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The simple Mk model provides a useful foundation for a number of innovative methods. These methods capture evolutionary processes that are more complicated than the original model, including models that vary through time or across clades. Modeling more than one discrete character at a time allows us to test for the correlated evolution of discrete characters.
Taken as a whole, chapters 7 through 9 provide a basis for the analysis of discrete characters on trees. One can test a variety of biologically relevant hypotheses about how these characters have changed along the branches of the tree of life.
References
Beaulieu, J. M., B. C. O’Meara, and M. J. Donoghue. 2013. Identifying hidden rate changes in the evolution of a binary morphological character: The evolution of plant habit in campanulid angiosperms. Syst. Biol. 62:725–737.
Felsenstein, J. 2012. A comparative method for both discrete and continuous characters using the threshold model. Am. Nat. 179:145–156.
Felsenstein, J. 1985. Phylogenies and the comparative method. Am. Nat. 125:1–15.
Felsenstein, J. 2005. Using the quantitative genetic threshold model for inferences between and within species. Philos. Trans. R. Soc. Lond. B Biol. Sci. 360:1427–1434.
FitzJohn, R. G. 2012. Diversitree: Comparative phylogenetic analyses of diversification in R. Methods Ecol. Evol. 3:1084–1092.
Fukuyama, K. 1991. Spawning behaviour and male mating tactics of a foam-nesting treefrog, Rhacophorus schlegelii. Anim. Behav. 42:193–199. Elsevier.
Gomez-Mestre, I., R. A. Pyron, and J. J. Wiens. 2012. Phylogenetic analyses reveal unexpected patterns in the evolution of reproductive modes in frogs. Evolution 66:3687–3700.
Hennig, W. 1966. Phylogenetic systematics. University of Illinois Press.
Huelsenbeck, J. P., B. Larget, and M. E. Alfaro. 2004. Bayesian phylogenetic model selection using reversible jump markov chain monte carlo. Mol. Biol. Evol. 21:1123–1133.
Liò, P., and N. Goldman. 1998. Models of molecular evolution and phylogeny. Genome Res. 8:1233–1244.
Maddison, W. P., P. E. Midford, S. P. Otto, and T. Oakley. 2007. Estimating a binary character’s effect on speciation and extinction. Syst. Biol. 56:701–710. Oxford University Press.
Marazzi, B., C. Ané, M. F. Simon, A. Delgado-Salinas, M. Luckow, and M. J. Sanderson. 2012. Locating evolutionary precursors on a phylogenetic tree. Evolution 66:3918–3930.
Marquez, R., and P. Verrell. 1991. The courtship and mating of the Iberian midwife toad Alytes cisternasii (Amphibia: Anura: Discoglossidae). J. Zool. 225:125–139. Blackwell Publishing Ltd.
O’Meara, B. C., C. Ané, M. J. Sanderson, and P. C. Wainwright. 2006. Testing for different rates of continuous trait evolution using likelihood. Evolution 60:922–933.
Pagel, M. 1999a. Inferring the historical patterns of biological evolution. Nature 401:877–884.
Pagel, M. 1999b. The maximum likelihood approach to reconstructing ancestral character states of discrete characters on phylogenies. Syst. Biol. 48:612–622.
Pagel, M., and A. Meade. 2006. Bayesian analysis of correlated evolution of discrete characters by reversible-jump markov chain monte carlo. Am. Nat. 167:808–825.
Revell, L. J. 2014. Ancestral character estimation under the threshold model from quantitative genetics. Evolution 68:743–759.
Rey, H. A. 2007. Curious George: Tadpole trouble. Houghton Mifflin Harcourt.
Steel, M., and D. Penny. 2000. Parsimony, likelihood, and the role of models in molecular phylogenetics. Mol. Biol. Evol. 17:839–850.
Thomas, G. H., R. P. Freckleton, and T. Székely. 2006. Comparative analyses of the influence of developmental mode on phenotypic diversification rates in shorebirds. Proc. Biol. Sci. 273:1619–1624.
Tuffley, C., and M. Steel. 1997. Links between maximum likelihood and maximum parsimony under a simple model of site substitution. Bull. Math. Biol. 59:581–607.
Tyler, M. J., and D. B. Carter. 1981. Oral birth of the young of the gastric brooding frog rheobatrachus silus. Anim. Behav. 29:280–282. Elsevier.
Uyeda, J. C., L. J. Harmon, and C. E. Blank. 2016. A comprehensive study of cyanobacterial morphological and ecological evolutionary dynamics through deep geologic time. PLoS One 11:e0162539. Public Library of Science.
Yang, Z. 2006. Computational molecular evolution. Oxford University Press.
Zamudio, K. R., R. C. Bell, R. C. Nali, C. F. B. Haddad, and C. P. A. Prado. 2016. Polyandry, predation, and the evolution of frog reproductive modes. Am. Nat. 188 Suppl 1:S41–61. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/09%3A_Beyond_the_Mk_Model/9.0S%3A_9.S%3A_Beyond_the_Mk_Model_%28Summary%29.txt |
This chapter introduces birth-death models and summarized their basic mathematical properties. Birth-death models predict patterns of species diversity over time intervals, and can also be used to model the growth of phylogenetic trees. We can visualize these patterns by measuring tree balance and creating lineage-through-time (LTT) plots.
• 10.1: Plant Diversity Imbalance
The diversity of flowering plants (the angiosperms) dwarfs the number of species of their closest evolutionary relatives. There are more than 260,000 species of angiosperms . The clade originated more than 140 million years ago, and all of these species have formed since then. One can contrast the diversity of angiosperms with the diversity of other groups that originated at around the same time. For example, gymnosperms, which are as old as angiosperms, include only around 1000 species.
• 10.2: The Birth-Death Model
A birth-death model is a continuous-time Markov process that is often used to study how the number of individuals in a population change through time. For macroevolution, these “individuals” are usually species, sometimes called "lineages" in the literature. In a birth-death model, two things can occur: births, where the number of individuals increases by one; and deaths, where the number of individuals decreases by one.
• 10.3: Birth-death models and phylogenetic trees
The main complication in phylogenetic studies of birth-death models is that we get a “censored” view of the process, in that we only observe lineages that survive to the present day.
• 10.4: Simulating Birth-Death Trees
We can use the statistical properties of birth-death models to simulate phylogenetic trees through time. We could begin with a single lineage at time 0. However, phylogenetic tree often start with the first speciation event in the clade, so one can also begin the simulation with two lineages at time 0 (this difference relates to the distinction between crown and stem ages of clades.
• 10.5: Tree topology, tree shape, and tree balance under a birth-death model
Tree topology summarizes the patterns of evolutionary relatedness among a group of species independent of the branch lengths of a phylogenetic tree. Two different trees have the same topology if they define the exact same set of clades. This is important because sometimes two trees can look very different and yet still have the same topology.
• 10.6: Lineage-through-Time Plots
The other main way to quantify phylogenetic tree shape is by making lineage-through-time plots. These plots have time along the x axis (from the root of the tree to the present day), and the reconstructed number of lineages on the y-axis. Since we are usually considering birth-death models, where the number of lineages is expected to grow (or shrink) exponentially through time, then it is typical practice to log-transform the y-axis.
• 10.S: Introduction to birth-death models (Summary)
10: Introduction to Birth-Death Models
The diversity of flowering plants (the angiosperms) dwarfs the number of species of their closest evolutionary relatives (Figure 10.1). There are more than 260,000 species of angiosperms (that we know; more are added every day). The clade originated more than 140 million years ago (Bell et al. 2005), and all of these species have formed since then. One can contrast the diversity of angiosperms with the diversity of other groups that originated at around the same time. For example, gymnosperms, which are as old as angiosperms, include only around 1000 species, and may even represent more than one clade. The diversity of angiosperms also dwarfs the diversity of familiar vertebrate groups of similar age (e.g. squamates - snakes and lizards - which diverged from their sister taxon, the tuatara, some 250 mya or more Hedges et al. 2006, include fewer than 8000 species).
The evolutionary rise of angiosperm diversity puzzled Darwin over his career, and the issues surrounding angiosperm diversification are often referred to as “Darwin’s abominable mystery” in the scientific literature (e.g. Davies et al. 2004). The main mystery is the tremendous variation in numbers of species across plant clades (see Figure 10.1). This variation even applies within angiosperms, where some clades are much more diverse than others.
At a global scale, the number of species in a clade can change only via two processes: speciation and extinction. This means that we must look to speciation and extinction rates – and how they vary through time and across clades – to explain phenomena like the extraordinary diversity of Angiosperms. It is to this topic that we turn in the next few chapters. Since Darwin’s time, we have learned a lot about the evolutionary processes that led to the diversity of angiosperms that we see today. These data provide an incredible window into the causes and effects of speciation and extinction over macroevolutionary time scales.
Comparative methods can be applied to understand patterns of species richness by estimating speciation and extinction rates, both across clades and through time. In this chapter, I will introduce birth-death models, by far the most common model for understanding diversification in a comparative framework. I will discuss the mathematics of birth-death models and how these models relate to the shapes of phylogenetic trees. I will describe how to simulate phylogenetic trees under a birth-death model. Finally, I will discuss tree balance and lineage-through-time plots, two common ways to measure the shapes of phylogenetic trees. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/10%3A_Introduction_to_Birth-Death_Models/10.01%3A_Plant_Diversity_Imbalance.txt |
A birth-death model is a continuous-time Markov process that is often used to study how the number of individuals in a population change through time. For macroevolution, these “individuals” are usually species, sometimes called "lineages" in the literature. In a birth-death model, two things can occur: births, where the number of individuals increases by one; and deaths, where the number of individuals decreases by one. We assume that no more than one new individual can form (or die) during any one event. In phylogenetic terms, that means that birth-death trees cannot have “hard polytomies” - each speciation event results in exactly two descendant species.
In macroevolution, we apply the birth-death model to species, and typically consider a model where each species has a constant probability of either giving birth (speciating) or dying (going extinct). We denote the per-lineage birth rate as λ and the per-lineage death rate as μ. For now we consider these rates to be constant, but we will relax that assumption later in the book.
We can understand the behavior of birth-death models if we consider the waiting time between successive speciation and extinction events in the tree. Imagine that we are considering a single lineage that exists at time t0. We can think about the waiting time to the next event, which will either be a speciation event splitting that lineage into two (Figure 10.2A) or an extinction event marking the end of that lineage (Figure 10.2B). Under a birth-death model, both of these events follow a Poisson process, so that the expected waiting time to an event follows an exponential distribution (Figure 10.2C). The expected waiting time to the next speciation event is exponential with parameter λ, and the expected waiting time to the next extinction event exponential with parameter μ. [Of course, only one of these can be the next event. The expected waiting time to the next event (of any sort) is exponential with parameter λ + μ, and the probability that that event is speciation is λ/(μ + λ), extinction μ/(μ + λ)].
When we have more than one lineage “alive” in the tree at any time point, then the waiting time to the next event changes, although its distribution is still exponential. In general, if there are N(t) lineages alive at time t, then the waiting time to the next event follows an exponential distribution with parameter N(t)(λ + μ), with the probability that that event is speciation or extinction the same as given above. You can see from this equation that the rate parameter of the exponential distribution gets larger as the number of lineages increases. This means that the expected waiting times across all lineages get shorter and shorter as more lineages accumulate.
Using this approach, we can grow phylogenetic trees of any size (Figure 10.2D).
We can derive some important properties of the birth-death process on trees. To do so, it is useful to define two additional parameters, the net diversification rate (r) and the relative extinction rate (ϵ):
$r = λ − μ \label{10.A}$
$\epsilon = \frac{\mu}{\lambda} \lable{10.1B}$
These two parameters simplify some of the equations below, and are also commonly encountered in the literature.
To derive some general properties of the birth-death model, we first consider the process over a small interval of time, Δt. We assume that this interval is so short that it contains at most a single event, either speciation or extinction (the interval might also contain no events at all). The probability of speciation and extinction over the time interval can be expressed as:
$Pr_{speciation} = N(t)λΔt \label{10.2A}$
$Pr_{extinction} = N(t)μΔt \label{10.2B}$
We now consider the total number of living species at some time t, and write this as N(t). It is useful to think about the expected value of N(t) under a birth-death model [we consider the full distribution of N(t) below]. The expected value of N(t) after a small time interval Δt is:
$N(t + Δt)=N(t)+N(t)λΔt − N(t)μΔt \label{10.3}$
We can convert this to a differential equation by subtracting N(t) from both sides, then dividing by Δt and taking the limit as Δt becomes very small:
$dN/dt = N(λ − μ) \label{10.4}$
We can solve this differential equation if we set a boundary condition that N(0)=n0; that is, at time 0, we start out with n0 lineages. We then obtain:
$N(t)=n_0e^{λ − μ}t = n_0e^{rt} \label{10.5}$
This deterministic equation gives us the expected value for the number of species through time under a birth-death model. Notice that the number of species grows exponentially through time as long as λ > μ, i.e. r > 0, and decays otherwise (Figure 10.3).
We are also interested in the stochastic behavior of the model – that is, how much should we expect N(t) to vary from one replicate to the next? We can calculate the full probability distribution for N(t), which we write as pn(t)=Pr[N(t)=n] for all n ≥ 0, to completely describe the birth-death model’s behavior. To derive this probability distribution, we can start with a set of equations, one for each value of n, to keep track of the probabilities of n lineages alive at time t. We will denote each of these as pn(t) (there are an infinite set of such equations, from p0 to p). We can then write a set of difference equations that describe the different ways that one can reach any state over some small time interval Δt. We again assume that Δt is sufficiently small that at most one event (a birth or a death) can occur. As an example, consider what can happen to make n = 0 at the end of a certain time interval. There are two possibilities: either we were already at n = 0 at the beginning of the time interval and (by definition) nothing happened, or we were at n = 1 and the last surviving lineage went extinct. We write this as:
(eq. 10.6)
p0(t + Δt)=p1(t)μΔt + p0(t)
Similarly, we can reach n = 1 by either starting with n = 1 and having no events, or going from n = 2 via extinction.
(eq. 10.7)
p1(t + Δt)=p1(t)(1 − (λ + μ))Δt + p2(t)2μΔt
Finally, any for n > 1, we can reach the state of n lineages in three ways: from a birth (from n − 1 to n), a death (from n + 1 to n), or neither (from n to n).
(eq. 10.8)
pn(t + Δt)=pn − 1(t)(n − 1)λΔt + pn + 1(t)(n + 1)μΔt + pn(t)(1 − n(λ + μ))Δt
We can convert this set of difference equations to differential equations by subtracting pn(t) from both sides, then dividing by Δt and taking the limit as Δt becomes very small. So, when n = 0, we use 10.6 to obtain:
(eq. 10.9)
dp0(t)/dt = μp1(t)
From 10.7:
(eq. 10.10)
dp1(t)/dt = 2μp2(t)−(λ − μ)p1(t)
and from 10.8, for all n > 1:
(eq. 10.11)
dpn(t)/dt = (n − 1)λpn − 1(t)+(n + 1)μpn + 1(t)−n(λ + μ)pn(t)
We can then solve this set of differential equations to obtain the probability distribution of pn(t). Using the same boundary condition, N(0)=n0, we have p0(t)=1 for n = n0 and 0 otherwise. Then, we can find the solution to the differential equations 10.9, 10.10, and 10.11. The derivation of the solution to this set of differential equations is beyond the scope of this book (but see Kot 2001 for a nice explanation of the mathematics). A solution was first obtained by Bailey (1964), but I will use the simpler equivalent form from Foote et al. (1999). For p0(t) – that is, the probability that the entire lineage has gone extinct at time t – we have:
(eq. 10.12)
p0(t)=α0n
And for all n ≥ 1:
(eq. 10.13)
$p_n(t) = \sum\limits_{j=1}^{min(n_0,n)} \binom{n_0}{j} \binom{n-1}{j-1} \alpha^{n_0 - j} \beta^{n-j} [(1-\alpha)(1-\beta)]^j$
Where α and β are defined as:
(eq. 10.14)
$\alpha=\frac{\epsilon (e^rt-1)}{(e^rt-\epsilon)}$
$\beta =\frac{(e^rt-1)}{(e^rt-\epsilon)}$
α is the probability that any particular lineage has gone extinct before time t.
Note that when n0 = 1 – that is, when we start with a single lineage - equations 10.12 and 10.13 simplify to (Raup 1985):
(eq. 10.15)
p0(t)=α
And for all n ≥ 1:
(eq. 10.16)
pn(t)=(1 − α)(1 − β)βi − 1
In all cases the expected number of lineages in the tree is exactly as stated above in equation (10.5), but now we have the full probability distribution of the number of lineages given n0, t, λ, and μ. A few plots capture the general shape of this distribution (Figure 10.4).
There are quite a few comparative methods that use clade species richness and age along with the distributions defined in 10.14 and 10.15 to make inferences about clade diversification rates (see chapter 11). | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/10%3A_Introduction_to_Birth-Death_Models/10.02%3A_The_Birth-Death_Model.txt |
The above discussion considered the number of lineages under a birth-death model, but not their phylogenetic relationships. However, just by keeping track of the parent-offspring relationships among lineages, we can consider birth-death models that result in phylogenetic trees (e.g. Figure 10.2D).
The main complication in phylogenetic studies of birth-death models is that we get a “censored” view of the process, in that we only observe lineages that survive to the present day. In the above example, if the true phylogenetic tree were the one plotted in 10.5A, we would only have a chance to observe the phylogenetic tree in figure 10.5B – and even then only if we sampled all of the species and reconstructed the tree with perfect accuracy! A partially sampled tree with only extant species can be seen in Figure 10.5C. I will cover the relationship between birth-death models and the branch lengths of phylogenetic trees in much more detail in the next chapter.
10.04: Simulating Birth-Death Trees
We can use the statistical properties of birth-death models to simulate phylogenetic trees through time. We could begin with a single lineage at time 0. However, phylogenetic tree often start with the first speciation event in the clade, so one can also begin the simulation with two lineages at time 0 (this difference relates to the distinction between crown and stem ages of clades; see also Chapter 11).
To simulate our tree, we need to draw waiting times between speciation and extinction events, connect new lineages to the tree, and prune lineages when they go extinct. We also need a stopping criterion, which can have to do with a particular number of taxa or a fixed time interval. We will consider the latter, and leave growing trees to a fixed number of taxa as an exercise for the reader. Our simulation algorithm is as follows.
simulation algorithm
Assume that we have a certain number of “living” lineages in our tree (1 or 2 initially), a current time (tc = 0 initially), and a stopping time tstop.
1. Draw a waiting time ti to the next speciation or extinction event. Waiting times are drawn from an exponential distribution with rate parameter Nalive * (λ + μ) where Nalive is the current number of living lineages in the tree.
2. Check to see if the simulation ends before the next event. That is, if tc + ti > tstop, end the simulation.
3. Decide whether the next event is a speciation event [with probability λ/(λ + μ)] or an extinction event [with probability μ/(λ + μ)]. This can be done by drawing a uniform random number ui from the interval [0, 1] and assigning speciation to the event if ui < λ/(λ + μ) and extinction otherwise.
4. If (3) is a speciation event, then choose a random living lineage in the tree. Attach a new branch to the tree at this point, and add one new living lineage to the simulation. Return to step 1.
5. If (3) is an extinction event, choose a random living lineage in the tree. That lineage is now dead. As long as there is still at least one living lineage in the tree, return to (1); otherwise, your whole clade has gone extinct, and you can stop the simulation.
This procedure returns a phylogenetic tree that includes both living and dead lineages. One can prune out any extinct taxa to return a birth-death tree of survivors, which is more in line with what we typically study using extant species. It is also worth noting that entire clades can – and often do – go extinct under this protocol before one reaches time tstop. Note also that there is a much more efficient way to simulate trees (Stadler 2011).
We can think about phylogenetic predictions of birth-death models in two ways: by considering tree topology, and by considering tree branch lengths. I will consider each of these two aspects of trees below. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/10%3A_Introduction_to_Birth-Death_Models/10.03%3A_Birth-death_models_and_phylogenetic_trees.txt |
Tree topology summarizes the patterns of evolutionary relatedness among a group of species independent of the branch lengths of a phylogenetic tree. Two different trees have the same topology if they define the exact same set of clades. This is important because sometimes two trees can look very different and yet still have the same topology (e.g. Figure 10.6 A, B, and C).
Tree shape ignores both branch lengths and tree tip labels. For example, the two trees in figure 10.7 A and B have the same tree shape even though they share no tips in common. What they do share is that their nodes have the same patterns in terms of the number of descendants on each “side” of the bifurcation. By contrast, the phylogenetic tree in 10.7 C has a different shape. (Note that what I am calling tree shape is sometimes referred to as “unlabeled” tree topology; e.g. Felsenstein 2004).
Finally, tree balance is a way of expressing differences in the number of descendants between pairs of sister lineages at different points in a phylogenetic tree. For example, consider the phylogenetic tree depicted in figure 10.7B. The deepest split in that tree separates a clade with five species (trout, hippo, bluejay, periwinkle snail, glass squid) from a clade with a single species (Shiitake mushroom), and so that node in the tree is unbalanced with a (5, 1) pattern. By contrast, the deepest split in 10.7C separates two clades of equal size. In that tree, the deepest node is balanced with a (3, 3) pattern. A number of approaches in macroevolution use balance at nodes and across whole trees to try to capture important evolutionary patterns.
We can start to understand these approaches by considering the balance of a single node n in a phylogenetic tree. There are two clades descended from this node; let’s call them a and b. We assume that the total number of species descended from the node Ntotal = Na + Nb is constant and that neither Na nor Nb is zero. An important result, first discussed by Farris (1976) for a pure-birth model, is that all possible numerical divisions of Ntotal into Na + Nb are equally probable. For example, if Ntotal = 10, then all possible divisions: 1 + 9, 2 + 8, 3 + 7, 4 + 6, 5 + 5, 6 + 4, 7 + 3, 8 + 2, and 9 + 1 are all equally probable, so that each will be predicted to occur with a probability 1/9. Formally,
$p(N_a \mid N_{total})=\frac{1}{N_{total}-1} \label{19.17}$
Note that there is a subtle difference between equation 10.2 above and some equations in the literature, e.g. Slowinski and Guyer (1993). This difference has to do with whether we label the two descendent clades, a and b, or not; if the clades are unlabeled, then there is no difference between 4+6 and 6+4, so that the probability that the largest clade, whichever it might be, has 6 species is twice what is given by my equation.
Equation 10.17 applies even if there is extinction, as long as both sister clades have the same speciation and extinction rates (Slowinski and Guyer 1993). This equation has been used to compare diversification rates between sister clades, either for a single pair or across multiple pairs (see Chapter 11).
Tree balance statistics provide a way of comparing numbers of taxa across all of the nodes in a phylogenetic tree simultaneously. There are a surprisingly large number of tree balance statistics, but all rely on summarizing information about the balance of each node across a whole tree. Colless’ index Ic (Colless 1982) is one of the simplest – and, perhaps, most commonly used – indices of tree balance. Ic is the sum of the difference in the number of tips subtended on each side of every node in the tree, standardized by the maximum that such a sum can achieve:
$I_C = \frac{\sum\limits_{all nodes} (N_L - N_R)}{(N-1)(N-2)/2} \label{10.18}$
If the tree is perfectly balanced (only possible when N is some power of 2, e.g. 2, 4, 8, 16, etc.), then IC = 0 (Figure 10.7C). By contrast, if the tree is completely pectinate, which means that each split in the tree contrasts a clade with 1 species with the rest of the species in the clade, then IC = 1 (Figure 10.7A). All phylogenetic trees have values of IC between 0 and 1 (Figure 10.7B).
There are a number of other indices of phylogenetic tree balance (reviewed in Mooers and Heard 1997). All of these indices are used in a similar way: one can then compare the value of the tree index to what one might expect under a particular model of diversification, typically birth-death. In fact, since these indices focus on tree topology and ignore branch lengths, one can actually consider their general behavior under a set of equal-rates Markov (ERM) models. This set includes any model where birth and death rates are equal across all lineages in a phylogenetic tree at a particular time. ERM models include birth-death models as described above, but also encompass models where birth and/or death rates change through time. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/10%3A_Introduction_to_Birth-Death_Models/10.05%3A_Tree_topology%2C_tree_shape%2C_and_tree_balance_under_a_birth-death_model.txt |
The other main way to quantify phylogenetic tree shape is by making lineage-through-time plots. These plots have time along the x axis (from the root of the tree to the present day), and the reconstructed number of lineages on the y-axis (Figure 10.8). Since we are usually considering birth-death models, where the number of lineages is expected to grow (or shrink) exponentially through time, then it is typical practice to log-transform the y-axis.
Lineage-through-time plots are effective ways to visualize patterns of lineage diversification through time. Under a pure-birth model on a semi-log scale, LTT plots follow a straight line on average (Figure 10.9A). By contrast, extinction should leave a clear signal in LTT plots because the probability of a lineage going extinct depends on how long it has been around; old lineages are much more likely to have been hit by extinction than relatively young lineages. We see this reflected in LTT plots as the “pull of the present” – an upturn in the slope of the LTT plot near the present day (Figure 10.9B). Incomplete sampling – that is, not sampling all of the living species in a clade – can also have a huge impact on the shape of LTT plots (Figure 10.9C). We will discuss LTT plots further in chapter 11, where we will use them to make inferences about patterns of lineage diversification through time.
10.0S: 10.S: Introduction to birth-death models (Summary)
In this chapter, I introduced birth-death models and summarized their basic mathematical properties. Birth-death models predict patterns of species diversity over time intervals, and can also be used to model the growth of phylogenetic trees. We can visualize these patterns by measuring tree balance and creating lineage-through-time (LTT) plots.
References
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Hedges, S. B., J. Dudley, and S. Kumar. 2006. TimeTree: A public knowledge-base of divergence times among organisms. Bioinformatics 22:2971–2972.
Kot, M. 2001. Stochastic birth and death processes. Pp. 25–42 in M. Kot, ed. Elements of mathematical ecology. Cambridge University Press, Cambridge.
Mooers, A. O., and S. B. Heard. 1997. Inferring evolutionary process from phylogenetic tree shape. Q. Rev. Biol. 72:31–54.
Raup, D. M. 1985. Mathematical models of cladogenesis. Paleobiology 11:42–52.
Slowinski, J. B., and C. Guyer. 1993. Testing whether certain traits have caused amplified diversification: An improved method based on a model of random speciation and extinction. Am. Nat. 142:1019–1024.
Stadler, T. 2011. Simulating trees with a fixed number of extant species. Syst. Biol. 60:676–684. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/10%3A_Introduction_to_Birth-Death_Models/10.06%3A_Lineage-through-Time_Plots.txt |
In this chapter, I describe how to estimate parameters from birth-death models using data on species diversity and ages, and how to use patterns of tree balance to test hypotheses about changing birth and death rates. I also describe how to calculate the likelihood for birth-death models on trees, which leads directly to both ML and Bayesian methods for estimating birth and death rates. In the next chapter, we will explore elaborations on birth-death models, and discuss models that go beyond constant-rates birth-death models to analyze the diversity of life on Earth.
• 11.1: Hotspots of Diversity
Some parts of the tree of life have more species than others. This imbalance in diversity tells us that speciation is much more common in some lineages than others. Likewise, numerous studies have argued that certain habitats are "hotbeds" of speciation.
• 11.2: Clade Age and Diversity
There are two different ways that one can measure the age of a clade: the stem age and the crown age. A clade's crown age is the age of the common ancestor of all species in the clade. By contrast, a clade's stem age measures the time that that clade descended from a common ancestor with its sister clade. The stem age of a clade is always at least as old, and usually older, than the crown age.
• 11.3: Tree Balance
Birth-death trees have a certain amount of "balance," perhaps a bit less than your intuition might suggest. We can look to real trees to see if the amount of balance matches what we expect under birth-death models. A less balanced pattern in real trees would suggest that speciation and/or extinction rate vary among lineages more than we would expect. By contrast, more balanced trees would suggest more even and predictable diversification across the tree of life than expected under birth-death mo
• 11.4: Fitting birth-death models to branching times
Most modern approaches to fitting birth-death models to phylogenetic trees use the intervals between speciation events on a tree - the "waiting times" between successive speciation - to estimate the parameters of birth-death models. Frequently, information about the pattern of species accumulation in a phylogenetic tree is summarized by a lineage-through-time (LTT) plot, which is a plot of the number of lineages in a tree against time
• 11.5: Sampling and birth-death models
It is important to think about sampling when fitting birth-death models to phylogenetic trees. If any species are missing from your phylogenetic tree, they will lead to biased parameter estimates. This is because missing species are disproportionally likely to connect to the tree on short, rather than long, branches. If we randomly sample lineages from a tree, we will end up badly underestimating both speciation and extinction rates (and wrongly inferring slowdowns).
• 11.S: Fitting Birth-Death Models (Summary)
11: Fitting Birth-Death Models
The number of species on the Earth remains highly uncertain, but our best estimates are around 10 million (Mora et al. 2011). This is a mind-boggling number, and far more than have been described. As far as we know, all of those species are descended from a single common ancestor that lived some 4.2 billion years ago (Hedges and Kumar 2009). All of these species formed by the process of speciation, the process by which one species splits into two (or more) descendants (Coyne and Orr 2004).
Some parts of the tree of life have more species than others. This imbalance in diversity tells us that speciation is much more common in some lineages than others (Mooers and Heard 1997). Likewise, numerous studies have argued that certain habitats are "hotbeds" of speciation (e.g. Hutter et al. 2017; Miller and Wiens 2017). For example, the high Andes ecosystem called the Páramo - a peculiar landscape of alien-looking plants and spectacled bears - might harbor the highest speciation rates on the planet (Madriñán et al. 2013).
Figure 11.1 Páramo ecosystem, Chingaza Natural National Park, Colombia. Photo taken by the author, can be reused under a CC-BY-4.0 license.
In this chapter we will explore how we can learn about speciation and extinction rates from the tree of life. We will use birth-death models, simple models of how species form and go extinct through time. Birth-death models can be applied to data on clade ages and diversities, or fit to the branching times in phylogenetic trees. We will explore both maximum likelihood and Bayesian methods to do both of these things. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/11%3A_Fitting_Birth-Death_Models/11.01%3A_Hotspots_of_Diversity.txt |
If we know the age of a clade and its current diversity, then we can calculate the net diversification rate for that clade. Before presenting equations, I want to make a distinction between two different ways that one can measure the age of a clade: the stem age and the crown age. A clade's crown age is the age of the common ancestor of all species in the clade. By contrast, a clade's stem age measures the time that that clade descended from a common ancestor with its sister clade. The stem age of a clade is always at least as old, and usually older, than the crown age.
Magallón and Sanderson (2001) give an equation for the estimate of net diversification rate given clade age and diversity:
$\hat{r} = \frac{ln(n)}{t_{stem}} \label{11.1}$
where n is the number of species in the clade at the present day and tstem is the stem group age. In this equation we also see the net diversification rate parameter, r = λ − μ (see Chapter 10). This is a good reminder that this parameter best predicts how species accumulate through time and reflects the balance between speciation and extinction rates.
Alternatively, one can use tcrown, the crown group age:
$\hat{r} = \frac{ln(n)-ln(2)}{t_{crown}} \label{11.2}$
The two equations differ because at the crown group age one is considering the clade's diversification starting with two lineages rather than one (Figure 11.2).
Even though these two equations reflect the balance of births and deaths through time, they give ML estimates for r only under a pure birth model where there is no extinction. If there has been extinction in the history of the clade, then our estimates using equations 11.1 and 11.2 will be biased. The bias comes from the fact that we see only clades that survive to the present day, and miss any clades of the same age that died out before they could be observed. By observing only the "winners" of the diversification lottery, we overestimate the net diversification rate. If we know the relative amount of extinction, then we can correct for this bias.
Under a scenario with extinction, one can define ϵ = μ/λ and use the following method-of-moments estimators from Rohatgi (1976, following the notation of Magallon and Sanderson (2001)):
$\hat{r} = \frac{log[n(1-\epsilon)+\epsilon]}{t_{stem}} \label{11.3}$
for stem age, and
$\hat{r} = \frac{ln[\frac{n(1-\epsilon^2 )}{2}+2 \epsilon+\frac{(1-\epsilon) \sqrt{n(n \epsilon^2-8\epsilon+2 n \epsilon+n)}}{2}]-ln(2)}{t_{crown}} \label{11.4}$
for crown age. (Note that eq. 11.3 and 11.4 reduce to 11.1 and 11.2, respectively, when ϵ = 0). Of course, we usually have little idea what ϵ should be. Common practice in the literature is to try a few different values for ϵ and see how the results change (e.g. Magallon and Sanderson 2001).
Magallón and Sanderson (2001), following Strathmann and Slatkin (1983), also describe how to use eq. 10.13 and 10.15 to calculate confidence intervals for the number of species at a given time.
As a worked example, let's consider the data in table 11.1, which give the crown ages and diversities of a number of plant lineages in the Páramo (from Madriñán et al. 2013). For each lineage, I have calculated the pure-birth estimate of speciation rate (from equation 11.2, since these are crown ages), and net diversification rates under three scenarios for extinction (ϵ = 0.1, ϵ = 0.5, and ϵ = 0.9).
Lineage n Age $\hat{r}_{pb}$ $\hat{r}_{\epsilon = 0.1}$ $\hat{r}_{\epsilon = 0.5}$ $\hat{r}_{\epsilon = 0.9}$
1 17 0.42 5.10 5.08 4.53 2.15
2 14 10.96 0.18 0.18 0.16 0.07
3 32 3.80 0.73 0.73 0.66 0.36
4 65 2.50 1.39 1.39 1.28 0.78
5 55 3.05 1.09 1.08 1.00 0.59
6 120 4.04 1.01 1.01 0.94 0.62
7 36 4.28 0.68 0.67 0.61 0.34
8 32 7.60 0.36 0.36 0.33 0.18
9 66 1.47 2.38 2.37 2.19 1.34
10 27 8.96 0.29 0.29 0.26 0.14
11 5 3.01 0.30 0.30 0.26 0.09
12 46 0.80 3.92 3.91 3.58 2.07
13 53 14.58 0.22 0.22 0.21 0.12
The table above shows estimates of net diversification rates for Páramo lineages (data from Madriñán et al. 2013) using equations 11.2 and 11.4. Lineages are as follows: 1: Aragoa, 2: Arcytophyllum, 3: Berberis, 4: Calceolaria, 5: Draba, 6: Espeletiinae, 7: Festuca, 8: Jamesonia + Eriosorus, 9: Lupinus, 10: Lysipomia, 11: Oreobolus, 12: Puya, 13: Valeriana.
Inspecting the last four columns of this table, we can make a few general observations. First, these plant lineages really do have remarkably high diversification rates (Madriñán et al. 2013). Second, the net diversification rate we estimate depends on what we assume about relative extinction rates (ϵ). You can see in the table that the effect of extinction is relatively mild until extinction rates are assumed to be quite high (ϵ = 0.9). Finally, assuming different levels of extinction affects the diversification rates but not their relative ordering. In all cases, net diversification rates for Aragoa, which formed 17 species in less than half a million years, is higher than the rest of the clades. This relationship holds only when we assume relative extinction rates are constant across clades, though. For example, the net diversification rate we calculate for Calceolaria with ϵ = 0 is higher than the calculated rate for Aragoa with ϵ = 0.9. In other words, we can't completely ignore the role of extinction in altering our view of present-day diversity patterns.
We can also estimate birth and death rates for clade ages and diversities using ML or Bayesian approaches. We already know the full probability distribution for birth-death models starting from any standing diversity N(0)=n0 (see equations 10.13 and 10.15). We can use these equations to calculate the likelihood of any particular combination of N and t (either tstem or tcrown) given particular values of λ and μ. We can then find parameter values that maximize that likelihood. Of course, with data from only a single clade, we cannot estimate parameters reliably; in fact, we are trying to estimate two parameters from a single data point, which is a futile endeavor. (It is common, in this case, to assume some level of extinction and calculate net diversification rates based on that, as we did in Table 11.1 above).
One can also assume that a set of clades have the same speciation and extinction rates and fit them simultaneously, estimating ML parameter values. This is the approach taken by Magallón and Sanderson (2001) in calculating diversification rates across angiosperms. When we apply this approach to the Paramo data, shown above, we obtain ML estimates of $\hat{r} = 0.27$ and $\hat{\epsilon} = 0$. If we were forced to estimate an overall average rate of speciation for all of these clades, this might be a reasonable estimate. However, the table above also suggests that some of these clades are diversifiying faster than others. We will return to the issue of variation in diversification rates across clades in the next chapter.
We can also use a Bayesian approach to calculate posterior distributions for birth and death rates based on clade ages and diversities. This approach has not, to my knowledge, been implemented in any software package, although the method is straightforward (for a related approach, see Höhna et al. 2016). To do this, we will modify the basic algorithm for Bayesian MCMC (see Chapter 2) as follows.
Note
1. Sample a set of starting parameter values, r and ϵ, from their prior distributions. For this example, we can set our prior distribution for both parameters as exponential with a mean and variance of λprior (note that your choice for this parameter should depend on the units you are using, especially for r). We then select starting r and ϵ from their priors.
2. Given the current parameter values, select new proposed parameter values using the proposal density Q(p′|p). For both parameter values, we can use a uniform proposal density with width wp, so that Q(p′|p) U(p − wp/2, p + wp/2). We can either choose both parameter values simultaneously, or one at a time (the latter is typically more effective).
3. Calculate three ratios:
• a. The prior odds ratio. This is the ratio of the probability of drawing the parameter values p and p′ from the prior. Since we have exponential priors for both parameters, we can calculate this ratio as: \[ R_{prior} = \frac{\lambda_{prior} e^{-\lambda_{prior} p'}}{\lambda_{prior} e^{-\lambda_{prior} p}}=e^{\lambda_{prior} (p-p')} \label{11.5}\[
• b. The proposal density ratio. This is the ratio of probability of proposals going from p to p′ and the reverse. We have already declared a symmetrical proposal density, so that Q(p′|p)=Q(p|p′) and Rproposal = 1.
• c. The likelihood ratio. This is the ratio of probabilities of the data given the two different parameter values. We can calculate these probabilities from equations 10.13 or 10.16 (depending on if the data are stem ages or crown ages).
1. Find Raccept as the product of the prior odds, proposal density ratio, and the likelihood ratio. In this case, the proposal density ratio is 1, so (eq. 11.6):
Raccept = Rprior ⋅ Rlikelihood | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/11%3A_Fitting_Birth-Death_Models/11.02%3A_Clade_Age_and_Diversity.txt |
As we discussed in Chapter 10, tree balance considers how "balanced" the branches of a phylogenetic tree are. That is, if we look at each node in the tree, are the two sister clades of the same size (balanced) or wildly different (imbalanced)?
Birth-death trees have a certain amount of "balance," perhaps a bit less than your intuition might suggest (see chapter 10). We can look to real trees to see if the amount of balance matches what we expect under birth-death models. A less balanced pattern in real trees would suggest that speciation and/or extinction rate vary among lineages more than we would expect. By contrast, more balanced trees would suggest more even and predictable diversification across the tree of life than expected under birth-death models. This approach traces back to Raup and colleagues, who applied stochastic birth-death models to paleontology in a series of influential papers in the 1970s (e.g. Raup et al. 1973, Raup and Gould (1974)). I will show how to do this for both individual nodes and for whole trees in the following sections.
Section 11.3a: Sister clades and the balance of individual nodes
For single nodes, we already know that the distribution of sister taxa species richness is uniform over all possible divisions of Nn species into two clades of size Na and Nb (Chapter 11). This idea leads to simple test of whether the distribution of species between two sister clades is unusual compared to the expectation under a birth-death model (Slowinski and Guyer 1993). This test can be used, for example, to test whether the diversity of exceptional clades, like passerine birds, is higher than one would expect when compared to their sister clade. This is the simplest measure of tree balance, as it only considers one node in the tree at a time.
Slowinsky and Guyer (1993) developed a test based on calculating a P-value for a division at least as extreme as seen in a particular comparison of sister clades. We consider Nn total species divided into two sister clades of sizes Na and Nb, where Na < Nb and Na + Nb = Nn. Then:
If Na ≠ Nb:
$P = \frac{2 N_a}{N_n - 1} \label{11.7}$
If Na = Nb or P > 1 then set P = 1
For example, we can assess diversification in the Andean representatives of the legume genus Lupinus (Hughes and Eastwood 2006). This genus includes one young radiation of 81 Andean species, spanning a wide range of growth forms. The likely sister clade to this spectacular Andean radiation is a clade of Lupinus species in Mexico that includes 46 species (Drummond et al. 2012). In this case Na = 81 − 46 = 35, and we can then calculate a P-value testing the null hypothesis that both of these clades have the same diversification rate:
$P = \frac{2 N_a}{N_n - 1} = \frac{2 \cdot 35}{81 - 1} = 0.875 \label{11.8}$
We cannot reject the null hypothesis. Indeed, later work suggests that the actual increase in diversification rate for Lupinus occurred deeper in the phylogenetic tree, in the ancestor of a more broadly ranging New World clade (Hughes and Eastwood 2006; Drummond et al. 2012).
Often, we are interested in testing whether a particular trait - say, dispersal into the Páramo - is responsible for the increase in species richness that we see in some clades. In that case, a single comparison of sister clades may be unsatisfying, as sister clades almost always differ in many characters, beyond just the trait of interest. Even if the clade with our putative "key innovation" is more diverse, we still might not be confident in inferring a correlation from a single observation. We need replication.
To address this problem, many studies have used natural replicates across the tree of life, comparing the species richnesses of many pairs of sister clades that differ in a given trait of interest. Following Slowinsky and Guyer (1993), we could calculate a p-value for each clade, and then combine those p-values into an overall test. In this case, one clade (with diversity N1) has the trait of interest and the other does not (N0), and our formula is half of equation 11.5 since we will consider this a one-tailed test:
$P = \frac{N_0}{N_n - 1} \label{11.9}$
When analyzing replicate clade comparisons - e.g. many sister clades, where in each case one has the trait of interest and the other does not - Slowinsky and Guyer (1993) recommended combining these p-values using Fisher's combined probability test, so that:
$χ^2){combined} = −2∑\ln (P_i) \label{11.10}$
Here, the Pi values are from i independent sister clade comparisons, each using equation 11.9. Under the null hypothesis where the character of interest does not increase diversification rates, the test statistic, χ2combined, should follow a chi-squared distribution with 2k degrees of freedom where k is the number of tests. But before you use this combined probability approach, see what happens when we apply it to a real example!
As an example, consider the following data, which compares the diversity of many sister pairs of plants. In each case, one clade has fleshy fruits and the other dry (data from Vamosi and Vamosi 2005):
Fleshy fruit clade nfleshy Dry fruit clade ndry
A 1 B 2
C 1 D 64
E 1 F 300
G 1 H 89
I 1 J 67
K 3 L 4
M 3 N 34
O 5 P 10
Q 9 R 150
S 16 T 35
U 33 V 2
W 40 X 60
Y 50 Z 81
AA 100 BB 1
CC 216 DD 3
EE 393 FF 1
GG 850 HH 11
II 947 JJ 1
KK 1700 LL 18
The clades in the above table are as follows: A: Pangium, B: Acharia+Kigellaria, C: Cyrilla, D: Clethra, E: Roussea, F: Lobelia, G: Myriophylum + Haloragis + Penthorum, H: Tetracarpaea, I: Austrobaileya, J: Illicium+Schisandra, K: Davidsonia, L: Bauera, M: Mitchella, N: Pentas, O: Milligania
, P: Borya, Q: Sambucus, R:
Viburnum, S: Pereskia, T: Mollugo, U: Decaisnea + Sargentodoxa + Tinospora + Menispermum +
Nandina Caulophyllum + Hydrastis + Glaucidium, V: Euptelea, W: Tetracera, X: Dillenia, Y: Osbeckia, Z: Mouriri, AA: Hippocratea, BB: Plagiopteron, CC: Cyclanthus + Sphaeradenia + Freycinetia, DD: Petrosavia + Japonlirion, EE: Bixa, FF: Theobroma + Grewia + Tilia + Sterculia + Durio, GG: Impatiens, HH: Idria, II: Lamium + Clerodendrum + Callicarpa + Phyla + Pedicularis + Paulownia, JJ: Euthystachys, KK: Callicarpa + Phyla + Pedicularis + Paulownia + Solanum, LL: Solanum.
The individual clades show mixed support for the hypothesis, with only 7 of the 18 comparisons showing higher diversity in the fleshy clade, but 6 of those 7 comparisons significant at P < 0.05 using equation 11.9. The combined probability test gives a test statistic of χ2combined = 72.8. Comparing this to a χ2 distribution with 36 degrees of freedom, we obtain P = 0.00027, a highly significant result. This implies that fleshy fruits do, in fact, result in a higher diversification rate.
However, if we test the opposite hypothesis, we see a problem with the combined probability test of equation 11.10 (Vamosi and Vamosi 2005). First, notice that 11 of 18 comparisons show higher diversity in the non-fleshy clade, with 4 significant at P < 0.05. The combined probability test gives χ2combined = 58.9 and P = 0.0094. So we reject the null hypothesis and conclude that non-fleshy fruits diversify at a higher rate! In other words, we can reject the null hypothesis in both directions with this example.
What's going on here? It turns out that this test is very sensitive to outliers - that is, clades with extreme differences in diversity. These clades are very different than what one would expect under the null hypothesis, leading to rejection of the null - and, in some cases with two characters, when there are outliers on both sides (e.g. the proportion of species in each state has a u-shaped distribution; Paradis 2012) we can show that both characters significantly increase diversity (Vamosi and Vamosi 2005)!
Fortunately, there are a number of improved methods that can be used that are similar in spirit to the original Slowinsky and Guyer test but more statistically robust (e.g Paradis 2012). For example, we can apply the "richness Yule test" as described in Paradis (2012), to the data from Vamosi et al. (2005). This is a modified version of the McConway-Sims test (McConway and Sims 2004), and compares the likelihood of a equal rate yule model applied to all clades to a model where one trait is associated with higher or lower diversification rates. This test requires knowledge of clade ages, which I don't have for these data, but Paradis (2012) shows that the test is robust to this assumption and recommends substituting a large and equal age for each clade. I chose 1000 as an arbitrary age, and found a significant likelihood ratio test (null model lnL = −215.6, alternative model lnL = −205.7, P = 0.000008). This method estimates a higher rate of diversification for fleshy fruits (since the age of the clade is arbitrary, the actual rates are not meaningful, but their estimated ratio λ1/λ0 = 1.39 suggests that fleshy fruited lineages have a diversification rate almost 40% higher).
Section 11.3b: Balance of whole phylogenetic trees
We can assess the overall balance of an entire phylogenetic tree using tree balance statistics. As discussed, I will describe just one common statistic, Colless' I, since other metrics capture the same pattern in slightly different ways.
To calculate Colless' I, we can use Equation 10.18. This result will depend strongly on tree size, and so is not comparable across trees of different sizes; to allow comparisons, Ic is usually standardized by subtracting the expected mean for trees of that size under an random model (see below), and dividing by the standard deviation. Both of these can be calculated analytically (Blum et al. 2006), and standardized Ic calculated using a small approximation (following Bortolussi et al. 2006) as:
$I^{'}_c = \frac{I_c-n*log(n)-n(\gamma-1-log(2))}{n} \label{11.11}$
Since the test statistics are based on descriptions of patterns in trees rather than particular processes, the relationship between imbalance and evolutionary processes can be difficult to untangle! But all tree balance indices allow one to reject the null hypothesis that the tree was generated under a birth-death model. Actually, the expected patterns of tree balance are absolutely identical under a broader class of models called "Equal-Rates Markov" (ERM) models (Harding 1971; Mooers and Heard 1997). ERM models specify that diversification rates (both speciation and extinction) are equal across all lineages for any particular point in time. However, those rates may or may not change through time. If they don't change through time, then we have a constant rate birth-death model, as described above - so birth-death models are ERM models. But ERM models also include, for example, models where birth rates slow through time, or extinction rates increase through time, and so on. As long as the changes in rates occur in exactly the same way across all lineages at any time, then all of these models predict exactly the same pattern of tree balance.
Typical steps for using tree balance indices to test the null hypothesis that the tree was generated under an ERM model are as follows:
1. Calculate tree balance using a tree balance statistic.
2. Simulate pure birth trees to general a null distribution of the test statistic. We are considering the set of ERM models as our null, but since pure-birth is simple and still ERM we can use it to get the correct null distribution.
3. Compare the actual test statistic to the null distribution. If the actual test statistic is in the tails of the null distribution, then your data deviates from an ERM model.
Step 2 is unnecessary in cases where we know null distributions for tree balance statistics analytically, true for some (but not all) balance metrics (e.g. Blum and François 2006). There are also some examples in the literature of considering null distributions other than ERM. For example, Mooers and Heard (1997) consider two other null models, PDA and EPT, which consider different statistical distributions of tree shapes (but both of these are difficult to tie to any particular evolutionary process).
Typically, phylogenetic trees are more imbalanced than expected under the ERM model. In fact, this is one of the most robust generalizations that one can make about macroevolutionary patterns in phylogenetic trees. This deviation means that diversification rates vary among lineages in the tree of life. We will discuss how to quantify and describe this variation in later chapters. These tests are all similar in that they use multiple non-nested comparisons of species richness in sister clades to calculate a test statistic, which is then compared to a null distribution, usually based on a constant-rates birth-death process (reviewed in Vamosi and Vamosi 2005; Paradis 2012).
As an example, we can apply the whole-tree balance approach to the tree of Lupinus (Drummond et al. 2012). For this tree, which has 137 tips, we calculate Ic = 1010 and Ic = 3.57. This is much higher than expected by chance under an ERM model, with P = 0.0004. That is, our tree is significantly more imbalanced than expected under a ERM model, which includes both pure birth and birth-death. We can safely conclude that there is variation in speciation and/or extinction rates across lineages in the tree. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/11%3A_Fitting_Birth-Death_Models/11.03%3A_Tree_Balance.txt |
Another approach that uses more of the information in a phylogenetic tree involves fitting birth-death models to the distribution of branching times. This approach traces all the way back to Yule (1925), who first applied stochastic process models to the growth of phylogenetic trees. More recently, a series of papers by Raup and colleagues (Raup et al. 1973; Raup and Gould 1974; Gould et al. 1977; Raup 1985) spurred modern approaches to quantitative macroevolution by simulating random clades, then demonstrating how variable such clades grown under simple birth-death models can be.
