RNases P, F, D
1. RNase P is an endonuclease that cleaves the precursor to generate the 5' end of the mature tRNA.
2. RNase F is an endonuclease that cleaves the precursor 3 nucleotides past the 3' end of the mature tRNA.
3. RNase D is an exonuclease that trims in a 3' to 5' direction to generate the 3' end or the mature tRNA.
The catalytic activity of RNase P is in the RNA component
1. RNAse P is composed of a 375 nt RNA and a 20 kDa protein.
2. The catalytic activity is in the RNA. The protein is thought to aid in the reaction, but is not required for catalysis. All enzymes are not proteins!
3. This was one of the first instances discovered of catalytic RNA, and Sidney Altman shared the Nobel Prize for this.
The enzyme tRNA nucleotidyl transferaseadds CCA to the 3' ends of pre-tRNAs.
1. Virtually all tRNAs end in CCA, forms the amino acceptor stem.
2. For most prokaryotic tRNA genes, the CCA is encoded at the 3' end of the gene.
3. No known eukaryotic tRNA gene encodes the CCA, but rather it is added posttranscriptionally by the enzyme tRNA nucleotidyl transferase. This enzyme is present in a wide variety of organisms, including bacteria, in the latter case presumably to add CCA to damaged tRNAs. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/12%3A_RNA_processing/12.0%3A_Overview_of_RNA_Processing.txt |
As discussed previously, eukaryotic mRNAs are capped at their 5' end and polyadenylated at their 3' end. In vitro assays for these reactions have been developed, and several of the enzymatic activities have been identified. These will be reviewed in this section. Polyadenylation is not limited to eukaryotes. Several mRNAs in E. coli are polyadenylated as well. This is a fairly new area of study.
Modification at the 5' end: Cap Structure
The "cap" is a methylated 5'‑GMP that is linked via its 5' phosphate to the b‑phosphoryl of the initiating nucleotide (usually A); see Figure 3.3.6. Capping occurs shortly after transcription has begun. It occurs in a series of enzymatic steps (Figure 3.3.7):
1. Remove the g‑phosphoryl of the initiating nucleotide (RNA triphosphatase)
2. Link a GMP to the b‑phosphoryl of the initiating nucleotide (mRNA guanylyl transferase). The GMP is derived from GTP, and is linked by its 5' phosphate to the 5' diphosphate of the initiating nucleotide. Pyrophosphate is released.
3. The N‑7 of the cap GMP is methylated (methyl transferase), donor is S‑adenosyl methionine.
4. Subsequent methylations occur on the 2' OH of the first two nucleotides of the mRNA.
Capping has been implicated in having a role in efficiency of translation and in mRNA stability.
Several proteins are required for cleavage and polyadenylation at the 3' end:
1. CPSF is a tetrameric specificity factor; it recognizes and binds to the AAUAAA polyadenylation signal.
2. CFI and CFII are cleavage factors.
3. PAP is the polyA polymerase.
4. CFI, CFII and PAP form a complex that binds to the nascent RNA at the cleavage site, directed by the CPSF specificity factor.
5. CstF is an additional protein implicated in this process in vitro, but its precise function is currently unknown.
12.3: Multiple Mechanisms are Used for Splicing Different Types of Introns
Different Types of Introns
At least four distinct classes of introns have been identified: Introns in nuclear protein-coding genes that are removed by spliceosomes (spliceosomal introns) Introns in nuclear and archaeal transfer RNA genes that are removed by proteins (tRNA introns) Self-splicing group I introns that are removed by RNA catalysis.
1. pre‑tRNA
2. group I, group II: Introns in fungal mitochondrial genes and in plastid (chloroplast) genes have been grouped into two different groups based on different consensus sequences found in the introns. As we will see below, the group II introns have a mechanism for splicing that is similar to that of pre‑mRNA.
3. pre‑mRNA
In all cases, splicing will remove the introns and join the exons to give the mature RNA.
Table. Features of splicing for different types of introns
Class Distribution Sequence Distinguishing feature Mechanism
pre-tRNA yeast to mammals very short (10-20 nucleotides) requires ATP cut, kinase, ligase
group I fungal mitochondria, plastids, pre-rRNA in Tetrahymena characteristic consensus self-splicing, G nucleot(s)ide to initiate phosphoester transfer
group II fungal mitochondria, plastids characteristic consensus can self-splice, internal A nucleotide to initiate phosphoester transfer
pre-mRNA yeast to mammals 5' GU...AG 3' spliceosome (ATP for assembly), internal A nucleotide to initiate phosphoester transfer
Splicing of pre‑tRNAs
Some precursor tRNAs contain short introns (only 10 to 20 nucleotides) with no apparent consensus sequences. These short introns are removed in a series of steps catalyzed by enzymes that include an endonuclease, a kinase and a ligase. Because the endonuclease generates a 2’, 3’ cyclic phosphodiester product, an additional phosphodiesterase is needed to open the cyclic phosphodiester to provide the 3’ hydroxyl for the ligase reaction. In addition, the 2’-phosphate (product of the phosphodiesterase) must be removed by a phosphatase. This process uses two ATPs for every splicing event. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/12%3A_RNA_processing/12.2%3A_Modifications_at_the_5%27_and_3%27_ends_of_mRNA.txt |
An in vitro reaction was established to examine the removal of an intron from the precursor to rRNA in Tetrahymena. Suprisingly, it was discovered that the splicing of the pre-RNA occurred in the absence of any added protein!
Further investigation revealed that the reaction requires a guanine nucleotide or nucleoside with a 3'‑OH, plus mono‑ and divalent cations. GTP, GDP, GMP or guanosine will work to initiate splicing. There is no requirement for protein or high energy bond cleavage
Self‑splicing occurs by a phosphoester transfer mechanism (Figure 3.3.11)
The 3'‑OH of the guanine nucleotide is the nucleophile that attacks and joins to the 5' phosphate of the first nucleotide of the intron. This leaves the 3'‑OH of the last nucleotide of the upstream exon available to attack and join the 5' phosphate of the first nucleotide of the downstream exon. These two phosphoester transfers result in a joining of the two exons and excision of the intron (with the initiating G nucleotide attached to the 5' end.) The excised intron is then circularized by attack of the 3'‑OH of the last nucleotide of the intron on the phosphate between the 15th and 16th nucleotides of the introns. Further degradation effectively removes the intron from the reaction and helps prevent the reverse reaction from occurring. Note that the phosphoester transfers are readily reversible unless the products (excised intron) are removed. There is no increase or decrease in the number of phosphoester bonds during this splicing.
The Intron is the Catalyst for Splicing in this System
RNA involvement in self‑splicing is stoichiometric, but the excised intron does have a catalytic activity in vitro. After a series of intramolecular cyclization and cleavage reactions, the linear excised intron lacking 19 nucleotides (called L-19 IVS) can be used catalytically to add and remove nucleotides to an artificial substrate. For instance, C5, which is complementary to the internal guide sequences of the intron, can be converted to C4 + C6 and other products (Figure 3.3.12).
The 3‑D structure of the folded RNA is responsible for the specificity and efficiency of the reaction (analogous to the general ideas about proteins with enzymatic activity). The specificity of splicing is caused, at least in part, by base‑pairing between the 3' end of the upstream exon and a region in the intron called the internal guide sequence. The initiating G nt also binds to a specific site in the RNA close to the 5' splice site. Thus two sites in the pre-rRNA intron are used sequentially in splicing (Figure 3.3.13 A and 3.3.13.B.).
Figure 3.3.13.A.
The internal guide sequence (IGS) is not not required for catalysis but does confer specificity. Thus one can design RNAs for exon exchange in cells. This potential avenue for therapy for genetic disorders is called "exon replacement therapy.
12.5: RNAs Can Function as Enzymes
Examples include the following:
• RNase P
• Group I introns (includes intron of pre‑rRNA in Tetrahymena)
• Group II introns
• RNA: peptide bond formation
Hammerhead ribozymes
Viroids and virusoids have a self-cleaving activity that localized to a 58 nucleotide structure illustrated in Figure 3.3.15. The mechanism differs in some respects from the phosphoester transfer. A divalent metal hydroxide binds in the active site, and abstracts a proton from the 2' OH of the nucleotide at the cleavage site. This now serves as a nucleophile to attack the 3' phosphate and cleave the phosphodiester bond, generating a 2',3' cyclic phosphate and a 5' OH on the ends of the cleaved RNA.
One application currently being explored is the use of designed hammerheads to cleave a particular mRNA, thereby turning off expression of a particular gene. If over-expression or ectopic expression of a defined gene were the cause of some pathology (e.g. some form of cancer), then reducing its expression could have therapeutic value.
Other RNAs possibly involved in catalysis, such as the snRNAs involved in splicing pre-mRNA.
Even though RNAs can be sufficient for catalysis, sometimes they are assisted by proteins to improve efficiency. For instance, group I introns are capable of splicing introns by themselves in a cell-free reaction. However, some are not very efficient in this process, and in the cell they are assisted by proteins that themselves are not catalytic but they enhance the reaction. Examples are maturases, which are proteins that assist in the splicing of some group I introns found in yeast mitochondria. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/12%3A_RNA_processing/12.4%3A_Selfsplicing_by_group_I_introns_%28prerRNA_of_Tetrahymena%29.txt |
1. Splice Sites
The sequence at the 5' and 3' ends of introns in pre-mRNAs is very highly conserved. Thus one can derive a consensus sequence for splice junctions.
5' exon...AG'GURAGU.................YYYYYYYYYYNCAG'G....exon
The GU is the 5' splice site (sometimes called the donor splice site) and the AG is the 3' splice site (or acceptor splice site). GU is invariant at the 5' splice site, and AG is (almost) invariant at the 3' splice site for the most prevalent class of introns in pre-mRNA.
Effects of mutations at the splice junctions demonstrate their importance in the splicing mechanism. Mutation of the GT at the donor site in DNA to an AT prevents splicing (this was seen in a mutation of the b‑globin gene that caused b0 thalassemia.) A different mutation of the b‑globin gene that generated a new splice site caused an aberrant RNA to be made, resulting in low levels of b‑globin being produced (b+ thalassemia).
2. The intron is excised as a lariat
The 2'‑OH of an A at the "branch" point forms a phosphoester with the first G of the intron to initiate splicing. Splicing occurs by a series of phosphoester transfers (also called trans‑esterifications). After the 2'-OH of the A at the branch has joined to the initial G of the intron, the 3'‑OH of the upstream exon is available to react with the first nucleotide of the downstream exon, thereby joining the two exons via the phosphoester transfer mechanism.
c. Intron lariat is the equivalent of a "circular" intermediate.
The sequence at the branch point is only moderately conserved in most species; examination of many branch points gives the consensus YNYYRAG. It lies 18 to 40 nucleotides upstream of the 3' splice site.
3. Small nuclear ribonucleoproteins (or snRNPs) form the functional splicesome on pre‑mRNA and catalyze splicing.
a. "U" RNAs and associated proteins. Small nuclear RNAs (snRNAs) are about 100 to 300 nts long and can be as abundant as 105 to 106 molecules per cell. They are named U followed by an integer. The major ones involved in splicing are U1, U2, U4/U6, and U5 snRNAs. They are conserved from yeast to human. The snRNAs are associated with proteins to form small nuclear ribonucleoprotein particles, or snRNPs. The snRNPs are named for the snRNAs they contain, hence the major ones involved in splicing are U1, U2, U4/U6, U5 snRNPs.
One class of proteins common to many snRNPs are the Sm proteins. There are 7 Sm proteins, called B/B’, D1, D2, D3, E, F, G. Each Sm protein has similar 3-D structure, consisting of an alpha helix followed by 5 beta strands. The Sm proteins interact via the beta strands, and may form circle around RNA.
A particular sequence common to many snRNAs is recognized by the Sm proteins, and is called the “Sm RNA motif”.
b. Use of antibodies from patients with SLE. Several of the common snRNPs are recognized by the autoimmune serum called anti‑Sm, initially generated by patients with the autoimmune disease Systemic Lupus Erythematosis. One of the critical early experiments showing the importance of snRNPs in splicing was the demonstration that anti-Sm antisera is a potent inhibitor of in vitrosplicing reactions. Thus the targets of the antisera, i.e. Sm proteins in snRNPs, are needed for splicing.
c. The snRNPs assemble onto the pre-mRNA to make a large protein-RNA complex called a spliceosome (Figure 3.3.17). Catalysis of splicing occurs within the spliceosome. Recent studies support the hypothesis that the snRNA components of the spliceosome actually catalyze splicing, providing another example of ribozymes.
d. U1 snRNP: Binds to the 5' splice site, and U1 RNA forms a base‑paired structure with the 5' splice site.
e. U2 snRNP: Binds to the branch point and forms a short RNA-RNA duplex. This step requires an auxiliary factor (U2AF) and ATP hydrolysis, and commits the pre-mRNA to the splicing pathway.
f. U5 snRNP plus the U4, U6 snRNP now bind to assemble the functional spliceosome. Evidence indicates that U4 snRNP dissociates from the U6 snRNP in the spliceosome. This then allows U6 RNA to form new base-paired structures with the U2 RNA and the pre-mRNA that catalyze the transesterification reaction (phosphoester transfers). One model is that U6 RNA pairs with the 5' splice site and with U2 RNA (which itself is paired to the branch point), thus bringing the branch point A close to the 5' splice site. U5 RNA may serve to hold close together the ends of the exons to be joined.
4. Trans‑splicing
All of the splicing we have discussed so far is between exons on the same RNA molecule, but in some cases exons can be spliced to other RNAs. This is very common in trypanosomes, in which a spliced leader sequence is found at the 5' ends of almost all mRNAs. A few examples of transsplicing have been described in mammalian cells. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/12%3A_RNA_processing/12.6%3A_Splicing_of_introns_in_premRNAs.txt |
1. Similar mechanism as that for nuclear pre‑mRNA splicing.
2. Can occur by self‑splicing, albeit under rather artificial conditions.
3. Reaction can be reversible (as can splicing of group I introns), leading to the idea that these introns can be transposable elements.
4. The group II self‑splicing may be the evolutionary ancestor to nuclear pre‑mRNA splicing
I. Mechanistic similarties for splicing group I, group II and pre‑mRNA introns
1. All involve transesterification = phosphoester transfers. No high energy bonds are utilized in the splicing process; the arrangement of phosphodiester bonds is reorganized, and as a result exons are joined together.
2. The initiating nucleophile is the 3' OH of a guanine nucleotide for Group I introns, whereas for Group II introns and introns in pre‑mRNA, it is the 2' OH of an internal adenine nucleotide in the intron.
3. In all cases, particular secondary structures in the RNAs are utilized to bring together the reactive components (e.g. ends of exons and introns). These secondary structures may be intramolecular in the case of self‑splicing Group I and Group II introns, or they may be intermolecular in the case of pre‑mRNA and the snRNAs, e.g. those in the U1, U2, perhaps U6 snRNPs.
12.8: Alternative Splicing
For many genes, all the introns in the mRNA are spliced out in a unique manner, resulting in one mRNA per gene. But there is a growing number of examples of other genes in which certain exons are included or excluded from the final mature mRNA, a process called alternative splicing. Some exons may be included in some tissues and not others, or may be sex‑specific, indicating some regulation over the selection of splice sites.
Alternative splicing of pre‑mRNA means that a single gene may encode more than one protein product (e.g., sex determination in Drosophila melanogaster).
Drosophila melanogaster
The X to autosome ratio (X:A ratio) in the zygote will determine which of two different developmental pathways along which the fly will develop. If the X:A ratio is high (e.g. the female is XX and the X:A ratio is 1.0), the fly will utilize the female pathway; if the ratio is low (e.g. 0.5 since the male is XY), it will develop as a male.
The X:A ratio is determined by "counting" certain genes (or their expression) on the X chromosome (e.g. sisterless a, sisterless b, and runt) for the numerator and counting other genes (such as deadpan) for the denominator. All of the products of these genes are homologous to various calsses of transcription factors, consistent with at least part of the regulation of sex determination being transcriptional. However, as discussed below, alternative splicing plays a key role as well, at least in Drosophila.
The pathways have at least four steps that were defined genetically by mutations that caused, e.g. genetically female flies (high X:A) to develop as males. In each case, the same gene encodes both male and female‑specific mRNAs (and proteins), but the sex‑specific mRNAs (and proteins) differ as a result of alternative splicing.
In all cases, the default state is male development, and some new activity has to be present to establish and maintain the female pathway.
1. The target of the X:A signal is the Sex-lethalgene (Sxl), which serves as a master switch gene. In early development, an X:A ratio of 1 in females leads to the activiation of an embryo-specific promoter of the Sxlgene, whereas Sxlis not transcribed in male embryos. Later in development, Sxlis transcribed in both sexes. Now the high X:A ratio leads to the skipping of an exon in the splicing of pre‑mRNA from the Sex‑lethalgene. This produces a functional Sxl protein in females. In males (default pathway), the mRNA has an early termination codon, and no functional Sxl protein is made.
1. A functional Sxl protein inhibits the default splicing of pre‑mRNA from the transformergene, to generate a functional Tra protein in female embryos. In the female‑specific splicing of trapre‑RNA, a 5' splice site (common to both male and female splicing) is connected to an alternative 3' splice site, thereby removing a termination codon and allowing function Tra protein to be made (Figure 3.3.15).
2. The Tra protein promotes female‑specific splicing of pre‑mRNA from the tra2gene, again generating a functional Tra2 protein only in females.
3. Tra and Tra2 proteins promote female‑specific splicing of pre‑mRNA from the doublesexgene (dsx). In this case, the male‑specific mRNA has skipped an exon (Figure 3.3.15). Skipping an exon requires an alteration in the splicing pattern at both the 3' splice site and the 5' splice sites surrounding the exon.
4. The male‑specific Dsx protein blocks female differentiation and leads to male development. The female‑specific Dsx protein represses expression of male genes and leads to female development.
Some clues about mechanism
1. Tra and Tra2 are RNA‑binding proteins related to Splicing Factor 2 (SF2). This latter protein has a domain rich in the dipeptide Arg‑Ser, which defines one type of RNA‑binding domain. SF2 is required for early steps in spliceosome assembly. The related Tra and Tra2 proteins are not required for viability, but they do regulate the specific splicing events for pre‑mRNA from dsx.
2. Tra2 binds in the female‑specific exon of the dsx transcript, and presumably regulates splice site selection. The binding site for Tra2 within the exon is an example of a splicing enhancer. The mechanisms by which the binding of splicing regulatory proteins (e.g. Tra, Tra2) to splicing enhancers is a very active area of research currently.
3. Sxl is another RNA‑binding protein that inhibits the default splicing pattern for tra pre‑mRNA.
Figure 3.3.20. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/12%3A_RNA_processing/12.7%3A_Splicing_of_group_II_introns.txt |
RNA editing refers to changing the sequence of RNA after transcription, either by adding nucleotides, taking them away, or substituting one for another. The extent of editing is dramatic in some mRNAs, e.g. in the mitochondria of trypansomes and Leishmania. For example, for some mRNAs 55% of the nucleotide sequence is added after transcription! In many of the cases characterized so far, a small number of U's are inserted at many places in the mRNA. Other examples of excising U's and adding C's are known for other mitochondrial genes from other organisms.
In at least some cases, the additional nucleotides are added under the direction of guide RNAs that are encoded elsewhere in the mitochondrial genome. A portion of the guide RNA is complementary to the mRNA in the vicinity of the position at which nucleotides will be added (Figure 3.3.16). The U at the 3' end of the guide RNA initiates a series of phosphoester transfer reactions to insert itself into the mRNA (see bottom of Figure 3.3.16). More U's at the 3' end of the guide RNA can be added, one at a time. Note the similarity in mechanism between these insertions of nucleotides (editing) and the self‑splicing of Group I intron.
For a situation in which one segment of DNA encodes the unedited mRNA and two other segments of DNA encode the guide RNAs required for editing, the "gene" is encoded in three portions, mutations in which would complement in trans! This is a counter‑example to one of our most powerful definitions of a gene.
In mammals, two different forms of apolipoprotein B are made, one in the liver and one in the intestine. The intestinal form is much shorter because of an earlier termination codon. Surprisingly, only one gene is found and it must encode both from of ApoB. A specifc enzyme must change one nucleotide of the mRNA for apolipoprotein B (a C in codon 2153, CAA) post‑transcriptionally from a C to a U to generate the termination codon (UAA) found in the intestinal form.
This enzymatic activity is present in a protein with no apparent RNA component, and hence no obvious guide RNA. Thus it appears to operate by a distinctly different mechanism from the editing in protist mitochondria (see. e.g. Greeve, J. et al., 1991, Nucleic Acids Research 19: 3569-3576).
12.E: RNA Processing (Exercises)
12.1 Nucleoside triphosphates labeled with [32P] at the a, b, or g position are useful for monitoring various aspects of transcription. For the specific process listed in a-c, give the position of the label that is appropriate for examining that step.
a) Initiation by E. coliRNA polymerase.
b) Forming the 5' end of eukaryotic mRNA.
c) Elongation by eukaryotic RNA polymerase II.
12.2 (POB) RNA posttranscriptional processing.
Predict the likely effects of a mutation in the sequence (5')AAUAAA in a eukaryotic mRNA transcript.
12.3 A phosphoester transfer mechanism (or transesterification) is observed frequently in splicing and other reactions involving RNA. Are the following statements about these mechanisms true or false?
a) The mechanism requires the cleavage of high-energy bonds from ATP.
b) The initiating nucleophile for splicing of Group I introns (including the intron of pre-rRNA from Tetrahymena) is the 3' hydroxyl of a guanine nucleotide.
c) The initiating nucleophile for splicing of nuclear pre-mRNA is the 2' hydroxyl of an internal adenine nucleotide.
d) The individual reactions in the phosphoester transfers are reversible, but the overall process is essentially irreversible because of circularization (includes lariat formation) of the excised intron.
12.4 What properties are shared by the splicing mechanism of Tetrahymena pre-rRNA and Group II fungal mitochrondrial introns?
12.5 Please answer these questions on splicing of precursors to mRNA.
a) What dinucleotides are almost invariably found at the 5’ and 3’ splice sites of introns?
b) Which splicing component binds at the 5' splice junction?
c) What nucleotides are joined by the branch structure in the intron during splicing?
d) What is ATP used for during splicing of precursors to mRNA?
12.6 (POB) RNA splicing.
What is the minimum number of transesterification reactions needed to splice an intron from an mRNA transcript? Why?
12.7 Match the following statements with the appropriate eukaryotic splicing process listed in parts a-c.
1) A guanine nucleoside or nucleotide initiates a concerted phosphotransfer reaction.
2) The consensus sequences at splice junctions are AG'GUAAGU...YYYAG'G (' is the junction, Y = any pyrimidine).
3) Splicing occurs in two separate steps, cutting to generate a 3'-phosphate followed by an ATP dependent ligation.
4) Splicing requires no protein factors.
5) Splicing requires U1 small nuclear ribonucleoprotein complexes.
a) Splicing of pre-mRNA.
b) Splicing of pre-tRNA in yeast
c) Splicing of pre-rRNA in Tetrahymena
12.8 The enzyme RNase H will cleave any RNA that is in a heteroduplex with DNA. Thus one can cleave a single-stranded RNA in any specific location by first annealing a short oligodeoxyribonucleotide that is complementary to that location and then treating with RNase H.
This approach is useful in determining the structure of splicing intermediates. Let's consider a hypothetical case shown in the figure below. After incubation of radiolabeled precursor RNA (exon1-intron-exon2) with a nuclear extract that is capable of carrying out splicing, the products were analyzed on a denaturing polyacrylamide gel. The results showed that the exons were joined as linear RNA, but the excised intron moved much slower than a linear RNA of the same size, indicative of some non-linear structure. The excised intron was annealed to a short oligodeoxyribonucleotide that is complementary to the region at the 5' splice site (labeled oligo 1 in the figure), treated with RNase H and analyzed on a denaturing polyacrylamide gel. The product ran as a linear RNA with the size of the excised intron (less the length of the RNase H cleavage site). As summarized in the figure, the excised intron was analyzed by annealing (separately) with three other oligodeoxyribonucleotides, followed by RNase H treatment and gel electrophoresis. Use of oligodeoxyribonucleotide number 2 generated a Y-shaped molecule, use of oligodeoxyribonucleotide number 3 generated a V-shaped molecule with one 5' end and 2 3' ends, and use of oligodeoxyribonucleotide number 4 generated a circle and a short linear RNA.
(a) What does the result with oligodeoxyribonucleotide 2 tell you?
(b) What does the result with oligodeoxyribonucleotide 4 tell you?
(c) What does the result with oligodeoxyribonucleotide 1 tell you?
(d) What does the result with oligodeoxyribonucleotide 3 tell you?
(e) What is the structure of the excised intron? Show the locations of the complementary oligos on your drawing. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/12%3A_RNA_processing/12.9%3A_RNA_editing.txt |
Once transcription and processing of rRNAs, tRNAs and snRNAs are completed, the RNAs are ready to be used in the cell ‑ assembled into ribosomes or snRNPs and used in splicing and protein synthesis. But the mature mRNA is not yet functional to the cell. It must be translated into the encoded protein. The rules for translating from the "language" of nucleic acids to that of proteins is the genetic code.
Introduction
Experiments testing the effects of frameshift mutations showed that the deletion or addition of 1 or 2 nucleotides caused a loss of function, whereas deletion or addition of 3 nucleotides allowed retention of considerable function. This demonstrated that the coding unit is 3 nucleotides. The nucleotide triplet that encodes an amino acid is called a codon. Each group of three nucleotides encodes one amino acid. Since there are 64 combinations of 4 nucleotides taken three at a time and only 20 amino acids, the code is degenerate (more than one codon per amino acid, in most cases). The adaptor molecule for translation is tRNA. A charged tRNA has an amino acid at one end, and at the other end it has an anticodon for matching a codon in the mRNA; ie. it "speaks the language" of nucleic acids at one end and the "language" of proteins at the other end. The machinery for synthesizing proteins under the direction of template mRNA is the ribosome.
Figure 3.4.1. tRNAs serve as an adaptor for translating from nucleic acid to protein
A. Size of a codon: 3 nucleotides
1. Three is the minimum number of nucleotides per codon needed to encode 20 amino acids.
a. 20 amino acids are encoded by combinations of 4 nucleotides
b. If a codon were two nucleotides, the set of all combinations could encode only
4x4 = 16 amino acids.
c. With three nucleotides, the set of all combinations can encode
4x4x4 = 64 amino acids (i.e. 64 different combinations of four nucleotides taken three at a time).
2. Results of combinations of frameshift mutations show that the code is in triplets. Length‑altering mutations that add or delete one or two nucleotides have severe defective phenotype (they change the reading frame, so the entire amino acid sequence after the mutation is altered.). But those that add or delete three nucleotides have little or no effect. In the latter case, the reading frame is maintained, with an insertion or deletion of an amino acid at one site. Combinations of three different single nucleotide deletions (or insertions), each of which has a loss-of-function phenotype individually, can restore substantial function to a gene. The wild-type reading frame is restored after the 3rd deletion (or insertion).
B. Experiments to decipher the code
1. Several different cell‑free systems have been developed that catalyze protein synthesis. This ability to carry out translation in vitro was one of the technical advances needed to allow investigators to determine the genetic code.
a. Mammalian (rabbit) reticulocytes: ribosomes actively making lots of globin.
b. Wheat germ extracts
c. Bacterial extracts
2. The ability to synthesize random polynucleotides was another key development to allow the experiments to decipher the code. S. Ochoa isolated the enzyme polynucleotide phosphorylase, and showed that it was capable of linking nucleoside diphosphates (NDPs) into polymers of NMPs (RNA) in a reversible reaction.
nNDP n + nPi
The physiological function of polynucleotide phosphorylase is to catalyze the reverse reaction, which is used in RNA degradation. However, in a cell-free system, the forward reaction is very useful for making random RNA polymers.
3. Homopolymers program synthesis of specfic homo‑polypeptides (Nirenberg and Matthei, 1961).
a. If you provide only UDP as a substrate for polynucleotide phosphorylase, the product will be a homopolymer poly(U).
b. Addition of poly(U) to an in vitro translation system (e.g. E. coli lysates), results in a newly synthesized polypeptide which is a polymer of polyphenylalanine.
c. Thus UUU encodes Phe.
d. Likewise, poly(A) programmed synthesis of poly‑Lys; AAA encodes Lys.
Poly(C) programmed synthesis of poly‑Pro; CCC encodes Pro.
Poly(G) programmed synthesis of poly‑Gly; GGG encodes Gly.
4. Use of mixed co‑polymers
a. If two NDPs are mixed in a known ratio, polynucleotide phosphorylase will make a mixed co‑polymer in which nucleotide is incorporated at a frequency proportional to its presence in the original mixture.
b. For example, consider a 5:1 mixture of A:C. The enzyme will use ADP 5/6 of the time, and CDP 1/6 of the time. An example of a possible product is:
AACAAAAACAACAAAAAAAACAAAAAACAAAC...
Table 3.4.1. Frequency of triplets in a poly(AC) (5:1) random copolymer
Composition
Number
Probability
Relative frequency
3 A
1
0.578
1.0
2 A, 1 C
3
3 x 0.116
3 x 0.20
1 A, 2 C
3
3 x 0.023
3 x 0.04
3 C
1
0.005
0.01
c. So the frequency that AAA will occur in the co‑polymer is
(5/6)(5/6)(5/6) = 0.578.
This will be the most frequently occurring codon, and can be normalized to 1.0 (0.578/0.578 = 1.0)
d. The frequency that a codon with 2 A's and 1 C will occur is
(5/6)(5/6)(1/6) = 0.116.
There are three ways to have 2 A's and 1 C, i.e. AAC, ACA and CAA. So the frequency of occurrence of all the A2C codons is 3 x 0.116.
Normalizing to AAA having a relative frequency of 1.0, the frequency of A2C codons is 3 x (0.116/0.578) = 3 x 0.2.
e. Similar logic shows that the expected frequency of AC2 codons is 3 x 0.04, and the expected fequency of CCC is 0.01.
Table 3.4.2. Amino acid incorporation with poly(AC) (5:1) as a template
Radioactive
Precipitable cpm
Observed
Theoretical
amino acid
- template
+ template
incorporation
incorporation
Lysine
60
4615
100.0
100
Threonine
44
1250
26.5
24
Asparagine
47
1146
24.2
20
Glutamine
39
1117
23.7
20
Proline
14
342
7.2
4.8
Histidine
282
576
6.5
4
These data are from Speyer et al. (1963) Cold Spring Harbor Symposium in Quantitative Biology, 28:559. The theoretical incorporation is the expected value given the genetic code as it was subsequently determined.
f. When this mixture of mixed copolymers is used to program in vitro translation, Lys is incorporated most frequently, which can be expressed as 100. This confirms that AAA encodes Lys.
g. Relative to Lys incorporation as 100, Thr, Asn, and Gln are incorporated with values of 24 to 26, very close to the expectation for amino acids encoded by one of the A2C codons. However, these data do not show which of the A2C codons encodes each specific amino acid. We now know that ACA encodes Thr, AAC encodes Asn, and CAA encodes Gln.
h. Pro and His are incorporated with values of 6 and 7, which is close to the expected 4 for amino acids encoded by AC2 codons. E.g. CCA encodes Pro, CAC encodes His. ACC encodes Thr, but this incorporation is overshadowed by the “26.5” units of incorporation at ACA. Or, more accurately, “26.5” @ 20 (ACA) + 4 (ACC) for Thr.
5. Defined trinucleotide codons stimulate binding of aminoacyl‑tRNAs to ribosomes
a. At high concentrations of Mg2+ cations, the normal initation mechanism, requiring f‑Met‑tRNAf, can be overriden, and defined trinucleotides can be used to direct binding of particular, labeled aminoacyl‑tRNAs to ribosomes.
b. E.g. If ribosomes are mixed with UUU and radiolabeled Phe‑tRNAphe, under these conditions, a ternary complex will be formed that will stick to nitrocellulose ("Millipore assay" named after the manufacturer of the nitrocellulose).
c. One can then test all possible combinations of triplet nucleotides.
Figure 3.4.2. Data from Nirenberg and Leder (1964) Science 145:1399.
6. Repeating sequence synthetic polynucleotides (Khorana)
a. Alternating copolymers: e.g. (UC)n programs the incorporation of Ser and Leu. So UCU and CUC encode Ser and Leu, but cannot tell which is which. But in combination with other data, e.g. the random mixed copolymers in section 4 above, one can make some definitive determinations. Such subsequent work showed that UCU encodes Ser and CUC encodes Leu.
b. poly(AUG) programs incorporation of poly‑Met and poly‑Asp at high Mg concentrations. AUG encodes Met, UGA is a stop, so GUA must encode Asp.
C. The genetic code
By compiling observations from experiments such as those outlined in the previous section, the coding capacity of each group of 3 nucleotides was determined. This is referred to as the genetic code. It is summarized in Table 3.4.4. This tells us how the cell translates from the "language" of nucleic acids (polymers of nucleotides) to that of proteins (polymers of amino acids).
Table 3.4.4. The Genetic Code
Position in Codon
1st
2nd .
3rd
U .
C .
A .
G .
U
UUU
Phe
UCU
Ser
UAU
Tyr
UGU
Cys
U
UUC
Phe
UCC
Ser
UAC
Tyr
UGC
Cys
C
UUA
Leu
UCA
Ser
UAA
Term
UGA
Term
A
UUG
Leu
UCG
Ser
UAG
Term
UGG
Trp
G
C
CUU
Leu
CCU
Pro
CAU
His
CGU
Arg
U
CUC
Leu
CCC
Pro
CAC
His
CGC
Arg
C
CUA
Leu
CCA
Pro
CAA
Gln
CGA
Arg
A
CUG
Leu
CCG
Pro
CAG
Gln
CGG
Arg
G
A
AUU
Ile
ACU
Thr
AAU
Asn
AGU
Ser
U
AUC
Ile
ACC
Thr
AAC
Asn
AGC
Ser
C
AUA
Ile
ACA
Thr
AAA
Lys
AGA
Arg
A
AUG*
Met
ACG
Thr
AAG
Lys
AGG
Arg
G
G
GUU
Val
GCU
Ala
GAU
Asp
GGU
Gly
U
GUC
Val
GCC
Ala
GAC
Asp
GGC
Gly
C
GUA
Val
GCA
Ala
GAA
Glu
GGA
Gly
A
GUG*
Val
GCG
Ala
GAG
Glu
GGG
Gly
G
* Sometimes used as initiator codons.
3. Degeneracy
The degeneracy of the genetic code refers to the fact that most amino acids are specified by more than one codon. The exceptions are methionine (AUG) and tryptophan (UGG). The degeneracy is found primarily the third position. Consequently, single nucleotide substitutions at the third position may not lead to a change in the amino acid encoded. These are called silent or synonymous nucleotide substitutions and do not alter the encoded protein. This is discussed in more detail below.
The pattern of degeneracy allows one to organize the codons into "families" and "pairs". In 9 groups of codons, the nucleotides at the first two positions are sufficient to specify a unique amino acid, and any nucleotide (abbreviated N) at the third position encodes that same amino acid. These comprise 9 codon "families". An example is ACN encoding threonine.
There are 13 codon "pairs", in which the nucleotides at the first two positions are sufficient to specify two amino acids. A purine (R) nucleotide at the third position specifies one amino acid, whereas a pyrimidine (Y) nucleotide at the third position specifies the other amino acid.
These examples add to more than 20 (the number of amino acids) because leucine (encoded by UUR and CUN), serine (encoded by UCN and AGY) and arginine (encoded by CGN and AGR) are encoded by both a codon family and a codon pair. The UAR codons specifying termination of translation were counted as a codon pair. The three codons encoding isoleucine (AUU, AUC and AUA) are half-way between a codon family and a codon pair.
5. The major codon specifying initiation of translation is AUG
Bacteria can also use GUG or UUG, and very rarely AUU and possibly CUG. Using data from the 4288 genes identified by the complete genome sequence of E. coli, the following frequency of use of codons in initiation was determined:
• AUG is used for 3542 genes.
• GUG is used for 612 genes.
• UUG is used for 130 genes.
• AUU is used for 1 gene.
• CUG may be used for 1 gene.
Regardless of which codon is used for initiation, the first amino acid incorporated during translation is f-Met in bacteria.
6. Three codons specify termination of translation: UAA, UAG, UGA.
Of these three codons, UAA is used most frequently in E. coli, followed by UGA. UAG is used much less frequently.
• UAA is used for 2705 genes.
• UGA is used for 1257 genes.
• UAG is used for 326 genes.
7. The genetic code is almost universal
In the rare exceptions to this rule, the differences from the genetic code are fairly small. For example, one exception is RNA from mitochondrial DNA, where both UGG and UGA encode Trp.
D. Differential codon usage
1. Various species have different patterns of codon usage: E.g. one may use 5' UUA to encode Leu 90% of the time (determined by nucleotide sequences of many genes). It may never use CUR, and the combination of UUG plus CUY may account for 10% of the codons.
2. tRNA abundance correlates with codon usage in natural mRNAs: In this example, the tRNALeu with 3' AAU at the anticodon will be the most abundant.
3. The pattern of codon usage may be a predictorof the level of expression of the gene: In general, more highly expressed genes tend to use codons that are frequently used in genes in the rest of the genome. This has been quantitated as a "codon adaptation index". Thus in analyzing complete genomes, a previously unknown gene whose codon usage profile matches the preferred codon usage for the organism would score high on the codon adaptation index, and one would propose that it is a highly expressed gene. Likewise, one with a low score on the index may encode a low abundance protein.
The observation of a gene with a pattern of codon usage that differs substantially from that of the rest of the genome indicates that this gene may have entered the genome by horizontal transfer from a different species.
4. The preferred codon usage is a useful consideration in "reverse genetics": If you know even a partial amino acid sequence for a protein and want to isolate the gene for it, the family of mRNA sequences that can encode this amino acid sequence can be determined easily. Because of the degeneracy in the code, this family of sequences can be very large. Since one will likely use these sequences as hybridization probes or as PCR primers, the larger the family of possible sequences is, the more likely that one can get hybridization to a target sequence that differs from the desired one. Thus one wants to limit the number of possible sequences, and by referring to a table of codon preferences (assuming they are known for the organism of interest), then one can use the preferred codons rather than all possible codons. This limits the number of sequences that one needs to make as hybridization probes or primers.
E. Wobble in the anticodon
This flexibility at the "wobble" position allows some tRNAs to pair with two or three codons, thereby reducing the number of tRNAs required for translation. The following “wobble” rules mean that the 61 codons (for 20 amino acids) can be read by as few as 31 anticodons (or 31 tRNAs).
In addition to the usual base pairs, one can have G‑U pairs and I in the anticodon 1st position can pair with U, C or A (wobble rules).
5' base of the anticodon = 3' base of the codon =
first position in the tRNA third position in the mRNA
C G
A U
U A or G
G C or U
I U, C or A
F. Types of mutations
Base substitutions
This has already been covered in Part Two, DNA Repair. Just as a reminder, there are two types of base substitutions.
1. Transitions: A purine substitutes for a purine or a pyrimidine substitutes for another pyrimidine. The same class of nucleotide remains. Examples are A substituting for G or C substituting for T.
2. Transversions: A purine substitutes for a pyrimidine or a pyrimidine substitutes for a purine. A different class of nucleotide is placed into the DNA, and the helix will be distorted (especially with a purine‑purine base pair). Examples are A substituting for T or C, or C substituting for A or G.
Over evolutionary time, the rate of accumulation of transitions exceeds the rate of accumulation of transversions.
Effect of mutations on the mRNA
1. Missense mutations cause the replacement of an amino acid. Depending on the particular replacement, it may or may not have a detectable phenotypic consequence. Some replacements, e.g. a valine for an leucine in a position that is important for maintaining an a‑helix, may not cause a detectable change in the structure or function of the protein. Other replacements, such as valine for a glutamate at a site that causes hemoglobin to polymerize in the deoxygenated state, cause significant pathology (sickle cell anemia in this example).
2. Nonsense mutations cause premature termination of translation. They occur when a substitution, insertion or deletion generates a stop codon in the mRNA within the region that encodes the polypeptide in the wild‑type mRNA. They almost always have serious phenotypic consequences.
3. Frameshift mutations are insertions or deletions that change the reading frame of the mRNA. They almost always have serious phenotypic consequences.
Not all base subsitutions alter the encoded amino acids
1. The base substitution may lead to an alteration in the encoded polypeptide sequence, in which case the substitution is called nonsynonymous or nonsilent.
2. If the base substitution occurs in a degenerate site in the codon, so that the encoded amino acid is not altered, it is called a synonymous or silent substitution.
Example:
ACU ‑> AAU is a nonsynonymous substitution that results in Thr → Asn
while,
ACU ‑> ACC is a synonymous substitution that results in Thr → Thr
1. Examination of the patterns of degeneracy in the genetic code shows that nonsynonymous substitutions occur mostly in the first and second positions of the codon, whereas synonymous substitutions occur mostly in the third position. However, there are several exceptions to this rule.
2. In general, the rate of fixation of synonymous substitutions in a population is significantly greater that the rate of fixation of nonsynonymous substitutions. This is one of the strongest supporting arguments in favor of model of neutral evolution, or evolutionary drift, as a principle cause of the substitutions seen in natural populations.
13: Genetic code
Questions for Chapter 13. Genetic Code
13.1 How does the enzyme polynucleotide phosphorylase differ from DNA and RNA polymerases?
13.2 A short oligopeptide is encoded in this sequence of RNA
5' GACUAUGCUCAUAUUGGUCCUUUGACAAG
a) Where does it start and stop, and how many amino acids are encoded?
b) Which codon position usually shows degeneracy?
The template strand of a sample of double-helical DNA contains the sequence:
(5')CTTAACACCCCTGACTTCGCGCCGTCG
a) What is the base sequence of mRNA that can be transcribed from this strand?
c) Suppose the other (nontemplate) strand of this DNA sample is transcribed and translated. Will the resulting amino acid sequence be the same as in (b)? Explain the biological significance of your answer.
13.5 The Basis of the Sickle-Cell Mutation.
b) Leu can be converted to either Ser, Val, or Met by a single nucleotide substitution (a different nucleotide substitution for each amino acid replacement). What is the codon for Leu?
b) valine?
b) 5'-G-A-U-3'
13.10 (POB) Identifying the Gene for a Protein with a Known Amino Acid Sequence.
H3N+-Ala-Pro-Met-Thr-Trp-Tyr-Cys-Met-Asp-Trp-Ile-Ala-Gly-Gly-Pro-Trp-Phe-Arg-Lys-Asn-Thr-Lys---
13.11 Let's suppose you are in a lab on the Starship Enterprise. One of the “away teams” has visited Planet Claire and brought back a fungus that is the star of this week's episode. While the rest of the crew tries to figure out if the fungus is friend or foe (and gets all the camera time), you are assigned to determine its genetic code. With the technologies of two centuries from now, you immediately discover that its proteins are composed of only eight amino acids, which we will call simply amino acids 1, 2, 3, 4, 5, 6, 7, and 8. Its genetic material is a nucleic acid containing only three nucleotides, called K, N and D, which are not found in earthly nucleic acids.
The results of frameshift mutations confirm your suspicion that the smallest possible coding unit is in fact used in this fungus. Insertions of a single nucleotide or three nucleotides into a gene cause a complete loss of function, but insertions or deletions of two nucleotides have little effect on the encoded protein.
You make synthetic polymers of the nucleotides K, N and D and use them to program protein synthesis. The amino acids incorporated into protein directed by each of the polynucleotide templates is shown below. Assume that the templates are read from left to right.
Template Amino acid(s) incorporated
Kn = KKKKKKKKKK 1
Nn = NNNNNNNNNN 2
Dn = DDDDDDDDDDD 3
(KN)n = KNKNKNKNKN 4 and 5
(KD)n = KDKDKDKDKD 6 and 7
(ND)n = NDNDNDNDND 8
(KND)n = KNDKNDKNDKND 4 and 6 and 8
Please report your results on the genetic code used in the fungus from Planet Claire.
a) What is size of a codon?
b) Is the code degenerate?
Amino acid Codon(s)
1
3
5
7
e) What is the mutation that will change a codon for amino acid 6 to a codon for amino acid 5? Show both the initial codon and the mutated codon.
f) What is the mutation that will change a codon for amino acid 8 to a codon for amino acid 7? Show both the initial codon and the mutated codon. | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/13%3A_Genetic_code/13.E%3A_Genetic_Code_%28Exercises%29.txt |
A reminder: mRNA encodes the polypeptide with each amino acid designated by a string of three nucleotides. tRNAs serve as the adaptors to translate from the language of nucleic acids to that of proteins. Ribosomes are the factories for protein synthesis.
A. tRNAs
1. The transfer RNAs, or tRNAs serve as adaptors to align the appropriate amino acids on the mRNA templates.
Figure 3.5.1.
2. Primary structure of tRNAs
a. tRNAs are short, being only 73 to 93 nts long.
b. All tRNAs have the trinucleotide CCA at the 3' end.
1. The amino acid is attached to the terminal A of the CCA.
2. In most prokaryotic tRNA genes, the CCA is encoded at the 3' end of the gene. No known eukaryotic tRNA gene encodes the CCA, but rather it is added posttranscriptionally by the enzyme tRNA nucleotidyl transferase.
3. The secondary structure of tRNA is a cloverleaf
a. tRNAs have 4 arms with 3 loops (see Figure 3. 5.2. for yeast phenylalanine tRNA)
b. Theamino acid acceptor arm is formed by complementary base‑pairing between the intial 7 nts of tRNA and a short segment near the 3' end. Again, the amino acid will be added to the terminal A.
c. The D arm ends in the D loop. It contains several dihydrouridines, which are abbreviated "D".
d. The anticodon arm ends in anticodon loop. The anticodon is located in the center of the loop. It will align 3' to 5' with the mRNA (reading 5' to 3').
e. The variable loop varies in size in different tRNAs. The difference in size between the 73 nt versus 93 nt tRNAs is found in the variable loop.
f. The TyC arm is named for this highly conserved motif found in the loop.
4. The tertiary structure of tRNA is a "fat L". (See Fig 3.5.3.)
F. The polarity of translation is from the amino (N) terminus to the caboxy (C) terminus.
This was demonstrated in a classic experiment by Dintzis.
1. Actively translating proteins were labeled with radioactive amino acids for a brief time (short relative to the time required to complete synthesis).
2. Completed polypeptides were collected, digested with trypsin, and the amount of radioactivity in tryptic fragments was determined.
3. Tryptic fragments from the C‑terminal end of the polypeptide had radioactivity at the earliest times of labeling.
4. As the period of labeling was increased (longer pulse), tryptic fragments closer to the N terminus were labeled.
5. This shows that the direction of polypeptide growth is from the N teminus to the C terminus, i.e. translation begins at the N terminal amino acid. This corresponds to mRNA chain growth in a 5' to 3' direction.
6. Note that this experimental protocol is also used to map origins of replication, as we covered in Part Two of the course.
b. IF3= Initiation Factor 3
1. An antiassociation factor; prevents association between the large and small ribosomal subunits.
2. It also must be associated with the small subunit for it to form an initiation complex, i.e. for the small subunit to correctly bind mRNA and fmet-tRNAf.
3. It dissociates prior to binding of the large subunit.
c. IF2
1. Brings fmet‑tRNAf to the partial P site on the small subunit.
2. At least in eukaryotes, it does this in a ternary complex with IF2, fmet‑tRNAf and GTP. In bacteria, the GTP may bind the initiation complex separately. [In some texts, such as MBOG, p. 412, the GTP-IF2 complex binds to the 30S subunit separately from fmet-tRNAf. How would you test the differences in these two models?]
3. IF2 activates a GTPase activity in the small subunit. The resulting change in conformation may allow the large subunit to bind.
3. Binding of 50S (large) subunit to initiation complex gives a complete ribosome ready for the elongation phase of translation. Note that f-met-tRNAfmet is positioned at the P site. It has recognized the initiator AUG in the mRNA.
4. Identification of initiator AUG in eukaryotes
a. Bases around AUG influence efficiency of initiation.
1. The most important effects are from a purine 3 nt before AUG and a G after it. The preferred context is RNNAUGG.
2. The consensus sequence for a large number of mRNAs is GCCRCCAUGG, but these other nucleotides have little effect in mutagenesis experiments.
a. Modified scanner model
(1) The mRNA is "prepared" for binding to the ribosome by the action of eukaryotic initiation factor 4, abbreviated eIF4 (Figure 3.5.16). eIF4 is a multisubunit factor; it includes a cap‑binding protein, eIF4F, that recognizes the 5' cap structure. It also includes proteins eIF4A and eIF4B. These are RNA helicases, which unwind secondary structures in the 5' untranslated region of the mRNA at the expense of ATP hydrolysis.
The mRNA then binds to the small ribosomal subunit. The met-tRNAi has already been brought to the small ribosomal subunit by eIF2, in a complex with GTP.
eIF3 keeps the small ribosomal subunit apart from the large subunit during the process of binding the mRNA.
(2) The small subunit, with associated factors, scans along the mRNA until it reaches (usually) the first AUG. Factors eIF1 and eIF1A help move the preinitiation complex to the AUG start.
Figure 3.5.16.
H. The elongation cycle during translation
1. Binding of aminoacyl‑tRNA to the A site
Recent review: Weijland, A. and A. Parmeggiani (1994) TIBS 19:188-193. Schroeder, R. (1994) Nature 370:597.
a. Elongation factor EF‑Tu
1. The ternary complex of aminoacyl‑tRNA, EF‑Tu, and GTP brings the aminoacyl‑tRNA to the A site on the 70S ribosome (fig. 3.5.17).
2. After the aminoacyl‑tRNA is deposited at the A site of the ribosome, the GTP is cleaved to GDP + Pi. The binary complex of EF‑Tu and GDP dissociates from the ribosome.
3. This is one of many examples of guanine‑nucleotide‑binding proteins that are active when GTP is bound and inactive when GDP is bound.
The general model is that the GTP-bound state of EF-Tu adopts a conformation with a high affinity for aminoacyl-tRNA. The conformation (shape, charge density, etc.) of the resulting ternary complex (containing EF-Tu,GTP, and aminoacyl-tRNA) then allows it to bind to the A site of the ribsosome. Hydrolysis of GTP to form GDP and inorganic phosphate causes the EF-Tu to adopt a different conformation. The aminoacyl-tRNA now has a lower affinity for EF-Tu in the GDP bound state, and presumably a higher affinity for the A site on the ribosome, so it stays on the ribosome when EF-Tu in the GDP bound state dissociates (both from aminoacyl-tRNA and from the ribosome).
Figure 3.5.17.
(4) EF‑Tu is one of the most abundant proteins in E. coli, at 70,000 copies per cell. This is almost equal to the number of aminoacyl‑tRNAs per cell, so most of the aminoacyl‑tRNAs are likely to be in the ternary complex when the concentration of GTP is sufficiently high.
b. GTP
1. Required for binding aminoacyl‑tRNA.
2. Hydrolysis promotes dissociation of the complex EF‑Tu plus GDP from the ribosome.
c. EF‑Ts
1. Aids in the recycling of EF‑Tu by GDP‑GTP exchange.
2. EF‑Ts binds to EF‑Tu complexed with GDP, causing dissociation of GDP. GTP can now bind to the EF‑Tu‑Ts complex, causing EF‑Ts to dissociate and leaving EF‑Tu complexed with GTP. This latter binary complex is ready to bind another aminoacyl‑tRNA.
d. The antibiotic kirromycin prevents release of EF‑Tu‑GDP, thereby blocking elongation. This demonstrates that one step must be completed before the next can take place, and illustrates the importance of the EF‑Tu‑GTP/GDP cycle.
2. Peptidyl transferase on the large ribosomal subunit
a. The peptidyl transferase reaction occurs via nucleophilic displacement. The amino group from aminoacyl‑tRNA (position n) attacks the "C‑terminal" carboxyl group of peptidyl‑tRNA (position n‑1in the mRNA). This results in cleavage of the high energy peptidyl‑tRNA ester linkage, thereby providing the free energy to drive the reaction. The resulting products of the reaction are deacylated tRNA at the P site and peptidyl‑tRNA at the A site.
Figure 3.5.18. Peptidyl transferase reaction
b. Role of rRNA in catalysis
It is likely that rRNA provides the catalytic center for the peptidyl transferase activity, with perhaps some ribosomal proteins aiding in holding the rRNA in the correct conformation for catalysis. This conclusion is supported by several lines of investigation, some of which are listed below.
1. No protein, singly or in combination with other proteins, has been shown to catalyze peptide bond formation.
2. Specific regions of 16S rRNA (in the small subunit) interact with the anticodon regions of tRNA in both the A and P sites. In contrast, 23S rRNA in the large subunit interacts with the CCA terminus of peptidyl‑tRNA, thus placing it in the right location for peptidyl transferase.
3. The antibiotics erythromycin and chloramphenicol block peptidyl transferase. Some mutations that confer resistance to them map to the 23S rRNA sequence (others map to some 50S ribosomal proteins).
4. A preparation consisting of 23S rRNA and some remnants of large subunit proteins retains peptidyl transferase activity. For more information, see Noller et al. (1992) Unusual resistance of peptidyl transferase to protein extraction procedures. Science 256: 1416-1419.
5. Ribozyme RNAs can be selected that catalyze peptide bond formation. In this experiment, the investigators started with a pool of 1.3 ´ 1015 different RNAs of 72 nucleotides, flanked by constant regions. They let this large population of RNAs catalyze a peptide bond formation that adds a biotinyl-labeled amino acid (in a chemical mimic of a P site) to an amino acid connected to the RNA (in a chemical mimic of an A site). The RNAs that successfully catalyzed the reaction were extremely rare, but were now covalently attached to a biotin label. Thus they could be selected from the population by binding to streptavidin. PCR was used to amplify the successful RNAs, and the procedure repeated 19 times. At this point, the investigators characterized 9 RNAs that catalyzed the reaction. They found that these RNAs increased the reaction rate by a factor of 106 over the uncatalyzed reaction.
6. The three-dimensional structure of the ribosome shows that the active site is comprised of RNA. The structure of a ribosome crystallized with an active site directed inhibitor has been determined, as well as the structure without the inhibitor. This allowed researchers to see precisely where the peptidyl transferase active site is within the structure. Only RNA is seen around this site. The nearest protein is 20 Angstroms away, too far to participate in catalysis.
3. Translocation
a. The translocation step moves the ribosome another 3 nucleotides downstream (one codon) and moves peptidyl‑tRNA to the P site (position n), deacylated tRNA exits through the E site, and the A site (position n+1) is vacant for another round of elongation.
b. Elongation Factor G = EF‑G
1. This is another very abundant protein, with about 20,000 copies per cell, which is equivalent to the number of ribosomes.
2. EF‑G‑GTP binds to the ribosome to aid translocation, and is released upon GTP hydrolysis (GTPase is from some ribosomal component).
3. Recent structural studies (from A. Dahlberg and colleagues) show that EF-G in the GTP-bound state has a shape similar to that of the ternary complex of EF-Tu, GTP and aminoacyl-tRNA. Like the latter ternary complex, EF-G in the GTP-bound state also has a high affinity for the A site on the ribosome. This may help drive the movement of the peptidyl-tRNA from the A site to the P site, replacing it with EF-G (GTP) in the A site.
c. Hydrolysis of GTP is required for dissociation of EF‑G after translocation. The GTPase is part of the ribosome, not EF-G.