Most modern approaches to fitting birth-death models to phylogenetic trees use the intervals between speciation events on a tree - the "waiting times" between successive speciation - to estimate the parameters of birth-death models. Figure 10.2 shows these waiting times. Frequently, information about the pattern of species accumulation in a phylogenetic tree is summarized by a lineage-through-time (LTT) plot, which is a plot of the number of lineages in a tree against time (see Figure 10.9). As I introduced in Chapter 10, the y-axis of LTT plots is log-transformed, so that the expected pattern under a constant-rate pure-birth model is a straight line. Note also that LTT plots ignore the relative order of speciation events. Stadler (2013b) calls models justifying such an approach "species-exchangable" models - we can change the identity of species at any time point without changing the expected behavior of the model. Because of this, approaches to understanding birth-death models based on branching times are different from - and complementary to - approaches based on tree topology, like tree balance.
As discussed in the previous chapter, even though we often have no information about extinct species in a clade, we can still (in theory) infer the presence of extinction from an LTT plot. The signal of extinction is an excess of young lineages, which is seen as the "pull of the recent" in our LTT plots (Figure 10.10). In the next chapter, I will demonstrate how statistical approaches can capture this pattern in a more rigorous way.
Section 11.4a: Likelihood of waiting times under a birth-death model
In order to use ML and Bayesian methods for estimating the parameters of birth-death models from comparative data, we need to write down the likelihoods of the waiting times between speciation events in a tree. There is a little bit of variation in notation in the literature, so I will follow Stadler (2013b) and Maddison et al. (2007), among others, to maintain consistency. We will assume that the clade begins at time t1 with a pair of species. Most analyses follow this convention, and condition the process as starting at the time t1, representing the node at the root of the tree. This makes sense because we rarely have information on the stem age of our clade. We will also condition on both of these initial lineages surviving to the present day, as this is a requirement to obtain a tree with this crown age (e.g. Stadler 2013a equation 5).
Speciation and extinction events occur at various times, and the process ends at time 0 when the clade has n extant species - that is, we measure time backwards from the present day. Extinction will result in species that do not extend all the way to time 0. For now, we will assume that we only have data on extant species. We will refer to the phylogenetic tree that shows branching times leading to the extant species as the reconstructed tree (Nee et al. 1994). For a reconstructed tree with n species, there are n − 1 speciation times, which we will denote as t1, t2, t3, ..., tn − 1. The leaves of our ultrametric tree all terminate at time 0.
Note that in this notation, t1 > t2 > … > tn − 1 > 0, that is, our speciation times are measured backwards from the tips, and as we increase the index the times are constantly decreasing [this is an important notational difference between Stadler (2013a), used here, and Nee (1994 and others), the latter of which considers the time intervals between speciation events, e.g. t1 − t2 in our notation]. For now, we will assume complete sampling; that is, all n species alive at the present day are represented in the tree.
We will now derive a likelihood of of observing the set of speciation times t1, t2, ..., tn − 1 given the extant diversity of the clade, n, and our birth-death model parameters λ and μ. To do this, we will follow very closely an approach based on differential equations introduced by Maddison et al. (2007).
The general idea is that we will assign values to these probabilities at the tips of the tree, and then define a set of rules to update them as we flow back through the tree from the tips to the root. When we arrive at the root of the tree, we will have the probability of observing the actual tree given our model - that is, the likelihood. This is another pruning algorithm.
To begin with, we need to keep track of two probabilities: DN(t), the probability that a lineage at some time t in the past will evolve into the extant clade N as observed today; and E(t), the probability that a lineage at some time t will go completely extinct and leave no descendants at the present day. (Later, we will redefine E(t) so that it includes the possibility that the lineages has descendants but none have been sampled in our data). We then apply these probabilities to the tree using three main ideas (Figure 11.4)
1. We define our starting points at the tips of the tree.
2. We define how the probabilities defined in (1) change as we move backwards along branches of the tree.
3. We define what happens to our probabilities at the tree nodes.
Then, starting at the tips of the tree, we make our way to the root. At each tip, we have a starting value for both DN(t) and E(t). We move backwards along the branches of the tree, updating both probabilities as we go using step 2. When two branches come togther at a node, we combine those probabilities using step 3.
In this way, we walk through the tree, starting with the tips and passing over every branch and node (Figure 11.4). When we get to the root we will have DN(troot), which is the full likelihood that we want.
You might wonder why we need to calculate both DN(t) and E(t) if the likelihood is captured by DN(t) at the root. The reason is that the probability of observing a tree is dependent on these extinction probabilities calculated back through time. We need to keep track of E(t) to know about DN(t) and how it changes. You will see below that E(t) appears directly in our differential equations for DN(t).
First, the starting point. Since every tip i represents a living lineage, we know it is alive at the present day - so we can define DN(t)=1. We also know that it will not go extinct before being included in the tree, so E(t)=0. This gives our starting values for the two probabilities at each tip in the tree (Figure 11.5).
Next, imagine we move backwards along some section of a tree branch with no nodes. We will consider an arbitrary branch of the tree. Since we are going back in time, we will start at some node in the tree N, which occurs at a time tN, and denote the time going back into the past as t (Figure 11.6).
Since that section of branch exists in our tree, we know two things: the lineage did not go extinct during that time, and if speciation occurred, the lineage that split off did not survive to the present day. We can capture these two possibilities in a differential equation that considers how our overall likelihood changes over some very small unit of time (Maddison et al. 2007).
$\frac{dD_N(t)}{dt} = -(\lambda + \mu) D_N(t) + 2 \lambda E(t) D_N(t) \label{11.12}$
Here, the first part of the equation, −(λ + μ)DN(t), represents the probability of not speciating nor going extinct, while the second part, 2λE(t)DN(t), represents the probability of speciation followed by the ultimate extinction of one of the two daughter lineages. The 2 in this equation appears because we must account for the fact that, following speciation from an ancestor to daughters A and B, we would see the same pattern no matter which of the two descendants survived to the present.
We also need to calculate our extinction probability going back through the tree (Maddison et al. 2007):
$\frac{dE(t)}{dt} = \mu - (\mu + \lambda) E(t) + \lambda E(t)^2 \label{11.13}$
The three parts of this equation represent the three ways a lineage might not make it to the present day: either it goes extinct during the interval being considered ( μ), it survives that interval but goes extinct some time later ( −(μ + λ)E(t)), or it speciates in the interval but both descendants go extinct before the present day ( λE(t)2) (Maddison et al. 2007). Unlike the DN(t) term, this probability depends only on time and not the topological structure of the tree.
We will also specify that λ > μ; it is possible to relax that assumption, but it makes the solution more complicated.
We can solve these equations so that we will be able to update the probability moving backwards along any branch of the tree with length t. First, solving equation 11.13 and using our initial condition E(0) = 0:
$E(t) = 1 - \frac{\lambda-\mu}{\lambda - (\lambda-\mu)e^{(\lambda - \mu)t}} \label{11.14}$
We can now substitute this expression for E(t) into Equation \ref{11.12} and solve
$D_N(t) = e^{-(\lambda - \mu)(t - t_N)} \frac{(\lambda - (\lambda-\mu)e^{(\lambda - \mu)t_N})^2}{(\lambda - (\lambda-\mu)e^{(\lambda - \mu)t})^2} \cdot D_N(t_N) \label{11.15}$
Remember that tN is the depth (measured from the present day) of node N (Figure 11.6).
Finally, we need to consider what happens when two branches come together at a node. Since there is a node, we know there has been a speciation event. We multiply the probability calculations flowing down each branch by the probability of a speciation event [Maddison et al. (2007); Figure 11.7].
So:
$D_{N′}(t)=D_N(t)D_M(t)λ \label{11.16}$
Where clade N' is the clade made up of the combination of two sister clades N and M.
To apply this approach across an entire phylogenetic tree, we multiply Equations \ref{11.15} and \ref{11.16} across all branches and nodes in the tree. Thus, the full likelihood is (Maddison et al. 2007; Morlon et al. 2011):
$L(t_1, t_2, \dots, t_n) = \lambda^{n-1} \big[ \prod_{k = 1}^{2n-2} e^{(\lambda-\mu)(t_{k,b} - t_{k,t})} \cdot \frac{(\lambda - (\lambda-\mu)e^{(\lambda - \mu)t_{k,t}})^2}{(\lambda - (\lambda-\mu)e^{(\lambda - \mu)t_{k,b}})^2} \big] \label{11.17}$
Here, n is the number of tips in the tree (note that the original derivation in Maddison uses n as the number of nodes, but I have changed it for consistency with the rest of the book).
The product in Equation \ref{11.17} is taken over all 2n − 2 branches in the tree. Each branch k has two times associated with it, one towards the base of the tree, tk, b, and one towards the tips, tk, t.
Most methods fitting birth-death models to trees condition on the existence of a tree - that is, conditioning on the fact that the whole process did not go extinct before the present day, and the speciation event from the root node led to two surviving lineages. To do this conditioning, we divide Equation \ref{11.17} by λ[1 − E(troot)]2 (Morlon et al. 2011; Stadler 2013a).
Additionally, likelihoods for birth-death waiting times, for example those in the original derivation by Nee, include an additional term, (n − 1)!. This is because there are (n − 1)! possible topologies for any set of n − 1 waiting times, all equally likely. Since this term is constant for a given tree size n, then leaving it out has no influence on the relative likelihoods of different parameter values - but it is necessary to know about this multiplier if comparing likelihoods across different models for model selection, or comparing the output of different programs (Stadler 2013a).
Accounting for these two factors, the full likelihood is:
$L(\tau) = (n-1)! \frac{\lambda^{n-2} \big[ \prod_{k = 1}^{2n-2} e^{(\lambda-\mu)(t_{k,b} - t_{k,t})} \cdot \frac{(\lambda - (\lambda-\mu)e^{(\lambda - \mu)t_{k,t}})^2}{(\lambda - (\lambda-\mu)e^{(\lambda - \mu)t_{k,b}})^2} \big]}{ [1-E(t_{root})]^2} \label{11.18}$
where:
$E(t_{root}) = 1 - \frac{\lambda-\mu}{\lambda - (\lambda-\mu)e^{(\lambda - \mu)t_{root}}}\label{11.19}$
Section 11.4b: Using maximum likelihood to fit a birth-death model
Given equation 11.19 for the likelihood, we can estimate birth and death rates using both ML and Bayesian approaches. For the ML estimate, we maximize equation 11.19 over λ and μ. For a pure-birth model, we can set μ = 0, and the maximum likelihood estimate of λ can be calculated analytically as:
$\lambda= \frac{n-2}{s_{branch}} \label{11.20}$
where sbranch is the sum of branch lengths in the tree,
$s_{branch} = \sum_{i=1}^{n-1} t_i + t_{n-1} \label{11.21}$
Equation \ref{11.21} is also called the Kendal-Moran estimator of the speciation rate (Nee 2006).
For a birth-death model, we can use numerical methods to maximize the likelihood over λ and μ.
For example, we can use ML to fit a birth-death model to the Lupinus tree (Drummond et al. 2012), which has 137 tip species and a total age of 16.6 million years. Doing so, we obtain ML parameter estimates of λ = 0.46 and μ = 0.20, with a log-likelihood of lnLbd = 262.3. Compare this to a pure birth model on the same tree, which gives λ = 0.35 and lnLpb = 260.4. One can compare the fit of these two models using AIC scores: AICbd = −520.6 and AICpb = −518.8, so the birth-death model has a better (lower) AIC score but by less than two AIC units. A likelihood ratio test, which gives Δ = 3.7 and P = 0.054. In other words, we estimate a non-zero extinction rate in the clade, but the evidence supporting that model over a pure-birth model is not particularly strong. Even if this model selection is a bit ambiguous, remember that we have also estimated parameters using all of the information that we have in the waiting times of the phylogenetic tree.
Section 11.4c: Using Bayesian MCMC to fit a Birth-Death Model
We can also estimate birth and death rates using a Bayesian MCMC. We can use exactly the method spelled out above for clade ages and diversities, but substitute Equation \ref{11.11} for the likelihood, thus using the waiting times derived from a phylogenetic tree to estimate model parameters.
Applying this to Lupines with the same priors as before, we obtain the posterior distributions shown in figure 11.5. The mean of the posterior for each parameter is λ = 0.48 and μ = 0.23, quite close to the ML estimates for these parameters. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/11%3A_Fitting_Birth-Death_Models/11.04%3A_Fitting_birth-death_models_to_branching_times.txt |
It is important to think about sampling when fitting birth-death models to phylogenetic trees. If any species are missing from your phylogenetic tree, they will lead to biased parameter estimates. This is because missing species are disproportionally likely to connect to the tree on short, rather than long, branches. If we randomly sample lineages from a tree, we will end up badly underestimating both speciation and extinction rates (and wrongly inferring slowdowns; see chapter 12).
Fortunately, the mathematics for incomplete sampling of reconstructed phylogenetic trees has also been worked out. There are two ways to do this, depending on how the tree is actually sampled. If we consider the missing species to be random with respect to the taxa included in the tree, then one can use a uniform sampling fraction to account for them. By contrast, we often are in the situation where we have tips in our tree that are single representatives of diverse clades (e.g. genera). We usually know the diversity of these unsampled clades in our tree of representatives. I will follow (Höhna et al. 2011; Höhna 2014) and refer to this approach as representative sampling (and the previous alternative as uniform sampling).
For the uniform sampling approach, we use the framework above of calculating backwards through time, but modify the starting points for each tip in the tree to reflect f, the probability of sampling a species (following Fitzjohn et al. (2009)):
(eq. 11.22)
DN(0)=1 − f
E(0)=f
Repeating the calculations above along branches and at nodes, but with the starting conditions above, we obtain the following likelihood (FitzJohn et al. 2009):
(eq. 11.23)
\begin{aligned} L(t_1, t_2, \dots, t_n) = \lambda^{n-1} \big[ \prod_{k = 1}^{2n-2} e^{(\lambda-\mu)(t_{k,b} - t_{k,t})} \cdot \ \frac{(f \lambda - (\mu - \lambda(1-f))e^{(\lambda - \mu)t_{k,t}})^2}{(f \lambda - (\mu - \lambda(1-f))e^{(\lambda - \mu)t_{k,b}})^2} \big] \end{aligned}
Again, the above formula is proportional to the full likelihood, which is:
(eq. 11.24)
$L(\tau) = (n-1)! \frac{\lambda^{n-2} \big[ \prod_{k = 1}^{2n-2} e^{(\lambda-\mu)(t_{k,b} - t_{k,t})} \cdot \frac{(f \lambda - (\mu - \lambda(1-f))e^{(\lambda - \mu)t_{k,t}})^2}{(f \lambda - (\mu - \lambda(1-f))e^{(\lambda - \mu)t_{k,b}})^2} \big]}{[1-E(t_{root})]^2}$
and:
(eq. 11.25)
$E(t_{root}) = 1 - \frac{\lambda-\mu}{\lambda - (\lambda-\mu)e^{(\lambda - \mu)t_{root}}}$
For representative sampling, one approach is to consider the data as divided into two parts, phylogenetic and taxonomic. The taxonomic part is the stem age and extant diversity of the unsampled clades, while the phylogenetic part is the relationships among those clades. Following Rabosky and Lovette (2007), we can then calculate:
(eq. 11.26)
Ltotal = Lphylogenetic ⋅ Ltaxonomic
Where Lphylogenetic can be calculated using equation 11.18 and Ltaxonomic calculated for each clade using equation 10.16 and then multiplied to get the overall likelihood.
There are two extensions to this approach that are worth mentioning. One is Hohna's (2011) diversified sampling ("DS") model. This model makes a different assumption: when sampling n taxa from an overall set of m, the deepest n − 1 nodes have been included. Hohna's approach allows users to fit a model with representative sampling but without requiring assignment of extant diversity to each clade. Another approach, by Stadler and Smrckova (2016), calculates likelihoods for representatively sampled trees and can fit models of time-varying speciation and extinction rates (see chapter 12). | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/11%3A_Fitting_Birth-Death_Models/11.05%3A_Sampling_and_birth-death_models.txt |
In this chapter, I described how to estimate parameters from birth-death models using data on species diversity and ages, and how to use patterns of tree balance to test hypotheses about changing birth and death rates. I also described how to calculate the likelihood for birth-death models on trees, which leads directly to both ML and Bayesian methods for estimating birth and death rates. In the next chapter, we will explore elaborations on birth-death models, and discuss models that go beyond constant-rates birth-death models to analyze the diversity of life on Earth.
References
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Höhna, S., M. R. May, and B. R. Moore. 2016. TESS: An R package for efficiently simulating phylogenetic trees and performing Bayesian inference of lineage diversification rates. Bioinformatics 32:789–791.
Höhna, S., T. Stadler, F. Ronquist, and T. Britton. 2011. Inferring speciation and extinction rates under different sampling schemes. Mol. Biol. Evol. 28:2577–2589.
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Hutter, C. R., S. M. Lambert, and J. J. Wiens. 2017. Rapid diversification and time explain amphibian richness at different scales in the tropical Andes, Earth’s most biodiverse hotspot. Am. Nat. 190:828–843.
Maddison, W. P., P. E. Midford, S. P. Otto, and T. Oakley. 2007. Estimating a binary character’s effect on speciation and extinction. Syst. Biol. 56:701–710. Oxford University Press.
Madriñán, S., A. J. Cortés, and J. E. Richardson. 2013. Páramo is the world’s fastest evolving and coolest biodiversity hotspot. Front. Genet. 4:192.
Magallon, S., and M. J. Sanderson. 2001. Absolute diversification rates in angiosperm clades. Evolution 55:1762–1780.
McConway, K. J., and H. J. Sims. 2004. A likelihood-based method for testing for nonstochastic variation of diversification rates in phylogenies. Evolution 58:12–23.
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In this chapter I discussed models that go beyond constant rate birth-death models. We can fit models where speciation rate varies across clades or through time (or both). In some cases, very different models predict the same pattern in phylogenetic trees, warranting some caution until direct fossil data can be incorporated. I also described a model of protracted speciation, where speciation takes some time to complete. This latter model is potentially better connected to microevolutionary models of speciation, and could point towards fruitful directions for the field. We know that simple birth-death models do not capture the richness of speciation and extinction across the tree of life, so these models that range beyond birth and death are critical to the growth of comparative methods.
• 12.1: Capturing Variable Evolution
Simple, constant-rate birth-death models are not adequate to capture the complexity and dynamics of speciation and extinction across the tree of life. Speciation and extinction rates vary through time, across clades, and among geographic regions. We can sometimes predict this variation based on what we know about the mechanisms that lead to speciation and/or extinction. This chapter explores some extensions to birth-death models that allow us to explore diversification in more detail.
• 12.2: Variation in Diversification Rates across Clades
We know from analyses of tree balance that the tree of life is more imbalanced than birth-death models predict. We can explore this variation in diversification rates by allowing birth and death models to vary along branches in phylogenetic trees. The simplest scenario is when one has a particular prediction about diversification rates to test. For example, we might wonder if diversification rates in one clade are higher than in the rest of the phylogenetic tree.
• 12.3: Variation in Diversification Rates through Time
In addition to considering rate variation across clades, we might also wonder whether birth and/or death rates have changed through time. For example, perhaps we think our clade is an adaptive radiation that experienced rapid diversification upon arrival to an island archipelago and slowed as this new adaptive zone got filled. This hypothesis is an example of density-dependent diversification, where diversification rate depends on the number of lineages that are present
• 12.4: Diversity-Dependent Models
Time-dependent models in the previous section are often used as a proxy to capture processes like key innovations or adaptive radiations. Many of these theories suggest that diversification rates should depend on the number of species alive in a certain time or place, rather than time . Therefore, we might want to define speciation rate in a truly diversity dependent manner rather than using time as a proxy.
• 12.5: Protracted Speciation
In all of the diversification models that we have considered so far, speciation happens instantly; one moment we have a single species, and then immediately two. But this is not biologically plausible. Speciation takes time, as evidenced by the increasing numbers of partially distinct populations that biologists have identified in the natural world. Furthermore, the fact that speciation takes time can have a profound impact on the shapes of phylogenetic trees.
• 12.S: Beyond Birth-Death Models (Summary)
12: Beyond Birth-Death models
As we discovered in Chapter 11, there are times and places where the tree of life has grown more rapidly than others. For example, islands and island-like habitats are sometimes described as hotspots of speciation (Losos and Schluter 2000; Hughes and Eastwood 2006), and diversification rates in such habitats can proceed at an extremely rapid pace. On a broader scale, many studies have shown that speciation rates are elevated and/or extinction rates depressed following mass extinctions (e.g. Sepkoski 1984). Finally, some clades seem to diversity much more rapidly than others. In my corner of the world, the Pacific Northwest of the United States, this variation is best seen in our local amphibians. We have species like the spotted frog and the Pacific tree frog, which represent two very diverse frog lineages with high diversification rates (Ranidae and Hylidae, respectively; Roelants et al. 2007). At the same time, if one drives a bit to the high mountain streams, you can find frogs with tiny tails. These Inland Tailed Frogs are members of Ascaphidae, a genus with only two species, one coastal and one inland. (As an aside, the tail, found only in males, is an intromittent organ used for internal fertilization - analogous to a penis, but different!) These two tailed frog species are the sister group to a small radiation of four species frogs in New Zealand (Leopelmatidae, which have no tails). These two clades together - just six species - make up the sister clade to all other frogs, nearly 7000 species (Roelants et al. 2007; Jetz and Pyron 2018). We seek to explain patterns like this contrast in the diversity of two groups which decend from a common ancestor and are, thus, the same age.
Simple, constant-rate birth-death models are not adequate to capture the complexity and dynamics of speciation and extinction across the tree of life. Speciation and extinction rates vary through time, across clades, and among geographic regions. We can sometimes predict this variation based on what we know about the mechanisms that lead to speciation and/or extinction.
In this chapter, I will explore some extensions to birth-death models that allow us to explore diversification in more detail. This chapter also leads naturally to the next, chapter 13, which will consider the case where diversification rates depend on species’ traits. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/12%3A_Beyond_Birth-Death_models/12.01%3A_Capturing_Variable_Evolution.txt |
We know from analyses of tree balance that the tree of life is more imbalanced than birth-death models predict. We can explore this variation in diversification rates by allowing birth and death models to vary along branches in phylogenetic trees. The simplest scenario is when one has a particular prediction about diversification rates to test. For example, we might wonder if diversification rates in one clade (clade A in Figure 12.2) are higher than in the rest of the phylogenetic tree. We can test this hypothesis by fitting a multiple-rate birth-death model.
The simplest method to carry out this test is by using model selection in a ML framework (Rabosky et al. 2007). To do this, we first fit a constant-rates birth-death model to the entire tree, and calculate a likelihood. We can then fit variable-rates birth-death models to the data, comparing the fit of these models using either likelihood ratio tests or AICC. The simplest way to fit a variable-rates model is to adapt the likelihood formula in equation 11.18 (or eq. 11.24 if species are unsampled). We calculate the likelihood in two parts, one for the background part of the tree (with rates λB and μB) and one for the focal clade that may have different diversification dynamics (with rates λA and μA). We can then compare this model to one where speciation and extinction rates are constant through time.
Consider the example in Figure 12.2. We would like to know whether clade A has speciation and extinction rates, λA and μA, that differ from the background rates, λB and μB – we will call this a “variable rates” model. The alternative is a “constant rates” model where the entire clade has constant rate parameters λT and μT. These two models are nested, since the constant-rates model is a special case of the variable rates model where λT = λA = λB and μT = μA = μB. Calculating the likelihood for these two models is reasonably straightforward - we simply calculate the likelihood for each section of the tree using the relevant equation from Chapter 11, and then multiply the likelihoods from the two parts of the tree (or add the log-likelihoods) to get the overall likelihood.
For a real example, let’s look at the phylogenetic tree of amphibians and evaluate the hypothesis that the tailed and New Zealand frogs, sister clade to the rest of frogs, diversified at a slower rate than other amphibians (Figure 12.3). We can use the phylogenetic "backbone" tree from Jetz and Pyron (Jetz and Pyron 2018), assigning diversities based on the classification associated with that publication. We can then calculate likelihoods based on Equation 11.24.
We can calculate the likelihood of the constant rates model, with two parameters λT and μT, to a variable rates model with four parameters λliop, μliop, λother, and μother. For this example, we obtain the following results.
Model Parameter estimates ln-Likelihood AICc
Constant rates λT = 0.30 -1053.9 2111.8
μT = 0.28
Variable rates lambdaliop = 0.010 -1045.4 2101.1
μliop = 0.007
lambdaother = 0.29
μother = 0.27
With a difference in AICc of more than 10, we see from these results that there is good reason to think that there is a difference in diversification rates in these "oddball" frogs compared to the rest of the amphibians.
Of course, more elaborate comparisons are possible. For example, one could compare the fit of four models, as follows: Model 1, constant rates; Model 2, speciation rate in clade A differs from the background; Model 3, extinction rate in clade A differs from the background; and Model 4, both speciation and extinction rates in clade A differ from the background. In this case, some of the pairs of models are nested – for example, Model 1 is a special case of Model 2, which is, in turn, a special case of Model 4 – but all four do not make a nested series. Here we benefit from using a model selection approach based on AICC. We can fit all four models and use their relative number of parameters to calculate AICC scores. We can then calculate AICC weights to evaluate the relative support for each of these four models. (As an aside, it might be difficult to differentiate among these four possibilities without a lot of data!)
But what if you do not have an a priori reason to predict differential diversification rates across clades? Or, what if the only reason you think one clade might have a different diversification rate than another is that it has more species? (Such reasoning is circular, and will wreak havoc with your analyses!) In such cases, we can use methods that allow us to fit general models where diversification rates are allowed to vary across clades in the tree. Available methods use stepwise AIC (MEDUSA, Alfaro et al. 2009; but see May and Moore 2016), or reversible-jump Bayesian MCMC (Rabosky 2014, 2017; but see Moore et al. 2016).
For example, running a stepwise-AIC algorithm on the amphibian data (Alfaro et al. 2009) results in a model with 11 different speciation and extinction regimes (Figure 12.4). This is good evidence that diversification rates have varied wildly through the history of amphibians.
One note: all current approaches fit a model where birth and death rates change at discrete times in the phylogenetic tree - that is, along certain branches in the tree leading to extant taxa. One might wish for an approach, then, that models such changes - using, for example, a Poisson process - and then locates the changes on the tree. However, we still lack the mathematics to solve for E(t) (e.g. Equation 11.19) under such a model (Moore et al. 2016). Given that, we can view current implementations of models where rates vary across clades as an approximation to the likelihood, and one that discounts the possibility of shifts in speciation and/or extinction rates among any clades that did not happen to survive until the present day (Rabosky 2017) - and we are stuck with that until a better alternative is developed! | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/12%3A_Beyond_Birth-Death_models/12.02%3A_Variation_in_Diversification_Rates_across_Clades.txt |
In addition to considering rate variation across clades, we might also wonder whether birth and/or death rates have changed through time. For example, perhaps we think our clade is an adaptive radiation that experienced rapid diversification upon arrival to an island archipelago and slowed as this new adaptive zone got filled (Schluter 2000). This hypothesis is an example of density-dependent diversification, where diversification rate depends on the number of lineages that are present (Rabosky 2013). Alternatively, perhaps our clade has been experiencing extinction rates that have changed through time, perhaps peaking during some time period of unfavorable climactic conditions (Benton 2009). This is another hypothesis that predicts variation in diversification (speciation and extinction) rates through time.
We can fit time-dependent diversification models using likelihood equations that allow arbitrary variation in speciation and/or extinction rates, either as a function of time or depending on the number of other lineages in the clade. To figure out the likelihood we can first make a simplifying assumption: though diversification rates might change, they are constant across all lineages at any particular time point. In particular, this means that speciation (and/or extinction) rates slow down (or speed up) in exactly the same way across all lineages in an evolving clade. This is again the Equal-Rates Markov (ERM) model for tree growth described in the previous chapter.
Our assumption about equal rates across lineages at any time means that we can consider time-slices through the tree rather than individual branches, i.e. we can get all the information that we need to fit these models from lineage through time plots.
The most general way to fit time-varying birth-death models to phylogenetic trees is described in Morlon et al. (2011). Consider the case where both speciation and extinction rates vary as a function of time, λ(t) and μ(t). Morlon et al. (2011) derive the likelihood for such a model as:
$L(t_1, t_2, \dots, t_{n-1}) = (n+1)! \dfrac{f^n \sum_{i=1}^{n-1} \lambda(t_i) \Psi(s_{i,1}, t_i) \Psi(s_{i,2}, t_i)}{\lambda [1-E(t_1)]^2} \label{12.1}$
where:
$E(t) = 1 - \frac{e^{\int_0^t [\lambda(u) - \mu(u)]du}}{\dfrac{1}{f} + \displaystyle \int_0^t e^{\int_0^s [\lambda(u) - \mu(u)]du} \lambda(s) ds}\label{12.2}$
and:
$\Psi(s, t) = e^{\int_s^t [\lambda(u) - \mu(u)]du} \left[ 1 + \frac{\displaystyle \int_s^t e^{\int_0^\tau [\lambda(\sigma) - \mu(\sigma)]d\sigma}\lambda(\tau)d\tau}{\frac{1}{f} + \displaystyle \int_0^s e^{\int_0^\tau \left[\lambda(\sigma) - \mu(\sigma) \right]d\sigma} \lambda(\tau) d\tau} \right]^{-2} \label{12.3}$
Following Chapter 11, n is the number of tips in the tree, and divergence times t1, t2, …, tn − 1 are defined as measured from the present (e.g. decreasing towards the present day). λ(t) and μ(t) are speciation and extinction rates expressed as an arbitrary function of time, f is the sampling fraction (under a uniform sampling model). For a node starting at time ti, si, 1 and si, 2 are the times when the two daughter lineages encounter a speciation event in the reconstructed tree1. E(t), as before, is the probability that a lineage alive at time t leaves no descendants in the sample. Finally, $Ψ(s, t)$ is the probability that a lineage alive at time s leaves exactly one descent at time t < s in the reconstructed tree. These equations look complex, and they are - but basically involve integrating the speciation and extinction functions (and their difference) along the branches of the phylogenetic tree.
Note that my equations here differ from the originals in Morlon et al. (2011) in two ways. First, Morlon et al. (2011) assumed that we have information about the stem lineage and, thus, used an index on ti that goes up to n instead of n − 1 and a different denominator conditioning on survival of the descendants of the single stem lineage (Morlon et al. 2011). Second, I also multiply by the total number of topological arrangements of n taxa, (n + 1)!.
If one substitutes constants for speciation and extinction (λ(t)=λc, μ(t)=μc) in Equation \ref{12.1}, then one obtains equation 11.24; if one additionally considers the case of complete sampling and substitutes f = 1 then we obtain equation 11.18. This provides a single unified framework for time-varying phylogenetic trees with uniform incomplete sampling (see also Höhna 2014 for independent but equivalent derivations that also extend to the case of representative sampling).
Equation \ref{12.1} requires that we define speciation rate as a function of time. Two types of time-varying models are currently common in the comparative literature: linear and expoential. If speciation rates change linearly through time (see Rabosky and Lovette 2008 for an early version of this model):
$λ(t)=λ_0 + α_λt \label{12.4}$
Where λ0 is the initial speciation rate at the present and alpha is the slope of speciation rate change as we go back through time. αλ must be chosen so that speciation rates do not become negative as we move back through the tree: αλ > −λ0/t1. Note that the interpretation of αλ is a bit strange since we measure time backwards: a positive αλ, for example, would mean that speciation rates have declined from the past to the present. Other time-dependent models published earlier (e.g. Rabosky and Lovette 2008, which considered a linearly declining pure-birth model) do not have this property.
We could also consider a linear change in extinction through time:
$μ(t)=μ_0 + α_μt \label{12.5}$
Again, αμ is the change in extinction rate through time, and must be interpreted in the same "backwards" way as αλ. Again, we must restrict our parameter to avoid a negative rate: αμ > −μ0/t1
One can then substitute either of these formulas into Equation \ref{12.1} to calculate the likelihood of a model where speciation rate declines through time. Many implementations of this approach use numerical approximations rather than analytic solutions (see, e.g., Morlon et al. 2011; Etienne et al. 2012).
Another common model has speciation and/or extinction rates changing exponentially through time:
$λ(t)=λ_0e^{β_λt} \label{12.6}$
and/or
$μ(t)=μ_0e^{β_μt} \label{12.7}$
We can again calculate likelihoods for this model numerically (Morlon et al. 2011, Etienne et al. (2012)).
As an example, we can test models of time-varying diversification rates across part of the amphibian tree of life from Jetz and Pyron (2018). I will focus on one section of the salamanders, the lungless salamanders (Plethodontidae, comprised of the clade that spans Bolitoglossinae, Spelerpinae, Hemidactylinae, and Plethodontinae). This interesting clade was already identified above as including both an increase in diversification rates (at the base of the clade) and a decrease (on the branch leading to Hemidactylinae; Figure 12.4). The tree I am using may be missing a few species; this section of the tree includes 440 species in Jetz and Pyron (2018) but 471 speices are listed on Amphibiaweb as of May 2018. Thus, I will assume random sampling with f = 440/471 = 0.934. Comparing the fit of a set of models, we obtain the following results:
Model Number of params. Param. estimates lnL AIC
CRPB 1 λ = 0.05111267 497.8 -993.6
CRBD 2 λ = 0.05111267 497.8 -991.6
SP-L μ = 0
3 λ0 = 0.035 513.0 -1019.9
αλ = 0.0011
μ = 0
SP-E 3 λ0 = 0.040 510.7 -1015.4
βλ = 0.016
μ = 0
EX-L 3 λ = 0.053 497.8 -989.6
μ0 = 0
αμ = 0.000036
EX-E 3 λ = 0.069 510.6 -1015.3
μ0 = 61.9
βμ = −111.0
Models are abbreviated as: CRPB = Constant rate pure birth; CRBD = Constant rate birth-death; SP-L = Linear change in speciation; SP-E = Exponential change in speciation; EX-L = Linear change in extinction; EX-E = Exponential change in extinction.
The model with the lowest AIC score has a linear decline in speciation rates, and moderate support compared to all other models. From this, we support the inference that diversification rates among these salamanders has slowed through time. Of course, there are other models I could have tried, such as models where both speciation and extinction rates are changing through time, or models where there are many more extant species of salamanders than currently recognized. The conclusion we make is only as good as the set of models being considered, and one should carefully consider any plausible models that are not in the candidate set. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/12%3A_Beyond_Birth-Death_models/12.03%3A_Variation_in_Diversification_Rates_through_Time.txt |
Time-dependent models in the previous section are often used as a proxy to capture processes like key innovations or adaptive radiations (Rabosky 2014). Many of these theories suggest that diversification rates should depend on the number of species alive in a certain time or place, rather than time (Phillimore and Price 2008; Etienne and Haegeman 2012; Etienne et al. 2012; Rabosky 2013; Moen and Morlon 2014). Therefore, we might want to define speciation rate in a truly diversity dependent manner rather than using time as a proxy:
$\lambda(t) = \lambda_0 (1 - \frac{N_t}{K}) \label{12.8}$
Since speciation rate now depends on number of lineages rather than time, we can't plug this expression into our general formula (Morlon et al. 2011). Instead, we can use the approach outlined by Etienne et al. (2012) and Etienne et al. (2016). This approach focuses on numerical solutions to differential equations moving forward through time in the tree. The overall idea of the approach is similar to Morlon, but details differ; likelihoods from Etienne et al. (2012) should be directly comparable to all the likelihoods presented in this book provided that the conditioning is the same and they are multiplied by the total number of topological arrangements, (n + 1)!, to get a likelihood for the tree rather than for the branching times. Etienne's approach can also deal with incomplete sampling under a uniform sampling model.
As an example, we can fit a basic model of diversity-dependent speciation to our phylogenetic tree of lungless salamanders introduced above. Doing so, we find a ML estimate of λ0 = 0.099, μ = 0, and K = 979.9, with a log-likelihood of 537.3 and an AIC of -1068.7. This is a substantial improvement over any of the time-varying models considered above, and evidence for diversity dependence among lungless salamanders.
Both density- and time-dependent approaches have become very popular, as time-dependent diversification models are consistent with many ecological models of how multi-species clades might evolve through time. For example, adaptive radiation models based on ecological opportunity predict that, as niches are filled and ecological opportunity “used up,” then we should see a declining rate of diversification through time (Etienne and Haegeman 2012; Rabosky and Hurlbert 2015). By contrast, some models predict that species create new opportunities for other species, and thus predict accelerating diversification through time (Emerson and Kolm 2005). These are reasonable hypotheses, but there is a statistical challenge: in each case, there is at least one conceptually different model that predicts the exact same pattern. In the case of decelerating diversification, the predicted pattern of a lineage-through-time plot that bends down towards the present day can also come from a model where lineages accumulate at a constant rate, but then are not fully sampled at the present day (Pybus and Harvey 2000). In other words, if we are missing some living species from our phylogenetic tree and we don’t account for that, then we would mistake a constant-rates birth death model for a signal of slowing diversification through time. Of course, methods that we have discussed can account for this. Some methods can even account for the fact that the missing taxa might be non-random, as missing taxa tend to be either rare or poorly differentiated from their sister lineages (e.g. often younger than expected by chance; Cusimano and Renner 2010; Brock et al. 2011). However, the actual number of species in a clade is always quite uncertain and, in every case, must be known for the method to work. So, an alternative explanation that is often viable is that we are missing species in our tree, and we don't know how many there are. Additionally, since much of the signal for these methods comes from the most recent branching events in the tree, some "missing" nodes may simply be too shallow for taxonomists to call these things "species." In other words, our inferences of diversity dependence from phylogenetic trees are strongly dependent on our understanding of how we have sampled the relevant taxa.
Likewise, a pattern of accelerating differentiation mimics the pattern caused by extinction. A phylogenetic tree with high but constant rates of speciation and extinction is nearly impossible to distinguish from a tree with no extinction and speciation rates that accelerate through time.
Both of the above caveats are certainly worth considering when interpreting the results of tests of diversification from phylogenetic data. In many cases, adding fossil information will allow investigators to reliably distinguish between the stated alternatives, although methods that tie fossils and trees together are still relatively poorly developed (but see Slater and Harmon 2013). And many current methods will give ambiguous results when multiple models provide equivalent explanations for the data - as we would hope!
12.05: Protracted Speciation
In all of the diversification models that we have considered so far, speciation happens instantly; one moment we have a single species, and then immediately two. But this is not biologically plausible. Speciation takes time, as evidenced by the increasing numbers of partially distinct populations that biologists have identified in the natural world (Coyne and Orr 2004; De Queiroz 2005). Furthermore, the fact that speciation takes time can have a profound impact on the shapes of phylogenetic trees (Losos and Adler 1995). Because of this, it is worth considering diversification models that explicitly account for the fact that the process of speciation has a beginning and an end.
The most successful models to tackle this question have been models of protracted speciation (Rosindell et al. 2010; Etienne and Rosindell 2012; Lambert et al. 2015). One way to set up such a model is to state that speciation begins by the formation of an incipient species at some rate λ1. This represents a “partial” species; one can imagine, for example, that this is a population that has split off from the main range of the species, but has not yet evolved full reproductive isolation. The incipient species only becomes a “full” species if it completes speciation, which occurs at a rate λ2. This represents the rate at which an incipient species evolves full species status (Figure 12.5).
Because speciation takes time, the main impact of this model is that we predict fewer very young species in our tree – that is, the nodes closest to the tips of the tree are not as young as they would be compared to pure-birth or birth-death models without protracted speciation (Figure 12.6). As a result, protracted speciation models produce lineage through time plots that can mimic the properties often attributed to diversity-dependence, even without any interactions among lineages (Etienne and Rosindell 2012)!
Likelihood approaches are available for this model of protracted speciation. Again, the likelihood must be calculated using numerical methods (Lambert et al. 2015). Fitting this model to the salamander tree, we obtain a maximum log-likeihood of 513.8 with parameter values λ1 = 0.059, λ2 = 0.44, and μ = 0.0. This corresponds to an AIC score of -1021.6; this model fits about as well as the best of the time-varying models but not as well as the diversity dependent model considered above. Again, though, I am not including plausible combinations of models, such as protracted speciation that varies through time.
So far, models of protracted speciation remain mostly in the realm of ecological neutral theory, and are just beginning to move into phylogenetics and evolutionary biology (see, e.g., Sukumaran and Lacey Knowles 2017). However, I think models that treat speciation as a process that takes time – rather than something instantaneous – will be an important addition to our macroevolutionary toolbox in the future. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/12%3A_Beyond_Birth-Death_models/12.04%3A_Diversity-Dependent_Models.txt |
In this chapter I discussed models that go beyond constant rate birth-death models. We can fit models where speciation rate varies across clades or through time (or both). In some cases, very different models predict the same pattern in phylogenetic trees, warranting some caution until direct fossil data can be incorporated. I also described a model of protracted speciation, where speciation takes some time to complete. This latter model is potentially better connected to microevolutionary models of speciation, and could point towards fruitful directions for the field. We know that simple birth-death models do not capture the richness of speciation and extinction across the tree of life, so these models that range beyond birth and death are critical to the growth of comparative methods.
Footnotes
1: Even though this approach requires topology, Morlon et al. (2011) show that their likelihood is equivalent to other approaches, such as Nee and Maddison, that rely only on branching times and ignore topology completely. This is because trees with the same set of branching times but different topologies have identical likelihoods under this model.
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References
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Benton, M. J. 2009. The red queen and the court jester: Species diversity and the role of biotic and abiotic factors through time. Science 323:728–732.
Brock, C. D., L. J. Harmon, and M. E. Alfaro. 2011. Testing for temporal variation in diversification rates when sampling is incomplete and nonrandom. Syst. Biol. 60:410–419. Oxford University Press.
Coyne, J. A., and H. A. Orr. 2004. Speciation. Sinauer, New York.
Cusimano, N., and S. S. Renner. 2010. Slowdowns in diversification rates from real phylogenies may not be real. Syst. Biol. 59:458–464.
De Queiroz, K. 2005. Ernst Mayr and the modern concept of species. Proceedings of the National Academy of Sciences 102:6600–6607. National Acad Sciences.
Emerson, B. C., and N. Kolm. 2005. Species diversity can drive speciation. Nature 434:1015–1017.
Etienne, R. S., and B. Haegeman. 2012. A conceptual and statistical framework for adaptive radiations with a key role for diversity dependence. Am. Nat. 180:E75–89.
Etienne, R. S., and J. Rosindell. 2012. Prolonging the past counteracts the pull of the present: Protracted speciation can explain observed slowdowns in diversification. Syst. Biol. 61:204–213.
Etienne, R. S., B. Haegeman, T. Stadler, T. Aze, P. N. Pearson, A. Purvis, and A. B. Phillimore. 2012. Diversity-dependence brings molecular phylogenies closer to agreement with the fossil record. Proc. Biol. Sci. 279:1300–1309.
Etienne, R. S., A. L. Pigot, and A. B. Phillimore. 2016. How reliably can we infer diversity-dependent diversification from phylogenies? Methods Ecol. Evol. 7:1092–1099.
Höhna, S. 2014. Likelihood inference of non-constant diversification rates with incomplete taxon sampling. PLoS One 9:e84184.
Hughes, C., and R. Eastwood. 2006. Island radiation on a continental scale: Exceptional rates of plant diversification after uplift of the Andes. Proc. Natl. Acad. Sci. U. S. A. 103:10334–10339.
Jetz, W., and R. A. Pyron. 2018. The interplay of past diversification and evolutionary isolation with present imperilment across the amphibian tree of life. Nat Ecol Evol 2:850–858.
Lambert, A., H. Morlon, and R. S. Etienne. 2015. The reconstructed tree in the lineage-based model of protracted speciation. J. Math. Biol. 70:367–397.
Losos, J. B., and F. R. Adler. 1995. Stumped by trees? A generalized null model for patterns of organismal diversity. Am. Nat. 145:329–342.
Losos, J. B., and D. Schluter. 2000. Analysis of an evolutionary species–area relationship. Nature 408:847. Macmillian Magazines Ltd.
May, M. R., and B. R. Moore. 2016. How well can we detect lineage-specific diversification-rate shifts? A simulation study of sequential AIC methods. Syst. Biol. 65:1076–1084.
Moen, D., and H. Morlon. 2014. Why does diversification slow down? Trends Ecol. Evol. 29:190–197.
Moore, B. R., S. Höhna, M. R. May, B. Rannala, and J. P. Huelsenbeck. 2016. Critically evaluating the theory and performance of Bayesian analysis of macroevolutionary mixtures. Proc. Natl. Acad. Sci. U. S. A. 113:9569–9574.
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Sukumaran, J., and L. Lacey Knowles. 2017. Multispecies coalescent delimits structure, not species. Proc. Natl. Acad. Sci. U. S. A. 114:1607–1612. National Academy of Sciences. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/12%3A_Beyond_Birth-Death_models/12.0S%3A_12.S%3A_Beyond_Birth-Death_Models_%28Summary%29.txt |
Many evolutionary models postulate a link between species characteristics and speciation, extinction, or both. These hypotheses can be tested using state-dependent diversification models, which explicitly consider the possibility that species’ characters affect their diversification rates. State-dependent models as currently implemented have some potential problems, but there are methods to deal with these critiques. The overall ability of state-dependent models to explain broad patterns of evolutionary change remains to be determined, but represents a promising avenue for future research.
• 13.1: The Evolution of Self-Incompatibility
Some species of angiosperms can avoid self-fertilization through self-incompatibility. In plants with self-incompatibility, the process by which the sperm meets the egg is interrupted at some stage if pollen grains have a genotype that is the same as the parent. This prevents self-fertilization – and also prevents sexual reproduction with plants that have the same genotype(s) at loci involved in the process.
• 13.2: A State-Dependent Model of Diversification
The models that we will consider in this chapter include trait evolution and associated lineage diversification. In the simplest case, we can consider a model where the character has two states, 0 and 1, and diversification rates depend on those states. We need to model the transitions among these states, which we can do in an identical way to what we did previously with a continuous-time Markov model.
• 13.3: Calculating Likelihoods for State-Dependent Diversification Models
To calculate likelihoods for state-dependent diversification models we use a pruning algorithm with calculations that progress back through the tree from the tips to the root. We have already used this approach to derive likelihoods for constant rate birth-death models on trees and this derivation is similar.
• 13.4: ML and Bayesian Tests for State-Dependent Diversification
Now that we can calculate the likelihood for state-dependent diversification models, formulating ML and Bayesian tests follows the same pattern we have encountered before. For ML, some comparisons are nested and so you can use likelihood ratio tests.
• 13.5: Potential Pitfalls and How to Avoid Them
The most serious limitation of state-dependent models as currently implemented is that they consider only a relatively small set of possible models. In particular, the approach we describe above compares two models: first, a model where birth and death rates are constant and do not depend on the state of the character; and second, a model where birth and death rates depend only on the character state.
• 13.S: Characters and Diversification Rates (Summary)
13: Characters and Diversification Rates
Most people have not spent a lot of time thinking about the sex lives of plants. The classic mode of sexual reproduction in angiosperms (flowering plants) involves pollen (the male gametophyte stage of the plant life cycle). Pollen lands on the pistil (the female reproductive structure) and produces a pollen tube. Sperm cells move down the pollen tube, and one sperm cell unites with the egg to form a new zygote in the ovule.
As you might imagine, plants have little control over what pollen grains land on their pistil (although plant species do have some remarkable adaptations to control pollination by animals; see Anders Nilsson 1992). In particular, this "standard" mode of reproduction leaves open the possibility of self-pollination, where pollen from a plant fertilizes eggs from the same plant (Stebbins 1950). Self-fertilization (sometimes called selfing) is a form of asexual reproduction, but one that involves meiosis; as such, there are costs to self-fertilization. The main cost is inbreeding depression, a reduction in offspring fitness associated with recessive deleterious alleles across the genome (Holsinger et al. 1984).
Some species of angiosperms can avoid self-fertilization through self-incompatibility (Bateman 1952). In plants with self-incompatibility, the process by which the sperm meets the egg is interrupted at some stage if pollen grains have a genotype that is the same as the parent (e.g. Schopfer et al. 1999). This prevents selfing – and also prevents sexual reproduction with plants that have the same genotype(s) at loci involved in the process.
Species of angiosperms are about evenly divided between these two states of self-compatibility and self-incompatibility (Igic and Kohn 2006). Furthermore, self-incompatible species are scattered throughout the phylogenetic tree of angiosperms (Igic and Kohn 2006).
The evolution of selfing is a good example of a trait that might have a strong effect on diversification rates by altering speciation, extinction, or both. One can easily imagine, for example, how incompatibility loci might facilitate the evolution of reproductive isolation among populations, and how lineages with such loci might diversify at a very different tempo than those without (Goldberg et al. 2010).
In this chapter, we will learn about a family of models where traits can affect diversification rates. I will also address some of the controversial aspects of these models and how we can improve these approaches in the future.
13.02: A State-Dependent Model of Diversification
The models that we will consider in this chapter include trait evolution and associated lineage diversification. In the simplest case, we can consider a model where the character has two states, 0 and 1, and diversification rates depend on those states. We need to model the transitions among these states, which we can do in an identical way to what we did in Chapter 7 using a continuous-time Markov model. We express this model using two rate parameters, a forward rate q01 and a backwards rate q10.
We now consider the idea that diversification rates might depend on the character state. We assume that species with character state 0 have a certain speciation rate (λ0) and extinction rate (μ0), and that species in 1 have potentially different rates of both speciation (λ1) and extinction (μ1). That is, when the character evolves, it affects the rate of speciation and/or extinction of the lineages. Thus, we have a six-parameter model (Maddison et al. 2007). We assume that parent lineages give birth to daughters with the same character state, that is that character states do not change at speciation.
It is straightforward to simulate evolution under our state-dependent model of diversification. We proceed in the same way as we did for birth-death models, by drawing waiting times, but these waiting times can be waiting times to the next character state change, speciation, or extinction event. In particular, imagine that there are n lineages present at time t, and that k of these lineages are in state 0 (and n − k are in state 1). The waiting time to the next event will follow an exponential distribution with a rate parameter of:
$ρ = k(q_{01} + λ_0 + μ_0)+(n − k)(q_{10} + λ_1 + μ_1) \label{13.1}$
This equation says that the total rate of events is the sum of the events that can happen to lineages with state 0 (state change to 1, speciation, or extinction) and the analogous events that can happen to lineages with state 1. Once we have a waiting time, we can assign an event type depending on probabilities. For example, the probability that the event is a character state change from 0 to 1 is:
$p_{q_{01}} = (n ⋅ q_{01})/ρ\label{13.2}$
And the probability that the event is the extinction of a lineage with character state 1 is:
$p_{μ_1} = \dfrac{(n − k)⋅μ_1}{ρ} \label{13.3}$
And so on for the other four possible events.