Figure 3.5.20.
d. Action of fusidic acid revealed the need for release of EF‑G‑GDP. In the presence of fusidic acid, EF‑G‑GTP binds the ribosome, GTP is hydrolyzed, and the ribosome moves three nucleotides. But the ribosome‑EF‑G‑GDP complex is stabilized by this compound, and translation is halted.
e. Ribosomes cannot bind EF‑Tu and EF‑G simultaneously. EF‑Tu must finish its action before EF‑G can act, and EF‑G must complete its cycle before EF‑Tu can act again to bring in another aminoacyl‑tRNA.
f. Effect of diptheria toxin
1. The eukaryotic analog to EF‑G is eEF2, which is also a translocase dependent on GTP hydrolysis. It is also is blocked by fusidic acid.
2. Diptheria toxin will catalyze the addition of ADP‑ribose (from substrate NAD+) to eEF2, thereby inactivating it. The target for ADP‑ribosylation is modified histidine found in eEF2 from many species.
c. The "down side" to nonsense suppression is that the suppressor tRNA can act at any amber codon, Therefore it competes with the releasing factors in recognizing the normal termination codons. When the suppressor tRNA is used instead of releasing factors, translation proceeds further down the mRNA than it is supposed to, leading to production of aberrant proteins. Suppressor strains of E. colican be pretty sick (i.e. they don't grow as well as wild type strains).
d. Two other amber suppressors are encoded by the supDgene, which encodes a tRNA that will insert Ser at a UAG, and supF, which will insert Tyr.
3. Missense suppressors: These are mutant tRNAs that lead to the insertion of an amino acid that is compatible with the wild type amino acid (altered by the original mutation).
4. Frameshift suppressors: These are mutant tRNAs whose anticodon has been expanded (or contracted?) to match the length‑altering mutation in the mRNA.
E.g. Consider an original mutation 5'GGG ‑> 5'GGGG (insert a G).
A frameshift suppressor would also have an additional C in the anticodon.
wt tRNA anticodon 3'CCC ‑‑> suppressor tRNA 3'CCCC.
14: Translation (Protein synthesis)
14.1 (POB) Methionine Has Only One Codon.
Methionine is one of the two amino acids having only one codon. Yet the single codon for methionine can specify both the initiating residue and interior Met residues of polypeptides synthesized by E. coli. Explain exactly how this is possible.
14.2 Are the following statements concerning aminoacyl‑tRNA synthetase true or false?
a) Two distinct classes of the enzymes have been defined that are not very related to each other.
b) The enzymes scan previously‑synthesized aminoacyl‑tRNAs and cleave off any amino acids that are linked to the incorrect tRNA.
c) Proofreading can occur at the formation of either the aminoacyl‑adenylate intermediate (in some synthetases) or at the aminoacyl‑tRNA (in other synthetases) to insure that the correct amino acid is attached to a given tRNA.
d) The product of the reaction has a high‑energy ester bond between the carboxyl of an amino acid and a hydroxyl on the terminal ribose of the tRNA.
14.3 A preparation of ribosomes in the process of synthesizing the polypeptide insulin was incubated in the presence of all 20 radiolabeled amino acids, tRNA's, aminoacyl-tRNA synthetases and other components required for protein synthesis. All the amino acids have the same specific radioactivity (counts per minute per nanomole of amino acid). It takes ten minutes to synthesize a complete insulin chain (from initiation to termination) in this system. After incubation for 1 minute, the completed insulin chains were cleaved with trypsin and the radioactivity of the fragments determined.
a) Which tryptic fragment has the highest specific activity?
b) In the same system described above, the insulin polypeptide chains still attached to the ribosomes after ten minutes were isolated, cleaved with trypsin, and the specific activity of each tryptic peptide determined. Which peptide has the highest specific activity?
14.4 Which component of the protein synthesis machinery of E. colicarries out the function listed for each statement.
a) Translocation of the peptidyl-tRNA from the A site to the P site of the ribosome.
b) Binding of f-Met-tRNA to the mRNA on the small ribosomal subunit.
c) Recognition of the termination codons UAG and UAA.
d) Holds the initiator AUG in register for formation of the initiation complex (via base pairing).
14.5 a) In the initiation of translation in E. coli, which ribosomal subunit does the mRNA initially bind to?
b) What nucleotide sequences in the mRNA are required to direct the mRNA to the initial binding site on the ribosome?
c) What other factors are required to form an initiation complex?
14.6 What steps in the activation of amino acids and elongation of a polypeptide chain require hydrolysis of high energy phosphate bonds? What enzymes catalyze these steps or which protein factors are required?
14.7(POB) Maintaining the Fidelity of Protein Synthesis
The chemical mechanisms used to avoid errors in protein synthesis are different from those used during DNA replication. DNA polymerases utilize a 3' ® 5' exonuclease proofreading activity to remove mispaired nucleotides incorrectly inserted into a growing DNA strand. There is no analogous proofreading function on ribosomes; and, in fact, the identity of amino acids attached to incoming tRNAs and added to the growing polypeptide is never checked. A proofreading step that hydrolyzed the last peptide bond formed when an incorrect amino acid was inserted into a growing polypeptide (analogous to the proofreading step of DNA polymerases) would actually be chemically impractical. Why? (Hint: Consider how the link between the growing polypeptide and the mRNA is maintained during the elongation phase of protein synthesis.)
14.8 (POB) Expressing a Cloned Gene.
You have isolated a plant gene that encodes a protein in which you are interested. What are the sequences or sites that you will need to get this gene transcribed, translated, and regulated in E. coli.)?
14.9 The three codons AUU, AUC, and AUA encode isoleucine. They correspond to "hybrid" between a codon family (4 codons) and a codon pair (2 codons). The single codon AUG encodes methionine. Given the prevalence of codon pairs and families for other amino acids, what are hypotheses for how this situation for isoleucine and methionine could have evolved?
14.10 Use the following processes to answer parts a-c:
1. synthesis of aminoacyl-tRNA from an amino acid and tRNA.
2. binding of aminoacyl-tRNA to the ribosome for elongation.
3. formation of the peptide bond between peptidyl-tRNA and aminoacyl-tRNA on the ribosome.
4. translocation of peptidyl-tRNA from the A site to the P site on the ribosome.
5. assembly of a spliceosome for removal of introns from nuclear pre-mRNA.
6. removal of introns from nuclear pre-mRNA after assembly of a spliceosome.
7. synthesis of a 5' cap on eukaryotic mRNA.
(a) Which of the above processes require ATP?
(b) Which of the above processes require GTP?
(c) For which of the above processes is there evidence that RNA is used as a catalyst? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_III%3A_The_Pathway_of_Gene_Expression/14%3A_Translation_(Protein_synthesis)/14.E%3A_Translation_-_Protein_synthesis_%28Exercises%29.txt |
Regulation is the controlled expression of gene functions. This can be done in many ways, but these can be grouped into two classes. The level of enzyme activity can be regulated by noncovalent or covalent modification of a protein. The amount of the protein can also be regulated. This latter class of regulation can be exerted at any step in the pathway of gene expression or during protein turnover. For many (perhaps most) genes, the principal level of regulation of expression is at transcription, and Part Four of this course will focus primarily on this. However, post-transcriptional control is also important in many genes, and this will also be discussed.
Unit IV: Regulation of Gene Expression
Operons
An operon is a cluster of coordinately regulated genes. It includes structural genes (generally encoding enzymes), regulatory genes (encoding, e.g. activators or repressors) and regulatory sites (such as promoters and operators). The type of control is defined by the response of the operon when no regulatory protein is present. In the case of negative control, the genes in the operon are expressed unless they are switched off by a repressor protein. Thus the operon will be turned on constitutively (the genes will be expressed) when the repressor in inactivated. In the case of positive control, the genes are expressed only when an active regulator protein, e.g. an activator, is present. Thus the operon will be turned off when the positive regulatory protein is absent or inactivated.
Table 4.1.1. Positive vs. negative control
Catabolic versus Biosynthetic Operons
Catabolic pathways catalyze the breakdown of nutrients (the substrate for the pathway) to generate energy, or more precisely ATP, the energy currency of the cell. In the absence of the substrate,there is no reason for the catabolic enzymes to be present, and the operon encoding them is repressed. In the presence of the substrate, when the enzymes are needed, the operon is induced or de-repressed.
Table 4.1.2. Comparison of catabolic and biosynthetic operons
Operon encodes Absence of Effect Presence of Effect
catabolic enzymes substrate repressed substrate derepressed (induced)
biosynthetic enzymes product induced product repressed
For example, the lac operon encodes the enzymes needed for the uptake (lactose permease) and initial breakdown of lactose (the disaccharide b-D-galactosyl-1->4-D-glucose) into galactose and glucose (catalyzed by b-galactosidase). These monosaccharides are broken down to lactate (principally via glycolysis, producing ATP), and from lactate to CO2 (via the citric acid cycle), producing NADH, which feeds into the electron-transport chain to produce more ATP (oxidative phosphorylation). This can provide the energy for the bacterial cell to live. However, the initial enzymes (lactose permease and b-galactosidase) are only needed, and only expressed, in the presence of lactose and in the absence of glucose. In the presence of the substrate lactose, the operon in turned on, and in its absence, the operon is turned off.
Anabolic, or biosynthetic, pathways use energy in the form of ATP and reducing equivalents in the form of NAD(P)H to catalyze the synthesis of cellular components (the product) from simpler materials, e.g. synthesis of amino acids from small dicarboxylic acids (components of the the citric acid cycle). If the cell has plenty of the product already (in the presence of the product), the the enzymes catalyzing its synthesis are not needed, and the operon encoding them is repressed. In the absence of the product, when the cell needs to make more, the biosynthetic operon is induced. E.g., the trpoperon encodes the enzymes that catalyze the conversion of chorismic acid to tryptophan. When the cellular concentration of Trp (or Trp-tRNAtrp) is high, the operon is not expressed, but when the levels are low, the operon is expressed.
Inducible versus repressible Operons
Inducible operons are turned on in reponse to a metabolite (a small molecule undergoing metabolism) that regulates the operon. E.g. the lac operon is induced in the presence of lactose (through the action of a metabolic by-product allolactose). Repressible operons are switched off in reponse to a small regulatory molecule. E.g., the trpoperon is repressed in the presence of tryptophan. Note that in this usage, the terms are defined by the reponse to a small molecule. Although lac is an inducible operon, we will see conditions under which it is repressed or induced (via derepression).
Table 4.1.3.
Map of the E. colilac operon
1. Promoters = p= binding sites for RNA polymerase from which it initiates transcription. There are separate promoters for the lacIgene and the lacZYAgenes.
2. Operator = o = binding site for repressor; overlaps with the promoter for lacZYA.
3. Repressor encoded by lacIgene
4. Structural genes: lacZYA
lacZ encodes b-galactosidase, which cleaves the disccharide lactose into galactose and glucose.
lacYencodes the lactose permease, a membrane protein that faciltitates uptake of lactose.
lacAencodes b-galactoside transacetylase; the function of this enzymes in catabolism of lactose is not understood (at least by me)
C. Negative control
The lac operon is under both negative and positive control. The mechanisms for these will be considered separately.
1. In negative control, the lacZYAgenes are switched off by repressor when the inducer is absent (signalling an absence of lactose). When the repressor tetramer is bound to o, lacZYAis not transcribed and hence not expressed.
2. When inducer is present (signalling the presence of lactose), it binds the repressor protein, thereby altering its conformation, decreasing its affinity for o, the operator. The dissociation of the repressor-inducer complex allows lacZYAto be transcribed and therefore expressed.
Inducers
The natural inducer (or antirepressor), is allolactose, an analog of lactose. It is made as a metabolic by-product of the reaction catalyzed by b-galactosidase. Usually this enzyme catalyzes the cleavage of lactose to galactose + glucose, but occasionally it will catalyze an isomerization to form allolactose, in which the galacose is linked to C6 of glucose instead of C4.
A gratuitous inducer will induce the operon but not be metabolized by the encoded enzymes; hence the induction is maintained for a longer time. One of the most common ones used in the laboratory is a synthetic analog of lactose called isopropylthiogalactoside (IPTG). In this compound the b-galactosidic linkage is to a thiol, which is not an efficient substrate for b-galactosidase.
E. Regulatory mutants
Regulatory mutations affect the amount of all the enzymes encoded by an operon, whereas mutations in a structural gene affects only the activity of the encoded (single) polypeptide.
Repressor mutants
• a. Wild-type strains (lacI+) are inducible.
• b. Most strains with a defective repressor (lacI-) are constitutive, i.e. they make the enzymes encoded by the lac operon even in the absence of the inducer.
• c. Strains with repressor that is not able to interact with the inducer (lacIS) are noninducible. Since the inducer cannot bind, the repressor stays on the operator and prevents expression of the operon even in the presence of inducer.
• d. Deductions based on phenotypes of mutants
Table 4.1.4. Phenotypes of repressor mutants
b-galactosidase transacetylase
Genotype -IPTG +IPTG -IPTG +IPTG Conclusion
I+Z+A+ <0.1> 100 <1> 100 Inducible
I+Z-A+ <0.1> <0.1> <1> 100 lacZencodes b-galactosidase
I-Z+A+ 100 100 90 90 Constitutive
I+Z-A+ /F' I-Z+A+ <0.1> 100 <1> 200 I+ >I- in trans
IsZ+A+ <0.1> <1> <1> <1> Noninducible
IsZ+A+ /F' I+Z+A+ <0.1> 1 <1> 1 Is>.I+ in trans
1. The wild-type operon is inducible by IPTG.
2. A mutation in lacZaffects only b-galactosidase, not the transacetylase (or other products of the operon), showing that lacZis a structural gene.
3. A mutation in lacIaffects both enzymes, hence lacIis a regulatory gene. Both are expressed in the absence of the inducer, hence the operon is constitutively expressed (the strain shows a constitutive phenotype).
4. In a merodiploid strain, in which one copy of the lac operon is on the chromosome and another copy is on an F' factor, one can test for dominance of one allele over another. The wild-type lacI+is dominant over lacI-intrans. In a situation where the only functional lacZgene is on the same chromosome as lacI-, the functional lacI still causes repression in the absence of inducer.
5. The lacISallele is noninducible.
6. In a merodiploid, the lacISallele is dominant over wild-type in trans.
e. The fact that the product of the lacIgene is trans-acting means that it is a diffusible molecule that can be encoded on one chromosome but act on another, such as the F' chromosome in example (d) above. In fact the product of the lacIgene is a repressor protein.
2. Operator mutants
a. Defects in the operator lead to constitutive expression of the operon, hence one can isolate operator constitutive mutations, abbreviated oc. The wild-type o+is inducible.
b. Mutations in the operator are cis-acting; they only affect the expression of structural genes on the same chromosome.
(1)The merodiploid I+ocZ+/I+o+Z- [this is an abbreviation for lacI+oclacZ+/lacI+o+lacZ-] expresses b-galactosidase constitutively. Thus oc is dominant to o+ when oc is in cisto lacZ+.
(2)The merodiploid I+ocZ-/I+o+Z+ is inducible for b-galactosidase expression. Thus o+ is dominant to oc when o+ is in cisto lacZ+.
(3)The allele of othat is in cisto the active reporter gene (i.e., on the same chromosome as lacZ+ in this case) is the one whose phenotype is seen. Thus the operator is cis-acting, and this property is referred to as cis-dominance. As in most cases of cis-regulatory sequences, these are sites on DNA that are required for regulation. In this case the operator is a binding site for the trans-acting repressor protein.
Interactions between Operator and Repressor
Sequence of operator
The operator overlaps the start the site of transcription and the promoter. It has a dyad symmetry centered at +11. Such a dyad symmetry is commonly found within binding sites for symmetrical proteins (the repressor is a homotetramer). The sequence of DNA that consititutes the operator was defined by the position of oC mutations, as well as the nucleotides protected from reaction with, e.g. DMS, upon binding of the repressor.
2. Repressor
a. Purification
(1)Increase the amount of repressor in the starting material by over-expression.
A wild-type cell has only about 10 molecules of the repressor tetramer. Isolation and purification of the protein was greatly aided by use of mutant strain with up-promoter mutations for lacI, so that many more copies of the protein were present in each cell. This general strategy of over-producing the protein is widely used in purification schemes. Now the gene for the protein is cloned in an expression vector, so that the host (bacteria in this case) makes a large amount of the protein - often a substantial fraction of the total bacterial protein.
(2)Assays for repressor
[1]Binding of radiolabeled IPTG (gratuitous inducer) to repressor
[2]Binding of radiolabeled operator DNA sequence to repressor. This can be monitored by the ability of the protein-DNA complex to bind to nitrocellulose (whereas a radiolabeled mutant operator DNA fragement, oc, plus repressor will not bind). Electrophoretic mobility shift assays would be used now in many cases.
[3]This ability of particular sequences to bind with high affinity to the desired protein is frequently exploited to rapidly isolate the protein. The binding site can be synthesized as duplex oligonucleotides. These are ligated together to form multimers, which are then attached to a solid substrate in a column. The desired DNA-binding protein can then be isolated by affinity chromatography, using the binding site in DNA as the affinity ligand.
b. The isolated, functional repressor is a tetramer; each of the four monomers is the product of the lacI gene (i.e. it is a homotetramer).
c. The DNA-binding domainof the lac repressor folds into a helix-turn-helixdomain. We will examine this structural domain in more in Chapter III. It is one of the most common DNA-binding domains in prokaryotes, and a similar structural domain (the homeodomain) is found in some eukaryotic transcriptional regulators.
3. Contact points between repressor and operator
a. Investigation of the contact points between repressor and the operator utiblized the same techniques that we discussed previously for mapping the binding site of RNA polymerase on the promoter, e.g. electrophoretic mobility shift assays (does the DNA fragment bind?), DNase footprints (where does the protein bind?) and methylation interference assays (methylation of which purines will prevent binding?). Alternative schemes will allow one to identify sites at which methylation is either prevented or enhanced by the binding of the repressor. These techniques provide a biochemical defintion of the operator = binding site for repressor.
b. The key contact points (see Figure 4.1.4.):
(1)are within the dyad symmetry.
(2)coincide (in many cases) with nucleotides that when mutated lead to constitutive expression. Note that the latter is a genetic definition of the operator, and it coincides with the biochemically-defined operator.
(3)tend to be distributed symmetrically around the dyad axis (+11).
(4)are largely on one face of the DNA double helix.
c. The partial overlap between the operator and the promoter initially suggested a model of steric interference to explain the mechanism of repression. As long a repressor was bound to the operator, the polymerase could not bind to the promoter. But, as will be explored in the next chapter, this is notthe case. RNA polymerase canbind to the lacpromoter even when repressor is boudn to the lac operator. However, the polymerase cannot initiatetranscription when juxtaposed to the repressor.
4. Conformational shift in repressor when inducer binds
a. The repressor has two different domains, one that binds to DNA ("headpiece" containing the helix-turn-helix domain) and another that binds to the inducer (and other subunits) (called the "core). These are connected by a "hinge" region.
b. These structural domains can be distinguished by the phenotypes of mutations that occur in them.
lacI-dprevents binding to DNA, leads to constitutive expression.
lacISprevents binding of inducer, leads to a noninducible phenotype.
c. Binding of inducer to the "core" causes an allosteric shift in the repressor so that the "headpiece" is no longer able to form a high affinity complex with the DNA, and the repressor can dissociate (go to one of the many competing nonspecific sites).
Positive control: "catabolite repression"
1. Catabolite repression
a. Even bacteria can be picky about what they eat. Glucose is the preferred source of carbon for E. coli; the bacterium will consume the available glucose before utilizing alternative carbon sources, such as lactose or amino acids.
b. Glucose leads to repression of expression of lacand some other catabolic operons. This phenomenon is called catabolite repression.
2. Two components are needed for this form of regulation
a. cAMP
[1]In the presence of glucose, the [cAMP] inside the cell decreases from 10-4 M to 10-7 M. A high [cAMP] will relieve catabolite repression.
[2]cAMP synthesis is catalyzed by adenylate cyclase (product of the cyagene)
ATP ® cAMP + PPi
b. Catabolite Activator Protein = CAP
[1]Product of the capgene, also called crp(cAMP receptor protein).
[2]Is a dimer
[3]Binds cAMP, and then the cAMP-CAP complex binds to DNA at specific sites
3. Binding site for cAMP-CAP
a. In the lac operon, the binding site is a region of about 20 bp located just upstream from the promoter, from -52 to -72.
b. The pentamer TGTGA is an essential element in recognition. For the lac operon, the binding site is a dyad with that sequence in both sides of the dyad.
c. Contact points betwen cAMP-CAP and the DNA are close to or coincident with mutations that render the lacpromoter no longer responsive to cAMP-CAP.
d. cAMP-CAP binds on one face of the helix.
Figure 4.1.5. Binding site for cAMP-CAP
4. Binding of cAMP-CAP to its site will enhance efficiency of transcription initiation at promoter
a. The lacpromoter is not a particularly strong promoter. The sequence at -10, TATGTT, does not match the consensus (TATAAT) at two positions.
b. In the presence of cAMP-CAP, the RNA polymerase will initiate transcription more efficiently.
c. The lacUV5 promoter is an up-promoter mutation in which the -10 region matches the consensus. The lac operon driven by the UV5 promoter will achieve high level induction without cAMP-CAP, but the wild-type promoter requires cAMP-CAP for high level induction.
5. Mode of action of cAMP-CAP
a. Direct positive interaction with RNA polymerase. The C-terminus of the a subunit is required for RNA polymerase to be activated by cAMP-CAP. This will be explored in more detail in Chapter 16.
b. cAMP-CAP bends the DNA about 90o.
Some generalities
• Repressors, activators and polymerases interact primarily with one face of the DNA double helix.
• Regulatory proteins, such as activators and repressors, are frequently symmetrical and bind symmetrical sequences in DNA.
• RNA polymerases are not symmetrical, and the promoters to which they bind also are asymmetrical. This confers directionality on transcription.
15: Positive and negative control of gene expression
Q15.1
Amber mutations are one class of nonsense mutations. They lead to premature termination of translation by alternation of an amino acid-encoding codon to a UAG terminator, e.g. CAG (Gln) may be changed to UAG (stop; amber). The phenotype of such amber mutants can be suppressed by amber-suppressor genes, which are mutant tRNA genes that encode tRNAs that recognize UAG codons and allow insertion of an amino acid during translation. Which genes or loci in the lac operon can give rise to amber-suppressible mutations?
15.2 (POB) Negative regulation.
In the lacoperon, describe the probable effect on lacZ gene expression of:
1. Mutations in the lacoperator
2. Mutations in the lacI gene
3. Mutations in the promoter
Q15.3
Consider a negatively controlled operon with two structural genes (A and B, for enzymes A and B) an operator gene (0) and a regulatory gene (R). In the wild-type haploid strain grown in the absence of inducer, the enzyme activities of A and B are both 1 unit. In the presence of an inducer, the enzyme activities of A and B are both 100 units. For parts a-d, choose the answer that best describes the enzyme activities in the designated strains.
Uninduced Induced
Enz A Enz B Enz A Enz B
a) R+0CA+B+ a) 1 1 100 100
b) 1 100 100 1
c) 50 50 100 100
b) R-0+A+B- a) 1 1 100 100
b) 100 100 100 100
c) 100 0 100 0
c) R+0CA+B+/R+0+A+B+ a) 2 2 200 200
b) 51 51 200 200
c) 200 2 2 200
d) R-0+A+B+/R+0+A+B+ a) 2 2 200 200
b) 2 101 2 101
c) 200 200 200 200
Q15.4 (POB) Positive regulation.
A new RNA polymerase activity is discovered in crude extracts of cells derived from an exotic fungus. The RNA polymerase initiates transcription only from a single, highly specialized promoter. As the polymerase is purified, its activity is observed to decline. The purified enzyme is completely inactive unless crude extract is added to the reaction mixture. Suggest an explanation for these observations.
Q15.5
Consider a hypothetical regulatory scheme in which citrulline induces the production of urea cycle enzymes. Four genes (citA, citB, citC, citD) affecting the activity or regulation of the enzymes were analyzed by assaying the wild-type and mutant strains for argininosuccinate lyase activity and arginase activity in the absence (-cit) or presence (+cit) of citrulline. In the following table, wild-type alleles of the genes are indicated by a + under the letter of the cit gene and mutant alleles are indicated by a - under the letter. The activities of the enzymes are given in units such that 1 = the uninduced wild-type activity, 100 = the induced activity of a wild-type gene, and 0 = no measurable activity. In the diploid analysis, one copy of each operon is present in each cell.
Strain lyase activity arginase act.
number genes - cit + cit - cit + cit
Haploid: A B C D
1 + + + + 1 100 1 100
2 - + + + 100 100 100 100
3 + - + + 0 0 1 100
4 + + - + 100 100 100 100
5 + + + - 1 100 0 0
Diploid: A B C D / A B C D
6 + + + - / + - + + 1 100 1 100
7 - + + + / + - + + 1 100 2 200
8 + + - + / + - + - 100 100 100 100
9 + - - + / + + + - 1 100 100 100
Use the data in the table to answer the following questions.
a) What is the phenotype of the following strains with respect to lyase and arginase activity? A single word will suffice for each phenotype.
Lyase activity Arginase activity
Strain 2 ______________________ ________________________
Strain 3 ______________________ ________________________
Strain 4 ______________________ ________________________
Strain 5 ______________________ ________________________
Strain 6 ______________________ ________________________
b) What can you conclude about the roles of citBand citDin the activity or regulation of the urea cycle in this organism? Brief answers will suffice.
c) What is the relationship (recessive or dominant) between wild-type and mutant alleles of citAand citC? Be as precise as possible in your answer.
d) What can you conclude about the roles of citAand citCin the activity or regulation of the urea cycle in this organism? Brief answers will suffice.
Q15.6
Consider a hypothetical operon responsible for synthesis of the porphyrin ring (the heterocyclic ring that is a precursor to heme, cytochromes and chlorophyll). Four genes or loci, porA, porB, porC, and porD that affect the activity or regulation of the biosynthetic enzymes were studied in a series of haploid and diploid strains. In the following table, wild-type alleles of the genes or loci are indicated by a + under the letter of the porgene or locus and mutant alleles are indicated by a — under the letter. The activities of two enzymes involved in porphyrin biosynthesis, d-aminolevulinic acid synthetase and d-aminolevulinic acid dehydrase (referred to in the table as ALA synthetase and ALA dehydrase), were assayed in the presence or absence of heme (one product of the pathway). The units of enzyme activity are 100 = non-repressed activity of the wild-type enzyme, 1 = repressed activity of the wild-type enzyme (in the presence of heme), and 0 = no measurable activity. In the diploid analysis, one copy of each operon is present in each cell.
Strain ALA synthetase ALA dehyd.
number por - heme +heme - heme + heme
Haploid: A B C D
1 + + + + 100 1 100 1
2 - + + + 100 100 100 100
3 + - + + 0 0 100 1
4 + + - + 100 1 0 0
5 + + + - 100 100 100 100
Diploid: A B C D / A B C D
6 + - + + / + + - + 100 1 100 1
7 - + + + / + + - + 200 101 100 100
8 + + + - / + + - + 200 2 100 1
9 - + - + / + - + - 100 100 100 1
Use the data in the table to answer the following questions.
a) Describe the phenotype of the following the strains with respect to ALA synthetase and ALA dehydrase activities. A single word will suffice for each phenotype.
ALA synthetase ALA dehydrase
Strain 2 ______________________ ________________________
Strain 3 ______________________ ________________________
Strain 4 ______________________ ________________________
Strain 5 ______________________ ________________________
Strain 6 ______________________ ________________________
b) What is the relationship (dominant or recessive) between wild-type and mutant alleles of the four genes, and which strain demonstrates this? Please answer in a sentence with the syntax in this example: "Strain 20 is repressible, which shows that mutant grk1 is dominant to wild-type."
porA Strain ___ is _____________, which shows that _________
porA is _______________________________________________.
porB Strain ___ is _____________, which shows that _________
porB is ________________________________________________.
porC Strain ___ is _____________, which shows that _________
porC is ________________________________________________.
porD Strain ___ is _____________, which shows that _________
porD is ________________________________________________.
c) What is the role of each of the genes in activity or regulation of porphyrin biosynthesis? Brief phrases will suffice.
d) Is this operon under positive or negative control? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_IV%3A_Regulation_of_Gene_Expression/15%3A_Positive_and_negative_control_of_gene_expression/15.E%3A_Positive_and_negative_control_of_gene_expression_%28Exercises%29.txt |
[Dr. Tracy Nixon made major contributions to this chapter.]
A. The multiple steps in initiation and elongation by RNA polymerase are targets for regulation
RNA Polymerase has to
• bind to promoters,
• form an open complex,
• initiate transcription,
• escape from the promoter,
• elongate , and
• terminate transcription.
See Figure 4.2.1.
Summarizing a lot of work, we know that strong promoters have high KB, high kf, low kr, and high rates of promoter clearance. Furthermore, weak promoters have low KB, low kf, high kr, and low rates of promoter clearance and moderate promoters have one or more "weak" spots.
To learn these facts, we need genetic data to identify which macromolecules (DNA and proteins) interact in a specific regulation event, and to determine which base pairs and amino acid residues are needed for that regulation event. We also need biochemical data to describe the binding events and chemical reactions that are affected by the specific regulation event. Ideally, we would determine all forward and reverse rate constants, or equilibrium constants (which are a function of the ratio of rate constants) if rates are inaccessible. Although, in reality, we cannot get either rates or equilibrium constants for many of the steps, some of the steps are amenable to investigation and have proved to be quite informative about the mechanisms of regulation.
B. Methods exist for measuring rate constants and equilibrium constants, and newer, more accurate methods are now being used
Classical methods of equilibrium studies and data analysis use low concentrations of enzymes and make assumptions that simplify complex reactions so that they can be treated by definite integrals of chemical flux equations. They also manipulate an equation into a form that can be plotted as a linear function, and derive parameter estimates by slope and intercept values
Driven by the success of recombinant DNA and protein purification technology, and by the increased computational power in desktop computers, the classical methods are being replaced by
• using of large amounts of enzymes to directly include them in kinetic studies. In this approach, the enzymes are used in substrate level quantities.
• numerical integrations of chemical flux equations (Kinetic Simulation)
• more rigorous methods based on NonLinear, Least Squares (NLLS) regression, and
• analyzing data from multiple experiments of different design simultaneously (global NLLS analysis).
These changes increase the steps in a reaction that can be examined experimentally, replace the limited set of simple mechanisms that can be analyzed with essentially any mechanism and increase knowledge of error, permitting conclusions to be drawn with more confidence
Box 1: The equations used in this chapter come from several different sources that use different names for the same thing. The following lists some of these synonyms.
C. Experimental approaches to macromolecular binding reactions
Several methods are available for measuring the amount of protein that binds specifically to a DNA molecule. We have already encountered these as methods for localizing protein-binding sites on DNA, and all are amenable to quantization. Major methods include nitrocellulose filter binding, electrophoretic mobility shift assays, and DNase protection assays.
Which Experimental Technique is Best?
• The kind of observations that can be made about the system differ for different experimental approaches.
• These differences lead to specific problems with each technique,
• Each technique depends on combining the analysis of more than one experiment to obtain enough information to resolve intrinsic binding free energy from cooperativity energy.
The most robust technique is DNase I footprinting. If you are studying the binding of multiple, interacting proteins, then it is possible that these proteins are showing cooperativity in their binding to DNA. When analyzing such cooperativity by DNase I footprinting, the resolution is limited to cooperativities >0.5 kcal/mole, and is subject to some critical assumptions. Gel-shifts (also called electrophoretic mobility shift assays, or EMSAs) are useful when there is no cooperativity, or when cooperativity is large relative to site heterogeneity. Filter binding studies require knowledge about filter retention efficiencies for the different protein-DNA complexes, which can only be empirically determined. And always keep in mind that flanking sequences do affect binding affinities, and even point mutations can have distant effects.
In any of these assays, we are devising a physical means for measuring a quantity that is related to fractional occupancy.
D. Measurement of equilibrium constants in macromolecular binding reactions
Classical methods with their linear transformation are not as accurate as the NonLinear, Least Squares (NLLS) regression analysis, but they can serve to show the general approach.
a.The binding constants can be determined by titrating labeled DNA binding sites with increasing amounts of the repressor, and measuring amount of protein-bound DNA and the amount of free DNA. Typical techniques are electrophoretic mobility shift assays or nitrocellulose filter binding.
Note that for a simple equilibrium of a single protein binding to a single site on the DNA, the equilibrium constant for binding (KB) is approximated by the inverse of the protein concentration at which the concentration of DNA bound to protein equals the concentration of free DNA (Figure 4.2.3).
If it were possible to reliably determine both the concentration of DNA bound to protein (i.e. [DP]) and the concentration of free DNA ([D]), then one could plot the ratio of bound DNA to free DNA at each concentration of repressor. If the results were linear, then the slope of the line would give the equilibrium binding constant, KB . See Figure 4.2.4.
However, the error associated with determining very low concentrations of free or bound DNA is substantial, and a more reliable measurement is that of the ratio of bound DNA to total DNA, i.e. [DP]/[D]tot , as illustrated in Figure 4.2.5. The equation describing this binding curve has a form equivalent to the Michelis-Menten equation for steady-state enyzme kinetics. Note that the concentration of protein at which half the DNA is bound to protein is the inverse of KB . You can show this for yourself by substituting 0.5 for [DP]/[D]tot in the equation. At this point, [P] = 1/KB .
2. Problems with the classical approach.
In this classical approach, experiments were designed such that
o one or more concentrations could be assumed to be unchanging, and
o observations were manipulated mathematically (transformed) to a linear equation so that one could
+ plot the transformed data,
+ decide where to draw a straight line, and
+ use the slope and intercepts to estimate the parameters in question. (Scatchard plots, Lineweaver-Burke plots, etc).
* Two problems are associated with the older technique
o Deciding where to draw the straight line is an arbitrary decision for each person doing the analysis (and using a linear regression to find the "best fit" line is not justified, as two of the assumptions about your data that are needed to justify such a regression are not true)
o There is no accurate estimate of the error in the estimate of the parameter value
3. These limitations have been overcome in the last 5 or so years, aided by the advent of recombinant DNA techniques that allow the production of large amounts of the proteins being analyzed, and the availability of powerful microcomputers that can carry out the large number of computations required for nonlinear, least squares regression analysis (NLLS).
a. We can model binding reactions by
• tabulating the different states that exist in a system,
• associating each state with a fractional probability based on the Boltzmann partition function and the Gibb's free energy for that state (DGs),
• and determine the probability of any observed measurement by the ratio of
• the sum of fractional probabilities that give the observation, and
• the sum of the fractional probabilities of all possible states.
Where jis the number of ligands bound, the fractional probability of a particular state is given by this equation for fs.
As an example, consider a one-site system, such as an operator that binds one protein. There are two states, the 0 state with no protein bound to the operator and the 1 state with one protein bound. Thus one can write the equation for f0 and for f1.
If we expand the fractional probabilities for each of these fractional occupancy equations, we derive equations relating fractional occupancy, , to a function of Gibb's free energies for binding (DG), protein concentration ([P2]), and complex stoichiometry (j).
For a single site system, we have the following equations:
Since Gibb's free energy is also related to the equilibrium constant for reactions:
$\Delta G= -RT\ln (K_{eq})$
these free energies can be re-cast as equilibrium constants, as follows.
A more complete presentation of this method, including a treatment of multiple binding sites, can be obtained at the BMB Courses web site (www.bmb.psu.edu/courses/default.htm) by clicking on BMB400 "Nixon Lectures."
b. Analyzing the data
After collecting the binding data, we are in a position to analyze the observed data to find out what values for DG or Kb make the function best predict the observations. Statisticians have developed Maximum Likelihood Theory to allow using the data to find, for each parameter, the value that is most likely to be correct. For biochemical data the approach that is most appropriate (most of the time) is global, nonlinear, least squares (NLLS) regression.
• Fortunately, desktop computers are now powerful enough to do these calculations in a few minutes, for one experiment, or even for many experiments combined in a global analysis. This method has several advantages. It gives you:
o the same parameter estimates, no matter what program or method you or someone else uses, provided that the program is written correctly and used correctly.
o much more rigorous estimates of error.
This last point is worth emphasizing:
• is it not true that $100 (minus$50) is much less attractive as a fee for your time than is $100 (minus$0.01)? The same can be true for estimates of binding free energies, or equilibrium constants .
• Moreover, when several experiments are required to estimate a parameter, the error in each experiment should be included in the estimate of the parameter. Without a global analysis that determines a conglomerate error, it is not possible to carefully carry forward the error of one experiment to the analysis of data from additional ones.
c. This analysis produces a plot of the variance of fit, or error, over a wide range of possible values for the parameter being measured, such as the DG for binding. The DG value with the smallest error is the most accurate value.
An example of this analysis is shown in Figure 4.2.6. The raw data shown in Figure 4.2.2 (left panel) produced the binding curves shown on right panel of that figure. These data were then subjected to non-linear least-squares analysis. The errors (or variance of fit) for each possible value of DG are plotted in Figure 4.2.6. For example, note that the lowest variance of fit for DG1 is about –9.5 kcal/mole.
dG1 = DG1 = Gibb's free energy for binding to the first site of a two-site system.
dG2 = DG2 = Gibb's free energy for binding to the second site of a two-site system.
The variance of fit for the DG for the cooperativity between proteins bound at the two sites is also plotted.
These data were kindly provided by Dr. Tracy Nixon.
As indicated above, once a value for DG is available, one can calculate Keq from
DG = -RT ln (Keq)
Figure 4.2.7.
Some key references for NLLS:
Senear and Bolen, 1992, Methods Enzymol. 210:463
Koblan et al, 1992, Methods Enzymol. 210:405.
Senear et al 1991, J. Biol. Chem. 266:13661
E. Insights into the mechanism of lac regulation by measuring binding constants
Having gone through both classical and non-linear least squares analysis for measuring binding constants, let’s look at an example of how one uses these measurements to better understand the mechanism of gene regulation. We know that transcription of the lacoperon is increased in the presence of the inducer, but how does this occur? One could list a number of possibilities, each with different predictions about how the inducer may affect the binding constant of repressor for operator, KB.
1. Does the inducer change the conformation of the lac repressor so that it now activates transcription? This could occur with no effect on KB.
2. Does inducer cause the repressor to dissociate from the operator DNA and remain free in solution? This predicts a decrease in KB for specific DNA, but no binding to nonspecific DNA.
3. Does inducer cause the repressor to dissociate from the operator and redistribute to nonspecific sites on the DNA? This predicts a decrease in KB for specific DNA, but proposes that most of the repressor is bound to non-operator sites.
Measurement of the equilibrium constants for lac repressor binding to operator and to nonspecific DNA, in the absence and presence of the inducer, shows that possibility c above is correct. This section of the chapter explores this result in detail.
In the absence of inducer, the repressor, or R4, will bind to specific sites (in this case the operator) with high affinity and to nonspecific sites (other DNA sequences) with lower affinity (Figure 4.2.8). This is stated quantitatively in the following values for the equilibrium association constant. Either equilibrium constant can be abbreviated Keq or KB. We will use the term KS to refer to KB at specific sites and KNS for the KB at nonspecific sites.
$KS = 2 \times 10^{13} M-1 KNS = 2 \times 10^6 M-1$
[A detailed presentation of some representative data and how to use them to determine these binding constants for the lac repressor is in Appendix A at the end of this chapter. This Appendix goes through the classic approach to measuring binding constants.]
3. The binding constant of lac repressor to its operator changes in the presence of inducer. (Figure 4.2.8)
Binding of the inducer to the repressor lowersthe affinity of the repressor for the operator1000 fold, but does not affect the affinity of repressor for nonspecific sites.
For R4 with inducer :
KS = 2 x 1010 M -1 KNS= 2 x 106 M -1
4. The difference in affinity for specific versus nonspecific sites can be described by the specificity parameter, which is the ratio between the equilibrium constant for specific binding and the equilibrium constant for nonspecific binding.
Specificity = in absenceof inducer
in presenceof inducer
Note the in the presence of the inducer, the specificity with which the lacrepressor binds to DNA is decreased 1000-fold.
Even though the repressor still has a higher affinity for specific DNA in the presence of the inducer, there are so many nonspecific sitesin the genome that the repressor stays bound to these nonspecific sites rather than finding the operator. Hence in the presence of the inducer, the operator is largely unoccupied by repressor, and the operon is actively transcribed.
The regulation of the lac operon via redistribution of the repressor to nonspecific sites in the genome is covered in more detail in the next two sections. They show the effect of having a large number of nonspecific, low affinity sites competing with a single, high affinity site for a small number of repressor molecules.
5.Distribution of repressor between operator and nonspecific sites
Although repressor has a much higher affinity for the operator than for nonspecific sites, there are so many more nonspecific sites (4.6 x 106, since essentially every nucleotide in the E. coligenome is the beginning of a nonspecific binding site) than specific sites (one operator per genome) that virtually all of the repressor is bound to DNA, even if only nonspecific sites are present.
1. We use the binding constants above, and couple them with a calculation that the concentration of repressor (10 molecules per cell) is 1.7 x 10-8 M and the concentration of nonspecific sites (4.6 x 106 per cell) is 7.64 x 10-3 M. These values for [R4] and [DNS] are essentially constant. With this information, we can compute that the ratio of free repressor to that bound to nonspecific sites is less that 1 x 10-4 ( it is about 6.6 x 10-5), as shown in the box below. Thus only about 1 in 15,000 repressor molecules is not bound to DNA.
2. This analysis shows that the lac repressor is partitioned between nonspecific sites and the operator. When it is not bound to the operator, it is bound elsewhere to any of about 4.6 million sites in the genome. Almost none of the repressor is unbound to DNA in the cell.
3. Box 2 (below) goes through these calculations in more detail.
Box 2. Effectively all repressor protein is bound to DNA.
6. Regulation of the lac operon via redistribution of the repressor to nonspecific sites in the genome
a. The high specificity of repressor for the operator means that in the absence of inducer, the operator is bound by the repressor virtually all the time. This is true despite the huge excess of nonspecific binding sites.
b. The specificity parameter described above (Ks/Kns) allows one to evaluate the simultaneous equilibria (repressor for operator and repressor for nonspecific sites on the DNA). We want to calculate the ratio of repressor-bound operators to free operators. Values for KS, KNS, and [DNS] are already known, and the concentration of repressor not bound to DNA is negligible.
Box 3. Specificity parameter is related to ratio of bound to free operator sites.
Specificity =
ratio of Bound:Free
operator sites
Now we need a value for [R4DNS]. This is obtained by realizing that under conditions that saturate specific sites, the concentration of repressor bound to nonspecific sites is closely approximated by [repressor]total - [operator], or [R4]total - [Ds]total in the equations in Box 4.
Box 4.
[R4]free is negligible (see above).
Under conditions that saturate specific sites,
Thus [R4DNS] = [R4]total - [Ds]total
c. After making these simplifying assumptions, we now have a value for every variable and constant in the equation, except the ratio of bound:free operator sites. Thus we can compute the desired ratio.
Box 5. Equation relating specificity to the ratio of bound to free operator and a set of constants.
Specificity =
already want to constants
measured determine
d. Now that we have the equation in Box 5, we can calculate the ratio of free operator to operator bound by repressor can be calculated in the absence and presence of inducer.
(1) In the absence of inducer:
Specificity =
i.e. the ratio of free operators to operators bound by repressor is 0.05. R4 is bound to the operator ~ 95% of the time. Thus the operon is not expressed.
(2) In the presence of inducer:
Specificity =
i.e. in the presence of inducer, only about 2% of the operators are bound by repressor, or R4 is bound to the operator ~ 2% of the time. Thus the operon is expressed.
In summary, these calculations show that in the absence of inducer, 95% of the operators are occupied (o is bound by R4 95% of the time). In the presence of inducer, the repressor re-distributes to nonspecific sites on the DNA, leaving only 2% of the operators bound by R4. Thus the operon is expressedin most of the cells.
An additional example of the use of the measured binding constants and the specificity parameter is in Appendix B at the end of this chapter. This example explores the effects of operator mutants.
F.Mechanism of repression and induction for the lacoperon
1.Effect of lac repressor on the ability of RNA polymerase to bind to the promoter
The analysis in the previous section showed how the inducer affects the partitioning of the repressor between specific and nonspecific sites. Now let’s examine the effect that repressor bound to the operator has on the function of the polymeraseat the promoter
a.Binding of repressor to the operator actually increases the affinityof the RNA polymerase for the promoter!
Consider the following equilibrium:
RNA polymerase + promoter RNA polymerase-promoter (closed complex)
In the absence of repressor on the operator, the affinity of RNA polymerase for the lacpromoter is
KB = 1.9 x 107 M-1
In the presence of repressor on the operator, the affinity is
KB = 2.5 x 109 M-1
b.Repressor bound to the operator increasesthe affinity of RNA polymerase for the lacpromoter about 100 fold, so the closed complex is formed much more readily. The repressor essentially holds the RNA polymerase in storage at the promoter, but transcription is not initiated.
c.Upon binding of the inducer to the repressor, the repressor dissociates and the RNA polymerase-promoter complex can shift to the open complex and initiate transcription, thus switching on the operon.
d.Thus the effect of repressor bound to the operator is not on Kb for the polymerase-promoter interaction, but rather is on kf for the conversion from closed to open complex.
G. Kinetic measurements of the abortive initiation reaction allow one to calculate kf.
1. Abortive Transcription Assay
The initial transcribing complex (ITC) that exists after open complex formation frequently fails to transform into the initial elongating complex (IEC). The RNA product is released, and the system initiates again. The rate at which the aborted transcripts accumulates can provide a measure of promoter strength, and experiments have been devised to use such an assay to estimate KB for polymerase binding to the promoter region, and kf for isomerization from closed to open complex form. Polymerase, promoter DNA, and nucleotides are mixed such that a radiolabeled phosphate will be introduced into transcripts that are made and aborted. The amount of radioactivity in the short transcripts is then counted as a function of time.
There is a lag between mixing reagents, and optimal rate of abortive transcript production. The length of this lag is inversely proportional to the [RNAP]. A plot of lag-time vs 1/[RNAP] gives a straight line plot, with slope equal to 1/[KB ´ kf] and y-intercept of 1/kf.
Figure 4.2.11.
H. Activation of transcription by the CAP protein of E. coli
1. Activation of transcription by the CAP protein of E. coliillustrates several general regulatory principles.
We will focus on the point that in different contexts (different promoters), a single protein can directly interact with RNAP via at least 2 distinct contact surfaces. Depending on the context, CAP can affect KB or kf for RNA polymerase-promoter interactions.
An additional discussion of the ability of CAP to affect the architecture of a protein-DNA complex which contains precise contacts between RNAP and an additional regulatory protein (MalT), by bending DNA, is at the BMB400 Web site, under "Nixon Lectures." This latter point will not be covered in detail here.
2. aSubunit of RNA polymerase
a. Recall from Part Three that the a subunit of RNA polymerase has two separate domains. The amino terminal domain (aNTD) is essential for dimerization and assembly of polymerase, and the carboxy terminal domain (aCTD) is needed for binding to DNA and for communication with many, but not all, transcription factors.
Most RNA polymerase (~60%) is associated with rRNA or tRNA genes. This is accomplished by a special sequence upstream of the promoter elements (i.e. the –35 and –10 boxes), called the UP element (-57)5'-AAAATTATTTT-3'(-47), which binds a2 dimers, and increases occupancy by polymerase by ~10-fold.
Figure 4.2.12.
b. Much of the communication between activators and E. coliRNA polymerase is mediated between the CTD of a and these factors.
(see Ebright and Busby, 1995, Curr. Opinion in Gen. & Dev. 5:197-203)
Figure 4.2.13.
3. Summary & Distinctions between Cap at Class I and Cap at Class II Promoters
(For reviews see Mol Micro 23:853-859 and Cur. Opin. Genet. Dev. 5:197-203).
Class I promoters have CAP binding sites centered at -62, -83, or -93.
At class II promoters, it is centered at -42 and overlaps the -35 determinant of the promoter.
Figure 4.2.14. CAP binding to class I and class II promoters.
4. CAP has at least two Activation Regions (ARs):
• AR1 (residues 156-164)
At class I promoters, AR1 in the downstream subunit of CAP "sees" residues 258-265 of CTD of a. This interaction increases KB for polymerase binding to the promoter.
At class II promoters, CAP displaces the aCTD (decreasing KB), which is overcome by increasing KB via upstream subunit AR1-aCTD interaction
• AR2 (residues 19, 21, 96, 101)
At class II promoters, the downstream subunit "sees" aNTD residues 162-165, increasing kf for isomerization from closed to open complexes.
Figure 4.2.15. Activation Regions on CAP
At both class I and class II promoters, CAP AR1 interacts with the CTD of a. It is clear that for class I promoters, residues 258-265 of the a subunit are the target of AR1 of CAP; it is not clear if these are the same residues needed for interaction at class II promoters. At class I promoters, this interaction provides "true" direct activation: the interaction is between the downstream subunit of CAP, and appears to only be used to increase KB for the binding of RNA polymerase to the promoter region (perhaps substituting for the lack of an UP sequence). At class II promoters, AR1 in the upstream subunit contacts the alpha subunit, but it does not appear to cause direct stimulation of transcription. Instead, it overcomes inhibition of polymerase that is hypothesized to arise from CAP displacing the alpha subunit from its preferred position near -45. This is evidenced by the following observations:
• aCTD binds to -40 to -55 region at class II promoters in the absence of CAP, but binds to the -58 to -74 region in its presence
• AR1 mutants in CAP decrease KB for RNA polymerase at class II promoters, but have no affect on kf .
• Removal of the a CDT eliminates the need for CAP AR1 in class II promoters, and has no negative affect.
• In contrast, removal of the a CDT prevents activation by CAP at class I promoters.
In addition to overcoming a decrease in KB by AR1, at class II promoters CAP also exerts a "direct" activation. This occurs between CAP residues 19, 21, 96 and 101 (AR2) in the downstream subunit of CAP, and residues 162-165 of the a subunit NTD. This interaction increases the kf and has no affect on KB. Region 162-165 is between regions 30-55 / 65-75 and 175-185 / 195-210 which are essential for contact with the b and b' subunits of polymerase, respectively. AR2 is not needed for CAP to work at class I promoters.
Appendix A for Chapter 17 (Part Four., section II)
Measurement of equilibrium constants for binding of lacrepressor to specific and nonspecific sites in DNA
R4 = Repressor
D S = Specific DNA site Þ operator
D NS = Nonspecific DNA site Þ all other sites in genome
R4+ DSR4DS R4+ DNSR4DNS
The lacrepressor will bind to its specific site, the operator, with very high affinity,
Keq = KS = 2 x 1013 M-1, where Ks is the equilibrium association constant for binding to a specific site
and it will bind to other DNA sequences, or nonspecific sites, with a lower affinity.
Keq = KNS = 2 x 106 M-1, where Kns is the equilibrium association constant for binding to a nonspecific site.
Measurements in the laboratory:
Since it can be difficult to measure the amount of bound or free probe at very low concentrations, it is more reliable to measure the fraction of probe bound as a function of [R4]. The fraction of probe bound is
= .
By substituting [Ds] = [Ds]total- [R4Ds] into the equation for Ks, you can derive the following relationship between the fraction of probe bound by repressor and the concentration of the repressor:
=
{Since the [R4] is usually much greater than the [Ds]total in these assays, the
[R4]free >> [R4Ds], and [R4] is well approximated by [R4]total .}
This equation has the form of the classic Michaelis-Menten equation for steady-state enzyme kinetics, and it is also useful in analysis of many binding assays. Once is plotted against [R4], one can do curve fitting to derive a value for Ks. One can also get a value for Ks by measuring the [R4] at which half the probe is bound. At this point, [R4] = . {This can be seen simply by substituting = 0.5 into the equation above. The algebra is exactly the same as is done for the determination of Km by the Michaelis-Menten analysis.}
Appendix B. Use of binding constants and the equations relating the specificity parameter to the ratio of bound to free operator sites to study the effects of operator mutants.
The same equations used in section E of this chapter also can be used to examine the effects of operator mutants. The following analysis shows that a mutation that decreases the affinity of the operator 20-fold for the repressor will result in about half the operators being free of repressor (or the operon being expressed about half the time).
Specificity =
This says that the operator is essentially equally distributed between the bound and free form.
50% of the operators are not occupied by repressor, thus only about half of the operons will be expressed (in a population of bacteria), or any particular operon will be expressed about half the time.
16: Transcription regulation via effects on RNA polymerases
16.1 The ratio [RDs]/[Ds] is the concentration of a hypothetical repressor (R) bound to its specific site on DNA divided by the concentration of unbound DNA, i.e. it is the ratio of bound DNA to free DNA. When the measured [RDs]/[Ds] is plotted versus the concentration of free repressor [R], the slope of the plot showed that the ratio [RDs]/[Ds] increased linearly by 60 for every increase of 1x10-11 M in [R]. What is the binding constant Ks for association of the repressor with its specific site?
16.2 The binding of the protein TBP to a labeled short duplex oligonucleotide containing a TATA box (the probe) was investigated quantitatively. The following table gives the fraction of total probe bound (column 2) and the ratio of bound to free probe (column 3) as a function of [TBP]. These data are provided courtesy of Rob Coleman and Frank Pugh.
[TBP]
nM
0.10
0.040
0.042
0.20
0.16
0.19
0.30
0.33
0.5
0.40
0.44
0.78
0.50
0.52
1.1
0.70
0.62
1.6
1.0
0.71
2.45
2.0
0.83
4.88
3.0
0.87
6.69
5.0
0.93
14
10
0.97
32.3
20
0.99
99
Plot the data for the two different measures of bound probe. Note that since the denominator for column 2 is a constant, the ratio of bound to total probe will level off, whereas the amount of free probe can continue to decrease with increasing [TBP], and thereby getting a continuing increase in the ration of bound to free probe.
What is the equilibrium constant for TBP binding to the TATA box?
16.3 What is the fate of the lac repressor after it binds the inducer?
16.4 How does the lac repressor prevent transcription of the lacoperon?
For the next two questions, let's imagine that you mixed increasing amounts of the DNA binding protein called AP1 with a constant amount of a labeled duplex oligonucleotide containing the binding site (TGACTCA). After measuring the fraction of DNA bound by AP1 (i.e. the fractional occupancy) as a function of [AP1], the data were analyzed by nonlinear, least squares regression analysis at a wide range of possible values for DG. The error associated with the fit of each of those values to experimental data is shown below; the higher the variance of fit, the larger the error.
16.5 What is the most accurate value of DG for binding of AP1 to this duplex oligonucleotide?
16.6 What is the most accurate measure of the equilibrium constant, Ks, for binding of AP1 to this duplex oligonucleotide?
For the next two problems, consider a hypothetical eubacterial operon in which the operator overlaps the -10 region of the promoter. Measurement of the lag time before production of abortive transcripts (in an abortive initiation assay) as a function of the inverse of the RNA polymerase concentration (1/[RNAP]) gave the results shown below. The filled circles are the results of the assay in the absence of repressor, and the open circles are the results in the presence of repressor bound to the operator.
16.7 What is the value of the forward rate constant (kf ) for closed to open complex formation under the two different conditions?
16.8 What is the value of the equilibrium constant (KB ) for binding of the RNA polymerase to the promoter under the 2 conditions? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_IV%3A_Regulation_of_Gene_Expression/16%3A_Transcription_regulation_via_effects_on_RNA_polymerases/16.E%3A_Transcription_regulation_via_effects_on_RNA_polymerases_%28E.txt |
Although regulation of the initiation of transcription appears to be a dominant factor in control of expression of many genes, the importance of regulation after initiation is becoming better appreciated in an increasing number and variety of systems. The classic systems in which these issues have been explored are antitermination in bacteriophage l and in attenuation of transcription in bacterial biosynthetic operons, in particular the trp operon in E. coli. Although some of the mechanistic details may be peculiar to bacteria, especially the need for coupled transcription and translation in the trpattenuation system, the phenomenon of regulation after initiation is seen in a wide variety of organisms, ranging from bacteria to humans. Some of this work was discussed in the sections on elongation of transcription in Chapter II of Part Three.
Introduction
Both systems discussed in this chapter control the frequency of termination of transcription. Antitermination in bacteriophage l can prevent RNA polymerase from stopping at r-dependent terminators, thus leading to transcription of downstream genes. Attenuation in the trp operon also controls the frequency at which RNA polymerase stops at an early terminator in the operon, hence regulating the transcription of downstream genes. In contrast to the system in l, attenuation in trp regulates termination at a r-independentterminator.
Antitermination in Bacteriophage l
Just to quickly review one of the points in Chapter III, antitermination occurs at two different times in the llife cycle. The N protein allows read-through transcription in the shift from immediate-early to early transcription, and the Q protein allows read-through transcription of the late genes.
Recall from Part Three of the text that r-dependent terminators do not have a well-conserved sequence or secondary structure. Also, the protein r tracks along protein-free regions of the RNA until it hits a paused transcriptional complex at a r-dependent site, at which point its RNA helicase activity can cause termination and dissociation of the polymerase and transcript from the template DNA.
Sites on the DNA needed for antitermination in bacteriophage l. nutsites (N utilization sites) for pN, qutsite for pQ and the nutsites are within the transcription unit, not at the promoter and not at the terminator.
• nutLis in the 5' untranslated region of the Ngene, and nutRis in the 3' untranslated region of the crogene.
• In both cases, the nutsite precedes the terminator at which pN will act.
Both nutLand nutR are 17 bp sequences with a dyad symmetry.
5' AGCCCTGAARAAGGGCA
TCGGGACTTYTTCCCGT 5'
The protein pN recognizes the nutsite and binds to RNA polymerase as it transcribes through the site. The complex of pN with the RNA polymerases is highly processive and overrides the efforts of r at the terminator.
E. coli (host) proteins needed for action of pN. These were isolated as host functions that when mutated prevented action of pN. NusA (encoded by nusA, for N utilization substance, complementation group A) is the best characterized.
1. Can form part of the transcription complex
2. Has been proposed to bind to the core RNA polymerase after s dissociates.
3. Can also bind pN.
4. Model:
NusA binds the core polymerase after s dissociates. As this complex transcribes through a nutsite, pN binds also. The complex a2bb'-NusA-pN prevents r-dependent termination at tR1, tR2, and tL1.
Several other nusgenes have been identified. NusG is the bacterial homolog of a family of conserved proteins involved in elongation. It is homologous to the large subunit of DSIF, which is an elongation factor in mammals. DSIF is the DRB-sensitivity inducing factor. Current studies implicate it in both negative and positive effects on elongation. It has two subunits, one of 160 kDa that is homologous to the yeast transcriptional regulatory protein Spt5, and one of 14 kDa that is homologous to the yeast Spt4 protein. Another nusgene encodes a ribosomal protein. Much more needs to be learned about both termination and antitermination. The nusphenotype of mutations in a gene encoding a ribosomal protein suggests that translation is also coupled to this process.
Components of the E. coli trp operon
The trpoperon encodes the enzymes required for biosynthesis of tryptophan. More specifically, its five genes (trpEDCBA) encode five subunits of proteins that in total catalyze five enzymatic steps, converting chorismic acid to tryptophan. However, there is not a 1:1 correspondence between a cistron and an enzyme. For example, trpBand trpAencode, respectively, the b and a subunits of tryptophan synthase, which catalyzes the replacement of glycerol-3-phosphate from indole-3-glycerol-phosphate with serine to form tryptophan, with glyceraldehyde-3-phosphate as the other product
A leader sequence separates the promoter and operator from the first structural gene of the operon, trpE. An attenuator of transcription follows the leader. As we will see in more detail below, the efficiency of "premature" termination at this attenuator is determined by the extent of translation of the leader, which in turn is determined by the availability of Trp-tRNAtrp. This is an important part of the regulation of the operon. Two terminators of transcription follow the structural genes, one dependent on r and one independent of r.
Modes of regulation: turn operon off in presence of Trp
Repressor-operator: requires a protein binding to a specific site in the presence of Trp to decrease the efficiency of initiation of transcription. Attenuation: the elongation (and termination) of transcription by RNA polymerase is linked to the progress of translation by a ribosome. In the presence of Trp, the translation by the ribosome causes transcription of the subsequent genes in the operon to terminate.
Repressor: apo-repressor and co-repressor (Trp)
The apo-repressor is encoded by trpRat a distant locus. The apo-repressor is a homo-tetramer. It has a high affinity for the operator only when it is bound by the amino acid Trp, which serves as a co-repressor. Thus the active repressor is a tetramer of (formerly apo-) repressor in complex with Trp. The active repressor binds to the operator to prevent initiation of transcription. The operator overlaps the promoter, including the -10 region of the promoter. It has a dyad axis of symmetry.
Attenuation
The attenuator is a conditional transcriptional terminator used to regulate expression of biosynthetic operons in bacteria. It is upstream of the structural genes trpEDCBA and is a r-independent termination site. Its ability to terminate transcription is dependent on its ability to form the stem of duplex RNA that is characteristic of r-independent termination sites.
The fraction of transcripts that read through the attenuator is determined by the [Trp-tRNAtrp]. The concentration of charged tRNAs is a measure of the amount of Trp available for protein synthesis. If most tRNAtrp is charged, there is an abundance of Trp, and the cell does not need to make more. Low [Trp-tRNAtrp] allows read-through transcription through the attenuator, so that trpEDCBA is expressed and high [Trp-tRNAtrp] causes termination of transcription at the attenuator.
The [Trp-tRNAtrp] determines the progress of ribosomes as they translate a short leader peptide. The leader peptide is a short 14 amino acid polypeptide encoded by trpL. Two codons for Trp are in the leader, and the progress of ribosomes past these Trp codons will be determined by the availability of Trp-tRNAtrp. When the concentration of tryptophanyl-tRNA is high, translation of the trpleader will be completed, but when it is low, translation will stall at the tryptophan codons.
The extent of progress of the ribosomes determines the secondary structures formed in the leader RNA. When the [Trp-tRNAtrp] is high, the ribosomes translate past the Trp codons to complete the synthesis leader of the peptide. This allows the nascent RNA to form the structure for r-independent terminator. Thus transcription terminates before the RNA polymerase reaches trpEDCBA. When the [Trp-tRNAtrp] is low, the ribosomes stall at the Trp codons, which prevents formation of the secondary structures in the RNA necessary for termination at the attenuator. Thus read-through transcription continues through trpEDCBAand the operon is expressed, so that more Trp is made.
Table 4.4.1: The basic components of regulation at the attenuator of the E. coli trpoperon are tabulated below.
[trp-tRNA] translation of trpL secondary structures formed in RNA Attenuator Operon
High complete 3-4 stem terminate transcription OFF
Low stalls at Trp codons 2-3 stem allow read-through transcription ON
Alternative base-paired structures in leader RNA. Four regions of the leader RNA can be involved in secondary structure formation, in particular base-paired stems; these are referred to simply as regions 1, 2, 3, and 4. Potentially, 1 can pair with 2, 2 can pair with 3, and 3 can pair with 4.
A stem formed by pairing between 3 and 4 makes a G+C rich stem followed by U's, which is sufficient for r-independent termination of transcription. When the [Trp-tRNAtrp] is high, the 3-4 base-paired structure forms, and transcription terminates at the attenuator. This turns the operon OFF. The formation of a base-paired stem between regions 2 and 3 precludes formation of the 3-4 terminator, and transcription will continue into the structural genes trpEDCBA. This turns the operon ON.
The choice between a 2-3 stem or a 3-4 stem is dictated by the progress of the ribosome. If the ribosome can translate past the Trp codons (when the [Trp-tRNAtrp] is high), then it will reach a natural translation termination codon. When the ribosome is in that position, region 2 of the leader RNA is covered by the ribosome, so the 2-3 stem cannot form but the 3-4 stem can. This generates the secondary structure needed for termination of transcription at the attenuator. In contrast, if the ribosome stalls at the Trp codons in the leader, because the [Trp-tRNAtrp] is low, then region 2 of the leader RNA is not covered by the ribosome. It can then base pair with region 3. This prevents formation of the 3-4 terminator, and RNA polymerase can continue elongation through trpEDCBA.
Mutational Analysis (selected examples)
• Translation of trpLis needed for regulation by attenuation. Mutation of the AUG for initiation of translation of the leader RNA prevents transcription past the attenuator. In the absence of translation, both the 1-2 and 3-4 stems can form. The latter 3-4 stem is the terminator.
• Charged tRNAtrp is required for regulation. Mutation of the genes for tRNAtrp or Trp-tRNAtrp synthetase leads to constitutive expression of trpEDCBA. In these mutants, translation will stall at Trp codons regardless of the intracellular [Trp], and no terminator will form at the attenuator.
• Specific secondary structures in the trpleader RNA are needed for regulation. E.g. mutations that decrease the number of base pairs between the 3 and 4 regions will decrease the amount of transcriptional termination (i.e. increase expression of the operon). Compensatory mutations that increase the number of base pairs between 3 and 4 will suppress the original mutations.
Attenuation requires coupled transcription and translation
Requires no regulatory proteins: charging of cognate tRNA is the regulatory signal. Need a transcriptional pause site at +90 to allow the ribosomes to catch up with the RNA polymerase and thereby affect the secondary structures in the nascent RNA.
Attenuation is a common mechanism for regulating biosynthetic operons
Many operons that encode the enzymes catalyzing biosynthesis of amino acids are regulated by attenuation. In each case, the leader polypeptide is rich in the amino acid that is the product of the pathway, e.g. his, phe, leu, thr, ilv.
Additional readings
• Friedman, D.I. and Count, D.L. (1995) Transcriptional antitermination: The lambda paradigm updated. Molecular Microbiology 18: 191-200.
• Henkin, T. (2000) Transcriptional termination in bacteria. Current Opinions in Microbiology 3: 149-153.
• Gusarov, I. and Nudler, E. (2001) Control of intrinsic transcriptional termination by N and NusA: The basic mechanism. Cell 107: 437-449.
18: Transcriptional regulation after initiation
18.1 Which of the following statements concerning the action of N protein are true?
1. N action requires sequences on thel DNA called nutL+ and nutR+.
2. N activity requires a host function encoded by nusA+.
3. N protein acts to promote rho-dependent termination.
4. N protein can relieve the polarity of certain amber mutations.
18.2 Antitermination at tL1 ofl by N protein allows read-through transcription through int, which encodes the integrase enzyme. However, large amounts of the Int protein are not produced lytic infection, because these transcripts continue past the r-dependent terminator tint. This allows the formation of a secondary structure in the RNA that serves as a signal for RNases to degrade the transcripts from the 3' end. Why are large amounts of Int made during lysogeny?
18.3 Sketch the RNA secondary structures in the trp leader/attenuator region being translated by a ribosome under conditions of low and high concentrations of tryptophan.
What determines the progress of the ribosome, and how does this affect trpexpression?
18.4 Which of the following events occur when E. coli is starved for the amino acid tryptophan?
1. No tryptophanyl-tRNA is made.
2. The ribosome translates the leader peptide completely (to the UGA stop codon).
3. A G+C rich stem-loop structure forms in the nascent RNA (regions 3 and 4) at the attenuator site.
4. A step-loop structure forms in the nascent RNA (regions 2 and 3) that precludes formation of the G+C rich stem-loop at the attenuator site.
5. Transcription reads through the attenuator into trp EDCBA.
18.5 (POB) Transcription attenuation.
In the leader region of the trpmRNA, what would be the effect of:
1. Increasing the distance (number of bases) between the leader peptide gene and sequence 2?
2. Increasing the distance between sequences 2 and 3?
3. Removing sequence 4? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_IV%3A_Regulation_of_Gene_Expression/18%3A_Transcriptional_regulation_after_initiation/18.E%3A_Transcriptional_regulation_after_initiation_%28Exercises%29.txt |
Promoters
1. Eukaryotic genes differ in their state of expression
a. Basal transcription
1. Is frequently studied by in vitrotranscription, using defined templates and either extracts from nuclei or purified components.
2. Requires RNA polymerase with general transcription factors (e.g. TFIID, TFIIA, TFIIB, TFIIE, TFIIF, and TFIIH for RNA polymerase II), as previously covered in Part Three.
b. Activated transcription
1. Occurs via transcriptional activators interacting directly or indirectly with the general transcription complex to increase the efficiency of initiation.
2. The transcriptional activators may bind to specific DNA sequences in the upstream promoter elements, or they may bind to enhancers (see Section B below).
3. The basic idea is to increase the local concentration of the general transcription factors so the initiation complex can be assembled more readily. The fact that the activators are bound to DNA that is close to the target (or becomes close because of looping of the DNA) means that the local concentration of that protein is high, and hence it can boost the local concentration of the interacting general transcription factors.
3. Stalled polymerases
RNA polymerase will transcribe about 20 to 40 nucleotides at the start of some genes and then stall at a pause site. The classic example are heat-shock genes in Drosophila, but other cases are also known. These genes are activated by release of stalled polymerases to elongate. In the case of the heat shock genes, this requires heat shock transcription factor (HSTF). The mechanism is still under study; some interesting ideas are:
1. Phosphorylation of the CTD of the large subunit of RNA polymerase II causes release to elongation ("promoter clearance"). One candidate (but not the only one) for the CTD kinase is TFIIH.
2. Addition of a processivity factor (analogous to E. coliNus A?), maybe TFIIS.
C. Enhancers
1. Enhancers are cis-acting regulatory sequences that increase level of expression of a gene, but they operate independentlyof position and orientation. These last two operational criteria distinguish enhancers from promoters.
2. Examples
a. SV40 control region
1. SV40 (simian virus 40) infects monkey kidney cells, and it will also cause transformation of rodent cells. It has a double stranded DNA genome of about 5 kb. Because of its involvement in tumorigenesis, it has been a favorite subject of molecular virologists. The early region encodes tumor antigens (T-Ag and t-Ag) with many functions, including stimulating DNA replication of SV40 and blocking the action of endogenous tumor suppressors like p53 (the 1993 "Molecule of the Year"). The late region encodes three capsid proteins called VP1, VP2 and VP3 (viral protein n). A region between the early and late genes controls both replication and transcription of both classes of genes.
2. The control region has an origin of replication with binding sites for T-Ag.
1. Wild type SV40 expresses T-Ag upon infection of monkey cells and lyses infected cells. However, a viral strain lacking the 72 bp repeats shows a highly reduced level of T-Ag and rarely lyses infected cells.
2. If the 72 bp repeats are added back to the mutant SV40 genome, except they are placed between the ends of the early and late genes (180° from their wild-type position), T-Ag is expressed at a high level and one obtains productive infections.
3. If the orientation of the 72 bp repeats is reversed, one still gets high level expression of viral genes and productive infection. In fact, it is needed for expression of the late genes in the wild-type, which are transcribed in the opposite direction from the early genes.
4. One concludes that the enhancer is needed for efficient transcription of the target promoters, but it can act in either orientation and at a variety of different positions and distances from the targets.
5. Work done virtually concurrently with that described above showed that the 72 bp repeats work on other "heterologous" genes, so that, for example b-globin genes could be expressed in nonerythroid cells. In fact this was one of the key observations in the discovery of the enhancer.
6. One copy of the 72 bp region will work as an enhancer, but two copies work better.
b. Immunoglobulin genes
1. This was the first enhancer of a cellular gene discovered. Researchers noted that a region of the intron was exceptionally well conserved among human, rabbit and mouse sequences, and subsequent deletion experiments showed that the intronic enhancer was required for expression.
2. After rearrangement of the immunoglobulin gene to fuse VDJ regions, one is left with a large intron between this combined variable region gene and the constant region. An enhancer is found in that intron, and another enhancer is found 3' to the polyA addition site.
(3) The enhancers have multiple binding sites for transcriptional regulatory proteins
(a) Several of these sites are named for the enhancer they were discovered in. E.g. mE1, mE2, etc. are binding sites for enhancer proteins identified in the gene for the immunoglobulin heavy chain m (mu).
The protein YY1 (ying yang 1) binds to the mE1 site (CCAT is the core of the consensus) and bends DNA there.
The octamer site (ATTTGCAT) is bound by two related proteins. Oct1 is found in all tissues examined, whereas Oct2 is lymphoid specific - the first example of a tissue-specific transcription factor. Transcriptional activators that do not have their own DNA binding sequence, like VP16 from Herpes virus, will bind to Oct proteins, which bind to DNA, and the complex can activate transcription.
(b) Some proteins will bind to sites both in the promoter and the enhancer, e.g. Oct proteins. Remember Oct1 also acts at the SV40 enhancer.
c. Summary
1. The position of the enhancer can be virtually anywhere relative to the gene, but the promoter is always at the 5' end.
2. Examples are known of enhancers 5' to the gene (upstream), adjacent to the promoter (like in SV40), downstream from the gene (some globin genes), within the gene (immunoglobulins) or far upstream within a locus control region (globin genes, see Chapter 20.)
3. Multiple binding sites for transcriptional activators
a. All enhancers characterized thus far have multiple binding sites for activator proteins.
b. Multiples of binding sites are needed for function of the enhancer.
1. In experiments with the SV40 enhancer, it was noted that some mutations that decreased the infectivity of the virus caused a mutation of one of the domains of the enhancer, e.g. domain A. When these mutants were then selected for pseudo-revertants to wild-type, with infectivity largely restored, it was found that the pseudo-revertants had duplicated one of the remaining domains. Subsequently, multimers of the various protein-binding sites were shown to be active, but monomers had little activity.
2. The domain (e.g. A, C and B in the SV40 enhancer) with at least two binding sites is called an enhanson. Multiple enhansons make up an enhancer.
The two-hybrid screening method is a rapid and sensitive way to test a large group of proteins for their ability to interact in vivowith a particular protein. For example, one component of a regulatory complex may be characterized and a cDNA available. This cDNA for the “bait” protein is fused to a DNA segments encoding a well-known DNA binding domain, such as that of LexA, which binds to lex o. When introduced into yeast cells with the lacZgene (encoding beta-galactosidase) under control of lex o, the lacZgene is not expressed because the hybrid bait protein has no activation domain. A library of cDNAs to be tested are fused to the DNA encoding the activation domain of GAL2. When these are transformed into yeast cells carrying the hybrid LexA_DBD-bait and the lex o - lacZ reporter, only the hybrid proteins that interact with the bait will stimulate expression of lacZ. Transformed cells that are positive in this assay are carrying a plasmid with a hybrid gene with the cDNA encoding a protein (the “trap”) that interacts with the protein of interest (bait).
D. DNA binding domains
Computer-assisted three-dimensional views of several transcription factors, illustrating many of the domains described here, can be viewed as Chime tutorials at
• www.bmb.psu.edu/pugh/514/mdls
• www.clunet.edu/BioDev/OMM/cro/cromast.htm
1. Helix-turn-helix, homeodomain
1. The sequence of the "homeodomain" forms three helices separated by tight turns.
2. Helix three occupies the major groove at the binding site on the DNA. It is the recognition helix, forming specific interactions (H-bonds and hydrophobic interactions) with the edges of the base pairs in the major groove.
3. Helices one and two are perpendicular to and above helix three, providing alignment with the phosphodiester backbone. The N-terminal tail of helix interacts with the minor groove of the DNA on the opposite face of the DNA.
4. Helix two + helix three is comparable to the helix-turn-helix motif first identified in the l Cro and repressor system.
(5) Examples
(a) Homeotic genes and their relatives.
All these are involved in regulating early developmental events in Drosophila. They are transcription factors (regulating the genes that determine the next developmental fate), and they have this same protein motif for their DNA binding domains.
Some specific examples are the products of these genes:
• the pair-rule gene eve= even skipped
• the segment polarity gene en= engrailed
• the homeotic gene Antp= antennapedia
(d) In a protein with 3 adjacent Zn fingers, e.g. Sp1 (remember this protein from the SV40 early promoter), each finger binds in the major groove to contact three adjacent base pairs. For the high affinity binding site, one finger contacts GGG, the next finger contacts GCG, and the remaining finger contacts GGG. So the three fingers curve along to contact the major groove for most of one turn of the helix.
(e) Members of this class of Zn finger proteins have multiple fingers, usually in a tandem array. Examples include TFIIIA (the motif was discovered in this protein) with 9 fingers, a CAC-binding protein (related to some extent to Sp1) with 3 fingers, and Drosophila ADR1 with 2 fingers.
(2)Cys2Cys2
(a) Consensus sequence:
Cys-X2-Cys-X1-3-Cys-X2-Cys
(b)Forms a distinctly different structure from the Cys2His2 Zn fingers.
1. Note that the number of amino acids between the 2 "halves" of the finger (1 to 3 in this case) is much less than the 12 that separate the two halves of a Cys2His2 Zn finger.
2. The Cys2Cys2 fingers are not interchangable with Cys2His2 Zn fingers in domain swap experiments.
3. The proteins do not have extensive repetitions of the motif, in contrast to proteins with Cys2His2 Zn fingers.
(3) Examples include heterodimers that can exchange partners
(a) MyoD is a key protein in committment of mesodermal tissues to muscle differentiation. Other relatives, such as myogenin and myf5, are equally important and provide redundant functions. All are muscle-specific and have a similar binding domain. MyoD is active when it has E12 or E47 as its heterodimeric partner; when active it will stimulate transcription of muscle specific genes such as the one encoding creatine kinase. E12 and E47 were initially discovered as proteins that bound to enhancers of immunoglobulin genes, but are found in virtually all cell-types. Another protein, called Id, can also bind to E12 or E47 by its HLH domain. However, Id lacks a basic domain, so heterodimers with Id are not active. So the activity of bHLH proteins can be regulated by exchange of partners.
(b)A developing theme is that one of partners of a bHLH heterodimer is ubiquitous (e.g. E12, E47 in mammals, da = daughterless in Drosophila) and the other is tissue-specific (MyoD or AC-S = achaete-scute, a regulator of neurogenesis in Drosophila). The ubiquitous components may be involved in regulating a variety of other tissue-specific proteins with bHLH domains.
(c) Myc, one of many regulators of the cell cycle, is a bHLH protein. It forms partners with Max, and it is possible that this is important in regulation of the cell cycle.
E. Transcriptional activation domains
1. Acidic
This domain has been postulated to be an "acid blob" or an amphipathic helix with acidic residues on one face. Recent physico-chemical studies of GAL4 have shown b-sheet structure. At this point no single structure has been established. Examples:
GAL4 protein, VP16, GCN4, glucocorticoid hormone receptor, AP1, and the l repressor (activation of PRM).
2. Gln-rich
This domain is rich in glutamine, as its name implies. Examples of proteins containing the domain are Sp1, Antp, Oct1 and Oct2
3. Pro-rich
Again, the domain is rich in proline. Examples include CTF/NF1 (involved in regulation of replication as nuclear factor 1, and proposed to be one of many proteins binding to CCAAT motifs).
4. Work so far has not established well-defined secondary or tertiary structures for these domains.
One possibility is that the activation domains assume their proper structure after binding to its target, i.e. an induced fit model.
Table 4.5.1. Selected eukarytoic transcription factors and their properties
Name
System
Binding site (top strand)
Quaternary structure
DNA binding domain
Activation domain
Other comments
Engrailed
early development
homeodomain
Sp1
SV40, cellular housekeeping genes
GGGGCGGGG
monomer
3 Zn fingers
Cys2His2
Gln-rich
phosphoprotein
AP1
SV40, cellular enhancers
TGASTCA
heterodimer, Jun-Fos, Jun2, others
basic region + Leu zipper
acidic
regulated by phosphorylation
Oct1
lymphoid and other genes
ATTTGCAT
monomer, but can bind VP16
POU domain + homeodomain (HTH)
Gln-rich, also binds VP16
Oct1 is ubiquitous, Oct2 is lymphoid specific
GAL4
yeast galactose regulon
CGGASGACWGTCSTCCG
homodimer
Zn2Cys6, binuclear cluster
acidic
Glucocorticoid receptor
glucocorticoid responsive genes
TGGTACAAATGTTCT
cytoplasm: with "heat shock" proteins;
nucleus: homodimer
2 Zn fingers,
Cys2Cys2
close to Zn finger
binding of hormone ligand changes conformation, move to nucleus and activate genes
MyoD
determination of myogenesis
CAGCTG
heterodimer with E12/E47: active;
heterodimer with ID: inactive
basic-helix-loop-helix
switch partners to activate or inactivate
HMG(I)Y
interferon gene and others
minor groove
monomer (?)
bends DNA to provide favorable interactions of other proteins
VP16
Herpes simplex virus
not bind tightly to DNA
binds to proteins like Oct1
acidic activation domain; very potent
binds to other proteins that themselves bind specifically to DNA
a. In looping models, the activators bound to the enhancer are brought in close proximity to their targets at the promoter by forming loops in the DNA.
1. The activators can make direct contact with their target (perhaps the pre-initiation complex), or they may operate through an intermediary called a co-activatoror mediator.
2. If a loop is formed, in principle it does not matter how large the loop is or if the activator binding site is 5' or 3' to the target. This could explain the ability of enhancers to operate independently of position.
c. The looping model is favored at this time. However, it has been difficult to design experiments that definitely rule out tracking. Several observations show that DNA can form loops in vitro, allowing contact between proteins at the enhancer and those at the promoter. For instance:
1. Using electron microscopy, one can visualize loops of DNA held together by interactions between enhancer-bound activator proteins and proteins bound to the promoter.
2. The biochemical approaches show that the activation domains of transcription factors canbind to components of the pre-initiation complex, such as TFIID (see Section H).
b. E.g. the enhancer for the interferon-bgene, which is located just upstream from the promoter, has binding sites for three dimeric "conventional" transcription factors: NFKB (p50 + p65), IRF, and a heterodimer of ATF2 + Jun (a relative of AP1). In addition, there are three specific binding sites for HMGI(Y).
1. HMGI(Y) is a member of the "high mobility group" of nonhistone chromosomal proteins. Most HMG proteins are abundant in the nucleus, albeit not as abundant as histones.
2. HMGI(Y) binds in the minor groove of DNA and bends the DNA.
3. It also makes specific protein-protein contacts with IRF, ATF2 and NFkB, even in the absence of DNA.
4. By bending the DNA at precise positions by a defined amount, and by aiding the binding of other proteins, HMGI(Y) seems to play a critical role in assembly of the enhancer complex in juxtaposition with the promoter.
5. In general, proteins that bend DNA can be the agents that cause the looping to bring the enhancer-binding proteins in proximity to their targets.
c. Other proteins that bend DNA
cAMP-CAP (recall this from catabolite repression in E. coli), IHF = integration host factor (required for integration of l DNA to form a prophage, via a large complex called an intasome), and YY1 (ying yang 1) which has either negative or positive effects on a large variety of genes in mammals.
19: Transcriptional regulation in eukaryotes
19.1 (POB) Specific DNA binding by regulatory proteins.
A typical prokaryotic repressor protein discriminates between its specific DNA binding site (operator) and nonspecific DNA by a factor of 105 to 106. About ten molecules of the repressor per cell are sufficient to ensure a high level of repression. Assume that a very similar repressor existed in a human cell and had a similar specificity for its binding site. How many copies of the repressor would be required per cell to elicit a level of repression similar to that seen in the prokaryotic cell? (Hint: The E. coligenome contains about 4.7 million base pairs and the human haploid genome contains about 2.4 billion base pairs).
Use the following information for the next 3 problems. Let's imagine that part of the regulation of expression of the OB gene is mediated by a protein we will call OBF1. There is one binding site for OBF1 in the OB gene, and let's assume that is the only specific binding site in the haploid genome, or 2 specific sites in a diploid genome. The haploid human genome has about 3 x 109 bp, or 6 x 109 bp in a diploid genome. If we assume that about 33.3% of the nuclear DNA is in an accessible chromatin conformation, that means that about 2 x 109 bp of DNA are available to bind OBF1 nonspecifically.
19.2 The diameter of a mammalian nucleus is about 10 mm. If you model a nucleus as a sphere, what is its volume? What is the molar concentration of specific and nonspecific binding sites in the nucleus?
Binding of OBF1 to a specific site and to nonspecific sites is described by the following equations.
Let P = OBF1
Ds = a specific binding site in DNA
Dns = a nonspecific binding site in the genomic DNA
P + Ds
Ks = = 1011 M-1 (eqn 2)
Kns = = 105 M-1 (eqn 3)
19.3 What fraction of the OBF1 (or P in the equations) is not bound to either specific or nonspecific sites in the DNA?
19.4 How many molecules of OBF1 are needed per nucleus to maintain 90% occupancy of the specific sites? This condition means
= 9
Use the following information for the next seven questions.
The agoutigene in mice controls the amount and distribution of pigments within coat hairs. Some mutations of this gene also lead to adult-onset obesity, a mild diabetes-like syndrome, tumor susceptibility and recessive embryonic lethality. The gene encodes a predicted protein of 131 amino acids that has the structural features of a secreted protein, but no striking homology to other known proteins has been recognized. This protein is likely to be a regulator of melanin pigment synthesis, and it may also be a more general metabolic regulator.
Let's suppose that you are investigating the regulation of the agoutigene, and have the capacity to transfect a melanocyte cell line, which transcribes the wild-type agouti gene, and an adipocyte cell line, which transcribes the wild-type agoutigene only at a very low level. Further, you already know that the basal promoter is in a DNA segment located between -100 and +50. You make progressive 5' deletions of a fragment that includes -300 to +50, link it to a luciferase reporter gene, and transfect the constructs into melanocyte and adipocyte cells, with the following results.
19.5. What do you conclude about the region between -250 and -200?
19.6. What do you conclude about the region between -200 and -150?
19.7. What do you conclude about the region between -150 and -100?
You also investigate the binding of nuclear proteins to these DNA segments located upstream of the agoutigene. Extracts containing nuclear proteins from melanocytes were tested for the ability to bind to the fragments delineated in the deletion series above.
The fragment from -150 to -100 was used as the labeled probe in a mobility shift assay. The mobility of the free probe is shown in lane 1, and the pattern after binding to melanocyte nuclear extract is shown in lane 2. Lanes 3-14 show the mobility shifts after addition of the competitors to the binding reaction; the triangle above the lanes indicates that an increasing amount of competitor is used in successive lanes. "Self" is the same -150 to -100 fragment that is used as a probe, but it is unlabeled and present in an excess over the labeled probe (lanes 3-5). A completely different DNA (sheared E. coli DNA) was used as a nonspecific competitor (lanes 6-8). Two different duplex oligonucleotides, one containing the binding site for AP1 (lanes 9-11) and the other containing the binding site for Sp1 (lanes 12-14) were also tested. Thinner, less densely filled boxes denote bands of less intensity than the darker, thicker bands. Use these results to answer the next two questions.
19.8. What do you conclude from these data?
19.9. What sequence within the -150 to -100 segment might you expect to be bound in melanocyte nuclei?
19.10. The fragment from -200 to -150 was also used as a labeled probe in a mobility shift assay similar to that described for the -150 to -100 segment, as shown below.
What do you conclude from these data?
19.11. Some mutant alleles of the agouti gene are expressed ectopically (i.e. in the wrong tissue). Just using the information on the 5' deletions above, what region is a likely candidate for the position of a loss-of-function mutation that leads to ectopic expression in adipose tissue?
19.12 (POB) Functional domains in regulatory proteins.
A biochemist replaces the DNA‑binding domain of the yeast GAL4 protein with the DNA‑binding domain from the lambda repressor (CI) and finds that the engineered protein no longer functions as a transcriptional activator (it no longer regulates transcription of the GALoperon in yeast). What might be done to the GAL4 binding site in the DNA to make the engineered protein functional in activating GALoperon transcription?
19.13 What is the DNA-binding domain of the transcription factor Sp1?
19.14 What is the dimerization domain of the transcription factor AP1?
19.15 (ASC) Describe three mechanisms for regulating the activity of transcription factors.
19.16 (ASC) You have constructed a plasmid set containing a series of nucleotide insertions spaced along the length of the glucocorticoid-receptor gene. Each insertion encodes three or four amino acids. The map positions of the various insertions in the coding sequence of the receptor gene is as follows:
0 Glucocorticoid-receptor coding sequence 783
| |
| |
| |
Insertion: A B C D E F G H I J K L M N O P Q R S
The plasmids containing the receptor gene can be functionally expressed in CV-1 and COS cells, which contain a steroid-responsive gene. Using these cells, you determine the effect of each of these insertions in the receptor on the induction of the steroid-responsive gene and on binding of the synthetic steroid dexamethasone. The results of these analyses are summarized in the tablebelow.
Insertion Induction Dexamethasone binding
A ++++ ++++
B ++++ ++++
C ++++ ++++
D 0 ++++
E 0 ++++
F 0 ++++
G ++++ ++++
H ++++ ++++
I + ++++
J ++++ ++++
K 0 ++++
L 0 ++++
M 0 ++++
N + ++++
O ++++ ++++
P ++++ ++++
Q 0 0
R 0 0
S 0 0
wild-type ++++ ++++
a) From this analysis, how many different functional domains does the glucocorticoid receptor have? Indicate the position of these domains relative to the insertion map.
b) Which domain is the steroid-binding domain?
c) How could you determine which of the domains is the DNA-binding domain? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_IV%3A_Regulation_of_Gene_Expression/19%3A_Transcriptional_regulation_in_eukaryotes/19.E%3A_Transcriptional_regulation_in_eukaryotes_%28Exercises%29.txt |
Nucleosome composition
• Nucleosomes are the repeating subunit of chromatin.
• Nucleosomes are composed of a nucleosome core, histone H1 (in higher eukaryotes) and variable length linker DNA (0-50bp).
• The nucleosome core contains an octamer of 2 each of the core histones (H2A, H2B, H3 and H4) and 146 bp of DNA wrapped 1.75 turns (Figure 4.6.1).
Histone interactions in the nucleosome
• Core histones dimerize through their histone fold motifs generating H3/H4 dimers and H2A H2B dimers (Figure 4.6.2.).
• Two H3/H4 dimers associate to form a tetramer, which binds DNA.
• Two H2A/H2B dimers associate with the tetramer to form the histone octamer.
• At physiological salt the octamer is not stable unless bound to DNA and dissociates into the H3/H4 tetramer and two H2A/H2B dimers.
Chromatin higher order structure
• Arrays of nucleosomes condense into higher order chromatin fibers (Figure 4.6.3.).
• Despite over 2 decades of investigation the structure of the “30nm” chromatin fiber is not known.
• This may be due to irregularity or instability of the structure.
• This level of structure has been implicated in mechanisms of chromatin repression; thus, the lack of structural information at this level is particularly troublesome
Biochemical investigation of different states of chromatin and gene activity in cells
Proposed sequence for gene activation
1. Open a chromatin domain
Relocate away from pericentromeric heterochromatin
Establish a locus-wide open chromatin configuration
General histone hyperacetylation
DNase I sensitivity
2. Activate transcription
Local hyperacetylation of histone H3
Promoter activation to initiate and elongate transcription
Summary of cis-regulatory elements that act in chromatin
Generate an open, accessible chromatin structure
Can extend over about hundreds of kb
Can be tissue specific
Enhance expression of individual genes
Can be tissue specific
Can function at specific stages of development.
Insulate genes from position effects.
Enhancer blocking assay
How is the structure of chromatin modified in cells to change transcriptional activity?
Competition vs. Replacement models for how transcription factors occupy their binding sites on a chromatin template.
The HAT complexes could be involved in other processes, or can affect them indirectly through their effects on transcription. For instance, one component of the SAGA HAT complex is Tra1, the yeast homolog of a human protein involved in cellular transformation. It may be a direct target of activator proteins.
Multiple nuclear HATs are found in yeast and in other species (Table 4.6.2). They are all large with many subunits. By comparison, their substrate, which is the nucleosome, is 0.2 MDa in mass. They have different substrate specificities. Some acetylated H3 preferentially, others acetylate H4. The reason for the diversity of HATs is a matter of current study.
Table 4.6.2.The four major nuclear HAT complexes in yeast
Complex
Mass (MDa=megadaltons)
SAGA
1.8
NuA4
1.4
ADA
0.8
NuA3
0.5
20: Transcriptional regulation via chromatin alterations
Use the following information to answer the next two questions.
DNase hypersensitive sites around a gene were mapped by treating nuclei from cells that express that gene with increasing amounts of DNaseI. The partially digested DNA was isolated, cut to completion with a restriction enzyme, and analyzed by Southern blot-hybridization using a radioactive probe that is located 3' to the gene. Cleavage of genomic DNA with the restriction enzyme generates an 8 kb fragment that contains the gene, and the probe for the blot hybridization is located at the right end of the fragment (left to right defined as the direction of transcription of the gene). The results of this indirect end-labeling assay shows a gradual fade-out of the 8 kb fragment with increasing [DNaseI], and the appearance of a new band at 6 kb with DNaseI treatment.
20.1 Where is the DNase I hypersensitive site?
20.2 If the start site for transcription is 5 kb from the right end of the restriction fragment, what is a likely possibility for the function of the region mapped by the DNase hypersensitive site?
For the next three questions, consider the following information about a protein called Gcn5p. [This problem is based on Brownell et al. (1996) Cell 84: 843-851.]
[1] Gcn5p is needed for activation of some, but not all, genes in yeast.
[2] Gcn5p does not bind with high affinity to any particular site on DNA.
[3] Gcn5p will interact with acidic transcriptional activators.
[4] When incubated with histones and the following substrates, Gcn5p will have the designated effects. A + in the column under "Effect" means that the histones move slower than unmodified histones on a polyacrylamide gel that separates on the basis of charge, with the histones moving toward the negatively charged electrode. A - means that the treatment has no effect on the histones. S-adenosylmethionine is a substrate for some methyl transfer reactions, and NADH is the substrate for ADPribosyl-transferases.
Mixture Effect
Gcn5p + histones -
Gcn5p + histones + ATP -
Gcn5p + histones + S-adenosylmethionine -
Gcn5p + histones + acetyl-coenzyme A +
Gcn5p + histones + NADH -
20.3 What conclusion is consistent with these observations?
20.4 What enzymatic activity is associated with Gcn5p?
20.5 Which step in the gene expression pathway is likely to be regulated by Gcn5p?
20.6 What functions have been ascribed to the locus control region of mammalian beta-globin genes?
20.7 Use the following information to answer the next 6 parts (a-f) of this question. The regulatory scheme is imaginary but illustrative of some of the models we have discussed.
The protein surfactin is produced in the lung to provide surface area for efficient gas exchange in the alveoli. Let's suppose that expression of the surfactin gene is induced in lung cells by a new polypeptide hormone called pulmonin. Induction by pulmonin requires a particular DNA sequence upstream of the surfactin gene; this is called PRE for pulmonin response element. Proteins that bind specifically to that site were isolated, and the most highly purified fraction that bound to the PRE contained two polypeptides. A cDNA clone was isolated that encoded one of the polypeptides called NFL2. Antisera that specifically recognizes NFL2 is available.
The mechanism of the induction by pulmonin was investigated by testing various cell fractions (nuclear or cytoplasmic) from uninduced or pulmonin‑induced lung cells in two assays. The presence or absence of NFL2 polypeptide was determined by reacting with the anti‑NFL2 antisera, and the ability to bind to the PRE DNA sequence was tested by an electrophoretic mobility shift assay. In a further series of experiments, the NFL2 polypeptide was synthesized in vitroby transcribing the cDNA clone and translating that artificial mRNA. The product has the same amino acid sequence as the native polypeptide and is referred to below as "expressed cDNA." The expressed cDNA (which is the polypeptide synthesized in vitro) was tested in the same assays, before and after treatment with the cytoplasmic and nuclear extracts and also with a protein kinase that will phosphorylate the expressed cDNA on a specific serine.
Line
Source of protein and Type of treatment
React with anti‑NFL2
Bind to PRE DNA
1
Uninduced cell cytoplasmic extract = unind. CE
+
2
Uninduced cell nuclear extract = unind. NE
3
Induced cell cytoplasmic extract = ind. CE
4
Induced cell nuclear extract = ind. NE
+
+
5
Induced cell nuclear extract + phosphatase
+
6
Expressed cDNA
+
7
Expressed cDNA + ind. CE
+
8
Expressed cDNA + unind. NE
+
9
Expressed cDNA + ind. CE + unind. NE
+
+
10
Expressed cDNA + unind. CE + unind. NE
+
11
Expressed cDNA + protein kinase + ATP
+
12
Expressed cDNA + protein kinase + ATP + unind. NE
+
+
13
Expressed cDNA + protein kinase + ATP + ind. CE
+
Based on these data, an affinity column was made with the expressed NFL2 cDNA as the ligand and used to test binding of proteins from nuclear extracts. When the column was pretreated with protein kinase + ATP (so that NFL2 was phosphorylated), a ubiquitous nuclear protein called UBF3 was bound from nuclear extracts from both induced and uninduced cells. If the NFL2 ligand was not phosphorylated, no binding of nuclear proteins was observed.
To confirm that NFL2 really was part of the protein complex on PRE, antibodies against NFL2 were shown to react with this protein‑DNA complex. Furthermore, antibodies against phosphoserine, but not antibodies against phosphotyrosine, reacted with the specific PRE‑protein complex.
Answer questions a to f based on the above observations.
a) Where is the NFL2 polypeptide? (Use data in lines 1‑5.)
b) Where is the activity that will bind to the PRE site in DNA? (Use data in lines 1‑5.)
c) From the data in lines 6‑13, what must happen to the in vitrosynthesized NFL2 (the expressed cDNA) in order to bind to the PRE site?
d) What proteins and covalent modifications of them are required to bind to the PRE site?
e) Which cell compartment has the protein kinase that acts on NFL2?
f) What model for pulmonin induction of the surfactin gene best fits the data given? | textbooks/bio/Genetics/Working_with_Molecular_Genetics_(Hardison)/Unit_IV%3A_Regulation_of_Gene_Expression/20%3A_Transcriptional_regulation_via_chromatin_alterations/20.E%3A_Transcriptional_regulation_via_chromatin_alterations_%28Exercis.txt |
WHAT YOU’LL LEARN TO DO: Describe and identify anatomical position and locate major organs.
Learning Objectives
• Identify anatomical position and be able to reference it when describing anatomical locations.
• Know where to cut to create each of the following and be able to recognize the views created by a cut in each of the following:
• mid-sagittal plane
• para-sagittal plane
• frontal plane
• transverse plane
• Be able to look up anatomical nouns and adjectives for external body areas and find what they refer to.
• Apply each pair of terms to locate a structure or direct someone to a structure:
• superior/inferior
• anterior/posterior
• medial/lateral
• proximal/distal
• superficial/deep
• dorsal/ventral
• cephalad/caudal
• supine/prone
• Locate each of the following:
• dorsal body cavity
• cranial cavity
• spinal cavity
• ventral body cavity
• thoracic body cavity
• diaphragm
• abdominopelvic cavity
• abdominal cavity
• pelvic cavity
• List the major organs and identify at least two physiological roles for each of the 11 human organ systems:
• integumentary
• skeletal
• muscular
• nervous
• endocrine
• cardiovascular
• lymphatic/immune
• respiratory
• digestive
• urinary
• reproductive
• Identify each organ in the anatomical model and know its location within the model body:
• cranial cavity
• palate
• parotid salivary gland
• sublingual salivary gland
• submandibular salivary gland
• larynx
• trachea
• esophagus
• aorta left
• lung (2 lobes)
• right lung (3 lobes)
• diaphragm
• kidney
• adrenal (suprarenal) glands
• liver
• gall bladder
• pancreas
• stomach
• spleen
• large intestine
• small intestine
• appendix
• ovaries
• uterus
• female urethra
• testes
• scrotum
01: Overview and the Microscope
Complete the following lab activities and exercises:
• Reading: Anatomical Position and Planes
• Exercises 1.1
• Reading: Anatomical Orientation and Directions
• Exercises 1.2
• Reading: The Human Body Cavities
• Exercises 1.3
• Reading: The Human Organ Systems
• Exercises 1.4
• Reading: Parts of a Compound Microscope and How To Handle Them Correctly
• Exercises 1.5, 1.6 and 1.7
• Reading: How The Virtual Image Differs From The Real Image
• Exercise 1.8
• Making Simple But Accurate Line Drawings of Magnified Specimens
• Exercise 1.9
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1.02: Anatomical Position and Planes
Information
When anatomists or health professionals identify the location of a structure in the human body, they do so in reference to a body in anatomical position. That is, they figure out the location based on the assumption that the body is starting out in anatomical position.
Anatomical position for a human is when the human stands up, faces forward, has arms extended, and has palms facing out.
When referencing a structure that is on one side of the body or the other, we use the terms “anatomical right” and “anatomical left.” Anatomical right means that the structure is on the side that a person in anatomical position would consider their right-hand side (not necessarily on the right of the viewer) and anatomical left means that the structure is the side that a person in anatomical position would consider their left-hand side (which likewise is not necessarily the left side of the viewer.)
Anatomical planes
Information
To view the interior of a body, we expose the organs and structures that are visible when that body is cut open along one of four commonly used sectional planes. These planes are the different directions a body is cut to reveal different views of its internal structures.
• Frontal plane—A vertical cut that separates the front from the back of the specimen. Also known as a coronal plane.
• Transverse plane—A horizontal cut that separates the top from the bottom of the specimen. Also known as a cross-sectional plane.
• Midsagittal plane—A vertical cut down the exact center line of the specimen that separates the left half from the right half.
• Parasagittal plane—A vertical cut that is off-center that separates the left of the specimen from the right in unequal portions. It does not matter whether it is the left side or the right side that is larger, as long as they are not equal.
Anatomical Vocabulary
Anatomical nouns and adjective for external body parts
Information
Like all areas of science, there is a lot of jargon associated with anatomy and physiology. Often terms are used within the field that differ from what we would name things in everyday conversation. Such jargon usually allows the specialist in the field to be more precise in what exactly they are referring to, but the jargon also can be intimidating and exclusionary. If you don’t know it, you are not in the club.
LAB 1 EXERCISE \(1\)
Here are a bunch of anatomical adjectives (followed in parentheses by the noun version of the same term). For each, use your smart phone or laptop or whatever is most convenient to you to find what body part the term refers to. (Shortcut hint: the Google search engine will return definitions for words if you type “define: word” in the search box, leaving out the quotation marks.)
Write down the body part or body region next to each term. Use Figure 1.4 to help you make sure you have the correct definition but look up each definition to make sure you are being accurate.
1. Find the body part or region indicated by each of the following terms. Use everyday language to describe the part or region. (Forearm, belly, etc.)
Abdominal (abdomen)
Acromial (acromion)
Antebrachial
(antebrachium)
Antecubital
(antecubitis)
Auricle (auris)
Axillary (axilla)
Brachial (brachium)
Buccal (bucca)
Carpal (carpus)
Cephalic (cephalus)
Cervical (cervicis)
Coxal (coxa)
Cranial (cranium)
Crural (crus)
Digital (digit)
Dorsal (dorsa)
Facial (facies)
Femoral (femur)
Frontal (frons)
Gluteal (gluteus)
Inguinal (inguen)
Lumbar (lumbus)
Mammary (mamma)
Manual (manus)
Mental (mentis)
Nasal (nasus)
Olecranal (olecranon)
Oral (oris)
Ocular (oculus)
Palmar (palma)
Patellar (patella)
Pelvic (pelvis)
Plantar (planta)
Popliteal (popliteus)
Pubic (pubis)
Sacrum (sacral)
Sural (sura)
Tarsal (tarsus)
Thoracic (thorax)
Umbilical (umbilicus)
Anatomical Orientation and Directions
Information
To be able to direct others to specific anatomical structures, or to find structures based on someone else’s directions, it is useful to have specific pairs of terms that allow you to orient your search with respect to the location of another, known structures. The following pairs of terms are used to make comparisons. Each term is used to orient a first structure or feature with respect to the position of a second structure or feature.
• Superior/Inferior–Equivalent to above and below when moving along the long axis of a body in anatomical position. The structure that is superior to another is above the second structure when the body is in anatomical position. A feature that is inferior to another is below the second feature when the body is in anatomical position.
• Proximal/Distal–Equivalent to near and far. Usually used to orient the positions of structures and features along the limbs with respect to the trunk of the body. A feature that is proximal to something else is closer to the limb’s point of attachment to the trunk. A structure that is distal to something else is farther away from the limb’s point of attachment. Less precisely but still occasionally used in the trunk of the body itself to indicate whether something is closer to (proximal) or farther away from (distal) something else.
• Medial/Lateral–Equivalent to towards the middle or towards the edge. Used with respect to the midline of the trunk of a body in anatomical position. A structure that medial to another is closer to the midline of the body’s trunk. A feature that is lateral to another is farther away from the midline of the trunk.
• Anterior/Posterior–Equivalent to the front and back of a body in anatomical position. A structure that is anterior to another is closer to the front of the body when the body is in anatomical position. A feature that is posterior to another is closer to the back of the body when the body is in anatomical position.
• Ventral/Dorsal–Equivalent to belly-side and back-side of a body in anatomical position. For a human in anatomical position, this pair of terms is equivalent to anterior and posterior. However, for four-legged animals in what is considered their anatomical position, the belly-side is not equivalent to the front of the animal. A structure that is ventral to another is closer to the belly-side of the body. A feature that is dorsal to another is closer to the back of the body.
• Superficial/Deep–Equivalent to closer to the surface and farther from the surface.
• Cephalic/Caudal–Equivalent to closer to the head and closer to the tail. This is more useful for four-legged animals with tails than for upright humans with only a vestigial tail.
LAB 1 EXERCISE \(2\)
1. Fill in the blank with the appropriate directional term to complete the following sentences. More than one answer may be correct.
• The heart is to the lungs.
• The knee is to the hip.
• The wrist is to the hand.
• The mouth is to the nose.
• The thorax is to the abdomen.
• The thumb is to the ring finger.
• The sternum is to the heart.
• The skull is to the scalp.
• The ears are to the nose.
• Dorsal refers to the of the human body, while ventral refers to the of the human body.
2. Find the indicated structures in the diagrams provided, based on the directional terms given. The structure to find will be one of those at the end of an unlabeled line.
A. Label the extensor digitorum (ED) muscle in the figure below. It is:
• Distal to the anconeus muscle
• Lateral to the extensor digiti minimi muscle
• Superficial to the Extensor pollicis brevis muscle
B. Label the Incus in the figure below. It is:
• Superior to the lateral end of the cochlear nerve
• Medial to the malleus
• Lateral to the stapes
3. Using your knowledge of the different body planes shown in Figure 1-2 (shown again below), fill in the blanks with the appropriate body plane for each of the following descriptions.
1. The plane that divides the body into anterior and posterior parts is the
plane.
2. A transverse plane divides the body into and
regions.
3. A or plane divides the body into right and left parts.
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Figure \(7\).. Anatomy of the human ear.. Authored by: Chittka L, Brockmann. Provided by: Wikipedia. Located at: https://commons.wikimedia.org/wiki/F...man_Ear_en.svg. License: CC BY-SA: Attribution- ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/01%3A_Overview_and_the_Microscope/1.01%3A_Learning_Objectives_and_Activities.txt |
Information
The major cavities of the human body are the spaces left over when internal organs are removed. There are additional body cavities which we will only discuss in lecture. These are the cavities created by serous membranes–the pleural cavities, the pericardial cavity, and the peritoneal cavity–and the mediastinum.
• Dorsal body cavity–the cranial cavity and the spinal cavity in combination.
• Cranial cavity–the space occupied by the brain, enclosed by the skull bones.
• Spinal cavity–the space occupied by the spinal cord enclosed by the vertebrae column making up the backbone. The spinal cavity is continuous with the cranial cavity.
• Ventral body cavity–the thoracic cavity, the abdominal cavity, and the pelvic cavity in combination.
• Thoracic cavity–the space occupied by the ventral internal organs superior to the diaphragm.
• Abdominopelvic cavity–the abdominal cavity and the pelvic cavity in combination.
• Abdominal cavity–the space occupied by the ventral internal organs inferior to the diaphragm and superior to the pelvic cavity.
• Pelvic cavity–the space occupied by the ventral internal organs that are bordered by the bones of the pelvic girdle.
LAB EXERCISE 1-3
Fill in the blank with the appropriate body cavity
1. The two main body cavities are the ________ __________and the _________ _________cavities.
2. The stomach is found in the _________ _________cavity.
3. The heart is found in the cavity, which is part of the larger cavity.
4. The brain is found within the cavity which is part of the larger cavity
5. The urinary bladder and reproductive organs are found within the _________ _________ cavity.
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Figure \(1\). The locations of the major body cavities of the human body.. Authored by: Mysid. Provided by: Wikipedia. Located at: https://commons.wikimedia.org/wiki/F...avities-en.svg. License: Public Domain: No Known Copyright
The Human Organ Systems
Information
Organ systems are groups of organs within the body that can be thought of as working together as a unit to carry out specific tasks or functions within the body. The human body is most commonly divided into eleven organ systems, the ones listed below.
It should be kept in mind that these divisions are somewhat arbitrary as to which organs are included and which are excluded. Skeletal muscles attached to bones are part of the muscular system, but the smooth muscles around soft tissues are not. Skeletal muscles are attached to bones, and serve to move the bones, but bones are part of the skeletal system, not the muscular system.
It also bears remembering that no one organ system ever functions independently of the others. The nervous system sends instructions to the muscular system as to when to move particular muscles. The cardiovascular system delivers nutrients and removes wastes from the muscle fibers of the muscular system to allow them to continue to function, etc. Dividing the human body into eleven organ systems is simply a way for the human mind to organize information about what parts do what. In the body itself, the parts that need to interact do interact, regardless of which system they have been grouped into.
The eleven organ systems are shown in Figure \(2\). The figure also lists the organs in each system and some roles for each system.
Identifying the major internal organs of the body
LAB 1 EXERCISE \(1\)
For each of the following organs, identify the organ system to which it belongs. There are three organs in this list which each belong to two organ systems; in those cases, list them both.
Brain
Ovaries
Cartilage
Pancreas
Skin
Spleen
Heart
Kidneys
Lungs
Testes
Mammary glands
Gall bladder
Thymus
Pituitary
gland
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Figure \(2\). Organ Systems, part 2.. Provided by: OpenStax College. Locate athttps://cnx.org/resources/9490de5610...ody(Page2).jpg. Licen se: CC BY-SA: Attribution-ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/01%3A_Overview_and_the_Microscope/1.03%3A_Human_Body_Cavities_and_Organ_Systems.txt |
Information
Many important anatomical features, especially those that function at the tissue or cellular levels, are too small to be seen by the unaided eye. The compound microscope is a valuable tool for magnifying small sections of biological material so that otherwise inaccessible details can be resolved.
There are many different types of microscopes. We shall only learn about the compound light microscope. It uses visible light to visualize the specimen, but passes that light through two separate lens to magnify the image. The compound microscopes we will use in this course are sturdy instruments but they still have a lot of moving parts. They can be damaged and broken through misuse and mishandling. A large part of learning how to use the microscopes properly involves learning how to avoid damaging it. To do that, you first have to know which parts are which. Figure 1.12 identifies the key parts of the microscope that you need to be familiar with.
In Figure 1.12, there are two compound microscopes shown. The one on the left is monocular and the one on the right is binocular. Many of the parts of the two microscopes are in slightly different locations. Get used to this. Different brands and different models of microscopes position the key parts differently.
When you first sit in front of a microscope, you should always take a second to find the key parts, especially the focus knobs, the condenser adjustment knob (if present), and the stage control knobs. When viewing a specimen, your eyes will be at the eyepieces, and if you grab the wrong knob by accident, you can lose your image at best, and damage the microscope at worst. Don’t assume you remember where the key knobs are. You may have a different microscope than last time.