Once we have picked an event in this way, we can randomly assign it to one of the lineages in the appropriate state, with each lineage equally likely to be chosen. We then proceed forwards in time until we have a dataset with the desired size or total time depth.
An example simulation is shown in Figure 13.1. As you can see, under these model parameters the impact of character states on diversification is readily apparent. In the next section we will figure out how to extract that information from our data. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/13%3A_Characters_and_Diversification_Rates/13.01%3A_The_Evolution_of_Self-Incompatibility.txt |
To calculate likelihoods for state-dependent diversification models we use a pruning algorithm with calculations that progress back through the tree from the tips to the root. We will follow the description of this algorithm in Maddison et al. (2007). We have already used this approach to derive likelihoods for constant rate birth-death models on trees (Chapter 12), and this derivation is similar.
We consider a phylogenetic tree with data on the character states at the tips. For the purposes of this example, we will assume that the tree is complete and correct – we are not missing any species, and there is no phylogenetic uncertainty. We will come back to these two assumptions later in the chapter.
We need to obtain the probability of obtaining the data given the model (the likelihood). As we have seen before, we will calculate that likelihood going backwards in time using a pruning algorithm (Maddison et al. 2007). The key principle, again, is that if we know the probabilities at some point in time on the tree, we can calculate those probabilities at some time point immediately before. By applying this method successively, we can move back towards the root of the tree. We move backwards along each branch in the tree, merging these calculations at nodes. When we get to the root, we have the probability of the data given the model and the entire tree – that is, we have the likelihood.
The other essential piece is that we have a starting point. When we start at the tips of the tree, we assume that our character states are fixed and known. We use the fact that we know all of the species and their character states at the present day as our starting point, and move backwards from there (Maddison et al. 2007). For example, for a species with character state 0, the likelihood for state 0 is 1, and for state 1 is zero. In other words, at the tips of the tree we can start our calculations with a probability of 1 for the state that matches the tip state, and 0 otherwise.
This discussion also highlights the fact that incorporating uncertainty and/or variation in tip states for these algorithms is not computationally difficult – we just need to start from a different point at the tips. For example, if we are completely unsure about the tip state for a certain taxa, we can begin with likelihoods of 0.5 for starting in state 0 and 0.5 for starting in state 1. However, such calculations are not commonly implemented in comparative methods software.
We now need to consider the change in the likelihood as we step backwards through time in the tree (Maddison et al. 2007). We will consider some very small time interval Δt, and later use differential equations to find out what happens in the limit as this interval goes to zero (Figure 13.2). Since we will eventually take the limit as Δt → 0, we can assume that the time interval is so small that, at most, one event (speciation, extinction, or character change) has happened in that interval, but never more than one. We will calculate the probability of the observed data given that the character is in each state at time t, again measuring time backwards from the present day. In other words, we are considering the probability of the observed data if, at time t, the character state were in state 0 [p0(t)] or state 1 [p1(t)]. For now, we can assume we know these probabilities, and try to calculate updated probabilities at some earlier time t + Δt: p0(t + Δt) and p1(t + Δt).
To calculate p0(t + Δt) and p1(t + Δt), we consider all of the possible things that could happen in a time interval Δt along a branch in a phylogenetic tree that are compatible with our dataset (Figure 13.2; Maddison et al. 2007). First, nothing at all could have happened; second, our character state could have changed; and third, there could have been a speciation event. This last event might seem incorrect, as we are only considering changes along branches in the tree and not at nodes. If we did not reconstruct any speciation events at some point along a branch, then how could one have taken place? The answer is that a speciation event could have occurred but all taxa descended from that branch have since gone extinct. We must also consider the possibility that either the right or the left lineage went extinct following the speciation event; that is why the speciation event probabilities appear twice in Figure 13.2 (Maddison et al. 2007).
We can write an equation for these updated probabilities. We will consider the probability that the character is in state 0 at time t + Δt; the equation for state 1 is similar (Maddison et al. 2007).
\begin{aligned} p_0 (t+\Delta t)=(1-\mu_0 )\Delta t \cdot [(1-q_{01} \Delta t)(1-\lambda_0 \Delta t) p_0 (t)+q_{01} \Delta t(1-\lambda_0 \Delta t) \ p_1 (t)+2 \cdot (1-q_{01} \Delta t) \lambda_0 \Delta t \cdot E_0 (t) p_0 (t)] \end{aligned} \label{13.4}
We can multiply through and simplify. We will also drop any terms that include [Δt]2, which become vanishingly small as Δt decreases. Doing that, we obtain (Maddison et al. 2007):
$p_0(t + Δt)=[1 − (λ_0 + μ_0 + q_{01})Δt]p_0(t)+(q_{01}Δt)p_1(t)+2(λ_0Δt)E_0(t)p_0(t) \label{13.5}$
Similarly,
$p_1(t + Δt)=[1 − (λ_1 + μ_1 + q_{10})Δt]p_1(t)+(q_{10}Δt)p_0(t)+2(λ_1Δt)E_1(t)p_1(t) \label{13.6}$
We can then find the instantaneous rate of change for these two equations by solving for p1(t + Δt)/[Δt], then taking the limit as Δt → 0. This gives (Maddison et al. 2007):
$\frac{dp_0}{dt} = -(\lambda_0 + \mu_0 + q_{01}) p_0(t) + q{01}p_1(t) + 2 \lambda_0 E_0(t) p_0(t) \label{13.7}$
and:
$\frac{dp_1}{dt} = -(\lambda_1 + \mu_1 + q_{10}) p_1(t) + q{10}p_1(t) + 2 \lambda_1 E_1(t) p_1(t) \label{13.8}$
We also need to consider E0(t) and E1(t). These represent the probability that a lineage with state 0 or 1, respectively, and alive at time t will go extinct before the present day. Neglecting the derivation of these formulas, which can be found in Maddison et al. (2007) and is closely related to similar terms in Chapters 11 and 12, we have:
$\frac{dE_0}{dt} = \mu_0-(\lambda_0+\mu_0+q_{01} ) E_0 (t)+q_{01} E_1 (t)+\lambda_0 [E_0 (t)]^2 \label{13.9}$
and:
$\frac{dE_1}{dt} = \mu_1-(\lambda_1+\mu_1+q_{10} ) E_1 (t)+q_{10} E_0 (t)+\lambda_1 [E_1 (t)]^2 \label{13.10}$
Along a single branch in a tree, we can sum together many such small time intervals. But what happens when we get to a node? Well, if we consider the time interval that contains the node, then we already know what happened – a speciation event. We also know that the two daughters immediately after the speciation event were identical in their traits (this is an assumption of the model). So we can calculate the likelihood for their ancestor for each state as the product of the likelihoods of the two daughter branches coming into that node and the speciation rate (Maddison et al. 2007). In this way, we merge our likelihood calculations along each branch when we get to nodes in the tree.
When we get to the root of the tree, we are almost done – but not quite! We have partial likelihood calculations for each character state – so we know, for example, the likelihood of the data if we had started with a root state of 0, and also if we had started at 1. To merge these we need to use probabilities of each character state at the root of the tree (Maddison et al. 2007). For example, if we do not know the root state from any outside information, we might consider root probabilities for each state to be equal, 0.5 for state 0 and 0.5 for state 1. We then multiply the likelihood associated with each state with the root probability for that state. Finally, we add these likelihoods together to obtain the full likelihood of the data given the model.
The question of which root probabilities to use for this calculation has been discussed in the literature, and does matter in some applications. Aside from equal probabilities of each state, other options include using outside information to inform prior probabilities on each state (e.g. Hagey et al. 2017), finding the calculated equilibrium frequency of each state under the model (Maddison et al. 2007), or weighting each root state by its likelihood of generating the data, effectively treating the root as a nuisance parameter (FitzJohn et al. 2009).
I have described the situation where we have two character states, but this method generalizes well to multi-state characters (the MuSSE method; FitzJohn 2012). We can describe the evolution of the character in the same way as described for multi-state discrete characters in chapter 9. We then can assign unique diversification rate parameters to each of the k character states: λ0, λ1, …, λk and μ0, μ1, …, μk (FitzJohn 2012). It is worth keeping in mind, though, that it is not too hard to construct a model where parameters are not identifiable and model fitting and estimation become very difficult. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/13%3A_Characters_and_Diversification_Rates/13.03%3A_Calculating_Likelihoods_for_State-Dependent_Diversification_Models.txt |
Now that we can calculate the likelihood for state-dependent diversification models, formulating ML and Bayesian tests follows the same pattern we have encountered before. For ML, some comparisons are nested and so you can use likelihood ratio tests. For example, we can compare the full BiSSe model (Maddison et al. 2007), with parameters q01, q10, λ0, λ1, μ0, μ1 with a restricted model with parameters q01, q10, λall, μall. Since the restricted model is a special case of the full model where λ0 = λ1 = λall and μ0 = μ1 = μall, we can compare the two using a likelihood ratio test, as described earlier in the book. Alternatively, we can compare a series of BiSSe-type models by comparing their AICc scores.
For example, I will apply this approach to the example of self-incompitability. I will use data from Goldberg and Igic (2012), who provide a phylogenetic tree and data for 356 species of Solanaceae. All species were classified as having any form of self incompatibility, even if the state is variable among populations. The data, along with a stochastic character map of state changes, are shown in Figure 13.4. Applying the BiSSe models to these data and assuming that q01 ≠ q10, we obtain the following results:
Model Number of parameters Parameter estimates lnL AIC
Character-independent model 4 λ = 0.65, μ = 0.57 -945.96 1899.9
q01 = 0.16, q10 = 0.09
Speciation rate depends on character 5 μ = 0.57 -945.57 1901.1
λ0 = 0.69, λ1 = 0.63
q01 = 0.17, q10 = 0.08
Extinction rate depends on character 5 λ = 0.65 -943.93 1897.9
μ0 = 0.45, μ1 = 0.67
q01 = 0.22, q10 = 0.06
Full character-dependent model 6 λ0 = 0.49, λ1 = 0.79 -941.94 1895.9
μ0 = 0.20, μ1 = 0.84
q01 = 0.29, q10 = 0.05
From this, we conclude that models where the character influences diversification fit best, with the full model receiving the most support. We can't discount the possibility that the character only influences extinction and not speciation, since that model is within 2 AIC units of the best model.
Alternatively, we can carry out a Bayesian test for state-dependent diversification.
Bayesian test for state-dependent diversification
Like other models in the book, this requires setting up an MCMC algorithm that samples the posterior distributions of our model parameters (FitzJohn 2012). In this case:
1. Sample a set of starting parameter values, q01, q10, λ0, λ1, μ0, μ1, from their prior distributions. For example, one could set prior distribution for all parameters as exponential with a mean and variance of λpriori (note that, as usual, the choice for this parameter should depend on the units of tree branch lengths you are using). We then select starting values for all parameters from the prior.
2. Given the current parameter values, select new proposed parameter values using the proposal density Q(p′|p). For all parameter values, we can use a uniform proposal density with width wp, so that Q(p′|p) U(p − wp/2, p + wp/2). We can either choose all parameter values simultaneously, or one at a time (the latter is typically more effective).
3. Calculate three ratios:
• The prior odds ratio. This is the ratio of the probability of drawing the parameter values p and p′ from the prior. Since we have exponential priors for all parameters, we can calculate this ratio as: \[ R_{prior} = \frac{\lambda_{prior_i} e^{-\lambda_{prior_i} p'}}{\lambda_{prior_i} e^{-\lambda_{prior_i} p}}=e^{\lambda_{prior_i} (p-p')} \label{13.11}\[
• The proposal density ratio. This is the ratio of probability of proposals going from p to p′ and the reverse. We have already declared a symmetrical proposal density, so that Q(p′|p)=Q(p|p′) and Rproposal = 1.
• The likelihood ratio. This is the ratio of probabilities of the data given the two different parameter values. We can calculate these probabilities from the approach described in the previous section.
4. Find Raccept as the product of the prior odds, proposal density ratio, and the likelihood ratio. In this case, the proposal density ratio is 1, so (eq. 13.12):
Raccept = Rprior ⋅ Rlikelihood | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/13%3A_Characters_and_Diversification_Rates/13.04%3A_ML_and_Bayesian_Tests_for_State-Dependent_Diversification.txt |
Recently, a few papers have been published that are critical of state-dependent diversification models (Rabosky and Goldberg 2015, Maddison and FitzJohn (2015)). These papers raise substantive critiques that are important to address when applying the methods described in this chapter to empirical data. In this section I will attempt to describe the critiques and their potential remedies.
The most serious limitation of state-dependent models as currently implemented is that they consider only a relatively small set of possible models. In particular, the approach we describe above compares two models: first, a model where birth and death rates are constant and do not depend on the state of the character; and second, a model where birth and death rates depend only on the character state (Maddison et al. 2007). But there is another possibility that might be (in general) more common than either of the models we consider: birth and death rates vary, but in a way that is not dependent on the particular character we have chosen to analyze. I say that this is probably a common pattern because we know that birth and death rates vary tremendously across lineages in the tree of life (Alfaro et al. 2009), and it seems probable to me that many of our hypotheses about which characters might contribute to that variation are, at this point, stabs in the dark.
This issue is a normal one for statistical analyses – after all, there are always other models outside of our set of considered possibilities. However, in this case, the fact that state-dependent diversification models fail to consider the possibility outlined above causes a very particular – and peculiar – problem: if we apply the tests to empirical phylogenetic trees, even with made-up data, we almost always find statistically significant results (Rabosky and Goldberg 2015). For example, Rabosky and Goldberg (2015) found that there is very often a statistically significant “signal” that the number of letters in a species name is significantly associated with speciation rates across a range of empirical datasets. This result might seem ridiculous and puzzling, as there is no way that species name length should be associated with the diversification processes. However, if we return to our alternative model above, then the results make sense. Rabosky and Goldberg (2015) simulated character evolution on real phylogenetic trees, and their results do not hold when the trees are simulated along with the characters (this is also why Rabosky and Goldberg’s (2015) results do not represent “type I errors,” contra their paper, because the data are not simulated under the null hypothesis). On these real trees, speciation and/or extinction rates vary across clades. Among the two models that the authors consider, both are wrong; speciation and extinction are independent of the character but not constant through time. Of the two alternatives, the state-dependent model tends to fit better because, from a statistical point of view, it is important for the model to capture some variation in birth and death rates across clades. Even a random character will pick up some of this variation, so that the alternative model tends to fit better than the null – even though, in this case, the character has nothing to do with diversification!
Fortunately, there are a number of ways to deal with this problem. First, one can compare the statistical support for the state-dependent model with the support that one obtains for random data. The random data could be simulated on the tree, or one could permute the tips or draw random data from a multinomial distribution (Rabosky and Goldberg 2015). One can then compare, for example, the distribution of ΔAICc scores obtained from these permutations to the ΔAICc for the original data. There are also semi-parametric methods based on permutations that have similar statistical properties (Rabosky and Goldberg 2017). Alternatively, we could explicitly consider the possibility that some unmeasured character is actually the thing that is influencing diversification rates (Beaulieu and O’Meara 2016). This latter approach is the most elegant as we can directly add the model described in this section to our list of candidates (see Beaulieu and O’Meara 2016).
A more general critique of state-dependent models of diversification was raised by Maddison and Fitzjohn (Maddison and FitzJohn 2015). This paper pointed out that statistically significant results for these tests can be driven by an event on a single branch of a tree, and therefore be unreplicated. This is a good criticism that applies equally well to a range of comparative methods. We can deal with this critique, in part, by making sure the events we test are replicated in our data. Together, both of these critiques argue for a stronger set of model adequacy approaches in comparative methods.
13.0S: 13.S: Characters and Diversification Rates (Summary)
Many evolutionary models postulate a link between species characteristics and speciation, extinction, or both. These hypotheses can be tested using state-dependent diversification models, which explicitly consider the possibility that species’ characters affect their diversification rates. State-dependent models as currently implemented have some potential problems, but there are methods to deal with these critiques. The overall ability of state-dependent models to explain broad patterns of evolutionary change remains to be determined, but represents a promising avenue for future research.
References
Alfaro, M. E., F. Santini, C. Brock, H. Alamillo, A. Dornburg, D. L. Rabosky, G. Carnevale, and L. J. Harmon. 2009. Nine exceptional radiations plus high turnover explain species diversity in jawed vertebrates. Proceedings of the National Academy of Sciences 106:13410–13414. National Acad Sciences.
Anders Nilsson, L. 1992. Orchid pollination biology. Trends Ecol. Evol. 7:255–259.
Bateman, A. J. 1952. Self-incompatibility systems in Angiosperms. Heredity 6:285. The Genetical Society of Great Britain.
Beaulieu, J. M., and B. C. O’Meara. 2016. Detecting hidden diversification shifts in models of Trait-Dependent speciation and extinction. Syst. Biol. 65:583–601.
FitzJohn, R. G. 2012. Diversitree: Comparative phylogenetic analyses of diversification in R. Methods Ecol. Evol. 3:1084–1092.
FitzJohn, R. G., W. P. Maddison, and S. P. Otto. 2009. Estimating trait-dependent speciation and extinction rates from incompletely resolved phylogenies. Syst. Biol. 58:595–611. sysbio.oxfordjournals.org.
Goldberg, E. E., and B. Igić. 2012. Tempo and mode in plant breeding system evolution. Evolution 66:3701–3709. Wiley Online Library.
Goldberg, E. E., J. R. Kohn, R. Lande, K. A. Robertson, S. A. Smith, and B. Igić. 2010. Species selection maintains self-incompatibility. Science 330:493–495.
Hagey, T. J., J. C. Uyeda, K. E. Crandell, J. A. Cheney, K. Autumn, and L. J. Harmon. 2017. Tempo and mode of performance evolution across multiple independent origins of adhesive toe pads in lizards. Evolution 71:2344–2358.
Holsinger, K. E., M. W. Feldman, and F. B. Christiansen. 1984. The evolution of self-fertilization in plants: A population genetic model. Am. Nat. 124:446–453.
Igic, B., and J. R. Kohn. 2006. Bias in the studies of outcrossing rate distributions. Evolution 60:1098–1103.
Maddison, W. P., and R. G. FitzJohn. 2015. The unsolved challenge to phylogenetic correlation tests for categorical characters. Syst. Biol. 64:127–136.
Maddison, W. P., P. E. Midford, S. P. Otto, and T. Oakley. 2007. Estimating a binary character’s effect on speciation and extinction. Syst. Biol. 56:701–710. Oxford University Press.
Rabosky, D. L., and E. E. Goldberg. 2017. FiSSE: A simple nonparametric test for the effects of a binary character on lineage diversification rates. Evolution 71:1432–1442.
Rabosky, D. L., and E. E. Goldberg. 2015. Model inadequacy and mistaken inferences of trait-dependent speciation. Syst. Biol. 64:340–355.
Schopfer, C. R., M. E. Nasrallah, and J. B. Nasrallah. 1999. The male determinant of self-incompatibility in Brassica. Science 286:1697–1700.
Stebbins, G. L. 1950. Variation and evolution in plants. Geoffrey Cumberlege.; London. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/13%3A_Characters_and_Diversification_Rates/13.05%3A_Potential_Pitfalls_and_How_to_Avoid_Them.txt |
Comparative methods occupy a central place in evolutionary biology. This is because phylogenies provide an accounting of the historical patterns of evolution and, in turn, give us a natural way to measure long-term evolutionary dynamics. The first phase of comparative methods was focused strongly on adaptation. As discussed in this book, we have now branched out into a wide number of new areas, including diversification, community ecology, quantitative genetics, and more. This expansion has involved new statistical approaches that increase the flexibility of comparative methods and their connection to biological processes. I expect this trend to continue, fueled by the creativity and energy of the next crop of young scientists.
• 14.1: The Lorax
Comparative methods have been rushing forward at breakneck speed to “speak for the trees” for more than 20 years now. At the same time, we have gained more information about the shape of the tree of life than any time in the history of the planet. So, what have we learned so far? And what can we learn moving forward? Perhaps most importantly, how can we overcome the perceived and actual limits of comparative approaches, and enable new breakthroughs in our understanding of evolutionary biology?
• 14.2: What We Have Learned so Far
The great success of comparative methods has been, I think, in testing hypotheses about adaptation. A variety of methods can be applied to test for evolutionary relationships between form and the environment – and, increasingly, organismal function. These methods applied to real data have shed great light onto the myriad ways that species can adapt. This has been a great boon to organismal biology, and comparative methods are now routinely used to analyze and test hypotheses of adaptation.
• 14.3: Where Can we go Next?
Comparative methods have proven to be an essential tool in identifying and describing adaptations. However, the scope of comparative methods has broadened, and now seeks to address long-standing theories of macroevolution. It is in this area that I think comparative methods has promise, but awaits new developments and ideas to really make progress towards the future.
• 14.4: A hint at the future of comparative methods
It is perilous to predict the future progress of science. Nonetheless, I will offer a few suggestions that I think might be productive avenues for work in comparative methods.
• 14.S: What have we learned from the trees? (Summary)
14: What have we learned from the trees
In Dr. Seuss’s “The Lorax” there is a quote that seems appropriate to begin my final chapter (Seuss 1971).
‘Mister,’ he said with a sawdusty sneeze,
‘I am the Lorax, I speak for the trees.
I speak for the trees, for the trees have no tongues.”
Comparative methods have been rushing forward at breakneck speed to “speak for the trees” for more than 20 years now. At the same time, we have gained more information about the shape of the tree of life than any time in the history of the planet. So, what have we learned so far? And what can we learn moving forward? Perhaps most importantly, how can we overcome the perceived and actual limits of comparative approaches, and enable new breakthroughs in our understanding of evolutionary biology?
14.02: What We Have Learned so Far
The great success of comparative methods has been, I think, in testing hypotheses about adaptation. A variety of methods can be applied to test for evolutionary relationships between form and the environment – and, increasingly, organismal function. These methods applied to real data have shed great light onto the myriad ways that species can adapt. This has been a great boon to organismal biology, and comparative methods are now routinely used to analyze and test hypotheses of adaptation across the tree of life. Methods for detecting adaptation using comparative approaches are growing increasingly sophisticated in terms of the types of data that they can handle, including massively multivariate gene expression data, function-valued trait data, and data from genome sequencing. One can only expect this trend to continue.
One thing seems certain after a few decades of comparative analysis: the tempo of evolution is incredibly variable. Rates of evolution vary both through time and across clades, with the quickest rates of both trait evolution and speciation thousands of times faster than the slowest rates. We can see this variation in analyses from relatively simple tree balance tests to sophisticated Bayesian analyses. So, evolution does not tick along like a clock; instead, rates of evolution depend strongly on lineage, time, and place. The details of these relationships, though, remain to be deciphered.
Comparative methods have played a critical role in our understanding of speciation. Studies using lineage-through-time plots have greatly enhanced our knowledge of diversification rates, and a wide range of results have shown increasing evidence for diversity-dependence in speciation (though this interpretation is not without controversy!). This set of studies provide a nice complement to paleobiological studies of diversification rates using the fossil record.
We can already gain some new biological insights as comparisons among clades start to hint at which factors are responsible for the fact that some species are so much more diverse than others. Perhaps for psychological reasons, most studies have tried to determine explanations for the fastest rates of speciation, as seen in young diverse clades like African cichlids and Andean plants. However, given the high potential for speciation and splitting to accumulate species in a geographic landscape, it might be true that the depauperate clades are really the mysterious parts of the tree of life. Many current research programs are aimed directly at explaining differences in diversity across both narrow and broad phylogenetic scales.
Overall, I think it is easy to see why comparative methods have risen to their current prominence in evolutionary biology. Phylogenetic trees provide a natural way to test evolutionary hypotheses over relatively long time scales without requiring any direct historical information. They have been applied across the tree of life to help scientists understand how species adapt and multiply over long time scales. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/14%3A_What_have_we_learned_from_the_trees/14.01%3A_The_Lorax.txt |
As emphasized by Harvey and Pagel (1991), comparative methods have proven to be an essential tool in identifying and describing adaptations. However, the scope of comparative methods has broadened, and now seeks to address long-standing theories of macroevolution. It is in this area that I think comparative methods has promise, but awaits new developments and ideas to really make progress towards the future.
The main challenge, I think, is in identifying and testing broad theories of macroevolution. Too many papers focus on “classic” verbal models of macroevolution – many of which have been defined in contradictory ways over the years and can never really be tested. At the same time, new quantitative theories of macroevolution are lacking.
Let me explore this in a bit more detail using the idea of adaptive radiation and the related concept of ecological opportunity. Perhaps, the theory goes, occasionally lineages enter a new adaptive zone full of niches just waiting to be occupied; the lineage then evolves rapidly to fill those niches. Based on this definition, there are several sets of criteria that one might apply to decide whether or not a particular lineage has experienced such an adaptive radiation. There are a few alternatives that are sometimes contrasted to this pattern, like nonadaptive radiation.
The concept of adaptive radiation has been very fruitful for inspiring creative and interesting work on model clades, but (in my opinion) we have mostly failed in terms of really predicting adaptive radiations and separating the phenomenon from ‘normal’ evolution. For example, most studies identify lineages undergoing adaptive radiation a priori. Even when the goal is to identify adaptive radiations, some criteria seem hard to pin down; for example, one can require evidence of adaptation, but surely every lineage on the planet has experienced selection and adapted in at least some way over its history. Likewise, we can require common descent, but there is only one tree of life on Earth (that we know!), so eventually one will find that as well. And authors differ dramatically on whether or not adaptive radiations need to be rapid relative to trait evolution and/or lineage diversification in other lineages. Finally, although a few studies have been able to characterize the unique features of adaptive radiations compared to their close relatives, comparisons across broader sections of the tree of life have mostly failed. We still do not know for certain if there is anything that links the “classic” adaptive radiations (e.g. anoles, Darwin’s finches, mammals) and distinguishes them from evolution in normal clades.
Comparative methods have cast doubt on another cornerstone of macroevolution, that of punctuated equilibrium. As we have argued, comparative methods have had varying success in tackling each of the parts of PE theory, but we can see little evidence to link them into a cohesive whole. For one thing, there is too much evidence that lineages adapt and evolve along branches of phylogenetic trees, rather than just at speciation. Quantitative tests do tend to find some statistical support for the idea that change depends on both anagenesis and speciation, but “pure” punctuated equilibrium is increasingly hard to defend.
As for other major macroevolutionary theories, some have received mixed support (e.g. Dollo’s law, escape-and-radiation, cospeciation, key innovations), while others have hardly been tested in a comparative framework, probably due to a lack of methods (the geographic mosaic theory, holey adaptive landscapes). | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/14%3A_What_have_we_learned_from_the_trees/14.03%3A_Where_Can_we_go_Next%3F.txt |
It is perilous to predict the future progress of science. Nonetheless, I will offer a few suggestions that I think might be productive avenues for work in comparative methods.
First, I think comparative methods can and should do a better job of integrating diverse data into a coherent framework. For example, despite clear connections, neither fossils nor contemporary data on the tempo and genetics of speciation typically can be integrated with phylogenetic studies of diversification (Rabosky and Matute 2013). Research projects with the same goal, like estimating when and why a lineage undergoes speciation, are better integrated than separate. There are a few hints about how to proceed: first, speciation models that we fit to both phylogenetic and fossil data must be better connected to the process of speciation; and second, analyses need to consider both paleontological and phylogenetic data simultaneously.
Second, it is absolutely essential to fully deal with uncertainty through entire pipelines of comparative analysis, from tree building to model fitting. The easiest way to do this is through a single integrated Bayesian framework, although using each step’s posterior as a prior is nearly as good. Even if one is not a Bayesian, I think it is critical to test how tree uncertainty might affect the results of our comparative analyses.
Third, comparative methods require a more diverse set of models that are better linked to biological processes. Current models like Brownian motion and OU have, at best, a weak and many-to-one connection to microevolutionary models. Other models are even more abstract; nothing we can measure about an evolving lineage from one generation to the next, for example, can inform us about the meaning of the lambda parameter from a PGLS analysis. This can be fine statistically, but I think we can do better. The easiest connections to make are between comparative methods and quantitative genetics. In this book I explore only the most basic aspects of this connection. More could, and should, be done. For example, no trait models that I know of deal with differences in abundance and range size among species, even though these vary tremendously among even very close relatives and are almost certain to affect the tempo and mode of trait evolution. Here we can look to other fields like ecology for inspiration.
14.0S: 14.S: What have we learned from the trees? (Summary)
Comparative methods occupy a central place in evolutionary biology. This is because phylogenies provide an accounting of the historical patterns of evolution and, in turn, give us a natural way to measure long-term evolutionary dynamics. The first phase of comparative methods was focused strongly on adaptation. As discussed in this book, we have now branched out into a wide number of new areas, including diversification, community ecology, quantitative genetics, and more. This expansion has involved new statistical approaches that increase the flexibility of comparative methods and their connection to biological processes. I expect this trend to continue, fueled by the creativity and energy of the next crop of young scientists.
References
Harvey, P. H., and M. D. Pagel. 1991. The comparative method in evolutionary biology. Oxford University Press.
Rabosky, D. L., and D. R. Matute. 2013. Macroevolutionary speciation rates are decoupled from the evolution of intrinsic reproductive isolation in drosophila and birds. Proc. Natl. Acad. Sci. U. S. A. 110:15354–15359. National Acad Sciences.
Seuss. 1971. The Lorax. Random House. | textbooks/bio/Evolutionary_Developmental_Biology/Phylogenetic_Comparative_Methods_(Harmon)/14%3A_What_have_we_learned_from_the_trees/14.04%3A_A_hint_at_the_future_of_comparative_methods.txt |
Thumbnail: Punnett Square. (CC BY-SA 3.0; Pbroks13 via Wikimedia Commons).
01: Introduction to heredity
Heredity and Classical Genetics. Dominant and recessive traits. Heterozygous and homozygous genotypes. Created by Sal Khan.
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• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
1.02: Alleles and genes
A gene as a stretch of DNA on a chromosome. Alleles as versions (sequence variants) of a gene.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
1.03: Worked example - Punnett squares
Learn how to use Punnett squares to calculate probabilities of different phenotypes. Includes worked examples of dihybrid crosses, independent assortment, incomplete dominance, codominance, and multiple alleles. Created by Sal Khan.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
1.04: Mendel and his peas
How can we study inheritance?
When spending time with your own family, friends, and neighbors, you may have noticed that many traits run in families. For instance, members of a family may share similar facial features, hair color (like the brother and sister below), or a predisposition to health problems such as diabetes. Characteristics that run in families often have a genetic basis, meaning that they depend on genetic information a person inherits from his or her parents.
What if you wanted to figure out how genetic information is transmitted between generations? For instance, you might be curious how traits can "skip" a generation, or why one child in a family may suffer from a genetic disease while another does not. How could you go about asking these kinds of questions scientifically?
An obvious first idea would be to study human inheritance patterns directly, but that turns out to be a tricky proposition (see the pop-up below for details). In this article, we'll see how a nineteenth-century monk named Gregor Mendel instead uncovered the key principles of inheritance using a simple, familiar system: the pea plant.
[Why didn't Mendel study humans?]
Many of us are curious about genetics because we want to understand human inheritance. If the burning questions of genetics are about humans, why would anyone study inheritance in something like the pea plant?
When it comes to basic principles of inheritance, humans may be the organisms we most want to learn about, but they aren't always the best organisms to study experimentally. For instance, it takes many years for a human being to grow, mature, and have children, making data collection slow. Also, when humans have children, they often have one or two (rather than, say, several thousand), making it harder to see mathematical patterns in the data. Finally, and perhaps most critically, it wouldn’t be possible (or ethical) to set up controlled experiments in human genetics – that is, you couldn’t ask a pair of people to have children just because you were curious what those children would look like.
Peas, on the other hand, grow quickly, make many seeds, and can be bred to one another in a simple, controlled way. In other words, they’re good organisms to study in many ways that humans aren’t. Using pea plants, Mendel was able to uncover fundamental principles of inheritance that apply to many different kinds of organisms. Although these principles would have been nearly impossible to deduce from human family trees alone, they form the core of the modern field of human genetics. Mendel’s laws, and later findings building on those laws, allow us to understand and predict the inheritance of many human traits, including genetic disorders.
The monk in the garden: Gregor Mendel
Johann Gregor Mendel (1822–1884), often called the “father of genetics,” was a teacher, lifelong learner, scientist, and man of faith. It would be fair to say that Mendel had a lot of grit: he persevered through difficult circumstances to make some of the most important discoveries in biology.
As a young man, Mendel had difficulty paying for his education due to his family's limited means, and he also suffered bouts of physical illness and depression; still, he persevered to graduate from high school and, later, university1. After finishing university, he joined the Augustinian Abbey of St. Thomas in Brno, in what is now the Czech Republic. At the time, the monastery was the cultural and intellectual hub of the region, and Mendel was immediately exposed to new teachings and ideas1.
His decision to join the order (against the wishes of his father, who expected him to carry on the family farm) appears to have been motivated in part by a desire to continue his education and pursue his scientific interests2. Supported by the monastery, he taught physics, botany, and natural science courses at the secondary and university levels.
Research on heredity
In 1856, Mendel began a decade-long research project to investigate patterns of inheritance. Although he began his research using mice, he later switched to honeybees and plants, ultimately settling on garden peas as his primary model system2. A model system is an organism that makes it easy for a researcher to investigate a particular scientific question, such as how traits are inherited. By studying a model system, researchers can learn general principles that apply to other, harder-to-study organisms or biological systems, such as humans.
Mendel studied the inheritance of seven different features in peas, including height, flower color, seed color, and seed shape. To do so, he first established pea lines with two different forms of a feature, such as tall vs. short height. He grew these lines for generations until they were pure-breeding (always produced offspring identical to the parent), then bred them to each other and observed how the traits were inherited.
In addition to recording how the plants in each generation looked, Mendel counted the exact number of plants that showed each trait. Strikingly, he found very similar patterns of inheritance for all seven features he studied:
• One form of a feature, such as tall, always concealed the other form, such as short, in the first generation after the cross. Mendel called the visible form the dominant trait and the hidden form the recessive trait.
• In the second generation, after plants were allowed to self-fertilize (pollinate themselves), the hidden form of the trait reappeared in a minority of the plants. Specifically, there were always about 3 plants that showed the dominant trait (e.g., tall) for every 1 plant that showed the recessive trait (e.g., short), making a 3:1 ratio.
• Mendel also found that the features were inherited independently: one feature, such as plant height, did not influence inheritance of other features, such as flower color or seed shape.
In 1865, Mendel presented the results of his experiments with nearly 30,000 pea plants to the local Natural History Society. Based on the patterns he observed, the counting data he collected, and a mathematical analysis of his results, Mendel proposed a model of inheritance in which:
• Characteristics such as flower color, plant height, and seed shape were controlled by pairs of heritable factors that came in different versions.
• One version of a factor (the dominant form) could mask the presence of another version (the recessive form).
• The two paired factors separated during gamete production, such that each gamete (sperm or egg) randomly received just one factor.
• The factors controlling different characteristics were inherited independently of one another.
We'll take a closer look at how Mendel reached these conclusions in the articles on the law of segregation and the law of independent assortment. In 1866, Mendel published his observations and his model of inheritance, under the title Experiments in Plant Hybridization3,4, in the Proceedings of the Natural History Society of Brünn.
Scientific legacy
Mendel's work went largely unnoticed by the scientific community during his lifetime. How could this have been the case?
In part, Mendel's contemporaries failed to recognize the importance of his work because his findings went against prevailing (popular) ideas about inheritance. In addition, although we now see Mendel's mathematical approach to biology as innovative and pioneering, it was new, unfamiliar, and perhaps confusing or unintuitive to other biologists of the time5.
In the mid-1800s, when Mendel was doing his experiments, most biologists subscribed to the idea of blending inheritance. Blending inheritance wasn't a formal, scientific hypothesis, but rather, a general model in which inheritance involved the permanent blending of parents' characteristics in their offspring (producing offspring with an intermediate form of a characteristic)6. The blending model fit well with some observations of human inheritance: for instance, children often look a bit like both of their parents.
But the blending model could not explain why Mendel crossed a tall and a short pea plant and got only tall plants, or why self-fertilization of one of those tall plants would produce a 3:1 ratio of tall to short plants in the next generation. Instead, if the blending model were correct, a tall plant crossed with a short plant should produce a medium plant, which would go on to produce more medium plants (see below).
As it turns out, both pea plant height and human height (along with many other characteristics in a wide range of organisms) are controlled by pairs of heritable factors that come in distinctive versions, just as Mendel proposed. In humans, however, there are many different factors (genes) that contribute fractionally to height and vary among individuals. This makes it difficult to see the contribution of any one factor and produces inheritance patterns that can resemble blending. In Mendel's experiments, in contrast, there was just one factor that differed between the tall and short pea plants, allowing Mendel to clearly see the underlying pattern of inheritance.
In 1868, Mendel became abbot of his monastery and largely set aside his scientific pursuits in favor of his pastoral duties. He was not recognized for his extraordinary scientific contributions during his lifetime. In fact, it was not until around 1900 that his work was rediscovered, reproduced, and revitalized. Its rediscoverers were biologists on the brink of discovering the chromosomal basis of heredity – that is, about to realize that Mendel's “heritable factors” were carried on chromosomes.
Mendel’s model system: The pea plant
Mendel carried out his key experiments using the garden pea, Pisum sativum, as a model system. Pea plants make a convenient system for studies of inheritance, and they are still studied by some geneticists today.
Useful features of peas include their rapid life cycle and the production of lots and lots of seeds. Pea plants also typically self-fertilize, meaning that the same plant makes both the sperm and the egg that come together in fertilization. Mendel took advantage of this property to produce true-breeding pea lines: he self-fertilized and selected peas for many generations until he got lines that consistently made offspring identical to the parent (e.g., always short).
Pea plants are also easy to cross, or mate in a controlled way. This is done by transferring pollen from the anthers (male parts) of a pea plant of one variety to the carpel (female part) of a mature pea plant of a different variety. To prevent the receiving plant from self-fertilizing, Mendel painstakingly removed all of the immature anthers from the plant’s flowers before the cross.
Because peas were so easy to work with and prolific in seed production, Mendel could perform many crosses and examine many individual plants, making sure that his results were consistent (not just a fluke) and accurate (based on many data points).
Mendel’s experimental setup
Once Mendel had established true-breeding pea lines with different traits for one or more features of interest (such as tall vs. short height), he began to investigate how the traits were inherited by carrying out a series of crosses.
First, he crossed one true-breeding parent to another. The plants used in this initial cross are called the P generation, or parental generation.
Mendel collected the seeds from the P generation cross and grew them up. These offspring were called the F1 generation, short for first filial generation. (Filius means “son” in Latin, so this name is slightly less weird than it seems!)
Once Mendel examined the F1 plants and recorded their traits, he let them self-fertilize naturally, producing lots of seeds. He then collected and grew the seeds from the F1 plants to produce an F2 generation, or second filial generation. Again, he carefully examined the plants and recorded their traits.
Mendel's experiments extended beyond the F2 generation to F3, F4, and later generations, but his model of inheritance was based mostly on the first three generations (P, F1, and F2).
Mendel didn’t just record what his plants looked like in each generation (e.g., tall vs. short). Instead, he counted exactly how many plants with each trait were present. This may sound tedious, but by recording numbers and thinking mathematically, Mendel made discoveries that eluded famous scientists of his time (such as Charles Darwin, who carried out similar experiments but didn’t grasp the significance of his results)5.
You can use the links below to learn more about Mendel's laws of inheritance:
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
[Attribution and references]
Attribution:
This article is a modified derivative of “Mendel’s experiments and the laws of probability,” by OpenStax College, Biology (CC BY 3.0). Download the original article for free at http://cnx.org/contents/[email protected].
The modified article is licensed under a CC BY-NC-SA 4.0 license.
Works cited:
1. Biography.com Editors. (2015). Gregor Mendel biography. In The Biography.com website. Retrieved from http://www.biography.com/people/gregor-mendel-39282.
2. Gregor Mendel. (2015, 1 September). Retrieved from Wikipedia on September 9, 2015: https://en.wikipedia.org/wiki/Gregor_Mendel.
3. Mendel, J. G. (1866). Versuche über Pflanzenhybriden. Verhandlungen des naturforschenden Vereines in Brünn, Bd. IV für das Jahr 1865, Abhandlungen, 3–47. English translation retrieved from http://www.esp.org/foundations/genetics/classical/gm-65.pdf.
4. Blumberg, R. B. (1997). Mendel's paper in English. In MendelWeb. Retrieved from http://www.mendelweb.org/Mendel.plain.html.
5. Purves, W. K., Sadava, D. E., Orians, G. H., and Heller, H.C. (2004). Genetics: Mendel and beyond. In Life: The science of biology (7th ed.). Sunderland, MA: Sinauer Associates, 189.
6. Blending inheritance. (2015, April 20). Retrieved November 17, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Blending_inheritance.
7. Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Mendel used the scientific approach to identify two laws of inheritance. In Campbell Biology (10th ed.). San Francisco, CA: Pearson, 268.
Additional references:
Blending inheritance. (2015, April 20). Retrieved November 17, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Blending_inheritance.
Blumberg, R. B. (1997). Mendel's paper in English. In MendelWeb. Retrieved from http://www.mendelweb.org/Mendel.plain.html.
Gregor Mendel. (2015, 1 September). Retrieved from Wikipedia on September 9, 2015: https://en.wikipedia.org/wiki/Gregor_Mendel.
Hybrid (biology). (2015, October 29). Retrieved November 16, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Hybrid_%28biology%29.
Mendel, J. G. (1866). Versuche über Pflanzenhybriden. Verhandlungen des naturforschenden Vereines in Brünn, Bd. IV für das Jahr 1865, Abhandlungen, 3–47. English translation retrieved from http://www.esp.org/foundations/genetics/classical/gm-65.pdf.
Nature Education. (2014). Gregor Mendel: A private scientist. In Scitable. Retrieved from http://www.nature.com/scitable/topicpage/gregor-mendel-a-private-scientist-6618227.
Purves, W. K., Sadava, D. E., Orians, G. H., and Heller, H.C. (2004). Genetics: Mendel and beyond. In Life: The science of biology (7th ed., pp. 187-212). Sunderland, MA: Sinauer Associates.
Raven, P. H., Johnson, G. B., Mason, K. A., Losos, J. B., and Singer, S. R. (2014). Patterns of inheritance. In Biology (10th ed., AP ed., pp. 221-238). New York, NY: McGraw-Hill.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Drawing from the deck of genes. In Campbell Biology (10th ed., pp. 267-268). San Francisco, CA: Pearson.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Mendel used the scientific approach to identify two laws of inheritance. In Campbell Biology (10th ed., pp. 268-269). San Francisco, CA: Pearson.
The origins of genetics. (n.d.). Retrieved from http://images.pcmac.org/SiSFiles/Schools/AL/JacksonCounty/NorthJacksonHigh/Uploads/Presentations/The%20Origins%20of%20Genetics%208-1.ppt. | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/01%3A_Introduction_to_heredity/1.01%3A_Introduction_to_heredity.txt |
Key points:
• Gregor Mendel studied inheritance of traits in pea plants. He proposed a model where pairs of "heritable elements," or genes, specified traits.
• Genes come in different versions, or alleles. A dominant allele hides a recessive allele and determines the organism's appearance.
• When an organism makes gametes, each gamete receives just one gene copy, which is selected randomly. This is known as the law of segregation.
• A Punnett square can be used to predict genotypes (allele combinations) and phenotypes (observable traits) of offspring from genetic crosses.
• A test cross can be used to determine whether an organism with a dominant phenotype is homozygous or heterozygous.
Introduction
Today, we know that many of people's characteristics, from hair color to height to risk of diabetes, are influenced by genes. We also know that genes are the way parents pass characteristics on to their children (including things like dimples, or—in the case of me and my father—a terrible singing voice). In the last hundred years, we've come to understand that genes are actually pieces of DNA that are found on chromosomes and specify proteins.
But did we always know those things? Not by a long shot! About 150 years ago, a monk named Gregor Mendel published a paper that first proposed the existence of genes and presented a model for how they were inherited. Mendel's work was the first step on a long road, involving many hard-working scientists, that's led to our present understanding of genes and what they do.
In this article, we’ll trace the experiments and reasoning that led Mendel to formulate his model for the inheritance of single genes.
Mendel's model: It started with a 3:1 ratio
Mendel studied the genetics of pea plants, and he traced the inheritance of a variety of characteristics, including flower color, flower position, seed color, and seed shape. To do so, he started by crossing pure-breeding parent plants with different forms of a characteristic, such as violet and white flowers. Pure-breeding just means that the plant will always make more offspring like itself, when self-fertilized over many generations.
[What is self-fertilization?]
In self-fertilization, sperm and eggs from the same pea plant combine inside a closed flower, producing seeds with a single plant as both mother and father.
What results did Mendel find in his crosses for flower color? In the parental, or P generation, Mendel crossed a pure-breeding violet-flowered plant to a pure-breeding white-flowered plant. When he gathered and planted the seeds produced in this cross, Mendel found that 100 percent of the plants in the next generation, or F1 generation, had violet flowers.
Conventional wisdom at that time would have predicted that the hybrid flowers should be pale violet—that is, that the parents' traits should blend in the offspring. Instead, Mendel’s results showed that the white flower trait had completely disappeared. He called the trait that was visible in the F1 generation (violet flowers) the dominant trait, and the trait that was hidden or lost (white flowers) the recessive trait.
Importantly, Mendel did not stop his experimentation there. Instead, he let the F1 plants self-fertilize. Among their offspring, called the F2 generation, he found that 705 plants had violet flowers and 224 had white flowers. This was a ratio of 3.15 violet flowers to one white flower, or approximately 3:1.
This 3:1 ratio was no fluke. For the other six characteristics that Mendel examined, both the F1 and F2 generations behaved in the same way they did for flower color. One of the two traits would disappear completely from the F1 generation, only to reappear in the F2 generation in a ratio of roughly 3:1.
[See Mendel's data for all seven characteristics]
Character Dominant trait Recessive trait F2 generation F2 ratio
Seed color Yellow Green 6022 yellow, 2001 green 3.01:1
Seed shape Round Wrinkled 5474 round, 1850 wrinkled 2.96:1
Pod color Green Yellow 428 green, 152 yellow 2.82:1
Pod shape Inflated Constricted 882 inflated, 299 constricted 2.95:1
Plant height Tall Short (dwarf) 787 tall, 277 short 2.84:1
Flower color Purple White 705 purple, 224 white 3.15:1
Flower position Axial (along branch) Terminal (end of branch) 651 axial, 207 terminal 3.14:1
Adapted from "The results of Mendel's garden pea hybridizations," by OpenStax College, Biology (CC BY 3.0).
As it turned out, the 3:1 ratio was a crucial clue that let Mendel crack the puzzle of inheritance. Let's take a closer look at what Mendel figured out.
Mendel's model of inheritance
Based on his results (including that magic 3:1 ratio), Mendel came up with a model for the inheritance of individual characteristics, such as flower color.
In Mendel's model, parents pass along “heritable factors," which we now call genes, that determine the traits of the offspring. Each individual has two copies of a given gene, such as the gene for seed color (Y gene) shown below. If these copies represent different versions, or alleles, of the gene, one allele—the dominant one—may hide the other allele—the recessive one. For seed color, the dominant yellow allele Y hides the recessive green allele y.
The set of alleles carried by an organism is known as its genotype. Genotype determines phenotype, an organism's observable features. When an organism has two copies of the same allele (say, YY or yy), it is said to be homozygous for that gene. If, instead, it has two different copies (like Yy), we can say it is heterozygous. Phenotype can also be affected by the environment in many real-life cases, though this did not have an impact on Mendel's work.
Mendel's model: The law of segregation
So far, so good. But this model alone doesn't explain why Mendel saw the exact patterns of inheritance he did. In particular, it doesn't account for the 3:1 ratio. For that, we need Mendel's law of segregation.
According to the law of segregation, only one of the two gene copies present in an organism is distributed to each gamete (egg or sperm cell) that it makes, and the allocation of the gene copies is random. When an egg and a sperm join in fertilization, they form a new organism, whose genotype consists of the alleles contained in the gametes. The diagram below illustrates this idea:
The four-squared box shown for the F2 generation is known as a Punnett square. To prepare a Punnett square, all possible gametes made by the parents are written along the top (for the father) and side (for the mother) of a grid. Here, since it is self-fertilization, the same plant is both mother and father.
The combinations of egg and sperm are then made in the boxes in the table, representing fertilization to make new individuals. Because each square represents an equally likely event, we can determine genotype and phenotype ratios by counting the squares.
[Why are the boxes all equally likely?]
A key point of the law of segregation is that a parent’s two gene copies are randomly distributed to its gametes. Thus, for a Yy heterozygote, Y and y gametes are equally likely to be made: 50% of the sperm and eggs will have a Y allele, 50% will have a y allele, and the same will be true for eggs.
Since each type of gamete is equally common, each fertilization event (meeting of gametes, corresponding to a square of the table) also has an equal chance of happening. Thus, the boxes of the table represent four equal-probability events.
Since the table contains 1 box with a YY genotype, 2 boxes with a Yy genotype, and 1 box with a yy genotype, we'd expect to see YY, Yy, and yy plants in a ratio of 1:2:1 in the F2 generation. Since both YY and Yy plants are yellow, this genotype ratio translates into a phenotype ratio of 3:1 yellow-seeded to green-seeded plants, almost exactly what Mendel observed.
YY and yy plants each appear in just one square, while Yy plants are found in two squares. This is because there are two different fertilization events that lead to a Yy plant: the fusion of a Y egg and a y sperm, or the fusion of a y egg and a Y sperm. Either event is equally likely, and has the same likelihood as a YY or yy fertilization event.
The test cross
Mendel also came up with a way to figure out whether an organism with a dominant phenotype (such as a yellow-seeded pea plant) was a heterozygote (Yy) or a homozygote (YY). This technique is called a test cross and is still used by plant and animal breeders today.
In a test cross, the organism with the dominant phenotype is crossed with an organism that is homozygous recessive (e.g., green-seeded):
If the organism with the dominant phenotype is homozygous, then all of the F1 offspring will get a dominant allele from that parent, be heterozygous, and show the dominant phenotype. If the organism with the dominant phenotype organism is instead a heterozygote, the F1 offspring will be half heterozygotes (dominant phenotype) and half recessive homozygotes (recessive phenotype).