When storing a microscope you should always follow this list:
1. Remove any slide found on the stage and return it to the slide box.
2. Rotate the smallest lens or no lens into place above the stage. Lower the stage a few turns.
3. Loosely coil the cord in your hand starting near the microscope and working toward the plug.
4. Hang the coiled cord over one ocular lens.
5. Look at the number on the back of the microscope, return that scope to its numbered box.
If there’s already a microscope in that numbered box, check its number and move it. If it is not numbered simply push it to the back of the box and place yours closer to the front. We have a few extra microscopes which we store in this fashion.
The eyepiece
This is where your eyes will be. If the microscope is binocular, use both eyepieces. With binocular microscopes, you almost always can adjust the width of the eyepieces to ensure they fit the spacing of your eyes. The eyepiece contains the eyepiece lens, one of the two lenses doing the actual magnifying in a compound microscope.
The carrying arm
When moving a microscope, even if it is just a few inches, always pick it up by the carrying arm. Do NOT drag the microscope: pick it up. The microscope will have rubber feet that prevent it from sliding, so if you try to drag it, it will shake and vibrate and possible damage parts. Never pick up the microscope by any part other than the carrying arm. The other parts are generally much more fragile and prone to breaking if you try.
The objective lenses
Most compound light microscopes will contain three to four objective lenses that can be rotated over the slide. Sometimes these lenses are just called objectives. When a particular objective has been fully rotated into position, you will hear or feel a click as that objective locks into place. The objective lens is the second of the two lenses doing the actual magnifying in a compound microscope, so if it is not snapped into proper position, you won’t see the proper image. Each objective lens can usually be unscrewed from its position in the rotating turret that houses it. Be careful you are rotating the turret, not unscrewing an objective. Do NOT unscrew the objectives from the turret. Each objective lens has a different magnifying power, so the image on your slide will be magnified to lesser or greater extents, depending on which objective lens you have chosen. Each objective’s magnification power will be written somewhere on the side of the objective, although sometimes it is hard to see the number. The magnification of an objective lens will always be a whole number. There will be other things written on the side of an objective, but the one that is a whole number greater than 1 will be the magnification. You can ignore everything else written there.
The stage
The stage is the platform that the slide will be clipped on to.
Stage clips
The slide will be held in place on the stage with stage clips. Most of the time, these will clip against the sides of the slide. They do not sit above or below the slide. They are spring-loaded to hold the slide edges and lock the slide in place so that the stage controls can move the position of the slide smoothly. If the slide is not clipped in place, you won’t be able to reposition the slide to find microscopic features of interest.
Stage Controls
These allow you to move your slide while you are viewing it, but only if the slide is properly clipped in with the stage clips. Always find where these are on your microscope before you start viewing your slide. They seem to never be in the same place in two different microscopes and if you just blindly grope for them while viewing your slide, you will likely do something unfortunate to your view or to the entire microscope. There are always two dials. One moves the slide left and right. The other moves the slide up and down. Sometimes they are on top of each other, as in the binocular microscope shown on the right in Figure \(1\) . Sometimes they are two separate dials, as in the monocular microscope shown on the left in Figure \(1\). Sometimes they are above the stage, as in the monocular microscope . Sometimes they are below the stage, as in the binocular scope . Spend a few seconds to find them every time before you sit down at a microscope.
Coarse focus
This is always the larger of the two focus knobs. You should usually only need to use the coarse focus knob once for each new slide. Use it with the lowest power objective to get the specimen approximately in focus. After that, only use the fine focus knob, even after you change to a higher-power objective. Sometimes the coarse focus knob is with the fine focus knob, as on the binocular microscope in Figure 1.12. Sometimes it is separate from the fine focus knob, as on the monocular microscope in Figure 1.12.
Fine focus
This is always the smaller of the two focus knobs. This is the focus know you will use over and over again in viewing slides. Don’t change the coarse focus after using it for the first time, only change the fine focus.
Condenser adjustment
Not all microscopes have a condenser adjustment knob. If there are only two knobs, as on the monocular microscope in Figure \(1\), those two are the coarse focus and the fine focus and you only have to keep those two separate. But if there is a third knob, it is the condenser adjustment knob. As a general rule, do NOT touch or adjust this knob. It controls how far the light condenser is from the slide, which should be properly adjusted before you use the microscope. If you move it, you will have it in the wrong position. If your scope has the knob, find out where it is and avoid it.
Diaphragm
This is directly under the hole in the stage where light passes through to the slide. It is controlled by a lever which opens and closes an iris to let more or less light through the slide. In some specimens there is not much contrast between the colors and shades of the different components being magnified. Changing how bright the view is by adjusting the diaphragm can allow you to better see some of the details you are trying to magnify.
1. Pick up your microscope and physically move it to a new location. Bring it close enough that you can look into it comfortably from where you are sitting. Arrange it so that the stage is facing you and the eyepiece is rotated towards you. What part of the microscope did you grab in order to pick it up and move it?
2. Where are the locations of the two stage adjustment knobs on your microscope?
3. Where is the location of the coarse focus knob?
4. Where is the location of the fine focus knob?
5. Is there a condenser adjustment knob? If so, where is it located?
6. Find the diaphragm lever. Looking in the hole in the center of the stage, what happens when you move the diaphragm leverclockwise?
7. Still looking down at the hole in the center of the stage, what happens when you slide the diaphragm lever counter-clockwise?
Lab 1 Exercise \(1\)
There are compound microscopes in the large wooden boxes in the front of the room. The boxes and the scopes are numbered. Use any scope you like but please return it to the correctly numbered box. Carry out the activities listed below and fill in the blanks as you do so.
1. Pick up your microscope and physically move it to a new location. Bring it close enough that you can look into it comfortably from where you are sitting. Arrange it so that the stage is facing you and the eyepiece is rotated towards you. What part of the microscope did you grab in order to pick it up and move it?
2. Where are the locations of the two stage adjustment knobs on your microscope?
3. Where is the location of the coarse focus knob?
4. Where is the location of the fine focus knob?
5. Is there a condenser adjustment knob? If so, where is it located?
6. Find the diaphragm lever. Looking in the hole in the center of the stage, what happens when you move the diaphragm lever clockwise?
7. Still looking down at the hole in the center of the stage, what happens when you slide the diaphragm lever counter- clockwise?
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Magnification and Resolution
The reason for using a microscope is to magnify features to the point where new details can be resolved.
Magnification is the factor by which an image appears to be enlarged. It will be a whole number greater than 1 and is usually followed by an “x”, as in 10x magnification.
When you look through microscope eyepieces, you are seeing a virtual image because in reality, what you are looking at is not as large as it appears through the eyepieces, and because there can be some distortion of the image.
Resolution is the shortest distance between two points that can still be visually distinguished as separate. The resolution of a typical unaided human eye is about 200 µm. Using a microscope decreases the resolution to distances as short as 0.2 µm. Resolution is a property of the eye.
Resolving power is the ability of a lens to show two adjacent objects as discrete. Resolving power is a property of a lens.
Each lens in a microscope has a numerical aperture, or NA, value. This has to do with the angles of light that enter and exit a lens. Its applications are beyond the scope of this lab, but numerical aperture does influence the resolution possible with a particular lens, and so the NA value for the lens is usually printed on each objective. It will be a number less than 1.0, and you can ignore it for our purposes.
Each lens in a microscope also has a magnifying factor. This is the degree to which that lens magnifies an image. It will be a number larger than 1.0. For instance a 10x objective magnifies the image ten-fold. The magnifying factor for each objective always printed on it, and the magnifying factor for each eyepiece is usually printed on it. (If the eyepiece is missing a printed magnifying factor, you can usually assume it is 10x.)
The total magnification for any image viewed under a compound microscope is calculated by using the formula:
Total Magnification = eyepiece magnifying factor * objective magnifying factor
So, each time you switch objectives, you change the total magnification. Total magnification does not have units, but is usually indicated by an “x”, as in “total magnification = 100x.”
Lab 1 Exercise \(2\)
There are compound microscopes in the large wooden boxes in the front of the room. The boxes and the scopes are numbered. Use any scope you like but please return it to the correctly numbered box. Carry out the activities listed below and fill in the blanks as you do so.
1. Write down the magnification factor for the eyepiece lenses on the microscope in front of you.
2. Using the microscope in front of you, write out all the words and numbers written on each objective on your microscope. There are probably three objectives, but some microscopes might have four. Start with the smallest objective and move through them in order of increasing size
Objective one:
Objective two:
Objective three:
Objective four:
3. In the above list, for each objective, circle just the magnification factor for that objective. Remember, the magnifying factor is a whole number, and differs for each different objective.
4. Write down the total magnification (eyepiece magnifying factor * objective magnifying factor) when using each objective on the microscope in front of you.
Objective one:
Objective two:
Objective three:
Objective four:
5. If you observed two features on a slide with your naked eye that were 0.5 mm apart, how far apart would they appear to be if you observed them with the microscope in front of you, using the second objective?
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All the microscopes in the lab are parfocals. That means that if the slide is in focus under one objective, it will stay largely in focus if the objective is changed. In practical terms, this means you should usually only need to use the coarse focus knob once per slide. You get the slide in focus under the lowest-power objective (where focusing is easiest), then, from that point onward, only make minor adjustments with the fine focus knobs even if you change objectives.
When you first get a new slide, you can usually determine the location of the specimen by looking at the slide while it is still in your hand. The specimen is usually a patch of color somewhere near the center of the coverslip. After you clip your slide securely onto the stage with the stage clips, use the stage control knobs to move the patch of color until it is directly over the hole in the center of the stage where the light comes through. Now when you look through the eyepieces using the lowest objective (always start with the lowest objective) you should be able to find the specimen and get it quickly in focus.
Occasionally the eyepiece or objective lenses will have specks of dirt or dust on them, making it difficult to focus on the specimen. To clean lenses, always use lens paper supplied by the lab instructor or a pure cotton swab. Do not use any other type of cloth or paper as they might scratch the lens. KimWipes are NOT lens paper, NEVER use KimWipes on glass lenses or slides. To remove dirt with lens paper, first roll up the lens paper and try to dry brush away the dirt in a spiraling motion that circles from the center of the lens out. If that doesn’t work, moisten the lens paper or pure cotton swab with blue lens cleaning solution (do not apply the water to the lens directly) and clean in a spiraling motion from the center of the lens out.
Below is a checklist for initially setting up a microscope. Every time you get a new slide, you should use this checklist.
Lab 1 Exercise \(1\)
1. Plug in the microscope & turn on light source.
2. Pick up microscope by carrying arm, position it so it is accessible to your seat, with open side of the stage facing you
3. Rotate the objectives so that the lowest power objective (smallest in size) clicks into place.
4. Look at the slide with your naked eye and find the location of the specimen.
5. Clip the slide into place with the stage clips. The cover slip on the slide must face up. Find the stage controls and make sure that, when they are turned, the slide moves smoothly left and right or up and down, depending on the knob.
6. Use the stage controls to move the slide so that the light source is shining directly on to the specimen to be magnified.
7. Find the coarse and fine focus knobs. Watching the stage and objective, use the coarse focus knob to bring the low power objective as close to the slide as it will go.
8. Put your eye to the eyepiece (or eyepieces, if the microscope is binocular) and rotate the coarse focus knob in the lowering direction until some aspect of the specimen comes into focus.
9. Move your hand to the fine focus knob and get the specimen into perfect focus for your eyes. Do NOT touch the coarse focus knob again.
10. Use the stage control knobs to move you specimen to close to the exact center of your field of view
11. Move to the next highest power objective (do not skip the individual objectives) and use only the fine focus to get your image into perfect focus for your eyes.
12. If you need further magnification, move to the next highest power objective and use only the fine focus to get your image into perfect focus for your eyes.
13. Do not use the 100x objective (if you have one) in this course. It must be used with immersion oil and we won’t have students doing that.
How The Virtual Image Differs From The Real Image
The virtual image you see when looking in your microscope is not quite the same as the real image you would see with your eye. For one thing, it is bigger. For another thing, the orientation of the image is different. The two lenses in a compound microscope reflect the original image two times, in two different planes, while magnifying it. So what you think of as the “top” of your image is really the bottom, and what you think of “right” is really left. Usually this is not an issue at the microscopic level, but it is important to understand how the microscope is rearranging your virtual image.
Lab 1 Exercise \(2\)
1. Get an “e” slide. If it is already under your microscope, rotate the lowest-power objective into place, use the coarse focus to lower the stage, and remove the slide.
2. Look at the unmagnified “e” on the slide by eye. Rotate the slide around in your hand so that the “e” is right side up. Now clip the slide onto the microscope stage with the stage clips so that the “e” is facing you right side up when you look at it with your unaided eye.
3. In the right-hand circle below, draw what the “e” looks like when you are looking at it right side up. Assume the circle below is the size of the entire coverslip. Draw the “e” you see unaided in the correct proportion to the coverslip. (The unmagnified “e” will take up a tiny portion of the coverslip area.)
4. Clip the slide into your microscope stage so that the “e” is still facing you right side up. Follow the checklist in Lab 1 Exercise 1-7 to ensure you are setting up the microscope properly for use, but stay on the lowest power objective. Get the “e” into your field of view and in focus.
5. In the left-hand circle above, draw what the “e” looks like when viewing it through the microscope under the lowest-power objective. Under the circle, write the total magnification of the image.
6. When viewed under a microscope, how is a specimen rotated?
7. Look at the stage and slide directly (not through the eyepieces.) Move the stage control knob that causes the slide to move away from you on the stage, then move it back to its original position.
8. Now move the stage control knob the exact same way you just did, but view the “e” through the eyepiece. When the stage is moving away from you, what direction does the virtual image appear to be moving?
9. Again, look at the stage and slide directly (not through the eyepiece.) This time, move the stage control knob that causes the slide to move to your right, then move it back to its original position.
10. Now move the stage control knob the exact same way you just did, but view the “e” through the eyepiece. When the stage is moving to your right, what direction does the virtual image appear to be moving?
11. The field of view is the entire area you can see when looking through an eyepiece. Use the stage control knobs to move the virtual image of your “e” to one side of the field of view. Keep most of the “e” in the field of view, but move it to one side or the other.
12. Now switch to the next-power objective. (Do not skip.) To get to the next-power objective, and not the highest-power objective, which way did you have to rotate the objectives, clockwise or counter-clockwise?
13. Using only the fine focus knob (you do NOT use the coarse focus knob on any objective other than the lowest objective), get the “e” in focus.
14. When moving to the next objective, which part of the field of view do you zoom in on?
15. Move away from the eyepieces and look at the distance between the slide and the bottom of the objective. Rotate back to the lowest power objective. Now rotate to the next objective (do not rotate to the highest-power objective by accident). Now rotate to the third-highest objective. What happens to the distance between the slide and the bottom of the objective as you rotate to higher power objectives?
16. With the third-highest power objective still in place, how much space is there between the slide and the bottom of the objective?
17. Notice there is a danger of smashing the objective lens into the slide if you were to use the coarse focus. Why are you instructed to only use the coarse focus with the lowest- power objective?
18. Only draw what you actually see. Even if you expect to see something, if it is not there you should not draw it. Do not base your drawings on what the textbook or some other source tells you should be there. Do not draw things in the shapes that texts or other sources tell you to expect unless you actually see those shapes.
Making Simple But Accurate Line Drawings of Magnified Specimens
You do not have to be a great artist to make a diagram of the cells and structures you see under a microscope. You only have to be careful to draw something that is approximately the same size and shape as what you see. Follow the following guidelines:
1. Only draw what you actually see. Even if you expect to see something, if it is not there you should not draw it. Do not base your drawings on what the textbook or some other source tells you should be there. Do not draw things in the shapes that texts or other sources tell you to expect unless you actually see those shapes.
2. Keep things as simple as possible. Draw strong unbroken lines. Avoid shading or cross- hatching unless there is a very good reason to add them.
3. Feel free to simplify reality by leaving out unnecessary details. Draw what is of interest, but leave out background material, debris, or any other distracting items. Just be careful that, if you are leaving something out, that it isn’t something that is an important part of what you are drawing.
You should always have a basic understanding of what you are looking for before looking in the microscope. Tissues and other microscopic specimens can be confusing and cluttered. If you know in general what you are looking for, and, sometimes more importantly, what you are not looking for, it will make it much easier to find what you want to draw and it will make it much easier to decide how to draw it.
Just remember, what you see under the microscope may look quite different from the perfect specimens that are usually found in the figures put into textbooks and websites. Use the idealized images to track down what you are looking for, but draw the specimen as it actually is, regardless of your expectations.
For instance, in most textbooks, neurons, the most common cell found in nervous tissue, are drawn to look like variations of the drawing in Figure \(1\)A.
In the typical diagram of a neuron that appears in texts and on websites, there usually is a clear nucleus, and often a nucleolus visible, too. Sometimes organelles such as mitochondria are visible (there are none in Figure \(1\)A.) The dendrites are typically short and branched. There almost always is a single, easily-identifiable axon that is longer than all the dendrites and that branches as it ends.
Figure \(1\)B shows an actual neuron as viewed through a microscope. If you view enough neurons through enough different types of microscopes, you can eventually create a composite diagram that incorporates features from many specimens to present a “typical” neuron, but it is unlikely that if you view a single neuron you will see everything in Figure \(1\)A. In fact, often actual specimens look very little like their textbook counterparts. Draw what you see, not what you think you are supposed to see. Just make sure you are looking at what you are supposed to be finding (for instance, a neuron and not a piece of dirt or cell debris), and then draw it as it is.
In the case of the actual neuron in Figure \(1\)B, there is no nucleus visible, there is maybe one large projection and one small projection you could call dendrites – but there aren’t many projections – and neither of the projections are branched. There is one long thin projection that is probably an axon, and it is not branched.
If you draw what you see, you end up with a drawing like the one in Figure 1.14B. It does not look like a textbook neuron, but it is a reasonable representation of what is there in this case.
Most students feel they “cannot draw” and are reluctant to sketch what they are seeing under a microscope. You don’t let your lack of artistic skills stop you.
1. Draw an outline that approximates the item you want to draw. Don’t obsess about making it match perfectly. Approximate is ok.
2. Try to get the proportions approximately right. If something is half as big, or as third as big, as something else, make it that way in the drawing, too.
3. Do not draw everything you see. Improve on reality by only drawing the parts of the specimen you are interested in. You do not have to draw every bit of debris or dirt. Decide what the important parts of your specimen are and draw only those.
4. Do not use shading or cross-hatching unless there is a very good reason to do so. It will actually make it easier to understand your drawing if you stick to drawing just outlines. It will also be easier and quicker to draw.
Lab 1 Exercise \(3\)
1. Get a human blood smear slide. Rotate your lowest power objective into place on your microscope.
2. Follow the checklist in Lab Exercise \(1\) until you are viewing the blood smear under your 40x objective.
3. You will see mostly red blood cells. They will probably be pinkish and they will be the circles without nuclei. Occasionally, some will appear to have blank circles in their centers, but these are not nuclei. If you search around your slide using your stage controls, you will find the rare circular cells with nuclei. These are white blood cells. There will be less than one white blood cell for every 100 red blood cells. These white blood cells will probably be light blue or grey and have purple or dark blue nuclei. Their nuclei will not always be round.
4. Find a section of your slide with two or more white blood cells among all the red blood cells.
5. In the circle below, draw four or five representative red blood cells (do not draw all the red blood cells you see) and draw all the white blood cells in your field of view. Pay careful attention to drawing the white blood cell nuclei as accurately as possible.
6 Do not remove or change the position of your slide until one of your lab partners has verified that your white blood cells are drawn accurately. Introduce yourself to a partner and ask for their help.
LICENSES AND ATTRIBUTIONS
CC LICENSED CONTENT, ORIGINAL
A&P Labs. Authored by: Ross Whitwam. Provided by: Mississippi University for Women. Located athttp://www.muw.edu/. License: CC BY-SA: Attribution-ShareAlike
Figure \(1\)B. A drawing of Figure Figure \(1\)A's neuron.. Authored by: Ross Whitwam. Provided by: Mississippi University for Women. Located athttp://www.muw.edu. License: CC BY-SA: Attribution-ShareAlike
CC LICENSED CONTENT, SHARED PREVIOUSLY
Figure \(\PageIndex{2a}\). A typical diagram of a neuron.. Authored by: Jonathan Haas. Located at: https://commons.wikimedia.org/w/inde...curid=18271454. License: CC BY-SA: Attribution-ShareAlike
CC LICENSED CONTENT, SPECIFIC ATTRIBUTION
Figure \(\PageIndex{2b}\)A. An actual neuron; Figure \(1\)B. An actual neuron. Authored by: W. Clay Spencer, Rebecca McWhirter, Tyne Miller, Pnina Strasbourger, Owen Thompson, LaDeana W. Hillier, Robert H. Waterston, David M. Miller III. Located athttp://dx.doi.org/10.1371/journal.pone.0112102. License: CC BY: Attribution
When storing a microscope, you should always follow this list:
1. Remove any slide found on the stage and return it to the slide box.
2. Rotate the smallest lens or no lens into place above the stage. Lower the stage a few turns.
3. Loosely coil the cord in your hand starting near the microscope and working toward the plug.
4. Hang the coiled cord over one ocular lens.
5. Look at the number on the back of the microscope, return that scope to its numbered box.
6. If there’s already a microscope in that numbered box, check its number and move it. If it is not numbered simply push it to the back of the box and place yours closer to the front. We have a few extra microscopes which we store in this fashion. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/01%3A_Overview_and_the_Microscope/1.05%3A_The_Parts_of_a_Compound_Microscope_and_How_To_Handle_Them_Correctly.txt |
In this lab students will learn to identify main external and internal cell structures. Use your textbook to identify and label the cell structures listed below. Watch the Cytology Lab Video associated with this lab.
• 2.1: Cell Structure
BIOL 2400: General Anatomy & Physiology http://www.stkate-las.com/uploads/7/0/2/5/70257827/cytology1_lab.pdf
• 2.2: Observing Mitosis
Cell Division: Open Lab City Tech CUNY https://openlab.citytech.cuny.edu/bio-oer/cell-division/
02: Cytology
Laab 2 Example \(1\)
1. Plasma membrane
2. Nucleus
3. Nuclear envelope
4. Rough endoplasmic reticulum
5. Smooth endoplasmic reticulum
6. Golgi complex
7. Golgi vesicles
8. Mitochondria
9. Lysosome
10. Centriole
There may be multiple pictures of the same structure
Lab 2 Exercise \(2\)
requested before 2.1 to help you complete this activity. Place the appropriate letter(s) on the line
between the numbers and the name. “NA” = not applicable, do not use.
1. _____ Plasma membrane
2. _____ Nucleus
3. _____ Nuclear envelope
4. _____ Rough endoplasmic reticulum
5. _____ Smooth endoplasmic reticulum
6. _____ Golgi complex
7. _____ Golgi vesicles
8. _____ Mitochondria
9. _____ Lysosome
10. _____ Centriole
There may be multiple pictures of the same structure
(CC-BY-SA-NC Cytology Laboratory Liberal Arts and Sciences in the College for Adults, St. Catherine University http://www.stkate-las.com/uploads/7/...ology1_lab.pdf)
2.02: Observing Mitosis
Mitosis In A Whitefish Embryo
The Cell Cycle and Mitosis:
The cell cycle refers to a series of events that describe the metabolic processes of growth and replication of cells. The bulk of the cell cycle is spent in the “living phase”, known as interphase. Interphase is further broken down in to 3 distinct phases: G1 (Gap 1), S (Synthesis) and G2 (Gap 2). G1 is the phase of growth when the cell is accumulating resources to live and grow. After attaining a certain size and having amassed enough raw materials, a checkpoint is reached where the cell uses biochemical markers to decide if the next phase should be entered. If the cell is in an environment with enough nutrients in the environment, enough space and having reached the appropriate size, the cell will enter the S phase. S phase is when metabolism is shifted towards the replication (or synthesis) of the genetic material. During S phase, the amount of DNA in the nucleus is doubled and copied exactly in preparation to divide. The chromosomes at the end of G1 consist of a single chromatid. At the end of S phase, each chromosome consists of two identical sister chromatids joined at the centromere.
When the DNA synthesis is complete, the cell continues on to the second growth phase called G2. Another checkpoint takes place at the end of G2 to ensure the fidelity of the replicated DNA and to re- establish the success of the cell’s capacity to divide in the environment. If conditions are favorable, the cell continues on to mitosis.
Follow this link above to view a mitosis video.
Mitosis is the process of nuclear division used in conjunction with cytokinesis to produce 2 identical daughter cells. Cytokinesis is the actual separation of these two cells enclosed in their own cellular membranes. Unicellular organisms utilize this process of division in order to reproduce asexually.
Prokaryotic organisms lack a nucleus, therefore they undergo a different process called binary fission. Multicellular eukaryotes undergo mitosis for repairing tissue and for growth. The process of mitosis is only a short period of the lifespan of cells. Mitosis is traditionally divided into four stages: prophase, metaphase, anaphase and telophase.
The actual events of mitosis are not discreet but occur in a continuous sequence—separation of mitosis into four stages is merely convenient for our discussion and organization. During these stages important cellular structures are synthesized and perform the mechanics of mitosis. For example, in animal cells two microtubule organizing centers called centrioles replicate. The pairs of centrioles move apart and form an axis of proteinaceous microtubules between them called spindle fibers.
These spindle fibers act as motors that pull at the centromeres of chromosomes and separate the sister chromatids into newly recognized chromosomes. The spindles also push against each other to stretch the cell in preparation of forming two new nuclei and separate cells. In animal cells, a contractile ring of actin fibers cinch together around the midline of the cell to coordinate cytokinesis. This cinching of the cell membrane creates a structure called the cleavage furrow.
Eventually, the cinching of the membrane completely separates into two daughter cells. Plant cells require the production of new cell wall material between daughter cells. Instead of a cleavage furrow, the two cells are separated by a series of vesicles derived from the Golgi. These vesicles fuse together along the midline and simultaneously secrete cellulose into the space between the two cells. This series of vesicles is called the cell plate.
Review Setting up Your Microscope
1. Plug in the microscope & turn on light source.
2. Pick up microscope by carrying arm, position it so it is accessible to your seat, with open side of the stage facing you
3. Rotate the objectives so that the lowest power objective (smallest in size) clicks into place.
4. Look at the slide with your naked eye and find the location of the specimen.
5. Clip the slide into place with the stage clips. The cover slip on the slide must face up. Find the stage controls and make sure that, when they are turned, the slide moves smoothly left and right or up and down, depending on the knob.
6. Use the stage controls to move the slide so that the light source is shining directly on to the specimen to be magnified.
7. Find the coarse and fine focus knobs. Watching the stage and objective, use the coarse focus knob to bring the low power objective as close to the slide as it will go.
8. Put your eye to the eyepiece (or eyepieces, if the microscope is binocular) and rotate the coarse focus knob in the lowering direction until some aspect of the specimen comes into focus.
9. Move your hand to the fine focus knob and get the specimen into perfect focus for your eyes. Do NOT touch the coarse focus knob again.
10. Use the stage control knobs to move you specimen to close to the exact center of your field of view
11. Move to the next highest power objective (do not skip the individual objectives) and use only the fine focus to get your image into perfect focus for your eyes.
12. If you need further magnification, move to the next highest power objective and use only the fine focus to get your image into perfect focus for your eyes.
13. Do not use the 100x objective (if you have one) in this course. It must be used with immersion oil and we won’t have students doing that.
Lab 2 Exercise \(1\)
1. Obtain a slide of a whitefish embryo (blastula) from the slide box at your table.
2. Follow the checklist above to set up your slide for viewing.
3. View the slide on the objective which provides the best view. Find the representative object.
4. In the circles below the name, draw a representative sample of the stage of mitosis, taking care to correctly and clearly draw the true shape in the slide. Draw your structures proportionately to their size in your microscope’s field of view.
5. Fill in the blanks next to your drawing.
6. Repeat this for each of the mitosis phases seen below.
A) Prophase
Function of this phase:
D) Telophase/cytokinesis
Function of this phase:
Storing the microscope
When storing a microscope you should always follow this list:
1. Remove any slide found on the stage and return it to the slide box.
2. Rotate the smallest lens or no lens into place above the stage. Lower the stage a few turns.
3. Loosely coil the cord in your hand starting near the microscope and working toward the plug.
4. Hang the coiled cord over one ocular lens.
5. Look at the number on the back of the microscope, return that scope to its numbered box.
6. If there’s already a microscope in that numbered box, check its number and move it. If it is not numbered simply push it to the back of the box and place yours closer to the front. We have a few extra microscopes which we store in this fashion. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/02%3A_Cytology/2.01%3A_Cell_Structure.txt |
Information
Epithelial tissue serves two main functions in the body.
1. It provides linings for external and internal surfaces that face harsh environments. The outer layer of the skin is epithelial tissue, as are the innermost layers of the digestive tract, the respiratory tract, and blood vessels.
2. It forms glands that secrete materials onto epithelial surfaces or into the blood. Sweat glands, salivary glands, mammary glands, adrenal glands, and pituitary glands are examples of glands made of epithelial tissue.
Epithelial tissue is often classified according to numbers of layers of cells present, and by the shape of the cells. See Figure 3.1.
simple epithelium is only one layer of cells thick. A stratified epithelium is more than one layer of cells thick. A pseudostratified epithelium is really a specialized form of a simple epithelium in which there appears at first glance to be more than one layer of epithelial cells, but a closer inspection reveals that each cell in the layer actually extends to the basolateral surface of the epithelium.
There are three basic shapes used to classify epithelial cells. A squamous epithelial cell looks flat under a microscope. A cuboidal epithelial cell looks close to a square.
columnar epithelial cell looks like a column or a tall rectangle. A few epithelial layers are constructed from cells that are said to have a transitional shape. Transitional epithelial cells are epithelial cells specialized to change shape if they are stretched laterally. They can transition from columnar- and cuboidal-looking shapes in their unstretched state to more squamous-looking shapes in their stretched state.
When classifying a stratified epithelial sheet, the sheet is named for the shape of the cells in its most superficial layers. So, a stratified squamous epithelium only necessarily has squamous-shaped cells in its highest layers and might have a different-shaped cell in its lower layers.
Under a microscope, epithelial cells are readily distinguished by the following features:
• The cells will usually be one of the three basic cell shapes – squamous, cuboidal, or columnar.
• The cells will be closely attached to one another, in either a single layer or in multiple layers, and usually will not have room for extracellular material between the attached cells.
The epithelial layer on one side will face an empty space (or, in some organs, it will face a secreted substance like mucus) and on the other side will usually be attached to connective tissue proper.
Usually, a slide will have a section of tissue cut out of a larger organ. Slides with epithelial tissues usually have some of the underlying tissue found beneath the epithelial tissue with them.
A layer of epithelial cells always serves as an outer layer for some structure, but, when looking at a tissue preparation on a slide, do not assume that just because you have found one end of the tissue sample you are automatically looking at epithelial tissue. Look for the cell characteristics listed above to be sure you are on the epithelial side of a tissue slice.
In Figure \(2\), only one edge of the tissue slice has epithelial cells. In Figure \(2\)A that edge is indicated with an arrow, but when looking at a specimen under a microscope, you have to figure out for yourself where the edge with the epithelial cells is.
In the tissue slice in Figure \(2\), there are three edges that are not epithelial cells. If you just mindlessly started viewing the first edge you find, you have a good chance of looking as something other than the epithelial cells in the preparation. Be sure what you are looking at has the three visual characteristics of epithelial tissue:
• The cells will usually be one of the three basic cell shapes – squamous, cuboidal, or columnar.
• The cells will be closely attached to one another, in either a single layer or in multiple layers, and usually will not have room for extracellular material between the attached cells.
• The epithelial layer on one side will face an empty space (or, in some organs, it will face a secreted substance like mucus) and on the other side will usually be attached to connective tissue proper.
In the following figures are some more epithelial layers. Note that whether the epithelial layer of the specimen is on the top, bottom, right, or left of the slice varies with how the specimen slice was positioned on the slide. In every case, you have to find which edge has the epithelial layer.
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Setting up a microscope
Lab 1 Exercise 1
1. Plug in the microscope & turn on light source.
2. Pick up microscope by carrying arm, position it so it is accessible to your seat, with open side of the stage facing you
3. Rotate the objectives so that the lowest power objective (smallest in size) clicks into place.
4. Look at the slide with your naked eye and find the location of the specimen.
5. Clip the slide into place with the stage clips. The cover slip on the slide must face up. Find the stage controls and make sure that, when they are turned, the slide moves smoothly left and right or up and down, depending on the knob.
6. Use the stage controls to move the slide so that the light source is shining directly on to the specimen to be magnified.
7. Find the coarse and fine focus knobs. Watching the stage and objective, use the coarse focus knob to bring the low power objective as close to the slide as it will go.
8. Put your eye to the eyepiece (or eyepieces, if the microscope is binocular) and rotate the coarse focus knob in the lowering direction until some aspect of the specimen comes into focus.
9. Move your hand to the fine focus knob and get the specimen into perfect focus for your eyes. Do NOT touch the coarse focus knob again.
10. Use the stage control knobs to move you specimen to close to the exact center of your field of view
11. Move to the next highest power objective (do not skip the individual objectives) and use only the fine focus to get your image into perfect focus for your eyes.
12. If you need further magnification, move to the next highest power objective and use only the fine focus to get your image into perfect focus for your eyes.
13. Do not use the 100x objective (if you have one) in this course. It must be used with immersion oil and we won’t have students doing that.
Lab 3 Exercise \(1\)
Simple columnar
Simple cuboidal
Simple squamous
Stratified squamous
Pseudostratified columnar
Transitional
1. Obtain a slide of one of the tissues listed below from the slide box at your table.
2. Follow the checklist above to set up your slide for viewing.
3. View the slide on the objective which provides the best view. Find the representative object.
4. In the circle below the name, draw a representative sample of the tissue, taking care to correctly and clearly draw their true shape in the slide. If it is a stratified epithelium draw all the layers. Draw your structures proportionately to their size in your microscope’s field of view.
5. Fill in the blanks next to your drawing.
A) Prophase
Repeat this for each of the tissue types seen below. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/03%3A_Histology/3.01%3A_Examining_epithelial_tissue.txt |
Information
Connective tissue is found throughout the body, usually in association with other tissues. As its name indicates, it often serves to connect different tissues together, but it also can serve as a wrapper (in locations where a tough epithelial wrapping is not required), a structural support, cushioning, a storage repository, a protective layer, or a transport medium.
Connective tissue has the most types of subcategories and the most varied functions of all the four major tissue types (epithelial, muscular, nervous, and connective tissues.) Bone and cartilage are connective tissues, as are blood and lymph, fat, ligaments, and tendons. Epimysium, the connective tissue wrapping around skeletal muscles, and periosteum, the connective tissue wrapping around bones, are both connective tissues.
The different types of connective tissue are so diverse, there is no one set of characteristics that encompasses all the different types. However, there are three characteristics that we consider diagnostic of most connective tissue types.
1. The cells are dispersed. Connective tissues generally have cells that are not tightly connected to each other, the way the cells in epithelial and muscular tissues usually are. There is usually a fair amount of space between the connective tissue cells. An exception to this is adipose tissue (also known as fat), the rare type of connective tissue in which the cells are packed tightly together.
2. The tissue has more extracellular material than cells. Most connective tissues are solid (blood and lymph are the exceptions) because all the volume between the dispersed cells is filled with an extracellular matrix of viscous ground substance and protein fibers.
3. An extensive network of protein fibers is found in the extracellular matrix. Protein fibers are complexes of millions of individual proteins threaded into long fibrous structures that provide strength and elasticity to the tissue as a whole. The protein fibers are so large, they are longer than the cells they surround and enmesh. Blood and lymph, being liquid connective tissues, do not have these enmeshing protein fibers, but they still have an extensive liquid extracellular matrix.
A common way of classifying the many different types of connective tissue is to subdivide it into three main sub-categories, and further divide those subcategories into specific types of connective tissue. The three main sub-categories of connective tissue are:
1. Connective tissue proper. These are the types of connective tissue that typically have all three of the defining characteristics listed above. It is further subdivided into dense connective tissue proper, in which the extracellular protein fibers predominate, and loose connective tissue proper, in which the extracellular protein fibers are not so densely woven.
2. Supporting connective tissue. Bones and cartilage are the two types of connective tissue in this sub-category. They both have all three of the defining characteristics listed above, but their extracellular matrix is tougher, denser, and more solid than the various types of connective tissue proper.
3. Fluid connective tissue.
Blood and lymph are the two types of connective tissue in this sub-category. Both are fluid, rather than solid, and both lack the network of extracellular protein fibers found in the other types of connective tissue.
A&P Labs. Authored by: Ross Whitwam. Provided by: Mississippi University for Women. Located athttp://www.muw.eduLicenseCC BY-SA: Attribution-ShareAlike
Connective Tissue Proper
Information
Connective tissue proper encompasses the types of connective tissue that usually show all three of the defining cellular characteristics of connective tissue with the fewest deviations from those characteristics.
Dispersed cells
More extracellular material than cells.
Extensive protein fibers in the extracellular matrix.
Nonetheless there is still a great variety among the subcategories of connective tissue proper. Some are classified as dense connective tissue proper and have a dense arrangement of extracellular protein fibers that give the tissue strength and toughness. Tendons connecting muscles to bone and ligaments connecting bone to bone are examples of dense connective tissue proper. Other tissues are classified loose connective tissue proper and have fewer extracellular protein fibers and more ground substance (the extracellular material surrounding the protein fibers), making the tissues spongier but more fragile. Areolar tissue, found in the hypodermis of the skin and below the epithelial layers of the digestive, respiratory, and urinary tracts, is a loose connective tissue proper, as is adipose tissue, also known as fat.
Table \(1\) lists some of the subcategories of connective tissue proper, along with some of their characteristics and properties.
In drawing images of connective tissue proper preparations seen under the microscope, it is important to simplify the visuals. Connective tissue preparations are often messy with a number of blotches and shapes irrelevant to the main components of the tissue, which are the cells and the extracellular protein fibers. Especially with connective tissue slides, it is important to make sure you know what you are looking for, find those components, and draw only them in as simple a form as possible, usually with just lines and with minimal shadings or hatchings. Leave out the unnecessary and irrelevant stuff on the slide.
For instance, Figure 3.7 A shows a section of connective tissue taken from just below (deep to) the epithelial layer of the stomach, magnified 40x. Figure 3.7 B illustrates how you should represent that view as a line drawing.
Notice that in the line drawing not every single thick collagen fiber, nor every single thin elastic fiber, not every single fibroblast was drawn. Also notice that some of the out-of-focus, blurry fibers were not drawn at all, rather than draw fuzzy blotches. The key is to get the important structures (once you know what those are) and leave out distracting, non-essential messiness.
LAB 3 EXERCISE \(2\)
1. Table 3.1 lists six categories of connective tissue proper. Use the information in that table to identify which category each of the following samples belong to.
2. Under each picture, list the evidence that points to the category you choose.
3. The key identifying features to look for in each picture are: the type(s) of protein fibers present, the type of cells present, and whether there are significant amounts of ground substance present.
4. Collagen protein fibers are thick. Elastic protein fibers are thin. Reticular protein fibers are thin but form a web-like arrangement.
5. If there is abundant space between protein fibers, the tissue is likely one of the loose connective tissues. If there is little space between protein fibers, the tissue is likely one of the dense connective tissues.
6. Adipose is mainly large adipocyte cells containing a large droplet of lipids and nucleus and cytoplasm crammed into one corner of the cell. There are more adipocytes than extracellular material in adipose.
7. The protein fibers in regular dense connective tissue proper will largely parallel each other, but they are often undulate in a wave-like arrangement while being parallel.
LAB 3 EXERCISE \(3\)
1. Obtain a slide of connective tissue proper/areolar CT from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
1. Obtain a slide of dense irregular connective tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
1. Obtain a slide of a spleen or lymph node with reticular connective tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
1. Obtain a slide of adipose connective tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
1. Obtain a slide of a large artery with elastic connective tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
1. Obtain a slide of a tendon with dense regular connective tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
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CC LICENSED CONTENT, ORIGINAL
• A&P Labs. Authored by: Ross Whitwam. Provided by: Mississippi University for Women. Located athttp://www.muw.eduLicenseCC BY-SA: Attribution-ShareAlike
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• Figure 3-1A photomicrograph, A, and a drawing from the photomicrograph, B, of the connective tissue in the wall of the stomach, just below the epithelial layer.. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located
at: 141.214.65.171/Histology/Basic%20Tissues/Epithelium%20and%20CT/160_HISTO_40X.svs/view.apml. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
• Exercise 3.1 A. Source: Stomach wall. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located
• at: 141.214.65.171/Histology/Basic%20Tissues/Epithelium%20and%20CT/160_HISTO_40X.svs/view.apml. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
• Exercise 3.1 B. Joint Capsule. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning
Resources. Located
at: 141.214.65.171/Histology/Basic%20Tissues/Epithelium%20and%20CT/033_HISTO_20X.svs/view.apml. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
• Exercise 3.1 C. Source: Lymph node. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located at141.214.65.171/Histology/Basi...0and%20CT/028- 2_HISTO_40X.svs/view.apmlLicenseCC BY-NC-SA: Attribution-NonCommercial-ShareAlike
• Exercise 3.1 D. Source: Fat. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located at141.214.65.171/Histology/Basi...0and%20CT/019- 2_HISTO_20X.svs/view.apmlLicenseCC BY-NC-SA: Attribution-NonCommercial-ShareAlike
• Exercise 3.1 E. Source: Aorta Wall. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located
at: 141.214.65.171/Histology/Cardiovascular%20System/036_HISTO_20X.svs/view.apml%20%20. License: CC BY-NC- SA: Attribution-NonCommercial-ShareAlike
• Exercise 3.1 F. Source: Tendon. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning
Resources. Located at: 141.214.65.171/Histology/Basic%20Tissues/Epithelium%20and%20CT/74.svs/view.apml. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/03%3A_Histology/3.02%3A_Examining_Connective_Tissue.txt |
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Supporting connective tissue comprises bone and cartilage. We will examine those tissues in greater detail in Lab 5 The Appendicular Skeleton & Lab 6 The Axial Skeleton.
In both bone and cartilage, as in the different types of connective tissue proper, there are extracellular protein fibers embedded in a viscous ground substance. However, in bone and cartilage, the ground substance is so viscous as to be very hard and tough solids. Both bone and cartilage use mainly collagen and elastic protein fibers in their extracellular matrix, but cartilage uses a ground substance rich in the carbohydrate hyaluronan and bone uses a ground substance rich in a mineralized calcium phosphate compound known as hydroxyapatite.
The carbohydrate hyaluronan (sometimes known as hyaluronic acid or hyaluronate) binds up huge numbers of water molecules in the extracellular matrix of cartilage. This helps solidify the ground substance around the collagen and elastic fibers of cartilage. As a result, it is often difficult to see the protein fibers in cartilage when viewing preparations under the microscope.
They hydroxyapatite that surrounds the mostly collagen protein fibers in the ground substance of bone is not soluble in water and forms a mineral solid in which both the bone cells and the collagen fibers are embedded. As with cartilage, it is usually difficult to see the collagen fibers in the extracellular matrix of bone due to the density of the ground substance that surrounds them.
There is only one type of cell in cartilage, chondrocytes. They secrete and maintain the extracellular matrix of the tissue. Chondrocytes arise from mesenchymal stem cells, just like the fibroblasts of connective tissue proper do, but chondrocytes are specialized to produce just cartilage. The extracellular matrix produced by the chondrocytes is so tough and durable, the chondrocytes are in danger of being crushed by it. This is why chondrocytes always leave a region around themselves free of the cartilaginous extracellular matrix that makes up the rest of the tissue. These non-cartilaginous pockets around each chondrocyte are called lacunae and are clearly visible when examining cartilage under the microscope.
There are four types of bone cells, osteoprogenitor cells, osteoblasts, osteoclasts, and osteocytes, but the osteocytes are the most abundant and the only ones found throughout the bone. Osteocytes are found in concentric circles of mineralized extracellular matrix. Each circle is called a lamella (plural: lamellae) and the osteocytes are found along the edges of each lamellae. In compact bone, groups of lamellae and osteocytes are arranged into individual osteons, the cylindrical arrangement of material that makes up the fundamental building block of the compact bone. Each osteon has a hollow central canal in its center that blood vessels and nerves can travel through. In spongy bone, groups of lamellae are arranged into trabeculae (singular: trabecula), which are the individual projections of spongy bone. Trabeculae do not have central canals.
Osteocytes, like chondrocytes, are protected from the extracellular matrix that surrounds them by being housed in lacunae, which are spaces free of mineralized extracellular matrix. Osteocytes, unlike chondrocytes, have numerous cytoplasmic extensions that project off of the main cell body. These extension connect up with the extensions from other near-by osteocytes. These projections, like the osteocyte cell body, are in tiny spaces free of the mineralized extracellular matrix. These spaces (but not the cytoplasmic projections themselves) are called canaliculi (singular: canaliculum) because, under the microscope, they look like tiny little canals.
LAB 3 EXERCISE \(1\)
1. In the photomicrograph below of cartilage tissue, find and label the indicated structures.
2. In the photomicrograph below of compact bone tissue, find and label the indicated structures.
1. Obtain a slide of hyaline cartilage connective tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
1. Obtain a slide of ground compact bone connective tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
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CC LICENSED CONTENT, ORIGINAL
A&P Labs. Authored by: Ross Whitwam. Provided by: Mississippi University for Women. Located athttp:www.muw.eduLicenseCC BY-SA: Attribution-ShareAlike
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Exercise 3.3. 1. In the photomicrograph below of cartilage tissue, find and label the indicated structures.. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located at: 141.214.65.171/Histology/Basic%20Tissues/Cartilage%20and%20Bone/044H_HISTO_20X.svs/view.apml. License: C C BY-NC-SA: Attribution-NonCommercial-ShareAlike
Exercise 3.3. 2. In the photomicrograph below of compact bone tissue, find and label the indicated structures.. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located at: 141.214.65.171/Histology/Basic%20Tissues/Cartilage%20and%20Bone/051xc_HISTO_40X.svs/view.apml. License: C C BY-NC-SA: Attribution-NonCommercial-ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/03%3A_Histology/3.03%3A_Examining_Connective_Tissue.txt |
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Muscular tissue is the third of the four major categories of animal tissue. Muscle tissue is subdivided into three broad categories: skeletal muscle, cardiac muscle, and smooth muscle. The three types of muscle can be distinguished by both their locations and their microscopic features.
Skeletal muscle is found attached to bones. It consists of long multinucleate fibers. The fibers run the entire length of the muscle they come from and so are usually too long to have their ends visible when viewed under the microscope. The fibers are relatively wide and very long, but unbranched. Fibers are not individual cells, but are formed from the fusion of thousands of precursor cells. This is why they are so long and why individual fibers are multinucleate (a single fiber has many nuclei). The nuclei are usually up against the edge of the fiber. There are striations in skeletal muscle. These are alternating dark and light bands perpendicular to the edge of the fiber that are present all along the fiber.
Cardiac muscle is only found in the heart. Its fibers are longer than they are wide, and they are striated, like skeletal muscle fibers. But, unlike skeletal muscle fibers, cardiac muscle
fibers have distinct ends to them, called intercalated discs. These are dark lines that run from one side of the fiber to the other. The intercalated discs are not much thicker than the striations, but they are usually darker and so distinct for that reason. One cardiac muscle fiber is the material between two intercalated discs. Cardiac muscle fibers are mononucleate, with only one nucleus per fiber, and they can sometimes be branched.
Smooth muscle is found in the walls of internal organs, such as the organs of the digestive tract, blood vessels, and others. It consists of mononucleate fibers with tapered edgesNo striations are visible in smooth muscle under the microscope. Because smooth muscle often is wrapping around the organ it is associated with, it can be hard to find an entire smooth muscle fiber in profile in a tissue slice on a microscope slide. Most of the fibers will be sectioned at angles or will be difficult to get into a single plane of focus, but a little bit of searching can usually turn up some with all of the defining characteristics visible.
LAB 3 EXERCISE \(1\)
In each of the three photomicrographs below, identify which type of muscle is present. List the defining visual characteristics of that type of muscle, and draw arrows to features on the photograph that illustrate each characteristic.
1. Obtain a slide of cardiac muscle tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
1. Obtain a slide of skeletal muscle tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
4. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
5. Obtain a slide of smooth muscle tissue from the slide box.
6. View the slide on an appropriate objective.
7. Fill out the blanks next to your drawing.
8. In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
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A&P Labs. Authored by: Ross Whitwam. Provided by: Mississippi University for Women. Located athttp://www.muw.eduLicenseCC BY-SA: Attribution-ShareAlike
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Exercise \(1\)AAuthored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located at: 141.214.65.171/Histology/Cardiovascular%20System/098HE_HISTO_40X.svs/view.apml. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
Exercise \(1\)B. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located at: 141.214.65.171/Histology/Basic%20Tissues/Muscle/058L_HISTO_40X.svs/view.apml. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
Exercise \(1\)C. Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located at: 141.214.65.171/Histology/Basic%20Tissues/Muscle/169_HISTO_40X.svs/view.apml. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/03%3A_Histology/3.04%3A_Distinguishing_Between_The_Three_Types_of_Muscle_Tissue.txt |
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Nervous tissue is the last of the four major categories of animal tissue. Nervous tissue comprises neurons, the cells specialized for the propagation of electrochemical signals, and neuroglia, the so-called “supporting cells” of nervous tissue. There are multiple types of neuroglia cells (sometimes called just glial cells), and most of them do assist neurons, but recent research has discovered that some of the neuroglia cells play an active role by themselves in the proper functioning of the central and peripheral nervous systems. It is turning out that neuroglia do more than just support neurons.
We will leave investigation of the neuroglia for later labs. In this lab, we will just focus on the structural features of neurons as they are visible under microscopic examination.
A neuron is typically represented as having the following features. A large cell body (sometimes known as the soma) in which the nucleus and other major organelles are found. Dendrites, which are usually represented as numerous small projections extending from the cell body. A single axon, which is usually represented as a large single projection extending from the cell body, much longer than any of the dendrites. Multiple axon terminals that branch off at the end of the axon. Figure \(1\) shows a typical drawn neuron showing all these features.
All of these features are important to the functioning of neurons. The cell body is where most of the cellular processes of the neuron take place: protein synthesis, metabolism, etc. The dendrites contact and pick up signals coming in from other cells (other neurons or sensory cells). The axon propagates the neuron’s electrical signal to its various targets. And the axon terminals actually contact (synapse with) those targets and send chemical signals to the targets to trigger changes in them. However, most neurons in the body do not look as clear-cut as the neuron in Figure \(1\). Some neurons have short axons. Others have dendrites that are almost as large as the axon. Others have two axons connected to the same cell body. Many have dendrites and axon terminals that are too thin and small too see clearly. Cell bodies tend to come in all sorts of sizes and shapes, and are not always bulbous lumps readily distinguished from the rest of the cell.
What’s more, nervous tissue is delicate and fragile. In creating preparations for slides to viewed under a microscope, some of the cells are smushed together, damaged, truncated, or torn. Other times, the nervous tissue cells are simply too densely packed together to make out individual details. Much of what is visible is not useful to the student looking to find an intact and well-defined neuron. However, a bit of searching will often reveal some intact neurons.
Often these are found at the edges of the preparation where the material is least dense. Keep in mind, you will often see neurons shaped distinctly different from those shown in textbooks and if Figure \(1\).
Figure \(2\) shows two different locations in the same slide. Figure \(2\) A shows a mass of neurons (stained black) in which it is difficult to distinguish the shape of even one neuron clearly. Figure 3.8 B shows a different area from the same slide, where the neurons are not so crowded, and where one neuron in particular (outlined in red) shows many of the structures we are interested in. So, keep in mind, it is important to know what you want to find in a particular slide, and it is important to search around the slide to find the best example of what you are looking for.
1. Obtain a slide of nervous tissue from the instructor. Use any nervous tissue except peripheral nerve, there are no nerve cell bodies in a peripheral nerve section.
2. View the slide on the second-highest objective. Search carefully until you find a clear, representative neuron in your field of view.
3. In the circle below, draw the neuron you found. Only draw the single neuron. Do not draw any of the other material. Draw your structures proportionately to their size in your microscope’s field of view.
4. Label any neural parts you can clearly recognize.
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Figure \(1\). A typical diagrammatic representation of neuron.. Authored by: Jonathan Haas. Located atcommons.wikimedia.org/w/inde...curid=18271454LicenseCC BY-SA: Attribution-ShareAlike
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Figure \(2\). Authored by: Kent Christensen, Ph.D., J. Matthew Velkey, Ph.D., Lloyd M. Stoolman, M.D., Laura Hessler, and Diedra Mosley-Brower. Provided by: University of Michigan Histology and Virtual Microscopy Learning Resources. Located at: 141.214.65.171/Histology/Central%20Nervous%20System/13270.svs/view.apml. License: CC BY-SA: Attribution- ShareAlike
3.06: Practice in Identifying Different Types of Tissues
Name 2 major components of Connective Tissue:
1. ____________________________________________________________
2. ____________________________________________________________
Name 2 components of Extracellular Matrix:
1. ____________________________________________________________
2. ____________________________________________________________
Name 3 major connective tissue fibers:
1. ____________________________________________________________
2. ____________________________________________________________
3. ____________________________________________________________
Name the 3 types of muscle tissue:
1. ____________________________________________________________
2. ____________________________________________________________
3. ____________________________________________________________
Name the 2 major cell types in neural tissue:
1. ____________________________________________________________
2. ____________________________________________________________
Epithelia can be divided into two basic groups based on the number of cell layers. These are:
1. ____________________________________________________________
2. ____________________________________________________________
Name two epithelia that can be considered an exception:
1. ____________________________________________________________
2. ____________________________________________________________
LAB 3 EXERCISE \(1\)
When storing a microscope, you should always follow this list:
1. Remove any slide found on the stage and return it to the slide box.
2. Rotate the smallest lens or no lens into place above the stage. Lower the stage a few turns.
3. Loosely coil the cord in your hand starting near the microscope and working toward the plug.
4. Hang the coiled cord over one ocular lens.
5. Look at the number on the back of the microscope, return that scope to its numbered box.
6. If there’s already a microscope in that numbered box, check its number and move it. If it is not numbered simply push it to the back of the box and place yours closer to the front. We have a few extra microscopes which we store in this fashion. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/03%3A_Histology/3.05%3A_Identifying_Neurons_Within_Nervous_Tissue.txt |
Skills to Develop
By the end of this section, you will be able to:
• Identify the components of the integumentary system
• Describe the layers of the skin and the functions of each layer
• Identify and describe the hypodermis and deep fascia
04: The Integumentary System
Although you may not typically think of the skin as an organ, it is in fact made of tissues that work together as a single structure to perform unique and critical functions. The skin and its accessory structures make up the integumentary system, which provides the body with overall protection. The skin is made of multiple layers of cells and tissues, which are held to underlying structures by connective tissue (Figure \(1\)). The deeper layer of skin is well vascularized (has numerous blood vessels). It also has numerous sensory, and autonomic and sympathetic nerve fibers ensuring communication to and from the brain.
The Epidermis
The epidermis is composed of keratinized, stratified squamous epithelium. It is made of four or five layers of epithelial cells, depending on its location in the body. It does not have any blood vessels within it (i.e., it is avascular). Skin that has four layers of cells is referred to as “thin skin.” From deep to superficial, these layers are the stratum basale, stratum spinosum, stratum granulosum, and stratum corneum. Most of the skin can be classified as thin skin. “Thick skin” is found only on the palms of the hands and the soles of the feet. It has a fifth layer, called the stratum lucidum, located between the stratum corneum and the stratum granulosum.