The fact that we get a 1:1 ratio in this second case is another confirmation of Mendel’s law of segregation.
Is that Mendel's complete model of inheritance?
Not quite! We've seen all of Mendel's model for the inheritance of single genes. However, Mendel's complete model also addressed whether genes for different characteristics (such as flower color and seed shape) influence each other's inheritance. You can learn more about Mendel's model for the inheritance of multiple genes in the law of independent assortment article.
One thing I find pretty amazing is that Mendel was able to figure out his entire model of inheritance simply from his observations of pea plants. This wasn't because he was some kind of crazy super genius, but rather, because he was very careful, persistent, and curious, and also because he thought about his results mathematically (for instance, the 3:1 ratio). These are some of the qualities of a great scientist—ones that anyone, anywhere, can develop!
Check your understanding
Query \(1\)
[Hint]
When two pure-breeding organisms with different forms of a characteristic are crossed, the offspring of the cross may show just one form of the characteristic, while the other form may be hidden.
The trait that is visible in the offspring of the cross is called the dominant trait, while the trait that is hidden is called the recessive trait.
In this example, tan fur is the dominant trait, because it is visible in the offspring of the cross. Black fur is the recessive trait, because it is hidden in the offspring of the cross.
Tan fur is dominant, while black fur is recessive.
Query \(2\)
[Hint]
Genotype, the set of alleles an organism carries, determines phenotype, its outward appearance or characteristics. However, environment influences how genotype is translated into phenotype, so an organism's phenotype is typically the product of both genotype and environment.
Genotype determines phenotype, but with influence from the environment.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
[Attribution and references]
Attribution:
This article is a modified derivative of the following articles:
The modified article is licensed under a CC BY-NC-SA 4.0 license.
Additional references:
Ding, Z. (2009). Eye color. In Stanford at the Tech: Understanding genetics. Retrieved from http://genetics.thetech.org/ask/ask316.
Miko, I. (2008). Gregor Mendel and the principles of inheritance. Nature Education, 11(1):134. Retrieved from http://www.nature.com/scitable/topicpage/gregor-mendel-and-the-principles-of-inheritance-593.
Monohybrid cross. (2015, September 18). Retrieved November 14, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Monohybrid_cross.
Phenotype. (2015, October 17). Retrieved November 14, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Phenotype.
Punnet. (2015, August 9). Retrieved November 18, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Punnet.
Punnett square. (2015, November 8). Retrieved November 14, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Punnett_square.
Purves, W. K., Sadava, D. E., Orians, G. H., and Heller, H.C. (2004). Genetics: Mendel and beyond. In Life: The science of biology (7th ed., pp. 187-212). Sunderland, MA: Sinauer Associates.
Raven, P. H., Johnson, G. B., Mason, K. A., Losos, J. B., and Singer, S. R. (2014). Patterns of inheritance. In Biology (10th ed., AP ed., pp. 221-238). New York, NY: McGraw-Hill.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Mendel and the gene idea. In Campbell biology (10th ed., pp. 267-291). San Francisco, CA: Pearson.
Starr, B. (2006, December 20). Eye color [answer]. In Stanford at the Tech: Understanding genetics. Retrieved from http://genetics.thetech.org/ask/ask203
Strehlow, A. T. (2005, March 9). Other traits [answer]. In Stanford at the Tech: Understanding genetics. Retrieved from http://genetics.thetech.org/ask/ask98.
White, D. and Rabago-Smith, M. (2011). Genotype-phenotype associations and human eye color. Journal of Human Genetics, 56, 5-7. http://dx.doi.org/10.1038/jhg.2010.126. | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/01%3A_Introduction_to_heredity/1.05%3A_The_law_of_segregation.txt |
Introduction
The law of segregation lets us predict how a single feature associated with a single gene is inherited. In some cases, though, we might want to predict the inheritance of two characteristics associated with two different genes. How can we do this?
[Refresher on the law of segregation]
The law of segregation states that each gamete (sperm or egg cell) made by an organism will get just one of the two gene copies present in a parent organism, and that the gene copies are randomly allocated to the gametes. For instance, if an organism has a genotype of Aa, half of its gametes will contain an A allele, and the other half will contain an a allele.
You can use the link at the start of the paragraph to learn more about the law of segregation.
To make an accurate prediction, we need to know whether the two genes are inherited independently or not. That is, we need to know whether they "ignore" one another when they're sorted into gametes, or whether they "stick together" and get inherited as a unit.
When Gregor Mendel asked this question, he found that different genes were inherited independently of one another, following what's called the law of independent assortment. In this article, we'll take a closer look at the law of independent assortment and how it is used to make predictions. We'll also see when and why the law of independent assortment does (or doesn't!) hold true.
Note: If you are not yet familiar with how individual genes are inherited, you may want to check out the article on the law of segregation or the introduction to heredity video before you dive into this article.
What is the law of independent assortment?
Mendel's law of independent assortment states that the alleles of two (or more) different genes get sorted into gametes independently of one another. In other words, the allele a gamete receives for one gene does not influence the allele received for another gene.
Example: Pea color and pea shape genes
Let's look at a concrete example of the law of independent assortment. Imagine that we cross two pure-breeding pea plants: one with yellow, round seeds (YYRR) and one with green, wrinkled seeds (yyrr). Because each parent is homozygous, the law of segregation tells us that the gametes made by the wrinkled, green plant all are ry, and the gametes made by the round, yellow plant are all RY. That gives us F1 offspring that are all RrYy.
The allele specifying yellow seed color is dominant to the allele specifying green seed color, and the allele specifying round shape is dominant to the allele specifying wrinkled shape, as shown by the capital and lower-case letters. This means that the F1 plants are all yellow and round. Because they are heterozygous for two genes, the F1 plants are called dihybrids (di- = two, -hybrid = heterozygous).
A cross between two dihybrids (or, equivalently, self-fertilization of a dihybrid) is known as a dihybrid cross. When Mendel did this cross and looked at the offspring, he found that there were four different categories of pea seeds: yellow and round, yellow and wrinkled, green and round, and green and wrinkled. These phenotypic categories (categories defined by observable traits) appeared in a ratio of approximately 9:3:3:1.
This ratio was the key clue that led Mendel to the law of independent assortment. That's because a 9:3:3:1 ratio is exactly what we'd expect to see if the F1 plant made four types of gametes (sperm and eggs) with equal frequency: YR, Yr, yR, and yr. In other words, this is the result we'd predict if each gamete randomly got a Y or y allele, and, in a separate process, also randomly got an R or r allele (making four equally probable combinations).
We can confirm the link between the four types of gametes and the 9:3:3:1 ratio using the Punnett square above. To make the square, we first put the four equally probable gamete types along each axis. Then, we join gametes on the axes in the boxes of the chart, representing fertilization events. The 16 equal-probability fertilization events that can occur among the gametes are shown in the 16 boxes. The offspring genotypes in the boxes correspond to a 9:3:3:1 ratio of phenotypes, just as Mendel observed.
[More about two-gene Punnett squares]
We can draw a Punnett square for a two-gene scenario by following the same basic rules as for a monohybrid cross, placing the gametes along the axes and combining them in the squares to represent fertilization events. However, since there are now more gamete types, there must also be more squares in the table: 4 possible types of maternal gametes x 4 possible types of paternal gametes = 16 squares total.
As with a single-gene Punnett square, we place all the possible types of gametes along the axes, then combine them in the squares where the columns and rows intersect to represent fertilization events (the formation of zygotes, or offspring).
To learn how you can use the rules of probability to predict the outcome of a dihybrid cross, see the probabilities in genetics article.
Independent assortment vs. linkage
The section above gives us Mendel's law of independent assortment in a nutshell, and lets us see how the law of independent assortment leads to a 9:3:3:1 ratio. But what was the alternative possibility? That is, what would happen if two genes didn't follow independent assortment?
In the extreme case, the genes for seed color and seed shape might have always been inherited as a pair. That is, the yellow and round alleles might always have stayed together, and so might the green and wrinkled alleles.
To see how this could work, imagine that the color and shape genes are physically stuck together and cannot be separated, as represented by the boxes around the alleles in the diagram below. For instance, this could happen if the two genes were located very, very close together on a chromosome (an idea we'll explore further at the end of the article).
Rather than giving a color allele and, separately, giving a shape allele to each gamete, the F1 dihybrid plant would simply give one “combo unit” to each gamete: a YR allele pair or a yr allele pair.
We can use a Punnett square to predict the results of self-fertilization in this case, as shown above. If the seed color and seed shape genes were in fact always inherited as a unit, or completely linked, a dihybrid cross should produce just two types of offspring, yellow/round and green/wrinkled, in a 3:1 ratio. Mendel's actual results were quite different from this (the 9:3:3:1 ratio we saw earlier), telling him that the genes assorted independently.
The reason for independent assortment
To see why independent assortment happens, we need to fast-forward half a century and discover that genes are physically located on chromosomes. To be exact, the two copies of a gene carried by an organism (such as a Y and a y allele) are located at the same spot on the two chromosomes of a homologous pair. Homologous chromosomes are similar but non-identical, and an organism gets one member of the pair from each of its two parents.
The physical basis for the law of independent assortment lies in meiosis I of gamete formation, when homologous pairs line up in random orientations at the middle of the cell as they prepare to separate. We can get gametes with different combos of "mom" and "dad" homologues (and thus, the alleles on those homologues) because the orientation of each pair is random.
To see what this means, compare chromosome arrangement 1 (top) and chromosome arrangement 2 (bottom) at the stage of metaphase I in the diagram below. In one case, the red "mom" chromosomes go together, while in the other, they split up and mix with the blue "dad" chromosomes. If meiosis happens many times, as it does in a pea plant, we will get both arrangements—and thus RY, Ry, rY, and ry classes of gametes—with equal frequency.
Genes that are on different chromosomes (like the Y and R genes) assort independently. The seed color and seed shape genes are on chromosomes 1 and 7 of the pea genome, respectively, in real life1. Genes that are far apart on the same chromosome also assort independently thanks to the crossing over, or exchange of homologous chromosome bits, that occurs early in meiosis I.
[See a picture]
There are, however, gene pairs that do not assort independently. When genes are close together on a chromosome, the alleles on the same chromosome tend to be inherited as a unit more frequently than not. Such genes do not display independent assortment and are said to be linked. We'll take a closer look at genetic linkage in other articles and videos.
[See a picture]
Check your understanding
Query \(1\)
[Hint]
Since all of the F1 dogs are black and straight-furred, we know that black fur color and straight fur texture are dominant over yellow fur color and curly fur texture. If we call the color gene B/b and the texture gene C/c, and use capital letters for the dominant form of each gene and lowercase letters for the recessive form, we can assign the two parental dogs genotypes of BBcc (black and curly) and bbCC (yellow and straight-furred). When the parental dogs are crossed, they produce black, straight-furred F1 dogs that are dihybrids: BbCc.
A cross between two F1 dihybrid dogs results in the Punnett square shown below. The F1 dogs can make four different types of gametes, which are represented along the two axes of the Punnett square. The squares of the table represent fertilization events in which the gametes on the axes combine. Since all of the gamete types are equally likely to be produced (because the genes assort independently, i.e., do not influence each other's inheritance), all the squares in the table represent equal-probability events, ones that occur 1/16 of the time.
Now, we need to find the squares that correspond to the outcome we are interested in: a puppy with yellow, straight fur. To have yellow, straight fur, the puppy must get two recessive alleles for fur color (bb genotype) and at least one dominant allele for fur texture (Cc or CC genotype). If we go through the table and circle the genotypes that match these requirements, we'll find that 3 out of the 16 boxes correspond to yellow, straight-furred puppies. Thus, we would expect 3/16 of the F2 puppies to have yellow, straight fur.
Answer: 3/16.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
[Attribution and references]
Attribution:
This article is a modified derivative of:
The modified article is licensed under a CC BY-NC-SA 4.0 license.
Works cited:
1. Reid, J. B., and Ross, J. J. (2011). Mendel's genes: Towards a full molecular characterization. Genetics 189(1), 3-10. http://dx.doi.org/10.1534/genetics.111.132118. Retrieved from www.ncbi.nlm.nih.gov/pmc/articles/PMC3176118/. | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/01%3A_Introduction_to_heredity/1.06%3A_The_law_of_independent_assortment.txt |
Introduction
The Punnett square is a valuable tool, but it's not ideal for every genetics problem. For instance, suppose you were asked to calculate the frequency of the recessive class not for an Aa x Aa cross, not for an AaBb x AaBb cross, but for an AaBbCcDdEe x AaBbCcDdEe cross. If you wanted to solve that question using a Punnett square, you could do it – but you'd need to complete a Punnett square with 1024 boxes. Probably not what you want to draw during an exam, or any other time, if you can help it!
The five-gene problem above becomes less intimidating once you realize that a Punnett square is just a visual way of representing probability calculations. Although it’s a great tool when you’re working with one or two genes, it can become slow and cumbersome as the number goes up. At some point, it becomes quicker (and less error-prone) to simply do the probability calculations by themselves, without the visual representation of a clunky Punnett square. In all cases, the calculations and the square provide the same information, but by having both tools in your belt, you can be prepared to handle a wider range of problems in a more efficient way.
In this article, we’ll review some probability basics, including how to calculate the probability of two independent events both occurring (event X and event Y) or the probability of either of two mutually exclusive events occurring (event X or event Y). We’ll then see how these calculations can be applied to genetics problems, and, in particular, how they can help you solve problems involving relatively large numbers of genes.
[Solution to the five-gene cross problem]
In this problem, we’re supposed to find the frequency of the recessive class among the offspring of an AaBbCcDdEe x AaBbCcDdEe cross – that is, the frequency of aabbccddee individuals. How do we get an aabbccddee individual? There’s only one way for that to happen: both parents must contribute an abcde gamete.
What, then, is the probability that one of the parents will make an abcde gamete? Both parents are heterozygous for all five genes, so there’s a 1/2 chance of getting the recessive (lowercase) allele for any one gene. To get our desired gamete, we need all five genes in recessive form (a and b and c and d and e). This is a case where we can apply the product rule, which states that the probability of event X and event Y happening is the product of their individual probabilities (probability of X times probability of Y), assuming that X and Y are independent events. Thus, the overall probability of one parent producing an abcde gamete is:
Probability of abcde gamete = (probability of a) x (probability of b) x (probability of c) x (probability of d) x (probability of e)
$P(abcde)=P(a)\cdot P(b)\cdot P(c)\cdot P(d)\cdot P(e)$
$P(abcde)=(1/2)\cdot (1/2)\cdot (1/2)\cdot (1/2)\cdot (1/2)=(1/2)^5=1/32$
If that’s the probability of one parent making an abcde gamete, what’s the likelihood of both parents doing so? Again, we can apply the "and" rule (product rule), since we need both parent 1 and parent 2 to make an abcde gamete in order to get our target recessive homozygote. Thus, the overall probability is:
Probability of aabbccddee individual = (probability of parent 1 making an abcde gamete) x (probability of parent 2 making an abcde gamete)
$P(aabbccddee)=P(abcde_\text{parent A})\cdot P(abcde_\text{parent B})$
$P(aabbccddee)=(1/32)\cdot (1/32)=1/1024$
That’s our overall probability for a recessive homozygote for all five genes.
The 1/1024 probability corresponds to 1 box out of the 1024 boxes of the Punnett square you’d have to draw to represent this cross. The probability calculation is the same calculation we’d implicitly do by drawing the Punnett square, just faster and with fewer chances for mistakes.
Probability basics
Probabilities are mathematical measures of likelihood. In other words, they’re a way of quantifying (giving a specific, numerical value to) how likely something is to happen. A probability of 1 for an event means that it is guaranteed to happen, while a probability of 0 for an event means that it is guaranteed not to happen. A simple example of probability is having a 1/2 chance of getting heads when you flip a coin, as Sal explains in this intro to probability video.
Probabilities can be either empirical, meaning that they are calculated from real-life observations, or theoretical, meaning that they are predicted using a set of rules or assumptions.
• The empirical probability of an event is calculated by counting the number of times that event occurs and dividing it by the total number of times that event could have occurred. For instance, if the event you were looking for was a wrinkled pea seed, and you saw it 1,850 times out of the 7,324 total seeds you examined, the empirical probability of getting a wrinkled seed would be 1,850/7,324 = 0.253, or very close to 1 in 4 seeds.
• The theoretical probability of an event is calculated based on information about the rules and circumstances that produce the event. It reflects the number of times an event is expected to occur relative to the number of times it could possibly occur. For instance, if you had a pea plant heterozygous for a seed shape gene (Rr) and let it self-fertilize, you could use the rules of probability and your knowledge of genetics to predict that 1 out of every 4 offspring would get two recessive alleles (rr) and appear wrinkled, corresponding to a 0.25 (1/4) probability. We’ll talk more below about how to apply the rules of probability in this case.
In general, the larger the number of data points that are used to calculate an empirical probability, such as shapes of individual pea seeds, the more closely it will approach the theoretical probability.
The product rule
One probability rule that's very useful in genetics is the product rule, which states that the probability of two (or more) independent events occurring together can be calculated by multiplying the individual probabilities of the events. For example, if you roll a six-sided die once, you have a 1/6 chance of getting a six. If you roll two dice at once, your chance of getting two sixes is: (probability of a six on die 1) x (probability of a six on die 2) = (1/6) ⋅ (1/6) = 1/36.
In general, you can think of the product rule as the “and” rule: if both event X and event Y must happen in order for a certain outcome to occur, and if X and Y are independent of each other (don’t affect each other’s likelihood), then you can use the product rule to calculate the probability of the outcome by multiplying the probabilities of X and Y.
We can use the product rule to predict frequencies of fertilization events. For instance, consider a cross between two heterozygous (Aa) individuals. What are the odds of getting an aa individual in the next generation? The only way to get an aa individual is if the mother contributes an a gamete and the father contributes an a gamete. Each parent has a 1/2 chance of making an a gamete. Thus, the chance of an aa offspring is: (probability of mother contributing a) x (probability of father contributing a) = (1/2) ⋅ (1/2) = 1/4.
This is the same result you’d get with a Punnett square, and actually the same logical process as well—something that took me years to realize! The only difference is that, in the Punnett square, we'd do the calculation visually: we'd represent the 1/2 probability of an a gamete from each parent as one out of two columns (for the father) and one out of two rows (for the mother). The 1-square intersect of the column and row (out of the 4 total squares of the table) represents the 1/4 chance of getting an a from both parents.
The sum rule of probability
In some genetics problems, you may need to calculate the probability that any one of several events will occur. In this case, you’ll need to apply another rule of probability, the sum rule. According to the sum rule, the probability that any of several mutually exclusive events will occur is equal to the sum of the events’ individual probabilities.
For example, if you roll a six-sided die, you have a 1/6 chance of getting any given number, but you can only get one number per roll. You could never get both a one and a six at the same time; these outcomes are mutually exclusive. Thus, the chances of getting either a one or a six are: (probability of getting a 1) + (probability of getting a 6) = (1/6) + (1/6) = 1/3.
You can think of the sum rule as the “or” rule: if an outcome requires that either event X or event Y occur, and if X and Y are mutually exclusive (if only one or the other can occur in a given case), then the probability of the outcome can be calculated by adding the probabilities of X and Y.
As an example, let's use the sum rule to predict the fraction of offspring from an Aa x Aa cross that will have the dominant phenotype (AA or Aa genotype). In this cross, there are three events that can lead to a dominant phenotype:
• Two A gametes meet (giving AA genotype), or
• A gamete from Mom meets a gamete from Dad (giving Aa genotype), or
• a gamete from Mom meets A gamete from Dad (giving Aa genotype)
In any one fertilization event, only one of these three possibilities can occur (they are mutually exclusive).
Since this is an “or” situation where the events are mutually exclusive, we can apply the sum rule. Using the product rule as we did above, we can find that each individual event has a probability of 1/4. So, the probability of offspring with a dominant phenotype is: (probability of A from Mom and A from Dad) + (probability of A from Mom and a from Dad) + (probability of a from Mom and A from Dad) = (1/4) + (1/4) + (1/4) = 3/4.
Once again, this is the same result we’d get with a Punnett square. One out of the four boxes of the Punnett square holds the dominant homozygote, AA. Two more boxes represent heterozygotes, one with a maternal A and a paternal a, the other with the opposite combination. Each box is 1 out of the 4 boxes in the whole Punnett square, and since the boxes don't overlap (they’re mutually exclusive), we can add them up (1/4 + 1/4 + 1/4 = 3/4) to get the probability of offspring with the dominant phenotype.
The product rule and the sum rule
Product rule Sum rule
For independent events X and Y, the probability ($P$) of them both occurring (X and Y) is $P(X)\cdot P(Y)$. For mutually exclusive events X and Y, the probability ($P$) that one will occur (X or Y) is $P(X)+P(Y)$.
Applying probability rules to dihybrid crosses
Direct calculation of probabilities doesn’t have much advantage over Punnett squares for single-gene inheritance scenarios. (In fact, if you prefer to learn visually, you may find direct calculation trickier rather than easier.) Where probabilities shine, though, is when you’re looking at the behavior of two, or even more, genes.
For instance, let’s imagine that we breed two dogs with the genotype BbCc, where dominant allele B specifies black coat color (versus b, yellow coat color) and dominant allele C specifies straight fur (versus c, curly fur). Assuming that the two genes assort independently and are not sex-linked, how can we predict the number of BbCc puppies among the offspring?
One approach is to draw a 16-square Punnett square. For a cross involving two genes, a Punnett square is still a good strategy. Alternatively, we can use a shortcut technique involving four-square Punnett squares and a little application of the product rule. In this technique, we break the overall question down into two smaller questions, each relating to a different genetic event:
1. What’s the probability of getting a Bb genotype?
2. What’s the probability of getting an Cc genotype?
In order for a puppy to have a BbCc genotype, both of these events must take place: the puppy must receive Bb alleles, and it must receive Cc alleles. The two events are independent because the genes assort independently (don't affect one another's inheritance). So, once we calculate the probability of each genetic event, we can multiply these probabilities using the product rule to get the probability of the genotype of interest (BbCc).
To calculate the probability of getting a Bb genotype, we can draw a 4-square Punnett square using the parents' alleles for the coat color gene only, as shown above. Using the Punnett square, you can see that the probability of the Bb genotype is 1/2. (Alternatively, we could have calculated the probability of Bb using the product rule for gamete contributions from the two parents and the sum rule for the two gamete combinations that give Bb.) Using a similar Punnett square for the parents' fur texture alleles, the probability of getting an Cc genotype is also 1/2. To get the overall probability of the BbCc genotype, we can simply multiply the two probabilities, giving an overall probability of 1/4.
[Let's check that with a Punnett square]
You can also use this technique to predict phenotype frequencies. Give it a try in the practice question below!
Check your understanding
Query $1$
[Hint]
We can break the question down into two smaller questions:
1. What fraction of offspring will have black coat color?
2. What fraction of offspring will have straight fur?
Since black coat color and straight fur are dominant traits, all BB and Bb puppies will have black coats, and all CC and Cc puppies will have straight fur, corresponding to 3/4 of puppies in each case. (You can draw out the individual Punnett squares for the color and texture genes to confirm these frequencies.)
To get the probability of a puppy having both black coat color and straight fur, you can multiply the probabilities of these two independent events: $(3/4)\cdot(3/4)=9/16$.
9/16 of the puppies will have black coats and straight fur.
Beyond dihybrid crosses
The probability method is most powerful (and helpful) in cases involving a large number of genes.
For instance, imagine a cross between two individuals with various alleles of four unlinked genes: AaBbCCdd x AabbCcDd. Suppose you wanted to figure out the probability of getting offspring with the dominant phenotype for all four traits. Fortunately, you can apply the exact same logic as in the case of the dihybrid crosses above. To have the dominant phenotype for all four traits, and organism must have: one or more copies of the dominant allele A and one or more copies of dominant allele B and one or more copies of the dominant allele C and one or more copies of the dominant allele D.
Since the genes are unlinked, these are four independent events, so we can calculate a probability for each and then multiply the probabilities to get the probability of the overall outcome.
• The probability of getting one or more copies of the dominant A allele is 3/4. (Draw a Punnett square for Aa x Aa to confirm for yourself that 3 out of the 4 squares are either AA or Aa.)
• The probability of getting one or more copies of the dominant B allele is 1/2. (Draw a Punnett square for Bb x bb: you’ll find that half the offspring are Bb, and the other half bb.)
• The probability of getting one or more copies of the dominant C allele is 1. (If one of the parents is homozygous CC, there’s no way to get offspring without a C allele!)
• The probability of getting one or more copies of the dominant D allele is 1/2, as for B. (Half the offspring will be Dd, and the other half will be dd.)
To get the overall probability of offspring with the dominant phenotype for all four genes, we can multiply the probabilities of the four independent events: $(3/4)\cdot(1/2)\cdot(1)\cdot(1/2)=3/16$.
Check your understanding
Query $2$
[Hint]
It’s not possible to get a quadruple homozygous recessive individual out of this cross. That’s because the probability of getting two recessive c alleles is zero. The first parent has only dominant alleles for this gene, ensuring that each of the offspring will receive at least one dominant C allele (and thus cannot display the recessive phenotype).
How does the zero probability of a cc genotype figure in mathematically? To get the overall probability of the aabbccdd genotype, we'd have to multiply the probabilities of the desired genotypes for the other three genes (aa, 1/4; bb, 1/2; and dd, 1/2) by the zero corresponding to the cc genotype, giving an overall probability of zero.
$P(aabbccdd)=P(aa) \cdot P(bb) \cdot P(cc) \cdot P(dd)$
$P(aabbccdd)=(1/4)\cdot(1/2)\cdot(0)\cdot(1/2)=0$
The probability of getting an individual with a recessive phenotype for all four genes is 0.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
[Attribution and references]
Attribution:
This article is a modified derivative of the following articles:
The modified article is licensed under a CC BY-NC-SA 4.0 license.
Additional references:
Griffiths, A. J. F., Miller, J. H., Suzuki, D. T., Lewontin, R. C., and Gelbart, W. M. (2000). Using genetic ratios. In An introduction to genetic analysis (7th ed.). New York, NY: W. H. Freeman. Retrieved from http://www.ncbi.nlm.nih.gov/books/NBK21812/.
Purves, W. K., Sadava, D., Orians, G. H., and Heller, H. C. (2003). Punnett squares or probability calculations: A choice of methods. In Life: The science of biology (7th ed., pp. 195-196). Sunderland, MA: Sinauer Associates.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Mendel and the gene idea. In Campbell Biology (10th ed., pp. 267-291). San Francisco, CA: Pearson.
Raven, P. H., Johnson, G. B., Mason, K. A., Losos, J. B., and Singer, S. R. (2014). Patterns of inheritance. In Biology (10th ed., AP ed., pp. 221-238). New York, NY: McGraw-Hill.
Staroscik, A. (2015). Punnett square calculator. In SciencePrimer.com. Retrieved from http://scienceprimer.com/punnett-square-calculator.
The Adapa Project. (2014, August 13). What are the laws of segregation and independent assortment and why are they so important? InBioBook. Retrieved from https://adapaproject.org/bbk_temp/tiki-index.php?page=Leaf%3A+What+are+the+laws+of+segregation+and+independent+assortment+and+why+are+they+so+important%3F. | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/01%3A_Introduction_to_heredity/1.07%3A_Probabilities_in_genetics.txt |
Key terms
Term Meaning
Genetics The study of biological inheritance
Trait A specific characteristic of an individual
Gene A unit of heredity that is passed from parent to offspring
Allele One of different forms of a gene
Genotype The genetic makeup of an organism (ex: TT)
Phenotype The physical characteristics of an organism (ex: tall)
Dominant allele Allele that is phenotypically expressed over another allele
Recessive allele Allele that is only expressed in absence of a dominant allele
Homozygous Having two identical alleles for a particular gene
Heterozygous Having two different alleles for a particular gene
Punnett square Diagram that can be used to predict the genotypes and phenotypes resulting from a genetic cross
Mendelian inheritance
Gregor Mendel's principles of heredity, observed through patterns of inheritance in pea plants, form the basis of modern genetics.
Mendel proposed that traits were specified by "heritable elements" called genes. Genes come in different versions, or alleles, with dominant alleles being expressed over recessive alleles. Recessive alleles are only expressed when no dominant allele is present.
In most sexually reproducing organisms, each individual has two alleles for each gene (one from each parent). This pair of alleles is called a genotype and determines the organism's appearance, or phenotype.
Mendel's laws
When an organism makes gametes, each gamete receives just one gene copy, which is selected randomly. This is known as the law of segregation.
Mendel's second law is the law of independent assortment, which states that the alleles for one gene sort into gametes independently of the alleles of another gene.
Punnett squares and probability
A Punnett square can be used to predict genotype and phenotypes of offspring from genetic crosses. A single-gene, or monohybrid cross is pictured below.
A test cross can be used to determine whether an organism with a dominant phenotype is homozygous or heterozygous.
Punnett squares can be used for two-gene crosses, or dihybrid crosses by following the same basic rules as for a monohybrid cross. However, since there are now more gamete types, there must also be more squares in the table.
Probabilities in genetics
The two probability rules that are most relevant to Punnett squares are the product rule and the sum rule.
The product rule states that the probability of two (or more) independent events occurring together can be calculated by multiplying the individual probabilities of the events.
In some genetics problems, you may need to calculate the probability that any one of several events will occur. In this case, you’ll need to apply another rule of probability, the sum rule. According to the sum rule, the probability that any of several mutually exclusive events will occur is equal to the sum of the events’ individual probabilities.
Common mistakes and misconceptions
• Dominant traits are not always the most common. Some people may think that dominant trait is the most likely to be found in the population, but the term "dominant" only refers to the fact that the allele is expressed over another allele. An example of this is Huntington's disease. Even though Huntington's is caused by a dominant allele, it only affects about 30,000 people in the United States1.
• Traits are not always the product of a single gene. For example, there are at least 3 different genes that are associated with eye color in humans. In addition, there are sometimes more than two alleles for each gene. For example, there are 3 different alleles of one gene that determine coat color of cats.
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• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org) | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/01%3A_Introduction_to_heredity/1.08%3A_Introduction_to_heredity_review.txt |
Query \(1\)
Step-by-step solution
1. Individuals with two of the same allele are considered homozygous, while individuals with two different alleles are heterozygous.
2. If the dominant allele is present, then the dominant trait will be shown:
WW - homozygous dominant (wide wings)
Ww - heterozygous (wide wings)
ww - homozygous recessive (thin wings)
3. The correct answer is
Ww - homozygous dominant
Query \(2\)
Step-by-step solution
1. The genotype of an organism is represented by two letters (ex: LL, Ll, or ll).
An organism's phenotype is its physical characteristic (ex: lactase persistence or lactose intolerance).
2. An individual with a homozygous dominant genotype (LL) would show the dominant phenotype (lactase persistence).
An individual with a heterozygous dominant genotype (Ll) would show the dominant phenotype (lactase persistence).
An individual with a homozygous recessive genotype (ll) would show the recessive phenotype (lactose intolerance).
3. The correct answer is
ll - homozygous recessive - lactose intolerant
Query \(3\)
Step-by-step solution
1. The only possible genotype for wacky wings is ww (homozygous recessive).
The possible genotypes for normal wings are WW (homozygous dominant) and Ww (heterozygous).
2. The only possible genotype for flashy feathers is ff (homozygous recessive).
The possible genotypes for normal feathers are FF (homozygous dominant) and Ff (heterozygous).
3. The correct answer is
wwff; WWFF
Query \(4\)
Step-by-step solution
1. Individuals that are heterozygous for a trait are called carriers. They have both the dominant and recessive allele in their genotype.
2. The presence of at least one dominant allele will result in the expression of the dominant phenotype in the individual.
3. The correct answer is
Dominant and recessive alleles can be found in the same individual
Dominant alleles mask the expression of recessive alleles
Query \(5\)
Step-by-step solution
1. A dominant allele is denoted by an uppercase letter such as H.
A recessive allele is denoted by a lowercase letter such as h.
2. A homozygous genotype contains two identical alleles. An individual can either be homozygous dominant (ex: HH) or homozygous recessive (ex: hh).
A heterozygous genotype contains two different alleles (ex: Hh).
3. The correct answer is
Heterozygous; one dominant allele and one recessive allele
Query \(6\)
Step-by-step solution
1. The genotype of an organism is represented by two letters (ex: AA, Aa, or aa).
An organism's phenotype is its physical characteristic (ex: black feathers or red feathers).
2. A chicken with a homozygous dominant genotype (AA) would show the dominant phenotype (black feathers).
A chicken with a heterozygous dominant genotype (Aa) would show the dominant phenotype (black feathers).
A chicken with a homozygous recessive genotype (aa) would show the recessive phenotype (red feathers).
3. The correct answer is
aa - genotype; red feathers - phenotype
Query \(7\)
Step-by-step solution
1. In order to have fancy fins, the possible genotypes are FF (homozygous) and Ff (heterozygous).
2. In order to have shiny scales, the possible genotypes are SS (homozygous) and Ss (heterozygous).
3. The correct answer is
FFSS; FfSs
Query \(8\)
Step-by-step solution
1. In a heterozygous organism, one of the alleles may completely determine the phenotype of the organism, masking the presence of the other allele.
2. The allele that masks the presence of the other allele is the dominant allele. The masked allele is the recessive allele.
3. In this example, both parents are heterozygous for brown coats (Bb x Bb).
When these two dogs are crossed, some of the puppies have black coats because they inherited two recessive (b) alleles. Others have brown coats because they have at least one dominant (B) allele.
4. The correct answer is
The allele for brown coat is dominant to the allele for black coat.
Query \(9\)
Step-by-step solution
1. Individuals with two of the same allele are considered homozygous, while individuals with two different alleles are heterozygous.
2. If the dominant allele is present, then the dominant trait will be shown:
TT - homozygous dominant (taster) Tt - heterozygous (taster)
tt - homozygous recessive (non-taster)
3. The correct answer is
TT - homozygous dominant
Query \(10\)
Step-by-step solution
1. The genotype of an organism is represented by two letters (ex: SS, Ss, or ss).
An organism's phenotype is its physical characteristic (ex: smooth pods or constricted pods).
2. A pea plant with a homozygous dominant genotype (SS) would show the dominant phenotype (smooth pods).
A pea plant with a heterozygous dominant genotype (Ss) would show the dominant phenotype (smooth pods).
A pea plant with a homozygous recessive genotype (ss) would show the recessive phenotype (constricted pods).
3. The correct answer is
ss - genotype; constricted pod - phenotype
Query \(11\)
Step-by-step solution
1. In a heterozygous organism, one of the alleles may completely determine the phenotype of the organism, masking the presence of the other allele.
2. The allele that masks the presence of the other allele is the dominant allele. The masked allele is the recessive allele.
3. In this example, one plant is homozygous for the red allele (RR), while the other is homozygous for the white allele (rr).
When these two plants are crossed, the offspring will all be heterozygous (Rr), with one of each allele. Because of the presence of the dominant R allele, the flowers will all be red.
4. The correct answer is
The allele for red flowers is dominant to the allele for white flowers.
Query \(12\)
Step-by-step solution
1. Individuals that are heterozygous for a trait are called carriers. They have both the dominant and recessive allele in their genotype.
2. The presence of at least one dominant allele will result in the expression of the dominant phenotype in the individual.
3. The correct answer is
An individual can have both dominant and recessive alleles
Recessive traits are expressed only if the dominant allele is absent
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org) | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/01%3A_Introduction_to_heredity/1.09%3A_Practice_-_Introduction_to_heredity.txt |
Query $1$
Step-by-step solution
1. Because there are two traits here, we must perform a dihybrid cross. We know the parent genotypes (rrss and RRSS), but we need to figure out what the possible gametes are by figuring out all the possible combinations of the two alleles.
For the rrss mother, the only possible gamete combination is rs.
For the RRSS father, the only possible gamete combination is RS.
2. Next we will perform the dihybrid cross.
Cross: rrss x RRSS
If we complete the cross, we find that the possible offspring can only be rSsR. Offspring that have at least one R and S allele must be rude and sneaky because the allele for rude (R) is dominant to the allele for respectful (r) and the allele for sneaky (S) is dominant to the allele for sincere (s).
3. After completing the Punnet square, we need to figure out how many of the offspring genotype combinations contain only the recessive r and s alleles because that is the only way to express the respectful and sincere phenotype combination.
Overall, there are 0 genotype combinations with only the recessive r and s alleles and thus, a 0/16 chance of offspring being respectful and sincere.
4. The correct answer is
0/16
Query $2$
Step-by-step solution
1. The father is homozygous recessive (bb), so he can only give the recessive b allele to the offspring, while the mother is homozygous dominant (BB), so she can only give the dominant B allele to the offspring. The cross for these parents is bb x BB.
2. We can complete a Punnett square to find the possible offspring genotypes.
Cross: bb x BB
If we complete the cross, we find that the possible offspring can only be Bb. Offspring that have at least one B allele must be boring because the allele for boring (B) is dominant to the allele for busy (b).
3. If we complete the cross, we find that there is a 100% chance of the offspring being Bb (boring). There are no other possible genotype combinations.
4. The correct answer is
Only Bb
Query $3$
Step-by-step solution
1. Because there are two traits here, we must perform a dihybrid cross. We know the parent genotypes (CcDd and CcDd), but we need to figure out what the possible gametes are by figuring out all the possible combinations of the two alleles.
For both parents, the CcDd genotype leads to the possible gamete combinations of CD, Cd, cD, or cd.
2. Next, we will perform the dihybrid cross.
Cross: CcDd x CcDd
After completing the cross, we need to figure out how many of the offspring genotype combinations contain at least one dominant C allele and one dominant D allele in order to get the clumsy and dazzling phenotype combination.
3. Because we are looking for the proportion of offspring that are both clumsy and dazzling, we can refer to the circled genotypes in ${\color{Red} red}$ on the Punnett square. These boxes all contain at least one dominant C allele and one dominant D allele.
Overall, there are 9 ${\color{Red} red}$ circled genotypes and thus, a 9/16 chance of offspring being clumsy and dazzling.
4. The correct answer is
9/16
Query $4$
Step-by-step solution
1. The mother is heterozygous, so she can possibly give one of either allele (P or p) to the offspring, while the father is homozygous dominant (PP), so he can only give the dominant P allele to the offspring. The cross for these parents is Pp x PP.
2. We can complete a Punnett square to find the possible offspring genotypes.
Cross: Pp x PP
If we complete the cross, we find that the possible offspring can be PP or Pp. Offspring that have at least one P allele must be playful because the allele for playful (P) is dominant to the allele for paranoid (p).
3. If we complete the cross, we find that there is a 50% chance of the offspring being PP (playful) and also a 50% chance of the offspring being Pp (playful). There are no other possible genotype combinations.
4. The correct answer is
PP and Pp
Query $5$
Step-by-step solution
1. Because there are two traits here, we must perform a dihybrid cross. We know the parent genotypes (Bbhh and bbHh), but we need to figure out what the possible gametes are by figuring out all the possible combinations of the two alleles.
For the Bbhh parent, the possible gamete combinations are Bh, Bh, bh, or bh.
For the bbHh parent, the possible gamete combinations are bH, bH, bh, or bh.
2. Next, we will perform the dihybrid cross.
Cross: Bbhh x bbHh
By completing the cross, we find that the possible offspring genotypes are BbHh, bbHh, Bbhh, or bbhh.
3. Because we are looking for the proportion of offspring that are brown with long fur, we need to identify offspring that are dominant for the fur color gene (either BB or Bb), and recessive for the fur length gene (hh).
If we look at our Punnett square, the only genotype that displays these phenotypes is Bbhh.
There is a 4/16 chance of offspring being brown with long fur, or Bbhh.
4/16 = 1/4 = 25%
4. The correct answer is
25%
Query $6$
Step-by-step solution
1. When a dominant allele is present, it will be expressed. The mother and father each have the genotype Aa, meaning that each will have pea combs because of the A allele.
The cross for these parents is Aa x Aa.
2. We can complete a Punnett square to find the possible offspring combinations.
Cross: Aa x Aa
If we complete the cross, we find that the possible offspring can be AA, Aa, or aa. Offspring that have an A allele must have pea combs, because the allele for pea combs (A) is dominant to the allele for single combs (a). Only aa offspring will have single combs.
3. If we look at our Punnett square, we find that there is a 1/4 chance of offspring having the genotype for single combs (aa). If we convert this to a percentage, it means there is a 25% chance these parents will have offspring with single combs.
4. The correct answer is
25%
Query $7$
Step-by-step solution
1. Because there are two traits here, we must perform a dihybrid cross. We know the parent genotypes (SsTt and SsTt), but we need to figure out what the possible gametes are by figuring out all the possible combinations of the two alleles.
For both parents, the SsTt genotype leads to the possible gamete combinations of ST, St, sT, or st.
2. Next we will perform the dihybrid cross.
Cross: SsTt x SsTt
After completing the cross, we need to figure out how many of the offspring genotype combinations contain two recessive s alleles and at least one dominant T allele in order to get the sassy and timely phenotype combination.
3. Because we are looking for the proportion of offspring that are both sassy and timely, we can refer to the circled genotypes in ${\color{Red} red}$ on the Punnett square. These boxes all contain two recessive s alleles and at least one dominant T allele.
Overall, there are 3 ${\color{Red} red}$ circled genotypes and thus, a 3/16 chance of offspring being sassy and timely. When converted to a percentage, 3/16 = 18.75%.
4. The correct answer is
Query $8$
Step-by-step solution
1. When a dominant allele is present, it will be expressed. Both parents are heterozygous (Kk), meaning that each will have smooth kernels because of the K allele.
The cross for these parents is Kk x Kk.
2. We can complete a Punnett square to find the possible offspring combinations.
Cross: Kk x Kk
If we complete the cross, we find that the possible offspring can be KK, Kk, or kk. Offspring that have a K allele must have smooth kernels because the allele for smooth kernels (K) is dominant to the allele for wrinkled kernels (k). Only kk offspring will have wrinkled kernels.
3. If we look at our Punnett square, we find that there is a 1/4 chance of offspring being homozygous dominant for smooth kernels (KK) and a 2/4 chance of that the offspring will be heterozygous (Kk) for smooth kernels.
Combined, we find that there is a 3/4 chance that the offspring will have smooth kernels.
4. The correct answer is
3/4
Query $9$
Step-by-step solution
1. Because there are two traits here, we must perform a dihybrid cross. We know the parent genotypes (AABb and aabb), but we need to figure out what the possible gametes are by figuring out all the possible combinations of the two alleles.
For the AABb parent, the possible gamete combinations are AB, AB, Ab, or Ab.
For the aabb parent, the possible gamete combinations are ab, ab, ab, or ab.
2. Next, we will perform the dihybrid cross.
Cross: AABb x aabb
By completing the cross, we find that the possible offspring genotypes are AaBb or Aabb.
3. Because we are looking for the proportion of offspring that are heterozygous for both traits, we need to identify offspring that are heterozygous for the awesome trait (Aa) and heterozygous for the bashful trait (Bb).
If we look at our Punnett square, the only dihybrid genotype that is heterozygous for both traits is AaBb.
There is an 8/16 chance of offspring being awesome and bashful, or heterozygous for both traits with the genotype AaBb.
8/16 = 1/2 = 50%
4. The correct answer is
50%
Query $10$
Step-by-step solution
1. The homozygous black horse has a genotype of BB. Homozygous means it has two of the same alleles, and it is black, so the allele is B.
The chestnut horse has a genotype of bb because it can only be chestnut if it has two recessive (b) alleles. If it had even one B allele, it would be black.
The cross for these parents is BB x bb.
2. We can complete a Punnett square to find the possible offspring combinations.
Cross: BB x bb
If we complete the cross, we find that the only possible offspring can only be Bb. Offspring that have a B allele must be black because the allele for black color (B) is dominant to the allele for chestnut color (b). So all of the offspring produced will be black.
3. If we look at our Punnett square, we find that there is a 0/4 chance of offspring being chestnut because Bb offspring will show the dominant trait (black). If we convert this to a percentage, it means there is a 0% chance these parents will have chestnut offspring.
4. The correct answer is
0%
Query $11$
Step-by-step solution
1. Because there are two traits here, we must perform a dihybrid cross. We know the parent genotypes (Ttgg and ttGg), but we need to figure out what the possible gametes are by figuring out all the possible combinations of the two alleles.
For the Ttgg parent, the possible gamete combinations are Tg, Tg, tg, or tg.
For the ttGg parent, the possible gamete combinations are tG, tG, tg, or tg.
2. Next, we will perform the dihybrid cross.
Cross: Ttgg x ttGg
By completing the cross, we find that the possible offspring are TtGg, ttGg, Ttgg or ttgg.
3. Because we are looking for the proportion of offspring that are short with yellow peas, we know that we are looking for a fully recessive offspring, or offspring with a ttgg genotype.
If we look at our Punnett square, we find that there is a 4/16 chance of offspring being short with yellow peas, or ttgg. We can then reduce this chance to 1/4.
4. The correct answer is
1/4
Query $12$
Step-by-step solution
1. The mother and father are both heterozygous, so they both can possibly give one of either allele (S or s) to the offspring. The cross for these parents is Ss x Ss.
2. We can complete a Punnett square to find the possible offspring genotypes.
Cross: Ss x Ss
If we complete the cross, we find that the possible offspring can be SS, Ss, or ss. Offspring that have at least one dominant S allele must be healthy (no sickle-cell anemia) because the allele for healthy (S) is dominant to the allele for sickle-cell anemia (s). Only ss offspring will have sickle-cell anemia.
3. If we complete the cross, we find that there are 2/4 possible offspring genotypes that are heterozygous (Ss). We can then reduce this chance to 1/2.
4. The correct answer is
1/2
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Thumbnail: Incomplete dominance in snapdragons. (CC BY-NC-SA / cropped from original; Khan Academy).
02: Non-Mendelian inheritance
In complete dominance, only one allele in the genotype is seen in the phenotype. In codominance, both alleles in the genotype are seen in the phenotype. In incomplete dominance, a mixture of the alleles in the genotype is seen in the phenotype. Created by Ross Firestone.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
2.02: Multiple alleles incomplete dominance and codominance
Introduction
Gregor Mendel knew how to keep things simple. In Mendel's work on pea plants, each gene came in just two different versions, or alleles, and these alleles had a nice, clear-cut dominance relationship (with the dominant allele fully overriding the recessive allele to determine the plant's appearance).
Today, we know that not all alleles behave quite as straightforwardly as in Mendel’s experiments. For example, in real life:
• Allele pairs may have a variety of dominance relationships (that is, one allele of the pair may not completely “hide” the other in the heterozygote).
• There are often many different alleles of a gene in a population.
In these cases, an organism's genotype, or set of alleles, still determines its phenotype, or observable features. However, a variety of alleles may interact with one another in different ways to specify phenotype.
As a side note, we're probably lucky that Mendel's pea genes didn't show these complexities. If they had, it’s possible that Mendel would not have understood his results, and wouldn't have figured out the core principles of inheritance—which are key in helping us understand the special cases!
Incomplete dominance
Mendel’s results were groundbreaking partly because they contradicted the (then-popular) idea that parents' traits were permanently blended in their offspring. In some cases, however, the phenotype of a heterozygous organism can actually be a blend between the phenotypes of its homozygous parents.
For example, in the snapdragon, Antirrhinum majus, a cross between a homozygous white-flowered plant (\(C^WC^W\)) and a homozygous red-flowered plant (\(C^RC^R\)) will produce offspring with pink flowers (\(C^RC^W\)). This type of relationship between alleles, with a heterozygote phenotype intermediate between the two homozygote phenotypes, is called incomplete dominance.
We can still use Mendel's model to predict the results of crosses for alleles that show incomplete dominance. For example, self-fertilization of a pink plant would produce a genotype ratio of \(1\, C^RC^R : 2\, C^RC^W : 1\, C^WC^W\) and a phenotype ratio of \(1:2:1\) red:pink:white. Alleles are still inherited according to Mendel's basic rules, even when they show incomplete dominance.
Codominance
Closely related to incomplete dominance is codominance, in which both alleles are simultaneously expressed in the heterozygote.
We can see an example of codominance in the MN blood groups of humans (less famous than the ABO blood groups, but still important!). A person's MN blood type is determined by his or her alleles of a certain gene. An \(L^M\) allele specifies production of an M marker displayed on the surface of red blood cells, while an \(L^N\) allele specifies production of a slightly different N marker.
Homozygotes (\(L^ML^M\) and \(L^NL^N\)) have only M or N markers, respectively, on the surface of their red blood cells. However, heterozygotes (\(L^ML^N\)) have both types of markers in equal numbers on the cell surface.
As for incomplete dominance, we can still use Mendel's rules to predict inheritance of codominant alleles. For example, if two people with \(L^ML^N\) genotypes had children, we would expect to see M, MN, and N blood types and \(L^ML^M\), \(L^ML^N\), and \(L^NL^N\) genotypes in their children in a \(1:2:1\) ratio (if they had enough children for us to determine ratios accurately!)
Multiple alleles
Mendel's work suggested that just two alleles existed for each gene. Today, we know that's not always, or even usually, the case! Although individual humans (and all diploid organisms) can only have two alleles for a given gene, multiple alleles may exist in a population level, and different individuals in the population may have different pairs of these alleles.
As an example, let’s consider a gene that specifies coat color in rabbits, called the \(C\) gene. The \(C\) gene comes in four common alleles: \(C\), \(c^{ch}\), \(c^h\), and \(c\):
• A \(CC\) rabbit has black or brown fur.
• A \(c^{ch}c^{ch}\) rabbit has chinchilla coloration (grayish fur).
• A \(c^hc^h\) rabbit has Himalayan (color-point) patterning, with a white body and dark ears, face, feet, and tail.
• A \(cc\) rabbit is albino, with a pure white coat.
Multiple alleles makes for many possible dominance relationships. In this case, the black \(C\) allele is completely dominant to all the others; the chinchilla \(c^{ch}\) allele is incompletely dominant to the Himalayan \(c^h\) and albino \(c\) alleles; and the Himalayan \(c^h\) allele is completely dominant to the albino \(c\) allele.
Rabbit breeders figured out these relationships by crossing different rabbits of different genotypes and observing the phenotypes of the heterozygous kits (baby bunnies).
[How do these alleles change the rabbit's color?]