Dermis
The dermis might be considered the “core” of the integumentary system (derma- = “skin”), as distinct from the epidermis (epi- = “upon” or “over”) and hypodermis (hypo- = “below”). It contains blood and lymph vessels, nerves, and other structures, such as hair follicles and sweat glands. The dermis is made of two layers of connective tissue that compose an interconnected mesh of elastin and collagenous fibers, produced by fibroblasts.
Hair
Hair is a keratinous filament growing out of the epidermis. It is primarily made of dead, keratinized cells. Strands of hair originate in an epidermal penetration of the dermis called the hair follicle. The hair shaft is the part of the hair not anchored to the follicle, and much of this is exposed at the skin’s surface. The rest of the hair, which is anchored in the follicle, lies below the surface of the skin and is referred to as the hair root. The hair root ends deep in the dermis at the hair bulb, and includes a layer of mitotically active basal cells called the hair matrix.
The hair bulb surrounds the hair papilla, which is made of connective tissue and contains blood capillaries and nerve endings from the dermis.
\
Nails
The nail bed is a specialized structure of the epidermis that is found at the tips of our fingers and toes. The nail body is formed on the nail bed, and protects the tips of our fingers and toes as they are the farthest extremities and the parts of the body that experience the maximum mechanical stress. In addition, the nail body forms a back-support for picking up small objects with the fingers. The nail body is composed of densely packed dead keratinocytes. The epidermis in this part of the body has evolved a specialized structure upon which nails can form. The nail body forms at the nail root, which has a matrix of proliferating cells from the stratum basale that enables the nail to grow continuously. The lateral nail fold overlaps the nail on the sides, helping to anchor the nail body. The nail fold that meets the proximal end of the nail body forms the nail cuticle, also called the eponychium. The nail bed is rich in blood vessels, making it appear pink, except at the base, where a thick layer of epithelium over the nail matrix forms a crescent-shaped region called the lunula (the “little moon”). The area beneath the free edge of the nail, furthest from the cuticle, is called the hyponychium. It consists of a thickened layer of stratum corneum.
4.03: Exercises
EPIDERMAL LAYERS
& Physiology Lab Homework by Laird C. Sheldahl, under a Creative Commons Attribution-ShareAlike License 4.0
Lab 4 Exercise \(1\)
Integument Layers
Label the following: *Hair follicle * Sebaceous gland * Epidermis * Dermis (papillary layer) *Dermis (reticular layer) * Hypodermis * Arrector pili muscle * Sweat gland.
1 Hair follicle
2 Sebaceous gland
3 Epidermis
4 Dermis (papillary layer)
5 Dermis (reticular layer)
6 Hypodermis
7 Arrector pili muscle
8 Sweat gland
9
Lab 4 Exercise \(2\)
Epidermal Layers
Label the following: Stratum corneum * Stratum lucidum * Stratum granulosum * Stratum spinosum * Stratum basale * Dermal papilla * Sweat duct * Sweat pore * Stratified squamous epithelium * Dendritic cell * Melanocyte
1 Stratum corneum
2 Stratum lucidum
3 Stratum granulosum
4 Stratum spinosum
5 Stratum basale
6 Dermal papilla
7 Sweat duct
8 Sweat pore
9 Stratified squamous epithelium
10 Dendritic cell
11 Melanocyte
Lab 4 Exercise \(3\)
Skin Accessories
Label the parts of a hair follicle: Sebaceous gland * Hair shaft * Hair bulb * Hair root * Hair papilla
1 Sebaceous gland
2 Hair shaft
3 Hair bulb
4 Hair root
5 Hair papilla
Label the parts of a nail: Eponychium * Free edge * Hyponychium * Lunula * Nail matrix * Nail root * Lateral nail fold * Nail bed * Nail plate.
1 Eponychium
2 Free edge
3 Hyponychium
4 Nail matrix
5 Nail root
6 Lateral nail fold
7 Nail bed
8 Nail plate.
Lab 4 Exercise \(4\)
Integument Layers
Histology: Hair * Hair follicle * Dermal papilla * Sebaceous gland * Eccrine sweat gland * Stratified squamous epithelium * Areolar CT * Dense irregular CT.
1 Hair follicle
2 Dermal papilla
3 Sebaceous gland
4 Eccrine sweat gland
5 Stratified squamous epithelium
6 Areolar CT
7 Dense irregular CT.
4.2: Skin Models
Thin skin
Thick skin
Layers of the skin:
Epidermis
• Stratum corneum
• Stratum lucidum
• Stratum granulosum
• Stratum spinosum
• Stratum basale
Dermis
• Papillary layer
• Reticular layer
• Layers beneath the skin (subcutaneous)
Hypodermis
• Skin structures:
Dermal papillae
Hair follicle
• Root
• Shaft
• Hair follicle
• Papilla
• Arrector pilli (or piloerector m.)
• Bulb
Blood vessels
Nerve endings:
• Lamellar corpuscle
• Tactile corpuscle
Glands:
• Eccrine (merocrine) sweat gland
• Pore
• Duct
• Apocrine sweat gland
• Sebaceous glands
Nail
• Eponychium
• Hyponychium
• Free edge
• Lateral nail fold
• Proximal nail fold
• Lunula
• Nail bed
• Nail plate (body)
• Nail matrix
• Skin cells:
Keratinocytes
Melanocytes
Fibroblasts
• Tissues types:
Stratified squamous epithelium
Areolar CT
Dense irregular CT
CT – adipose
Glandular epithelium (simple or stratified cuboidal) | textbooks/bio/Human_Biology/Human_Anatomy_Lab/04%3A_The_Integumentary_System/4.01%3A_Layers_of_the_Skin.txt |
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The bones of the human body can be divided into two broad groups, the axial skeleton and the appendicular skeleton. The axial skeleton comprises the bones found along the central axis traveling down the center of the body. The appendicular skeleton comprises the bones appended to the central axis.
The axial skeleton consists of the bones of the skull, the bones of the inner ear (known as ossicles), the hyoid bone in the throat, and the bones of the vertebral column, including the sacrum and coccyx bones in the center of the pelvic girdle.
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Figure \(1\) The axial skeleton highlighted in blue.. Authored by: Axial_skeleton_diagram.svg: LadyofHats Mariana Ruiz Villarreal. Located at: https://commons.wikimedia.org/wiki/F...gram_blank.svg. License: Public Domain: No Known Copyright
The Bones of the Skull
There is only one movable joint in the skull. That is the joint connecting the lower jaw, or mandible, to the rest of the skull. All the other bones in the skull are firmly attached to one another by sutures.
Sutures are rigid immovable connections holding bones tightly to one another. Some of the sutures in the skull take a few months-to-years after birth to completely form.
The brain is encased in the cranium of the skull. The bones that make up the cranium are called the cranial bones. The remainder of the bones in the skull are the facial bones.
Figure \(2\) and Figure \(3\) show all the bones of the skull, as they appear from the outside. In Figure \(5\), some of the bones of the hard palate forming the roof of the mouth are visible because the mandible is not present. Figure \(9\) also shows the foramen magnum, the large hole at the base of the skull that allows the spinal cord to attach to the brain
The sphenoid bone, from the outside, appears to contribute to only a small portion of the cranium, but when the parietal bones are removed and the interior of the cranial cavity (where the brain would be housed) is viewed, you can see the butterfly-like shape of the sphenoid bone makes a large contribution to the floor of the cranial cavity. The ethmoid bone, which from the outside is only visible in the eye sockets and as the upper conchae (internal bumps) of the nasal cavity, also contributes to the floor of the cranial cavity. The contributions of these two bones to the floor of the cranial cavity are shown in Figure \(5\).
What is commonly referred to as the “cheekbone” is really the processes of two bones connected together: the zygomatic process of the temporal bone is sutured to the temporal process of the zygomatic bone to produce the zygomatic arch.
There are three prominent bone markings on the temporal bones. The external acoustic meatus is the opening that leads to the organs of the inner ear. The styloid process is a thin, pen-like projection where muscles and ligaments of the neck are attached. The mastoid process is a wide and rough projection that serves as another attachment point for neck muscles.
While all the bones of the skull, other than the mandible, are sutured to one another, the flat bones of the cranium are visibly sutured where they articulate to another. There are four different cranial sutures.
The coronal suture is the articulation point of the frontal bone with the two parietal bones. The sagittal suture is the articulation point between the two parietal bones.
The squamous sutures are the articulation points between the each temporal bone and the parietal bone superior to it.
The lambdoid suture is the articulation point between the occipital bone and the two parietal bones.
Lab 5 Exercise \(1\)
Models of human skulls may be found in the cabinets. On the models identify all of the following and on the diagrams below be able to label and/or color in all the following bones, processes, and foramina:
Bones
Sutures
Foramina
Processes
B1 – frontal
S1 – coronal
F1 – supraorbital
P1 – mastoid
B2 – parietal
S2 – squamous
F2 – infraorbital
P2 – styloid
B3 – occipital
S3 – lambdoid
F3 – mental
P3 – zygomatic
B4 – temporal
P4 temporal
B5 – sphenoid
B6 – ethmoid
B7 – lacrimal
B8 – nasal
B9 – maxilla
B10 – zygomatic
B11 – mandible
B12 – vomer
B13 – palatine (not visible here)
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Figure \(4\). The bones of the skull, inferior view, looking up. Mandible removed.. Authored by: Made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en. Located at: http://lifesciencedb.jp/bp3d/?lng=en. License: CC BY-SA: Attribution-ShareAlike
Figure \(5\). The interior of the cranial cavity, viewed from above and behind, with the parietal bones removed.. Authored by: Made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en. Located at: http://lifesciencedb.jp/bp3d/?lng=en. License: CC BY-SA: Attribution-ShareAlike
Figure \(6\). The cranial sutures.. Authored by: Made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en. Located at: http://lifesciencedb.jp/bp3d/?lng=en. License: Public Domain: No Known Copyright
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Figure \(2\). The bones of the skull, left lateral view.. Authored by: LadyofHats Mariana Ruiz Villarreal. Located at: https://en.wikipedia.org/wiki/File:H...ed_(bones).svg. License: Public Domain: No Known Copyright
Figure \(3\). The bones of the skull, anterior view.. Authored by: LadyofHats Mariana Ruiz Villarreal. Located at: https://commons.wikimedia.org/wiki/F...ront_bones.svg. License: Public Domain: No Known Copyright
Lab exercises \(1\). Located at: https://pixabay.com/en/skull-diagram...anatomy-41553/. License: Public Domain: No Known Copyright | textbooks/bio/Human_Biology/Human_Anatomy_Lab/05%3A_The_Axial_Skeleton/5.01%3A_Bones_of_the_Skull.txt |
The Hyoid Bone
The hyoid bone in the neck is the only bone in the body that does not articulate directly with at least one other bone. It is U-shaped and is held in place by, and helps anchor, muscles that connect to the floor of the mouth and the tongue. It helps provide greater movement of the tongue and larynx, and so is crucial to human speech.
In about 50% of strangulations, the hyoid bone is fractured.
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Figure \(1\). The hyoid bone. Authored by: Was a bee. Located at: https://commons.wikimedia.org/wiki/File:Hyoid_bone_animation.gif. License: CC BY-SA: Attribution-ShareAlike
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Figure \(1\). The hyoid bone. Authored by: Henry Grey, Henry Vandyke Carter. Provided by: Anatomy of the Human Body, 20th Edition (1918). Located at: https://commons.wikimedia.org/wiki/File:Gray1194.png. License: Public Domain: No Known Copyright
The Vertebral Column
The vertebral column is more colloquially called the backbone or the spine. It consists of 24 vertebrae bones, and two bones from the axial section of the pelvic girdle, the sacrum and the coccyx.
The vertebrae are divided into three groups. There are seven cervical vertebrae (names C1 through C7), twelve thoracic vertebrae (named T1 through T12), and five lumbar vertebrae (named L1 through L5).
You can use a meal-related mnemonic to remember them – imagine a crunchy breakfast at 7 am (7 cervical vertebrae), a tasty lunch at 12 noon (12 thoracic vertebrae), and a light dinner at 5 pm (5 lumbar vertebrae).
The first two cervical vertebrae have alternate names to C1 and C2. C1 is also called the atlas. In Greek mythology, Atlas was a titan who held the entire world on his shoulders. As the first vertebra in the column, Atlas in a sense holds up the skull. C2 is also called the axis. The axis allows both the skull and the atlas to rotate, so the head can be turned from side to side by neck muscles.
All three types of vertebrae have some structural features in common and some features that should allow you to readily distinguish vertebrae in one category from another.
All vertebrae, except C1 and C2, the atlas and axis, have a solid round portion on their anterior side called the body of the vertebra. The body is what allows the vertebrae in the vertebral column to be stacked upon one another, separated by pads of fibrocartilage called the intervertebral discs. The lower you go in the vertebral column, the larger the vertebrae’s bodies become.
The axis, or C2 vertebra, also has a bulbous vertical process not found in any of the other vertebrae. This is called the dens and it is what allows the axis vertebra above it to rotate.
Posterior to the vertebral body is a large opening in each vertebra called the vertebral foramen. This is the hole through which the spinal cord passes.
Posterior to the vertebral foramen there is a central process jutting out of each vertebra. This is the spinous process. It points more or less downwards when the vertebrae are correctly stacked into their column.
Surrounding the central spinous processes, there are other processes whose position and number vary depending on whether a vertebra is cervical, thoracic, or lumbar. Some of these processes allow a vertebra to articulate with the vertebrae superior and inferior to it, others, found only on thoracic vertebrae, allow a pair of ribs to articulate with the vertebra.
Figure \(2\) The three types of vertebrae. All are being viewed from behind. Note that C1 and C2 vertebrae, the atlas and axis, do not have vertebral bodies.
The sacrum is part of both the vertebral column and the pelvic girdle. It articulates with the intervertebral disc under the L5 vertebra above it, and with two coxal bones lateral to it.
The sacrum starts out as five vertebrae that fuse to form the one structure. This fusion is not complete until somewhere between the 18th and 30th year.
The coccyx is a vestigial tailbone. It is the evolutionary remnant of an ancestral species to humans that did have tails. It no longer serves as a functional tail, but some muscles, tendons, and ligaments do attach to it, making it useful. It forms from the fusion of usually three vertebrae, but a small proportion of the population have four or even five vertebrae in their coccyx.
Lab 5 Exercise \(1\)
1. Real or replica vertebrae bones may be found in the cabinets and on the back counters. Identify which in your group are the atlas, axis, cervical vertebrae, thoracic vertebrae, and lumbar vertebrae.
2. Stack your vertebrae in the correct order and in the correct orientation.
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Figure \(3\). The three types of vertebrae. All are being viewed from behind. Note that C1 and C2 vertebrae, the atlas and axis, do not have vertebral bodies. . Authored by: Anatomist90. Located at: https://commons.wikimedia.org/wiki/F..._vertebrae.jpg. License: CC BY- SA: Attribution-ShareAlike
Figure \(4\). The sacrum and coccyx. Authored by: OpenStax College. Located at: http://cnx.org/resources/e394fc66a09810ea99d0165de1078c8af2d4dc46/720_Sacrum_and_Coccyx.jpg. License: CC BY-SA: Attribution-ShareAlike
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Figure \(2\). The bones of the vertebral column. Authored by: Henry Grey, Henry Vandyke Carter. Located at: https://upload.wikimedia.org/wikiped...red_Ro_mod.JPG. Project: Anatomy of the Human Body, 20th Edition (1918). License: Public Domain: No Known Copyright | textbooks/bio/Human_Biology/Human_Anatomy_Lab/05%3A_The_Axial_Skeleton/5.02%3A_The_Vertebral_Column.txt |
Information
There is one last component of the axial skeleton we did not cover last lab: the thoracic cage, also called the rib cage. The thoracic cage surrounds and protects the heart and lungs in the thoracic cavity. It consists of the ribs, the sternum, and the thoracic vertebrae, to which the ribs articulate.
We examined the thoracic vertebrae last lab, so here we will only examine the ribs and sternum.
There are twelve pairs of ribs. The number is the same in both males and females. Each pair articulates with a different thoracic vertebra on the posterior side of the body. The most superior rib is designated rib 1 and it articulates with the T1 thoracic vertebrae. The rib below that is rib 2, and it connects to the T2 thoracic vertebra, and so on. Ten of the twelve ribs connect to strips of hyaline cartilage on the anterior side of the body. The cartilage strips are called costal cartilage (“costal” is the anatomical adjective that refers to the rib) and connect on their other end to the sternum.
On an individual rib, one end has various processes, facets, and bumps. This is the end that articulates with the vertebra. The other end is blunt and smooth. This is the end that connects to costal cartilage (unless it is a floating rib. See below.)
Ribs 1-7 are called the true ribs. Each true rib connects to its own strip of costal cartilage, which in turn connects to the sternum. Ribs 8-12 are called the false ribs. Ribs 8, 9, and 10 do connect to the sternum, but the costal cartilage of each of these ribs connects to the costal cartilage of the rib above it, rather than directly to the sternum. Ribs 11 and 12 do not have any costal cartilage connected to them at all, and in addition to being grouped in the false ribs, these two are also called floating ribs, to reflect that fact.
The sternum has three parts. The manubrium, at the superior end of the sternum, and wider than the rest of the bone, provides articulation points for the clavicles and for the costal cartilage extending from rib 1. The central, thin body provides articulation points for costal cartilage from ribs 2 through 7. The xiphoid process which hangs down at the inferior end of the process (“xiphoid” is from the Greek for sword), starts out as cartilage, and does not typically ossify into bone until an individual is about 40 years old.
Lab 5 Exercise \(1\)
1. Ribs may be found in the cabinets. On an individual rib, identify which end is the head and which is the anterior end.
2. On one of the intact skeletons in the lab, identify all the following components of the thoracic cage:
the true ribs the false ribs the floating ribs
costal cartilage sternum xiphoid process
manubrium sternal body
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Figure \(1\). Ribs and sternum.. Authored by: OpenStax College. Provided by: Anatomy & Physiology, Connexions Web site. https://cnx.org/contents/[email protected] athttps://cnx.org/resources/e1cbee95a6...1_Rib_Cage.jpgLicenseCC BY-SA: Attribution- ShareAlike
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Figure \(2\). True, false, and floating ribs.. Authored by: Cristobal Carrasco. Located athttps://commons.wikimedia.org/wiki/File:Costillas.pngLicenseCC BY-SA: Attribution-ShareAlike
Figure \(3\). The parts of a rib.. Authored by: Henry Vandyke Carter. Located atcommons.wikimedia.org/wiki/File:Image122.gifLicenseCC BY-SA: Attribution-ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/05%3A_The_Axial_Skeleton/5.03%3A_The_Thoracic_Cage__Ribs_and_Sternum.txt |
Lab 5 Exercise \(1\)
1. The instructor will provide you with a single rib from the human body. On that individual rib, identify which end is the head, and which is the anterior end.
2. On one of the intact skeletons in the lab, identify all the following components of the thoracic cage:
the true ribs the false ribs the floating ribs
costal cartilag sternum xiphoid process
manubrium sternal body
E
Lab 5 Exercise \(1\)
1. The instructor will provide you with a single rib from the human body. On that individual rib, identify which end is the head, and which is the anterior end.
2. On one of the intact skeletons in the lab, identify all the following components of the thoracic cage:
1 Zygomatic bones 6 Inferior orbital foramen
2 Lacrimal bones 7 Glabella
3 Coronal suture 8 Superior orbital fissure
4 Sagittal suture 9 Inferior orbital fissure
5 Superior orbital notch/foramen
1 Occipital bone 7 Mastoid process
2 Temporal bone 8 Glabella
3 Sphenoid bone 9 Styloid process
4 Maxilla 10 Lambdoid suture
5 Mandible 11 Zygomatic process of the temporal bone
6 External acoustic meatus 12 Temporal process of the zygomatic bone
13 Sutural bones
1 Zygomatic arch8 8 Internal acoustic meatus
2 Foramen ovale 9 Occipital condyle
3 Foramen spinosum 10 Mandibular fossa
4 Foramen lacerum 11 Ethmoid bone
5 Jugular foramen 12 Anterior cranial fossa
6 Carotid canal 13 Middle cranial fossa
7 Foramen magnum 14 Posterior cranial fossa.
1 Cervical vertebrae 8 Lumbar curvature
2 Cervical curvature 9 Atlas
3 Thoracic vertebrae 10 C1
4 Thoracic curvature 11 C2
5 5 12 12
6 Carotid canal 13 Axis
7 7 14 Dens
1 True ribs 3 False ribs
2 Floating ribs 4 Costal cartilages.
Lab 5 Exercise \(3\)
1. Using one of the full skeletons in the room, fill out the tables below with three or four steps to determine whether each individual upper limb bone comes from the anatomical left or anatomical right.
2. You can describe any features on that bone and which direction it has to face to allow you to determine whether that particular bone came from anatomical left or anatomical right.
3. You don't have to use anatomical jargon if you don't want. Use terms which will make sense to you when you read it again. Use as many steps as you need, not necessarily four.
Humerus – Anatomical left from anatomical right
1.
2.
3.
4.
Ulna – Anatomical left from anatomical right
1.
2.
3.
4.
Radius – Anatomical left from anatomical right
1.
2.
3.
4.
Lab 5 Exercise \(4\)
1. Using one of the full skeletons in the room, fill out the tables below with three or four steps to determine whether an individual coxal bone comes from the anatomical left or the anatomical right.
2. You can describe any features on the bone and which direction it has to face to allow you to determine whether that particular bone came from anatomical left or anatomical right.
3. You don't have to use anatomical jargon if you don't want. Use terms which will make sense to you when you read it again. Use as many steps as you need, not necessarily four.
Coxal bone – Anatomical left from anatomical right
1.
2.
3.
4.
e}\
e}
Exercise \(5\)
1. There will be an intact pelvis set up at the instructor’s station. Using the criteria in this image determine if the pelvis came from a male or female. Give three lines of evidence to support your conclusion.
Male or female?
Evidence 1
Evidence 2
Evidence 2
Exercise \(1\)
1. Using one of the full skeletons in the room, fill out the tables below with three or four steps to determine whether each individual lower limb bone comes from the anatomical left or anatomical right.
2. You can describe any features on that bone and which direction it has to face to allow you to determine whether that particular bone came from anatomical left or anatomical right.
3. You don't have to use anatomical jargon if you don't want. Use terms which will make sense to you when you read it again. Use as many steps as you need, not necessarily four.
Femur – Anatomical left from anatomical right
1.
2.
3.
4.
Tibia – Anatomical left from anatomical right
1.
2.
3.
4.
Fibula – Anatomical left from anatomical right
1.
2.
3.
4.
)
5.05: MODELS- Bones
Lab List Skull
Easy difficulty
Medium difficulty
Hard difficulty
Bones:
• Identify from external views:
• Parietal, Ethmoid, Sphenoid, Temporal,
• Occipital, Frontal
• Vomer, Inferior nasal concha, Nasal, Maxilla, Mandible, Palatine, Zygomatic, Lacrimal
Bones:
• identify bones than can be seen from the internal view (occipital, temporal, ethmoid, sphenoid, frontal)
• identify disarticulated skull bones
• (sphenoid and ethmoid)
Bones :
Landmarks:
• Mastoid Process, Mandibular Fossa,
• Sphenoid: Greater & lesser wings
• Sutures: sagittal, coronal, squamous, lambdoid.
Landmarks:
• Cranium: Glabella. Occipital condyles. Styloid process.
• Face: Alveolar processes of maxilla & mandible. lacrimal fossa
• Understand the difference between the zygomatic bone, zygomatic arch & zygomatic process (of the temporal bone).
• Temporal process of zygomatic bone
• Sphenoid: sella turcica, pterygoid plates.
• Mandible: ramus, condyle, body, coronoid process.
Landmarks:
• Sinuses: frontal, sphenoid, maxillary.
• Petrous & squamous part of temporal bone
• Palatine process of maxilla
• orbital margins
• Sphenoid: Hypophyseal fossa
• Ethmoid: lateral masses, superior & middle nasal conchae, crista galli, perpendicular plate, cribiform plates.
Foramen
• Foramen magnum
• External acoustic meatus
Foramen:
• mental, optic canal, superior orbital notch/foramen, infraorbital , olfactory foramina, carotid canal, jugular, ovale, lacerum, spinosum.
Foramen:
• stylomastoid, internal acoustic meatus, rotundum, hypoglossal, condylar canal, incisive, greater palatine.
Areas:
• Cranium (mnemonic: Pest of 6)
• Face (mnemonic: Virgil cannot make
• my pet zebra laugh)
Areas:
• Cranial fossa: anterior, middle, posterior.
Lab List Spine / Axial Skeleton
Bones:
• 7 cervical, 12 thoracic, 5 lumbar vertebrae
• Atlas & axis
• Sacrum & coccyx
• Sternum
• Hyoid
Bones:
• Identify disarticulated vertebrae as
• atlas/axis/cervical/thoracic or lumbar
Vertebrae landmarks:
Body
• Transverse process, spinous process
• Transverse foramen, Intervertebra
• foramenl, Vertebral foramen,
• Superior & inferior articular processes
Vertebrae landmarks:
• Costal facets
• Costal demifacets: inferior & superior.
Dens of axis
Vertebrae landmarks:
• Pedicles, Arch, Lamina, Superior & inferior notch
Sternum landmarks
• manubrium, body, xiphoid process
Sternum landmarks
• Jugular notch, sternal angle.
Sacrum landmarks
• Sacral foramen, Ala, Body, Auricular surface
Misc:
• Identify true/false/floating ribs on articulated spine.
• Know number of ribs.
• List 1 identifying characteristic of the 3 different types of vertebrae
Misc:
Misc:
• Spinal curves: cervical, thoracic, lumbar, sacral | textbooks/bio/Human_Biology/Human_Anatomy_Lab/05%3A_The_Axial_Skeleton/5.04%3A_Exercises.txt |
The appendicular skeleton is made up by the bones attached or appended to the axial skeleton. These are the bones of the limbs, hands, and feet, the bones of the pectoral (shoulder) girdles, and the coxal bones of the pelvic girdle.
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Figure \(1\. The appendicular skeleton highlighted in blue.. Authored by: LadyofHats Mariana Ruiz Villarreal. Located at: https://commons.wikimedia.org/wiki/F...gram_blank.svg. License: CC BY-SA: Attribution- ShareAlike
06: The Appendicular Skeleton
There is a pectoral girdle above each arm, consisting of the scapula (shoulder blade) and clavicle (collar bone.)
The scapula has a triangular body with three main processes coming off the body. On the posterior side is the spine of the scapula that ends in the acromion process, which articulates with the lateral end of the scapula. On the anterior side, facing into the body, is a right-angled process called the coracoid process. While at the superior lateral edge of the scapula the glenoid cavity, which serves as the socket for the head of the humerus bone. The glenoid cavity always points laterally, while the spine and acromion are posterior. This will help you determine whether a particular scapula comes from the anatomical left or right.
The clavicle is a vaguely S-shaped bone which articulates with the scapula in the lateral and articulates with the sternum at the medial end. The flatter broader end is the acromial extremity (end), which articulates with the acromion process of the scapula. The superior surface of the acromial extremity is smoother than its inferior surface, which can help you determine whether a particular clavicle bone comes from the left or right. The more circular end is the sternal extremity (end), which articulates with the manubrium of the sternum.
LAB 6 EXERCISE \(1\)
1. Using one of the full skeletons in the room, fill out the table below with three or four steps to determine whether an individual scapula comes from the anatomical left or anatomical right. You must use any features that are already filled in. The clavicle was done as an example.
2. You can describe any features on that bone and which direction it has to face to allow you to determine whether that particular bone came from anatomical left or anatomical right.
3. Sample instructions for the clavicle are provided as an example for all subsequent exercises. Use proper anatomical terminology. Use terms which will make sense to anyone schooled in anatomy if they read it. Use as many steps as you need, not necessarily four.
Clavicle – Anatomical left from anatomical right
1. Put the smooth side of the acromial end facing superiorly.
2. Put the curve in the middle of the bone facing anteriorly.
3. If the acromial end is on the right, the clavicle is from the anatomical right.
4. If the acromial end is on the left, the clavicle is from the anatomical left
Scapula – Fill in four steps to determine anatomical left from anatomical right
1. Glenoid cavity -
2.
3.
4. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/06%3A_The_Appendicular_Skeleton/6.01%3A_The_Pectoral_girdles.txt |
Information
For anatomists, the upper limb consists of the arm (the upper arm), the forearm (the lower arm), and the hand. The arm consists of a single bone, the humerus. The forearm consists of two bones, the ulna and radius. And the hand consists of 27 bones, which are grouped into the phalanges, metacarpals, and carpals.
The major processes and markings of the humerus, ulna, and radius bones are shown in Figures \(2\), \(3\), and \(4\), respectively.
The bones of the hands are divided into three groups. The carpals articulate with the ulna and radius bones of the forearm and are named after the carpus, or wrist. There are 8 carpal bones and each has its own name. In Figure \(5\) they are numbered so that 1-trapezium, 2-trapezoid, 3-capitate, 4-hamate, 5-pisiform, 6-triquetrum, 7-lunate, 8- scaphoid.
The metacarpals are five individual bones that are wrapped in muscle and collective tissue to create a single solid mass that serves as the palm of the hand. They are numbered I – V, starting with the metacarpal under the thumb and moving sequentially to the little finger.
Metacarpals are individually named according to the hand they come from and their number. So the metacarpal under the little finger in Figure 6.9 is named left metacarpal V.
There are 14 bones that make up the fingers. All are called phalanges (singular is phalanx). Each finger is made up of three phalanges, labelled the proximal, middle and distal phalanges as you move farther out from the metacarpals. Each thumb only has two phalanges, labelled proximal and distal. The phalanges are numbered I through V, like the metacarpals. Each phalanx then is named according to which hand it comes from, which number it is, and whether it is proximal, middle, or distal. So the second phalanx on the pointing finger in Figure is the left medial phalanx II.
LAB 6 EXERCISE \(1\):
1. Using one of the full skeletons in the room, fill out the table below with three or four steps to determine whether an individual bone comes from the anatomical left or anatomical right. You must use any features that are already filled in.
2. You can describe any features on that bone and which direction it has to face to allow you to determine whether that particular bone came from anatomical left or anatomical right.
3. Sample instructions for the clavicle are provided as an example for all subsequent exercises. Use proper anatomical terminology. Use terms which will make sense to anyone schooled in anatomy if they read it. Use as many steps as you need, not necessarily four.
Humerus – Anatomical left from anatomical right.
1. Olecranon fossa -
2.
3.
4.
Ulna – Anatomical left from anatomical right.
1. Radial notch -
2.
3.
4.
Radius – Anatomical left from anatomical right.
1. Styloid process -
2.
3.
4.
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Fig \(1\):. The bones of the left upper limb.. Authored by: Images in Figure 7-8 were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en. Located athttp://lifesciencedb.jp/bp3d/?lng=en. License: CC BY-SA: Attribution-ShareAlike
Figure \(2\):. The left humerus and its various processes and markings.. Authored by: Images in Figure 7-9 were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en. Located athttp://lifesciencedb.jp/bp3d/?lng=en. License: CC BY-SA: Attribution-ShareAlike
Figure \(3\):. The left ulna (in brown) and its major markings and processes.. Authored by: Images in Figure 7-10 were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=enLocated athttp://lifesciencedb.jp/bp3d/?lng=enLicenseCC BY-SA: Attribution- ShareAlike
Figure \(4\):. The left radius (in brown) and its major markings and processes.. Authored by: Images in Figure 7-11 were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=enLocated athttp://lifesciencedb.jp/bp3d/?lng=enLicenseCC BY-SA: Attribution- ShareAlike
Figure \(5\):. The bones of the left hand.. Authored by: Images in Figure 7-12 were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en. Located athttp://lifesciencedb.jp/bp3d/?lng=en. License: CC BY-SA: Attribution-ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/06%3A_The_Appendicular_Skeleton/6.02%3A_The_Upper_Limbs.txt |
Information
The pelvic girdle is name given the left and right coxal bones. Colloquially, these are known as the “hip bones”. The pelvic girdle is just the two coxal bones, but the pelvis itself is the bowl- like structure created by the two coxal bones joined in the anterior by the sacrum, and coccyx. (“Pelvis “comes from the Latin word for “basin”.)
At birth, each coxal bone starts out as three separate bones – the ilium, (ILL-ee-um),
the ischium, (ISH-ee-um) and the pubis (PYOO-bus) bones – joined by hyaline cartilage. Figure 6.10 shows what these bones look like initially. By the age of 25, these three bones have fully fused into a single coxal bone. We still subdivide the fully-formed coxal bone into three regions based on the positions of the three bones that fused to form it, each region named after the bone that gave rise to that region.
In a fully fused coxal bone, the ilium is the most superior portion, forming the “wing” that makes up the most prominent part of the coxal bone. The interior-facing side of this “wing” is called the iliac fossa. The ilium is where the sacrum attaches to each coxal bone to complete the pelvic bowl. This attachment point is alternatively called the sacroiliac joint, the sacroiliac articulation, or the iliac tuberosity, and is a rough surface. In anatomical position, the rough sacroiliac joint is always facing anterior.
In anatomical position, the ischium is posterior to the pubis. You “sit on the ischium” portion of the coxal bone. It is thicker and stronger than the pubis, allowing it to support your weight.
The large socket of the coxal bone is called the acetabulum (“ah-set-TAB-you-lum”). It faces laterally and is where the ball-like head of the femur bone articulates with the pelvis. Its name is derived from the Latin for vinegar cup, because of its cup-like shape. Inferior to the acetabulum is a large opening called the obturator foramen (“OB-tur-aye-tor for-AY-men”).
The major processes and markings of the coxal bone are shown in Figure \(2\).
LAB 6 EXERCISE \(1\):
Coxal bone – Anatomical left from anatomical right
1. Using one of the full skeletons in the room, fill out the table below with three or four steps to determine whether an individual bone comes from the anatomical left or anatomical right. You must use any features that are already filled in.
2. You can describe any features on that bone and which direction it has to face to allow you to determine whether that particular bone came from anatomical left or anatomical right.
3. Sample instructions for the clavicle are provided as an example for all subsequent exercises. Use proper anatomical terminology. Use terms which will make sense to anyone schooled in anatomy if they read it. Use as many steps as you need, not necessarily four.
1. Pubis -
2.
3.
4.
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Figure\(2\). The right coxal bone.. Authored by: OpenStax College. Located athttp://cnx.org/contents/[email protected] BY-SA: Attribution- ShareAlike
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Figure \(1\). The three bones of an unfused coxal bone.. Authored by: Henry Vandyke Carter. Located athttps://commons.wikimedia.org/wiki/File:Gray237.pngLicenseCC BY-SA: Attribution-ShareAlike
6.04: The Pelvis
Information
The pelvis is the bowl-shaped structure generated by the two coxal bones articulated with the sacrum and coccyx bones. On the anterior side of the pelvis, the pubis portions of the two coxal bones do not articulate with each other, but instead are joined with a small piece of cartilage called the pubic symphysis (“PYOO-bick SIM-fiss-is”)
Anatomists divide the pelvis into two regions. The false pelvis is superior and is surrounded by iliac fossa portions of the coxal bones and the upper portion of the sacrum. The true pelvis is inferior and is surrounded by the pubis and ischium portions of the coxal bones, in addition to the lower sections of the ilium and the sacrum. In women, the true pelvis defines the space babies must squeeze through during childbirth.
One of the few ways a skeleton stripped of all flesh can be reliably established as either male or female comes from examining the pelvis. The female pelvis can be distinguished from the male pelvis by a number or criterion, three of which are shown in Figure\(3\).
Most of these anatomical differences between the pelvises of males and females reflect the fact that only female pelvises have to serve as part of the birth canal, and these sex differences are not as pronounced in the skeletons of children who have not finished puberty.
In female pelvises, both the pelvic inlet and the pelvic outlet (not shown in Figure \(3\)) are wider and more oval-shaped than those in male pelvises. The pelvic inlet in males tends to be more heart-shaped (narrower on the dorsal side) and the pelvic outlet tends to be more narrow. The pubic arch, found immediately inferior to the pubic symphysis, tends to form an angle closer to 90° in females, but forms an angle closer to 60° in males. The sacrum in female pelvises tends to be less curved; in male pelvises, the sacrum is more curved and tends to impinge upon the space of the pubic outlet.
LAB 6 EXERCISE \(1\)
Obtain an intact pelvis from the cabinet. Using the criteria in Figure 6.14 determine if the pelvis came from a male or female. Give three lines of evidence to support your conclusion.
Male or female?
Evidence 1.
Evidence 2.
Evidence 3.
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Figure \(1\). The bones of the pelvis.. Authored by: BruceBlaus.. Located athttps://commons.wikimedia.org/wiki/F...723_Pelvis.pngLicensePublic Domain: No Known Copyright
Figure \(2\). The true pelvis vs. the false pelvis.. Located atcommons.wikimedia.org/wiki/F...is_(male)_03_-_superior_view.pngLicenseCC BY-SA: Attribution-ShareAlike
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Figure \(3\). Major differences that distinguish the adult male pelvis from the adult female pelvis.. Authored by: Henry Vandyke Carter. Located atcommons.wikimedia.org/wiki/F...lvis_LT.PNG%20LicenseCC BY-SA: Attribution-ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/06%3A_The_Appendicular_Skeleton/6.03%3A_The_Pelvic_Girdle.txt |
Information
For anatomists, the lower limb consists of the thigh (the upper leg), the leg (the lower leg), and the foot. The thigh consists of a single bone, the femur. The leg consists of two long bones, the tibia and fibula, and the sesamoid bone, the patella, that serves as the knee cap. The foot consists of 26 bones, which are grouped into the tarsalsmetatarsals, and phalanges.
The major processes and markings of the femur, patella, and tibia & fibula bones are shown in Figures \(2\), \(3\), and \(4\), respectively. The interosseous membrane connecting the tibia and fibula bones is shown in Figure \(5\)
Lateral condyle
Medial condyle
The bones of the foot are shown in Figure 6.20 below. The calcaneus is the heel bone, and the talus bone forms the ankle joint with the tibia and fibula. The calcaneus and tarsus are two of the seven tarsal bones that are posterior to the first long bones of the foot, the metatarsal bones. The bones of the toes are phalanges, the same name used for finger bones.
LAB 6 EXERCISE \(1\)
1. Using one of the full skeletons in the room, fill out the table below with three or four steps to determine whether an individual bone comes from the anatomical left or anatomical right. You must use any features that are already filled in.
2. You can describe any features on that bone and which direction it has to face to allow you to determine whether that particular bone came from anatomical left or anatomical right.
3. Sample instructions for the clavicle are provided as an example for all subsequent exercises. Use proper anatomical terminology. Use terms which will make sense to anyone schooled in anatomy if they read it. Use as many steps as you need, not necessarily four.
Femur – Anatomical left from anatomical right.
1 Head
2
3
4
Tibia – Anatomical left from anatomical right.
1 Tibial tuberosity
2
3
4
Fibula – Anatomical left from anatomical right.
1 Lateral malleolus
2
3
4
1. Using one of the full skeletons again, fill out the tables below with three or four steps to determine how to distinguish the calcaneus bone from the talus bone.
2. Use proper anatomical terminology. Use terms which will make sense to anyone schooled in anatomy if they read it. Use as many steps as you need, not necessarily three.
Calcaneus bone
1
2
3
Talus bone
1
2
3
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CC LICENSED CONTENT, ORIGINALA&P Labs. Authored by: Ross Whitwam. Provided by: Mississippi University for Women. Located athttp://www.muw.eduLicenseCC BY-SA: Attribution-ShareAlike
Figure \(1\). The bones of the left lower limb.. Authored by: Images in Figure 7-18 were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en.. Located at: . License: CC BY-SA: Attribution-ShareAlike
Figure \(2\). The left femur and its various processes and markings.. Authored by: Images in Figure 7-19 were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en.. Located at: . License: CC BY-SA: Attribution-ShareAlike
Figure \(4\). The left tibia and femur and their various processes and markings.. Authored by: Images in Figure 7-21 were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en.. Located at: . LicenseCC BY-SA: Attribution-ShareAlike
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Figure \(5\). The interosseous membrane of the left leg.. Authored by: Berichard. Located athttps://commons.wikimedia.org/wiki/F...ia_Fibula1.pngLicenseCC BY-SA: Attribution-ShareAlike
Figure \(6\). The bones of the left foot.. Authored by: Anatomist90. Located atupload.wikimedia.org/wikiped...7/Foot_bones_-_tarsus%2C_metatarsus_and_phalanges.jpgLicenseCC BY-SA: Attribution-ShareAlike
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Figure \(\PageIndex3}\). The Right patella.. Authored by: Palica . Located at: commons.wikimedia.org/wiki/F...e:Patella_post.jpgLicensePublic Domain: No Known Copyright | textbooks/bio/Human_Biology/Human_Anatomy_Lab/06%3A_The_Appendicular_Skeleton/6.05%3A_The_Lower_Limbs.txt |
1 Scapula Spine 10 Trochlea
2 Coracoid process 11 Capitulum
3 Acromion 12 Olecranon fossa
4 Supra-spinous fossa 13 Coronoid fossa
5 Infra-spinous fossa 14 Radial fossa
6 Sub-scapular fossa 15 Medial epicondyle
7 Glenoid cavity Humerus Head 16 Lateral epicondyle
8 Neck 17 Greater tubercle
9 (surgical and anatomical) 18 Lesser tubercle
1 Femur 9 Lateral epicondyle
2 Head 10 Linea aspera
3 Neck 11 Olecranon fossa
4 Greater trochanter 12 Tibia
5 Lesser trochanter 13 Medial malleolus
6 Medial condyle 14 Tibial tuberosity
7 Lateral condyle 15 Fibula
8 Medial epicondyl 16 Lateral malleolus
17 Patella
1 Radius 13 Triquetral
2 Styloid process 14 Pisiform
3 Radial head 15 Trapezium
4 Radial tuberosity 16 Trapezoid
5 Ulna 17 Capitate
6 Olecranon 18 Hamate
7 Trochlear notch 19 Metacarpals
8 Coronoid process 20 Lateral malleolus
9 Styloid process 21 Phalanges proximal
10 Carpals 22 middle
11 Scaphoid 23 distal
12 Lunate 24 Pollex
Tarsals * Talus * Calcaneus * Cuboid * Navicular * Medial,intermediate & lateral cuneiforms * Metatarsals Phalanges * proximal * middle * distal. Hallux Clavicle * sternal end * acromial end
1 Tarsals 10 proximal
2 Talus 11 middle
3 Calcaneus 12 distal
4 Cuboid 13 Hallux
5 Navicular 14 Clavicle
6 Medial cuneiforms 15 sternal end
7 intermediate cuneiforms lateral 16 acromial end
8 lateral cuneiforms
9 Metatarsals Phalanges
6.07: MODELS- Upper and Lower Limbs
Upper Limb
Easy difficulty
Medium difficulty
Hard difficulty
Bones:
• Clavicle, Scapula
• Humerus, Ulna, Radius,
• Metacarpals, Carpals (group),
• Phalanges (distal, medial, proximal)
Bones
• Scaphoid, Lunate, Triquetral, Pisiform,
• Trapezium, Trapezoid, Capitate, Hamate.
• Pollex
Humerus landmarks:
• Head
• Medial & Lateral epidondyles.
Humerus landmarks :
• Anatomical neck, Surgical neck
• Greater & lesser tubercles
• Deltoid tuberosity
• Trochlea, Capitulum.
Humerus landmarks:
Coronoid fossa, Olecrannon fossa, Radial fossa.
Ulna landmarks:
• Olecranon process.
Ulna landmarks :
• Coronoid process, Trochlear notch, Styloid process.
Radius landmarks :
• Head
• Styloid process.
Ulna landmarks:
• Radial notch.
Radius landmarks
• Radial tuberosity.
Clavicle landmarks:
• Acromial end, Sternal end.
Clavicle landmarks:
• Conoid tubercle.
Scapula landmarks :
• Acromion, Coracoid process
• Glenoid cavity
• Spine.
Scapula landmarks
• Supraspinous fossa, Infraspinous,
• fossa, Subscapular fossa.
Lower Limb
Easy difficulty
Medium difficulty
Hard difficulty
Bones:
• Ilium, Ischium, Pubis,
• Femur, Patella, Tibia, Fibula,
• Tarsals (group), Metatarsals.
Bones
• Talus
• Calcaneus
Bones
• Cuboid
• Navicular
• Cuneiforms: medial, intermediate, lateral
• Hallux
Hip landmarks:
• Pubis symphysis
Hip landmarks
• Acetabulum
• Obturator foramen
• Iliac crest
Hip landmarks:
• Anterior Superior Iliac spine
• Anterior Inferior Iliac spine
• Posterior Superior Iliac spine
• Posterior Inferior Iliac spine
• Greater sciatic notch
• Iliac fossa
• Ischial tuberosity
• Ischial spine
Femur landmarks:
• Head
• Neck
• Greater trochanter , Lesser
• trochanter.
Femur landmarks :
• Lateral & medial condyles
• Lateral & medial epicondyles
Femur landmarks:
• Fovea capitis
• Gluteal tuberosity
• Linea aspera
Tibia landmarks :
• Medial malleolus.
Fibula landmarks:
• Lateral malleolus.
Tibia landmarks:
• Tibial tuberosity | textbooks/bio/Human_Biology/Human_Anatomy_Lab/06%3A_The_Appendicular_Skeleton/6.06%3A_Exercises.txt |
The adult human body has 206 bones, and with the exception of the hyoid bone in the neck, each bone is connected to at least one other bone. Joints are the location where bones come together. Many joints allow for movement between the bones. At these joints, the articulating surfaces of the adjacent bones can move smoothly against each other. However, the bones of other joints may be joined to each other by connective tissue or cartilage. These joints are designed for stability and provide for little or no movement. Importantly, joint stability and movement are related to each other. This means that stable joints allow for little or no mobility between the adjacent bones. Conversely, joints that provide the most movement between bones are the least stable. Understanding the relationship between joint structure and function will help to explain why particular types of joints are found in certain areas of the body.
The articulating surfaces of bones at stable types of joints, with little or no mobility, are strongly united to each other. For example, most of the joints of the skull are held together by fibrous connective tissue and do not allow for movement between the adjacent bones. This lack of mobility is important, because the skull bones serve to protect the brain. Similarly, other joints united by fibrous connective tissue allow for very little movement, which provides stability and weight-bearing support for the body. For example, the tibia and fibula of the leg are tightly united to give stability to the body when standing. At other joints, the bones are held together by cartilage, which permits limited movements between the bones. Thus, the joints of the vertebral column only allow for small movements between adjacent vertebrae, but when added together, these movements provide the flexibility that allows your body to twist, or bend to the front, back, or side. In contrast, at joints that allow for wide ranges of motion, the articulating surfaces of the bones are not directly united to each other. Instead, these surfaces are enclosed within a space filled with lubricating fluid, which allows the bones to move smoothly against each other. These joints provide greater mobility, but since the bones are free to move in relation to each other, the joint is less stable. Most of the joints between the bones of the appendicular skeleton are this freely moveable type of joint. These joints allow the muscles of the body to pull on a bone and thereby produce movement of that body region. Your ability to kick a soccer ball, pick up a fork, and dance the tango depend on mobility at these types of joints.
7.01: Knee Joint
The knee joint is the largest joint of the body (Figure \(1\)). It actually consists of three articulations. The femoropatellar joint is found between the patella and the distal femur. The medial tibiofemoral joint and lateral tibiofemoral joint are located between the medial and lateral condyles of the femur and the medial and lateral condyles of the tibia. All of these articulations are enclosed within a single articular capsule. The knee functions as a hinge joint, allowing flexion and extension of the leg. This action is generated by both rolling and gliding motions of the femur on the tibia. In addition, some rotation of the leg is available when the knee is flexed, but not when extended. The knee is well constructed for weight bearing in its extended position, but is vulnerable to injuries associated with hyperextension, twisting, or blows to the medial or lateral side of the joint, particularly while weight bearing.
At the femoropatellar joint, the patella slides vertically within a groove on the distal femur. The patella is a sesamoid bone incorporated into the tendon of the quadriceps femoris muscle, the large muscle of the anterior thigh. The patella serves to protect the quadriceps tendon from friction against the distal femur. Continuing from the patella to the anterior tibia just below the knee is the patellar ligament. Acting via the patella and patellar ligament, the quadriceps femoris is a powerful muscle that acts to extend the leg at the knee. It also serves as a “dynamic ligament” to provide very important support and stabilization for the knee joint.
The medial and lateral tibiofemoral joints are the articulations between the rounded condyles of the femur and the relatively flat condyles of the tibia. During flexion and extension motions, the condyles of the femur both roll and glide over the surfaces of the tibia. The rolling action produces flexion or extension, while the gliding action serves to maintain the femoral condyles centered over the tibial condyles, thus ensuring maximal bony, weight-bearing support for the femur in all knee positions. As the knee comes into full extension, the femur undergoes a slight medial rotation in relation to tibia. The rotation results because the lateral condyle of the femur is slightly smaller
than the medial condyle. Thus, the lateral condyle finishes its rolling motion first, followed by the medial condyle. The resulting small medial rotation of the femur serves to “lock” the knee into its fully extended and most stable position. Flexion of the knee is initiated by a slight lateral rotation of the femur on the tibia, which “unlocks” the knee. This lateral rotation motion is produced by the popliteus muscle of the posterior leg.
Located between the articulating surfaces of the femur and tibia are two articular discs, the medial meniscus and lateral meniscus (see Figure \(1\)b). Each is a C-shaped fibrocartilage structure that is thin along its inside margin and thick along the outer margin. They are attached to their tibial condyles, but do not attach to the femur. While both menisci are free to move during knee motions, the medial meniscus shows less movement because it is anchored at its outer margin to the articular capsule and tibial collateral ligament. The menisci provide padding between the bones and help to fill the gap between the round femoral condyles and flattened tibial condyles.
Some areas of each meniscus lack an arterial blood supply and thus these areas heal poorly if damaged.
The knee joint has multiple ligaments that provide support, particularly in the extended position (see Figure \(1\)c). Outside of the articular capsule, located at the sides of the knee, are two extrinsic ligaments. The fibular collateral ligament (lateral collateral ligament) is on the lateral side and spans from the lateral epicondyle of the femur to the head of the fibula. The tibial collateral ligament (medial collateral ligament) of the medial knee runs from the medial epicondyle of the femur to the medial tibia. As it crosses the knee, the tibial collateral ligament is firmly attached on its deep side to the articular capsule and to the medial meniscus, an important factor when considering knee injuries. In the fully extended knee position, both collateral ligaments are taut (tight), thus serving to stabilize and support the extended knee and preventing side-to-side or rotational motions between the femur and tibia.
The articular capsule of the posterior knee is thickened by intrinsic ligaments that help to resist knee hyperextension. Inside the knee are two intracapsular ligaments, the anterior cruciate ligament and posterior cruciate ligament. These ligaments are anchored inferiorly to the tibia at the intercondylar eminence, the roughened area between the tibial condyles. The cruciate ligaments are named for whether they are attached anteriorly or posteriorly to this tibial region. Each ligament runs diagonally upward to attach to the inner aspect of a femoral condyle. The cruciate ligaments are named for the X-shape formed as they pass each other (cruciate means “cross”). The posterior cruciate ligament is the stronger ligament. It serves to support the knee when it is flexed and weight bearing, as when walking downhill. In this position, the posterior cruciate ligament prevents the femur from sliding anteriorly off the top of the tibia. The anterior cruciate ligament becomes tight when the knee is extended, and thus resists hyperextension. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/07%3A_Joints/07.00%3A_Introduction.txt |
The elbow joint is a uniaxial hinge joint formed by the humeroulnar joint, the articulation between the trochlea of the humerus and the trochlear notch of the ulna. Also associated with the elbow are the humeroradial joint and the proximal radioulnar joint. All three of these joints are enclosed within a single articular capsule (Figure \(1\)).
The articular capsule of the elbow is thin on its anterior and posterior aspects, but is thickened along its outside margins by strong intrinsic ligaments. These ligaments prevent side-to-side movements and hyperextension. On the medial side is the triangular ulnar collateral ligament. This arises from the medial epicondyle of the humerus and attaches to the medial side of the proximal ulna. The strongest part of this ligament is the anterior portion, which resists hyperextension of the elbow. The ulnar collateral ligament may be injured by frequent, forceful extensions of the forearm, as is seen in baseball pitchers. Reconstructive surgical repair of this ligament is referred to as Tommy John surgery, named for the former major league pitcher who was the first person to have this treatment.
The lateral side of the elbow is supported by the radial collateral ligament. This arises from the lateral epicondyle of the humerus and then blends into the lateral side of the annular ligament. The annular ligament encircles the head of the radius. This ligament supports the head of the radius as it articulates with the radial notch of the ulna at the proximal radioulnar joint. This is a pivot joint that allows for rotation of the radius during supination and pronation of the forearm.
7.03: Shoulder Joint
The shoulder joint is called the glenohumeral joint. This is a ball-and-socket joint formed by the articulation between the head of the humerus and the glenoid cavity of the scapula (Figure 7.3). This joint has the largest range of motion of any joint in the body. However, this freedom of movement is due to the lack of structural support and thus the enhanced mobility is offset by a loss of stability.
The large range of motions at the shoulder joint is provided by the articulation of the large, rounded humeral head with the small and shallow glenoid cavity, which is only about one third of the size of the humeral head. The socket formed by the glenoid cavity is deepened slightly by a small lip of fibrocartilage called the glenoid labrum, which extends around the outer margin of the cavity. The articular capsule that surrounds the glenohumeral joint is relatively thin and loose to allow for large motions of the upper limb. Some structural support for the joint is provided by thickenings of the articular capsule wall that form weak intrinsic ligaments. These include the coracohumeral ligament, running from the coracoid process of the scapula to the anterior humerus, and three ligaments, each called a glenohumeral ligament, located on the anterior side of the articular capsule. These ligaments help to strengthen the superior and anterior capsule walls.
However, the primary support for the shoulder joint is provided by muscles crossing the joint, particularly the four rotator cuff muscles. These muscles (supraspinatus, infraspinatus, teres minor, and subscapularis) arise from the scapula and attach to the greater or lesser tubercles of the humerus. As these muscles cross the shoulder joint, their tendons encircle the head of the humerus and become fused to the anterior, superior, and posterior walls of the articular capsule. The thickening of the capsule formed by the fusion of these four muscle tendons is called the rotator cuff. Two bursae, the subacromial bursa and the subscapular bursa, help to prevent friction between the rotator cuff muscle tendons and the scapula as these tendons cross the glenohumeral joint. In addition to their individual actions of moving the upper limb, the rotator cuff muscles also serve to hold the head of the humerus in position within the glenoid cavity. By constantly adjusting their strength of contraction to resist forces acting on the shoulder, these muscles serve as “dynamic ligaments” and thus provide the primary structural support for the glenohumeral joint.
Injuries to the shoulder joint are common. Repetitive use of the upper limb, particularly in abduction such as during throwing, swimming, or racquet sports, may lead to acute or chronic inflammation of the bursa or muscle tendons, a tear of the glenoid labrum, or degeneration or tears of the rotator cuff. Because the humeral head is strongly supported by muscles and ligaments around its anterior, superior, and posterior aspects, most dislocations of the humerus occur in an inferior direction. This can occur when force is applied to the humerus when the upper limb is fully abducted, as when diving to catch a baseball and landing on your hand or elbow.
Inflammatory responses to any shoulder injury can lead to the formation of scar tissue between the articular capsule and surrounding structures, thus reducing shoulder mobility, a condition called adhesive capsulitis (“frozen shoulder”)
7.04: Execises
LAB 7 EXERCISES \(1\)
Match the following
1 Joint cavity
2 Articular cartilages
3 Fibrous capsule
4 Synovial membrane
Organize the following types of joints into a diagram or table
1 Ball & socket 7 Plane
2 Cartilaginous 8 Saddle
3 Condylar 9 Suture
4 Fibrous 10 Synovial
5 Hinge 11 Synchondrosis
6 Pivot 12 Symphysis
LAB 7 EXERCISES \(2\)
Identify the components of the shoulder and hip joints
1 Bursa
2 Glenoid labrum
3 Synovial membrane
4 Head (of humerus)
5 Glenoid cavity of scapula
Draw and Label the following on this hip socket cross-section
1 Articular cartilage
2 Acetabular labrum
3 Ligamentum teres
4 Synovial membrane
5 Synovial cavity
6 Joint capsule
LAB 7 EXERCISES \(3\)
Identify the components of the knee joint
1 Medial condyle
2 Lateral condyle
3 Medial collateral ligament
4 Medial meniscus
5 Lateral meniscus
6 Anterior cruciate ligament
7 Posterior cruciate ligamen
1 Anterior cruciate ligament
2 Posterior cruciate ligament
3 Medial collateral ligament
4 Lateral collateral ligament
5 Medial meniscus
6 Lateral meniscus
7 Medial condyle
8 Lateral condyle
LAB 7 EXERCISES \(4\)
Identify the components of the knee joint
1 Quadriceps femoris tendon
2 Patella
3 Bursae
4 Fat pad
5 Joint cavity
6 Synovial membrane
7 Meniscus
8 Tibia
9 Articular cartilage
1 Nucleus pulposis
2 Annulus fibrosus
3 Superior articular process
4 Inferior articular process.
7.05: Models- Knee Shoulder Hip and Spine
Knee joint:
• Anterior cruciate ligament
• Posterior cruciate ligament
• Fibular (Lateral) collateral ligament
• Tibial (Medial) collateral ligament
• Lateral meniscus
• Medial meniscus
• Quadriceps tendon
Elbow joint:
• Fibrous capsule
• Ulnar collateral ligament
• Annular ligament of radius
• Radial collateral ligament
Shoulder joint
• Fibrous capsule
• Coraco-humeral ligament
• Coraco-acromial ligament
• Gleno-humeral ligaments
• Transverse humeral ligament
Vertebral column
• Vertebral disc
Skeletal structures to review:
• Acromion
• Acetabulum
• Capitulum
• Clavicle
• Coracoid process
• Femur
• Fibula
• Glenoid cavity
• Head of humerus
• Head of femur
• Humerus
• Lateral condyle of femur
• Medial condyle of femur
• Olecranon
• Olecranon fossa
• Patella
• Popliteal surface
• Radius
• Radial fossa
• Scapula
• Trochlea
• Ulna | textbooks/bio/Human_Biology/Human_Anatomy_Lab/07%3A_Joints/7.02%3A_Elbow_Joint.txt |
Skeletal muscle is found attached to bones. It consists of long multinucleate fibers. The fibers run the entire length of the muscle they come from and so are usually too long to have their ends visible when viewed under the microscope. The fibers are relatively wide and very long, but unbranched. Fibers are not individual cells, but are formed from the fusion of thousands of precursor cells. This is why they are so long and why individual fibers are multinucleate (a single fiber has many nuclei). The nuclei are usually up against the edge of the fiber. There are striations in skeletal muscle. These are alternating dark and light bands perpendicular to the edge of the fiber that are present all along the fiber.
08: The Axial Muscles
Figure \(1\) lists the muscles of the head and neck that you will need to know. A single platysma muscle is only shown in the lateral view of the head muscles in Figure 8.1. There are two platysma muscles, one on each side of the neck. Each is a broad sheet of a muscle that covers most of the anterior neck on that side of the body. The other anterior neck muscles are below them, and most models have the platysma muscles cut away to show the deeper muscles. The platysma muscles help pull down the lower jaw (mandible.)
Under the platysma are two sternocleidomastoid muscles. One on each side of the neck. These muscles have two origins, one on the sternum and the other on the clavicle. They insert on the mastoid process of the temporal bone. They can flex or extend the head, or can rotate the towards the shoulders.
The epicranius muscle is also very broad and covers most of the top of the head. The epicranius muscle includes a middle section which is all aponeurosis. The actual muscle tissue is only found over the forehead (the portion of the muscle called the epicranius frontalis; sometimes called the frontal belly of the epicranius) and the back of the head (the portion of the muscle called the epicranius occipitalis; sometimes called the occipital belly of the epicranius).
The buccinator muscles, one on each side of the face, compress the cheeks when contracted. The name is from the Latin for trumpet, which requires blowing air out of the cheeks to play, and also reflects the anatomical adjective for the cheek, buccal.
The two masseter muscles are also on each side of the face. They close the jaw when contracted. Its name is derived from the same Greek root as mastication, which means to chew.
The zygomaticus major muscles and the zygomaticus minor muscles are found on each side of the face both have their origins on the zygomatic bone. They both can change the shape of the mouth by elevating it.
LAB 8 EXERCISE 8-1
1. The following are muscles of facial expression. For each, give its location and describe its action when it contracts. Complete figure \(1\): above by adding in any muscles found in the table below.
Muscle
Location
Action when contracted
Epicranius frontalis
Epicranius occipitalis
Orbicularis oculi
Zygomaticus major
Zygomaticus minor
Buccinator
Orbicularis oris
Platysma
Levator labii sup.
Depressor labii inf.
Levator anguli oris
Depressor anguli oris
2. The following are muscles of mastication. For each, give its location and describe its action when it contracts.
Muscle
Location
Action when contracted
Masseter
Temporalis
Medial pterygoid
Lateral pterygoid
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Figure \(1\). The muscles of the head.. Authored by: OpenStax College. Located at: https://cnx.org/resources/9b369a7466..._the_Muscles_o f_Facial_Expressions.jpg. License: CC BY-SA: Attribution-ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/08%3A_The_Axial_Muscles/8.01%3A_The_Muscles_of_the_Head_and_Neck.txt |
Information
Figures \(1\) and \(2\) shows many of the muscles of the body’s trunk that you need to know, as well as some of the muscles of the arms and legs you will learn about in the next lab.
The deltoid muscles are the triangular muscles over each shoulder.
Some of the trunk muscles have been given nicknames by gym rats. For instance, the pecs are the pectoralis major muscles at each breast.
Lats are the latissimus dorsi muscle that covers most of the lower back with its lateral fibers.
The upper back is covered by the large trapezius muscle that is almost diamond-shaped as it extends from the neck, out to the shoulders, then tapers in midway down the back.
Obliques are the external oblique muscle whose fibers angle down as it covers both sides of the abdominal region. The external oblique muscle has two sets of fibers, which cover the left and right abdomen, that are connected by a wide aponeurosis sheet in the center of the abdomen. In most muscle models that aponeurosis sheet is cut away to reveal the rectus abdominis muscles below it.
What gym rats call the core muscles are three layers of muscle that sit over the abdomen. These layers are shown in Figure \(3\). The outer layer is the external oblique muscle, with its aponeurosis covering the medial abdomen. Under the external oblique are the internal obliques on the sides of the abdomen and the rectus abdominis muscle in-between the internal obliques. The fibers of the internal obliques run up at an angle, opposite in direction to the fiber angle of the external obliques. The rectus abdominis muscle is also known as the abs. The deepest layer has the transverse abdominis muscle, whose fibers run laterally. Its fibers are concentrated at the sides of the abdomen and, like the external oblique, has an aponeurosis covering the medial abdomen under the rectus abdominis.
Extending from the back and wrapping around the sides of the rib cage is the serratus anterior muscle. This muscle’s anterior edges are serrated like the teeth of a saw because this muscle’s origins are on ribs 1 through 8 and each serration is the attachment point to another rib. This muscle is shown in Figure \(4\).
LAB 8 EXERCISE \(1\)
1. The following are muscles that move the pectoral girdle. For each, give its location and describe its action when it contracts.
Muscle
Location
Action when contracted
Trapezius
Pectoralis minor
Serratus anterior
2. The following are muscles that move the arm. For each, give its location and describe its action when it contracts.
Muscle
Location
Action when contracted
Pectoralis major
Latissimus dorsi
Deltoid
3. The following are muscles of the abdominal wall. For each, give its location and describe its action when it contracts.
Muscle
Location
Action when contracted
Rectus abdominis
External oblique
Internal oblique
Transversus abdominis
4. Label the indicated facial muscles in Figure \(5\). (CC-BY-SA, Wikimedia)
1
2
3
4
5
6
7
8
9
10
11
12
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Figure \(4\). The external muscles of the body, lateral view.. Authored by: This image was made out of, or made from, content published in a BodyParts3D/Anatomography web site. . Provided by: BodyParts3D, u00a9 The Database Center for Life Science licensed under CC Attribution-Share Alike 2.1 Japan.. Located at: https://commons.wikimedia.org/wiki/F...es_lateral.png. License: CC BY-SA: Attribution- ShareAlike
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Figure \(1\) The major muscles of the body, anterior view. Anatomical right shows superficial muscles. Anatomical left shows deep muscles.. Authored by: OpenStax College. Located at: https://cnx.org/resources/8a3b1231f3...s_of_Muscles.j pg. License: CC BY-SA: Attribution-ShareAlike
Figure \(2\). The major muscles of the body, posterior view. Anatomical right shows superficial muscles. Anatomical left shows deep muscles.. Authored by: OpenStax College. Located at: https://cnx.org/resources/8a3b1231f3...s_of_Muscles.j pg. License: CC BY-SA: Attribution-ShareAlike
Figure \(4\). The external muscles of the body, lateral view.. Authored by: OpenStax College. Located at: https://cnx.org/resources/33fa36d780...he_Abdomen.jpg. License: CC BY-SA: Attribution-ShareAlike
Figure \(5\). Facial muscles.. Authored by: Patrick J. Lynch. Located at: https://commons.wikimedia.org/wiki/F...ad_anatomy.jpg. License: CC BY-SA: Attribution-ShareAlike
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Figure \(3\) The three layers of muscles in the abdomen.. Authored by: Arcadian. Located at: https://commons.wikimedia.org/wiki/F...nk_muscles.jpg. License: Public Domain: No Known Copyright | textbooks/bio/Human_Biology/Human_Anatomy_Lab/08%3A_The_Axial_Muscles/8.02%3A_The_Muscles_of_the_Trunk.txt |
LAB 8 EXERCISE \(1\)
Identify the following
1 epimysium
2 perimysium
3 endomysium
4 muscle fascicle
5 muscle fiber
1 origin
2 insertion
3 extensor
4 flexor
List the defining visual characteristics of this muscle and draw arrows to features on the photograph that illustrate each characteristic.
Muscle Type
Visual characteristics
1. _____________________________________________
2. _____________________________________________
3. _____________________________________________
4. _____________________________________________
1. Obtain a slide of skeletal muscle tissue from the slide box.
2. View the slide on an appropriate objective.
3. Fill out the blanks next to your drawing.
In the circle below, draw a representative sample of key features you identified, taking care to correctly and clearly draw their true shapes and directions. Draw your structures proportionately to their size in your microscope’s field of view.
Total Magnification: _________________
Type of eptihelium: _________________
Source of tissue: ____________________
Function of this phase: _______________
__________________________________
Type of muscle tissue:
8.04: MODELS- Head and Neck and Torso
Name
Action
Origin
Insertion
Muscles of facial expression
Frontalis (epicranius)
Raises eyebrows , wrinkles forehead
Frontal Bone
Skin of the brow
Occipitalis (epicranius)
Pull scalp posteriorly
Occipital bone
Aponeurosis connecting to frontalis
Orbicularis oris
Closes mouth
Maxillae and Mandible
Lips
Zygomaticus (major/minor)
smile
Zygomatic Bone
mouth
Orbicularis oculi
Closes eye
Margin of Orbit
Eyelid
Masseter
Elevates mandible
Zygomatic Arch
Mandible
Temporalis
Elevates mandible
Temporal Bone
Mandible
Buccinator
Presses cheek inward
Maxillae and Mandible
orbicularis oris
Muscles of the head, vertebral column and abdominal wall
Splenius capitis
extend + laterally flex head
upper spine
Temporal & occipital
Sternocleidomastoid
Flexes + rotates neck (also elevates ribs)
Sternum & Clavicle
Mastoid Process
Scalenes
Flexes neck (also elevates ribs)
Cervical Vertebrae
1st Two Ribs
Rectus abdominus
Flexes vertebral column , compresses abdomen
Pubis
Lower Ribs and Xiphoid
External oblique
Flexes + rotates vertebral column,
compresses abdomen
Lower Ribs
Linea alba and Ilium
Internal oblique
Flexes + rotates vertebral column, compresses abdomen
Lumbar Vertebrae & Ilium
Lower Ribs, Linea alba, Sternum
Transverse abdominus
Compresses abdomen
Lower Ribs, Ilium, Lumbar
Vertebrae
Linea Alba & Pubis
Erector spinae group
Extends vertebral column
Ilium, Sacrum, Ribs, Vertebrae
Ribs, Vertebrae, Base of Skull
Levator scapulae
Extends neck (also elevates scapula)
Cervical Vertebrae
Scapula
Name
Action
Origin
Insertion
Thoracic & shoulder muscles
Pectoralis major
Flexes, adducts + medially rotates arm at shoulder
Sternum & Clavicle
Humerus
Pectoralis minor
Elevates ribs (also moves scapula anterior and inferior)
Ribs
Scapula
External intercostals
Elevates ribs
Inferior Rib
Superior rib
Internal intercostals
Depresses ribs
Superior rib
Inferior Rib
Diaphragm
Increases thoracic volume
Xiphoid Process, Ribs, Lumbar Vertebrae
Central Tendinous Sheet
Arm movers
Serratus anterior
Moves and fixes scapula anteriorly
Ribs
Scapula
Trapezius
Elevates, retracts, depresses+ rotates scapula upward (also extends neck)
Occipital Bone & Thoracic Vertebrae
Scapula
Rhomboids (major & minor)
Elevates + adducts scapula
Cervical & Thoracic
Vertebrae
Scapula
Latissimus dorsi
Extends, adducts + medially rotates arm at shoulder
Thoracic, Lumbar Vertebrae, Ribs
Humerus
Deltoid
Abducts arm at shoulder (also anterior fibers flex + posterior fibers extend arm at shoulder)
Clavicle & Scapula
Humerus
Teres major
Medially rotate arm
Scapula
Humerus
Rotator cuff (SITS)
Infraspinatus
Laterally rotates shoulder
Scapula
Humerus
Supraspinatus
Abducts shoulder
Scapula
Humerus
Subscapularis
Medially rotates shoulder
Scapula
Humerus
Teres minor
Laterally rotates shoulder
Scapula
Humerus | textbooks/bio/Human_Biology/Human_Anatomy_Lab/08%3A_The_Axial_Muscles/8.03%3A_Exercises.txt |
Information
The rotator cuff is the name given to the group of four muscles that are largely responsible for the ability to rotate the arm. Three of the four rotator cuff muscles are deep to the deltoid and trapezius muscles and cannot be seen unless those muscles are first removed and one is on the anterior side of the scapula bone and cannot be seen from the surface.
On the anterior side of scapula bone is a single muscle, the subscapularis. It is triangular in shape and covers the entire bone. Its origin is along the fossa that makes up most of the “wing” of the scapula and it inserts on the lesser tubercle of the humerus bone. The subscapularis muscle is shown in Figure \(1\).
On the posterior side of the scapula bone are the other three muscles of the rotator cuff. All three insert on the greater tubercle of the humerus, allowing them, in combination with the subscapularis, to control rotation of the arm. The supraspinatus muscle is above the spine of the scapula. The infraspinatus muscle is below the spine of the scapula. The relatively thin teres minor muscle is the most inferior of the rotator cuff muscles. The three posteriorly- positioned muscles of the rotator cuff are shown in Figure \(2\).
The teres major muscle has its origin on the scapula, like the rotator cuff muscles, but is not involved in rotating the arm. It inserts lower on the humerus than the rotator cuff muscles and is involved in adducting the arm (bringing it closer to the midline of the body.)
LAB 9 EXERCISES \(1\)
1. Using the full-scale arm model, locate and identify all four muscles of the rotator cuff, as well as the deltoid muscle and the teres major muscle.
2. The following are muscles of arm rotation and adduction. For each, give its origin(s) and insertion(s) and whether or not it is part of the rotator cuff.
Muscle
Origin(s)
Insertion(s)
Part of rotator cuff?
Subscapularis
Supraspinatus
Infraspinatus
Teres major
Teres minor
Deltoid
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Figure \(1\). The subscapularis muscle of the rotator cuff, in red, anterior view.. Authored by: Was a bee. Provided by: Images in Figure \(1\) were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en.. Located at: https://commons.wikimedia.org/wiki/F...le_frontal.png. License: CC BY-SA: Attribution-ShareAlike
Figure \(2\). The muscles of the rotator cuff and arm, posterior view.. Authored by: OpenStax College. Located at: http://cnx.org/resources/e5ba9b5bb7343a347f55336ebd7a61f3b35b0cdc/1119_Muscles_that_Move_the_Humerus.jpg. Lice nse: CC BY-SA: Attribution-ShareAlike
9.02: Muscles of the Upper Arm
Anatomists refer to the upper arm as just the arm or the brachium. (The lower arm is the forearm or antebrachium.) There are three muscles on the upper arm that are parallel to the long axis of the humerus, the biceps brachii, the brachialis, and the triceps brachii.
The biceps brachii is on the anterior side of the humerus and is the prime mover (agonist) responsible for flexing the forearm. It has two origins (hence the “biceps” part of its name), both of which attach to the scapula bone. It inserts on the radius bone. The biceps brachii has two synergist muscles that assist it in flexing the forearm. Both are found on the anterior side of the arm and forearm. One of these is the brachioradialis muscle which is largely on the forearm (see the next section) and the other is the brachialis, which is largely on the upper arm. The brachialis muscle is deep to the biceps brachii and both its origin and its insertion are more distal to the shoulder than its equivalents on the biceps brachii. Like the biceps brachii the origin of the brachialis is on the humerus. Parts on the brachialis can be seen peeking out from under the biceps brachii, especially lower on the arm. The locations of these three muscles are shown in Figure 9.3.
On the posterior side of the arm is the triceps brachii muscle. It the antagonist to the biceps brachii. When the triceps brachii contracts it extends the forearm, undoing any flexing brought about by contractions of the biceps brachii. As a result, when the triceps brachii is contracted, the biceps brachii and its synergists must be relaxed, and vice versa. The triceps brachii has three origins, called the long head, the lateral head, and the medial head. Figure 9.4 shows the three origins of the triceps brachii in different colors. It is easiest to view the triceps brachii from the posterior, but the medial head and its origin are deep to the lateral head and the long head, and so is the medial head of the triceps brachii is partially obscured from the posterior.
LAB 9 EXERCISES 9-2
1. Using the full-scale arm model, locate and identify the biceps brachii, brachialis, and triceps brachii muscles.
2. The following are muscles of arm rotation and adduction. For each, give its origin(s) and insertion(s).
Muscle
Origin(s)
Insertion(s)
Biceps brachii
Brachialis
Triceps brachii
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Figure \(1\):. The muscles of the arm.. Authored by: OpenStax College. Located at: http://cnx.org/resources/6669b272a691b9377071de429a1336fec0469a5c/1120_Muscles_that_Move_the_Forearm.jpg. Lice nse: CC BY-SA: Attribution-ShareAlike
Figure \(2\):. The three heads of the triceps brachii color-coded to distinguish them. Keep in mind, despite the different colors all three are parts of the same one muscle.. Authored by: Was a bee. Provided by: Images in Figure 9.4 were made out of, or made from, content published in a BodyParts3D/Anatomography web site. The content of their website is published under the Creative Commons Attribution 2.1 Japan license. The author and licenser of the contents is http://lifesciencedb.jp/bp3d/?lng=en. Located at: https://commons.wikimedia.org/wiki/F...nimation02.gif. License: CC BY-SA: Attribution- ShareAlike | textbooks/bio/Human_Biology/Human_Anatomy_Lab/09%3A_The_Appendicular_Muscles/9.01%3A_Muscles_of_the_Rotator_Cuff.txt |
Information
Anatomists refer to the lower arm as the forearm or antebrachium. The musculature of the forearm is complicated. Figure \(1\) shows the muscles of the forearm.
Figure \(2\) shows the muscles of the hands.
LAB 9 EXERCISES \(1\)
1. Using the full-scale arm model, locate and identify the muscles of the forearm selected by your instructor.
2. Write down the muscles of the forearm selected by your instructor and, for each, give the location of that muscle and what effect contracting that muscle has.
Muscle
Location & description
Action(s)
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Figure \(1\). Muscles of the forearm.. Authored by: OpenStax College. Located at: https://cnx.org/resources/6669b272a6...he_Forearm.jpg. Lic ense: CC BY-SA: Attribution-ShareAlike
Figure \(2\). The muscles of the hands.. Authored by: OpenStax College. Located at: https://cnx.org/resources/49b609261e...f_the_Hand.jpg. Licen se: Public Domain: No Known Copyright
9.04: Muscles of the Hips and Thighs
Information
There are three layers of gluteal muscles on the posterior hips, just like there are three layers of muscles in the abdominal trunk. The largest of them is the most superficial muscle, the gluteus maximus. Its origin is on the ilium of the coxal bone, and it inserts part-way down the shaft of the femur. It helps maintain erect posture, abducts the thigh, and rotates the thigh outward.
Below the gluteus maximus is the smaller gluteus medius. The gluteus medius muscle helps abducts the thigh along with the gluteus maximus, but can rotate the thigh inward where the gluteus maximus rotates the thigh outward.
Below the gluteus medius are several muscles, one of which is the gluteus minimus, the smallest of the gluteal muscles. It is a synergist for the gluteus medius.
Like the forearm, the upper leg, or thigh, has a dense arrangement of many muscles. On the anterior side, the most prominent of the muscles are the sartorius muscle and the four muscles that make up quadriceps muscle group (the “quads”.)
The quadriceps sounds like it should be just one muscle, akin to the triceps brachii, but it is a group of four muscles, three visible on the surface, and the fourth obscured. The three surface muscles of the quadriceps are the rectus femoris in the center, the vastus medialis on the medial side, and the vastus lateralis on the lateral side. These three muscles are visible in Figure \(2\). Below the rectus femoris and largely hidden by it is the vastus intermedius. This muscle’s position can be seen in Figure 9.9. The four muscle of the quadriceps all extend the lower leg, and the rectus femoris additionally can flex the thigh at the hip.
The sartorius muscle is a distinctively long and thin muscle that crosses the thigh diagonally. It is visible in Figure 9.9. Sartorius comes from the Latin for tailor, and this is sometimes called the tailor’s muscle, although the reasons for the nickname are obscure. It may be because the shape of the muscle is thin and long, like a tailor’s measuring tape; it may be because it is close to the inseam a tailor measures when tailoring pants, or it may be because it helps bring about the cross-legged position that tailors often adopt when working.
In the posterior thigh the bulk of the musculature is made up of three long muscles that are collectively called the hamstrings. The origin of this nickname is obscure, but it may have to do with the practice of butchers of hanging the thighs of butchered animals such as pig (the “hams”) by the tendons of these three muscles. Move from the medial edge to the lateral edge of the posterior thigh, the hamstring muscles are the semimembranous muscle, the semitendinosus muscle, and the biceps femorismuscle. Notice the upper leg has a “biceps” muscle just like the upper arm does. This is why you have to indicate which biceps you are taking about when discussing one or other of these muscles. On the medial edge of the posterior thigh is the gracilis muscle. It is also visible on the medial edge of the thigh from the anterior.
LAB 9 EXERCISES \(1\)
1. Using the full-scale leg model, locate and identify the muscles of the thigh listed in the table below.
2. Write down the muscles of the thigh in the table below and, for each, give the location of that muscle and what effect contracting that muscle has.
Muscle
Location & description
Action(s)
Rectus femoris
Vastus intermedius
Vastus medialis
Vastus lateralis
Sartorius
Gracilis
Semimembranosus
Semitendinosus
Biceps femoris
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Figure \(1\). The three layers of gluteal muscles, gluteus maximus, gluteus medius, gluteus minimus. Authored by: Beth ohara~commonswiki. Located at: https://commons.wikimedia.org/wiki/F..._Muscles_3.PNG. License: CC BY- SA: Attribution-ShareAlike
Figure \(2\). The superficial muscles of the thigh.. Authored by: OpenStax College. Located at: https://cnx.org/resources/49a26b0c63...e_the_Femur.jp g. License: CC BY-SA: Attribution-ShareAlike
Figure \(3\). The quadriceps group of four muscles. The view on the left has the rectus femoris cut away to show the vastus intermedius which is below it.. Authored by: Athikhun.suw. Located at: https://commons.wikimedia.org/wiki/F...ius_muscle.jpg. License: CC BY-SA: Attribution- ShareAlike
Figure \(4\) The muscles of the posterior thigh.. Authored by: OpenStax College. Located at: http://cnx.org/resources/49a26b0c6351a2052a16c4fcd339bc092505e492/1122_Gluteal_Muscles_that_Move_the_Femur.jpg. License: CC BY-SA: Attribution-ShareAlike
Figure \(5\). The hamstring group of muscles of the posterior thigh.. Authored by: BruceBlaus. Located at: https://commons.wikimedia.org/wiki/F..._Hamstring.png. License: CC BY-SA: Attribution-ShareAlike
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Figure \(1\). The three layers of gluteal muscles, gluteus maximus, gluteus medius, gluteus minimus.. Authored by: Dr. Johannes Sobotta. Provided by: Sobotta's Human Anatomy 1909. Located at: https://commons.wikimedia.org/wiki/F...Sobo_1909_575- 576.png. License: Public Domain: No Known Copyright | textbooks/bio/Human_Biology/Human_Anatomy_Lab/09%3A_The_Appendicular_Muscles/9.03%3A_Muscles_of_the_Lower_Arm_and_Hand.txt |
Information
The muscles of the lower leg, called simply the leg by anatomists, largely move the foot and toes. The major muscles of the lower leg, other than the gastrocnemius which is cut away, are shown in Figure 9.12. The gastrocnemius muscle has two large bellies, called the medial head and the lateral head, and inserts into the calcaneus bone of the foot via its calcaneal tendon (also known as the Achilles tendon.) The soleus muscle is deep to the gastrocnemius, and the two muscles serve together as the calf of the leg. The gastrocnemius muscle is shown in Figure 9.13.
Figure \(2\) shows the muscles of the feet.
LAB 9 EXERCISES 9-5
1. Using the full-scale leg model, locate and identify the muscles of the lower leg listed in the table below.
2. For each of the muscles of the leg, give the location of that muscle and what effect contracting that muscle has.
Muscle
Location & description
Action
Gastrocnemius
Soleus
Tibialis anterior
Fibularis longus
LICENSES AND ATTRIBUTIONS
CC LICENSED CONTENT, ORIGINAL
A&P Labs. Authored by: Ross Whitwam. Provided by: Mississippi University for Women. Located at: http://www.muw.edu. License: CC BY-SA: Attribution-ShareAlike
CC LICENSED CONTENT, SPECIFIC ATTRIBUTION
Figure \(1\). The muscles of the lower leg. Authored by: OpenStax College. Located at: http://cnx.org/resources/adeb0a7399faaf57fe6dd39c017a6d25a03e6916/1123_Muscles_of_the_Leg_that_Move_the_Foot_ and_Toes.jpg. License: CC BY-SA: Attribution-ShareAlike
Figure\(2\). The gastrocnemius muscle. Authored by: Nikai . Located at: https://commons.wikimedia.org/wiki/F...trocnemius.png. License: CC BY-SA: Attribution-ShareAlike
Figure \(3\). The muscles of the feet.. Authored by: OpenStax College. Located at: https://cnx.org/resources/747c45c729...f_the_Foot.jpg. Licens e: CC BY-SA: Attribution-ShareAlike
9.07: MODELS- Arm and Leg
Name
Action
Origin
Insertion
Forearm movers
Biceps brachii
Flexes arm at elbow
Scapula
Radius
Brachialis
Flexes arm at elbow
Humerus
Ulna
Triceps brachii
Extends arm at elbow
Humerus & Scapula
Ulna
Pronator teres
Pronates forearm
Humerus & Ulna
Radius
Supinator
Supinates forearm
Humerus & Ulna
Radius
Brachioradialis
Flexes arm at elbow (beer raising)
Humerus
Radius
Hand & finger movers
Name
Action
Origin
Insertion
Flexor carpi radialis
Flexes & abducts hand at wrist
Humerus
Carpal Bones
Palmaris longus
Flexes hand at wrist
Humerus
Carpal Bones
Flexor carpi ulnaris
Flexes & adducts hand at wrist
Humerus
Carpal Bones
Extensor carpi radialis
Extends & abducts hand at wrist
Humerus
Carpal Bones
Extensor digitorum
Extends fingers
Humerus
Digits
Extensor carpi ulnaris
Extends & adducts hand at wrist
Humerus
Carpal Bones
Muscles of the thigh & leg
Name
Action
Origin
Insertion
Flexor digitorum longus
Flexes toes
Tibia
Digits
Flexor hallucis longus
Flexes hallux
Fibula
big toe
Tibialis anterior
Dorsiflex foot at ankle, inverts foot
Tibia
1st metatarsal
Extensor digitorum longus
Extends toes
Tibia & fibula
digits
Fibularis (peroneus) longus
plantarflexes
Tibia & fibula
1st metatarsal
Gastrocnemius
plantarflexes
Femur
calcaneous
Soleus
Plantarflexes
Femur & tibia
calcaneous
Name
Action
Origin
Insertion
Muscles of the leg & thigh
Psoas major
Flexes thigh at hip (also flexes vertebral column)
Thoracic and Lumbar Vertebrae
Femur
Iliacus
Flexes thigh at hip
Ilium
Femur
Adductor longus
Adducts thigh at hip
Pubis
Femur
Adductor magnus
Adducts thigh at hip
Pubis and Ischium
Femur
Gracilis
Adducts thigh at hip
Pubis
Tibia
Gluteus maximus
Extends thigh at hip, rotates thigh laterally
Ilium, Sacrum and Coccyx
Femur, ITB
Gluteus medius
Abducts thigh at hip, rotates thigh medially
Ilium
Femur
Sartorius
Flexes leg at knee (also flexes + rotates thigh at hip)
Ilium
Tibia
Tensor fascia latae
tenses the ITB, supports knee
Ilium
ITB
Iliotibial band (IT band/tract)
Not a muscle
n/a
n/a
Quadriceps femoris
Name
Action
Origin
Insertion
Rectus femoris
Extends leg at knee (also flexes thigh at hip)
Ilium
Tibia
Vastus lateralis
Extends leg at knee
Femur
Tibia
Vastus medialis
Extends leg at knee
Femur
Tibia
Vastus intermedius
Extends leg at knee
Femur
Tibia
Hamstrings
Name
Action
Origin
Insertion
Biceps femoris
Flexes leg at knee (also extends thigh at hip)
Ischium & Femur
Tibia & Fibula
Semitendinosus
Flexes leg at knee (also extends thigh at hip)
Ischium
Tibia
Semimembranosus
Flexes leg at knee (also extends thigh at hip)
Ischium
Tibia | textbooks/bio/Human_Biology/Human_Anatomy_Lab/09%3A_The_Appendicular_Muscles/9.05%3A_Muscles_of_the_Lower_Leg_and_Foot.txt |
Nervous tissue is composed of two types of cells, neurons and glial cells. Neurons are the primary type of cell that most anyone associates with the nervous system. They are responsible for the computation and communication that the nervous system provides. They are electrically active and release chemical signals to target cells. Glial cells, or glia, are known to play a supporting role for nervous tissue. Ongoing research pursues an expanded role that glial cells might play in signaling, but neurons are still considered the basis of this function. Neurons are important, but without glial support they would not be able to perform their function.
Neurons
Neurons are the cells considered to be the basis of nervous tissue. They are responsible for the electrical signals that communicate information about sensations, and that produce movements in response to those stimuli, along with inducing thought processes within the brain. An important part of the function of neurons is in their structure, or shape. The three- dimensional shape of these cells makes the immense numbers of connections within the nervous system possible.
Glial Cells
Glial cells, or neuroglia or simply glia, are the other type of cell found in nervous tissue. They are considered to be supporting cells, and many functions are directed at helping neurons complete their function for communication. The name glia comes from the Greek word that means “glue,” and was coined by the German pathologist Rudolph Virchow, who wrote in 1856: “This connective substance, which is in the brain, the spinal cord, and the special sense nerves, is a kind of glue (neuroglia) in which the nervous elements are planted.” Today, research into nervous tissue has shown that there are many deeper roles that these cells play. And research may find much more about them in the future.
10.02: Exercises
Lab 10 Exercise \(2\)
Label the following:
1 Axon
2 Synaptic cleft
3 Receptor,
4 Neurotransmitter
5 Synaptic terminal
6 Vesicles
7
Voltage-gated ion channels.
Lab 10 Exercise \(3\)
Match these items or actions to their locations on a neuron: graded potentials, action
1 graded potentials 7 EPSPs
2 action action 8 IPSPs
3 neurotransmitter-filled vesicles 9 pre-synaptic membrane
4 neurotransmitter receptors 10 post-synaptic membrane
5 voltage-gated sodium channels 11 Genes (that encode receptor proteins)
6 ligand-gated ion channels
Lab 10 Exercise \(4\)
Label the following glia & related items:
1 Astrocyte 6 Cerebrospinal fluid (CSF)
2 Perivascular feet 7 Extracellular fluid (ECF)
3 Oligodendrocyte 8 Neuron
4 Microglia 9 Capillary
5 Ependymal cell
Lab 10 Exercise \(6\)
Obtain slides of each of the following tissues, observe them, draw and label the significant features. SLIDES: Spinal cord, cerebral cortex, cerebellum, dorsal root ganglion, peripheral nerve, neuromuscular junction:
1. Obtain a slide of nervous tissue from the slide box. Use any nervous tissue except peripheral nerve, there are no nerve cell bodies in a peripheral nerve section.
2. View the slide on the second-highest objective. Search carefully until you find a clear, representative neuron in your field of view.
3. In the circle below, draw the neuron you found. Only draw the single neuron. Do not draw any of the other material. Draw your structures proportionately to their size in your microscope’s field of view.
4. Label any neural parts you can clearly recognize.
10.03: MODELS- Nerve Tissue
Anatomical divisions:
• Brain
• Spine
Peripheral nervous system
• Nerves
• Functional divisions:
Autonomic nervous system
• Sympathetic nervous system
• Parasympathetic nervous system
Histology:
Neurons
• Cell body
• Nuclei
• Axon
• Axon hillock
• Myelin sheath
• Nodes of Ranvier
• Telodendria
• Axon terminal
• Dendrite
• Synaptic cleft
Neuroglia
• Schwann cells
• Oligodendrocytes
• Satellite cells
• Astrocytes
• Microglial cells
• Ependymal cells | textbooks/bio/Human_Biology/Human_Anatomy_Lab/10%3A_Nervous_Tissue/10.01%3A_Neurons_and_Glial_Cells.txt |
The brain and the spinal cord are the central nervous system, and they represent the main organs of the nervous system. The spinal cord is a single structure, whereas the adult brain is described in terms of four major regions: the cerebrum, the diencephalon, the brain stem, and the cerebellum. A person’s conscious experiences are based on neural activity in the brain. The regulation of homeostasis is governed by a specialized region in the brain. The coordination of reflexes depends on the integration of sensory and motor pathways in the spinal cord.
11: The Central Nervous System (Brain)
The iconic gray mantle of the human brain, which appears to make up most of the mass of the brain, is the cerebrum (Figure 11.1). The wrinkled portion is the cerebral cortex, and the rest of the structure is beneath that outer covering. There is a large separation between the two sides of the cerebrum called the longitudinal fissure. It separates the cerebrum into two distinct halves, a right and left cerebral hemisphere. Deep within the cerebrum, the white matter of the corpus callosum provides the major pathway for communication between the two hemispheres of the cerebral cortex.
Cerebral Cortex
The cerebrum is covered by a continuous layer of gray matter that wraps around either side of the forebrain—the cerebral cortex. This thin, extensive region of wrinkled gray matter is responsible for the higher functions of the nervous system. A gyrus (plural = gyri) is the ridge of one of those wrinkles, and a sulcus (plural = sulci) is the groove between two gyri. The pattern of these folds of tissue indicates specific regions of the cerebral cortex.
The head is limited by the size of the birth canal, and the brain must fit inside the cranial cavity of the skull. Extensive folding in the cerebral cortex enables more gray matter to fit into this limited space. If the gray matter of the cortex were peeled off of the cerebrum and laid out flat, its surface area would be roughly equal to one square meter.
The folding of the cortex maximizes the amount of gray matter in the cranial cavity. During embryonic development, as the telencephalon expands within the skull, the brain goes through a regular course of growth that results in everyone’s brain having a similar pattern of folds. The surface of the brain can be mapped on the basis of the locations of large gyri and sulci. Using these landmarks, the cortex can be separated into four major regions, or lobes (Figure 11.2). The lateral sulcus that separates the temporal lobe from the other regions is one such landmark. Superior to the lateral sulcus are the parietal lobe and frontal lobe, which are separated from each other by the central sulcus. The posterior region of the cortex is the occipital lobe, which has no obvious anatomical border between it and the parietal or temporal lobes on the lateral surface of the brain. From the medial surface, an obvious landmark separating the parietal and occipital lobes is called the parieto- occipital sulcus.
The fact that there is no obvious anatomical border between these lobes is consistent with the functions of these regions being interrelated.
11.02: The Diencephalon
The diencephalon is the one region of the adult brain that retains its name from embryologic development. The etymology of the word diencephalon translates to “through brain.” It is the connection between the cerebrum and the rest of the nervous system, with one exception. The rest of the brain, the spinal cord, and the PNS all send information to the cerebrum through the diencephalon. Output from the cerebrum passes through the diencephalon. The single exception is the system associated with olfaction, or the sense of smell, which connects directly with the cerebrum. In the earliest vertebrate species, the cerebrum was not much more than olfactory bulbs that received peripheral information about the chemical environment (to call it smell in these organisms is imprecise because they lived in the ocean).
The diencephalon is deep beneath the cerebrum and constitutes the walls of the third ventricle. The diencephalon can be described as any region of the brain with “thalamus” in its name. The two major regions of the diencephalon are the thalamus itself and the hypothalamus (Figure 11.3). There are other structures, such as the epithalamus, which contains the pineal gland, or the subthalamus, which includes the subthalamic nucleus that is part of the basal nuclei.
11.03: The Brain Stem
The midbrain and hindbrain (composed of the pons and the medulla) are collectively referred to as the brain stem (Figure \(1\)). The structure emerges from the ventral surface of the forebrain as a tapering cone that connects the brain to the spinal cord. Attached to the brain stem, but considered a separate region of the adult brain, is the cerebellum. The midbrain coordinates sensory representations of the visual, auditory, and somatosensory perceptual spaces. The pons is the main connection with the cerebellum. The pons and the medulla regulate several crucial functions, including the cardiovascular and respiratory systems and rates.
11.04: The Cerebellum
The cerebellum, as the name suggests, is the “little brain.” It is covered in gyri and sulci like the cerebrum, and looks like a miniature version of that part of the brain (Figure \(1\)). The cerebellum is largely responsible for comparing information from the cerebrum with sensory feedback from the periphery through the spinal cord. It accounts for approximately 10 percent of the mass of the brain. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/11%3A_The_Central_Nervous_System_(Brain)/11.01%3A_The_Cerebrum.txt |
LAB 11 EXERCISES \(1\)
Label the (visible) lobes of the cerebrum
1
2
3
Label the following
1 a Gyrus
2 a Sulcus
3 Cerebrum
4 Cerebellum
5 Brainstem
LAB 11 EXERCISES \(2\)
Lobes and Limbic System
Label the following
1 Pre-central gyrus
2 Post-central gyrus
3 Central sulcus
4 Pre-frontal cortex
5 Temporal lobe
6 Occipital lobe
Label the following
1 Amygdala
2 Hippocampus
3 Pituitary gland
4 Medulla Oblongata
5 Pons
6 Thalamus
7 Superior & Inferior colliculi
LAB 11 EXERCISES \(3\)
Diencephalon and Brain Stem
Midbrain, Lateral ventricle, 3rd ventricle, Cerebral Aqueduct, 4th ventricle,
Central canal
Label the following
1 Arbor vitae
2 Corpus callosu
3 Septum pellucidum
4 Fornix
5 Optic chiasm
6 Pons
LAB 11 EXERCISES \(4\)
Ventricular System
Label the following
1 Midbrain
2 Lateral ventricle
3 3rd ventricle
4 Cerebral Aqueduct
5 4th ventricle
6 Central canal
LAB 11 EXERCISES \(6\)
Cortex and CSF Circulation
1) Label the ventricles, aqueduct and canal.
2) Draw two choroid plexuses (in red, if you have a
red pen handy).
Label CSF-related areas:
1 Lateral ventricle,
2 Lateral ventricle
3 3rd ventricle
4 4th ventricle
5 Superior sagittal sinus
6 Straight sinus,
7 Subarachnoid space,
8 Cerebral aqueduct,
9 Arachnoid granulation,
10 Choroid plexus,
11 Interthalamic adhesion
LAB 11 EXERCISES \(7\)
Cerebral Cortex
Label the following regions of the cortex
1 Primary motor cortex
2 Primary sensory cortex
3 Primary auditory cortex
4 Primary visual cortex
5 Motor association area
6 Pre-frontal cortex
7 Broca’s area
8 Wernicke’s area
Match the following symptoms to the brain
region they are associated with:
A Ataxia
B Inability to form words
C Nonsensical speech
D Hearing loss
E Vision loss
F Poor decision making
G Drooping facial muscles
H Paralysis of fingers
I Phantom limb
LAB 11 EXERCISES \(8\)
CNS
Label the following
1 Falx cerebri
2 Tentorum cerebellum
3 Superior sagittal sinus
4 Straight sinus
Label the following
1 Pituitary gland
2 Mamillary bodies
3 Pons
4 Medulla oblongata
5 Temporal lobe
6 Frontal lobe
7 Cerebellum
Identify the following regions:
1 Corpus callosum
2 Anterior commissure
3 Posterior commissure
4 Fornix
5 Interthalamic adhesion
6 Choroid plexus
7 Corpora quadrigemina
LAB 11 EXERCISES \(9\)
Coronal View and Miscellaneous
Identify the following regions:
1 Septum pellucidum
2 Grey matter
3 Insula
4 Basal ganglia
5 Corpus callosum
6 Lateral ventricle
7 Longitudinal fissure
11.06: MODELS- 3D Brains Ventricles Sagittal and Coronal Brain Slices Torsos Sagittal Head
3D Brains, Ventricles, Sagittal & Coronal Brain Slices, Torsos, Sagittal Head
Cerebrum
• Gyrus (plural = gyri)
• Pre-central gyrus
• Post-central gyrus
• Sulcus (plural = sulci or sulcusses)
• Central sulcus
• Lateral sulcus
• Fissures
• Longitudinal fissure
• Lobes of cerebrum
• Frontal lobe
• Parietal lobe
• Temporal lobe
• Occipital lobe
• Insula
• Cerebral cortex (gray matter)
• Basal nuclei
• White matter (tracts)
• Corpus callosum
• Septum pellucidum
• Fornix
• Choroid Plexuses
• Olfactory bulb
• Olfactory tract
• Optic chiasm
• Optic tract
Brain stem
• Midbrain
• Pons
• Medulla oblongata
Cerebellum
• Vermis
• Cerebellar cortex
• Arbor vitae
Ventricular system
• Lateral ventricles
• Interventricular foramen
• 3rd ventricle
• Cerebral aqueduct
• 4th ventricle
• Central canal
Meningeal layers
• Dura mater
• Falx cerebri
• Falx cerebelli
• Tentorium cerebella
• Dural venous sinuses
• Straight sinus
• Superior sagittal sinus
• Arachnoid mater
• Arachnoid granulations
• Pia mater
Diencephalon
• Thalamus
• Interthalamic adhesion
• Anterior commissure
• Posterior commissure
• Hypothalamus
• Pituitary gland
• Infundibulum
• Mamillary bodies
• Pineal gland
• Copora Quadrigemina (superior & inferior colliculi) | textbooks/bio/Human_Biology/Human_Anatomy_Lab/11%3A_The_Central_Nervous_System_(Brain)/11.05%3A_Exercises.txt |
Background Information:
The sheep brain is remarkably similar to the human brain. One major difference, however, is in proportion. For example, the sheep brain has a proportionately smaller cerebrum. Another difference is in orientation of the spinal cord. The sheep spinal cord is orientated anterior to posterior, as in any four-legged animal. The human spinal cord is orientated superior to inferior. This orientation difference has a major affect on the location of the brain stem. The sheep brain stem is located more towards the rear (posteriorly). The sheep skull, in order to compensate for this, has the foramen magnum located more towards the rear of the skull. In humans, since we walk upright (bipedalism), the spinal cord is in a more vertical plane, thus the foramen magnum is located centrally on the bottom of the skull (inferiorly). By observing the movement of this major skull feature from the rear (as in very early human ancestors) to the current location (modern humans), scientists have been able to determine when the human species began to walk upright on two legs.
Dissection Instructions:
1. Obtain a preserved sheep brain from the bucket in the front of the classroom. Place this on your dissection tray.
2. You will need the following dissection tools to properly perform this lab:
• scalpel
• scissors
• probes
3. The sheep brain is enclosed in a tough outer covering called the dura mater. You can still see some structures on the brain before you remove the dura mater. Take special note of the pituitary gland and the optic chiasma. These two structures will likely be pulled off when you remove the dura mater.
4. This image shows the ventral surface of the sheep's brain with most of the dura mater removed. The pituitary gland and the optic chiasma are still intact. (A = pituitary gland, B = optic chiasma, C = olfactory bulb)
5. On this image, the dura mater has been completely removed, you can still see the optic chiasma but the pituitary gland is missing. The infundibulum (pituitary stalk) is now visible in the center. Careful dissection also reveals two other large nerves: the oculomotor nerves (C.2). Often these two nerves are removed with the dura mater, but in this image they are still intact.
These two figures show the fissures located on the surface of the brain. Longitudinal fissure Transverse fissure
6. If you flip the brain over to the other side, you can see the cerebellum, it will be loosely attached to the cerebrum in most cases. If you did not carefully remove the dura mater you may have accidentally pulled the entire cerebellum away from the brain. The lobes of the brain are visible, as well as the transverse fissure, which separates the cerebrum from the cerebellum. The convolutions of the brain are also visible as bumps (gyri) and grooves (sulci).
7. The gap between the cerebrum and the cerebellum at the transverse fissure can reveal some internal parts of the brain. In this image, a student is bending the cerebellum down to show the superior and inferior colliculi. Just behind the colliculi, the pineal gland is just barely visible.
8. Using a scalpel and the longitudinal fissure as a guide, the brain is separated into the left and the right hemispheres. Sharp scalpels work best for this procedure. Always leave the specimen in the dissecting tray when cutting it, do NOT hold it in your hand!
If you are very careful, you will cleanly cut the brain into two halves and can see the internal structures, the most visible of them being the corpus callosum, which divides the left and right hemispheres. The cerebrum will still be visible as a wrinkled structrure, and you can even locate the "bumps" of the superior and inferior colliculi. Remember, you located those structures by pulling down the cerebellum.
The cerebellum, when cut will have a very distinct tree-like white area within it. This is called the arbor vitae, or the tree of life.
9. In the image below, a probe indicates the location of the lateral ventricle.
10. Once the brain is cut this way, the colliculi can also be seen from the inside and the pineal gland is revealed only if you made a very careful incision.
11. Other major structures are visible, here the probe indicates the arbor vitae (tree of life) found within the cerebellum. The fissure between the cerebrum and the cerebellum is called the transverse fissure. The cerebellum only loosely connects to the rest of the brain when the dura is removed.
12. This brain is pinned to show the pineal gland (blue pin), thalamus (red pin) and lateral ventricle (green pin).
13. The image below shows a cleanly separated brain with the major internal structures visible and labeled.
14. Finally, a section of the brain is cut to examine the difference between white matter and gray matter.
15. Once you have made the cut like in the above diagram, you should be able to see the difference between the white and gray matter, just like the diagram below.
16. The diagram below will help you understand the difference between the gyri and the sulcus.
Sheep brain dissection maybe downloaded at the following site: https://jb004.k12.sd.us/my%20website...P%20BRAIN%20DI SSECTION/SHEEP%20BRAIN%20LAB.htm
LAB 11 EXERCISES \(1\):
Sheep Brain Dissection Analysis:
Match the structure to the description
a looks like a butt Arbor Vitae
b leathery covering over the entire brain Lateral Ventricle
c cauliflower, the area toward the back of the brain Optic Chiasma
d behind the colliculi, looks like a little nub Superior Colliculi
e looks like a "tree" Dura Mater
f the rounded part of the brain stem Cerebellum
g shaped like an X Pineal Gland
h large area under the corpus callosum Thalamus
i space for fluid between the corpus callosum and the fornix Pons
j contains nerves, connects to the far front of the brain Olfactory Bulb | textbooks/bio/Human_Biology/Human_Anatomy_Lab/11%3A_The_Central_Nervous_System_(Brain)/11.07%3A_Sheep_Brain_Dissection.txt |
The description of the CNS is concentrated on the structures of the brain, but the spinal cord is another major organ of the system. Whereas the brain develops out of expansions of the neural tube into primary and then secondary vesicles, the spinal cord maintains the tube structure and is only specialized into certain regions. As the spinal cord continues to develop in the newborn, anatomical features mark its surface. The anterior midline is marked by the anterior median fissure, and the posterior midline is marked by the posterior median sulcus. Axons enter the posterior side through the dorsal (posterior) nerve root, which marks the posterolateral sulcus on either side. The axons emerging from the anterior side do so through the ventral (anterior) nerve root. Note that it is common to see the terms dorsal (dorsal = “back”) and ventral (ventral = “belly”) used interchangeably with posterior and anterior, particularly in reference to nerves and the structures of the spinal cord. You should learn to be comfortable with both.
Gray Horns
In cross-section, the gray matter of the spinal cord has the appearance of an ink-blot test, with the spread of the gray matter on one side replicated on the other—a shape reminiscent of a bulbous capital “H.” As shown in Figure \(1\), the gray matter is subdivided into regions that are referred to as horns.
The posterior horn is responsible for sensory processing. The anterior horn sends out motor signals to the skeletal muscles. The lateral horn, which is only found in the thoracic, upper lumbar, and sacral regions, is the central component of the sympathetic division of the autonomic nervous system.
Cranial Nerves
The nerves attached to the brain are the cranial nerves, which are primarily responsible for the sensory and motor functions of the head and neck (one of these nerves targets organs in the thoracic and abdominal cavities as part of the parasympathetic nervous system). There are twelve cranial nerves, which are designated CNI through CNXII for “Cranial Nerve,” using Roman numerals for 1 through 12. They can be classified as sensory nerves, motor nerves, or a combination of both, meaning that the axons in these nerves originate out of sensory ganglia external to the cranium or motor nuclei within the brain stem. Sensory axons enter the brain to synapse in a nucleus. Motor axons connect to skeletal muscles of the head or neck. Three of the nerves are solely composed of sensory fibers; five are strictly motor; and the remaining four are mixed nerves.
12.02: Exercises
LAB 12 EXERCISES \(1\)
Label the following
1 Dura mater
2 Arachnoid mater
3 Pia mater
4 Epidural space
5 Subarachnoid space
6 Spinal cord
7 Denticulate ligament
8 Dorsal Root Ganglion
9 Body (of vertebrae)
10 Spinous process (of vertebrae)
11 Dorsal Ramus
12 Ventral ramus
LAB 12 EXERCISES \(2\)
Label the following
1 Dorsal horn
2 Lateral horn
3 Ventral horn
4 Dorsal column
5 Lateral column
6 Ventral column
7 Central column
8 Grey commisure
LAB 12 EXERCISES \(3\)
Spinal Cord and Nerve Anatomy
Label the following
1 Dorsal root
2 Ventral root
3 Dorsal root ganglion
4 Sympathetic chain ganglion
5 Spinal nerve
6 Rami communicantes
7 Dorsal ramus
8 Ventral ramus
Label the parts of a nerve
1 Blood vessels
2 Schwann cell
3 Fascicle
4 Epineurium
5 Perineurium
6 Endoneurium
7 Axon
LAB 12 EXERCISES \(4\)
Vertebral Column
A
B
Identify the following
1 Cervical plexus
2 Brachial plexus
3 Lumbar plexus
4 Sacral plexus
5 Sciatic nerve
6 Cervical enlargement
7 Lumbar enlargement
Match these items to their relative position: Cervical enlargement, Lumbar enlargement*, Conus medullaris, Cauda equina, Filum terminale (* trickier than you might expect)
Cervical region:
Thoracic region:
Lumbo-sacral region:
Peripheral nerves and plexuses
Label the nerves
1 Axillary nerve
2 Median nerve
3 Radial nerve
4 Ulnar nerve
5 Musculocutaneous nerve
LAB 12 EXERCISES \(6\)
Draw your own cranial nerve study sheet
Mnemonic Cranial nerve name
I
II
III
IV
V
VI
VII
VIII
IX
X
XI
XII
12.03: MODELS Spinal Cord Spinal Columns (Flat) Spinal Cord in Vertebral Column Blank Brain Picture
Spinal cord anatomy
Meningeal layers Dura mater Epidural space
Arachnoid mater Subarachnoid space
Pia mater Denticulate ligaments
Conus medullaris
Cauda equina Filum terminale
Cervical enlargement
Lumbar enlargement
Spinal nerve Ventral root
Dorsal root
Dorsal root ganglion
Ventral Ramus
Dorsal Ramus
Spinal cord cross sectional anatomy: Dorsal horn
Lateral horn
Ventral horn
Dorsal column
Lateral column
Ventral column
Gray commissure
Central canal
Dorsal median sulcus
Ventral median fissure
Cranial nerves
I Olfactory
II Optic
III Occulomotor
IV Trochlear
V Trigeminal
VI Abducens
VII Facial
VIII Acoustic / Vestibulocochlear
IX Glossopharyngeal
X Vagus
XI (spinal) Accessory
XII Hypoglossal
Peripheral Nerves
Nerve anatomy: Epineurium
Perineurium
Endoneurium
Fascicle
Cervical Plexus Phrenic n.
Brachial Plexus Axillary n.
Radial n.
Ulnar n.
Median n.
Musculocutaneous n.
Intercostal nerves
Lumbar plexus Femoral n.
Sacral plexus Sciatic n.
Pudendal n.
Autonomic NS Sympathetic chain ganglion
Sympathetic chain
Rami communicantes
Lateral horn
Note: remember that dorsal/posterior, and ventral/anterior are interchangeable.
Histology
Spinal cord & dorsal root ganglia See cross sectional anatomy
See cross sectional anatomy | textbooks/bio/Human_Biology/Human_Anatomy_Lab/12%3A_Cranial_and_Spinal_Nerves/12.01%3A_The_Spinal_Cord.txt |
Gustation is the special sense associated with the tongue. The surface of the tongue, along with the rest of the oral cavity, is lined by a stratified squamous epithelium. Raised bumps called papillae (singular = papilla) contain the structures for gustatory transduction. There are four types of papillae, based on their appearance (Figure \(1\)): circumvallate, foliate, filiform, and fungiform. Within the structure of the papillae are taste buds that contain specialized gustatory receptor cells for the transduction of taste stimuli. These receptor cells are sensitive to the chemicals contained within foods that are ingested, and they release neurotransmitters based on the amount of the chemical in the food.