The \(C\) gene in rabbits encodes an enzyme that’s needed to make a type of pigment called melanin in hairs1,2.
• The \(C\) allele of this gene encodes a fully functional enzyme that makes lots of pigment and results in black fur.
• The \(c^{ch}\) allele encodes an enzyme that is less effective at making pigment, resulting in lighter, more grayish fur.
• The \(c^h\) allele encodes a defective enzyme, where the defect makes the enzyme very sensitive to temperature: it works fine at low temperatures, but doesn’t work at all at higher temperatures. The rabbit's extremities (paws, ears, etc.) are cooler, so the enzyme functions there and makes pigment. The rabbit’s main body is warmer, so the enzyme does not function and no pigment is made.
• The \(c\) allele encodes a completely nonfunctional enzyme, leading to an albino rabbit (one that does not produce any pigment in its hairs).
Note to rabbit fanciers: Rabbit coat color is determined by a number of genes, not just by the \(C\) gene. The allelic series described here for \(C\) assumes a certain genetic background for the other genes, one in which a \(CC\) genotype results in a black rabbit.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
[Attribution and references]
Attribution:
This article is a modified derivative of “Characteristics and traits”,” by OpenStax College, Biology (CC BY 3.0). Download the original article for free at http://cnx.org/contents/[email protected].
The modified article is licensed under a CC BY-NC-SA 4.0 license.
Works cited:
1. Coat color in the Himalayan rabbit. (n.d.). In Cengage learning. Retrieved from http://www.cengage.com/biology/discipline_content/animations/himalayan_rabbit_m.html.
2. Hinkle, Amy. (n.d.). Rabbit coat color biochemistry. In Amy's rabbit ranch. Retrieved from http://www.amysrabbitranch.com/Color%26Genetics/Biochemistry-CoatColor.pdf.
Additional references:
Blumberg, R. B. (1997). Mendel's paper in English. In MendelWeb. Retrieved from http://www.mendelweb.org/Mendel.plain.html.
Coat color photo matrix. (2013, September 29).In Green barn farm. Retrieved from http://www.gbfarm.org/rabbit/holland-colors-matrix.shtml.
Color genetics: The C series. (n.d.) In The nature trail. Retrieved from http://www.thenaturetrail.com/rabbit-genetics/color-c-series-chinchilla-sable-himalayan-rew/.
Cystic fibrosis transmembrane conductance regulator. (2015, November 12). Retrieved November 22, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Cystic_fibrosis_transmembrane_conductance_regulator.
Dominance (genetics). (2015, November 3). Retrieved November 21, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Dominance_%28genetics%29.
Fox, Richard R. (1974). Taxonomy and genetics. In S. H. Weisbroth, R. E. Flatt, and A. L. Kraus (Eds.), The biology of the laboratory rabbit (6-22). New York, NY: Academic Press.
Orfano, Finn. (2010, December 15). Codominance in genetics: An overview. In Bright hub. Retrieved from http://www.brighthub.com/science/genetics/articles/99400.aspx.
Purves, W. K., Sadava, D., Orians, G. H., and Heller, H. C. (2003). Alleles and their interactions. In Life: The science of biology (7th ed., pp. 197-199). Sunderland, MA: Sinauer Associates.
Rabbit color genotypes chart. (n.d.). In The nature trail. http://www.thenaturetrail.com/rabbit-genetics/rabbit-color-genotypes-chart/.
Raven, P. H., Johnson, G. B., Mason, K. A., Losos, J. B., and Singer, S. R. (2014). Patterns of inheritance. In Biology (10th ed., AP ed., pp. 221-238). New York, NY: McGraw-Hill.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Mendel and the gene idea. In Campbell Biology (10th ed., pp. 267-291). San Francisco, CA: Pearson.
Sweat test. (2015, Septebmer 24). Retrieved November 22, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Sweat_test. | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/02%3A_Non-Mendelian_inheritance/2.01%3A_Co-dominance_and_Incomplete_Dominance.txt |
Introduction
From Mendel’s experiments, you might imagine that all genes control a single characteristic and affect some harmless aspect of an organism’s appearance (such as color, height, or shape). Those predictions are true for some genes, but definitely not all of them! For example:
• A human genetic disorder called Marfan syndrome is caused by a mutation in one gene, yet it affects many aspects of growth and development, including height, vision, and heart function. This is an example of pleiotropy, or one gene affecting multiple characteristics.
• A cross between two heterozygous yellow mice produces yellow and brown mice in a ratio of 2:1, not 3:1. This is an example of lethality, in which a particular genotype makes an organism unable to survive.
In this article, we’ll take a closer look at pleiotropic genes and lethal alleles, seeing how these variations on Mendel's rules fit into our modern understanding of inheritance.
Pleiotropy
When we mentioned Mendel’s experiments with purple-flowered and white-flowered plants, we didn’t discuss any other phenotypes associated with the two flower colors. However, Mendel noticed that the flower colors were always correlated with two other features: the color of the seed coat (covering of the seed) and the color of the axils (junctions where the leaves met the stem)1,2.
In plants with white flowers, the seed coats and axils were colorless. In plants with purple flowers, on the other hand, the seed coats were brown-gray and the axils were reddish. Thus, rather than affecting just one characteristic, the flower color gene actually affected three.
Genes like this, which control multiple, seemingly unrelated features, are said to be pleiotropic (pleio- = many, -tropic = effects)1. We now know that Mendel’s flower color gene specifies a protein that causes colored particles, or pigments, to be made2. This protein works in several different parts of the pea plant (flowers, seed coat, and leaf axils). In this way, the seemingly unrelated phenotypes can be traced back to a defect in one gene with several jobs.
Importantly, alleles of pleiotropic genes are transmitted in the same way as alleles of genes that affect single traits. Although the phenotype has multiple elements, these elements are specified as a package, and the dominant and recessive versions of the package would appear in the offspring of two heterozygotes in a ratio of 3:1.
Pleiotropy in human genetic disorders
Genes affected in human genetic disorders are often pleiotropic. For example, people with a hereditary disorder called Marfan syndrome may have a set of seemingly unrelated symptoms, including the following1,3:
• Unusually tall height
• Thin fingers and toes
• Dislocation of the lens of the eye
• Heart problems (in which the aorta, the large blood vessel carrying blood away from the heart, bulges or ruptures).
These symptoms don’t seem directly related, but as it turns out, they can all be traced back to the mutation of a single gene. This gene encodes a protein that assembles into chains, making elastic fibrils that give strength and flexibility to the body’s connective tissues4. Mutations that cause Marfan syndrome reduce the amount of functional protein made by the body, resulting in fewer fibrils.
How does the identity of this gene explain the range of symptoms? Our eyes and the aortas normally contain many fibrils that help maintain structure, which is why these two organs are affected in Marfan syndrome5. In addition, the fibrils serve as “storage shelves” for growth factors. When there are fewer of them in Marfan syndrome, the growth factors cannot be shelved and thus cause excess growth (leading to the characteristic tall, thin Marfan build)4.
Lethality
For the alleles that Mendel studied, it was equally possible to get homozygous dominant, homozygous recessive, and heterozygous genotypes. That is, none of these genotypes affected the survival of the pea plants. However, this is not the case for all genes and all alleles.
Many genes in an organism’s genome are needed for survival. If an allele makes one of these genes nonfunctional, or causes it to take on an abnormal, harmful activity, it may be impossible to get a living organism with a homozygous (or, in some cases, even a heterozygous) genotype.
Example: The yellow mouse
A classic example of an allele that affects survival is the lethal yellow allele, a spontaneous mutation in mice that makes their coats yellow. This allele was discovered around the turn of the 20th century by the French geneticist Lucien Cuenót, who noticed that it was inherited in an unusual pattern6,7.
When yellow mice were crossed with normal agouti (brown) mice, they produced half yellow and half brown offspring. This suggested that the yellow mice were heterozygous, and that the yellow allele, \(A^Y\), was dominant to the agouti allele, \(A\). But when two yellow mice were crossed with each other, they produced yellow and brown offspring in a ratio of 2:1, and the yellow offspring did not breed true (were heterozygous). Why was this the case?
As it turned out, this unusual ratio reflected that some of the mouse embryos (homozygous \(A^YA^Y\) genotype) died very early in development, long before birth. In other words, at the level of eggs, sperm, and fertilization, the color gene segregated normally, resulting in embryos with a 1:2:1 ratio of \(A^YA^Y\), \(A^YA\), and \(AA\) genotypes. However, the \(A^YA^Y\) mice died as tiny embryos, leaving a 2:1 genotype and phenotype ratio among the surviving mice7,8.
Alleles like \(A^Y\), which are lethal when they're homozygous but not when they're heterozygous, are called recessive lethal alleles.
[Wait, isn't yellow color dominant?]
The lethal recessive \(A^Y\) allele also happens to give a dominant, non-lethal phenotype in the heterozygote (yellow coloration, along with health problems such as obesity, diabetes, and tumor formation). The trick is that the lethality is recessive, even though the color phenotype is dominant.
The color of the yellow mouse heterozygote is helpful because it lets us keep track of genotypes, but in many other cases, recessive lethal genes don’t have a dominant phenotype in the heterozygote. Instead, they are purely recessive to the normal allele, resulting in a normal phenotype for the heterozygote and an embryonic lethal phenotype for the homozygote.
Lethal alleles and human genetic disorders
Some alleles associated with human genetic disorders are recessive lethal. For example, this is true of the allele that causes achondroplasia, a form of dwarfism. A person heterozygous for this allele will have shortened limbs and short stature (achondroplasia), a condition that is not lethal. However, homozygosity for the same allele causes death during embryonic development or the first months of life, an example of recessive lethality7,9.
Some human disorders are also caused by dominant lethal alleles. These are alleles that cause death when they are present in just a single copy. If an allele leads to death of heterozygotes before birth, we’ll never see that allele in the living human population (but rather, as an implantation failure or miscarriage). However, if a dominant lethal allele allows heterozygotes to survive past birth, it can be seen in the population as a genetic disorder.
In fact, if a dominant lethal allele lets a person survive to reproductive age, it may even be passed on to children. This is the case in Huntington’s disease, a fatal genetic disorder affecting the nervous system. People with a Huntington allele inevitably develop the disease, but they may not show any symptoms until age 40 and can unknowingly pass the allele on to their children.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
[Attribution and references]
Attribution:
This article is a modified derivative of “Characteristics and traits,” by OpenStax College, Biology (CC BY 3.0). Download the original article for free at http://cnx.org/contents/[email protected].
The modified article is licensed under a CC BY-NC-SA 4.0 license.
Works cited:
1. Lobo, I. (2008). Pleiotropy: One gene can affect multiple traits. Nature Education, 1(1), 10. Retrieved from www.nature.com/scitable/topicpage/pleiotropy-one-gene-can-affect-multiple-traits-569.
2. Reid, J. B. and Ross, J. J. (2011). Mendel's genes: Towards a full molecular characterization. Genetics 189(1), 3-10. Retrieved from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3176118/#s4title.
3. Marfan syndrome. (2012). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/condition/marfan-syndrome.
4. FBN1. (2015). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/gene/FBN1.
5. Marfan syndrome. (2015, November 3). Retrieved November 21, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Marfan_syndrome.
6. Lethal allele. (2016, July 13). Retrieved July 26, 2016 from Wikipedia: https://en.wikipedia.org/wiki/Lethal_allele#History.
7. Lobo, I. (2008). Mendelian ratios and lethal genes. Nature Education, 1(1), 138. Retrieved from http://www.nature.com/scitable/topicpage/mendelian-ratios-and-lethal-genes-557.
8. Griffiths, A. J. F., Gelbart, W. M., Miller, J. H., and Lewontin, R. C. (1999). Interactions between alleles of one gene. In Modern Genetic Analysis. New York, NY: W. H. Freeman. Retrieved from http://www.ncbi.nlm.nih.gov/books/NBK21226/#_A824_.
9. Achrondroplasia. (2015, September 12). Retrieved November 21, 2015 from WIkipedia: https://en.wikipedia.org/wiki/Achondroplasia.
Additional references:
Achrondroplasia. (2015, September 12). Retrieved November 21, 2015 from WIkipedia: https://en.wikipedia.org/wiki/Achondroplasia.
Bergmann, D. C. (2011). Genetics lecture notes. Biosci 41, Stanford.
FBN1. (2015). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/gene/FBN1.
Griffiths, A. J. F., Gelbart, W. M., Miller, J. H., and Lewontin, R. C. (1999). Interactions between alleles of one gene. In Modern Genetic Analysis. New York, NY: W. H. Freeman. Retrieved from http://www.ncbi.nlm.nih.gov/books/NBK21226.
Lethal allele. (2015). In The free dictionary. Retrieved from http://medical-dictionary.thefreedictionary.com/lethal+allele.
Lethal allele. (2016, July 13). Retrieved July 26, 2016 from Wikipedia: https://en.wikipedia.org/wiki/Lethal_allele.
Lobo, I. (2008). Mendelian ratios and lethal genes. Nature Education, 1(1), 138. Retrieved from http://www.nature.com/scitable/topicpage/mendelian-ratios-and-lethal-genes-557.
Lobo, I. (2008). Pleiotropy: One gene can affect multiple traits. Nature Education, 1(1), 10. Retrieved from www.nature.com/scitable/topicpage/pleiotropy-one-gene-can-affect-multiple-traits-569.
Lucien Cuénot. (2016, February 13). Retrieved July 26, 2016 from Wikipedia: https://en.wikipedia.org/wiki/Lucien_Cuénot.
Marfan syndrome. (2012). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/condition/marfan-syndrome.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Mendel and the gene idea. In Campbell biology (10th ed., pp. 267-291). San Francisco, CA: Pearson.
Reid, J. B. and Ross, J. J. (2011). Mendel's genes: Towards a full molecular characterization. Genetics 189(1), 3-10. Retrieved from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3176118/ | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/02%3A_Non-Mendelian_inheritance/2.03%3A_Pleiotropy_and_lethal_alleles.txt |
How is height inherited?
If what you're really interested in is human genetics, learning about Mendelian genetics can sometimes be frustrating. You'll often hear a teacher use a human trait as an example in a genetics problem, but then say, "that's an oversimplification" or "it's much more complicated than that." So, what's actually going on with those interesting human traits, such as eye color, hair and skin color, height, and disease risk?
As an example, let's consider human height. Unlike a simple Mendelian characteristic, human height displays:
• Continuous variation. Unlike Mendel's pea plants, humans don’t come in two clear-cut “tall” and “short” varieties. In fact, they don't even come in four heights, or eight, or sixteen. Instead, it’s possible to get humans of many different heights, and height can vary in increments of inches or fractions of inches1.
• A complex inheritance pattern. You may have noticed that tall parents can have a short child, short parents can have a tall child, and two parents of different heights may or may not have a child in the middle. Also, siblings with the same two parents may have a range of heights, ones that don't fall into distinct categories.
Simple models involving one or two genes can't accurately predict all of these inheritance patterns. How, then, is height inherited?
Height and other similar features are controlled not just by one gene, but rather, by multiple (often many) genes that each make a small contribution to the overall outcome. This inheritance pattern is sometimes called polygenic inheritance (poly- = many). For instance, a recent study found over 400 genes linked to variation in height2.
When there are large numbers of genes involved, it becomes hard to distinguish the effect of each individual gene, and even harder to see that gene variants (alleles) are inherited according to Mendelian rules. In an additional complication, height doesn’t just depend on genetics: it also depends on environmental factors, such as a child’s overall health and the type of nutrition he or she gets while growing up.
In this article, we’ll examine how complex traits such as height are inherited. We'll also see how factors like genetic background and environment can affect the phenotype (observable features) produced by a particular genotype (set of gene variants, or alleles).
Polygenic inheritance
Human features like height, eye color, and hair color come in lots of slightly different forms because they are controlled by many genes, each of which contributes some amount to the overall phenotype. For example, there are two major eye color genes, but at least 14 other genes that play roles in determining a person’s exact eye color3.
Looking at a real example of a human polygenic trait would get complicated, largely because we’d have to keep track of tens, or even hundreds, of different allele pairs (like the 400 involved in height!). However, we can use an example involving wheat kernels to see how several genes whose alleles "add up" to influence the same trait can produce a spectrum of phenotypes1,4.
In this example, there are three genes that make reddish pigment in wheat kernels, which we’ll call A, B, and C. Each comes in two alleles, one of which makes pigment (the capital-letter allele) and one of which does not (the lowercase allele). These alleles have additive effects: the aa genotype would contribute no pigment, the Aa genotype would contribute some amount of pigment, and the AA genotype would contribute more pigment (twice as much as Aa). The same would hold true for the B and C genes1,4.
Now, let’s imagine that two plants heterozygous for all three genes (AaBbCc) were crossed to one another. Each of the parent plants would have three alleles that made pigment, leading to pinkish kernels. Their offspring, however, would fall into seven color groups, ranging from no pigment whatsoever (aabbcc) and white kernels to lots of pigment (AABBCC) and dark red kernels. This is in fact what researchers have seen when crossing certain varieties of wheat1,4.
This example shows how we can get a spectrum of slightly different phenotypes (something close to continuous variation) with just three genes. It’s not hard to imagine that, as we increased the number of genes involved, we’d be able to get even finer variations in color, or in another trait such as height.
Environmental effects
Human phenotypes—and phenotypes of other organisms—also vary because they are affected by the environment. For instance, a person may have a genetic tendency to be underweight or obese, but his or her actual weight will depend on diet and exercise (with these factors often playing a greater role than genes). In another example, your hair color may depend on your genes—until you dye your hair purple!
One striking example of how environment can affect phenotype comes from the hereditary disorder phenylketonuria (PKU)6. People who are homozygous for disease alleles of the PKU gene lack activity of an enzyme that breaks down the amino acid phenylalanine. Because people with this disorder cannot get rid of excess phenylalanine, it rapidly builds up to toxic levels in their bodies7.
If PKU is not treated, the extra phenylalanine can keep the brain from developing normally, leading to intellectual disability, seizures, and mood disorders. However, because PKU is caused by the buildup of too much phenylalanine, it can also be treated in a very simple way: by giving affected babies and children a diet low in phenylalanine8.
If people with phenylketonuria follow this diet strictly from a very young age, they can have few, or even no, symptoms of the disorder. In many countries, all newborns are screened for PKU and similar genetic diseases shortly after birth through a simple blood test, as shown in the image below.
Variable expressivity, incomplete penetrance
Even for characteristics that are controlled by a single gene, it’s possible for individuals with the same genotype to have different phenotypes. For example, in the case of a genetic disorder, people with the same disease genotype may have stronger or weaker forms of the disorder, and some may never develop the disorder at all.
In variable expressivity, a phenotype may be stronger or weaker in different people with the same genotype. For instance, in a group of people with a disease-causing genotype, some might develop a severe form of the disorder, while others might have a milder form. The idea of expressivity is illustrated in the diagram below, with the shade of green representing the strength of the phenotype.
[Example]
The genetic disorder retinoblastoma causes cancerous tumors of the eyes, but the disease varies in severity and speed of onset. Children with retinoblastoma may develop tumors in just one eye or in both eyes, and the tumors may appear more quickly or slowly after birth9.
In incomplete penetrance, individuals with a certain genotype may or may not develop a phenotype associated with the genotype. For example, among people with the same disease-causing genotype for a hereditary disorder, some might never actually develop the disorder. The idea of penetrance is illustrated in the diagram below, with green or white color representing the presence or absence of a phenotype.
[Example]
Some mutations in the retinoblastoma gene have high penetrance, but others have lower (incomplete) penetrance. For mutations in this latter category, some family members with the mutation are normal, while others develop retinoblastoma tumors6,11.
What causes variable expressivity and incomplete penetrance? Other genes and environmental effects are often part of the explanation. For example, disease-causing alleles of one gene may be suppressed by alleles of another gene elsewhere in the genome, or a person's overall health may influence the strength of a disease phenotype11.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
[Attribution and references]
This article is licensed under a CC BY-NC-SA 4.0 license.
Works cited:
1. Kimball, J. W. (2011, March 8). Continuous variation: Quantitative traits. Retrieved from http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/Q/QTL.html.
2. Wood, A. R., Esko, T., Yang, J., Vedantam, S., Pers, T. H., Gustafsson, S., ... Frayling, T. M. (2014). Defining the role of common variation in the genomic and biological architecture of adult human height. Nature Genetics, 46, 1173-1186. http://dx.doi.org/10.1038/ng.3097.
3. White, D. and Rabago-Smith, M. (2011). Genotype-phenotype associations and human eye color. Journal of Human Genetics, 56, 5-7. http://dx.doi.org/10.1038/jhg.2010.126.
4. Department of Agronomy, Iowa State University. (2016). Inheritance of a quantitative trait. Retrieved July 26, 2016 from https://masters.agron.iastate.edu/classes/527/lesson07/detail/kernelColor.html.
5. Armstrong, W. P. (n.d.). Continuous variation and Rh blood factor. In Wayne's Word. Retrieved July 26, 2016 from http://waynesword.palomar.edu/lmexer5.htm.
6. Bergmann, D. C. (2011). Variations on and complications of the basic Mendelian rules. In Genetics lecture notes. Biosci41, Stanford University, 10.
7. PAH. (2008). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/gene/PAH.
8. Phenylketonuria. (2012). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/condition/phenylketonuria.
9. National Cancer Institute. (2015, August 14). Retinoblastoma treatment – for health professionals. In Cancer types. Retrieved from http://www.cancer.gov/types/retinoblastoma/hp/retinoblastoma-treatment-pdq.
10. Carr, S. M. (2014). Penetrance versus expressivity. Retrieved from https://www.mun.ca/biology/scarr/Penetrance_vs_Expressivity.html.
11. Kelly, Jane. (2015, September 14). Retinoblastoma. In OMIM. Retrieved from http://www.omim.org/entry/180200.
12. Griffiths, A. J. F., Miller, J. H., Suzuki, D. T., Lewontin, R. C., and Gelbart, W. M. (2000). Penetrance and expressivity. In An introduction to genetic analysis (7th ed.). Retrieved from http://www.ncbi.nlm.nih.gov/books/NBK22090/. | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/02%3A_Non-Mendelian_inheritance/2.04%3A_Polygenic_inheritance_and_environmental_effects.txt |
Key terms
Term Meaning
Incomplete dominance Pattern of heredity in which one allele is not completely dominant over another
Codominance Pattern of heredity in which both alleles are simultaneously expressed in the heterozygote
Multiple alleles A gene that is controlled by more than two alleles
Pleiotropy When one gene affects multiple characteristics
Lethal allele Allele that results in the death of an individual
Polygenic trait Traits that are controlled by multiple genes
Variations involving single genes
Some of the variations on Mendel’s rules involve single genes.
• Incomplete dominance. Two alleles may produce an intermediate phenotype when both are present, rather than one fully determining the phenotype.
An example of this is the snapdragon plant. A cross between a homozygous white-flowered plant (\(C^WC^W\)) and a homozygous red-flowered plant (\(C^RC^R\)) will produce offspring with pink flowers (\(C^RC^W\)).
• Codominance. Two alleles may be simultaneously expressed when both are present, rather than one fully determining the phenotype.
In some varieties of chickens, the allele for black feathers is codominant with the allele for white feathers. A cross between a black chicken and a white chicken will result in a chicken with both black and white feathers.
• Multiple alleles. Mendel studied just two alleles of his pea genes, but real populations often have multiple alleles of a given gene.
An example of this is the gene for coat color in rabbits (the \(C\) gene) which comes in four common alleles: \(C\), \(c^{ch}\), \(c^h\), and \(c\).
• Pleiotropy. Some genes affect many different characteristics, not just a single characteristic.
An example of this is Marfan syndrome, which results in several symptoms (unusually tall height, thin fingers and toes, lens dislocation, and heart problems). These symptoms don’t seem directly related, but as it turns out, they can all be traced back to the mutation of a single gene.
• Lethal alleles. Some genes have alleles that prevent survival when homozygous or heterozygous.
A classic example of an allele that affects survival is the lethal yellow allele, a spontaneous mutation in mice that makes their coats yellow. Mice that are homozygous (\(A^YA^Y\)) genotype die early in development. Although this particular allele is dominant, lethal alleles can be dominant or recessive, and can be expressed in homozygous or heterozygous conditions.
Polygenic inheritance and environmental effects
Many characteristics, such as height, skin color, eye color, and risk of diseases, are controlled by many factors. These factors may be genetic, environmental, or both.
• Polygenic inheritance. Some characteristics are polygenic, meaning that they’re controlled by a number of different genes. In polygenic inheritance, traits often form a phenotypic spectrum rather than falling into clear-cut categories.
An example of this is skin pigmentation in humans, which is controlled by several different genes.
• Environmental effects. Most real-world characteristics are determined not just by genotype, but also by environmental factors that influence how genotype is translated into phenotype.
An example of this is the hydrangea flower. Hydrangea of the same genetic variety may vary in color from blue to pink depending on the pH of the soil they are in.
Common mistakes and misconceptions
• Some people confuse pleiotropy and polygenic inheritance. The major difference between the two is that pleiotropy is when one gene affects multiple characteristics (e.g. Marfan syndrome) and polygenic inheritance is when one trait is controlled by multiple genes (e.g. skin pigmentation).
• Codominance and incomplete dominance are not the same. In codominance, neither allele is dominant over the other, so both will be expressed equally in the heterozygote. In incomplete dominance, there is an intermediate heterozygote (such as a pink flower when the parents' phenotypes are red and white).
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org) | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/02%3A_Non-Mendelian_inheritance/2.05%3A_Non-Mendelian_inheritance_review.txt |
Query \(1\)
Step-by-step solution
1. In incomplete dominance, heterozygotes show an intermediate phenotype, such as blue-tipped feathers.
The blue-tipped parent's genotype is Bb and the white parent's genotype is bb, so the cross for these parents is Bb x bb.
We can complete a Punnett square to find the possible offspring combinations.
If we complete the cross, we find that the possible offspring can be either Bb or bb.
2. Bb offspring will have blue-tipped feathers, and bb offspring will have white feathers.
In order to have blue feathers, the offspring's genotype must be BB. Because this is not a possible genotype of this cross, 0/4 or 0% of the offspring can be blue.
3. The correct answer is
0%
Query \(2\)
Step-by-step solution
1. The lethal yellow allele is a spontaneous mutation in mice that makes their coats yellow. Mice that have a homozygous (\(A^YA^Y\)) genotype die early in development.
2. Some genes have alleles that prevent survival when homozygous or heterozygous. These are known as lethal alleles.
3. The correct answer is
Lethal allele
Query \(3\)
Step-by-step solution
1. In codominance, both traits are dominant and will be expressed equally if present.
The cross for these parents is WR x WR.
We can complete a Punnett square to find the possible offspring combinations.
If we complete the cross, we find that the possible offspring can be WW, WR, or RR.
2. 1/4 of the offspring will be white (WW), 1/4 of the offspring will be red (RR), and 2/4 of the offspring will be roan (WR).
2/4 WR = 1/2 = 50%
3. The correct answer is
50%
Query \(4\)
Step-by-step solution
1. Lethal alleles result in the death of affected individuals.
Multiple alleles refers to a gene that is controlled by more than two alleles.
2. Codominance refers to traits that are both expressed at the same time in heterozygotes.
If this was an example of codominance, the offspring would have both black and white hairs.
3. Incomplete dominance is the blending of alleles, resulting in a phenotype that is in between the two extremes.
In this example, the gray color is an intermediate between the black and the white coats.
4. The correct answer is
Incomplete dominance
Query \(5\)
Step-by-step solution
1. Many characteristics, such as height, skin color, eye color, and risk of diseases, are controlled by many factors. These factors may be genetic, environmental, or both.
2. Some characteristics are polygenic, meaning that they’re controlled by a number of different genes. In polygenic inheritance, traits often form a phenotypic spectrum rather than falling into clear-cut categories.
3. The correct answer is
Polygenic inheritance; it involves multiples genes coding for a trait that falls within a wide spectrum
Query \(6\)
Step-by-step solution
1. The cross for these parents is AB x AO.
We can complete a Punnett square to find the possible offspring combinations.
If we complete the cross, we find that the possible offspring can be AA, AB, AO or BO.
2. The offspring that are AA or AO will have A type blood and the offspring that are BO will have B type blood.
In order to have AB type blood, offspring must be AB because of codominance. Therefore, 1/4 or 25% of the offspring will have AB type blood.
3. The correct answer is
25%
Query \(7\)
Step-by-step solution
1. In incomplete dominance, heterozygotes show an intermediate phenotype, such as wavy hair.
The cross for these parents is Hh x Hh.
We can complete a Punnett square to find the possible offspring combinations.
If we complete the cross, we find that the possible offspring can be HH, Hh, or hh.
2. HH offspring will have curly hair, hh offspring will have straight hair, and heterozygous offspring (Hh) will have the intermediate phenotype, wavy hair.
Based on the Punnett square, the overall chance of having wavy haired (Hh) children is:
2/4 Hh = 1/2 wavy hair
3. The correct answer is
1/2
Query \(8\)
Step-by-step solution
1.
This diagram shows how the human ABO blood type system works.
2. Codominance is a pattern of heredity in which both alleles are simultaneously expressed in the heterozygote (e.g. AB blood type).
Multiple allele inheritance is when a gene that is controlled by more than two alleles (e.g. A, B, and O alleles).
3. The correct answers are
• The A and B alleles are codominant because both alleles are simultaneously expressed in the heterozygote.
• Human blood type is an example of multiple allele inheritance.
Query \(9\)
Step-by-step solution
1. In codominance, both traits are dominant and will be expressed equally if present.
The cross for these parents is RR x RW.
We can complete a Punnett square to find the possible offspring combinations.
If we complete the cross, we find that the possible offspring can be either RR or RW.
2. 2/4, or 50%, of the offspring will be red (RR) and 2/4, or 50%, of the offspring will be red-and-white (RW).
3. The correct answer is
50% of the offspring will be red and 50% of the offspring will be red-and-white.
Query \(10\)
Step-by-step solution
1.
This diagram describes how pleiotropy works.
2.
These human skin color charts exemplify polygenic inheritance.
3. The major difference between the two is that pleiotropy is when one gene affects multiple characteristics (e.g. Marfan syndrome) and polygenic inheritance is when one trait is controlled by multiple genes (e.g. skin pigmentation).
4. The correct answer is
Pleiotropy is when one gene affects multiple characteristics (e.g. Marfan syndrome).
Query \(11\)
Step-by-step solution
1. Pleiotropy occurs when one gene is controlled by multiple alleles and polygenic inheritance occurs when one trait is controlled by multiple genes.
2. Incomplete dominance is the blending of alleles, resulting in a phenotype that is in between the two extremes.
If this was an example of incomplete dominance, we would see corn with an intermediate color (such as light yellow) rather than corn that has mixed kernels.
3. Codominance refers to traits that are both expressed at the same time in heterozygotes.
In this example, both the yellow kernels and the white kernels are expressed in the phenotype.
4. The correct answer is
Codominance
Query \(12\)
Step-by-step solution
1. The cross for these parents is cbc x Ccb.
We can complete a Punnett square to find the possible offspring combinations.
If we complete the cross, we find that the possible offspring can be Ccb, Cc, cbcb, or cbc.
2. Because C (agouti) is dominant to both other alleles, it will be displayed if present. This means the offspring that are Ccb or Cc will have agouti coats (2/4 = 50%).
cb (black) is recessive to C but dominant to c, so the cbcb and cbc will have black fur (2/4 = 50%).
3. The correct answer is
50% of the offspring will have agouti coats and 50% will have black coats.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org) | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/02%3A_Non-Mendelian_inheritance/2.06%3A_Practice_-_Non-Mendelian_inheritance.txt |
Thumbnail: Tortoiseshell cat. (Public domain; Michael Bodega via Wikimedia Commons).
03: Sex linkage
Example punnet square for sex-linked recessive trait.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
3.02: X-linked inheritance
Key points:
• In humans and other mammals, biological sex is determined by a pair of sex chromosomes: XY in males and XX in females.
• Genes on the X chromosome are said to be X-linked. X-linked genes have distinctive inheritance patterns because they are present in different numbers in females (XX) and males (XY).
• X-linked human genetic disorders are much more common in males than in females due to the X-linked inheritance pattern.
Introduction
If you’re a human being (which seems like a good bet!), most of your chromosomes come in homologous pairs. The two chromosomes of a homologous pair contain the same basic information – that is, the same genes in the same order – but may carry different versions of those genes.
Are all of your chromosomes organized in homologous pairs? The answer depends on whether you’re (chromosomally) male.
• A human male has two sex chromosomes, the X and the Y. Unlike the 44 autosomes (non-sex chromosomes), the X and Y don’t carry the same genes and aren’t considered homologous.
• Instead of an X and a Y, a human female has two X chromosomes. These X chromosomes do form a bona fide homologous pair.
Because sex chromosomes don’t always come in homologous pairs, the genes they carry show unique, distinctive patterns of inheritance.
Sex chromosomes in humans
Human X and Y chromosomes determine the biological sex of a person, with XX specifying female and XY specifying male. Although the Y chromosome contains a small region of similarity to the X chromosome so that they can pair during meiosis, the Y chromosome is much shorter and contains many fewer genes.
To put some numbers to it, the X chromosome has about 800-900 protein-coding genes with a wide variety of functions, while the Y chromosome has just 60-70 protein-coding genes, about half of which are active only in the testes (sperm-producing organs)1,2,3,4.
The human Y chromosome plays a key role in determining the sex of a developing embryo. This is mostly due to a gene called SRY (“sex-determining region of Y”). SRY is found on the Y chromosome and encodes a protein that turns on other genes required for male development5,6.
• XX embryos don't have SRY, so they develop as female.
• XY embryos do have SRY, so they develop as male.
In rare cases, errors during meiosis may transfer SRY from the Y chromosome to the X chromosome. If an SRY-bearing X chromosome fertilizes a normal egg, it will produce a chromosomally female (XX) embryo that develops as a male7. If an SRY-deficient Y chromosome fertilizes a normal egg, it will produce a chromosomally male embryo (XY) that develops as a female8.
[Do all species with X and Y chromosomes have the SRY gene?]
No, not all of them. The molecular mechanisms of sex determination vary greatly among species, even those that use an X-Y sex determination system. In most placental mammals, SRY is found on the Y chromosome and is used for sex determination. However, the SRY gene is not present in other species with X-Y sex determination systems, such as fruit flies and other insects9.
X and Y chromosomes have evolved independently many times
To understand how this is possible, it's useful to keep in mind that “X” and “Y” are just generic labels applied to the dimorphic (di- = two, -morphe = form), or dissimilar, chromosomes found in species with X-Y sex determination systems10. X is whatever chromosome the female has two of, while Y is the chromosome paired with a single X in the male. The nature of the chromosomes — what genes are on them, and how they determine maleness or femaleness — may be quite different between species.
Sex chromosomes have evolved independently many times11. Thus, although fruit flies and humans both have X and Y chromosomes that match the above definition, and although XY and XX chromosome combinations correspond to maleness and femaleness, respectively, in both species, the sex determination mechanism in flies is very different from that of humans.
Example: Humans vs. fruit flies
In humans, as described above, the Y chromosome specifies maleness through the action of the SRY gene, a master regulator of male development.
In flies, the presence of a single X chromosome specifies maleness, while the presence of two Xs specifies femaleness, regardless of the presence or absence of a Y chromosome. (More specifically, it appears that sex is determined by the ratio of X chromosomes to autosomes, with a 1:2 ratio specifying maleness and a 2:2 ratio specifying femaleness. These ratios are "measured" through the levels of proteins produced by specific genes on the X chromosome and on autosomes.)12
Because of these differences in sex determination mechanisms, an XXY fly will develop as a female (see article on Thomas Hunt Morgan's experiments), while an XXY human will develop as a male (in a condition known as Klinefelter syndrome; see article on large-scale chromosomal changes).
X-linked genes
When a gene is present on the X chromosome, but not on the Y chromosome, it is said to be X-linked. X-linked genes have different inheritance patterns than genes on non-sex chromosomes (autosomes). That's because these genes are present in different copy numbers in males and females.
[What about genes on the Y?]
Genes on either the X or the Y have unusual inheritance patterns and are called sex-linked, but X-linkage is much more common than Y-linkage (since there are many more genes on the X than the Y).
Since a female has two X chromosomes, she will have two copies of each X-linked gene. For instance, in the fruit fly Drosophila (which, like humans, has XX females and XY males), there is an eye color gene called white that's found on the X chromosome, and a female fly will have two copies of this gene. If the gene comes in two different alleles, such as $\text{X}^W$ (dominant, normal red eyes) and $\text{X}^w$ (recessive, white eyes), the female fly may have any of three genotypes: $\text{X}^W\text{X}^W$ (red eyes), $\text{X}^W\text{X}^w$ (red eyes), and $\text{X}^w\text{X}^w$ (white eyes).
A male has different genotype possibilities than a female. Since he has only one X chromosome (paired with a Y), he will have only one copy of any X-linked genes. For instance, in the fly eye color example, the two genotypes a male can have are $\text{X}^W\text{Y}$ (red eyes) and $\text{X}^w\text{Y}$ (white eyes). Whatever allele the male fly inherits for an X-linked gene will determine his appearance, because he has no other gene copy—even if the allele is recessive in females. Rather than homozygous or heterozygous, males are said to be hemizygous for X-linked genes.
We can see how sex linkage affects inheritance patterns by considering a cross between two flies, a white-eyed female ($\text{X}^w\text{X}^w$) and a red-eyed male ($\text{X}^W\text{Y}$). If this gene were on a non-sex chromosome, or autosome, we would expect all of the offspring to be red-eyed, because the red allele is dominant to the white allele. What we actually see is the following:
However, because the gene is X-linked, and because it was the female parent who had the recessive phenotype (white eyes), all the male offspring—who get their only X from their mother—have white eyes ($\text{X}^w\text{Y}$). All the female offspring have red eyes because they received two Xs, with the $\text{X}^W$ from the father concealing the recessive $\text{X}^w$ from the mother.
X-linked genetic disorders
The same principles we see at work in fruit flies can be applied to human genetics. In humans, the alleles for certain conditions (including some forms of color blindness, hemophilia, and muscular dystrophy) are X-linked. These diseases are much more common in men than they are in women due to their X-linked inheritance pattern.
Why is this the case? Let's explore this using an example in which a mother is heterozygous for a disease-causing allele. Women who are heterozygous for disease alleles are said to be carriers, and they usually don't display any symptoms themselves. Sons of these women have a 50% chance of getting the disorder, but daughters have little chance of getting the disorder (unless the father also has it), and will instead have a 50% chance of being carriers.
Why is this the case? Recessive X-linked traits appear more often in males than females because, if a male receives a "bad" allele from his mother, he has no chance of getting a "good" allele from his father (who provides a Y) to hide the bad one. Females, on the other hand, will often receive a normal allele from their fathers, preventing the disease allele from being expressed.
Case study: Hemophilia
Let's look at a Punnett square example using an X-linked human disorder: hemophilia, a recessive condition in which a person's blood does not clot properly13. A person with hemophilia may have severe, even life-threatening, bleeding from just a small cut.
Hemophilia is caused by a mutation in either of two genes, both of which are located on the X chromosome. Both genes encode proteins that help blood clot14. Let's focus on just one of these genes, calling the functional allele $\text{X}^H$ and the disease allele $\text{X}^h$.
In our example, a woman who is heterozygous for normal and hemophilia alleles ($\text{X}^H\text{X}^h$) has children with a man who is hemizygous for the normal form ($\text{X}^H\text{Y}$). Both parents have normal blood clotting, but the mother is a carrier. What is the chance of their sons and daughters having hemophilia?
Since the mother is a carrier, she will pass on the hemophilia allele ($\text{X}^h$) on to half of her children, both boys and girls.
• None of the daughters will have hemophilia (zero chance of the disorder). That's because, in order to have the disorder, they must get a $\text{X}^h$ allele from both their mother and their father. There is 0 chance of the daughters getting an $\text{X}^h$ allele from their father, so their overall chance of having hemophilia is zero.
• The sons get a Y from their father instead of an X, so their only copy of the blood clotting gene comes from their mother. The mother is heterozygous, so half of the sons, on average, will get an $\text{X}^h$ allele and have hemophilia (1/2 chance of the disorder).
[Can a woman ever have hemophilia?]
Yes, it's possible for a woman to have a recessive, X-linked condition such as hemophilia. However, she must get two recessive copies of the X-linked gene (one from each parent) in order to have the condition. The odds of this are much lower than the odds of a man getting just one recessive disease allele from his mother.
Check your understanding
Query $1$
[Hint]
Because hemophilia is a recessive disorder, a woman must receive two disease alleles (one on each X chromosome) in order to display the disease. Thus, she must get a disease allele from both her mother and her father.
Since hemophilia is an X-linked disorder, males are hemizygous for the hemophilia-related gene (have only one allele and display the phenotype associated with that allele). In order to have a hemophilia allele that can be passed on to offspring, a male must himself be hemophiliac. Pairs of parents in which the male is not hemophiliac cannot produce a hemophiliac daughter, barring rare events such as spontaneous mutations in the the germline or during development of the embryo.
Women who are either homozygous for the hemophilia allele (hemophiliac) or heterozygous for the hemophilia allele (unaffected carriers) may pass on a hemophilia allele to their offspring. Pairs of parents in which the female is neither hemophiliac nor a carrier cannot produce a hemophiliac daughter (again, barring rare spontaneous mutation events).
Of the pairs above, a carrier mother and a hemophiliac father are the most likely to have a hemophiliac daughter.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
[Attribution and references]
Attribution:
This article is a modified derivative of "Characteristics and traits," by OpenStax College, Biology, CC BY 4.0. Download the original article for free at http://cnx.org/contents/[email protected].
This article is licensed under a CC BY-NC-SA 4.0 license.
Works cited:
1. X chromosome. (2015, December 11). Retrieved December 11, 2015 from Wikipedia: https://en.wikipedia.org/wiki/X_chromosome.
2. Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). The chromosomal basis of sex. In Campbell biology (10th ed., pp. 296-297). San Francisco, CA: Pearson.
3. X chromosome. (2012). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/chromosome/X.
4. Y chromosome. (2010). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/chromosome/Y.
5. Meiosis. (2015). In HHMI biointeractive. Retrieved from http://www.hhmi.org/biointeractive/meiosis.
6. SRY. (2015). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/gene/SRY.
7. 46,XX testicular disorder of sex development. (2008). In Genetics home reference. Retrieved from https://ghr.nlm.nih.gov/condition/46xx-testicular-disorder-of-sex-development.
8. Swyer syndrome. (2015). In Genetics home reference. Retrieved from https://ghr.nlm.nih.gov/condition/swyer-syndrome.
9. King, V., Goodfellow, P. N., Pearks Wilkerson, A. J., Johnson, W. E., O'Brien, S. J., and Pecon-Slattery, J. Evolution of the male-determining gene SRY within the cat family Felidae. Genetics, 175(4), 1855-1867. http://dx.doi.org/10.1534/genetics.106.066779.
10. Dimorphism. (2015). In Fine dictionary. Retrieved from http://www.finedictionary.com/dimorphism.html.
11. Bachtrog, Doris. (2013). Y chromosome evolution: Emerging insights into processes of Y chromosome degeneration. Nature Reviews Genetics, 14(2), 113-124. http://dx.doi.org/10.1038/nrg3366. Retrieved from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4120474/.
12. Gilbert, S. F. (2000). Chromosomal sex determination in Drosophila. In _Developmental Biology (6th ed.). Sunderland, MA: Sinauer Associates. Retrieved from www.ncbi.nlm.nih.gov/books/NBK10025/.
13. Hemophilia. (2015). In Genetics home reference. Retrieved from https://ghr.nlm.nih.gov/condition/hemophilia.
14. Kimball, J. W. (2015, December 23). Sex chromosomes. In Kimball's biology pages. Retrieved July 27, 2016 from http://www.biology-pages.info/S/SexChromosomes.html.
Additional references:
Bachtrog, Doris. (2013). Y chromosome evolution: Emerging insights into processes of Y chromosome degeneration. Nature Reviews Genetics, 14(2), 113-124. http://dx.doi.org/10.1038/nrg3366. Retrieved from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4120474/.
Bergmann, D. C. (2011). Genetics lecture notes. Biosci41, Stanford University.
Bourke, A. F. G. and Franks, N. R. (1995). Relatedness and reproductive value in the social Hymenoptera. In Social evolution in ants (p. 78). Princeton, NJ: Princeton University Press.
Bowen, R. A. (2000, August 17). Preparing a karyotype. In General and medical genetics. Retrieved from http://www.vivo.colostate.edu/hbooks/genetics/medgen/chromo/cytotech.html.
Carvalho, A. B., Koerich, L. B., and Clark, A. G. (2009). Origin and evolution of Y chromosomes: Drosophila tales. Trends in Genetics, 25(6), 270-277. http://dx.doi.org/10.1016/j.tig.2009.04.002. Retrieved from http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2921885/.
Deeb, S. S. and Motulsky, A. G. (2015, February 5). Red-green color vision defects. In GeneReviews. Retrieved from http://www.ncbi.nlm.nih.gov/books/NBK1301/.
Genetic factors and hormones that determine gender. (2007, June 27). In Human embryology: Organogenesis. Retrieved from http://www.embryology.ch/anglais/ugenital/molec02.html.
Haplodiploidy. (2015, November 7). Retrieved December 11, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Haplodiploidy.
Kimball, J. W. (2015, December 23). Sex chromosomes. In Kimball's biology pages. Retrieved July 27, 2016 from http://www.biology-pages.info/S/SexChromosomes.html.
Krempels, D. M. (n.d.). The genetics of calico cats. Retrieved from www.bio.miami.edu/dana/dox/calico.html.
OpenStax College, Biology. (2015, May 13). Chromosomal basis of inherited disorders. In OpenStax CNX. Retrieved from http://cnx.org/contents/[email protected]:65/Chromosomal-Basis-of-Inherited.
OpenStax College, Biology. (2015, May 13). Chromosomal theory and genetic linkage. In OpenStax CNX. Retrieved from http://cnx.org/contents/[email protected]:64/Chromosomal-Theory-and-Genetic
OPN1LW. (2015). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/gene/OPN1LW.
Pseudoautosomal region. (2015, September 2). Retrieved December 16, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Pseudoautosomal_region.
Purves, W. K., Sadava, D. E., Orians, G. H., and Heller, H.C. (2003). Sex determination and sex-linked inheritance. In Life: The science of biology (7th ed., pp. 125-144). Sunderland, MA: Sinauer Associates.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Sex-linked genes exhibit unique patterns of inheritance. In Campbell biology (10th ed., pp. 205-209). San Francisco, CA: Pearson.
Testis determining factor. (2015, November 10). Retrieved December 11, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Testis_determining_factor.
X0 sex-determination system (2015, July 20). Retrieved December 11, 2015 from Wikipedia: https://en.wikipedia.org/wiki/X0_sex-determination_system.
XY sex-determination system. (2015, November 19). Retrieved December 11, 2015 from Wikipedia: https://en.wikipedia.org/wiki/XY_sex-determination_system.
ZW sex-determination system. (2015, November 16). Retrieved December 11, 2015 from Wikipedia: https://en.wikipedia.org/wiki/ZW_sex-determination_system.
Zygosity. (2015, November 29). Retrieved December 11, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Zygosity. | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/03%3A_Sex_linkage/3.01%3A_Example_punnet_square_for_sex-linked_recessive_trait.txt |
Introduction
Having extra or missing chromosomes is not usually a good thing. In fact, for most chromosomes, having an extra or missing copy is lethal to humans (causing an embryo to die early in development).
Yet, human females have two X chromosomes (XX), while human males have just one (XY). Why doesn't it cause problems for men to have just one copy of the X chromosome, while women have two?
X-inactivation
As it turns out, the level of gene activity produced by a single X chromosome is the normal "dosage" for a human. Men have this dosage because, well, they only have one X chromosome! Women have the same dosage for a different reason: they shut down one of their two X chromosomes in a process called X-inactivation.
In X-inactivation, an X chromosome is compacted (or, as my intro bio professor liked to say, "crumpled up into a ball"), to make a small, dense structure called a Barr body. Most of the genes on the Barr body are inactive, meaning that they are not transcribed. The process of X-inactivation was discovered by the British geneticist Mary F. Lyon and is sometimes called lyonization in her honor1.
A woman has two X chromosomes, one from each parent. Which one will she inactivate? X-inactivation is a random process that happens separately in individual cells during embryonic development. One cell might shut down the paternal X, while its next-door neighbor might shut down the maternal X instead. All the cells descended from each of these original cells will maintain the same pattern of X-inactivation.
Interesting note: if you were a kangaroo, what I just said would not be true! In kangaroos and other marsupials, it is always the paternal X chromosome that undergoes X-inactivation2.
X-inactivation example: Calico cat
A classic example of X-inactivation is seen in cats. If a female cat is heterozygous for black and tan alleles of a coat color gene found on the X, she will inactivate her two Xs (and thus, the two alleles of the coat color gene) at random in different cells during development.
The result is a tortoiseshell coat pattern, made up of alternating patches of black and tan fur. The black patches come from groups of cells in which the X with the black allele is active, while the tan patches come from cells in which the X with the tan allele is active.
Although it's rarely as easy to see as in the case of the tortoiseshell cat, human females are also "mosaic" for any genes that are present in different alleles on their two X chromosomes.
[If that's true, why don't female carriers show X-linked disorders?]
At this point, you may be wondering why women heterozygous for a recessive, X-linked allele don’t display the associated condition. After all, roughly half of their cells will inactivate the normal allele, leaving only the disorder-causing allele active.
In many cases, it appears that having 50% of cells with a normal copy of a gene is sufficient to produce a normal, or close to normal, phenotype.
A good example of this comes from the X-linked condition of red-green color vision deficiency (colorblindness). Women who are carriers for red-green color vision deficiency alleles have a mixture of working and non-working photoreceptor cells in their eyes. However, the working cells allow them to perceive red and green well enough to function normally in daily life. In more sensitive tests in a laboratory setting, these women actually do show subtle color vision deficiencies relative to women who are not carriers3.
Sex chromosome aneuploidies
When an organism has an extra or missing copy of a chromosome, it is said to be aneuploid. Aneuploidies involving autosomes (non-sex chromosomes), especially large ones, are usually so harmful to development that an aneuploid embryo can't survive to birth.
Aneuploidies of X chromosomes, however, tend to be much less harmful, despite the fact that the X is a large chromosome. This is mostly thanks to X inactivation. Although the purpose of the X-inactivation system is to shut down the second X of an XX female, it can also do a pretty good job of shutting down more X chromosomes if they are present.
Examples of X chromosome aneuploidies include:
• Triple X syndrome, in which a woman has an XXX genotype, which occurs in about 1 out of every 1,000 female newborns4. Women with an XXX genotype have female sex characteristics and are fertile (able to have children). In some cases, triple X syndrome may be associated with learning difficulties, late development of motor skills in infants, and problems with muscle tone4.
• Klinefelter syndrome, in which males have an extra X chromosome, leading to a genotype of XXY. (In rarer cases, Klinefelter syndrome can involve several extra Xs, leading to an XXXY or XXXXY genotype.) Affected men may be infertile or develop less dense body and facial hair than other men. Klinefelter syndrome is thought to affect 1 out of every 500 to 1,000 male newborns5.
Like females, XXY males with Klinefelter syndrome will convert one X to a Barr body in each cell. Triple X females (as well as Klinefelter males with more than two X chromosomes) neutralize their extra Xs by forming additional Barr bodies. For example, there would be two Barr bodies in a cell from an XXX female or XXXY male.
In Turner syndrome, a woman lacks part or all of one of her X chromosomes (leaving her with just one functional X). People with this disorder develop as females, but often have short stature and may exhibit symptoms like infertility and learning difficulties. Turner syndrome is thought to occur in about 1 out of every 2,500 female births6. It has relatively mild effects because humans normally have only one X active in the cells of their body anyway.
Check your understanding
Query $1$
[Hint]
Tortoiseshell cats typically have a pattern of mixed black and orange patches on their coats as a result of X-inactivation. In cats, a gene borne on the X chromosome controls coat color, and heterozygous female cats with one orange allele ($\text{X}^O$) and one black allele ($\text{X}^o$) will randomly inactivate one of the two X chromosomes in each cell during embryonic development. All the cells descended from each of these early cells will retain the same inactivation state, giving rise to patches of cells with the same active X and, thus, the same active gene for coat pigmentation.
Normally, male cats do not show a tortoiseshell coat pattern because they have only one X chromosome, and thus do not carry out X-inactivation. Instead, they simply display the solid coat color corresponding to the allele of the coat color gene found on their single X (with $\text{X}^O$ giving orange and $\text{X}^o$ giving black). However, the X-inactivation system is still present in males, even though they don't typically use it, and can come into play if a male receives an unusually high number of X chromosomes.
It's possible to get a tortoiseshell pattern in a male cat if the cat receives two X chromosomes as well as a Y, and if his two Xs bear different alleles of the coat color gene (genotype $\text{X}^O\text{X}^o\text{Y}$). An XXY genotype can arise through nondisjunction, the failure of sex chromosomes to separate correctly from one another during meiosis. One example of how nondisjunction could lead to an XXY genotype and a tortoiseshell phenotype is shown in the Punnett square below.
In cats (as in humans), male development is specified by the SRY gene on the Y chromosome, so an XXY individual will develop as male. However, only one X can be active in each cell, so one of the two X chromosomes will be randomly inactivated in each of the XXY male's cells during early embryonic development. This results in a patchy tortoiseshell coat pattern like that of a heterozygous female cat.
Male tortoiseshells with an XXY genotype are outwardly male, but are usually sterile (unable to have offspring), with a feline version of the disorder known as Klinefelter syndrome in humans. Not all male tortoiseshell kittens have an abnormal number of X chromosomes, however. Instead, some are chimeras, or genetic mosaics, that result from the fusion of two embryos into one during early development. If the two embryos that fuse have different alleles of the coat color gene, the chimeric embryo will have a mixture of cells with different genotypes, and the mature cat will have patchy fur similar to that of a female tortoiseshell.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
[Attribution and references]
Attribution:
This article is a modified derivative of "Chromosomal basis of inherited disorders," by OpenStax College, Biology, CC BY 4.0. Download the original article for free at http://cnx.org/contents/[email protected].
This article is licensed under a CC BY-NC-SA 4.0 license.
Works cited:
1. Mary F. Lyon. (2015, November 22). Retrieved December 12, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Mary_F._Lyon.
2. Kimball, J. W. (2015, December 23). Sex chromosomes. In Kimball's biology pages. Retrieved from http://www.biology-pages.info/S/SexChromosomes.html.
3. Chen, Shuai. (2010, March 4). Since only one of a woman's X chromosomes works in a cell, why aren't more women colorblind? [answer]. In Stanford at the Tech: Understanding genetics. Retrieved from http://genetics.thetech.org/ask/ask349.
4. Triple X syndrome. (2014). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/condition/triple-x-syndrome.
5. Klinefelter syndrome. (2013). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/condition/klinefelter-syndrome.
6. Turner syndrome. (2012). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/condition/turner-syndrome.
Additional references:
Bergmann, D. C. (2011). Mutations, aneuploidy and sex determination. In Genetics lecture notes (pp. 36-44). Biosci41, Stanford University.
Chen, Shuai. (2010, March 4). Since only one of a woman's X chromosomes works in a cell, why aren't more women colorblind? [answer]. In Stanford at the Tech: Understanding genetics. Retrieved from http://genetics.thetech.org/ask/ask349.
Deeb, S. S. and Motulsky, A. G. (2015, February 5). Red-green color vision defects. In GeneReviews. Retrieved from http://www.ncbi.nlm.nih.gov/books/NBK1301/.
Genetic factors and hormones that determine gender. (2007, June 27). In Human embryology: Organogenesis. Retrieved from http://www.embryology.ch/anglais/ugenital/molec02.html.
Kimball, J. W. (2015, December 23). Sex chromosomes. In Kimball's biology pages. Retrieved from http://www.biology-pages.info/S/SexChromosomes.html.
Klinefelter syndrome. (2013). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/condition/klinefelter-syndrome.
Krempels, D. M. (n.d.). The genetics of calico cats. Retrieved from www.bio.miami.edu/dana/dox/calico.html.
Mary F. Lyon. (2015, November 22). Retrieved December 12, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Mary_F._Lyon.
OpenStax College, Biology. (2015, May 13). Chromosomal theory and genetic linkage. In OpenStax CNX. Retrieved from http://cnx.org/contents/[email protected]:64/Chromosomal-Theory-and-Genetic
O'Neil, Dennis. (2013). Sex chromosome abnormalities. In Human chromosomal abnormalities. Retrieved from http://anthro.palomar.edu/abnormal/abnormal_5.htm.
OPN1LW. (2015). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/gene/OPN1LW.
Purves, W. K., Sadava, D. E., Orians, G. H., and Heller, H.C. (2003). Sex determination and sex-linked inheritance. In Life: The science of biology (7th ed., pp. 125-144). Sunderland, MA: Sinauer Associates.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Alterations of chromosome number or structure cause some genetic disorders. In Campbell biology (10th ed., pp. 304-307). San Francisco, CA: Pearson.
Reece, J. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. (2011). Sex-linked genes exhibit unique patterns of inheritance. In Campbell biology (10th ed., pp. 205-209). San Francisco, CA: Pearson.
Testis determining factor. (2015, November 10). Retrieved December 11, 2015 from Wikipedia: https://en.wikipedia.org/wiki/Testis_determining_factor.
Triple X syndrome. (2014). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/condition/triple-x-syndrome.
Turner syndrome. (2012). In Genetics home reference. Retrieved from http://ghr.nlm.nih.gov/condition/turner-syndrome.
XY sex-determination system. (2015, November 19). Retrieved December 11, 2015 from Wikipedia: https://en.wikipedia.org/wiki/XY_sex-determination_system. | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/03%3A_Sex_linkage/3.03%3A_X-inactivation.txt |
Key terms
Term Meaning
Sex chromosome One of two chromosomes that determines an organism's biological sex
Autosome Chromosome that is not a sex chromosome
Sex-linked gene Gene that is located on one of the two sex chromosomes
Carrier Heterozygous individual that inherited a recessive allele for a genetic disorder but does not display symptoms of that disorder
Barr body A condensed region in the nucleus of a cell, consisting of an inactivated X chromosome
Aneuploidy Condition of having too many or too few chromosomes
Sex linkage
In humans, biological sex is determined by a pair of sex chromosomes: XX in females and XY in males. The other 44 chromosomes are autosomes.
Genes on either the X or Y chromosome are sex-linked traits. Genes found on the X chromosome can be found in either males or females, while genes found on the Y chromosome can only be found in males.
X-linked inheritance
There are many more X-linked traits than Y-linked traits because the Y chromosome is much shorter and fewer genes than the X chromosome.
X-linked genes have distinctive inheritance patterns because they are present in different numbers in females (XX) and males (XY).
Females have two X chromosomes, so she will have two copies of each X-linked gene, giving her the opportunity to be either homozygous or heterozygous for each sex-linked gene.
X-linked disorders
X-linked human genetic disorders are much more common in males than in females. Since males only have one X chromosome, and therefore one copy of any X-linked genes, whatever allele the male inherits for an X-linked gene will be expressed.
An example of this is the blood-clotting disorder, hemophilia. Women who are heterozygous for hemophilia are carriers, and they usually don't display any symptoms themselves.
Sons of these women have a 550, percent chance of having hemophilia. Daughters have little chance of having hemophilia (unless the father also has it), and will instead have a 550, percent chance of being carriers.
X-inactivation
If males can survive with only one X chromosome, why doesn't it cause problems for women who have two X chromosomes?
As it turns out, for females, most of the genes in one of the X chromosomes is inactivated, forming a Barr body. This inactivation happens randomly during embryonic development.
Example:
A common example of X-inactivation is seen in cats. If a female cat is heterozygous for black and tan alleles of a coat color gene found on the X, two Xs (and thus, the two alleles of the coat color gene) will be inactivated at random in different cells during development.
The result of this is a tortoiseshell coat pattern, made up of alternating patches of black and tan fur.
Sex chromosome aneuploidy
Aneuploidy, or disorders of chromosome number, are generally caused by nondisjunction. This occurs when pairs of homologous chromosomes or sister chromatids fail to separate during cell division.
Individuals that have autosomal aneuploidy rarely survive to birth. However, due to the size of the X chromosome and because of X-inactivation, X chromosome aneuploidies tend to be much less harmful.
In Klinefelter syndrome, males have one or more extra X chromosomes, leading to a genotype of XXY. (Or in rare cases, XXXY or XXXXY!) Affected men may be infertile or develop less dense body and facial hair than other men.
Women affected with Triple X syndrome have an XXX genotype. Women with Triple X syndrome have female sex characteristics and are fertile (able to have children).
Women with Turner syndrome lack part or all of one of their X chromosomes (leaving her with just one functional X). People with this disorder develop as females, but often have short stature and may experience infertility and learning difficulties.
Common mistakes and misconceptions
• Some people think that a recessive X-linked trait will show up more often in women because they have two X chromosomes. However, women are less likely to express recessive X-linked traits because there is potential for a "good" allele to mask a "bad" allele. On the other hand, if a male receives a "bad" allele from his mother, he has no chance of getting a "good" allele from his father (who provides a Y) to hide the bad one.
• Codominance and X-inactivation are not the same. Although these two concepts may result in similar looking organisms, a heterozygous individual expressing a codominant trait will express both alleles fully and separately.
In X-inactivation, females express only one X chromosome in each cell, meaning that genes on the X chromosome are expressed singly instead of in a pair. Because the inactivated X chromosome is not the same in every cell, neighboring cells may express different proteins if different X chromosomes carry different alleles.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org) | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/03%3A_Sex_linkage/3.04%3A_Sex_linkage_review.txt |
Query $1$
Step-by-step solution
1. Cells that have missing or extra chromosomes are said to be aneuploid.
Most aneuploidies, including those of sex chromosomes, occur during the production of gametes (sperm or eggs)
2. When pairs of homologous chromosomes or sister chromatids fail to separate during meiosis I or II, gametes with too many or too few chromosomes can be produced.
3. When an aneuploid sperm or egg combines with a normal sperm or egg in fertilization, it makes a zygote that is also aneuploid.
For instance, if a sperm containing an extra $\text{X}$ chromosome, (making it $\text{XY}$ instead of $\text{Y}$) combines with a normal egg cell ($\text{X}$), the resulting zygote will have a sex chromosome composition of $\text{XXY}$ rather than the normal $\text{XY}$.
4. The correct answer is
Nondisjunction occurred during meiosis
Query $2$
Step-by-step solution
1. If the woman has the trait, then her genotype must be $\text{X}^a\text{X}^a$.
A male who does not have the trait must have a genotype of $\text{X}^A\text{Y}$
2.
In order for a son to be affected he must be $\text{X}^a\text{Y}$.
All sons produced by this cross will have this genotype.
3. The correct answer is
There is a 100% chance he will be affected.
Query $3$
Step-by-step solution
1. Aneuploidy refers to an abnormal number of chromosomes - either an additional chromosome or missing chromosomes.
2. Each human gamete should have one sex chromosome.
Male gametes can have either an X or a Y chromosome, and female gametes should have one X chromosome.
3. If a cell has too many chromosomes, such as an egg having two X-chromosomes, the cell has likely undergone nondisjunction.
This means the cell did not properly divide its material during meiosis.
4. The correct answer is
An egg contains two X chromosomes.
Query $4$
Step-by-step solution
1. The cross for these parents is $\text{X}^H\text{Y}$ and $\text{X}^H\text{X}^h$.
2. If we look at the Punnett square, we see that the possible offspring produced by this cross are:
• $\text{X}^H\text{X}^H$
• daughter, unaffected
• $\text{X}^H\text{X}^h$
• daughter, unaffected carrier
• $\text{X}^H\text{Y}$
• son, unaffected
• $\text{X}^h\text{Y}$
• son, affected
3. The correct answer is
No daughter they have will be affected.
Query $5$
Step-by-step solution
1. Normally, male cats do not show a calico coat pattern because they have only one X chromosome, and cannot carry out X-inactivation.
Instead, they simply display the solid coat color corresponding to the allele of the coat color gene found on their single X.
However, the X-inactivation system is still present in males, even though they don't typically use it, and can come into play if a male receives an unusually high number of X chromosomes.
2. The X inactivation system can occur if a male receives an unusually high number of X chromosomes.
XXY individuals develop as outwardly male, but since only one X can be active in each cell, one of the two X chromosomes will be randomly inactivated during his development, resulting in the same calico pattern seen in heterozygous females.
3. The correct answer is
The male kitten has a genotype of XXY.
Query $6$
Step-by-step solution
1. The cross for these parents is $\text{X}^H\text{Y}$ and $\text{X}^h\text{X}^h$.
2. While you can complete a Punnett square for this cross, the simplest way to figure out the answer to this problem is by recognizing that a male offspring always inherits his X-chromosome from his mother.
Because the mother's genotype is $\text{X}^h\text{X}^h$, she can only contribute an $\text{X}^h$ to her sons.
This means that all of the sons will be $\text{X}^h\text{Y}$ after getting the hemophiliac gene from the mother and the Y-chromosome from the father.
[Let me see the Punnett square!]
The question asks specifically for the probability of males having hemophilia. If we look at our Punnett square, we find that there are two boxes with male offspring.
In order for a male to have the trait, he must have a genotype of $\text{X}^h\text{Y}$. Of the two male offspring boxes, 2/2 have this genotype.
This means that 2/2, or 100%, of the males will be hemophiliacs.
3. The correct answer is
100%
Query $7$
Step-by-step solution
1. Because all of the male offspring have white eyes, and white eyes are a recessive sex-linked trait, the male offspring's genotypes must be $\text{X}^w\text{Y}$.
The only option for the father to have a normal phenotype is if his genotype is $\text{X}^W\text{Y}$.
2. In order to have a normal phenotype, the mother must be either $\text{X}^W\text{X}^W$ or $\text{X}^W\text{X}^w$.
However, because a male offspring always inherits his X chromosome from his mother, she must be able to donate a $\text{X}^w$ allele to her male offspring for them to have white eyes.
This leaves only one option for the mother's genotype: $\text{X}^W\text{X}^w$.
3. The correct answer is
$\text{X}^W\text{X}^w$ and $\text{X}^W\text{Y}$
Query $8$
Step-by-step solution
1. Because of X-inactivation, X chromosome aneuploidies are generally less harmful than autosome aneuplodies.
This is because the X-inactivation that shuts down the second X of an XX female can also shut down excess X chromosomes if they are present.
2. Male sex chromosome aneuploidy occurs as either multiples of X or Y chromosomes or both. Common affected genotypes may be XXY or XYY.
The presence or absence of a Y chromosome determines biological sex in humans, despite aneuploidy.
3. The correct answer is
Male sex chromosome aneuploidy can affect either the X or Y chromosome.
Query $9$
Step-by-step solution
1. Females have two X-chromosomes. Thus, their sex chromosomes can be
• $\text{X}^D\text{X}^D$
• normal
• $\text{X}^D\text{X}^d$
• carrier
• $\text{X}^d\text{X}^d$
• DMD
Since a female must have two $\text{X}^d$ alleles to have DMD, it is less likely that females will express the trait.
2. Because males only have one X-chromosome, whatever allele is present on that single X-chromosome will be expressed.
3. The correct answer is
Males only have one copy of the X chromosome.
Query $10$
Step-by-step solution
1. Because red-green colorblindness is a recessive disorder, a woman must receive two colorblindness alleles (one on each X chromosome) in order to be colorblind.
Therefore, a daughter can only be colorblind if she gets a colorblind allele from each her mother and her father.
2. Males only have one X chromosome, so in order to have a colorblind allele that can be passed on to offspring, the male must be colorblind ($\text{X}^b\text{Y}$).
Women who are either homozygous for colorblindess ($\text{X}^b\text{X}^b$) or a carrier ($\text{X}^B\text{X}^b$) may pass on the colorblind allele to their offspring.
3. The correct answer is
A carrier mother and a colorblind father
Query $11$
Step-by-step solution
1. If the woman is a carrier, but does not have the trait, then her genotype must be $\text{X}^B\text{X}^b$.
A male who does not have the trait must have a genotype of $\text{X}^B\text{Y}$
2.
In order for a son to be affected he must be $\text{X}^b\text{Y}$.
There is a 1 in 4 chance that they will produce a child with this genotype.
3. The correct answer is
They have a 25% chance of having an affected son.
Query $12$
Step-by-step solution
1. The cross for these parents is $\text{X}^a\text{Y}$ and $\text{X}^A\text{X}^A$.
2. In order for a daughter to have the trait, she must have a genotype of $\text{X}^a\text{X}^a$. Of the two daughter offspring boxes, 0/2 have this genotype.
This means that 0/2, or 0%, of the daughters will have ocular albinism.
3. The correct answer is
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org) | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/03%3A_Sex_linkage/3.05%3A_Practice_-_Sex_linkage.txt |
Thumbnail: Pedigree for an autosomal dominant trait. (CC BY-NC-SA; Khan Academy).
04: Pedigrees
An introduction to reading and analyzing pedigrees.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
4.02: Pedigree for determining probability of exhibiting sex linked recessive trait
Pedigree for determining probability of exhibiting sex linked recessive trait.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org)
4.03: Pedigrees review
Key terms
Term Meaning
Pedigree Chart that shows the presence or absence of a trait within a family across generations
Genotype The genetic makeup of an organism (ex: TT)
Phenotype The physical characteristics of an organism (ex: tall)
Dominant allele Allele that is phenotypically expressed over another allele
Recessive allele Allele that is only expressed in absence of a dominant allele
Autosomal trait Trait that is located on an autosome (non-sex chromosome)
Sex-linked trait Trait that is located on one of the two sex chromosomes
Homozygous Having two identical alleles for a particular gene
Heterozygous Having two different alleles for a particular gene
Pedigrees
Pedigrees are used to analyze the pattern of inheritance of a particular trait throughout a family. Pedigrees show the presence or absence of a trait as it relates to the relationship among parents, offspring, and siblings.
Reading a pedigree
Pedigrees represent family members and relationships using standardized symbols.
By analyzing a pedigree, we can determine genotypes, identify phenotypes, and predict how a trait will be passed on in the future. The information from a pedigree makes it possible to determine how certain alleles are inherited: whether they are dominant, recessive, autosomal, or sex-linked.
To start reading a pedigree:
1. Determine whether the trait is dominant or recessive. If the trait is dominant, one of the parents must have the trait. Dominant traits will not skip a generation. If the trait is recessive, neither parent is required to have the trait since they can be heterozygous.
2. Determine if the chart shows an autosomal or sex-linked (usually X-linked) trait. For example, in X-linked recessive traits, males are much more commonly affected than females. In autosomal traits, both males and females are equally likely to be affected (usually in equal proportions).
Example: Autosomal dominant trait
The diagram shows the inheritance of freckles in a family. The allele for freckles (F) is dominant to the allele for no freckles (f).
At the top of the pedigree is a grandmother (individual I-2) who has freckles. Two of her three children have the trait (individuals II-3 and II-5) and three of her grandchildren have the trait (individuals III-3, III-4, and III-5).
[What is the genotype of individual I-2?]
Since freckles are dominant to no freckles, an affected individual such as I-2 must at least have one F allele.
The trait shows up in all generations and affects both males and females equally. This suggests that it is an autosomal dominant trait.
Unaffected individuals must have two recessive alleles (ff) in order to not have freckles. If we notice, I-2 has some children who do not have freckles. In order to produce children with a genotype of ff, I-2 must be able to donate a f allele.
We can therefore conclude that her genotype is Ff.
Example: X-linked recessive trait
The diagram shows the inheritance of colorblindness in a family. Colorblindness is a recessive and X-linked trait $(\text{X}^b)$. The allele for normal vision is dominant and is represented by $\text{X}^B$.
In generation I, neither parent has the trait, but one of their children (II-3) is colorblind. Because there are unaffected parents that have affected offspring, it can be assumed that the trait is recessive. In addition, the trait appears to affect males more than females (in this case, exclusively males are affected), suggesting that the trait may be X-linked.
[What is the genotype of individual III-2?]
We can determine the genotype of III-2 by looking at her children. Since she is an unaffected female, she must have at least one normal vision allele $(\text{X}^B)$. Her two genotype options are then $\text{X}^{B}\text{X}^{B}$ or $\text{X}^{B}\text{X}^{b}$.
However, her son (IV-1) is colorblind, meaning that he has a genotype of $\text{X}^{b}\text{Y}$. Because males always get their $\text{X}$ chromosome from their mothers (and their $\text{Y}$ from their fathers), his colorblind allele must come from III-2.
We can then determine that III-2's genotype is $\text{X}^{B}\text{X}^{b}$, so she can pass the $\text{X}^{b}$ on to her son.
Common mistakes and misconceptions
• The presence of many affected individuals in a family does not always mean that the trait is dominant. The terms dominant and recessive refer to the way that a trait is expressed, not by how often it shows up in a family. In fact, although it is uncommon, a trait may be recessive but still show up in all generations of a pedigree.
• You may not always be able to determine the genotype of an individual based on a pedigree. Sometimes an individual can either be homozygous dominant or heterozygous for a trait. Often, we can use the relationships between an individual and their parents, siblings, and offspring to determine genotypes. However, not all carriers are always explicitly indicated in a pedigree, and it may not be possible to determine based on the information provided.
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org) | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/04%3A_Pedigrees/4.01%3A_Pedigrees.txt |
Autosomal recessive trait
Query $1$
Step-by-step solution
1. Because the trait we are tracking (attached earlobes) is autosomal recessive, shaded individuals, like III-6, will have a homozygous recessive genotype (ee).
2. If III-6 (ee) were to have a child with a man who was homozygous for unattached earlobes (EE), then all of the children would be heterozygous - getting one E from their father and one e from their mother.
Attached earlobes is a recessive trait and will only occur in ee genotypes. Heterozygotes (Ee) will have unattached earlobes, as that is the dominant condition.
3. The correct answer is
All of their children would have unattached earlobes.
Query $2$
Step-by-step solution
1. Because the trait we are tracking (attached earlobes) is autosomal recessive, shaded individuals, like I-2, will have a homozygous recessive genotype (ee).
I-1 must have a heterozygous genotype because he is able to pass on a recessive allele to some of his offspring (II-2 and II-4).
2. If I-1 and I-2 had another child, the cross would be:
Only offspring with ee genotypes will have attached earlobes (2/4 boxes).
$2\div 4=0.5=50\%$
3. The correct answer is
50%
Query $3$
Step-by-step solution
1. Individual I-1 is represented by a non-shaded square, indicating that it is a male with unattached earlobes.
2. Because the trait we are tracking, attached earlobes, is autosomal recessive, shaded individuals will have a homozygous recessive genotype (ee).
Individuals that are non-shaded will have at least one E allele.
3. I-1 has children with attached earlobes (II-2 and II-4 are ee), meaning he must be able to pass on at least e allele. However, he shows the dominant condition, so he must also have one E allele.
Therefore, his genotype is Ee.
4. The correct answer is
Ee
Query $4$
Step-by-step solution
1. Individual II-3 is represented by a non-shaded square, indicating that it is a male with unattached earlobes.
2. Because the trait we are tracking, attached earlobes, is autosomal recessive, shaded individuals will have a homozygous recessive genotype (ee).
Individuals that are non-shaded will have at least one E allele.
3. II-3 has a mother with attached earlobes (ee), meaning he must get one e allele from her. However, he shows the dominant condition, so he must also have one E allele.
Therefore, his genotype is Ee.
4. The correct answer is
Ee
X-linked recessive trait
Query $5$
Step-by-step solution
1. Individual I-2 is represented by a shaded circle, indicating that it is an affected female.
Therefore, she must have a homozygous recessive genotype of $\text{X}^{d}\text{X}^{d}$.
2. Because males always get their $\text{X}$ chromosome from their mother, all of the sons that individual 2 has will receive a recessive $\text{X}^{d}$ allele.
Males will also receive their $\text{Y}$ chromosome from their father, giving any son of individuals I-1 and I-2 a genotype of $\text{X}^{d}\text{Y}$.
3. The correct answer is
100%
Query $6$
Step-by-step solution
1. Unaffected males, such as individual II-1 have a genotype of $\text{X}^{D}\text{Y}$.
On the other hand, affected males, such as individual II-3, have a genotype of $\text{X}^{d}\text{Y}$.
Since males only have one $\text{X}$ chromosome, they cannot be carriers.
2. Individuals II-4 and II-5 are both shaded in, indicating that they are affected.
In order to be affected, they must have the recessive genotypes $\text{X}^{d}\text{Y}$ and $\text{X}^{d}\text{X}^{d}$. This means that any child they have will have DMD because each parent can only pass on a recessive DMD allele.
3. Individual III-1 is an unaffected male, meaning that he has a genotype of $\text{X}^{D}\text{Y}$.
If he mates with an unaffected, non-carrier female ($\text{X}^{D}\text{X}^{D}$), there is no chance that the children will inherit the DMD allele.
4. The correct answer is
If individual III-1 marries an unaffected, non-carrier female, none of their offspring will have DMD.
Query $7$
Step-by-step solution
1. Individual II-2 is represented by a non-shaded circle, indicating that it is an unaffected female.
2. In order for individual II-2 to have a normal phenotype, but also produce an affected son, she must be a carrier for DMD. This means that she has one of each allele, $\text{X}^{D}\text{X}^{d}$.
3. The correct answer is
$\text{X}^{D}\text{X}^{d}$
Query $8$
Step-by-step solution
1. Individual I-3 is represented by a shaded square, indicating that it is an affected male. Therefore, he must have a genotype of $\text{X}^{d}\text{Y}$.
If he has a child with a DMD carrier ($\text{X}^{D}\text{X}^{d}$), the cross would be:
2. In order for a daughter to be affected, her genotype must be $\text{X}^{d}\text{X}^{d}$.
Only one box has this genotype, so the chances of having an affected daughter is:
$\dfrac{1}{4}=25\%$
3. The correct answer is
25%
Autosomal dominant trait
Query $9$
Step-by-step solution
1. Because the trait we are tracking, dimples, is autosomal dominant, any shaded individuals will have at least one dominant allele (D).
Any unshaded individuals will have the recessive genotype (dd).
2. II-2 has dimples, meaning she must have at least one D allele. In addition, she has a recessive parent, and one of her children has no dimples (dd), so she must also have at least one d allele. This makes her genotype Dd.
Individual II-1 has no dimples, meaning that he must have a homozygous recessive genotype (dd).
3. Now that we know their genotypes, if we perform a cross between individuals II-1 and II-2 we find:
Only offspring with a D allele will have dimples (2/4 boxes).
$2\div 4=50\%$
4. The correct answer is
50%
Query $10$
Step-by-step solution
1. Phenotype is the physical characteristic that we see (ex: dimples).
A genotype is the allele combination (ex: DD)
2. Because the trait we are tracking is having dimples, shaded individuals, like III-4, have dimples. Unshaded individuals, like III-1, do not have dimples.
3. The correct answer is
Dimples
Query $11$
Step-by-step solution
1. The trait that we are tracking, dimples, appears to be dominant, as all offspring who have the trait have an affected parent.
Having dimples also does not skip a generation, which suggests that it is likely dominant.
2. Shaded individuals have dimples, meaning that they must have at least one D allele.
Unshaded individuals, like I-1, do not have dimples, meaning that they must have the homozygous recessive genotype dd.
3. An individual with dimples can be either DD or Dd.
Individuals like II-2, II-3, and III-2 all have at least one recessive parent. Since the recessive parent can only donate a d, each of them must have a d in their genotype.
Because they all have dimples, we know they must have one D allele as well, giving them all the genotype of Dd.
4. The correct answer is
III-2 → Dd
Query $12$
Step-by-step solution
1. Because the trait we are tracking (dimples) is autosomal dominant, any shaded individuals have at least one dominant allele (D).
Any unshaded individuals have the recessive genotype (dd).
2. Individual III-3 must be heterozygous (Dd) because he has an unshaded father (dd).
If he was to have a child with a woman who was heterozygous for dimples (Dd), then the cross would be:
3. Dimples is a dominant trait and will occur whenever a dominant allele is present in the genotype.
Homozygous recessive individuals (dd) will have no dimples, as that is the recessive condition.
Looking at the Punnett square, we find that $\dfrac{3}{4}=75\%$ offspring will have at least one D allele.
4. The correct answer is
75%
Contributors and Attributions
• Khan Academy (CC BY-NC-SA 3.0; All Khan Academy content is available for free at www.khanacademy.org) | textbooks/bio/Genetics/Classical_Genetics_(Khan_Academy)/04%3A_Pedigrees/4.04%3A_Practice_-_Pedigrees.txt |
These are homework exercises to accompany Hardison's "Working with Molecular Genetics" TextMap. Genetics is the study of genes, genetic variation, and heredity in living organisms.
Exercises: Genetics (Hardison)
Question 1.5. Calculating recombination frequencies
Corn kernels can be colored or white, determined by the alleles C(colored, which is dominant) or c(white, which is recessive) of the coloredgene. Likewise, alleles of the shrunkengene determine whether the kernels are nonshrunken (Sh, dominant) or shrunken (sh, recessive). The geneticist Hutchison crossed a homozygous colored shrunken strain (CC shsh) to a homozygous white nonshrunken strain (cc ShSh) and obtained the heterozygous colored nonshrunken F1. The F1 was backcrossed to a homozygous recessive white shrunken strain (cc shsh). Four phenotypes were observed in the F2 progeny, in the numbers shown below.
Phenotype Number of plants
colored shrunken 21,379
white nonshrunken 21,096
colored nonshrunken 638
white shrunken 672
a) What are the predicted frequencies of these phenotypes if the coloredand shrunkengenes are not linked?
b) Are these genes linked, and if so, what is the recombination frequency between them?
Question 1.6. Constructing a linkage map:
Consider three genes, A, B and C, that are located on the same chromosome. The arrangement of the three genes can be determined by a series of three crosses, each following two of the genes (referred to as two-factor crosses). In each cross, a parental strain that is homozygous for the dominant alleles of the two genes (e.g. AB/AB) is crossed with a strain that is homozygous for the recessive alleles of the two genes (e.g. ab/ab), to yield an F1 that is heterozygous for both of the genes (e.g. AB/ab). In this notation, the slash (/) separates the alleles of genes on one chromosome from those on the homologous chromosome. The F1 (AB/ab) contains one chromosome from each parent. It is then backcrossed to a strain that is homozygous for the recessive alleles (ab/ab) so that the fates of the parental chromosomes can be easily followed. Let's say the resulting progeny in the F2 (second) generation showed the parental phenotypes (AB and ab) 70% of the time. That is, 70% of the progeny showed only the dominant characters (AB) or only the recessive characters (ab), which reflect the haploid genotypes AB/aband ab/ab, respectively, in the F2 progeny. The remaining 30% of the progeny showed recombinant phenotypes (Aband aB) reflecting the genotypes Ab/aband aB/abin the F2 progeny. Similar crosses using F1's from parental AC/ACand ac/acbackcrossed to a homozygous recessive strain (ac/ac) generated recombinant phenotypes Acand aCin 10% of the progeny. And finally, crosses using F1's from parental BC/BCand bc/bcbackcrossed to a homozygous recessive strain (bc/bc) generated recombinant phenotypes Bcand bCin 25% of the progeny.
a. What accounts for the appearance of the recombinant phenotypes in the F2 progeny?
b. Which genes are closer to each other and which ones are further away?
c. What is a linkage map that is consistent with the data given?
Question 1.7
Why are the distances in the previous problem not exactly additive, e.g. why is the distance between the outside markers (A and B) not 35 map units (or 35% recombination)? There are several possible explanations, and this problem explores the effects of multiple crossovers. The basic idea is that the further apart two genes are, the more likely that recombination can occur multiple times between them. Of course, two (or any even number of) crossover events between two genes will restore the parental arrangement, whereas three (or any odd number of) crossover events will give a recombinant arrangement, thereby effectively decreasing the observed number of recombinants in the progeny of a cross.
For the case examined in the previous problem, with genes in the order A___C_______B, let the term abrefer to the frequency of recombination between genes Aand B, and likewise let acrefer to the frequency of recombination between genes Aand C, and cbrefer to the frequency of recombination between genes Cand B.
a) What is the probability that when recombination occurs in the interval between Aand C, an independent recombination event also occurs in the interval between Cand B?
b) What is the probability that when recombination occurs in the interval between Cand B, an independent recombination event also occurs in the interval between Aand C?
c) The two probabilities, or frequencies, in a and b above will effectively lower the actual recombination between the outside markers Aand Bto that observed in the experiment. What is an equation that expresses this relationship, and does it fit the data in problem 3?
d. What is the better estimate for the distance between genes Aand Bin the previous problem?
Question 1.8 Complementation and recombination in microbes.
The State College Bar Association has commissioned you to study an organism, Alcophila latrobus, which thrives on Rolling Rock beer and is ruining the local shipments. You find three mutants that have lost the ability to grow on Rolling Rock (RR).
a) Recombination between the mutants can restore the ability to grow on RR. From the following recombination frequencies, construct a linkage map for mutations 1, 2, and 3.
Recombination between Frequency
1- and 2- 0.100
1- and 3- 0.099
2- and 3- 0.001
b) The following diploid constructions were tested for their ability to grow on RR. What do these data tell you about mutations 1, 2, and 3?
Grow on RR?
1) 1- 2+ / 1+ 2- yes
2) 1- 3+ / 1+ 3- yes
3) 2- 3+ / 2+ 3- no
Question 1.9 Using recombination frequencies and complementation to deduce maps and pathways in phage.
A set of four mutant phage that were unable to grow in a particular bacterial host (lets call it restrictive) were isolated; however, both mutant and wild type phage will grow in another, permissive host. To get information about the genes required for growth on the restrictive host, this host was co-infected with pairs of mutant phage, and the number of phage obtained after infection was measured. The top number for each co-infection gives the total number of phage released (grown on the permissive host) and the bottom number gives the number of wild-type recombinant phage (grown on the restrictive host). The wild-type parental phage gives 1010 phage after infecting either host. The limit of detection is 102 phage.
Phenotypes of phage, problem 1.9:
Assays after co-infection with mutant phage:
Results of assays, problem 1.9:
Number of phage
mutant 1 mutant 2 mutant 3 mutant 4
mutant 1 total <102
recombinants <102
mutant 2 total 1010 <102
recombinants 5x106 <102
mutant 3 total 1010 1010 <102
recombinants 107 5x106 <102
mutant 4 total 105 1010 1010 <102
recombinants 105 5x106 107 <102
a) Which mutants are in the same complementation group? What is the minimum number of genes in the pathway for growth on the restrictive host?
b) Which mutations have the shortest distance between them?
c) Which mutations have the greatest distance between them?
d) Draw a map of the genes in the pathway required for growth on the restrictive host. Show the positions of the genes, the positions of the mutations and the relative distances between them.
Question 1.10
One of the classic experiments in bacterial genetics is the fluctuation analysisof Luria and Delbrück (1943, Mutations of bacteria from virus sensitivity to virus resistance, Genetics 28: 491-511). These authors wanted to determine whether mutations arose spontaneouslywhile bacteria grew in culture, or if the mutations were inducedby the conditions used to select for them. They knew that bacteria resistant to phage infection could be isolated from infected cultures. When a bacterial culture is infected with a lytic phage, initially it “clears” because virtually all the cells are lysed, but after several hours phage-resistant bacteria will start to grow.
Luria and Delbrück realized that the two hypothesis for the source of the mutations could be distinguished by a quantitative analysis of the number of the phage-resistant bacteria found in many infected cultures. The experimental approach is outlined in the figure below. Many cultures of bacteria are grown, then infected with a dose of phage T1 that is sufficient to kill all the cells, except those that have acquired resistance. These resistant bacteria grow into colonies on plates and can be counted.
a. What are the predictions for the distribution of the number of resistant bacteria in the two models? Assume that on average, about 1 in 107 bacteria are resistant to infection by phage T1.
b. What do results like those in the figure and table tell you about which model is correct?
Figure for question 1.10.
The actual results from Luria and Delbrück are summarized in the following table. They examined 87 cultures, each with 0.2 ml of bacteria, for phage resistant colonies.
Number of resistant bacteria
Number of cultures
0
29
1
17
2
4
3
3
4
3
5
2
6-10
5
11-20
6
21-50
7
51-100
5
101-200
2
201-500
4
501-1000
0 | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/01.E%3A_Fundamental_Properties_of_Genes_%28Exercises%29.txt |
Question 1.5. Calculating recombination frequencies
Corn kernels can be colored or white, determined by the alleles C(colored, which is dominant) or c(white, which is recessive) of the coloredgene. Likewise, alleles of the shrunkengene determine whether the kernels are nonshrunken (Sh, dominant) or shrunken (sh, recessive). The geneticist Hutchison crossed a homozygous colored shrunken strain (CC shsh) to a homozygous white nonshrunken strain (cc ShSh) and obtained the heterozygous colored nonshrunken F1. The F1 was backcrossed to a homozygous recessive white shrunken strain (cc shsh). Four phenotypes were observed in the F2 progeny, in the numbers shown below.
Phenotype Number of plants
colored shrunken 21,379
white nonshrunken 21,096
colored nonshrunken 638
white shrunken 672
a) What are the predicted frequencies of these phenotypes if the coloredand shrunkengenes are not linked?
b) Are these genes linked, and if so, what is the recombination frequency between them?
Question 1.6. Constructing a linkage map:
Consider three genes, A, B and C, that are located on the same chromosome. The arrangement of the three genes can be determined by a series of three crosses, each following two of the genes (referred to as two-factor crosses). In each cross, a parental strain that is homozygous for the dominant alleles of the two genes (e.g. AB/AB) is crossed with a strain that is homozygous for the recessive alleles of the two genes (e.g. ab/ab), to yield an F1 that is heterozygous for both of the genes (e.g. AB/ab). In this notation, the slash (/) separates the alleles of genes on one chromosome from those on the homologous chromosome. The F1 (AB/ab) contains one chromosome from each parent. It is then backcrossed to a strain that is homozygous for the recessive alleles (ab/ab) so that the fates of the parental chromosomes can be easily followed. Let's say the resulting progeny in the F2 (second) generation showed the parental phenotypes (AB and ab) 70% of the time. That is, 70% of the progeny showed only the dominant characters (AB) or only the recessive characters (ab), which reflect the haploid genotypes AB/aband ab/ab, respectively, in the F2 progeny. The remaining 30% of the progeny showed recombinant phenotypes (Aband aB) reflecting the genotypes Ab/aband aB/abin the F2 progeny. Similar crosses using F1's from parental AC/ACand ac/acbackcrossed to a homozygous recessive strain (ac/ac) generated recombinant phenotypes Acand aCin 10% of the progeny. And finally, crosses using F1's from parental BC/BCand bc/bcbackcrossed to a homozygous recessive strain (bc/bc) generated recombinant phenotypes Bcand bCin 25% of the progeny.
a. What accounts for the appearance of the recombinant phenotypes in the F2 progeny?
b. Which genes are closer to each other and which ones are further away?
c. What is a linkage map that is consistent with the data given?
Question 1.7
Why are the distances in the previous problem not exactly additive, e.g. why is the distance between the outside markers (A and B) not 35 map units (or 35% recombination)? There are several possible explanations, and this problem explores the effects of multiple crossovers. The basic idea is that the further apart two genes are, the more likely that recombination can occur multiple times between them. Of course, two (or any even number of) crossover events between two genes will restore the parental arrangement, whereas three (or any odd number of) crossover events will give a recombinant arrangement, thereby effectively decreasing the observed number of recombinants in the progeny of a cross.
For the case examined in the previous problem, with genes in the order A___C_______B, let the term abrefer to the frequency of recombination between genes Aand B, and likewise let acrefer to the frequency of recombination between genes Aand C, and cbrefer to the frequency of recombination between genes Cand B.
a) What is the probability that when recombination occurs in the interval between Aand C, an independent recombination event also occurs in the interval between Cand B?
b) What is the probability that when recombination occurs in the interval between Cand B, an independent recombination event also occurs in the interval between Aand C?
c) The two probabilities, or frequencies, in a and b above will effectively lower the actual recombination between the outside markers Aand Bto that observed in the experiment. What is an equation that expresses this relationship, and does it fit the data in problem 3?
d. What is the better estimate for the distance between genes Aand Bin the previous problem?
Question 1.8 Complementation and recombination in microbes.
The State College Bar Association has commissioned you to study an organism, Alcophila latrobus, which thrives on Rolling Rock beer and is ruining the local shipments. You find three mutants that have lost the ability to grow on Rolling Rock (RR).
a) Recombination between the mutants can restore the ability to grow on RR. From the following recombination frequencies, construct a linkage map for mutations 1, 2, and 3.
Recombination between Frequency
1- and 2- 0.100
1- and 3- 0.099
2- and 3- 0.001
b) The following diploid constructions were tested for their ability to grow on RR. What do these data tell you about mutations 1, 2, and 3?
Grow on RR?
1) 1- 2+ / 1+ 2- yes
2) 1- 3+ / 1+ 3- yes
3) 2- 3+ / 2+ 3- no
Question 1.9 Using recombination frequencies and complementation to deduce maps and pathways in phage.
A set of four mutant phage that were unable to grow in a particular bacterial host (lets call it restrictive) were isolated; however, both mutant and wild type phage will grow in another, permissive host. To get information about the genes required for growth on the restrictive host, this host was co-infected with pairs of mutant phage, and the number of phage obtained after infection was measured. The top number for each co-infection gives the total number of phage released (grown on the permissive host) and the bottom number gives the number of wild-type recombinant phage (grown on the restrictive host). The wild-type parental phage gives 1010 phage after infecting either host. The limit of detection is 102 phage.
Phenotypes of phage, problem 1.9:
Assays after co-infection with mutant phage:
Results of assays, problem 1.9:
Number of phage
mutant 1 mutant 2 mutant 3 mutant 4
mutant 1 total <102
recombinants <102
mutant 2 total 1010 <102
recombinants 5x106 <102
mutant 3 total 1010 1010 <102
recombinants 107 5x106 <102
mutant 4 total 105 1010 1010 <102
recombinants 105 5x106 107 <102
a) Which mutants are in the same complementation group? What is the minimum number of genes in the pathway for growth on the restrictive host?
b) Which mutations have the shortest distance between them?
c) Which mutations have the greatest distance between them?
d) Draw a map of the genes in the pathway required for growth on the restrictive host. Show the positions of the genes, the positions of the mutations and the relative distances between them.
Question 1.10
One of the classic experiments in bacterial genetics is the fluctuation analysisof Luria and Delbrück (1943, Mutations of bacteria from virus sensitivity to virus resistance, Genetics 28: 491-511). These authors wanted to determine whether mutations arose spontaneouslywhile bacteria grew in culture, or if the mutations were inducedby the conditions used to select for them. They knew that bacteria resistant to phage infection could be isolated from infected cultures. When a bacterial culture is infected with a lytic phage, initially it “clears” because virtually all the cells are lysed, but after several hours phage-resistant bacteria will start to grow.
Luria and Delbrück realized that the two hypothesis for the source of the mutations could be distinguished by a quantitative analysis of the number of the phage-resistant bacteria found in many infected cultures. The experimental approach is outlined in the figure below. Many cultures of bacteria are grown, then infected with a dose of phage T1 that is sufficient to kill all the cells, except those that have acquired resistance. These resistant bacteria grow into colonies on plates and can be counted.
a. What are the predictions for the distribution of the number of resistant bacteria in the two models? Assume that on average, about 1 in 107 bacteria are resistant to infection by phage T1.
b. What do results like those in the figure and table tell you about which model is correct?
Figure for question 1.10.
The actual results from Luria and Delbrück are summarized in the following table. They examined 87 cultures, each with 0.2 ml of bacteria, for phage resistant colonies.
Number of resistant bacteria
Number of cultures
0
29
1
17
2
4
3
3
4
3
5
2
6-10
5
11-20
6
21-50
7
51-100
5
101-200
2
201-500
4
501-1000
0 | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/01.E%3A_Fundamental_Properties_of_Genes_(Exercises).txt |
3.2 Altering the ends of DNA fragments for ligation into vectors.
(Adapted from POB)
a) Draw the structure of the end of a linear DNA fragment that was generated by digesting with the restriction endonuclease EcoRI. Include those sequences remaining from the EcoRI recognition sequence.
b) Draw the structure resulting from the reaction of this end sequence with DNA polymerase I and the four deoxynucleoside triphosphates.
c) Draw the sequence produced at the junction if two ends with the structure derived in (b) are ligated.
d) Design two different short synthetic DNA fragments that would permit ligation of structure (a) with a DNA fragment produced by a PstI restriction digest. In one of these synthetic fragments, design the sequence so that the final junction contains the recognition sequences for both EcoRI and PstI. Design the sequence of the other fragment so that neither the EcoRI nor the PstI sequence appears in the junction.
3.3. What properties are required of vectors used in molecular cloning of DNA?
3.4. A student ligated a BamHI fragment containing a gene of interest to a pUC vector digested with BamHI, transformed E. coliwith the mixture of ligation products and plated the cells on plates containing the antibiotic ampicillin and the chromogenic substrate X‑gal. Which colonies should the student pick to find the ones containing the recombinant plasmid (with the gene of interest in pUC)?
3.5. Starting with an isolated mRNA, one wishes to make a double stranded copy of the mRNA and insert it at the PstI site of pBR322 via G-C homopolymer tailing. One then transforms E. coliwith this recombinant plasmid, selecting for tetracycline resistance. What are the four enzymatic steps used in preparing the cDNA insert? Name the enzymes and describe the intermediates.
3.6 A researcher needs to isolate a cDNA clone of giraffe actin mRNA, and she knows the size (Mr = 42,000) and partial amino acid sequence of giraffe actin protein and has specific antibodies against giraffe actin. After constructing a bank of cDNA plasmids from total mRNA of giraffe fibroblasts (dG-dC tailed into the PstI site of pBR322), what methods of screening the bank could be used to identify the actin cDNA clone?
3.7 The restriction map of pBR322 is
The distance in base pairs between restriction sites is as follows:
PstI to EcoRI 750 bp
EcoRI to HindIII 50 bp
HindIII to BamHI 260 bp
BamHI to PstI 3300 bp
A recombinant cDNA plasmid, pAlc-1, has double-stranded cDNA inserted at the PstI site of pBR322, using a technique that retains this cleavage site at both ends of the insert. Digestion of pBR322 and pAlc-1 with restriction endonucleases gives the following pattern after gel electrophoresis (left). The sizes of the fragments are given in base pairs. The DNA fragments were transferred out of the gel onto nitrocellulose and hybridized with radiolabeled cDNA from wild-type A. latrobus(a Southern blot-hybridizaton). Hybridizing fragments are shown in the autoradiogam diagram on the right.
a) What is the size of the cDNA insert?
b) What two restriction endonucleases cleave within the cDNA insert?
c) For those two restriction endonucleases, each DNA fragment in the single digest is cut by PstI into two DNA fragments in the double digest (i.e. the restriction endonuclease plus PstI). Determine which fragments each single digest fragment is cut into, and use this information to construct a map.
d) Draw a restriction map for pAlc-1, showing sites for PstI, EcoRI, BamHI and HindIII. Indicate the distance between sites and show the cDNA insert clearly.
3.8. You isolate and clone a KpnI fragment from A. latrobusgenomic DNA that encodes the mRNA cloned in pAlc-1 (as analyzed in question 3.7). The restriction map of the genomic fragment is
Each fragment that hybridizes to pAlc-1 is indicated by an asterisk. What does this map, especially when compared to that in problem 3.7, tell you about the structure of the gene? Be as quantitative as possible.
3.9. Some particular enzyme is composed of a polypeptide chain of 192 amino acids. The gene that encodes it has 1,440 nucleotide pairs. Explain the relationship between the number of amino acids in this polypeptide and the number of nucleotide pairs in its gene.
3.10. When viewed in the electron microscope, a hybrid between a cloned giraffe actin gene (genomic DNA) and mature actin mRNA looks like this:
What can you conclude about actin gene structure in the giraffe?