13.02: Olfaction (Smell)
Like taste, the sense of smell, or olfaction, is also responsive to chemical stimuli. The olfactory receptor neurons are located in a small region within the superior nasal cavity (Figure \(1\) ). This region is referred to as the olfactory epithelium and contains bipolar sensory neurons. Each olfactory sensory neuron has dendrites that extend from the apical surface of the epithelium into the mucus lining the cavity. As airborne molecules are inhaled through the nose, they pass over the olfactory epithelial region and dissolve into the mucus. These odorant molecules bind to proteins that keep them dissolved in the mucus and help transport them to the olfactory dendrites. The odorant–protein complex binds to a receptor protein within the cell membrane of an olfactory dendrite. These receptors are G protein–coupled, and will produce a graded membrane potential in the olfactory neurons.
The axon of an olfactory neuron extends from the basal surface of the epithelium, through an olfactory foramen in the cribriform plate of the ethmoid bone, and into the brain. The group of axons called the olfactory tract connect to the olfactory bulb on the ventral surface of the frontal lobe. From there, the axons split to travel to several brain regions. Some travel to the cerebrum, specifically to the primary olfactory cortex that is located in the inferior and medial areas of the temporal lobe. Others project to structures within the limbic system and hypothalamus, where smells become associated with long-term memory and emotional responses. This is how certain smells trigger emotional memories, such as the smell of food associated with one’s birthplace. Smell is the one sensory modality that does not synapse in the thalamus before connecting to the cerebral cortex. This intimate connection between the olfactory system and the cerebral cortex is one reason why smell can be a potent trigger of memories and emotion.
The nasal epithelium, including the olfactory cells, can be harmed by airborne toxic chemicals. Therefore, the olfactory neurons are regularly replaced within the nasal epithelium, after which the axons of the new neurons must find their appropriate connections in the olfactory bulb. These new axons grow along the axons that are already in place in the cranial nerve.
13.03: Audition (Hearing)
Hearing, or audition, is the transduction of sound waves into a neural signal that is made possible by the structures of the ear (Figure \(1\)). The large, fleshy structure on the lateral aspect of the head is known as the auricle. Some sources will also refer to this structure as the pinna, though that term is more appropriate for a structure that can be moved, such as the external ear of a cat. The C-shaped curves of the auricle direct sound waves toward the auditory canal. The canal enters the skull through the external auditory meatus of the temporal bone. At the end of the auditory canal is the tympanic membrane, or ear drum, which vibrates after it is struck by sound waves. The auricle, ear canal, and tympanic membrane are often referred to as the external ear. The middle ear consists of a space spanned by three small bones called the ossicles. The three ossicles are the malleus, incus, and stapes, which are Latin names that roughly translate to hammer, anvil, and stirrup. The malleus is attached to the tympanic membrane and articulates with the incus. The incus, in turn, articulates with the stapes. The stapes is then attached to the inner ear, where the sound waves will be transduced into a neural signal. The middle ear is connected to the pharynx through the Eustachian tube, which helps equilibrate air pressure across the tympanic membrane. The tube is normally closed but will pop open when the muscles of the pharynx contract during swallowing or yawning.
A cross-sectional view of the cochlea shows that the scala vestibuli and scala tympani run along both sides of the cochlear duct (Figure \(2\)). The cochlear duct contains several organs of Corti, which tranduce the wave motion of the two scala into neural signals. The organs of Corti lie on top of the basilar membrane, which is the side of the cochlear duct located between the organs of Corti and the scala tympani. As the fluid waves move through the scala vestibuli and scala tympani, the basilar membrane moves at a specific spot, depending on the frequency of the waves. Higher frequency waves move the region of the basilar membrane that is close to the base of the cochlea. Lower frequency waves move the region of the basilar membrane that is near the tip of the cochlea. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/13%3A_The_Somatic_Nervous_System_(Special_Senses)/13.01%3A_Gustation_%28Taste%29.txt |
Vision is the special sense of sight that is based on the transduction of light stimuli received through the eyes. The eyes are located within either orbit in the skull. The bony orbits surround the eyeballs, protecting them and anchoring the soft tissues of the eye (Figure \(1\)). The eyelids, with lashes at their leading edges, help to protect the eye from abrasions by blocking particles that may land on the surface of the eye. The inner surface of each lid is a thin membrane known as the palpebral conjunctiva. The conjunctiva extends over the white areas of the eye (the sclera), connecting the eyelids to the eyeball. Tears are produced by the lacrimal gland, located beneath the lateral edges of the nose. Tears produced by this gland flow through the lacrimal duct to the medial corner of the eye, where the tears flow over the conjunctiva, washing away foreign particles.
Movement of the eye within the orbit is accomplished by the contraction of six extraocular muscles that originate from the bones of the orbit and insert into the surface of the eyeball (Figure \(2\)). Four of the muscles are arranged at the cardinal points around the eye and are named for those locations. They are the superior rectus, medial rectus, inferior rectus, and lateral rectus. When each of these muscles contract, the eye to moves toward the contracting muscle. For example, when the superior rectus contracts, the eye rotates to look up. The superior oblique originates at the posterior orbit, near the origin of the four rectus muscles. However, the tendon of the oblique muscles threads through a pulley-like piece of cartilage known as the trochlea. The tendon inserts obliquely into the superior surface of the eye. The angle of the tendon through the trochlea means that contraction of the superior oblique rotates the eye medially. The inferior oblique muscle originates from the floor of the orbit and inserts into the inferolateral surface of the eye. When it contracts, it laterally rotates the eye, in opposition to the superior oblique. Rotation of the eye by the two oblique muscles is necessary because the eye is not perfectly aligned on the sagittal plane. When the eye looks up or down, the eye must also rotate slightly to compensate for the superior rectus pulling at approximately a 20-degree angle, rather than straight up. The same is true for the inferior rectus, which is compensated by contraction of the inferior oblique. A seventh muscle in the orbit is the levator palpebrae superioris, which is responsible for elevating and retracting the upper eyelid, a movement that usually occurs in concert with elevation of the eye by the superior rectus (see Figure \(3\)).
Light falling on the retina causes chemical changes to pigment molecules in the photoreceptors, ultimately leading to a change in the activity of the RGCs. Photoreceptor cells have two parts, the inner segment and the outer segment (Figure \(4\)). The inner segment contains the nucleus and other common organelles of a cell, whereas the outer segment is a specialized region in which photoreception takes place. There are two types of photoreceptors—rods and cones—which differ in the shape of their outer segment. The rod-shaped outer segments of the rod photoreceptor contain a stack of membrane-bound discs that contain the photosensitive pigment rhodopsin. The cone-shaped outer segments of the cone photoreceptor contain their photosensitive pigments in infoldings of the cell membrane. There are three cone photopigments, called opsins, which are each sensitive to a particular wavelength of light. The wavelength of visible light determines its color. The pigments in human eyes are specialized in perceiving three different primary colors: red, green, and blue. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/13%3A_The_Somatic_Nervous_System_(Special_Senses)/13.04%3A_Vision.txt |
Lab 13 Exercise \(2\)
Basic Ear Anatomy
Label the following on this stylized diagram: Auricle, Auditory canal, Pharyngotympanic tube, Stapes, Incus, Malleus, Tympanic membrane, Cochlea, Outer ear, Middle ear, Inner ear
Lab 13 Exercise \(3\)
Inner Ear Anatomy
Label the following: Auricle, External Acoustic meatus, Pharyngotympanic tube, Cochlea, Vestibule, Semicircular canals, Tympanic membrane, Middle ear
Lab 13 Exercise \(5\)
Label the following: Lens, Suspensory ligaments, Ciliary body, Cornea, Iris, Choroid, Vitreous humor
Lab 13 Exercise \(9\)
Optical Chiasm
Label the following:
1. Conscious visual sensation;
2. Beginning of axons of retinal ganglion cells;
3. 3rd order neuron;
4. synapse between 2nd & 3rd order neuron;
5. Half of nerve fibers decussate here;
6. First region with axons from only the left side of both eyes
Lab 13 Exercise \(10\)
Retina Histology
Label the following
1 Rod 6 Rods and cones
2 Cone 7 Bipolar cells
3 Horizontal Cell 8 Ganglion cells
4 Bipolar Cell 9 Optic nerve axons
5 Ganglion cell 10 Pigmented epithelium
11 Vitreous humor
13.06: MODELS- Eye Ear Torso Mid-Sagittal Head and Cochlea
Olfaction(sagittal head model)
• Nare
• Vestibule
• Nasal conchae (superior, middle, inferior)
• Nasal meatuses
• Nasal cavity
Eye(various models)
External Structures
• Palpebrae
• Tarsal plates
• Levator Palpebrae m.
• Orbicularis Oculi m.
• Medial and lateral canthus
• Lacrimal caruncle
• Lacrimal gland
• Lacrimal ducts
• Lacrimal sac
Extra-ocular muscles:
• Lateral rectus m.
• Medial rectus m.
• Superior rectus m.
• Inferior rectus m.
• Superior oblique m.
• Inferior oblique m.
Internal structures
• Fibrous tunic:
• Sclera
• Cornea
• Vascular tunic:
• Choroid
• Ciliary body / muscle
• Suspensory ligaments
• Iris
• Sensory (neural) tunic:
• Retina
• Macula Lutea
• Fovea centralis
• Optic Nerve
• Optic disc
• Lens
• Posterior cavity/Vitreous Humor
• Anterior cavity/ Aqueous Humor
Structures of the ear (various models)
• Outer ear
• Auricle
• External auditory canal
• Middle ear
• Tympanic membrane
• Auditory ossicles:
• Malleus
• Incus
• Stapes
• Pharyngotympanic tube (Eustachian, auditory)
• Inner ear
• Bony labyrinth
• Membranous labyrinth
• Vestibule
• Utricle
• Saccule
• Semicircular canals
• Semicircular ducts
• Ampulla
• Cochlea
• Oval window
• Round window
• Vestibulocochlear nerve
Cochlea (cross-sectional model)
• Vestibular duct (scala vestibuli)
• Vestibular duct (scala vestibuli)
• Cochlear duct (scala media)
• Tympanic duct (scala tympani)
• Vestibular membrane
• Basilar membrane
• Organ of Corti (& hair cells)
• Tectorial membrane
• Spiral ganglion
• Tongue (tongue model)
• Papillae:
• Filiform
• Fungiform
• Foliate
• Circumvallate.
• Sulcus terminalis
• Median sulcus
• Taste buds
• Taste pore
• Support & taste cells
• Histology:
• Eye – retina:
• Rods & Cones
• Bipolar cell layer
• Ganglion cell layer
• Axons of ganglion cells
• Retinal pigment epithelium
• Choroid layer
• Sclera
• Ear
• Basilar membrane
• Scala media (cochlear duct)
• Scala tympani (tympanic duct)
• Scala vestibuli (vestibular duct)
• Spiral ganglion
• Spiral organ (Organ of Corti)
• Tectorial membrane
• Vestibular membrane
• Tongue
• Taste bud
• Taste pore
• Support and Taste cells
13.07: Cow Eye Dissection
1. Examine the outside of the eye. You should be able to find the sclera, or the whites of the eye. This tough, outer covering of the eyeball has fat and muscle attached to it
2. Locate the covering over the front of the eye, the cornea. When the cow was alive, the cornea was clear. In your cow’s eye, the cornea may be cloudy or blue in color.
2. Cut away the fat and muscle, this may only be necessary if fat is covering the cornea of the eye and is in your way. Fat around the backside of the eye can be left alone. Flip the eye over to find the optic nerve where it exits out the back of the eye. It will be stronger and more rope-like than the surrounding fat tissue.
4. Use a scalpel or scissors to make an incision in the cornea. The cornea is tougher than it appears and may require some force to puncture, be careful when using the scalpel. Once the cornea is broken, clear liquid will leak (or squirt) out – this liquid is the aqueous humor.
5. Quick Check: Outer Tunic
A. The white of the eye is the ______________________________________
B. The front surface of the eye,continuous with (A) is the ______________________________________
C. The liquid found in the front of the eye is the ______________________________________
D. What is the name of the nerve found on the back of the eye? ______________________________________
6. Use a scalpel or scissors to make an incision in the sclera so that you can cut around the outside of the eye. Your goal is to separate the eye into a front and a back half.
7. Separate the inner parts of the eye.
The gelatinous liquid in the middle of the eye is the vitreous humor , which will also contain a hard, sphere-shaped lens
Find the cornea (which you punctured in step 1) and then the disk-shaped iris behind it. The iris will be dark in color and contain a center opening, the pupil
8. The image below shows how each part of the eye appears when it has been separated. Take a photo of your own eye and share it on social media. #coweye (optional)
9. The back of the eye has two layers, a very thin layer of cells that is easy to scrape off (and may fall off on its own), which is the retina . Behind, the retina is a blue, reflective layer known as the tapetum
10. The retina will converge at a point on the eye where it connects with the optic nerve. This is the optic disk. It may be easiest to find by scraping off the retina and locating the spot where it remains closely attached. Flipping the eye over will also show how that spot is directly in front of the optic nerve.
Cow Eye Dissection Downloaded at: https://www.biologycorner.com/worksheets/cow_eye_dissection.html
This work is licensed under a Creative Commons Attribution-NonCommercial- ShareAlike
LAB 13 EXERCISE \(1\)
Label the Cow Eye (use your book or other resources) | textbooks/bio/Human_Biology/Human_Anatomy_Lab/13%3A_The_Somatic_Nervous_System_(Special_Senses)/13.05%3A_Exercises.txt |
The endocrine system consists of cells, tissues, and organs that secrete hormones as a primary or secondary function. The endocrine gland is the major player in this system. The primary function of these ductless glands is to secrete their hormones directly into the surrounding fluid. The interstitial fluid and the blood vessels then transport the hormones throughout the body. The endocrine system includes the pituitary, thyroid, parathyroid, adrenal, and pineal glands (Figure \(1\)). Some of these glands have both endocrine and non-endocrine functions. For example, the pancreas contains cells that function in digestion as well as cells that secrete the hormones insulin and glucagon, which regulate blood glucose levels. The hypothalamus, thymus, heart, kidneys, stomach, small intestine, liver, skin, female ovaries, and male testes are other organs that contain cells with endocrine function. Moreover, adipose tissue has long been known to produce hormones, and recent research has revealed that even bone tissue has endocrine functions.
The ductless endocrine glands are not to be confused with the body’s exocrine system, whose glands release their secretions through ducts. Examples of exocrine glands include the sebaceous and sweat glands of the skin. As just noted, the pancreas also has an exocrine function: most of its cells secrete pancreatic juice through the pancreatic and accessory ducts to the lumen of the small intestine
Contributed by
14.02: Exercises
Lab 14 Exercise \(1\)
Label the following:
On the person, draw and label the following endocrine organs:
1 Thyroid
2 Adrenals
3 Pancreas
4 Ovaries
5 Hypothalmus
6 Pituitary
Lab 14 Exercise \(2\)
Endocrine system - CNS and histology
Identify the organ on the right and it's related structures. For the figure on the left label the Thyroid and parathyroid
Lab 14 Exercise \(3\)
Identify the organs and their related structures
Lab 14 Exercise \(4\)
1. Obtain a slide of each of the tissues listed below from the slide box at your table.
2. Follow the checklist above to set up your slide for viewing.
3. View the slide on the objective which provides the best view. Find the representative object.
4. In the circle below the name, draw a representative sample of the tissue, taking care to correctly and clearly draw their true shape in the slide. If it is a stratified epithelium draw all the layers. Draw your structures proportionately to their size in your microscope’s field of view.
5. Fill in the blanks next to your drawing and identify the structures listed on the last page under, “Histology”
Repeat this for each of the tissue types seen below
Lab 14 Exercise \(4\)
Hormone action
Label the following
1 Target cell
2 Endocrine cells x2
3 Water soluble hormone
4 Steroid hormone
5 Steroid hormone receptor
6 Cell surface hormone receptor
7 Second messenger system
14.03: MODELS- Torso Mid-Sagittal Head
Found on models:
• Hypothalamus
• Pituitary gland
• Infundibulum
• Pancreas
• Pineal gland
• Adrenal glands
• Testis
• Ovary
• Thyroid gland
• Ithsmus
Histology:
• Pituitary
• Anterior pituitary
• Posterior pituitary
• Infundibulum
• Pancreas
• Endocrine pancreas
• Pancreatic islet of Langerhans
• Capillary in pancreatic islet
• Exocrine pancreas
• acini
• duct
• Thyroid gland
• Follicular cell
• Parafollicular cell
• Colloid
• Thyroid follicle
• Parathyroid gland
• Adrenal glands
• Capsule of suprarenal gland
• Suprarenal cortex
• Zona fasciculata
• Zona glomerulosa
• Zona reticularis
• Suprarenal medulla | textbooks/bio/Human_Biology/Human_Anatomy_Lab/14%3A_The_Endocrine_System/14.01%3A_Structures_of_the_Endocrine_System.txt |
An Overview of Blood
Name the fluid component of blood and the three major types of formed elements, and identify their relative proportions in a blood sample
15: Blood
You have probably had blood drawn from a superficial vein in your arm, which was then sent to a lab for analysis. Some of the most common blood tests—for instance, those measuring lipid or glucose levels in plasma—determine which substances are present within blood and in what quantities. Other blood tests check for the composition of the blood itself, including the quantities and types of formed elements.
One such test, called a hematocrit, measures the percentage of RBCs, clinically known as erythrocytes, in a blood sample. It is performed by spinning the blood sample in a specialized centrifuge, a process that causes the heavier elements suspended within the blood sample to separate from the lightweight, liquid plasma (Figure 15.1). Because the heaviest elements in blood are the erythrocytes, these settle at the very bottom of the hematocrit tube. Located above the erythrocytes is a pale, thin layer composed of the remaining formed elements of blood. These are the WBCs, clinically known as leukocytes, and the platelets, cell fragments also called thrombocytes. This layer is referred to as the buffy coat because of its color; it normally constitutes less than 1 percent of a blood sample. Above the buffy coat is the blood plasma, normally a pale, straw- colored fluid, which constitutes the remainder of the sample.
The volume of erythrocytes after centrifugation is also commonly referred to as packed cell volume (PCV). In normal blood, about 45 percent of a sample is erythrocytes. The hematocrit of any one sample can vary significantly, however, about 36–50 percent, according to gender and other factors. Normal hematocrit values for females range from 37 to 47, with a mean value of 41; for males, hematocrit ranges from 42 to 52, with a mean of 47. The percentage of other formed elements, the WBCs and platelets, is extremely small so it is not normally considered with the hematocrit. So the mean plasma percentage is the percent of blood that is not erythrocytes: for females, it is approximately 59 (or 100 minus 41), and for males, it is approximately 53 (or 100 minus 47).
15.02: Exercises
LAB 15 EXERCISES \(1\)
Identify the 3 major components of a hematocrit:
LAB 15 EXERCISES \(2\)
Blood Cell Identification
Identify the formed elements by their picture below:
Label the image below
Fill in the Following Table
Blood Type Can Safely Receive Can Safely Donate T0
O+
A-
B+
AB-
AB+
LAB 15 EXERCISES \(3\)
1. Obtain a slide of blood from the slide box at your table.
2. Follow the checklist above to set up your slide for viewing.
3. View the slide on the objective which provides the best view (400X). Find the representative object.
4. In the circle below the name, draw a representative sample of the blood cell, taking care to correctly and clearly draw their true shape in the slide. Draw your structures proportionately to their size in your microscope’s field of view.
Repeat this for each of the blood cell types seen below
15.03: MODELS- Blood Histology
Hematocrit:
• Plasma
• Buffy coat
• RBCs
• (including average ranges for men and women)
Blood histology
• Erythrocytes
• Leukocytes:
• Granulocytes:
• Neutrophil
• Basophil
• Eosinophil
• Agranulocytes:
• Lymphocyte
• Monocyte
• Platelets
• Plasma
Blood types – students should understand the following terms and how they fit into the two blood typing mechanisms listed.
Know the antigens and antibodies present for a given blood type.
Students must know which blood types can be safely donated to or received from another given blood type.
Antigens
Antibodies
ABO system
• Type A
• Type B
• Type AB
• Type O
Rh system
• Rh+
• Rh- | textbooks/bio/Human_Biology/Human_Anatomy_Lab/15%3A_Blood/15.01%3A_Composition_of_Blood.txt |
Learning Objectives
• Describe the location and position of the heart within the body cavity
• Describe the internal and external anatomy of the heart
• Identify the tissue layers of the heart
• Identify the veins and arteries of the coronary circulation system
• Trace the pathway of oxygenated and deoxygenated blood thorough the chambers of the heart
16: The Heart
The human heart is located within the thoracic cavity, medially between the lungs in the space known as the mediastinum. Figure \(1\) shows the position of the heart within the thoracic cavity. Within the mediastinum, the heart is separated from the other mediastinal structures by a tough membrane known as the pericardium, or pericardial sac, and sits in its own space called the pericardial cavity. The dorsal surface of the heart lies near the bodies of the vertebrae, and its anterior surface sits deep to the sternum and costal cartilages. The great veins, the superior and inferior venae cavae, and the great arteries, the aorta and pulmonary trunk, are attached to the superior surface of the heart, called the base. The base of the heart is located at the level of the third costal cartilage, as seen in Figure \(1\). The inferior tip of the heart, the apex, lies just to the left of the sternum between the junction of the fourth and fifth ribs near their articulation with the costal cartilages. The right side of the heart is deflected anteriorly, and the left side is deflected posteriorly. It is important to remember the position and orientation of the heart when placing a stethoscope on the chest of a patient and listening for heart sounds, and also when looking at images taken from a midsagittal perspective. The slight deviation of the apex to the left is reflected in a depression in the medial surface of the inferior lobe of the left lung, called the cardiac notch.
16.02: Chambers and Circulation through the Heart
The human heart consists of four chambers: The left side and the right side each have one atrium and one ventricle. Each of the upper chambers, the right atrium (plural = atria) and the left atrium, acts as a receiving chamber and contracts to push blood into the lower chambers, the right ventricle and the left ventricle. The ventricles serve as the primary pumping chambers of the heart, propelling blood to the lungs or to the rest of the body.
There are two distinct but linked circuits in the human circulation called the pulmonary and systemic circuits. Although both circuits transport blood and everything it carries, we can initially view the circuits from the point of view of gases. The pulmonary circuit transports blood to and from the lungs, where it picks up oxygen and delivers carbon dioxide for exhalation. The systemic circuit transports oxygenated blood to virtually all of the tissues of the body and returns relatively deoxygenated blood and carbon dioxide to the heart to be sent back to the pulmonary circulation.
The right ventricle pumps deoxygenated blood into the pulmonary trunk, which leads toward the lungs and bifurcates into the left and right pulmonary arteries. These vessels in turn branch many times before reaching the pulmonary capillaries, where gas exchange occurs: Carbon dioxide exits the blood and oxygen enters. The pulmonary trunk arteries and their branches are the only arteries in the post-natal body that carry relatively deoxygenated blood. Highly oxygenated blood returning from the pulmonary capillaries in the lungs passes through a series of vessels that join together to form the pulmonary veins—the only post-natal veins in the body that carry highly oxygenated blood. The pulmonary veins conduct blood into the left atrium, which pumps the blood into the left ventricle, which in turn pumps oxygenated blood into the aorta and on to the many branches of the systemic circuit. Eventually, these vessels will lead to the systemic capillaries, where exchange with the tissue fluid and cells of the body occurs. In this case, oxygen and nutrients exit the systemic capillaries to be used by the cells in their metabolic processes, and carbon dioxide and waste products will enter the blood.
The blood exiting the systemic capillaries is lower in oxygen concentration than when it entered. The capillaries will ultimately unite to form venules, joining to form ever-larger veins, eventually flowing into the two major systemic veins, the superior vena cava and the inferior vena cava, which return blood to the right atrium. The blood in the superior and inferior venae cavae flows into the right atrium, which pumps blood into the right ventricle. This process of blood circulation continues as long as the individual remains alive. Understanding the flow of blood through the pulmonary and systemic circuits is critical to all health professions (Figure 16.2).
16.03: Membranes Surface Features and Layers
Our exploration of more in-depth heart structures begins by examining the membrane that surrounds the heart, the prominent surface features of the heart, and the layers that form the wall of the heart. Each of these components plays its own unique role in terms of function.
Membranes
The membrane that directly surrounds the heart and defines the pericardial cavity is called the pericardium or pericardial sac. It also surrounds the “roots” of the major vessels, or the areas of closest proximity to the heart. The pericardium, which literally translates as “around the heart,” consists of two distinct sublayers: the sturdy outer fibrous pericardium and the inner serous pericardium. The fibrous pericardium is made of tough, dense connective tissue that protects the heart and maintains its position in the thorax. The more delicate serous pericardium consists of two layers: the parietal pericardium, which is fused to the fibrous pericardium, and an inner visceral pericardium, or epicardium, which is fused to the heart and is part of the heart wall. The pericardial cavity, filled with lubricating serous fluid, lies between the epicardium and the pericardium
In most organs within the body, visceral serous membranes such as the epicardium are microscopic. However, in the case of the heart, it is not a microscopic layer but rather a macroscopic layer, consisting of a simple squamous epithelium called a mesothelium, reinforced with loose, irregular, or areolar connective tissue that attaches to the pericardium. This mesothelium secretes the lubricating serous fluid that fills the pericardial cavity and reduces friction as the heart contracts. Figure \(1\) illustrates the pericardial membrane and the layers of the heart.
Surface Features of the Heart
Inside the pericardium, the surface features of the heart are visible, including the four chambers. There is a superficial leaf-like extension of the atria near the superior surface of the heart, one on each side, called an auricle—a name that means “ear like”—because its shape resembles the external ear of a human (Figure \(2\)). Auricles are relatively thin-walled structures that can fill with blood and empty into the atria or upper chambers of the heart. You may also hear them referred to as atrial appendages. Also prominent is a series of fat-filled grooves, each of which is known as a sulcus (plural = sulci), along the superior surfaces of the heart. Major coronary blood vessels are located in these sulci. The deep coronary sulcus is located between the atria and ventricles. Located between the left and right ventricles are two additional sulci that are not as deep as the coronary sulcus. The anterior interventricular sulcus is visible on the anterior surface of the heart, whereas the posterior interventricular sulcus is visible on the posterior surface of the heart. Figure \(2\)illustrates anterior and posterior views of the surface of the heart. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/16%3A_The_Heart/16.01%3A_Location_of_the_Heart.txt |
Recall that the heart’s contraction cycle follows a dual pattern of circulation—the pulmonary and systemic circuits—because of the pairs of chambers that pump blood into the circulation. In order to develop a more precise understanding of cardiac function, it is first necessary to explore the internal anatomical structures in more detail.
Septa of the Heart
The word septum is derived from the Latin for “something that encloses;” in this case, a septum (plural = septa) refers to a wall or partition that divides the heart into chambers. The septa are physical extensions of the myocardium lined with endocardium. Located between the two atria is the interatrial septum. Normally in an adult heart, the interatrial septum bears an oval-shaped depression known as the fossa ovalis, a remnant of an opening in the fetal heart known as the foramen ovale. The foramen ovale allowed blood in the fetal heart to pass directly from the right atrium to the left atrium, allowing some blood to bypass the pulmonary circuit. Within seconds after birth, a flap of tissue known as the septum primum that previously acted as a valve closes the foramen ovale and establishes the typical cardiac circulation pattern.
Between the two ventricles is a second septum known as the interventricular septum. Unlike the interatrial septum, the interventricular septum is normally intact after its formation during fetal development. It is substantially thicker than the interatrial septum, since the ventricles generate far greater pressure when they contract.
The septum between the atria and ventricles is known as the atrioventricular septum. It is marked by the presence of four openings that allow blood to move from the atria into the ventricles and from the ventricles into the pulmonary trunk and aorta. Located in each of these openings between the atria and ventricles is a valve, a specialized structure that ensures one-way flow of blood. The valves between the atria and ventricles are known generically as atrioventricular valves. The valves at the openings that lead to the pulmonary trunk and aorta are known generically as semilunar valves. The interventricular septum is visible in Figure \(1\). In this figure, the atrioventricular septum has been removed to better show the bicupid and tricuspid valves; the interatrial septum is not visible, since its location is covered by the aorta and pulmonary trunk. Since these openings and valves structurally weaken the atrioventricular septum, the remaining tissue is heavily reinforced with dense connective tissue called the cardiac skeleton, or skeleton of the heart. It includes four rings that surround the openings between the atria and ventricles, and the openings to the pulmonary trunk and aorta, and serve as the point of attachment for the heart valves. The cardiac skeleton also provides an important boundary in the heart electrical conduction system.
16.05: Exercises
LAB 16 EXERCISES \(1\)
Visible human project, nlm.nih.gov
Draw the pathway of blood through the heart
Label the following
1 Aorta
2 Inferior vena cava
3 L. ventricle
4 R. ventricle
5 Lungs
6 Interventricular septum
Draw the pathway of blood through the heart
LAB 16 EXERCISES \(2\)
Label the following
1 Apex
2 Ascending aorta
3 Aortic arch
4 Anterior interventricular sulcus
5 L. Pulmonary artery
6 L. auricle
7 L. ventricle
8 Pulmonary trunk
9 Superior vena cava
10 R. ventricle
11 Coronary Sulcus
LAB 16 EXERCISES \(3\)
Label the following
1 Aorta
2 Coronary sulcus
3 L. atrium
4 L. ventricle
5 Inferior Vena Cava
6 R. Pumonary a
7 R. Pulmonary v.
8 Posterior interventricular sulcus
9 R. ventricle
LAB 16 EXERCISES \(4\)
Label the following
1 R. Coronary a. (RCA)
2 R. Marginal branch of RCA
3 Anterior interventricular a
4 Great cardiac vein
5 Circumflex A
Label the following
1 Great cardiac vein
2 Posterior interventricular a
3 Middle cardiac vein
4 Right coronary a.
5 Coronary sinus
6 Circumflex a.
7 L. marginal vein
LAB 16 EXERCISES \(5\)
Label the following
1 L. Pulmonary a.
2 L. Pulmonary v.
3 Interventricular septum
4 Mitral valve
5 Tricuspid valve
6 Pulmonary semilunar valve
7 Chordae tendinae
8 Papillary muscle
9 Endocardium
10 Myocardium
11 Epicardium
12 L. coronary a.
13 Pulmonary trunk
14 Conus arteriosis
LAB 16 EXERCISES \(6\)
Valves and Conduction System
Label the following
1 R. coronary a.
2 L. coronary a.
3 Mitral valve
4 Tricuspid valve
5 Pulmonary SL valve
6 Aortic SL valve
LAB 16 EXERCISES \(7\)
On this picture, draw and label the following:
1 SA Node
2 AV node
3 Bundle of His
4 Bundle branches
5 Purkinje fibers
16.06: MODELS- Heart and Torsos
Inferior vena cava
Superior vena cava
Right atrium
• Pectinate muscles
• Fossa ovalis
Right atrioventricular valve (tricuspid)
• Chordae tendineae
• Papillary muscles
Right ventricle
• Trabeculae carneae
Conus arteriosus
Pulmonary SL valve
Pulmonary trunk
Pulmonary arteries
Pulmonary veins
Left atrium
Left atrioventricular valve (mitral, bicuspid)
Left ventricle Aortic SL valve
Aorta
• Ascending, arch, descending, abdominal
• R. Brachiocephalic, L. Common Carotid a, L subclavian a.
• Apex
Auricles
Anterior interventricular sulcus
Posterior interventricular sulcus
Coronary sulcus
Inter-atrial septum
Interventricular septum
Electrical conduits: (clear heart model)
• Sinoatrial (SA) node
• Atrioventricular (AV) node
• Atrioventricular (AV) bundle (Bundle of His)
• Right and left bundle branches
• Purkinje fibers
• Moderator band
Tissue layers:
• Parietal pericardium
• Visceral pericardium (epicardium)
• Myocardium
• Endocardium
Coronary circulation:
• R. coronary a.
• Anterior interventricular a.
• Posterior interventricular a.
• Circumflex a.
• L. coronary a.
• Middle cardiac v.
• Great cardiac v.
• Coronary sinus
• Marginal arteries | textbooks/bio/Human_Biology/Human_Anatomy_Lab/16%3A_The_Heart/16.04%3A_Internal_Structure_of_the_Heart.txt |
The heart dissection is probably one of the most difficult dissections you will do. Part of the reason it is so difficult to learn is that the heart is not perfectly symmetrical, but it is so close that it becomes difficult to discern which side you are looking at (dorsel, ventral, left or right). Finding the vessels is directly related to being able to orient the heart correctly and figuring out which side you are looking at.
The heart is also difficult because the fatty tissue that surrounds the heart can obscure the openings to the vessels. This means that you really must experience the heart with your hands and feel your way to find the openings. Many people will be squeamish about this, and because the heart is slippery, it is easy to drop. Don't be shy with the heart, use your fingers to feel your way through the dissection.
1. Step One: Orientation
When you first remove your heart from the bag, you will see a lot of fatty tissue surrounding it. It is usually a waste of time to try to remove this tissue. Grab some colored pencils to help you identify and mark the vessels you find.
There are a few clues to help you figure out the left and the right side, but often the packaging and preserving process can cause the heart to be misshapen. If you are lucky, the heart will be nicely preserved and you will see that the front (ventral) side of the heart has a couple of key features: 1) a large pulmonary trunk that extends off the top of it 2) the flaps of the auricles covering the top of the atria. 3) the curve of the entire front side, whereas the backside is much flatter.
The first image shows the front side of the heart, often identified by the anterior interventricular artery that runs cross it at an angle (yellow).
The auricle is the flap that covers the atrium, it looks like an ear. The pulmonary trunk is the located at the front of the heart and enters at an angle.
Step 2: Locate the Aorta
Use your fingers to probe around the top of the heart. Four major vessels can be found entering the heart: the pulmonary trunk, aorta, superior vena cava, and the pulmonary vein. Remember that if you are looking at the back of the heart, then the right and left sides are the same as your right and left hand. This picture was on the board the day of the dissection so that you could glance up and recall which vessel entered which part of the heart.
If you find the pulmonary vein, the aorta should be situated a little bit behind it. It may be covered by fat, so use your fingers to poke around until you find the opening. Push your finger all the way in and you will feel inside of the left ventricle. The left ventricle has a very thick wall, unlike the right ventricle. Insert your finger through the pulmonary vessel to feel the left ventricle and you will notice and feel that it is much thinner than the left side of the heart.
With your fingers or probes in the aorta and the pulmonary trunk you should notice that they criss-cross each other, with the pulmonary trunk in the front. At this point, you may want to use your colored pencils to mark these vessels so that you don't get them confused when you are searching for the other two openings that top of the heart.
Step 3: Locate the Veins
The two major veins that enter the heart can be found on the backside, as both enter the atria. On the left side, you should be able to find the opening of the pulmonary vein as it enters the left atrium. The superior vena cava enters the right atrium. In many preserved hearts, the heart was cut at these points, so you won't see the vessels themselves, you will just find the openings. Again, use your fingers to feel around the heart to find the openings. If you've marked the aorta and pulmonary then you won't mistake them for the veins you are looking for. This picture shows all of the vessels labeled.
Sometimes, the aorta still has its branches attached to it. There are three vessels that branch from the aorta: the brachiocephalic, left common carotid and the left subclavian. The majority of the time, these vessels are not visible because the aorta was cut too close to the main part of the heart when the heart was removed from the animal. Occassionally, you can find the brachiocephalic artery attached, as it is in this photo.
Step 4: Make the Incisions
Now that you have all of the vessels located and marked, you can now open the heart to view the inner chambers. Use the superior vena cava and pulmonary vein as guides for where to cut. You are basically going to be cutting each side of the heart so that you can look inside. (Some dissections will ask you to make a coronal cut where a single cut opens the entire back side of the heart). The heart below is marked to show you where the two incisions should be made.
Cut the heart in half to expose the chambers. My students affectionally call these two variations the "hot dog cut" as pictured above because it looks like a hot dog bun, or the "hamburger cut, where the heart is cut into the front and the back half, as shown below.Figure \(2\): Copy and Paste Caption here. (Copyright; author via source)
Step 5: Viewing the Chambers
At this point it is helpful to have two hands, one to hold the heart apart so you can take a peak inside of it and another to use a probe to locate the specific parts. Your colored pencils you used to mark the heart in step 2 can also now be used to see where those vessels connect within the heart. For instance, the aorta pencil can now be seen ending in the left ventricle.
You can also now see how much thicker the walls of the left ventricle are compared to the right ventricle.
The other obvious structures seen within the heart are the chordae tendinae which are attached to papillary muscles. These tendons hold the heart valves in place, sometimes they are called the "heartstrings". The valves were probably cut when the heart was opened, but if you follow the "cords" they should lead you to a thin flap that is the atrioventricular (bicuspid) valve. You can find a similar valve on the right side of the heart (tricuspid).
Image shows the left atrioventricular valve (bicuspid) and the chordae tendinae. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/16%3A_The_Heart/16.07%3A_Heart_Dissection.txt |
Skills to Develop
• Compare and contrast the three tunics that make up the walls of most blood vessels
• Distinguish between elastic arteries, muscular arteries, and arterioles on the basis of structure and location
• Describe the basic structure of a capillary bed, from the supplying metarteriole to the venule into which it drains
• Explain the structure of venous valves in the large veins of the extremities
17: Blood Vessels and Circulation
Different types of blood vessels vary slightly in their structures, but they share the same general features. Arteries and arterioles have thicker walls than veins and venules because they are closer to the heart and receive blood that is surging at a far greater pressure (Figure 17.1). Each type of vessel has a lumen—a hollow passageway through which blood flows. Arteries have smaller lumens than veins, a characteristic that helps to maintain the pressure of blood moving through the system. Together, their thicker walls and smaller diameters give arterial lumens a more rounded appearance in cross section than the lumens of veins.
Comparison of Tunics in Arteries and Veins
Arteries Veins
General Appearance
Thick Walls with small lumens
Generally appear round
Thin walls with large lumens
Generally appear flattened
Tunica Intima Endothelium usually appears wavy due to constriction of smooth muscle
Internal elastic membrane presents in larger vessels
Endolithelium appears smooth
Internal elastic membrane absent
Tunica media Normally the thickest layer in arteries
Smooth muscle cells and elastic fibers predominate (the proportions of these vary with distance from the heart"
External elastic membrane present in larger vessels
Normally thinner than the tunica externa
Smooth muscle cells and collagenous fibers predeominate
Esternal elastic membrane absent
Tunica externa Normally thinner than the tunica media in all but the largest arteries
Collagenous and elastic fibers
Nervi vasorum and vasa vasorum present
Normally the thickest layer in veins
Collagenous and smooth fibers predominate
Some smooth muscle fibers
Nervi vasoum and vasa vasorum present
17.02: Exercises
LAB 17 EXERCISES \(1\)
Label the following
1 Tunica interna
2 Tunica media
3 Tunica externa
4 Valve
5 Vasa vasorum
6 Artery
7 Vein.
Label the following (include Left/right on all pages)
1 Aorta
2 L common carotid a
3 Brachiocephalic trunk
4 L subclavian a.
LAB 17 EXERCISES \(2\)
Arteries of the Abdomen
Label the following (include Left/right where appropriate)
1 Celiac trunk
2 Common hepatic a.
3 Common hepatic a.
4 Superior mesenteric a. s
5 Inferior mesenteric a.
6 Internal iliac a.
7 External iliac a.
8 Common Hepatic a.
9 Splenic a.
10 Abdominal aorta
LAB 17 EXERCISES \(3\)
Arteries of the limbs
Label the following (include Left/right where appropriate)
1 Popliteal a.
2 Posterior tibial a
3 Anterior tibial a
4 Fibular a.
5 Femoral a.
Label the following (include Left/right where appropriate)
1 Ulnar a
2 Radial a
3 Brachial a
4 Subclavian a.
5 Axillary a
LAB 17 EXERCISES \(4\)
Arteries of the head
Identify the following:
1 Superficial temporal a.
2 Facial a
3 Common carotid a
4 Vertebral a.
5 Internal carotid a.
6 External carotid a
7 Occipital a
Identify the following:
1 Circle of Willis
2 Basilar a.
3 Vertebral a.
4 Internal carotid a.
LAB 17 EXERCISES \(5\)
Veins
Identify the veins (Include Left/Right where appropriate)
1 Basilic v
2 Cephalic v.
3 Brachial v
4 Axillary v.
5 Median cubital v
6 Radial v
7 Ulnar v
Label the veins (Include Left/Right where appropriate):
1 Femoral v
2 Common iliac v.
3 External iliac v.
4 Great saphenous v.
5 Inferior vena cava.
LAB 17 EXERCISES \(6\)
Veins
Label the veins:
1 R. Hepatic v
2 Hepatic Portal v
3 Hepatic Portal v
4 Splenic v.
5 Inferior mesenteric v.
6 Superior mesenteric v.
7 L Gonadal v.
8 * L Renal v.
LAB 17 EXERCISES \(7\)
Veins and pressure points
Label the following (Include Left/Right where appropriate):
1 Internal iliac v.
2 Great saphenous v
3 Small saphenous v.
4 Popliteal v
5 Anterior tibial v
6 Posterior tibial v
7 Fibular v
Label 4 common pulse pressure points with names and arrows (places where you could find a pulse on a patient)
17.03: MODELS- Flat Body Skull with Arteries and Nerves Torso Leg Arm and Sagittal Head
Blood vessel histology:
• Tunica externa
• Tunica media
• Tunica interna
• Vasa vasorum
• Valves
• Varicose vein
Arteries (All arteries should have the word arteries at the end)
• Aorta
• Ascending * arch * descending * thoracic * abdominal
• Brachiocephalic trunk
• Common carotid (skull)
• External carotid (skull)
• Superior thyroid (torso)
• Facial (skull)
• Superficial temporal (skull)
• Occipital (skull)
• Internal carotid
• Circle of Willis (skull)
• Subclavian
• Vertebral artery
• Basilar artery (skull)
• Axillary artery
• Brachial artery
• Radial artery
• Ulnar artery
• Celiac trunk
• Left & right gastric artery
• Splenic
• Common hepatic a.
• Hepatic
• Superior mesenteric
• Suprarenal
• Renal
• Gonadal
• Inferior mesenteric
• Common iliac
• Internal iliac
• External iliac
• Femoral
• Popliteal
• Anterior tibial
• Posterior tibial
Veins (All veins except the Vena Cava should have the word vein at the end)
• Superior vena cava
• Brachiocephalic
• Internal jugular
• Superior sagittal sinus
• Straight sinus
• External jugular
• Subclavian
• Axillary
• Cephalic
• Brachial
• Radial
• Ulnar
• Basilic
• Median cubital
• Inferior vena cava
• Hepatic
• Suprarenal
• Renal
• Gonadal
• Common iliac
• Internal iliac
• External iliac
• Great saphenous
• Femoral
• Popliteal
• Anterior tibial
• Posterior tibial
• Fibular
• Small saphenous
• Inferior mesenteric
• Splenic
• Superior mesenteric
• Hepatic portal vein | textbooks/bio/Human_Biology/Human_Anatomy_Lab/17%3A__Blood_Vessels_and_Circulation/17.01%3A_Shared_Structures.txt |
Skills to Develop
• Describe the structure of the lymphatic tissue (lymph fluid, vessels, ducts, and organs)
• Describe the structure of the primary and secondary lymphatic organs
The lymphatic system is the system of vessels, cells, and organs that carries excess fluids to the bloodstream and filters pathogens from the blood. The swelling of lymph nodes during an infection and the transport of lymphocytes via the lymphatic vessels are but two examples of the many connections between these critical organ systems.
18: The Lymphatic System
The lymphatic vessels begin as open-ended capillaries, which feed into larger and larger lymphatic vessels, and eventually empty into the bloodstream by a series of ducts. Along the way, the lymph travels through the lymph nodes, which are commonly found near the groin, armpits, neck, chest, and abdomen. Humans have about 500– 600 lymph nodes throughout the body (Figure \(1\)).
Lymphatic Capillaries
Lymphatic capillaries, also called the terminal lymphatics, are vessels where interstitial fluid enters the lymphatic system to become lymph fluid. Located in almost every tissue in the body, these vessels are interlaced among the arterioles and venules of the circulatory system in the soft connective tissues of the body (Figure \(2\)).
Larger Lymphatic Vessels, Trunks, and Ducts
The lymphatic capillaries empty into larger lymphatic vessels, which are similar to veins in terms of their three- tunic structure and the presence of valves. These one-way valves are located fairly close to one another, and each one causes a bulge in the lymphatic vessel, giving the vessels a beaded appearance (see Figure \(2\)).
The superficial and deep lymphatics eventually merge to form larger lymphatic vessels known as lymphatic trunks. On the right side of the body, the right sides of the head, thorax, and right upper limb drain lymph fluid into the right subclavian vein via the right lymphatic duct (Figure \(3\)). On the left side of the body, the remaining portions of the body drain into the larger thoracic duct, which drains into the left subclavian vein. The thoracic duct itself begins just beneath the diaphragm in the cisterna chyli, a sac-like chamber that receives lymph from the lower abdomen, pelvis, and lower limbs by way of the left and right lumbar trunks and the intestinal trunk.
The overall drainage system of the body is asymmetrical (see Figure \(3\)). The right lymphatic duct receives lymph from only the upper right side of the body. The lymph from the rest of the body enters the bloodstream through the thoracic duct via all the remaining lymphatic trunks. In general, lymphatic vessels of the subcutaneous tissues of the skin, that is, the superficial lymphatics, follow the same routes as veins, whereas the deep lymphatic vessels of the viscera generally follow the paths of arteries.
18.02: Lymphoid Organs
The thymus gland is a bilobed organ found in the space between the sternum and the aorta of the heart (Figure \(1\)). Connective tissue holds the lobes closely together but also separates them and forms a capsule. The connective tissue capsule further divides the thymus into lobules via extensions called trabeculae. The outer region of the organ is known as the cortex and contains large numbers of thymocytes with some epithelial cells, macrophages, and dendritic cells (two types of phagocytic cells that are derived from monocytes). The cortex is densely packed so it stains more intensely than the rest of the thymus (see Figure \(1\)). The medulla, where thymocytes migrate before leaving the thymus, contains a less dense collection of thymocytes, epithelial cells, and dendritic cells.
Lymph Nodes
Lymph nodes function to remove debris and pathogens from the lymph, and are thus sometimes referred to as the “filters of the lymph” (Figure \(2\)). Any bacteria that infect the interstitial fluid are taken up by the lymphatic capillaries and transported to a regional lymph node. Dendritic cells and macrophages within this organ internalize and kill many of the pathogens that pass through, thereby removing them from the body. The lymph node is also the site of adaptive immune responses mediated by T cells, B cells, and accessory cells of the adaptive immune system. Like the thymus, the bean- shaped lymph nodes are surrounded by a tough capsule of connective tissue and are separated into compartments by trabeculae, the extensions of the capsule. In addition to the structure provided by the capsule and trabeculae, the structural support of the lymph node is provided by a series of reticular fibers laid down by fibroblasts.
The major routes into the lymph node are via afferent lymphatic vessels (see Figure \(2\)). Cells and lymph fluid that leave the lymph node may do so by another set of vessels known as the efferent lymphatic vessels. Lymph enters the lymph node via the subcapsular sinus, which is occupied by dendritic cells, macrophages, and reticular fibers. Within the cortex of the lymph node are lymphoid follicles, which consist of germinal centers of rapidly dividing B cells surrounded by a layer of T cells and other accessory cells. As the lymph continues to flow through the node, it enters the medulla, which consists of medullary cords of B cells and plasma cells, and the medullary sinuses where the lymph collects before leaving the node via the efferent lymphatic vessels.
Spleen
In addition to the lymph nodes, the spleen is a major secondary lymphoid organ (Figure \(3\)). It is about 12 cm (5 in) long and is attached to the lateral border of the stomach via the gastrosplenic ligament. The spleen is a fragile organ without a strong capsule, and is dark red due to its extensive vascularization.
The spleen is sometimes called the “filter of the blood” because of its extensive vascularization and the presence of macrophages and dendritic cells that remove microbes and other materials from the blood, including dying red blood cells. The spleen also functions as the location of immune responses to blood- borne pathogens.
The spleen is also divided by trabeculae of connective tissue, and within each splenic nodule is an area of red pulp, consisting of mostly red blood cells, and white pulp, which resembles the lymphoid follicles of the lymph nodes. Upon entering the spleen, the splenic artery splits into several arterioles (surrounded by white pulp) and eventually into sinusoids. Blood from the capillaries subsequently collects in the venous sinuses and leaves via the splenic vein. The red pulp consists of reticular fibers with fixed macrophages attached, free macrophages, and all of the other cells typical of the blood, including some lymphocytes.
The white pulp surrounds a central arteriole and consists of germinal centers of dividing B cells surrounded by T cells and accessory cells, including macrophages and dendritic cells. Thus, the red pulp primarily functions as a filtration system of the blood, using cells of the relatively nonspecific immune response, and white pulp is where adaptive T and B cell responses are mounted.
Lymphoid Nodules
Tonsils are lymphoid nodules located along the inner surface of the pharynx and are important in developing immunity to oral pathogens (Figure \(4\)). The tonsil located at the back of the throat, the pharyngeal tonsil, is sometimes referred to as the adenoid when swollen. Such swelling is an indication of an active immune response to infection. Histologically, tonsils do not contain a complete capsule, and the epithelial layer invaginates deeply into the interior of the tonsil to form tonsillar crypts. These structures, which accumulate all sorts of materials taken into the body through eating and breathing, actually “encourage” pathogens to penetrate deep into the tonsillar tissues where they are acted upon by numerous lymphoid follicles and eliminated. This seems to be the major function of tonsils—to help children’s bodies recognize, destroy, and develop immunity to common environmental pathogens so that they will be protected in their later lives. Tonsils are often removed in those children who have recurring throat infections, especially those involving the palatine tonsils on either side of the throat, whose swelling may interfere with their breathing and/or swallowing. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/18%3A_The_Lymphatic_System/18.01%3A_Structure_of_the_Lymphatic_System.txt |
In addition to the lymph nodes, the spleen is a major secondary lymphoid organ ( Figure \(1\) ). It is about 12 cm (5 in) long and is attached to the lateral border of the stomach via the gastrosplenic ligament. The spleen is a fragile organ without a strong capsule, and is dark red due to its extensive vascularization.
The spleen is sometimes called the “filter of the blood” because of its extensive vascularization and the presence of macrophages and dendritic cells that remove microbes and other materials from the blood, including dying red blood cells. The spleen also functions as the location of immune responses to blood- borne pathogens.
The spleen is also divided by trabeculae of connective tissue, and within each splenic nodule is an area of red pulp, consisting of mostly red blood cells, and white pulp, which resembles the lymphoid follicles of the lymph nodes. Upon entering the spleen, the splenic artery splits into several arterioles (surrounded by white pulp) and eventually into sinusoids. Blood from the capillaries subsequently collects in the venous sinuses and leaves via the splenic vein. The red pulp consists of reticular fibers with fixed macrophages attached, free macrophages, and all of the other cells typical of the blood, including some lymphocytes.
The white pulp surrounds a central arteriole and consists of germinal centers of dividing B cells surrounded by T cells and accessory cells, including macrophages and dendritic cells. Thus, the red pulp primarily functions as a filtration system of the blood, using cells of the relatively nonspecific immune response, and white pulp is where adaptive T and B cell responses are mounted.
18.05: Lymphoid Nodules
Tonsils are lymphoid nodules located along the inner surface of the pharynx and are important in developing immunity to oral pathogens (Figure Figure \(1\)). The tonsil located at the back of the throat, the pharyngeal tonsil, is sometimes referred to as the adenoid when swollen. Such swelling is an indication of an active immune response to infection. Histologically, tonsils do not contain a complete capsule, and the epithelial layer invaginates deeply into the interior of the tonsil to form tonsillar crypts. These structures, which accumulate all sorts of materials taken into the body through eating and breathing, actually “encourage” pathogens to penetrate deep into the tonsillar tissues where they are acted upon by numerous lymphoid follicles and eliminated. This seems to be the major function of tonsils—to help children’s bodies recognize, destroy, and develop immunity to common environmental pathogens so that they will be protected in their later lives. Tonsils are often removed in those children who have recurring throat infections, especially those involving the palatine tonsils on either side of the throat, whose swelling may interfere with their breathing and/or swallowing.
18.06: Exercises
LAB 17 EXERCISES \(1\)
Match the following
1 Cisterna Chyli, R.
2 Lymphatic duct
3 Thoracic duct
4 Axillary lymph nodes
5 Abdominal lymph nodes
6 Cervical lymph nodes
7 Inguinal lymph nodes
Match the following
1 Capsule
2 Afferent vessel
3 Efferent vessel
4 Hilus
5 Cortex
6 Medulla
7 Nodules
8 Trabeculae
LAB 17 EXERCISES \(3\)
Tonsils
Label tonsils
Identify the lymphatic structure found in the ileum and colon
LAB 17 EXERCISES \(1\)
Abdominal Region
Label the following
1 Tunica interna
2 Tunica media
3 Tunica externa
4 Valve
5 Vasa vasorum
6 Artery
7 Vein.
Fill in and identify all the boxes
1
2
3
4
5
6
Organ
LAB 17 EXERCISES \(4\)
Peyer’s patches (Mucosal associated lymphatic tissue, MALT) and Tonsil
1. Obtain a slide of each of the gland listed below from the slide box at your table.
2. Follow the checklist above to set up your slide for viewing.
3. View the slide on the objective which provides the best view. Find the representative object.
4. In the circle below the name, draw a representative sample of the tissue, taking care to correctly and clearly draw their true shape in the slide. Draw your structures proportionately to their size in your microscope’s field of view.
5. Fill in the blanks next to your drawing and identify the structures listed on the last page under, “Histology”
Repeat this for each of the tissue types seen below.
F
18.07: MODELS- Torsos and Mid-Sagittal Head
• Appendix (vermiform)
• Lymph node clusters:
• Cervical
• Inguinal
• Thoracic
• Axillary
• Thoracic duct
• Right lymphatic duct
• Cisterna Chyli
• Spleen
• Tonsils:
• Pharyngeal tonsils
• Palatine tonsils
• Lingual tonsil
• Tubal tonsils
• Histology:
• Lymph node
• Medulla
• Cortex
• Hilus
• Afferent & efferent vessels
• Spleen
• White pulp
• Red pulp
• Peyer’s patches (Mucosal associated lymphatic tissue, MALT)
• Tonsil | textbooks/bio/Human_Biology/Human_Anatomy_Lab/18%3A_The_Lymphatic_System/18.03%3A_Spleen.txt |
Skills to Develop
• List the structures that make up the respiratory system
19: The Respiratory System
The major entrance and exit for the respiratory system is through the nose. When discussing the nose, it is helpful to divide it into two major sections: the external nose, and the nasal cavity or internal nose.
The external nose consists of the surface and skeletal structures that result in the outward appearance of the nose and contribute to its numerous functions (Figure \(1\)). The root is the region of the nose located between the eyebrows. The bridge is the part of the nose that connects the root to the rest of the nose. The dorsum nasi is the length of the nose. The apex is the tip of the nose. On either side of the apex, the nostrils are formed by the alae (singular = ala). An ala is a cartilaginous structure that forms the lateral side of each naris (plural = nares), or nostril opening. The philtrum is the concave surface that connects the apex of the nose to the upper lip.