3.11. DNA complementary to pepper mRNA was synthesized using oligo (dT) as a primer for first strand synthesis. The second strand (synonymous with the mRNA) was then synthesized, and the population of double stranded cDNAs were ligated into a plasmid vector using a procedure that leaves PstI sites flanking the cDNA insert (i.e. the terminal PstI sites for each clone are not part of the cDNA). This cDNA library was screened for clones made from the mRNA from the pepper yellow gene. One clone was isolated, and subsequent analysis of the pattern of restriction endonuclease cleavage patterns showed it had the following structure:
The map shows the positions of restriction endonuclease cleavage sites and the distance between them in kilobases (kb). The map of the cDNA insert is shown with solid lines, and plasmid vector DNA flanking the cDNA is shown as dotted lines. The top strand is oriented 5' to 3' from left to right, and the bottom strand is oriented 5' to 3' from right to left. The positions and orientations of two oligonucleotides to prime synthesis for sequence determination are shown, and are placed adjacent to the strand that will be synthesized in the sequencing reaction.
a) Oligonucleotides that anneal to the plasmid vector sequences that flank the duplex cDNA insert were used to prime synthesis of DNA for sequencing by the Sanger dideoxynucleotide procedure. A primer that annealed to the vector sequences to the left of the map shown above generated the sequencing gel pattern shown below on the left. A primer that annealed to the vector sequences to the right of the map shown above generated the sequencing gel pattern shown below on the right. The gels were run from the negative electrode at the top to the positive electrode at the bottom, and the segment presented is past the PstI site (i.e. do not look for a PstI recognition site).
Left primer Right primer
G
A
T
C
G
A
T
C
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
____
a) What is the DNA sequence of the left and right ends of the insert in the cDNA clone? Be sure to specify the 5' to 3' orientation, and the strand (top or bottom) whose sequence is reported. The terms left, right, top and bottom all refer to the map shown above for the cDNA clone.
b) Which end of the cDNA clone (left or right in the map above) is most likely to include the sequence synonymous with the 3' end of the mRNA?
c) What restriction endonuclease cleavage sites do you see in the sequencing data given?
3.12. Genomic DNA from the pepper plant was ligated into EcoRI sites in a l phage vector to construct a genomic DNA library. This library was screened by hybridization to the yellowcDNA clone. The pattern of EcoRI cleavage sites for one clone that hybridized to the yellowcDNA clone was analyzed in two experiments.
In the first experiment, the genomic DNA clone was digested to completion with EcoRI, the fragments separated on an agarose gel, transferred to a nylon filter, and hybridized with the radioactive yellowcDNA clone. The digest pattern (observed on the agarose gel) is shown in lane 1, and the pattern of hybridizing fragments (observed on an autoradiogram after hybridization) is shown in lane 2. Sizes of the EcoRI fragments are indicated in kb. The right arm of this l vector is 6 kb long, and the left arm is 30 kb.
In the second experiment, the genomic DNA clone was digested with a range of concentrations of EcoRI, so that the products ranged from a partial digest to a complete digest. The cleavage products were annealed to a radioactive oligonucleotide that hybridized only to the right cohesive end (cossite) of the l vector DNA. This simply places a radioactive tag at the right end of all the products of the reaction that extend to the right end of the l clone (partial or complete); digestion products that do not include the right end of the l clone will not be seen. The results of the digestion are shown above, on the right. Lane 1 is the clone of genomic DNA in l that has not been digested, lane 5 is the complete digest with EcoRI, and lanes 2, 3 and 4 are partial digests using increasing amounts of EcoRI. The sizes of the radioactive DNA fragments (in kb) are given, and the density of the fill in the boxes is proportional to the intensity of the signal on the autoradiogram.
a) What is the map of the EcoRI fragments in the genomic DNA clone, and which fragments encode mRNA for the yellowgene? You may wish to fill in the figure below; the left and right arms of the l vector are given. Show positions of the EcoRI cleavage sites, distances between them (in kb) and indicate the fragments that hybridize to the cDNA clone.
EcoRI EcoRI
Left arm ___| |_ Right arm
(30 kb) (6 kb)
In a third experiment, the pepper DNA from the genomic DNA clone was excised, hybridized with yellowmRNA under conditions that favor RNA-DNA duplexes and examined in the electron microscope to visualize R-loops. A pattern like the following was observed. The lines in the figure can be duplex DNA, RNA-DNA duplexes and single-stranded DNA.
b) What do the R-loop data indicate? Please draw an interpretation of the R-loops, showing clearly the two DNA strands and the mRNA and distinguishing between the template (bottom, or message complementary) and nontemplate (top, or message synonymous) strands.
The EcoRI fragments that hybridize to the yellowcDNA clone were isolated and digested with SalI (S in the figure below), HindIII (H), and the combination of SalI plus HindIII (S+H). The resulting patterns of DNA fragments are shown below; all will hybridize to the yellowcDNA clone. Cleavage of the 5 kb EcoRI fragment with SalI generates two fragments of 2.5 kb.
c) What are the maps of the SalI and HindIII site(s) in each of the EcoRI fragments? Show positions of the cleavage sites and distances between them on the diagram below.
5 kb EcoRI fragment: 4 kb EcoRI fragment:
EcoRI EcoRI EcoRI EcoRI
|___________________________| |___________________________|
d) Compare these restriction maps with that of the cDNA clone (problem 1.38) and the R-loops shown above. Assuming that the SalI and HindIII sites in the genomic DNA correspond to those in the cDNA clone, what can you deduce about the intron/exon structure of the yellowgene(s) contained within the 5 kb and 4 kb EcoRI fragments? Please diagram the exon-intron structure in as much detail as the data permit (i.e. show the size of the intron(s) and positions of intron/exon junctions as precisely as possible).
5 kb EcoRI fragment: 4 kb EcoRI fragment:
EcoRI EcoRI EcoRI EcoRI
|___________________________| |___________________________|
e) Considering all the data (maps of cDNA and genomic clones and R-loop analysis), what can you conclude about the number and location(s) of yellowgene(s) in this genomic clone?
3.13 You have isolated an 1100 base pair (bp) cDNA clone for a gene called azure that when mutated causes blue eyes in frogs. You also isolate a 3000 bp SalI genomic DNA fragment that hybridizes to the azurecDNA. The map of the azure cDNA is as follows, with sizes of fragments given in bp.
Digestion of the 3000 bp SalI fragment of genomic DNA with the indicated restriction endonucleases yields the following pattern of fragments, all of which hybridize to the azurecDNA. Remember that the starting fragment has SalI sites at each end. Sizes of fragments are in bp.
Restriction enzymes
BamHI Bam+Pst PstI Pst+Eco EcoRI Bam+Eco
2700
2300
2000
1900 1900
1200
1100 1100
800
700 700 700
300 300 300
The SalI to SalI (3000 bp) genomic fragment was hybridized to the 1100 bp cDNA fragment, and the heteroduplexes were examined in the electron microscope. Measurements on a large number of molecules resulted in the determination of the sizes indicated in the structure on the left, i.e. duplex regions of 400 and 600 bp are interrupted by a single stranded loop of 1500 nucleotides and are flanked by single stranded regions of 500 and 100 nucleotides. When the same experiment is carried out with the 2700 bp SalI to EcoRI genomic DNA fragment hybridized to the cDNA fragment, the structure on the right is observed.
a) What is the restriction map of the 3000 bp SalI to SalI genomic DNA fragment from the azuregene? Specify distances between sites in base pairs.
b) How many introns are present in the azuregenomic DNA fragment?
c) Where are the exons in the azuregenomic DNA fragment? Draw the exons as boxes on the restriction map of the 3000 bp SalI to SalI genomic DNA fragment? Specify (in base pairs) the distances between restriction sites and the intron/exon boundaries.
3.14 The T-cell receptor is present only on T-lymphocytes, not on B-lymphocytes or other cells. Describe a strategy to isolate the T-cell receptor by subtractive hybridization, using RNA from T-lymphocytes and from B-lymphocytes.
3.15.How many exons are in the human insulin (INS) gene, how big are they, and how large are the introns that separate them? Use three different bioinformatic approaches to answer this.
a. Align the available genomic sequence containing INS(encoding insulin) with the sequence of the mRNA to find exons and introns in the INSgene. The sequence files are:
INSmRNA: accession number NM_000207
INS gene (includes part of THand IGF2in addition to INS): accession number L15440
Files can be obtained from NCBI (http://www.ncbi.nlm.nih.gov), or from the course web site (www.bmb.psu.edu/Courses/bmb400/default.htm)
Align the mRNA (cDNA) and genomic sequence using the BLAST2sequences server at
http://www.ncbi.nlm.nih.gov/blast/
and the sim4server at
pbil.univ-lyon1.fr/sim4.html
Sim4is designed to take into account terminal redundancy at the exon/intron junctions, whereas BLAST2does not. Do you see this effect in the output?
b. Use the ab initioexon finding program Genscan, available at
genes.mit.edu/GENSCAN.html
to predict exons in the INSgenomic sequence (L15440).
How does this compare with the results of analyzing with the program genscan?
c. What do you see for INSat the Human Genome Browser and Ensembl? They are accessed at:
http://genome.ucsc.edu/goldenPath/hgTracks.html
http://www.ensembl.org/ | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/03.E%3A_Isolating_and_Analyzing_Genes_%28Exercises%29.txt |
3.2 Altering the ends of DNA fragments for ligation into vectors.
(Adapted from POB)
a) Draw the structure of the end of a linear DNA fragment that was generated by digesting with the restriction endonuclease EcoRI. Include those sequences remaining from the EcoRI recognition sequence.
b) Draw the structure resulting from the reaction of this end sequence with DNA polymerase I and the four deoxynucleoside triphosphates.
c) Draw the sequence produced at the junction if two ends with the structure derived in (b) are ligated.
d) Design two different short synthetic DNA fragments that would permit ligation of structure (a) with a DNA fragment produced by a PstI restriction digest. In one of these synthetic fragments, design the sequence so that the final junction contains the recognition sequences for both EcoRI and PstI. Design the sequence of the other fragment so that neither the EcoRI nor the PstI sequence appears in the junction.
3.3. What properties are required of vectors used in molecular cloning of DNA?
3.4. A student ligated a BamHI fragment containing a gene of interest to a pUC vector digested with BamHI, transformed E. coliwith the mixture of ligation products and plated the cells on plates containing the antibiotic ampicillin and the chromogenic substrate X‑gal. Which colonies should the student pick to find the ones containing the recombinant plasmid (with the gene of interest in pUC)?
3.5. Starting with an isolated mRNA, one wishes to make a double stranded copy of the mRNA and insert it at the PstI site of pBR322 via G-C homopolymer tailing. One then transforms E. coliwith this recombinant plasmid, selecting for tetracycline resistance. What are the four enzymatic steps used in preparing the cDNA insert? Name the enzymes and describe the intermediates.
3.6 A researcher needs to isolate a cDNA clone of giraffe actin mRNA, and she knows the size (Mr = 42,000) and partial amino acid sequence of giraffe actin protein and has specific antibodies against giraffe actin. After constructing a bank of cDNA plasmids from total mRNA of giraffe fibroblasts (dG-dC tailed into the PstI site of pBR322), what methods of screening the bank could be used to identify the actin cDNA clone?
3.7 The restriction map of pBR322 is
The distance in base pairs between restriction sites is as follows:
PstI to EcoRI 750 bp
EcoRI to HindIII 50 bp
HindIII to BamHI 260 bp
BamHI to PstI 3300 bp
A recombinant cDNA plasmid, pAlc-1, has double-stranded cDNA inserted at the PstI site of pBR322, using a technique that retains this cleavage site at both ends of the insert. Digestion of pBR322 and pAlc-1 with restriction endonucleases gives the following pattern after gel electrophoresis (left). The sizes of the fragments are given in base pairs. The DNA fragments were transferred out of the gel onto nitrocellulose and hybridized with radiolabeled cDNA from wild-type A. latrobus(a Southern blot-hybridizaton). Hybridizing fragments are shown in the autoradiogam diagram on the right.
a) What is the size of the cDNA insert?
b) What two restriction endonucleases cleave within the cDNA insert?
c) For those two restriction endonucleases, each DNA fragment in the single digest is cut by PstI into two DNA fragments in the double digest (i.e. the restriction endonuclease plus PstI). Determine which fragments each single digest fragment is cut into, and use this information to construct a map.
d) Draw a restriction map for pAlc-1, showing sites for PstI, EcoRI, BamHI and HindIII. Indicate the distance between sites and show the cDNA insert clearly.
3.8. You isolate and clone a KpnI fragment from A. latrobusgenomic DNA that encodes the mRNA cloned in pAlc-1 (as analyzed in question 3.7). The restriction map of the genomic fragment is
Each fragment that hybridizes to pAlc-1 is indicated by an asterisk. What does this map, especially when compared to that in problem 3.7, tell you about the structure of the gene? Be as quantitative as possible.
3.9. Some particular enzyme is composed of a polypeptide chain of 192 amino acids. The gene that encodes it has 1,440 nucleotide pairs. Explain the relationship between the number of amino acids in this polypeptide and the number of nucleotide pairs in its gene.
3.10. When viewed in the electron microscope, a hybrid between a cloned giraffe actin gene (genomic DNA) and mature actin mRNA looks like this:
What can you conclude about actin gene structure in the giraffe?
3.11. DNA complementary to pepper mRNA was synthesized using oligo (dT) as a primer for first strand synthesis. The second strand (synonymous with the mRNA) was then synthesized, and the population of double stranded cDNAs were ligated into a plasmid vector using a procedure that leaves PstI sites flanking the cDNA insert (i.e. the terminal PstI sites for each clone are not part of the cDNA). This cDNA library was screened for clones made from the mRNA from the pepper yellow gene. One clone was isolated, and subsequent analysis of the pattern of restriction endonuclease cleavage patterns showed it had the following structure:
The map shows the positions of restriction endonuclease cleavage sites and the distance between them in kilobases (kb). The map of the cDNA insert is shown with solid lines, and plasmid vector DNA flanking the cDNA is shown as dotted lines. The top strand is oriented 5' to 3' from left to right, and the bottom strand is oriented 5' to 3' from right to left. The positions and orientations of two oligonucleotides to prime synthesis for sequence determination are shown, and are placed adjacent to the strand that will be synthesized in the sequencing reaction.
a) Oligonucleotides that anneal to the plasmid vector sequences that flank the duplex cDNA insert were used to prime synthesis of DNA for sequencing by the Sanger dideoxynucleotide procedure. A primer that annealed to the vector sequences to the left of the map shown above generated the sequencing gel pattern shown below on the left. A primer that annealed to the vector sequences to the right of the map shown above generated the sequencing gel pattern shown below on the right. The gels were run from the negative electrode at the top to the positive electrode at the bottom, and the segment presented is past the PstI site (i.e. do not look for a PstI recognition site).
Left primer Right primer
G
A
T
C
G
A
T
C
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a) What is the DNA sequence of the left and right ends of the insert in the cDNA clone? Be sure to specify the 5' to 3' orientation, and the strand (top or bottom) whose sequence is reported. The terms left, right, top and bottom all refer to the map shown above for the cDNA clone.
b) Which end of the cDNA clone (left or right in the map above) is most likely to include the sequence synonymous with the 3' end of the mRNA?
c) What restriction endonuclease cleavage sites do you see in the sequencing data given?
3.12. Genomic DNA from the pepper plant was ligated into EcoRI sites in a l phage vector to construct a genomic DNA library. This library was screened by hybridization to the yellowcDNA clone. The pattern of EcoRI cleavage sites for one clone that hybridized to the yellowcDNA clone was analyzed in two experiments.
In the first experiment, the genomic DNA clone was digested to completion with EcoRI, the fragments separated on an agarose gel, transferred to a nylon filter, and hybridized with the radioactive yellowcDNA clone. The digest pattern (observed on the agarose gel) is shown in lane 1, and the pattern of hybridizing fragments (observed on an autoradiogram after hybridization) is shown in lane 2. Sizes of the EcoRI fragments are indicated in kb. The right arm of this l vector is 6 kb long, and the left arm is 30 kb.
In the second experiment, the genomic DNA clone was digested with a range of concentrations of EcoRI, so that the products ranged from a partial digest to a complete digest. The cleavage products were annealed to a radioactive oligonucleotide that hybridized only to the right cohesive end (cossite) of the l vector DNA. This simply places a radioactive tag at the right end of all the products of the reaction that extend to the right end of the l clone (partial or complete); digestion products that do not include the right end of the l clone will not be seen. The results of the digestion are shown above, on the right. Lane 1 is the clone of genomic DNA in l that has not been digested, lane 5 is the complete digest with EcoRI, and lanes 2, 3 and 4 are partial digests using increasing amounts of EcoRI. The sizes of the radioactive DNA fragments (in kb) are given, and the density of the fill in the boxes is proportional to the intensity of the signal on the autoradiogram.
a) What is the map of the EcoRI fragments in the genomic DNA clone, and which fragments encode mRNA for the yellowgene? You may wish to fill in the figure below; the left and right arms of the l vector are given. Show positions of the EcoRI cleavage sites, distances between them (in kb) and indicate the fragments that hybridize to the cDNA clone.
EcoRI EcoRI
Left arm ___| |_ Right arm
(30 kb) (6 kb)
In a third experiment, the pepper DNA from the genomic DNA clone was excised, hybridized with yellowmRNA under conditions that favor RNA-DNA duplexes and examined in the electron microscope to visualize R-loops. A pattern like the following was observed. The lines in the figure can be duplex DNA, RNA-DNA duplexes and single-stranded DNA.
b) What do the R-loop data indicate? Please draw an interpretation of the R-loops, showing clearly the two DNA strands and the mRNA and distinguishing between the template (bottom, or message complementary) and nontemplate (top, or message synonymous) strands.
The EcoRI fragments that hybridize to the yellowcDNA clone were isolated and digested with SalI (S in the figure below), HindIII (H), and the combination of SalI plus HindIII (S+H). The resulting patterns of DNA fragments are shown below; all will hybridize to the yellowcDNA clone. Cleavage of the 5 kb EcoRI fragment with SalI generates two fragments of 2.5 kb.
c) What are the maps of the SalI and HindIII site(s) in each of the EcoRI fragments? Show positions of the cleavage sites and distances between them on the diagram below.
5 kb EcoRI fragment: 4 kb EcoRI fragment:
EcoRI EcoRI EcoRI EcoRI
|___________________________| |___________________________|
d) Compare these restriction maps with that of the cDNA clone (problem 1.38) and the R-loops shown above. Assuming that the SalI and HindIII sites in the genomic DNA correspond to those in the cDNA clone, what can you deduce about the intron/exon structure of the yellowgene(s) contained within the 5 kb and 4 kb EcoRI fragments? Please diagram the exon-intron structure in as much detail as the data permit (i.e. show the size of the intron(s) and positions of intron/exon junctions as precisely as possible).
5 kb EcoRI fragment: 4 kb EcoRI fragment:
EcoRI EcoRI EcoRI EcoRI
|___________________________| |___________________________|
e) Considering all the data (maps of cDNA and genomic clones and R-loop analysis), what can you conclude about the number and location(s) of yellowgene(s) in this genomic clone?
3.13 You have isolated an 1100 base pair (bp) cDNA clone for a gene called azure that when mutated causes blue eyes in frogs. You also isolate a 3000 bp SalI genomic DNA fragment that hybridizes to the azurecDNA. The map of the azure cDNA is as follows, with sizes of fragments given in bp.
Digestion of the 3000 bp SalI fragment of genomic DNA with the indicated restriction endonucleases yields the following pattern of fragments, all of which hybridize to the azurecDNA. Remember that the starting fragment has SalI sites at each end. Sizes of fragments are in bp.
Restriction enzymes
BamHI Bam+Pst PstI Pst+Eco EcoRI Bam+Eco
2700
2300
2000
1900 1900
1200
1100 1100
800
700 700 700
300 300 300
The SalI to SalI (3000 bp) genomic fragment was hybridized to the 1100 bp cDNA fragment, and the heteroduplexes were examined in the electron microscope. Measurements on a large number of molecules resulted in the determination of the sizes indicated in the structure on the left, i.e. duplex regions of 400 and 600 bp are interrupted by a single stranded loop of 1500 nucleotides and are flanked by single stranded regions of 500 and 100 nucleotides. When the same experiment is carried out with the 2700 bp SalI to EcoRI genomic DNA fragment hybridized to the cDNA fragment, the structure on the right is observed.
a) What is the restriction map of the 3000 bp SalI to SalI genomic DNA fragment from the azuregene? Specify distances between sites in base pairs.
b) How many introns are present in the azuregenomic DNA fragment?
c) Where are the exons in the azuregenomic DNA fragment? Draw the exons as boxes on the restriction map of the 3000 bp SalI to SalI genomic DNA fragment? Specify (in base pairs) the distances between restriction sites and the intron/exon boundaries.
3.14 The T-cell receptor is present only on T-lymphocytes, not on B-lymphocytes or other cells. Describe a strategy to isolate the T-cell receptor by subtractive hybridization, using RNA from T-lymphocytes and from B-lymphocytes.
3.15.How many exons are in the human insulin (INS) gene, how big are they, and how large are the introns that separate them? Use three different bioinformatic approaches to answer this.
a. Align the available genomic sequence containing INS(encoding insulin) with the sequence of the mRNA to find exons and introns in the INSgene. The sequence files are:
INSmRNA: accession number NM_000207
INS gene (includes part of THand IGF2in addition to INS): accession number L15440
Files can be obtained from NCBI (http://www.ncbi.nlm.nih.gov), or from the course web site (www.bmb.psu.edu/Courses/bmb400/default.htm)
Align the mRNA (cDNA) and genomic sequence using the BLAST2sequences server at
http://www.ncbi.nlm.nih.gov/blast/
and the sim4server at
pbil.univ-lyon1.fr/sim4.html
Sim4is designed to take into account terminal redundancy at the exon/intron junctions, whereas BLAST2does not. Do you see this effect in the output?
b. Use the ab initioexon finding program Genscan, available at
genes.mit.edu/GENSCAN.html
to predict exons in the INSgenomic sequence (L15440).
How does this compare with the results of analyzing with the program genscan?
c. What do you see for INSat the Human Genome Browser and Ensembl? They are accessed at:
http://genome.ucsc.edu/goldenPath/hgTracks.html
http://www.ensembl.org/ | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/03.E%3A_Isolating_and_Analyzing_Genes_(Exercises).txt |
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
04.E: Genomes and Chromosomes (Exercises)
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 | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/04.E%3A_Genomes_and_Chromosomes_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/05.E%3A_DNA_replication_I%3A_Enzymes_and_Mechanism_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/05.E%3A_DNA_replication_I%3A_Enzymes_and_Mechanism_(Exercises).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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/06.E%3A_DNA_replication_II%3A_Start_stop_and_control_%28Exercises%29.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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/06.E%3A_DNA_replication_II%3A_Start_stop_and_control_(Exercises).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?
07.E : Mutation and Repair of DNA (Exercises)
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? | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/07.E_%3A_Mutation_and_Repair_of_DNA_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/08.E%3A_Recombination_of_DNA_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/08.E%3A_Recombination_of_DNA_(Exercises).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?
09.E: Transposition of DNA (Exercises)
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? | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/09.E%3A_Transposition_of_DNA_%28Exercises%29.txt |
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?
10.E: Transcription: RNA polymerases (Exercises)
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/10.E%3A_Transcription%3A_RNA_polymerases_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/11.E%3A_Transcription%3A_Promoters_terminators_and_mRNA_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/11.E%3A_Transcription%3A_Promoters_terminators_and_mRNA_(Exercises).txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/12.E%3A_RNA_Processing_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/12.E%3A_RNA_Processing_(Exercises).txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/13.E%3A_Genetic_Code_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/13.E%3A_Genetic_Code_(Exercises).txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/14.E%3A_Translation_-_Protein_synthesis_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/14.E%3A_Translation_-_Protein_synthesis_(Exercises).txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/15.E%3A_Positive_and_negative_control_of_gene_expression_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/15.E%3A_Positive_and_negative_control_of_gene_expression_(Exercises).txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/16.E%3A_Transcription_regulation_via_effects_on_RNA_polymerases_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/16.E%3A_Transcription_regulation_via_effects_on_RNA_polymerases_(Exercises).txt |
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?
18.E: Transcriptional regulation after initiation (Exercises)
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/18.E%3A_Transcriptional_regulation_after_initiation_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/19.E%3A_Transcriptional_regulation_in_eukaryotes_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/19.E%3A_Transcriptional_regulation_in_eukaryotes_(Exercises).txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/20.E%3A_Transcriptional_regulation_via_chromatin_alterations_%28Exercises%29.txt |
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/Exercise_-_Genetics/Exercises%3A_Genetics_(Hardison)/20.E%3A_Transcriptional_regulation_via_chromatin_alterations_(Exercises).txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Exercises: Online Open Genetics (Nickle and Barrette-Ng)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
1.1 How would the results of the cross in Figure 1.11 have been different if heredity worked through blending inheritance rather than particulate inheritance?
1.2 Imagine that astronauts provide you with living samples of multicellular organisms discovered on another planet. These organisms reproduce with a short generation time, but nothing else is known about their genetics.
a) How could you define laws of heredity for these organisms?
b) How could you determine what molecules within these organisms contained genetic information?
c) Would the mechanisms of genetic inheritance likely be similar for all organisms from this planet?
d) Would the mechanisms of genetic inheritance likely be similar to organisms from earth?
1.3 It is relatively easy to extract DNA and protein from cells; biochemists had been doing this since at least the 1800’s. Why then did Hershey and Chase need to use radioactivity to label DNA and proteins in their experiments?
1.4 Compare Watson and Crick’s discovery with Avery, MacLeod and McCarty’s discovery.
a) What did each discover, and what was the impact of these discoveries on biology?
b) How did Watson and Crick’s approach generally differ from Avery, MacLeod and McCarty’s?
c) Briefly research Rosalind Franklin on the internet. Why is her contribution to the structure of DNA controversial?
1.5 Starting with mice and R and S strains of S. pneumoniae, what experiments in additional to those shown in Figure 1.3 to demonstrate that DNA is the genetic material?
1.6 List the information that Watson and Crick used to deduce the structure of DNA.
1.7 Refer to Watson and Crick’
a) List the defining characteristics of the structure of a DNA molecule.
b) Which of these characteristics are most important to replication?
c) Which characteristics are most important to the Central Dogma?
1.8 Compare Figure 1.13 and Table 1.1. Which of the mutants (#1, #2, #3) shown in Figure 1.13 matches each of the phenotypes expected for mutations in genes A, B,C?
1.9 Refer to Table 1.2
a) What is the relationship between DNA content of a genome, number of genes, gene density, and chromosome number?
b) What feature of genomes explains the c-value paradox?
c) Do any of the numbers in Table 1.2 show a correlation with organismal complexity?
1.10 a) List the characteristics of an ideal model organism.
b) Which model organism can be used most efficiently to identify genes related to:
i) eye development
ii) skeletal development
iii) photosynthesis
iii) cell division
iv) cell differentiation
v) cancer
1.11 Refer to Figure 1.8
a) Identify the part of the DNA molecule that would be radioactively labeled in the manner used by Hershey & Chase
b) DNA helices that are rich in G-C base pairs are harder to separate (e.g. by heating) than A-T rich helices. Why? | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/01.E%3A_Overview_DNA_and_Genes_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
1.1 How would the results of the cross in Figure 1.11 have been different if heredity worked through blending inheritance rather than particulate inheritance?
1.2 Imagine that astronauts provide you with living samples of multicellular organisms discovered on another planet. These organisms reproduce with a short generation time, but nothing else is known about their genetics.
a) How could you define laws of heredity for these organisms?
b) How could you determine what molecules within these organisms contained genetic information?
c) Would the mechanisms of genetic inheritance likely be similar for all organisms from this planet?
d) Would the mechanisms of genetic inheritance likely be similar to organisms from earth?
1.3 It is relatively easy to extract DNA and protein from cells; biochemists had been doing this since at least the 1800’s. Why then did Hershey and Chase need to use radioactivity to label DNA and proteins in their experiments?
1.4 Compare Watson and Crick’s discovery with Avery, MacLeod and McCarty’s discovery.
a) What did each discover, and what was the impact of these discoveries on biology?
b) How did Watson and Crick’s approach generally differ from Avery, MacLeod and McCarty’s?
c) Briefly research Rosalind Franklin on the internet. Why is her contribution to the structure of DNA controversial?
1.5 Starting with mice and R and S strains of S. pneumoniae, what experiments in additional to those shown in Figure 1.3 to demonstrate that DNA is the genetic material?
1.6 List the information that Watson and Crick used to deduce the structure of DNA.
1.7 Refer to Watson and Crick’
a) List the defining characteristics of the structure of a DNA molecule.
b) Which of these characteristics are most important to replication?
c) Which characteristics are most important to the Central Dogma?
1.8 Compare Figure 1.13 and Table 1.1. Which of the mutants (#1, #2, #3) shown in Figure 1.13 matches each of the phenotypes expected for mutations in genes A, B,C?
1.9 Refer to Table 1.2
a) What is the relationship between DNA content of a genome, number of genes, gene density, and chromosome number?
b) What feature of genomes explains the c-value paradox?
c) Do any of the numbers in Table 1.2 show a correlation with organismal complexity?
1.10 a) List the characteristics of an ideal model organism.
b) Which model organism can be used most efficiently to identify genes related to:
i) eye development
ii) skeletal development
iii) photosynthesis
iii) cell division
iv) cell differentiation
v) cancer
1.11 Refer to Figure 1.8
a) Identify the part of the DNA molecule that would be radioactively labeled in the manner used by Hershey & Chase
b) DNA helices that are rich in G-C base pairs are harder to separate (e.g. by heating) than A-T rich helices. Why? | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/01.E%3A_Overview_DNA_and_Genes_(Exercises).txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
2.1 Define chromatin. What is the difference between DNA, chromatin and chromosomes?
2.2 Species A has n=4 chromosomes and Species B has n=6 chromosomes. Can you tell from this information which species has more DNA? Can you tell which species has more genes?
2.3 The answer to question 2 implies that not all DNA within a chromosome encodes genes. Can you name any examples of chromosomal regions that contain relatively few genes?
2.4 a) How many centromeres does a typical chromosome have?
b) What would happen if there was more than one centromere per chromosome?
c) What if a chromosome had zero centromeres?
2.5 For a diploid with 2n=16 chromosomes, how many chromosomes and chromatids are per cell present in the gamete, and zygote and immediately following G1, S, G2, mitosis, and meiosis?
2.6 Bread wheat (Triticum aestivum) is a hexaploid. Using the nomenclature presented in class, an ovum cell of wheat has n=21 chromosomes. How many chromosomes in a zygote of bread wheat?
2.7 For a given gene:
a) What is the maximum number of alleles that can exist in a 2n cell of a given diploid individual?
b) What is the maximum number of alleles that can exist in a 1n cell of a tetraploid individual?
c) What is the maximum number of alleles that can exist in a 2n cell of a tetraploid individual?
d) What is the maximum number of alleles that can exist in a population?
2.8 a) Why is aneuploidy more often lethal than polyploidy?
b) Which is more likely to disrupt gene balance: polyploidy or duplication?
2.9 For a diploid organism with 2n=4 chromosomes, draw a diagram of all of the possible configurations of chromosomes during normal anaphase I, with the maternally and paternally derived chromosomes labelled.
2.10 For a triploid organism with 2n=3x=6 chromosomes, draw a diagram of all of the possible configurations of chromosomes at anaphase I (it is not necessary label maternal and paternal chromosomes).
2.11 For a tetraploid organism with 2n=4x=8 chromosomes, draw all of the possible configurations of chromosomes during a normal metaphase.
2.12 A simple mnemonic for leptotene, zygotene, pachytene, diplotene, & diakinesis is Lame Zebras Pee Down Drains. Make another one yourself.
02.E: Chromosomes Mitosis and Meiosis (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
2.1 Define chromatin. What is the difference between DNA, chromatin and chromosomes?
2.2 Species A has n=4 chromosomes and Species B has n=6 chromosomes. Can you tell from this information which species has more DNA? Can you tell which species has more genes?
2.3 The answer to question 2 implies that not all DNA within a chromosome encodes genes. Can you name any examples of chromosomal regions that contain relatively few genes?
2.4 a) How many centromeres does a typical chromosome have?
b) What would happen if there was more than one centromere per chromosome?
c) What if a chromosome had zero centromeres?
2.5 For a diploid with 2n=16 chromosomes, how many chromosomes and chromatids are per cell present in the gamete, and zygote and immediately following G1, S, G2, mitosis, and meiosis?
2.6 Bread wheat (Triticum aestivum) is a hexaploid. Using the nomenclature presented in class, an ovum cell of wheat has n=21 chromosomes. How many chromosomes in a zygote of bread wheat?
2.7 For a given gene:
a) What is the maximum number of alleles that can exist in a 2n cell of a given diploid individual?
b) What is the maximum number of alleles that can exist in a 1n cell of a tetraploid individual?
c) What is the maximum number of alleles that can exist in a 2n cell of a tetraploid individual?
d) What is the maximum number of alleles that can exist in a population?
2.8 a) Why is aneuploidy more often lethal than polyploidy?
b) Which is more likely to disrupt gene balance: polyploidy or duplication?
2.9 For a diploid organism with 2n=4 chromosomes, draw a diagram of all of the possible configurations of chromosomes during normal anaphase I, with the maternally and paternally derived chromosomes labelled.
2.10 For a triploid organism with 2n=3x=6 chromosomes, draw a diagram of all of the possible configurations of chromosomes at anaphase I (it is not necessary label maternal and paternal chromosomes).
2.11 For a tetraploid organism with 2n=4x=8 chromosomes, draw all of the possible configurations of chromosomes during a normal metaphase.
2.12 A simple mnemonic for leptotene, zygotene, pachytene, diplotene, & diakinesis is Lame Zebras Pee Down Drains. Make another one yourself. | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/02.E%3A_Chromosomes_Mitosis_and_Meiosis_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Q3.1
hat is the maximum number of alleles for a given locus in a normal gamete of a diploid species?
Q3.2
Wirey hair (W) is dominant to smooth hair (w) in dogs.
1. If you cross a homozygous, wirey-haired dog with a smooth-haired dog, what will be the genotype and phenotype of the F1 generation?
2. If two dogs from the F1 generation mated, what would be the most likely ratio of hair phenotypes among their progeny?
3. When two wirey-haired Ww dogs actually mated, they had alitter of three puppies, which all had smooth hair. How do you explain this observation?
4. Someone left a wirey-haired dog on your doorstep. Without extracting DNA, what would be the easiest way to determine the genotype of this dog?
5. Based on the information provided in question 1, can you tell which, if either, of the alleles is wild-type?
Q3.3
An important part of Mendel’s experiments was the use of homozygous lines as parents for his crosses. How did he know they were homozygous, and why was the use of the lines important?
Q3.4
Does equal segregation of alleles into daughter cells happen during mitosis, meiosis, or both?
Q3.5
If your blood type is B, what are the possible genotypes of your parents at the locus that controls the ABO blood types?
Q3.6
In the table below, match the mouse hair color phenotypes with the term from the list that best explains the observed phenotype, given the genotypes shown. In this case, the allele symbols do not imply anything about the dominance relationships between the alleles. List of terms: haplosufficiency, haploinsufficiency, pleiotropy, incomplete dominance, co-dominance, incomplete penetrance, broad (variable) expressivity.
Table for Question 3.6
A1A1 A1A2 A2A2
1 all hairs black on the same individual: 50% of hairs are all black and 50% of hairs are all white all hairs white
2 all hairs black all hairs are the same shade of grey all hairs white
3 all hairs black all hairs black 50% of individuals have all white hairs and 50% of individuals have all black hairs
4 all hairs black all hairs black mice have no hair
5 all hairs black all hairs white all hairs white
6 all hairs black all hairs black all hairs white
7 all hairs black all hairs black hairs are a wide range of shades of grey
Q3.7
A rare dominant mutation causes a neurological disease that appears late in life in all people that carry the mutation. If a father has this disease, what is the probability that his daughter will also have the disease?
Q3.8
Make Punnett Squares to accompany the crosses shown in Figure 3.10.
Q3.9
Another cat hair colour gene is called White Spotting. This gene is autosomal. Cats that have the dominant S allele have white spots. What are the possible genotypes of cats that are:
1. entirely black
2. entirely orange
3. black and white
4. orange and white
5. orange and black (tortoiseshell)
6. orange, black, and white (calico)
Q3.10
Draw reciprocal crosses that would demonstrate that the turkey E-gene is on the Z chromosome.
Q3.11
Mendel’s First Law (as stated in class) does not apply to alleles of most genes located on sex chromosomes. Does the law apply to the chromosomes themselves?
Q3.12
What is the relationship between the O0 and OB alleles of the Orange gene in cats?
Q3.13
Make a diagram similar to those in Figures 3.9, 3.11, and 3.13 that shows the relationship between genotype and phenotype for the F8 gene in humans.
Answers
3.1 There is a maximum of two alleles for a normal autosomal locus in a diploid species.
3.2 a) In the F1 generation, the genotype of all individuals will be Ww and all of the dogs will have wirey hair.
b) In the F2 generation, there would be an expected 3:1 ratio of wirey-haired to smooth-haired dogs.
c) Although it is expected that only one out of every four dogs in the F2 generation would have smooth hair, large deviations from this ratio are possible, especially with small sample sizes. These deviations are due to the random nature in which gametes combine to produce offspring. Another example of this would be the fairly common observation that in some human families, all of the offspring are either girls, or boys, even though the expected ratio of the sexes is essentially 1:1.
d) You could do a test cross, i.e. cross the wirey-haired dog to a homozygous recessive dog (ww). Based on the phenotypes among the offspring, you might be able to infer the genotype of the wirey-haired parent.
e) From the information provided, we cannot be certain which, if either, allele is wild-type. Generally, dominant alleles are wild-type, and abnormal or mutant alleles are recessive.
3.3 Even before the idea of a homozygous genotype had really been formulated, Mendel was still able to assume that he was working with parental lines that contained the genetic material for only one variant of a trait (e.g. EITHER green seeds of yellow seeds), because these lines were pure-breeding. Pure-breeding means that the phenotype doesn’t change over several generations of self-pollination. If the parental lines had not been pure-breeding, it would have been very hard to make certain key inferences, such as that the F1 generation could contain the genetic information for two variants of a trait, although only one variant was expressed. This inference led eventually to Mendel’s First Law.
3.4 Equal segregation of alleles occurs only in meiosis. Although mitosis does produce daughter cells that are genetically equal, there is no segregation (i.e. separation) of alleles during mitosis; each daughter cell contains both of the alleles that were originally present.
3.5 If your blood type is B, then your genotype is either IBi or IBIB. If your genotype is IBi, then your parents could be any combination of genotypes, as long as one parent had at least one i allele, and the other parent had at least one IB allele. If your genotype was IB IB, then both parents would have to have at least one IB allele.
3.6 case 1 co-dominance
case 2 incomplete-dominance
case 3 incomplete penetrance
case 4 pleiotropy
case 5 haplosufficiency
case 6 haploinsufficiency
case 7 broad (variable) expressivity
3.7 If the gene is autosomal, the probability is 50%. If it is sex-linked, 100%. In both situations the probability would decrease if the penetrance was less than 100%.
3.8
3.9 Note that a semicolon is used to separate genes on different chromosomes.
Phenotype Genotype(s)
a entirely black OB / OB ; s / s OB / Y ; s / s
b entirely orange O0 / O0 ; s / s O0 / Y ; s / s
c black and white OB / OB ; S / _ OB / Y ; S / _
d orange and white O0 / O0 ; S / _ O0 / Y ; S / _
e orange and black (tortoiseshell) O0 / OB ; s / s
f orange, black, and white (calico) O0 / OB ; S / _
3.10
3.11 Because each egg or sperm cell receives exactly one sex chromosome (even though this can be either an X or Y, in the case of sperm), it could be argued that the sex chromosomes themselves do obey the law of equal segregation, even though the alleles they carry may not always segregate equally. However, this answer depends on how broadly you are willing to stretch Mendel’s First Law.
3.12 Co-dominance
3.13 People with hemophilia A use injections of recombinant Factor VIII proteins on demand (to control bleeding) or regularily (to limit damage to joints). | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/03.E%3A_Genetic_Analysis_of_Single_Genes_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Q3.1
hat is the maximum number of alleles for a given locus in a normal gamete of a diploid species?
Q3.2
Wirey hair (W) is dominant to smooth hair (w) in dogs.
1. If you cross a homozygous, wirey-haired dog with a smooth-haired dog, what will be the genotype and phenotype of the F1 generation?
2. If two dogs from the F1 generation mated, what would be the most likely ratio of hair phenotypes among their progeny?
3. When two wirey-haired Ww dogs actually mated, they had alitter of three puppies, which all had smooth hair. How do you explain this observation?
4. Someone left a wirey-haired dog on your doorstep. Without extracting DNA, what would be the easiest way to determine the genotype of this dog?
5. Based on the information provided in question 1, can you tell which, if either, of the alleles is wild-type?
Q3.3
An important part of Mendel’s experiments was the use of homozygous lines as parents for his crosses. How did he know they were homozygous, and why was the use of the lines important?
Q3.4
Does equal segregation of alleles into daughter cells happen during mitosis, meiosis, or both?
Q3.5
If your blood type is B, what are the possible genotypes of your parents at the locus that controls the ABO blood types?
Q3.6
In the table below, match the mouse hair color phenotypes with the term from the list that best explains the observed phenotype, given the genotypes shown. In this case, the allele symbols do not imply anything about the dominance relationships between the alleles. List of terms: haplosufficiency, haploinsufficiency, pleiotropy, incomplete dominance, co-dominance, incomplete penetrance, broad (variable) expressivity.
Table for Question 3.6
A1A1 A1A2 A2A2
1 all hairs black on the same individual: 50% of hairs are all black and 50% of hairs are all white all hairs white
2 all hairs black all hairs are the same shade of grey all hairs white
3 all hairs black all hairs black 50% of individuals have all white hairs and 50% of individuals have all black hairs
4 all hairs black all hairs black mice have no hair
5 all hairs black all hairs white all hairs white
6 all hairs black all hairs black all hairs white
7 all hairs black all hairs black hairs are a wide range of shades of grey
Q3.7
A rare dominant mutation causes a neurological disease that appears late in life in all people that carry the mutation. If a father has this disease, what is the probability that his daughter will also have the disease?
Q3.8
Make Punnett Squares to accompany the crosses shown in Figure 3.10.
Q3.9
Another cat hair colour gene is called White Spotting. This gene is autosomal. Cats that have the dominant S allele have white spots. What are the possible genotypes of cats that are:
1. entirely black
2. entirely orange
3. black and white
4. orange and white
5. orange and black (tortoiseshell)
6. orange, black, and white (calico)
Q3.10
Draw reciprocal crosses that would demonstrate that the turkey E-gene is on the Z chromosome.
Q3.11
Mendel’s First Law (as stated in class) does not apply to alleles of most genes located on sex chromosomes. Does the law apply to the chromosomes themselves?
Q3.12
What is the relationship between the O0 and OB alleles of the Orange gene in cats?
Q3.13
Make a diagram similar to those in Figures 3.9, 3.11, and 3.13 that shows the relationship between genotype and phenotype for the F8 gene in humans.
Answers
3.1 There is a maximum of two alleles for a normal autosomal locus in a diploid species.
3.2 a) In the F1 generation, the genotype of all individuals will be Ww and all of the dogs will have wirey hair.
b) In the F2 generation, there would be an expected 3:1 ratio of wirey-haired to smooth-haired dogs.
c) Although it is expected that only one out of every four dogs in the F2 generation would have smooth hair, large deviations from this ratio are possible, especially with small sample sizes. These deviations are due to the random nature in which gametes combine to produce offspring. Another example of this would be the fairly common observation that in some human families, all of the offspring are either girls, or boys, even though the expected ratio of the sexes is essentially 1:1.
d) You could do a test cross, i.e. cross the wirey-haired dog to a homozygous recessive dog (ww). Based on the phenotypes among the offspring, you might be able to infer the genotype of the wirey-haired parent.
e) From the information provided, we cannot be certain which, if either, allele is wild-type. Generally, dominant alleles are wild-type, and abnormal or mutant alleles are recessive.
3.3 Even before the idea of a homozygous genotype had really been formulated, Mendel was still able to assume that he was working with parental lines that contained the genetic material for only one variant of a trait (e.g. EITHER green seeds of yellow seeds), because these lines were pure-breeding. Pure-breeding means that the phenotype doesn’t change over several generations of self-pollination. If the parental lines had not been pure-breeding, it would have been very hard to make certain key inferences, such as that the F1 generation could contain the genetic information for two variants of a trait, although only one variant was expressed. This inference led eventually to Mendel’s First Law.
3.4 Equal segregation of alleles occurs only in meiosis. Although mitosis does produce daughter cells that are genetically equal, there is no segregation (i.e. separation) of alleles during mitosis; each daughter cell contains both of the alleles that were originally present.
3.5 If your blood type is B, then your genotype is either IBi or IBIB. If your genotype is IBi, then your parents could be any combination of genotypes, as long as one parent had at least one i allele, and the other parent had at least one IB allele. If your genotype was IB IB, then both parents would have to have at least one IB allele.
3.6 case 1 co-dominance
case 2 incomplete-dominance
case 3 incomplete penetrance
case 4 pleiotropy
case 5 haplosufficiency
case 6 haploinsufficiency
case 7 broad (variable) expressivity
3.7 If the gene is autosomal, the probability is 50%. If it is sex-linked, 100%. In both situations the probability would decrease if the penetrance was less than 100%.
3.8
3.9 Note that a semicolon is used to separate genes on different chromosomes.
Phenotype Genotype(s)
a entirely black OB / OB ; s / s OB / Y ; s / s
b entirely orange O0 / O0 ; s / s O0 / Y ; s / s
c black and white OB / OB ; S / _ OB / Y ; S / _
d orange and white O0 / O0 ; S / _ O0 / Y ; S / _
e orange and black (tortoiseshell) O0 / OB ; s / s
f orange, black, and white (calico) O0 / OB ; S / _
3.10
3.11 Because each egg or sperm cell receives exactly one sex chromosome (even though this can be either an X or Y, in the case of sperm), it could be argued that the sex chromosomes themselves do obey the law of equal segregation, even though the alleles they carry may not always segregate equally. However, this answer depends on how broadly you are willing to stretch Mendel’s First Law.
3.12 Co-dominance
3.13 People with hemophilia A use injections of recombinant Factor VIII proteins on demand (to control bleeding) or regularily (to limit damage to joints). | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/03.E%3A_Genetic_Analysis_of_Single_Genes_(Exercises).txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Q4.1
How are polymorphisms and mutations alike? How are they different?
Q4.2
What are some of the ways a substitution can occur in a DNA sequence?
Q4.3
What are some of the ways a deletion can occur in a DNA sequence?
Q4.4
What are all of the ways an insertion can occur in a DNA sequence?
Q4.5
In the context of this chapter, explain the health hazards of smoking tobacco.
Q4.6
You have a female fruit fly, whose father was exposed to a mutagen (she, herself, wasn’t). Mating this female fly with another non-mutagenized, wild type male produces offspring that all appear to be completely normal, except there are twice as many daughters as sons in the F1 progeny of this cross.
1. Propose a hypothesis to explain these observations.
2. How could you test your hypothesis?
Q4.7
You decide to use genetics to investigate how your favourite plant makes its flowers smell good.
1. What steps will you take to identify some genes that are required for production of the sweet floral scent? Assume that this plant is a self-pollinating diploid.
2. One of the recessive mutants you identified has fishy-smelling flowers, so you name the mutant (and the mutated gene) fishy. What do you hypothesize about the normal function of the wild-type fishy gene?
3. Another recessive mutant lacks floral scent altogether, so you call it nosmell. What could you hypothesize about the normal function of this gene?
Q4.8
Suppose you are only interested in finding dominant mutations that affect floral scent.
1. What do you expect to be the relative frequency of dominant mutations, as compared to recessive mutations, and why?
2. How will you design your screen differently than in the previous question, in order to detect dominant mutations specifically?
3. Which kind of mutagen is most likely to produce dominant mutations, a mutagen that produces point mutations, or a mutagen that produces large deletions?
Q4.9
hich types of transposable elements are transcribed?
Q4.10
You are interested in finding genes involved in synthesis of proline (Pro), an amino acid that is normally synthesizes by a particular model organism.
1. How would you design a mutant screen to identify genes required for Pro synthesis?
2. Imagine that your screen identified ten mutants (#1 through #10) that grew poorly unless supplemented with Pro. How could you determine the number of different genes represented by these mutants?
3. If each of the four mutants represents a different gene, what will be the phenotype of the F1 progeny if any pair of the four mutants are crossed?
4. If each of the four mutants represents the same gene, what will be the phenotype of the F1 progeny if any pair of the four mutants are crossed?
Chapter 4 - Answers
4.1 Polymorphisms and mutations are both variations in DNA sequence and can arise through the same mechanisms. We use the term polymorphism to refer to DNA variants that are relatively common in populations. Mutations affect the phenotype.
4.2 Misreading of bases during replication can lead to substitution and can be caused by things like tautomerism, DNA alkylating agents, and irradiation.
4.3 Looping out of DNA on the template strand during replication; strand breakage, due to radiation and other mutagens; and (discussed in earlier chapters) chromosomal aberrations such as deletions and translocations.
4.4 Looping out of DNA on the growing strand during replication; transposition; and (discussed in earlier chapters) chromosomal aberrations such as duplications, insertions, and translocation.
4.5 Benzopyrene is one of many hazardous compounds present in smoke. Benzopyrene is an intercalating agent, which slides between the bases of the DNA molecule, distorting the shape of the double helix, which disrupts transcription and replication and can lead to mutation.
4.6 a) One possible explanation is that original mutagenesis resulted in a loss-of-function mutation in a gene that is essential for early embryonic development, and that this mutation is X-linked recessive in the female. Because half of the sons will inherit the X chromosome that bears this mutation, half of the sons will fail to develop beyond very early development and will not be detected among the F1 progeny. The proportion of male flies that were affected depends on what fraction of the female parent’s gametes carried the mutation. In this case, it appears that half of the female’s gametes carried the mutation.
b) To test whether a gene is X-linked, you can usually do a reciprocal cross. However, in this case it would be impossible to obtain adult male flies that carry the mutation; they are dead. If the hypothesis proposed in a) above is correct, then half of the females, and none of the living males in the F1 should carry the mutant allele. You could therefore cross F1 females to wild type males, and see whether the expected ratios were observed among the offspring (e.g. half of the F1 females should have a fewer male offspring than expected, while the other half of the F1 females and all of the males should have a roughly equal numbers of male and female offspring).
4.7 a) Treat a population of seeds with a mutagen such as EMS. Allow these seeds to self-pollinate, and then allow the F1 generation to also self-pollinate. In the F2 generation, smell each flower to find individuals with abnormal scent.
b) The fishy gene appears to be required to make the normal floral scent. Because the flowers smell fishy in the absence of this gene, one possibility explanation of this is that fishy makes an enzyme that converts a fishy-smelling intermediate into a chemical that gives flowers their normal, sweet smell.