The nares open into the nasal cavity, which is separated into left and right sections by the nasal septum (Figure \(2\)). The nasal septum is formed anteriorly by a portion of the septal cartilage (the flexible portion you can touch with your fingers) and posteriorly by the perpendicular plate of the ethmoid bone (a cranial bone located just posterior to the nasal bones) and the thin vomer bones (whose name refers to its plough shape). Each lateral wall of the nasal cavity has three bony projections, called the superior, middle, and inferior nasal conchae. The inferior conchae are separate bones, whereas the superior and middle conchae are portions of the ethmoid bone.
Conchae serve to increase the surface area of the nasal cavity and to disrupt the flow of air as it enters the nose, causing air to bounce along the epithelium, where it is cleaned and warmed. The conchae and meatuses also conserve water and prevent dehydration of the nasal epithelium by trapping water during exhalation. The floor of the nasal cavity is composed of the palate. The hard palate at the anterior region of the nasal cavity is composed of bone. The soft palate at the posterior portion of the nasal cavity consists of muscle tissue. Air exits the nasal cavities via the internal nares and moves into the pharynx.
The conchae, meatuses, and paranasal sinuses are lined by respiratory epithelium composed of pseudostratified ciliated columnar epithelium (Figure \(3\)). The epithelium contains goblet cells, one of the specialized, columnar epithelial cells that produce mucus to trap debris. The cilia of the respiratory epithelium help remove the mucus and debris from the nasal cavity with a constant beating motion, sweeping materials towards the throat to be swallowed. Interestingly, cold air slows the movement of the cilia, resulting in accumulation of mucus that may in turn lead to a runny nose during cold weather. This moist epithelium functions to warm and humidify incoming air. Capillaries located just beneath the nasal epithelium warm the air by convection. Serous and mucus-producing cells also secrete the lysozyme enzyme and proteins called defensins, which have antibacterial properties. Immune cells that patrol the connective tissue deep to the respiratory epithelium provide additional protection. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/19%3A_The_Respiratory_System/19.01%3A_The_Nose_and_its_Adjacent_Structures.txt |
The pharynx is a tube formed by skeletal muscle and lined by mucous membrane that is continuous with that of the nasal cavities (see Figure 19.2). The pharynx is divided into three major regions: the nasopharynx, the oropharynx, and the laryngopharynx (Figure \(1\)).
19.03: Larynx
The larynx is a cartilaginous structure inferior to the laryngopharynx that connects the pharynx to the trachea and helps regulate the volume of air that enters and leaves the lungs (Figure \(1\)). The structure of the larynx is formed by several pieces of cartilage. Three large cartilage pieces—the thyroid cartilage (anterior), epiglottis (superior), and cricoid cartilage (inferior)—form the major structure of the larynx. The thyroid cartilage is the largest piece of cartilage that makes up the larynx. The thyroid cartilage consists of the laryngeal prominence, or “Adam’s apple,” which tends to be more prominent in males. The thick cricoid cartilage forms a ring, with a wide posterior region and a thinner anterior region. Three smaller, paired cartilages—the arytenoids, corniculates, and cuneiforms—attach to the epiglottis and the vocal cords and muscle that help move the vocal cords to produce speech.
19.04: Trachea
The trachea (windpipe) extends from the larynx toward the lungs (\(1\)). The trachea is formed by 16 to 20 stacked, C-shaped pieces of hyaline cartilage that are connected by dense connective tissue. The trachealis muscle and elastic connective tissue together form the fibroelastic membrane, a flexible membrane that closes the posterior surface of the trachea, connecting the C-shaped cartilages. The fibroelastic membrane allows the trachea to stretch and expand slightly during inhalation and exhalation, whereas the rings of cartilage provide structural support and prevent the trachea from collapsing. In addition, the trachealis muscle can be contracted to force air through the trachea during exhalation. The trachea is lined with pseudostratified ciliated columnar epithelium, which is continuous with the larynx. The esophagus borders the trachea posteriorly.
19.05: Bronchial Tree
The trachea branches into the right and left primary bronchi at the carina. These bronchi are also lined by pseudostratified ciliated columnar epithelium containing mucus-producing goblet cells ( \(1\)b). The carina is a raised structure that contains specialized nervous tissue that induces violent coughing if a foreign body, such as food, is present. Rings of cartilage, similar to those of the trachea, support the structure of the bronchi and prevent their collapse. The primary bronchi enter the lungs at the hilum, a concave region where blood vessels, lymphatic vessels, and nerves also enter the lungs. The bronchi continue to branch into bronchial a tree. A bronchial tree (or respiratory tree) is the collective term used for these multiple-branched bronchi. The main function of the bronchi, like other conducting zone structures, is to provide a passageway for air to move into and out of each lung. In addition, the mucous membrane traps debris and pathogens.
A bronchiole branches from the tertiary bronchi. Bronchioles, which are about 1 mm in diameter, further branch until they become the tiny terminal bronchioles, which lead to the structures of gas exchange. There are more than 1000 terminal bronchioles in each lung. The muscular walls of the bronchioles do not contain cartilage like those of the bronchi. This muscular wall can change the size of the tubing to increase or decrease airflow through the tube. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/19%3A_The_Respiratory_System/19.02%3A_Pharynx.txt |
In contrast to the conducting zone, the respiratory zone includes structures that are directly involved in gas exchange. The respiratory zone begins where the terminal bronchioles join a respiratory bronchiole, the smallest type of bronchiole (Figure \(1\)), which then leads to an alveolar duct, opening into a cluster of alveoli.
Figure 19.7 Respiratory Zone Bronchioles lead to alveolar sacs in the respiratory zone, where gas exchange occurs.
Alveoli
An alveolar duct is a tube composed of smooth muscle and connective tissue, which opens into a cluster of alveoli. An alveolus is one of the many small, grape-like sacs that are attached to the alveolar ducts.
An alveolar sac is a cluster of many individual alveoli that are responsible for gas exchange. An alveolus is approximately 200 μm in diameter with elastic walls that allow the alveolus to stretch during air intake, which greatly increases the surface area available for gas exchange. Alveoli are connected to their neighbors by alveolar pores, which help maintain equal air pressure throughout the alveoli and lung (Figure \(2\)).
The alveolar wall consists of three major cell types: type I alveolar cells, type II alveolar cells, and alveolar macrophages. A type I alveolar cell is a squamous epithelial cell of the alveoli, which constitute up to 97 percent of the alveolar surface area. These cells are about 25 nm thick and are highly permeable to gases. A type II alveolar cell is interspersed among the type I cells and secretes pulmonary surfactant, a substance composed of phospholipids and proteins that reduces the surface tension of the alveoli. Roaming around the alveolar wall is the alveolar macrophage, a phagocytic cell of the immune system that removes debris and pathogens that have reached the alveoli.
The simple squamous epithelium formed by type I alveolar cells is attached to a thin, elastic basement membrane. This epithelium is extremely thin and borders the endothelial membrane of capillaries. Taken together, the alveoli and capillary membranes form a respiratory membrane that is approximately 0.5 mm thick. The respiratory membrane allows gases to cross by simple diffusion, allowing oxygen to be picked up by the blood for transport and CO2 to be released into the air of the alveoli.
19.07: Exercises
LAB 19 EXERCISES \(1\)
Identify the following
1 Nasal vestibule
2 Nasal cavity
3 Esophagus
4 Pharynx
5 Larynx.
Match the following
1 Naris
2 Ala (of nose)
3 Bridge (of nose)
4 Philtrum.
LAB 19 EXERCISES \(2\)
Identify the following components of the upper repiratory tract
1 Superior nasal conchae 8 Oropharynx
2 Middle nasal conchae 9 Laryngopharynx
3 Inferior nasal conchae 10 Epiglottis
4 Hard palate 11 Uvula
5 Soft palate 12 Sphenoid sinus
6 Eustachian tube 13 Frontal sinus
7 Nasopharynx
LAB 19 EXERCISES \(2\)
Upper Respiratory Tract
Label the following
1 All of the lobes of the lungs
2 All the fissures of the lungs
3 Apex
4 Base
5 Diaphragm.
7 Vein.
Label these components of the larynx
1 Epiglottis
2 Tunica media
3 Tunica externa
4 Valve
5 Vasa vasorum
6 Artery
7 Vein.
LAB 19 EXERCISES \(4\)
Upper --> Lower Respiratory Tract
Label the following
1 Vestibular folds
2 Epiglottis
3 Glottis
4 Arytenoid cartilage.
Label the following
1 Thyroid cartilage
2 Cricoid cartilage
3 Cricoid cartilage
4 Carina
5 R. primary bronchus
6 L. primary bronchus
7 Secondary bronchi
8 Tertiary bronchi
LAB 19 EXERCISES \(5\)
Lower Respiratory Tract
Label the following
1 Cardiac notch
2 Pulmonary artery
3 Pulmonary vein
4 Abdominal aorta
Label the following
1 Bronchiole
2 Alveoli
3 Alveolar sac
4 Pulmonary venule
5 Pulmonary arteriole
6 Capillary network
LAB 19 EXERCISES \(6\)
Histology
Label the tracheal histology
1 Pseudostratified ciliated columnar epithelium
2 Goblet cell
3 Cilia
Lower Respiratory Tract
Label the tracheal cartilage histology
1 Hyaline cartilage
2 Mucosa
3 Submucosa
4 Lumen
5 Mucous gland.
Label the following lung histology components
1 Simple squamous epithelium
2 Smooth muscle
3 Alveloi
4 Aveolar sac
5 Bronchiole
6 Arteriole
19.08: MODELS- Bronchial Tree Lungs and Heart Trachea Torsos and Mid-Sagittal Head
• Pleura
• parietal & visceral
• Pleural cavity
• Right lung
• Lobes: upper * middle * lower
• Horizontal fissure
• Oblique fissure
• Hilus
• Left Lung:
• lobes: upper * lower
• Oblique fissure
• Cardiac notch
• Hilus
• Philtrum
• Nose
• Bridge
• Ala
• Nares
• Nasal cavity
• Vestibule
• Nasal conchae:
• superior * middle * & inferior
• Nasal meatuses:
• superior * middle * & inferior
• Nasal septum
• Pharyngotympanic tube (Eustachian)
• Sinuses (Sphenoid * frontal)
• Hard palate
• Soft palate
• Uvula
• Hyoid bone
• Pharynx:
• Nasopharynx
• Oropharynx
• Laryngopharynx
• Larynx:
• Epiglottis
• Thyroid cartilage
• Cricoid cartilage
• Crico-thyroid ligament
• Crico-tracheal ligament
• Hyo-thyroid ligament
• Vestibular folds (false vocal cords)
• True vocal cords
• Arytenoid cartilage
• Corniculate cartilage
• Trachea:
• Tracheal cartilages
• Carina
• Bronchial tree:
• L * R primary (main) bronchi
• Secondary (lobar) bronchi
• Tertiary (segmental) bronchi
• Lung tissue:
• Bronchioles
• Alveolar sac
• Alveoli
• Muscles:
• Diaphragm
• External intercostals
• Internal intercostals
• Vasculature
• Pulmonary arteries
• Pulmonary veins
•
Lung Histology: (2 slides)
• Trachea
• Adventitia
• Tracheal cartilage (hyaline)
• Trachealis muscle
• Mucosa (pseudostratified ciliated columnar epithelium)
• Lungs
• Alveoli (Simple squamous epithelia)
• Bronchioles
• Pulmonary arterioles * venules * & capillaries | textbooks/bio/Human_Biology/Human_Anatomy_Lab/19%3A_The_Respiratory_System/19.06%3A_Respiratory_Zone.txt |
Skills to Develop
• Identify the organs of the alimentary canal from proximal to distal
• Identify the accessory digestive organs
20: The Digestive System
The function of the digestive system is to break down the foods you eat, release their nutrients, and absorb those nutrients into the body. Although the small intestine is the workhorse of the system, where the majority of digestion occurs, and where most of the released nutrients are absorbed into the blood or lymph, each of the digestive system organs makes a vital contribution to this process (Figure \(1\)).
Digestive System Organs
The easiest way to understand the digestive system is to divide its organs into two main categories. The first group is the organs that make up the alimentary canal. Accessory digestive organs comprise the second group and are critical for orchestrating the breakdown of food and the assimilation of its nutrients into the body.
Accessory digestive organs, despite their name, are critical to the function of the digestive system.
20.02: Alimentary Canal Organs
Also called the gastrointestinal (GI) tract or gut, the alimentary canal (aliment- = “to nourish”) is a one-way tube about 7.62 meters (25 feet) in length during life and closer to 10.67 meters (35 feet) in length when measured after death, once smooth muscle tone is lost. The main function of the organs of the alimentary canal is to nourish the body. This tube begins at the mouth and terminates at the anus. Between those two points, the canal is modified as the pharynx, esophagus, stomach, and small and large intestines to fit the functional needs of the body. Both the mouth and anus are open to the external environment; thus, food and wastes within the alimentary canal are technically considered to be outside the body. Only through the process of absorption do the nutrients in food enter into and nourish the body’s “inner space.”
Accessory Structures
Each accessory digestive organ aids in the breakdown of food (Figure \(1\)). Within the mouth, the teeth and tongue begin mechanical digestion, whereas the salivary glands begin chemical digestion. Once food products enter the small intestine, the gallbladder, liver, and pancreas release secretions—such as bile and enzymes— essential for digestion to continue. Together, these are called accessory organs because they sprout from the lining cells of the developing gut (mucosa) and augment its function; indeed, you could not live without their vital contributions, and many significant diseases result from their malfunction. Even after development is complete, they maintain a connection to the gut by way of ducts.
The Mouth
The cheeks, tongue, and palate frame the mouth, which is also called the oral cavity (or buccal cavity). The structures of the mouth are illustrated in Figure \(2\).
The Pharynx
The pharynx (throat) is involved in both digestion and respiration. It receives food and air from the mouth, and air from the nasal cavities. When food enters the pharynx, involuntary muscle contractions close off the air passageways.
The Esophagus
The esophagus is a muscular tube that connects the pharynx to the stomach. It is approximately 25.4 cm (10 in) in length, located posterior to the trachea, and remains in a collapsed form when not engaged in swallowing. As you can see in Figure \(4\), the esophagus runs a mainly straight route through the mediastinum of the thorax. To enter the abdomen, the esophagus penetrates the diaphragm through an opening called the esophageal hiatus. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/20%3A__The_Digestive_System/20.01%3A_Overview.txt |
There are four main regions in the stomach: the cardia, fundus, body, and pylorus (Figure \(1\)). The cardia (or cardiac region) is the point where the esophagus connects to the stomach and through which food passes into the stomach. Located inferior to the diaphragm, above and to the left of the cardia, is the dome-shaped fundus.
Below the fundus is the body, the main part of the stomach. The funnel-shaped pylorus connects the stomach to the duodenum. The wider end of the funnel, the pyloric antrum, connects to the body of the stomach. The narrower end is called the pyloric canal, which connects to the duodenum. The smooth muscle pyloric sphincter is located at this latter point of connection and controls stomach emptying. In the absence of food, the stomach deflates inward, and its mucosa and submucosa fall into a large fold called a ruga.
Histology
The wall of the stomach is made of the same four layers as most of the rest of the alimentary canal, but with adaptations to the mucosa and muscularis for the unique functions of this organ. In addition to the typical circular and longitudinal smooth muscle layers, the muscularis has an inner oblique smooth muscle layer (Figure 20.7). As a result, in addition to moving food through the canal, the stomach can vigorously churn food, mechanically breaking it down into smaller particles.
20.04: Small Intestine
The coiled tube of the small intestine is subdivided into three regions. From proximal (at the stomach) to distal, these are the duodenum, jejunum, and ileum (Figure \(1\)).
Histology
The wall of the small intestine is composed of the same four layers typically present in the alimentary system. However, three features of the mucosa and submucosa are unique. These features, which increase the absorptive surface area of the small intestine more than 600-fold, include circular folds, villi, and microvilli (Figure 20.9). These adaptations are most abundant in the proximal two-thirds of the small intestine, where the majority of absorption occurs.
20.05: Large Intestine
The large intestine runs from the appendix to the anus. It frames the small intestine on three sides. Despite its being about one-half as long as the small intestine, it is called large because it is more than twice the diameter of the small intestine, about 3 inches.
Three features are unique to the large intestine: teniae coli, haustra, and epiploic appendages (Figure \(2\)).
Histology
There are several notable differences between the walls of the large and small intestines (Figure \(3\)). For example, few enzyme-secreting cells are found in the wall of the large intestine, and there are no circular folds or villi. Other than in the anal canal, the mucosa of the colon is simple columnar epithelium made mostly of enterocytes (absorptive cells) and goblet cells. In addition, the wall of the large intestine has far more intestinal glands, which contain a vast population of enterocytes and goblet cells. These goblet cells secrete mucus that eases the movement of feces and protects the intestine from the effects of the acids and gases produced by enteric bacteria. The enterocytes absorb water and salts as well as vitamins produced by your intestinal bacteria.
* Enteric neuron. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/20%3A__The_Digestive_System/20.03%3A_Stomach.txt |
LAB 20 EXERCISES \(1\)
Label the parts of a hollow organ:
1 Mucosa
2 Submucosa
3 Muscularis externa (circular and longitudinal layers) Musculari
4 Serosa
5 Muscularis mucosa
6 Peyer’s patch
Label the parts of a hollow organ:
1 Villus
2 Lacteal
3 Goblet cell
4 Intestinal crypt
5 Intestinal gland
6 Enteric neuron
LAB 20 EXERCISES \(2\)
Oral Cavity
Label the components of the oral cavity
1 Uvula
2 Gingiva
3 Oral vestibule
4 Hard palate
5 * Soft palate
Label the parts of a tooth
1 Enamel
2 Dentin
3 Cementum
4 Pulp cavity
5 Apical foramen
6 Root canal
7 Crown
8 Root
LAB 20 EXERCISES \(3\)
Oral Cavity and Stomach
Label the 3 major salivary glands
Label the parts of the stomach
1 Fundus 6 Pyloric sphincter
2 Body 7 Cardiac sphincter
3 Cardiac region 8 Lesser curvature
4 Pyloric canal 9 Greater curvature
5 Pyloric antrum
LAB 20 EXERCISES \(4\)
Abdominal Cavity
Label the 4 types of teeth: in one quadrant
Label the anatomy of the liver
1 Right lobe
2 Left lobe
3 Gall bladder
4 Hepatic portal vein
5 Hepatic artery
6 Inferior vena cava
7 Falciform ligament
LAB 20 EXERCISES \(5\)
Bile Pancreas and Intestines
Label the path of bile and pancreatic secretions
1 L. Hepatic duct 7 * Accessory pancreatic duct
2 R. Hepatic duct 8 Hepato-pancreatic ampulla
3 Cystic duct 9 Pancreas
4 * Common Bile duct 10 Duodenum
5 Common Hepatic duct 11 Gall bladder
6 Pancreatic duct
Label the following
1 Stomach
2 Duodenum
3 Jejunum
4 * Ileum
5 Colon
LAB 20 EXERCISES \(6\)
Colon and Rectium
Label the parts of the colon
1 Cecum 7 Mesocolon
2 Ascending colon 8 Hepatic flexure
3 Transverse colon 9 Splenic flexure
4 Descending colon 10 Haustra
5 Sigmoid colon 11 Teniae Coli
6 Appendix
Label the following
1 Rectum
2 Anus
3 Rectal valve
4 External Anal sphincter
5 Anal columns
LAB 20 EXERCISES \(7\)
Histology
Histology of the liver
1 Hepatic triad
2 Central vein
3 Hepatic artery branch
4 Hepatic Portal vein branch
5 Bile ductule
6 Hepatocyte
7 * Sinusoid
Histology of the liver
1 Hepatic triad
2 Central vein
3 Hepatocyte
4 Sinusoid
Identify this tissue and its relevant components: (see lab list)
1 Tunica interna
2 Tunica media
3 Tunica externa
4 Valve
5 Vasa vasorum
6 Artery
7 Vein.
Label the parts of this gland: (see lab list)
1 Tunica interna
2 Tunica media
3 Tunica externa
4 Valve
5 Vasa vasorum
6 Artery
7 Vein.
20.07: MODELS- Hollow Organs Teeth Live Gall Bladder Upright Torsos Mid-Sagittal Head
• Oral vestibule
• Philtrum
• Labia
• Oral cavity
• Hard & Soft palate
• Uvula
• Pharynx (oro- and laryngo-)
• Esophagus
• Stomach
• Cardiac sphincter
• Cardiac region
• Fundus
• Body
• Pyloric antrum
• Pyloric canal
• Pyloric sphincter
• Rugae
• Lesser & greater curvature
• Small intestine
• Duodenum
• Jejunum
• Ileum
• Ileocecal valve
• Large intestine
• Haustra
• Taeniae coli
• Epiploic appendages
• Cecum
• Vermiform appendix
• Ascending colon
• Hepatic flexure (R )
• Transverse colon
• Splenic flexure (L)
• Descending colon
• Sigmoid colon
• Rectum
• Rectal valves
• Anal canal
• Anal columns
• Pancreas
• Pancreatic duct
• Hepato-pancreatic ampulla (or papilla)
• Accessory pancreatic duct
• Teeth
• Crown
• Root
• Enamel
• Cementum
• Dentin
• Pulp
• Apical foramen
• Dental Formula: 2 *1 *2 *3
• Incisors * canines * pre-molar (bicuspids) * molars (tricuspids)
• Tongue
• Filiform * Fungiform * Circumvallate * Foliate papillae
• Sulcus terminalis
• Median lingual sulcus
• Foramen cecum
• Taste buds
• Minor salivary glands
• Body
• Base
• Salivary glands (torso)
• Parotid (gland & duct)
• Submandibular
• Sublingual
• Liver/ Gall bladder
• Lobes: left * right * caudate * quadrate
• Left/ * Right and Common hepatic ducts
• Hepatic portal vein
• Hepatic vein
• Hepatic arteries
• Cystic duct
• Common bile duct
• Inferior vena cava
• Hollow Organ models:
• Mucosa
• Muscularis mucosa
• Lacteal
• goblet cell
• crypt
• Intestinal gland
• Submucosa
• Lymphatic nodule (Peyer’s patch)
• Lymphatic vessels
• Enteric neurons
• Muscularis Externa:
• Circular *
• Longitudinal
• Adventitia or serosa
• Peritoneal cavity
• Visceral & parietal peritoneum
• Mesentaries
• Greater omentum
• Mesocolon
Histology:
• Liver (also see liver histology model)
• Lobules
• Triad
• Central vein
• Hepatocytes
• Sinusoids
• Kupffer cells (macrophages)
• Esophagus
• Stratified squamous epithelium
• Tongue
• Skeletal muscle
• Taste buds
• Taste pore
• Receptor & support cells
• Intestines
• Brush border (microvilli)
• Simple columnar epithelium
• Goblet cells
• Villi
• Intestinal crypt
• Intestinal glands
• MALT
• Salivary glands
• Acini
• Ducts
• Pancreas
• Acini
• Ducts
• Pancreatic islets (islet of Langerhans) | textbooks/bio/Human_Biology/Human_Anatomy_Lab/20%3A__The_Digestive_System/20.06%3A_Exercises.txt |
Learning Objectives
Gross Anatomy of Urine Transport
• Identify the ureters, urinary bladder, and urethra, as well as their location, structure, and histology
• Compare and contrast male and female urethras
21: The Urinary System
The urethra transports urine from the bladder to the outside of the body for disposal. The urethra is the only urologic organ that shows any significant anatomic difference between males and females; all other urine transport structures are identical (Figure \(1\)). The urethra in both males and females begins inferior and central to the two ureteral openings forming the three points of a triangular-shaped area at the base of the bladder called the trigone (Greek tri- = “triangle” and the root of the word “trigonometry”). The urethra tracks posterior and inferior to the pubic symphysis (see Figure \(1\)). In both males and females, the proximal urethra is lined by transitional epithelium, whereas the terminal portion is a nonkeratinized, stratified squamous epithelium. In the male, pseudostratified columnar epithelium lines the urethra between these two cell types. Voiding is regulated by an involuntary autonomic nervous system-controlled internal urinary sphincter, consisting of smooth muscle and voluntary skeletal muscle that forms the external urinary sphincter below it
21.02: Bladder
The urinary bladder collects urine from both ureters (Figure \(1\)). The bladder lies anterior to the uterus in females, posterior to the pubic bone and anterior to the rectum. During late pregnancy, its capacity is reduced due to compression by the enlarging uterus, resulting in increased frequency of urination. In males, the anatomy is similar, minus the uterus, and with the addition of the prostate inferior to the bladder. The bladder is partially retroperitoneal (outside the peritoneal cavity) with its peritoneal-covered “dome” projecting into the abdomen when the bladder is distended with urine.
The bladder is a highly distensible organ comprised of irregular crisscrossing bands of smooth muscle collectively called the detrusor muscle. The interior surface is made of transitional cellular epithelium that is structurally suited for the large volume fluctuations of the bladder. When empty, it resembles columnar epithelia, but when stretched, it “transitions” (hence the name) to a squamous appearance (see Figure \(1\)). Volumes in adults can range from nearly zero to 500–600 mL.
The detrusor muscle contracts with significant force in the young. The bladder’s strength diminishes with age, but voluntary contractions of abdominal skeletal muscles can increase intra-abdominal pressure to promote more forceful bladder emptying. Such voluntary contraction is also used in forceful defecation and childbirth.
21.03: Ureters
The kidneys and ureters are completely retroperitoneal, and the bladder has a peritoneal covering only over the dome. As urine is formed, it drains into the calyces of the kidney, which merge to form the funnel-shaped renal pelvis in the hilum of each kidney. The renal pelvis narrows to become the ureter of each kidney. As urine passes through the ureter, it does not passively drain into the bladder but rather is propelled by waves of peristalsis. As the ureters enter the pelvis, they sweep laterally, hugging the pelvic walls. As they approach the bladder, they turn medially and pierce the bladder wall obliquely. This is important because it creates an one-way valve (a physiological sphincter rather than an anatomical sphincter) that allows urine into the bladder but prevents reflux of urine from the bladder back into the ureter. Children born lacking this oblique course of the ureter through the bladder wall are susceptible to “vesicoureteral reflux,” which dramatically increases their risk of serious UTI. Pregnancy also increases the likelihood of reflux and UTI.
The ureters are approximately 30 cm long. The inner mucosa is lined with transitional epithelium (Figure \(1\)) and scattered goblet cells that secrete protective mucus. The muscular layer of the ureter consists of longitudinal and circular smooth muscles that create the peristaltic contractions to move the urine into the bladder without the aid of gravity. Finally, a loose adventitial layer composed of collagen and fat anchors the ureters between the parietal peritoneum and the posterior abdominal wall. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/21%3A_The_Urinary_System/21.01%3A_Urethra.txt |
External Anatomy
The left kidney is located at about the T12 to L3 vertebrae, whereas the right is lower due to slight displacement by the liver. Upper portions of the kidneys are somewhat protected by the eleventh and twelfth ribs (Figure \(1\)). Each kidney weighs about 125–175 g in males and 115–155 g in females. They are about 11–14 cm in length, 6 cm wide, and 4 cm thick, and are directly covered by a fibrous capsule composed of dense, irregular connective tissue that helps to hold their shape and protect them. This capsule is covered by a shock-absorbing layer of adipose tissue called the renal fat pad, which in turn is encompassed by a tough renal fascia. The fascia and, to a lesser extent, the overlying peritoneum serve to firmly anchor the kidneys to the posterior abdominal wall in a retroperitoneal position.
Internal Anatomy
A frontal section through the kidney reveals an outer region called the renal cortex and an inner region called the medulla (Figure \(2\)). The renal columns are connective tissue extensions that radiate downward from the cortex through the medulla to separate the most characteristic features of the medulla, the renal pyramids and renal papillae. The papillae are bundles of collecting ducts that transport urine made by nephrons to the calyces of the kidney for excretion. The renal columns also serve to divide the kidney into 6–8 lobes and provide a supportive framework for vessels that enter and exit the cortex. The pyramids and renal columns taken together constitute the kidney lobes.
Nephrons and Vessels
The renal artery first divides into segmental arteries, followed by further branching to form interlobar arteries that pass through the renal columns to reach the cortex (Figure \(3\)). The interlobar arteries, in turn, branch into arcuate arteries, cortical radiate arteries, and then into afferent arterioles. The afferent arterioles service about 1.3 million nephrons in each kidney.
Nephrons are the “functional units” of the kidney; they cleanse the blood and balance the constituents of the circulation. The afferent arterioles form a tuft of high-pressure capillaries about 200 μm in diameter, the glomerulus. The rest of the nephron consists of a continuous sophisticated tubule whose proximal end surrounds the glomerulus in an intimate embrace—this is Bowman’s capsule. The glomerulus and Bowman’s capsule together form the renal corpuscle. As mentioned earlier, these glomerular capillaries filter the blood based on particle size. After passing through the renal corpuscle, the capillaries form a second arteriole, the efferent arteriole (Figure \(4\)). These will next form a capillary network around the more distal portions of the nephron tubule, the peritubular capillaries and vasa recta, before returning to the venous system. As the glomerular filtrate progresses through the nephron, these capillary networks recover most of the solutes and water, and return them to the circulation. Since a capillary bed (the glomerulus) drains into a vessel that in turn forms a second capillary bed, the definition of a portal system is met. This is the only portal system in which an arteriole is found between the first and second capillary beds. (Portal systems also link the hypothalamus to the anterior pituitary, and the blood vessels of the digestive viscera to the liver.) | textbooks/bio/Human_Biology/Human_Anatomy_Lab/21%3A_The_Urinary_System/21.04%3A_The_Kidney.txt |
Nephrons: The Functional Unit
Nephrons take a simple filtrate of the blood and modify it into urine. Many changes take place in the different parts of the nephron before urine is created for disposal. The term forming urine will be used hereafter to describe the filtrate as it is modified into true urine. The principle task of the nephron population is to balance the plasma to homeostatic set points and excrete potential toxins in the urine. They do this by accomplishing three principle functions—filtration, reabsorption, and secretion. They also have additional secondary functions that exert control in three areas: blood pressure (via production of renin), red blood cell production (via the hormone EPO), and calcium absorption (via conversion of calcidiol into calcitriol, the active form of vitamin D).
Renal Corpuscle
As discussed earlier, the renal corpuscle consists of a tuft of capillaries called the glomerulus that is largely surrounded by Bowman’s (glomerular) capsule. The glomerulus is a high-pressure capillary bed between afferent and efferent arterioles. Bowman’s capsule surrounds the glomerulus to form a lumen, and captures and directs this filtrate to the PCT. The outermost part of Bowman’s capsule, the parietal layer, is a simple squamous epithelium. It transitions onto the glomerular capillaries in an intimate embrace to form the visceral layer of the capsule. Here, the cells are not squamous, but uniquely shaped cells (podocytes) extending finger-like arms (pedicels) to cover the glomerular capillaries (Figure \(1\)). These projections interdigitate to form filtration slits, leaving small gaps between the digits to form a sieve. As blood passes through the glomerulus, 10 to 20 percent of the plasma filters between these sieve-like fingers to be captured by Bowman’s capsule and funneled to the PCT. Where the fenestrae (windows) in the glomerular capillaries match the spaces between the podocyte “fingers,” the only thing separating the capillary lumen and the lumen of Bowman’s capsule is their shared basement membrane (Figure \(2\)). These three features comprise what is known as the filtration membrane. This membrane permits very rapid movement of filtrate from capillary to capsule though pores that are only 70 nm in diameter.
21.06: Exercises
LAB 21 EXERCISES \(1\)
Identify the following: L. and R. Kidneys
1 L. and R. Kidneys
2 Adrenal glands
3 Ureters
4 Bladder
Identify all of the blood vessels shown.
1
2
3
4
5
6
LAB 21 EXERCISES \(2\)
Gross Anatomy Juxtaglomellular
Identify the following:
1 Cortex 7 Minor Calyx
2 Medulla 8 Major Calyx
3 Capsule 9 Renal Pelvis
4 Renal Pyramid 10 Ureter
5 * Renal Column 11 Papilla
6 Renal sinus 12 Hilus
Label the following:
1 Juxtaglomerular apparatus
2 JG cells
3 Macula densa
4 Distal convoluted tubule
5 Proximal convoluted tubule
LAB 21 EXERCISES \(3\)
Glomerulusses
Label the following:
1 Afferent arteriole
2 Efferent arteriole
3 Bowman’s capsule
4 Podocyte
5 Fenestrated endothelium
Label the following:
1 Filtration membrane
2 Fenestrated endothelia
3 Podocyte foot process
4 Filtration slit
5 Basement membrane
6 Plasma
7 * Filtrate
LAB 21 EXERCISES \(4\)
Nephron and Bladder
Label the parts of a nephron:
1 Bowman’s capsule 6 Afferent arteriole
2 Proximal convoluted tubule 7 Efferent arteriol
3 Distal convoluted tubule 8 Peritubular capillaries
4 Loop of Henle 9 Glomerulus
5 Collecting duct
Label the following:
1 Detrusor muscle
2 Ureters
3 Ureteral slits
4 Urethra
5 Internal urethral sphincter
6 External urethral sphincter
7 Trigone
8 Rugae.
LAB 21 EXERCISES \(5\)
Histology
Name this organ
Identify 3 tissue types
1
2
3
Name this organ
Identify 3 tissue types
1
2
3
LAB 21 EXERCISES \(6\)
Histology
Label the following:
1 Afferent arteriole
2 Glomerulus
3 Peritubular capillaries
Name this organ
Label the following:
1 Glomerulus
2 Bowman’s capsule
3 Convoluted tubule.
21.07: MODELS- Kidney Nephron Torso and Hemi-Pelvis
• Blood Supply:
• Renal artery
• Segmental
• Interlobar
• Arcuate
• Cortical radiate
• Afferent arteriole
• Glomerular capillaries
• Efferent arteriole
• Peritubular capillaries
• Cortical radiate venules (interlobular)
• Arcuate veins
• Interlobar veins (segmental veins)
• Renal vein
• Nephron: (model)
• Renal corpuscle
• Glomerulus
• Afferent + efferent arteriole
• Peritubular capillaries
• Vasa recta
• Glomerular (Bowman’s) capsule
• Podocytes
• Renal tubules:
• Proximal Convoluted Tubule (PCT)
• Loop of Henle
• Descending & ascending limbs
• Thick & thin segments
• Distal Convoluted Tubule (DCT)
• Collecting Duct
• Papillary duct
• Juxtaglomerular Apparatus (JGA)
• JG cells
• Macula Densa
• Ureters (torso)
• Urinary Bladder
• Trigone (2 ureters * 1 urethra)
• Rugae
• Detrusor muscle
• Urethra
• Kidney anatomy (model)
• Renal capsule
• Hilum
• Nephrons
• Juxtamedullary nephron
• Cortical nephron
• Renal cortex
• Renal medulla:
• Columns
• Renal sinus
• Renal Pyramids
• Papilla
• Minor Calyx (calyces)
• Major Calyx (calyces)
• Pelvis
Histology:
• Kidney:
• Bowman’s capsule
• Glomerulus
• Afferent & efferent arterioles
• simple cuboidal epithelium
• convoluted tubules
• Adipose capsule
• Bladder/Ureters
• Mucosa:
• Transitional epithelium
• Submucosa
• Areolar CT (& other CT)
• Muscularis externa (detrusor muscle)
• Smooth muscle | textbooks/bio/Human_Biology/Human_Anatomy_Lab/21%3A_The_Urinary_System/21.05%3A_Microscopic_Anatomy_of_the_Kidney.txt |
Learning Objectives
Anatomy and Physiology of the Male Reproductive System
• Describe the structure of the organs of the male reproductive system
• Describe the structure of the sperm cell
22: The Reproductive System (Male)
Scrotum
The testes are located in a skin-covered, highly pigmented, muscular sack called the scrotum that extends from the body behind the penis (see Figure \(1\)). This location is important in sperm production, which occurs within the testes, and proceeds more efficiently when the testes are kept 2 to 4°C below core body temperature. The dartos muscle makes up the subcutaneous muscle layer of the scrotum (Figure \(2\)).
22.02: Testes
The testes (singular = testis) are the male gonads—that is, the male reproductive organs. They produce both sperm and androgens, such as testosterone, and are active throughout the reproductive lifespan of the male.
Paired ovals, the testes are each approximately 4 to 5 cm in length and are housed within the scrotum (see this figure). They are surrounded by two distinct layers of protective connective tissue (Figure \(1\): Anatomy of the Testis). The outer tunica vaginalis is a serous membrane that has both a parietal and a thin visceral layer. Beneath the tunica vaginalis is the tunica albuginea, a tough, white, dense connective tissue layer covering the testis itself. The tightly coiled seminiferous tubules form the bulk of each testis.
22.03: Sperm
Sperm are smaller than most cells in the body; in fact, the volume of a sperm cell is 85,000 times less than that of the female gamete. Approximately 100 to 300 million sperm are produced each day, whereas women typically ovulate only one oocyte per month. As is true for most cells in the body, the structure of sperm cells speaks to their function. Sperm have a distinctive head, mid-piece, and tail region (Figure \(1\)). The head of the sperm contains the extremely compact haploid nucleus with very little cytoplasm. These qualities contribute to the overall small size of the sperm (the head is only 5 μm long). A structure called the acrosome covers most of the head of the sperm cell as a “cap” that is filled with lysosomal enzymes important for preparing sperm to participate in fertilization. Tightly packed mitochondria fill the mid- piece of the sperm. ATP produced by these mitochondria will power the flagellum, which extends from the neck and the mid-piece through the tail of the sperm, enabling it to move the entire sperm cell. The central strand of the flagellum, the axial filament, is formed from one centriole inside the maturing sperm cell during the final stages of spermatogenesis.
Sperm Transport
To fertilize an egg, sperm must be moved from the seminiferous tubules in the testes, through the epididymis, and—later during ejaculation—along the length of the penis and out into the female reproductive tract.
Role of the Epididymis
From the lumen of the seminiferous tubules, the immotile sperm are surrounded by testicular fluid and moved to the epididymis (plural = epididymides), a coiled tube attached to the testis where newly formed sperm continue to mature (see Figure \(1\)).
Duct System
During ejaculation, sperm exit the tail of the epididymis and are pushed by smooth muscle contraction to the ductus deferens (also called the vas deferens). The ductus deferens is a thick, muscular tube that is bundled together inside the scrotum with connective tissue, blood vessels, and nerves into a structure called the spermatic cord (see Figure \(1\) and Figure \(2\)).
Seminal Vesicles
As sperm pass through the ampulla of the ductus deferens at ejaculation, they mix with fluid from the associated seminal vesicle (Figure \(2\)).
Prostate Gland
As shown in Figure \(2\), the centrally located prostate gland sits anterior to the rectum at the base of the bladder surrounding the prostatic urethra (the portion of the urethra that runs within the prostate). About the size of a walnut, the prostate is formed of both muscular and glandular tissues.
Bulbourethral Glands
The final addition to semen is made by two bulbourethral glands (or Cowper’s glands) that release a thick, salty fluid that lubricates the end of the urethra and the vagina, and helps to clean urine residues from the penile urethra. The fluid from these accessory glands is released after the male becomes sexually aroused, and shortly before the release of the semen. It is therefore sometimes called pre-ejaculate. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/22%3A_The_Reproductive_System_(Male)/22.01%3A_Scrotum.txt |
The penis is the male organ of copulation (sexual intercourse). It is flaccid for non-sexual actions, such as urination, and turgid and rod-like with sexual arousal. When erect, the stiffness of the organ allows it to penetrate into the vagina and deposit semen into the female reproductive tract.
The shaft of the penis surrounds the urethra (Figure \(1\)). The shaft is composed of three column-like chambers of erectile tissue that span the length of the shaft. Each of the two larger lateral chambers is called a corpus cavernosum (plural = corpora cavernosa). Together, these make up the bulk of the penis. The corpus spongiosum, which can be felt as a raised ridge on the erect penis, is a smaller chamber that surrounds the spongy, or penile, urethra. The end of the penis, called the glans penis, has a high concentration of nerve endings, resulting in very sensitive skin that influences the likelihood of ejaculation (see ). The skin from the shaft extends down over the glans and forms a collar called the prepuce (or foreskin). The foreskin also contains a dense concentration of nerve endings, and both lubricate and protect the sensitive skin of the glans penis. A surgical procedure called circumcision, often performed for religious or social reasons, removes the prepuce, typically within days of birth.
22.05: Exercises
LAB 23 EXERCISES \(1\)
Label the following:
1 Deep artery
2 Dorsal vein
3 Urethra
4 Corpora cavernosa
5 Copora spongiosa
6 Median septum of the penis
7 Skin
Label the following:
1 Spermatic cord
2 Epididymis
3 Seminiferous tubule
4 Tunica albuginea
5 Tunica vaginalis
6 Rete testis
7 Vas deferens
LAB 23 EXERCISES \(2\)
Label the following:
1 Fascia of spermatic cord
2 Glans
3 Cremaster muscle
4 Pampiniform plexus
5 Tunica vaginalis
6 Ureter
LAB 22 EXERCISES \(3\)
Label the following:
1 Testis
2 Epididymis
3 R. vas deferensi
4 L. vas deferens
5 Ampulla (of vas deferens)
6 Seminal vesicle
7 Prostate
8 Urinary bladder
LAB 22 EXERCISES \(4\)
Label the following:
1 Rectum 6 Spongy urethra
2 Prostate 7 Shaft of penis
3 Ejaculatory duct 8 Root of penis
4 Prostatic urethra 9 Urinary bladder
5 Membranous urethra 10 Urogenital diaphragm
LAB 22 EXERCISES \(5\)
Histology
Label the following:
1 Head
2 Acrosome
3 Midpiece
4 Tail
1) Name this organ sample
2) Label the following:
1 Spermatogonia
2 Spermatozoa
3 Interstitial cells
4 Nurse cells
5 Lumen
LAB 22 EXERCISES \(1\)
1) Name this organ sample
2) Label the following:
1 Lumen
2 Mucosa
3 Submucosa
4 Muscularis externa
LAB 22 EXERCISES \(1\)
Label the following:
1 Mucosa
2 Submucosa
3 Muscularis externa (circular and longitudinal layers) Musculari
4 Serosa
5 Muscularis mucosa
6 Peyer’s patch
22.06: MODELS- Male Hemi-Pelvis and Torso
• Scrotum
• Dartos muscle
• Raphe
• Testes:
• Tunica albuginea
• Tunica vaginalis
• Seminiferous tubules
• Rete testis
• Epididymis
• Inguinal Canal
• Spermatic cord
• Testicular arteries
• Pampiniform plexus (blood vessels)
• Vas deferens (ductus deferens)
• Cremasteric muscle
• Ejaculatory duct
• Urethra:
• Prostatic * membranous * spongy
• External urethral orifice
• Penis
• Shaft
• Root
• Crus (Crura)
• Corpora cavernosa (x2)
• Corpora spongiosa
• Glans penis
• Dorsal artery/vein Deep artery
• Accessory glands:
• Seminal Vesicles
• Prostate
• Bulbourethral (Cowpers) Glands
• Rectum
• Bladder
• Urogenital diaphragm
• Pubis Symphisis
Histology:
• Penis:
• corpora cavernosa
• corpora spongiosa
• Erectile tissue
• Urethra
• Central artery
• Deep dorsal vein
• Testis
• tunica albuginea
• tunica vaginalis
• seminiferous tubules
• lumen *
• interstitial cells
• spermatids
• spermatogonia
• Sperm
• Acrosome
• head
• midpiece
• tail/flagellum
• Vas deferens & urethra
• Mucosa
• Submucosa
• Muscularis externa
• Prostate gland
• Glandular epithelium
• concretion | textbooks/bio/Human_Biology/Human_Anatomy_Lab/22%3A_The_Reproductive_System_(Male)/22.04%3A_Penis.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe the structure of the organs of the female reproductive system
• Trace the path of an oocyte from ovary to fertilization
23: The Reproductive System (Female)
(CC-BY-4.0, OpenStax, Human Anatomy)
The female reproductive system functions to produce gametes and reproductive hormones, just like the male reproductive system; however, it also has the additional task of supporting the developing fetus and delivering it to the outside world. Unlike its male counterpart, the female reproductive system is located primarily inside the pelvic cavity (Figure \(1\)). Recall that the ovaries are the female gonads. The gamete they produce is called an oocyte.
External Female Genitals
The external female reproductive structures are referred to collectively as the vulva (Figure \(2\)). The mons pubis is a pad of fat that is located at the anterior, over the pubic bone. After puberty, it becomes covered in pubic hair. The labia majora (labia = “lips”; majora = “larger”) are folds of hair-covered skin that begin just
posterior to the mons pubis. The thinner and more pigmented labia minora (labia = “lips”; minora = “smaller”) extend medial to the labia majora. Although they naturally vary in shape and size from woman to woman, the labia minora serve to protect the female urethra and the entrance to the female reproductive tract.
The superior, anterior portions of the labia minora come together to encircle the clitoris (or glans clitoris), an organ that originates from the same cells as the glans penis and has abundant nerves that make it important in sexual sensation and orgasm. The hymen is a thin membrane that sometimes partially covers the entrance to the vagina. An intact hymen cannot be used as an indication of “virginity”; even at birth, this is only a partial membrane, as menstrual fluid and other secretions must be able to exit the body, regardless of penile–vaginal intercourse. The vaginal opening is located between the opening of the urethra and the anus. It is flanked by outlets to the Bartholin’s glands (or greater vestibular glands).
Vagina
The vagina, shown at the bottom of \(1\) and \(2\), is a muscular canal (approximately 10 cm long) that serves as the entrance to the reproductive tract. It also serves as the exit from the uterus during menses and childbirth. The outer walls of the anterior and posterior vagina are formed into longitudinal columns, or ridges, and the superior portion of the vagina—called the fornix—meets the protruding uterine cervix. The walls of the vagina are lined with an outer, fibrous adventitia; a middle layer of smooth muscle; and an inner mucous membrane with transverse folds called rugae. Together, the middle and inner layers allow the expansion of the vagina to accommodate intercourse and childbirth. The thin, perforated hymen can partially surround the opening to the vaginal orifice. The hymen can be ruptured with strenuous physical exercise, penile–vaginal intercourse, and childbirth. The Bartholin’s glands and the lesser vestibular glands (located near the clitoris) secrete mucus, which keeps the vestibular area moist.
The vagina is home to a normal population of microorganisms that help to protect against infection by pathogenic bacteria, yeast, or other organisms that can enter the vagina. In a healthy woman, the most predominant type of vaginal bacteria is from the genus Lactobacillus. This family of beneficial bacterial flora secretes lactic acid, and thus protects the vagina by maintaining an acidic pH (below 4.5). Potential pathogens are less likely to survive in these acidic conditions. Lactic acid, in combination with other vaginal secretions, makes the vagina a self- cleansing organ. However, douching—or washing out the vagina with fluid—can disrupt the normal balance of healthy microorganisms, and actually increase a woman’s risk for infections and irritation. Indeed, the American College of Obstetricians and Gynecologists recommend that women do not douche, and that they allow the vagina to maintain its normal healthy population of protective microbial flora. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/23%3A_The_Reproductive_System_(Female)/23.01%3A_External_Female_Genitals.txt |
The ovaries are the female gonads (see Figure \(1\)). Paired ovals, they are each about 2 to 3 cm in length, about the size of an almond. The ovaries are located within the pelvic cavity, and are supported by the mesovarium, an extension of the peritoneum that connects the ovaries to the broad ligament. Extending from the mesovarium itself is the suspensory ligament that contains the ovarian blood and lymph vessels. Finally, the ovary itself is attached to the uterus via the ovarian ligament.
The ovary comprises an outer covering of cuboidal epithelium called the ovarian surface epithelium that is superficial to a dense connective tissue covering called the tunica albuginea. Beneath the tunica albuginea is the cortex, or outer portion, of the organ. The cortex is composed of a tissue framework called the ovarian stroma that forms the bulk of the adult ovary. Oocytes develop within the outer layer of this stroma, each surrounded by supporting cells. This grouping of an oocyte and its supporting cells is called a follicle. The growth and development of ovarian follicles will be described shortly. Beneath the cortex lies the inner ovarian medulla, the site of blood vessels, lymph vessels, and the nerves of the ovary. You will learn more about the overall anatomy of the female reproductive system at the end of this section.
23.03: Uterine Tubes
The uterine tubes (also called fallopian tubes or oviducts) serve as the conduit of the oocyte from the ovary to the uterus (Figure \(1\)). Each of the two uterine tubes is close to, but not directly connected to, the ovary and divided into sections. The isthmus is the narrow medial end of each uterine tube that is connected to the uterus. The wide distal infundibulum flares out with slender, finger-like projections called fimbriae. The middle region of the tube, called the ampulla, is where fertilization often occurs. The uterine tubes also have three layers: an outer serosa, a middle smooth muscle layer, and an inner mucosal layer. In addition to its mucus-secreting cells, the inner mucosa contains ciliated cells that beat in the direction of the uterus, producing a current that will be critical to move the oocyte.
23.04: Uterus and Cervix
The uterus is the muscular organ that nourishes and supports the growing embryo (see Figure \(1\)). Its average size is approximately 5 cm wide by 7 cm long (approximately 2 in by 3 in) when a female is not pregnant. It has three sections. The portion of the uterus superior to the opening of the uterine tubes is called the fundus. The middle section of the uterus is called the body of uterus (or corpus). The cervix is the narrow inferior portion of the uterus that projects into the vagina. The cervix produces mucus secretions that become thin and stringy under the influence of high systemic plasma estrogen concentrations, and these secretions can facilitate sperm movement through the reproductive tract.
The Breasts
Whereas the breasts are located far from the other female reproductive organs, they are considered accessory organs of the female reproductive system. The function of the breasts is to supply milk to an infant in a process called lactation. The external features of the breast include a nipple surrounded by a pigmented areola (Figure 23.4), whose coloration may deepen during pregnancy. The areola is typically circular and can vary in size from 25 to 100 mm in diameter. The areolar region is characterized by small, raised areolar glands that secrete lubricating fluid during lactation to protect the nipple from chafing. When a baby nurses, or draws milk from the breast, the entire areolar region is taken into the mouth.
Breast milk is produced by the mammary glands, which are modified sweat glands. The milk itself exits the breast through the nipple via 15 to 20 lactiferous ducts that open on the surface of the nipple. These lactiferous ducts each extend to a lactiferous sinus that connects to a glandular lobe within the breast itself that contains groups of milk-secreting cells in clusters called alveoli (see Figure 23.4). The clusters can change in size depending on the amount of milk in the alveolar lumen. Once milk is made in the alveoli, stimulated myoepithelial cells that surround the alveoli contract to push the milk to the lactiferous sinuses. From here, the baby can draw milk through the lactiferous ducts by suckling. The lobes themselves are surrounded by fat tissue, which determines the size of the breast; breast size differs between individuals and does not affect the amount of milk produced. Supporting the breasts are multiple bands of connective tissue called suspensory ligaments that connect the breast tissue to the dermis of the overlying skin. | textbooks/bio/Human_Biology/Human_Anatomy_Lab/23%3A_The_Reproductive_System_(Female)/23.02%3A_Ovaries.txt |
Whereas the breasts are located far from the other female reproductive organs, they are considered accessory organs of the female reproductive system. The function of the breasts is to supply milk to an infant in a process called lactation. The external features of the breast include a nipple surrounded by a pigmented areola ( Figure \(1\) ), whose coloration may deepen during pregnancy. The areola is typically circular and can vary in size from 25 to 100 mm in diameter. The areolar region is characterized by small, raised areolar glands that secrete lubricating fluid during lactation to protect the nipple from chafing. When a baby nurses, or draws milk from the breast, the entire areolar region is taken into the mouth.
Breast milk is produced by the mammary glands, which are modified sweat glands. The milk itself exits the breast through the nipple via 15 to 20 lactiferous ducts that open on the surface of the nipple. These lactiferous ducts each extend to a lactiferous sinus that connects to a glandular lobe within the breast itself that contains groups of milk-secreting cells in clusters called alveoli (see Figure \(1\)). The clusters can change in size depending on the amount of milk in the alveolar lumen. Once milk is made in the alveoli, stimulated myoepithelial cells that surround the alveoli contract to push the milk to the lactiferous sinuses. From here, the baby can draw milk through the lactiferous ducts by suckling. The lobes themselves are surrounded by fat tissue, which determines the size of the breast; breast size differs between individuals and does not affect the amount of milk produced. Supporting the breasts are multiple bands of connective tissue called suspensory ligaments that connect the breast tissue to the dermis of the overlying skin.
23.06: Exercises
LAB 23 EXERCISES \(1\)
Label the following:
1 Tunica albuginea 7 Secondary follicle
2 Cortex 8 Vesicular follicle
3 Medulla 9 Ovulated oocyte
4 Ovarian ligament 10 Corpus luteum
5 Primordial follicles 11 Corpus albicans
6 Primary follicle
LAB 23 EXERCISES \(2\)
Label the following:
1 Body 8 Ovary
2 Fundus 9 Ovarian ligament
3 Cervix 10 Endometrium
4 Fornix 11 Myometrium
5 Fallopian tube 12 Perimetrium
6 Ampulla 13 Internal os
7 Fimbriae 14 External os
LAB 23 EXERCISES \(3\)
Label the following:
1 Fallopian tube 7 Clitoris
2 Ovary 8 Labia minora
3 Recto-uterine pouchi 9 Labia Majora
4 Rectum 10 Uterus
5 Urinary bladder 11 Vagina
6 Urethra 12 Pubis Symphysis
LAB 23 EXERCISES \(4\)
Label the following:
1 Urogenital diaphragm
2 Round ligament
3 Infundibulum (of the fallopian tube)
Label the following:
1 Uterus
2 Fallopian tube
3 Round ligamenti
4 Ureter
LAB 23 EXERCISES \(5\)
Label the structures:
1 Glans Clitoris
2 Greater vestibular gland
3 Skene’s (para-urethral) glands
4 Urethral orifice
Label the following:
1 FSH 6 Luteal phase
2 LH 7 Menstrual phase
3 Estrogen (E2 ) 8 Proliferative phase
4 Progesterone (P) 9 Secretory phase
5 Follicular phase 10 Ovulation
LAB 23 EXERCISES \(6\)
Label the following:
1 Lactiferous ducts
2 Suspensory ligaments
3 Lactiferous sinus
4 Pectoralis minor
5 Pectoralis major
6 Intercostal muscles
7 Fascia
8 Fascia
23.07: MODELS- Female Hemi-Pelvises Torso Breast and Ovary
• Fallopian Tubes (uterine tubes)
• Ampulla
• Fimbriae
• Infundibulum
• Isthmus
• Uterus
• Fundus
• Body
• Cervix
• Cervical os (internal & external)
• Endometrium
• Myometrium
• Perimetrium
• Round ligament
• Vagina
• Rugae
• Fornix
• Vulva
• Mons pubis
• Vestibule
• Urethral orifice
• Labia minora
• Labia majora
• Clitoris
• Glans clitoris
• Clitoral hood
• Ovary (model)
• Primordial * Primary * Secondary * Vesicular/Graafian follicles
• Corpus luteum
• Corpus albicans
• Ovarian ligament
• Suspensory ligament (of the ovary)
• Recto-vaginal (recto-uterine) pouch
• Breast (Mammary Glands)
• Areola
• Nipple
• Lobe
• Lactiferous duct
• Lactiferous sinus
• Adipose tissue
• Pectoralis major muscle
• Suspensory (Cooper’s) ligaments
Histology:
• Ovary
• Follicles: graafian (secretory) * secondary * primary.
• corpus luteum
• corpus albicans
• oocyte
• follicular cells
• antrum
• Uterus
• Endometrium & myometrium
• Fallopian tubes
• Mucosa * Muscularis * Lumen | textbooks/bio/Human_Biology/Human_Anatomy_Lab/23%3A_The_Reproductive_System_(Female)/23.05%3A_Breasts.txt |
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