Note that although we show this biochemical pathway as leading from the fishy-smelling chemical to the sweet-smelling chemical in one step, it is likely that there are many other enzymes that act after the fishy enzyme to make the final, sweet-smelling product. In either case, blocking the pathway at the step catalyzed by the fishy enzyme would explain the fishy smell.
c) In nosmell plants, the normal sweet smell disappears. Unlike fishy, the sweet smell is not replaced by any intermediate chemical that we can easily detect. Thus, we cannot conclude where in the biochemical pathway the nosmell mutant is blocked; nosmell may normally therefore act either before or after fishy normally acts in the pathway:
Alternatively, nosmell may not be part of the biosynthetic pathway for the sweet smelling chemical at all. It is possible that the normal function of this gene is to transport the sweet-smelling chemical into the cells from which it is released into the air, or maybe it is required for the development of those cells in the first place. It could even be something as general as keeping the plants healthy enough that they have enough energy to do things like produce floral scent.
4.8 a) Dominant mutations are generally much rarer than recessive mutations. This is because mutation of a gene tends to cause a loss of the normal function of this gene. In most cases, having just one normal (wt) allele is sufficient for normal biological function, so the mutant allele is recessive to the wt allele. Very rarely, rather than destroying normal gene function, the random act of mutation will cause a gene to gain a new function (e.g. to catalyze a new enzymatic reaction), which can be dominant (since it performs this new function whether the wt allele is present or not). This type of gain-of-function dominant mutation is very rare because there are many more ways to randomly destroy something than by random action to give it a new function (think of the example given in class of stomping on an iPod).
b) Dominant mutations should be detectable in the F1 generation, so the F1 generation, rather than the F2 generation can be screened for the phenotype of interest.
c) Large deletions, such as those caused by some types of radiation, are generally less likely than point mutations to introduce a new function into a protein: it is hard for a protein to gain a new function if the entire gene has been removed from the genome by deletion.
4.9 Class I. see Figure 4.5 on Transposable Elements.
4.10 a) Mutagenize a wild type (auxotrophic) strain and screen for mutations that fail to grow on minimal media, but grow well on minimal media supplemented with proline.
b) Take mutants #1-#10) and characterize them, based on (1) genetic mapping of the mutants (different locations indicate different genes); (2) different response to proline precursors (a different response suggests different genes); (3) complementation tests among the mutations (if they complement then they are mutations in different genes).
c) If the mutations are in different genes then the F1 progeny would be wild type (able to grow on minimal medium without proline).
d) If the mutations are in the same gene then the F1 progeny would NOT be wild type (unable to grow on minimal medium without proline). | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/04.E%3A_Mutation_and_Variation_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Q4.1
How are polymorphisms and mutations alike? How are they different?
Q4.2
What are some of the ways a substitution can occur in a DNA sequence?
Q4.3
What are some of the ways a deletion can occur in a DNA sequence?
Q4.4
What are all of the ways an insertion can occur in a DNA sequence?
Q4.5
In the context of this chapter, explain the health hazards of smoking tobacco.
Q4.6
You have a female fruit fly, whose father was exposed to a mutagen (she, herself, wasn’t). Mating this female fly with another non-mutagenized, wild type male produces offspring that all appear to be completely normal, except there are twice as many daughters as sons in the F1 progeny of this cross.
1. Propose a hypothesis to explain these observations.
2. How could you test your hypothesis?
Q4.7
You decide to use genetics to investigate how your favourite plant makes its flowers smell good.
1. What steps will you take to identify some genes that are required for production of the sweet floral scent? Assume that this plant is a self-pollinating diploid.
2. One of the recessive mutants you identified has fishy-smelling flowers, so you name the mutant (and the mutated gene) fishy. What do you hypothesize about the normal function of the wild-type fishy gene?
3. Another recessive mutant lacks floral scent altogether, so you call it nosmell. What could you hypothesize about the normal function of this gene?
Q4.8
Suppose you are only interested in finding dominant mutations that affect floral scent.
1. What do you expect to be the relative frequency of dominant mutations, as compared to recessive mutations, and why?
2. How will you design your screen differently than in the previous question, in order to detect dominant mutations specifically?
3. Which kind of mutagen is most likely to produce dominant mutations, a mutagen that produces point mutations, or a mutagen that produces large deletions?
Q4.9
hich types of transposable elements are transcribed?
Q4.10
You are interested in finding genes involved in synthesis of proline (Pro), an amino acid that is normally synthesizes by a particular model organism.
1. How would you design a mutant screen to identify genes required for Pro synthesis?
2. Imagine that your screen identified ten mutants (#1 through #10) that grew poorly unless supplemented with Pro. How could you determine the number of different genes represented by these mutants?
3. If each of the four mutants represents a different gene, what will be the phenotype of the F1 progeny if any pair of the four mutants are crossed?
4. If each of the four mutants represents the same gene, what will be the phenotype of the F1 progeny if any pair of the four mutants are crossed?
Chapter 4 - Answers
4.1 Polymorphisms and mutations are both variations in DNA sequence and can arise through the same mechanisms. We use the term polymorphism to refer to DNA variants that are relatively common in populations. Mutations affect the phenotype.
4.2 Misreading of bases during replication can lead to substitution and can be caused by things like tautomerism, DNA alkylating agents, and irradiation.
4.3 Looping out of DNA on the template strand during replication; strand breakage, due to radiation and other mutagens; and (discussed in earlier chapters) chromosomal aberrations such as deletions and translocations.
4.4 Looping out of DNA on the growing strand during replication; transposition; and (discussed in earlier chapters) chromosomal aberrations such as duplications, insertions, and translocation.
4.5 Benzopyrene is one of many hazardous compounds present in smoke. Benzopyrene is an intercalating agent, which slides between the bases of the DNA molecule, distorting the shape of the double helix, which disrupts transcription and replication and can lead to mutation.
4.6 a) One possible explanation is that original mutagenesis resulted in a loss-of-function mutation in a gene that is essential for early embryonic development, and that this mutation is X-linked recessive in the female. Because half of the sons will inherit the X chromosome that bears this mutation, half of the sons will fail to develop beyond very early development and will not be detected among the F1 progeny. The proportion of male flies that were affected depends on what fraction of the female parent’s gametes carried the mutation. In this case, it appears that half of the female’s gametes carried the mutation.
b) To test whether a gene is X-linked, you can usually do a reciprocal cross. However, in this case it would be impossible to obtain adult male flies that carry the mutation; they are dead. If the hypothesis proposed in a) above is correct, then half of the females, and none of the living males in the F1 should carry the mutant allele. You could therefore cross F1 females to wild type males, and see whether the expected ratios were observed among the offspring (e.g. half of the F1 females should have a fewer male offspring than expected, while the other half of the F1 females and all of the males should have a roughly equal numbers of male and female offspring).
4.7 a) Treat a population of seeds with a mutagen such as EMS. Allow these seeds to self-pollinate, and then allow the F1 generation to also self-pollinate. In the F2 generation, smell each flower to find individuals with abnormal scent.
b) The fishy gene appears to be required to make the normal floral scent. Because the flowers smell fishy in the absence of this gene, one possibility explanation of this is that fishy makes an enzyme that converts a fishy-smelling intermediate into a chemical that gives flowers their normal, sweet smell.
Note that although we show this biochemical pathway as leading from the fishy-smelling chemical to the sweet-smelling chemical in one step, it is likely that there are many other enzymes that act after the fishy enzyme to make the final, sweet-smelling product. In either case, blocking the pathway at the step catalyzed by the fishy enzyme would explain the fishy smell.
c) In nosmell plants, the normal sweet smell disappears. Unlike fishy, the sweet smell is not replaced by any intermediate chemical that we can easily detect. Thus, we cannot conclude where in the biochemical pathway the nosmell mutant is blocked; nosmell may normally therefore act either before or after fishy normally acts in the pathway:
Alternatively, nosmell may not be part of the biosynthetic pathway for the sweet smelling chemical at all. It is possible that the normal function of this gene is to transport the sweet-smelling chemical into the cells from which it is released into the air, or maybe it is required for the development of those cells in the first place. It could even be something as general as keeping the plants healthy enough that they have enough energy to do things like produce floral scent.
4.8 a) Dominant mutations are generally much rarer than recessive mutations. This is because mutation of a gene tends to cause a loss of the normal function of this gene. In most cases, having just one normal (wt) allele is sufficient for normal biological function, so the mutant allele is recessive to the wt allele. Very rarely, rather than destroying normal gene function, the random act of mutation will cause a gene to gain a new function (e.g. to catalyze a new enzymatic reaction), which can be dominant (since it performs this new function whether the wt allele is present or not). This type of gain-of-function dominant mutation is very rare because there are many more ways to randomly destroy something than by random action to give it a new function (think of the example given in class of stomping on an iPod).
b) Dominant mutations should be detectable in the F1 generation, so the F1 generation, rather than the F2 generation can be screened for the phenotype of interest.
c) Large deletions, such as those caused by some types of radiation, are generally less likely than point mutations to introduce a new function into a protein: it is hard for a protein to gain a new function if the entire gene has been removed from the genome by deletion.
4.9 Class I. see Figure 4.5 on Transposable Elements.
4.10 a) Mutagenize a wild type (auxotrophic) strain and screen for mutations that fail to grow on minimal media, but grow well on minimal media supplemented with proline.
b) Take mutants #1-#10) and characterize them, based on (1) genetic mapping of the mutants (different locations indicate different genes); (2) different response to proline precursors (a different response suggests different genes); (3) complementation tests among the mutations (if they complement then they are mutations in different genes).
c) If the mutations are in different genes then the F1 progeny would be wild type (able to grow on minimal medium without proline).
d) If the mutations are in the same gene then the F1 progeny would NOT be wild type (unable to grow on minimal medium without proline). | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/04.E%3A_Mutation_and_Variation_(Exercises).txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
1. What are some of the modes of inheritance that are consistent with this pedigree?
2. In this pedigree in question 1, the mode of inheritance cannot be determined unamibguously.What are some examples of data (e.g. from other generations) that, if added to the pedigree would help determine the mode of inheritance?
3. For each of the following pedigrees, name the most likely mode of inheritance (AR=autosomal recessive, AD=autosomal dominant, XR=X-linked recessive, XD=X-linked dominant).(These pedigrees were obtained from various external sources).
a)
b)
c)
d)
4. The following pedigree represents a rare, autosomal recessive disease. What are the genotypes of the individuals who are indicated by letters?
5. If individual #1 in the following pedigree is a heterozygote for a rare, AR disease, what is the probability that individual #7 will be affected by the disease? Assume that #2 and the spouses of #3 and #4 are not carriers.
6. You are studying a population in which the frequency of individuals with a recessive homozygous genotype is 1%. Assuming the population is in Hardy-Weinberg equilibrium, calculate:
a) The frequency of the recessive allele.
b) The frequency of dominant allele.
c) The frequency of the heterozygous phenotype.
d) The frequency of the homozygous dominant phenotype.
7. Determine whether the following population is in Hardy-Weinberg equilibrium.
genotype number of individuals
AA
Aa
aa
432
676
92
8. Out of 1200 individuals examined, 432 are homozygous dominant (AA)for a particular gene. What numbers of individuals of the other two genotypic classes (Aa, aa) would be expected if the population is in Hardy-Weinberg equilibrium?
9. Propose an explanation for the deviation between the genotypic frequencies calculated in question 8 and those observed in the table in question 7.
Chapter 5 - Answers
5.1 The pedigree could show an AD, AR or XR mode of inheritance. It is most likely AD. It could be AR if the mother was a carrier, and the father was a homozygote. It could be XR if the mother was a carrier, and the father was a hemizygote. It cannot be XD, since the daughter (#2) would have necessarily inherited the disease allele on the X chromosome she received from her father.
5.2 There are many possible answers. Here are some possibilities: if neither of the parents of the father were affected (i.e. the paternal grandparents of children 1, 2, 3), then the disease could not be dominant. If only the paternal grandfather was affected, then the disease could only be X-linked recessive if the paternal grandmother was a heterozygote (which would be unlikely given that this is a rare disease allele).
5.3 a) The mode of inheritance is most likely AD, since every affected individual has an affected parent, and the disease is inherited even in four different matings to unrelated, unaffected individuals. It is very unlikely that it is XD or XR, in part because affected father had an affected son.
b) The mode of inheritance cannot be AD or XD, because affected individuals must have an affected parent when a disease allele is dominant. Neither can it be XR, because there is an affected daughter of a normal father. Therefore, it must be AR, and this is consistent with the pedigree.
c) The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It is not XR, because there are unaffected sons of an affected mother. It is therefore likely AR, but note that the recessive alleles for this condition appear to be relatively common in the population (note that two of the marriages were to unrelated, affected individuals).
d) The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It could be either XR or AR, but because all of the affected individuals are male, and no affected males pass the disease to their sons, it is likely XR.
5.4 If a represents the disease allele, individuals a, d, f (who all married into this unusual family) are AA, while b, c, e, g, h, i, j are all Aa, and k is aa.
5.5 There is a ½ chance that an offspring of any mating Aa x AA will be a carrier (Aa). So, there is a ½ chance that #3 will be Aa, and likewise for #4. If #3 is a carrier, there is again a ½ chance that #5 will be a carrier, and likewise for #6. If #5 and #6 are both Aa, then there is a ¼ chance that this monohybrid cross will result in #7 having the genotype aa, and therefore being affected by the disease. Thus, the joint probability is 1/2 x 1/2 x 1/2 x 1/2 x 1/4 =1/64.
5.6 a) q = -0.01. = 0.1
b) 1-q = p; 1-0.1 = 0.9
c) 2pq = 2(0.1)(0.9) = 0.18
d) p2 = 0.81
5.7 First, calculate allele frequencies:
$\mathrm{p = \dfrac{2(AA) + (Aa)}{total\: number\: of\: alleles\: scored} = \dfrac{2(432) + 676}{2(432+676+92)} = 0.6417}$
$\mathrm{q = \dfrac{2(aa) + (Aa)}{total\: number\: of\: alleles\: scored} = \dfrac{2(92) + 676}{2(432+676+92)} = 0.3583}$
Next, given these observed allele frequencies, calculate the genotypic frequencies that would be expected if the population was in Hardy-Weinberg equilibrium.
$\mathrm{p^2 = 0.6417^2 = 0.4118}$
$\mathrm{2pq = 2(0.6417)(0.3583) = 0.4598}$
$\mathrm{q^2 = 0.3583^2 = 0.1284}$
Finally, given these expected frequencies of each class, calculate the expected numbers of each in your sample of 1200 individuals, and compare these to your actual observations.
expected observed (reported in the original question)
AA 0.4118 × 1200 = 494
Aa 0.4598 × 1200 = 552
aa 0.1284 × 1200 = 154
432
676
92
The population does not appear to be at Hardy-Weinberg equilibrium, since the observed genotypic frequencies do not match the expectations. Of course, you could do a chi-square test to determine how significant the discrepancy is between observed and expected.
5.8 If in this theoretical question, the frequency of genotype of AA is set at 432/1200 and we are asked what frequencies of the other classes would fit a Hardy-Weinberg equilibrium. So, given that p2 = 432/1200, then p=0.6, and q=0.4. Given these allele frequencies and a sample size of 1200 individuals, then there should be 576 Aa individuals (2pq × 1200 = 2(0.6)(0.4) × 1200=576) and 192 aa individuals (q2 × 1200 = 0.42 × 1200 = 192), if the population was at Hardy-Weinberg equilibrium with 432 AA individuals.
5.9 The actual population appear to have more heterozygotes and few recessive homozygotes than would be expected for Hardy-Weinberg equilibrium. There are many possible reasons that a population may not be in equilibrium (see Table 5.1). In this case, there is possibly some selection against homozygous recessive genotypes, in favour of heterozygotes in particular. Perhaps the heterozygotes have some selective advantage that increases their fitness.
It is also worth noting the discrepancies between the allele frequencies calculated in 5.8 and 5.7. In question 5.7, we calculated the frequencies directly from the genotypes (this is the most accurate method, and does not require the population to be in equilibrium). In 5.8, we essentially estimated the frequency base on one of the phenotypic classes. The discrepancy between these calculations shows the limitations of using phenotypes to estimate allele frequencies, when a population is not in equilibrium. | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/05.E%3A_Pedigrees_and_Populations_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
1. What are some of the modes of inheritance that are consistent with this pedigree?
2. In this pedigree in question 1, the mode of inheritance cannot be determined unamibguously.What are some examples of data (e.g. from other generations) that, if added to the pedigree would help determine the mode of inheritance?
3. For each of the following pedigrees, name the most likely mode of inheritance (AR=autosomal recessive, AD=autosomal dominant, XR=X-linked recessive, XD=X-linked dominant).(These pedigrees were obtained from various external sources).
a)
b)
c)
d)
4. The following pedigree represents a rare, autosomal recessive disease. What are the genotypes of the individuals who are indicated by letters?
5. If individual #1 in the following pedigree is a heterozygote for a rare, AR disease, what is the probability that individual #7 will be affected by the disease? Assume that #2 and the spouses of #3 and #4 are not carriers.
6. You are studying a population in which the frequency of individuals with a recessive homozygous genotype is 1%. Assuming the population is in Hardy-Weinberg equilibrium, calculate:
a) The frequency of the recessive allele.
b) The frequency of dominant allele.
c) The frequency of the heterozygous phenotype.
d) The frequency of the homozygous dominant phenotype.
7. Determine whether the following population is in Hardy-Weinberg equilibrium.
genotype number of individuals
AA
Aa
aa
432
676
92
8. Out of 1200 individuals examined, 432 are homozygous dominant (AA)for a particular gene. What numbers of individuals of the other two genotypic classes (Aa, aa) would be expected if the population is in Hardy-Weinberg equilibrium?
9. Propose an explanation for the deviation between the genotypic frequencies calculated in question 8 and those observed in the table in question 7.
Chapter 5 - Answers
5.1 The pedigree could show an AD, AR or XR mode of inheritance. It is most likely AD. It could be AR if the mother was a carrier, and the father was a homozygote. It could be XR if the mother was a carrier, and the father was a hemizygote. It cannot be XD, since the daughter (#2) would have necessarily inherited the disease allele on the X chromosome she received from her father.
5.2 There are many possible answers. Here are some possibilities: if neither of the parents of the father were affected (i.e. the paternal grandparents of children 1, 2, 3), then the disease could not be dominant. If only the paternal grandfather was affected, then the disease could only be X-linked recessive if the paternal grandmother was a heterozygote (which would be unlikely given that this is a rare disease allele).
5.3 a) The mode of inheritance is most likely AD, since every affected individual has an affected parent, and the disease is inherited even in four different matings to unrelated, unaffected individuals. It is very unlikely that it is XD or XR, in part because affected father had an affected son.
b) The mode of inheritance cannot be AD or XD, because affected individuals must have an affected parent when a disease allele is dominant. Neither can it be XR, because there is an affected daughter of a normal father. Therefore, it must be AR, and this is consistent with the pedigree.
c) The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It is not XR, because there are unaffected sons of an affected mother. It is therefore likely AR, but note that the recessive alleles for this condition appear to be relatively common in the population (note that two of the marriages were to unrelated, affected individuals).
d) The mode of inheritance cannot be AD or XD, because, again, there are affected individuals with unaffected parents. It could be either XR or AR, but because all of the affected individuals are male, and no affected males pass the disease to their sons, it is likely XR.
5.4 If a represents the disease allele, individuals a, d, f (who all married into this unusual family) are AA, while b, c, e, g, h, i, j are all Aa, and k is aa.
5.5 There is a ½ chance that an offspring of any mating Aa x AA will be a carrier (Aa). So, there is a ½ chance that #3 will be Aa, and likewise for #4. If #3 is a carrier, there is again a ½ chance that #5 will be a carrier, and likewise for #6. If #5 and #6 are both Aa, then there is a ¼ chance that this monohybrid cross will result in #7 having the genotype aa, and therefore being affected by the disease. Thus, the joint probability is 1/2 x 1/2 x 1/2 x 1/2 x 1/4 =1/64.
5.6 a) q = -0.01. = 0.1
b) 1-q = p; 1-0.1 = 0.9
c) 2pq = 2(0.1)(0.9) = 0.18
d) p2 = 0.81
5.7 First, calculate allele frequencies:
$\mathrm{p = \dfrac{2(AA) + (Aa)}{total\: number\: of\: alleles\: scored} = \dfrac{2(432) + 676}{2(432+676+92)} = 0.6417}$
$\mathrm{q = \dfrac{2(aa) + (Aa)}{total\: number\: of\: alleles\: scored} = \dfrac{2(92) + 676}{2(432+676+92)} = 0.3583}$
Next, given these observed allele frequencies, calculate the genotypic frequencies that would be expected if the population was in Hardy-Weinberg equilibrium.
$\mathrm{p^2 = 0.6417^2 = 0.4118}$
$\mathrm{2pq = 2(0.6417)(0.3583) = 0.4598}$
$\mathrm{q^2 = 0.3583^2 = 0.1284}$
Finally, given these expected frequencies of each class, calculate the expected numbers of each in your sample of 1200 individuals, and compare these to your actual observations.
expected observed (reported in the original question)
AA 0.4118 × 1200 = 494
Aa 0.4598 × 1200 = 552
aa 0.1284 × 1200 = 154
432
676
92
The population does not appear to be at Hardy-Weinberg equilibrium, since the observed genotypic frequencies do not match the expectations. Of course, you could do a chi-square test to determine how significant the discrepancy is between observed and expected.
5.8 If in this theoretical question, the frequency of genotype of AA is set at 432/1200 and we are asked what frequencies of the other classes would fit a Hardy-Weinberg equilibrium. So, given that p2 = 432/1200, then p=0.6, and q=0.4. Given these allele frequencies and a sample size of 1200 individuals, then there should be 576 Aa individuals (2pq × 1200 = 2(0.6)(0.4) × 1200=576) and 192 aa individuals (q2 × 1200 = 0.42 × 1200 = 192), if the population was at Hardy-Weinberg equilibrium with 432 AA individuals.
5.9 The actual population appear to have more heterozygotes and few recessive homozygotes than would be expected for Hardy-Weinberg equilibrium. There are many possible reasons that a population may not be in equilibrium (see Table 5.1). In this case, there is possibly some selection against homozygous recessive genotypes, in favour of heterozygotes in particular. Perhaps the heterozygotes have some selective advantage that increases their fitness.
It is also worth noting the discrepancies between the allele frequencies calculated in 5.8 and 5.7. In question 5.7, we calculated the frequencies directly from the genotypes (this is the most accurate method, and does not require the population to be in equilibrium). In 5.8, we essentially estimated the frequency base on one of the phenotypic classes. The discrepancy between these calculations shows the limitations of using phenotypes to estimate allele frequencies, when a population is not in equilibrium. | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/05.E%3A_Pedigrees_and_Populations_(Exercises).txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
Answer questions 6.1 -6.3 using the following biochemical pathway for fruit color.Assume all mutations (lower case allele symbols) are recessive, and that either precursor 1 or precursor 2 can be used to produce precursor 3.If the alleles for a particular gene are not listed in a genotype, you can assume that they are wild-type.
6.1 If1 and 2 and 3 are all colorless, and 4 is red, what will be the phenotypes associated with the following genotypes?
1. aa
2. bb
3. dd
4. aabb
5. aadd
6. bbdd
7. aabbdd
8. What will be the phenotypic ratios among the offspring of a cross AaBb × AaBb?
9. What will be the phenotypic ratios among the offspring of a cross BbDd× BbDd?
10. What will be the phenotypic ratios among the offspring of a cross AaDd × AaDd?
6.2 If1 and 2 are both colorless, and 3 is blue and 4 is red, what will be the phenotypes associated with the following genotypes?
1. aa
2. bb
3. dd
4. aabb
5. aadd
6. bbdd
7. aabbdd
8. What will be the phenotypic ratios among the offspring of a cross AaBb × AaBb?
9. What will be the phenotypic ratios among the offspring of a cross BbDd× BbDd?
10. What will be the phenotypic ratios among the offspring of a cross AaDd × AaDd?
6.3 If1 is colorless, 2 is yellow and 3 is blue and 4 is red, what will be the phenotypes associated with the following genotypes?
1. aa
2. bb
3. dd
4. aabb
5. aadd
6. bbdd
7. aabbdd
8. What will be the phenotypic ratios among the offspring of a cross AaBb × AaBb?
9. What will be the phenotypic ratios among the offspring of a cross BbDd× BbDd?
10. What will be the phenotypic ratios among the offspring of a cross AaDd × AaDd?
6.4 Which of the situations in questions 6.1 – 6.3 demonstrate epistasis?
6.5 If the genotypes written within the Punnett Square are from the F2 generation, what would be the phenotypes and genotypes of the F1 and P generations for:
1. Figure 6.6
2. Figure 6.8
3. Figure 6.10
4. Figure 6.12
6.6 To better understand how genes control the development of three‐dimensional structures, you conducted a mutant screen in Arabidopsis and identified a recessive point mutation allele of a single gene (g) that causes leaves to develop as narrow tubes rather than the broad flat surfaces that develop in wild‐type (G). Allele g causes a complete loss of function. Now you want to identify more genes involved in the same process. Diagram a process you could use to identify other genes that interact with gene g. Show all of the possible genotypes that could arise in the Fgeneration.
6.7 With reference to question 6.6, if the recessive allele, g is mutated again to make allele g*, what are the possible phenotypes of a homozygous g* g* individual?
6.8 Again, in reference to question 6.7, what are the possible phenotypes of a homozygous aagg individual, where a is a recessive allele of a second gene? In each case, also specify the phenotypic ratios that would be observed among the F1 progeny of a cross of AaGg x AaGg
6.9 Calculate the phenotypic ratios from a dihybrid cross involving the two loci shown in Figure 6.13.There may be more than one possible set of ratios, depending on the assumptions you make about the phenotype of allele b.
6.10 Use the product rule to calculate the phenotypic ratios expected from a trihybrid cross.Assume independent assortment and no epistasis/gene interactions.
Chapter 6 - Answers
6.1 If 1 and 2 and 3 are all colorless, and 4 is red, what will be the phenotypes associated with the following genotypes? All of these mutations are recessive. As always, if the genotype for a particular gene is not listed, you can assume that alleles for that gene are wild-type.
1. red(because A and B are redundant, so products 3 and then 4 can be made)
2. red(because A and B are redundant, so products 3 and then 4 can be made)
3. white (because product3 will accumulate and it is colorless)
4. white (because only product 1 and 2 will be present and both are colorless)
5. white (because only product 1 and 3 will be present and both are colorless)
6. white (because only product 2 and 3 will be present and both are colorless)
7. white (because only product 1 and 2 will be present and both are colorless)
8. 15 red : 1 white
9. 12 red : 4 white
10. 12 red :4 white
6.2
1. red(because A and B are redundant, so products 3 and then 4 can be made)
2. red(because A and B are redundant, so products 3 and then 4 can be made)
3. blue (because product 3 will accumulate, and it is blue)
4. white (because only product 1 and 2 will be present and both are colorless)
5. blue (because only product 1 and 3 will be present and 1 is colorless and 3 is blue)
6. blue(because only product 2 and 3 will be present and 2 is colorless and 3 is blue)
7. white (because only product 1 and 2 will be present and both are colorless)
8. 15 red : 1 white
9. 12 red : 4 blue
10. 12 red : 4 blue
6.3
1. red(because A and B are redundant, so products 3 and then 4 can be made)
2. red(because A and B are redundant, so products 3 and then 4 can be made)
3. blue (because product 3 will accumulate, and it is blue)
4. yellow (because only product 1 and 2 will be present and 1 is colorless and 2 is yellow)
5. blue (because only product 1 and 3 will be present and 1 is colorless and 3 is blue)
6. green? (because only product 2 and 3 will be present and 2 is yellow and 3 is blue, so probably the fruit will be some combination of those two colors)
7. yellow (because only product 1 and 2 will be present and 1 is colorless and 2 is yellow)
8. 15 red : 1 yellow
9. 12 red : 3 blue:1 green
10. 12 red : 4 blue
6.4 Epistasis is demonstrated when the phenotype for a homozygous mutant in one gene is the same as the phenotype for a homozygous mutant in two genes. So, the following situations from questions 6.1-6.3 demonstrated epistasis:
6.1 No epistasis is evident from the phenotypes, even though we know from the pathway provided that D is downstream of A and B.
6.2 The phenotypes show that D is epistatic to A, because aadd looks like AAdd or Aadd. Also D is epistatic to B, because bbdd looks like BBdd or Bbdd.
6.3 The phenotypes show that D is epistatic to A, because aadd looks like AAdd or Aadd. The phenotypes do not provide evidence for epistasis between B and D.
6.5 The answer is the same for a) – d)
P could have been either: AABB xaabboraaBB x AAbb;
F1 was : AaBb x AaBb
6.6 Conduct an enhancer/suppressor screen (which can also result in the identification of revertants, as well)
allow the plants to self‐pollinate in order to make any new, recessive mutations homozygous
6.7 Depending which amino acids were altered, and how they were altered, a second mutation in g*g* could either have no effect (in which case the phenotype would be the same as gg), or it could possibly cause a reversion of the phenotype to wild‐type, so that g*g* and GG have the same phenotype.
6.8 Depending on the normal function of gene A, and which amino acids were altered in allele a, there are many potential phenotypes for aagg:
Case 1: If the normal function of gene A is in an unrelated process (e.g. A is required for
root development, but not the development of leaves), then the phenotype of aagg will
be: short roots and narrow leaves. The phenotypic ratios among the progeny of a
dihybrid cross will be:
9
3
3
1
A_G_
A_gg
aaG_
aagg
wild-type
tubular leaves
normal roots
short roots
normal leaves
tubular leaves
short roots
Case 2: If the normal function of gene A is in the same process as G, such that a is a recessive allele that increases the severity of the gg mutant (i.e. a is an enhancer of g) then the phenotype of aagg could be : no leaves. The phenotypic ratios among the progeny of a dihybrid cross depend on whether aa mutants have a phenotype independent of gg, in other words, do aaG_ plants have a phenotype that is different from wild‐type or from A_gg. There is no way to know this without doing the experiment, since it depends on the biology of the particular gene, mutation and pathway involved, so there are three possible outcomes:
Case 2a) If aa is an enhancer of gg, and aaG_ plants have a mutant phenotype that differs from wild‐type or (A_gg) then the phenotypic ratios among the progeny of a dihybrid cross will be:
9
3
3
1
A_G_
A_gg
aaG_
aagg
wild-type
tubular leaves
(some phenotype that differs from gg; maybe small twisted leaves)
abnormal leaves
no leaves
Case 2b) If aa is an enhancer of gg, and aaG_ plants have a mutant phenotype that is the same as A_gg , the phenotypic ratios among the progeny of a dihybrid cross will be:
9
6
1
A_G_
A_gg aaG_
aagg
wild-type
tubular leaves
no leaves
Case 2c) If aa is an enhancer of gg, and aaG_ do not have a phenotype that differs from wild‐type then the phenotypic ratios among the progeny of a dihybrid cross will be:
12
3
1
A_G_ aaG_
A_gg
aagg
wild-type
tubular leaves
no leaves
Case 3: If the normal function of gene A is in the same process as G, such that a is a recessive allele that decreases the severity of the gg mutant (i.e. a is an suppressor of g) then the phenotype of aagg could be : wild‐type. The phenotypic ratios among the progeny of a dihybrid cross depend on whether aa mutants have a phenotype independent of gg, in other words, do aaG_ plants have a phenotype that is different from wild‐type or from A_gg. There is no way to know this without doing the experiment, since it depends on the biology of the particular gene, mutation and pathway involved, so there are three possible outcomes:
Case 3a) If aa is a suppressor of gg, and aaG_ plants have a mutant phenotype that differs from wild‐type or (A_gg) then the phenotypic ratios among the progeny of a dihybrid cross will be:
10
3
3
A_G_ aagg
A_gg
aaG_
wild-type
tubular leaves (some phenotype that differs from gg)
no leaves
Case 3b) If aa is an suppressor of gg, and aaG_ plants have a mutant phenotype that is the same as A_gg the phenotypic ratios among the progeny of a dihybrid cross will be:
10
6
A_G_ aagg
A_gg aaG_
wild-type
tubular leaves
Case 3c) If aa is an suppressor of gg, and aaG_ plants do not have a phenotype that differs from wild‐type then the phenotypic ratios among the progeny of a dihybrid cross will be:
13
3
A_G_ aaG_ aagg
A_gg
wild-type
tubular leaves
Case 4: If the normal function of gene A is in the same process as G, such that a is a
recessive allele that with a phenotype that is epistatic to the gg mutant then the
phenotype of both aaG_ and aagg could be : no leaves. The phenotypic ratios among
the progeny of a dihybrid cross will be:
9
4
3
A_G_
aaG_ aagg
A_gg
wild-type
no leaves
tubular leaves
Case … ?: There are many more phenotypes and ratios that could be imagined (e.g.
different types of dominance relationships, different types of epistasis, lethality…etc).
Isn’t genetics wonderful? It is sometimes shocking that more people don’t want to
become geneticists.
The point of this exercise is to show that many different ratios can be generated,
depending on the biology of the genes involved. On an exam, you could be asked to
calculate the ratio, given particular biological parameters. So, this exercise is also meant
to demonstrate that it is better to learn how to calculate ratios than just trying to
memorize which ratios match which parameters. In a real genetic screen, you would
observe the ratios, and then try to deduce something about the biology from those
ratios.
6.9 Calculate the phenotypic ratios from a dihybrid cross involving the two loci shown in Figure 6.13. There may be more than one possible set of ratios, depending on the assumptions you make about the phenotype of allele b.
Assuming that bb has no phenotype on its own (i.e. A_bb looks like A_B_), then aaB_ will have the mutant phenotype, and A_bb, A_B_, and aabb will appear phenotypically wild-type. The phenotypic ratio will be 13 wild-type : 3 mutant.
6.10 For a dihybrid cross, there are 4 classes, 9:3:3:1. In a trihybrid cross without gene interactions, each of these 4 classes will be further split into a 3:1 ratio based on the phenotype at the third locus. For example, 9 x 3 =27 and 9 x 1 = 9. This explains the first two terms of the complete ratio: 27:9:9:9:3:3:3:1. | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/06.E%3A_Genetic_Analysis_of_Multiple_Genes_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
Answer questions 6.1 -6.3 using the following biochemical pathway for fruit color.Assume all mutations (lower case allele symbols) are recessive, and that either precursor 1 or precursor 2 can be used to produce precursor 3.If the alleles for a particular gene are not listed in a genotype, you can assume that they are wild-type.
6.1 If1 and 2 and 3 are all colorless, and 4 is red, what will be the phenotypes associated with the following genotypes?
1. aa
2. bb
3. dd
4. aabb
5. aadd
6. bbdd
7. aabbdd
8. What will be the phenotypic ratios among the offspring of a cross AaBb × AaBb?
9. What will be the phenotypic ratios among the offspring of a cross BbDd× BbDd?
10. What will be the phenotypic ratios among the offspring of a cross AaDd × AaDd?
6.2 If1 and 2 are both colorless, and 3 is blue and 4 is red, what will be the phenotypes associated with the following genotypes?
1. aa
2. bb
3. dd
4. aabb
5. aadd
6. bbdd
7. aabbdd
8. What will be the phenotypic ratios among the offspring of a cross AaBb × AaBb?
9. What will be the phenotypic ratios among the offspring of a cross BbDd× BbDd?
10. What will be the phenotypic ratios among the offspring of a cross AaDd × AaDd?
6.3 If1 is colorless, 2 is yellow and 3 is blue and 4 is red, what will be the phenotypes associated with the following genotypes?
1. aa
2. bb
3. dd
4. aabb
5. aadd
6. bbdd
7. aabbdd
8. What will be the phenotypic ratios among the offspring of a cross AaBb × AaBb?
9. What will be the phenotypic ratios among the offspring of a cross BbDd× BbDd?
10. What will be the phenotypic ratios among the offspring of a cross AaDd × AaDd?
6.4 Which of the situations in questions 6.1 – 6.3 demonstrate epistasis?
6.5 If the genotypes written within the Punnett Square are from the F2 generation, what would be the phenotypes and genotypes of the F1 and P generations for:
1. Figure 6.6
2. Figure 6.8
3. Figure 6.10
4. Figure 6.12
6.6 To better understand how genes control the development of three‐dimensional structures, you conducted a mutant screen in Arabidopsis and identified a recessive point mutation allele of a single gene (g) that causes leaves to develop as narrow tubes rather than the broad flat surfaces that develop in wild‐type (G). Allele g causes a complete loss of function. Now you want to identify more genes involved in the same process. Diagram a process you could use to identify other genes that interact with gene g. Show all of the possible genotypes that could arise in the Fgeneration.
6.7 With reference to question 6.6, if the recessive allele, g is mutated again to make allele g*, what are the possible phenotypes of a homozygous g* g* individual?
6.8 Again, in reference to question 6.7, what are the possible phenotypes of a homozygous aagg individual, where a is a recessive allele of a second gene? In each case, also specify the phenotypic ratios that would be observed among the F1 progeny of a cross of AaGg x AaGg
6.9 Calculate the phenotypic ratios from a dihybrid cross involving the two loci shown in Figure 6.13.There may be more than one possible set of ratios, depending on the assumptions you make about the phenotype of allele b.
6.10 Use the product rule to calculate the phenotypic ratios expected from a trihybrid cross.Assume independent assortment and no epistasis/gene interactions.
Chapter 6 - Answers
6.1 If 1 and 2 and 3 are all colorless, and 4 is red, what will be the phenotypes associated with the following genotypes? All of these mutations are recessive. As always, if the genotype for a particular gene is not listed, you can assume that alleles for that gene are wild-type.
1. red(because A and B are redundant, so products 3 and then 4 can be made)
2. red(because A and B are redundant, so products 3 and then 4 can be made)
3. white (because product3 will accumulate and it is colorless)
4. white (because only product 1 and 2 will be present and both are colorless)
5. white (because only product 1 and 3 will be present and both are colorless)
6. white (because only product 2 and 3 will be present and both are colorless)
7. white (because only product 1 and 2 will be present and both are colorless)
8. 15 red : 1 white
9. 12 red : 4 white
10. 12 red :4 white
6.2
1. red(because A and B are redundant, so products 3 and then 4 can be made)
2. red(because A and B are redundant, so products 3 and then 4 can be made)
3. blue (because product 3 will accumulate, and it is blue)
4. white (because only product 1 and 2 will be present and both are colorless)
5. blue (because only product 1 and 3 will be present and 1 is colorless and 3 is blue)
6. blue(because only product 2 and 3 will be present and 2 is colorless and 3 is blue)
7. white (because only product 1 and 2 will be present and both are colorless)
8. 15 red : 1 white
9. 12 red : 4 blue
10. 12 red : 4 blue
6.3
1. red(because A and B are redundant, so products 3 and then 4 can be made)
2. red(because A and B are redundant, so products 3 and then 4 can be made)
3. blue (because product 3 will accumulate, and it is blue)
4. yellow (because only product 1 and 2 will be present and 1 is colorless and 2 is yellow)
5. blue (because only product 1 and 3 will be present and 1 is colorless and 3 is blue)
6. green? (because only product 2 and 3 will be present and 2 is yellow and 3 is blue, so probably the fruit will be some combination of those two colors)
7. yellow (because only product 1 and 2 will be present and 1 is colorless and 2 is yellow)
8. 15 red : 1 yellow
9. 12 red : 3 blue:1 green
10. 12 red : 4 blue
6.4 Epistasis is demonstrated when the phenotype for a homozygous mutant in one gene is the same as the phenotype for a homozygous mutant in two genes. So, the following situations from questions 6.1-6.3 demonstrated epistasis:
6.1 No epistasis is evident from the phenotypes, even though we know from the pathway provided that D is downstream of A and B.
6.2 The phenotypes show that D is epistatic to A, because aadd looks like AAdd or Aadd. Also D is epistatic to B, because bbdd looks like BBdd or Bbdd.
6.3 The phenotypes show that D is epistatic to A, because aadd looks like AAdd or Aadd. The phenotypes do not provide evidence for epistasis between B and D.
6.5 The answer is the same for a) – d)
P could have been either: AABB xaabboraaBB x AAbb;
F1 was : AaBb x AaBb
6.6 Conduct an enhancer/suppressor screen (which can also result in the identification of revertants, as well)
allow the plants to self‐pollinate in order to make any new, recessive mutations homozygous
6.7 Depending which amino acids were altered, and how they were altered, a second mutation in g*g* could either have no effect (in which case the phenotype would be the same as gg), or it could possibly cause a reversion of the phenotype to wild‐type, so that g*g* and GG have the same phenotype.
6.8 Depending on the normal function of gene A, and which amino acids were altered in allele a, there are many potential phenotypes for aagg:
Case 1: If the normal function of gene A is in an unrelated process (e.g. A is required for
root development, but not the development of leaves), then the phenotype of aagg will
be: short roots and narrow leaves. The phenotypic ratios among the progeny of a
dihybrid cross will be:
9
3
3
1
A_G_
A_gg
aaG_
aagg
wild-type
tubular leaves
normal roots
short roots
normal leaves
tubular leaves
short roots
Case 2: If the normal function of gene A is in the same process as G, such that a is a recessive allele that increases the severity of the gg mutant (i.e. a is an enhancer of g) then the phenotype of aagg could be : no leaves. The phenotypic ratios among the progeny of a dihybrid cross depend on whether aa mutants have a phenotype independent of gg, in other words, do aaG_ plants have a phenotype that is different from wild‐type or from A_gg. There is no way to know this without doing the experiment, since it depends on the biology of the particular gene, mutation and pathway involved, so there are three possible outcomes:
Case 2a) If aa is an enhancer of gg, and aaG_ plants have a mutant phenotype that differs from wild‐type or (A_gg) then the phenotypic ratios among the progeny of a dihybrid cross will be:
9
3
3
1
A_G_
A_gg
aaG_
aagg
wild-type
tubular leaves
(some phenotype that differs from gg; maybe small twisted leaves)
abnormal leaves
no leaves
Case 2b) If aa is an enhancer of gg, and aaG_ plants have a mutant phenotype that is the same as A_gg , the phenotypic ratios among the progeny of a dihybrid cross will be:
9
6
1
A_G_
A_gg aaG_
aagg
wild-type
tubular leaves
no leaves
Case 2c) If aa is an enhancer of gg, and aaG_ do not have a phenotype that differs from wild‐type then the phenotypic ratios among the progeny of a dihybrid cross will be:
12
3
1
A_G_ aaG_
A_gg
aagg
wild-type
tubular leaves
no leaves
Case 3: If the normal function of gene A is in the same process as G, such that a is a recessive allele that decreases the severity of the gg mutant (i.e. a is an suppressor of g) then the phenotype of aagg could be : wild‐type. The phenotypic ratios among the progeny of a dihybrid cross depend on whether aa mutants have a phenotype independent of gg, in other words, do aaG_ plants have a phenotype that is different from wild‐type or from A_gg. There is no way to know this without doing the experiment, since it depends on the biology of the particular gene, mutation and pathway involved, so there are three possible outcomes:
Case 3a) If aa is a suppressor of gg, and aaG_ plants have a mutant phenotype that differs from wild‐type or (A_gg) then the phenotypic ratios among the progeny of a dihybrid cross will be:
10
3
3
A_G_ aagg
A_gg
aaG_
wild-type
tubular leaves (some phenotype that differs from gg)
no leaves
Case 3b) If aa is an suppressor of gg, and aaG_ plants have a mutant phenotype that is the same as A_gg the phenotypic ratios among the progeny of a dihybrid cross will be:
10
6
A_G_ aagg
A_gg aaG_
wild-type
tubular leaves
Case 3c) If aa is an suppressor of gg, and aaG_ plants do not have a phenotype that differs from wild‐type then the phenotypic ratios among the progeny of a dihybrid cross will be:
13
3
A_G_ aaG_ aagg
A_gg
wild-type
tubular leaves
Case 4: If the normal function of gene A is in the same process as G, such that a is a
recessive allele that with a phenotype that is epistatic to the gg mutant then the
phenotype of both aaG_ and aagg could be : no leaves. The phenotypic ratios among
the progeny of a dihybrid cross will be:
9
4
3
A_G_
aaG_ aagg
A_gg
wild-type
no leaves
tubular leaves
Case … ?: There are many more phenotypes and ratios that could be imagined (e.g.
different types of dominance relationships, different types of epistasis, lethality…etc).
Isn’t genetics wonderful? It is sometimes shocking that more people don’t want to
become geneticists.
The point of this exercise is to show that many different ratios can be generated,
depending on the biology of the genes involved. On an exam, you could be asked to
calculate the ratio, given particular biological parameters. So, this exercise is also meant
to demonstrate that it is better to learn how to calculate ratios than just trying to
memorize which ratios match which parameters. In a real genetic screen, you would
observe the ratios, and then try to deduce something about the biology from those
ratios.
6.9 Calculate the phenotypic ratios from a dihybrid cross involving the two loci shown in Figure 6.13. There may be more than one possible set of ratios, depending on the assumptions you make about the phenotype of allele b.
Assuming that bb has no phenotype on its own (i.e. A_bb looks like A_B_), then aaB_ will have the mutant phenotype, and A_bb, A_B_, and aabb will appear phenotypically wild-type. The phenotypic ratio will be 13 wild-type : 3 mutant.
6.10 For a dihybrid cross, there are 4 classes, 9:3:3:1. In a trihybrid cross without gene interactions, each of these 4 classes will be further split into a 3:1 ratio based on the phenotype at the third locus. For example, 9 x 3 =27 and 9 x 1 = 9. This explains the first two terms of the complete ratio: 27:9:9:9:3:3:3:1. | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/06.E%3A_Genetic_Analysis_of_Multiple_Genes_(Exercises).txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
7.1 Compare recombination and crossover. How are these similar? How are they different?
7.2 Explain why it usually necessary to start with pure-breeding lines when measuring genetic linkage by the methods presented in this chapter.
7.3 If you knew that a locus that affected earlobe shape was tightly linked to a locus that affected susceptibility to cardiovascular disease human, under what circumstances would this information be clinically useful?
7.4 In a previous chapter, we said a 9:3:3:1 phenotypic ratio was expected among the progeny of a dihybrid cross, in absence of gene interaction.
a) What does this ratio assume about the linkage between the two loci in the dihybrid cross?
b) What ratio would be expected if the loci were completely linked? Be sure to consider every possible configuration of alleles in the dihybrids.
7.5 Given a dihybrid with the genotype CcEe:
a) If the alleles are in coupling (cis) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?
b) If the alleles are in repulsion (trans) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?
7.6 Imagine the white flowers are recessive to purple flowers, and yellow seeds are recessive to green seeds. If a green-seeded, purple-flowered dihybrid is testcrossed, and half of the progeny have yellow seeds, what can you conclude about linkage between these loci? What do you need to know about the parents of the dihybrid in this case?
7.7 In corn (i.e. maize, a diploid species), imagine that alleles for resistance to a particular pathogen are recessive and are linked to a locus that affects tassel length (short tassels are recessive to long tassels). Design a series of crosses to determine the map distance between these two loci. You can start with any genotypes you want, but be sure to specify the phenotypes of individuals at each stage of the process. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.
7.8 In a mutant screen in Drosophila, you identified a gene related to memory, as evidenced by the inability of recessive homozygotes to learn to associate a particular scent with the availability of food. Given another line of flies with an autosomal mutation that produces orange eyes, design a series of crosses to determine the map distance between these two loci. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.
7.9 Image that methionine heterotrophy, chlorosis (loss of chlorophyll), and absence of leaf hairs (trichomes) are each caused by recessive mutations at three different loci in Arabidopsis. Given a triple mutant, and assuming the loci are on the same chromosome, explain how you would determine the order of the loci relative to each other.
7.10 If the progeny of the cross aaBB x AAbb is testcrossed, and the following genotypes are observed among the progeny of the testcross, what is the frequency of recombination between these loci?
AaBb 135
Aabb 430
aaBb 390
aabb 120
7.11 Three loci are linked in the order B-C-A. If the A-B map distance is 1cM, and the B-C map distance is 0.6cM, given the lines AaBbCc and aabbcc, what will be the frequency of Aabb genotypes among their progeny if one of the parents of the dihybrid had the genotypes AABBCC?
7.12 Genes for body color (B black dominant to b yellow) and wing shape (C straight dominant to c curved) are located on the same chromosome in flies. If single mutants for each of these traits are crossed (i.e. a yellow fly crossed to a curved-wing fly), and their progeny is testcrossed, the following phenotypic ratios are observed among their progeny.
black, straight
yellow, curved
black, curved
yellow, straight
17
12
337
364
a) Calculate the map distance between B and C.
b) Why are the frequencies of the two smallest classes not exactly the same?
7.13 Given the map distance you calculated between B-C in question 12, if you crossed a double mutant (i.e. yellow body and curved wing) with a wild-type fly, and testcrossed the progeny, what phenotypes in what proportions would you expect to observe among the F2 generation?
7.14 In a three-point cross, individuals AAbbcc and aaBBCC are crossed, and their F1 progeny is testcrossed. Answer the following questions based on these F2 frequency data.
aaBbCc
480
AaBbcc
15
AaBbCc
10
aaBbcc
1
aabbCc
13
Aabbcc
472
AabbCc
1
aabbcc
8
a) Without calculating recombination frequencies, determine the relative order of these genes.
b) Calculate pair-wise recombination frequencies (without considering double cross overs) and produce a genetic map.
c) Recalculate recombination frequencies accounting for double recombinants.
7.15 Wild-type mice have brown fur and short tails. Loss of function of a particular gene produces white fur, while loss of function of another gene produces long tails, and loss of function at a third locus produces agitated behaviour. Each of these loss of function alleles is recessive. If a wild-type mouse is crossed with a triple mutant, and their F1 progeny is test-crossed, the following recombination frequencies are observed among their progeny. Produce a genetic map for these loci.
Chapter 7 - Answers
7.1
Crossovers are defined cytologically; they are observed directly under the microscope.
Recombination is defined genetically; it is calculated from observed phenotypic proportions.
Some crossovers lead to recombination, but not all crossovers result in recombination.
Some recombinations involve crossovers, but not all recombinations result from crossovers.
Crossovers happen between sister and non-sister chromatids. If the chromatids involved the crossover have identical alleles, there will not be any recombination.
Crossovers can also happen without causing recombination when there are two crossovers between the loci being scored for recombination.
Recombination can occur without crossover when loci are on different chromosomes.
7.2
The use of pure breeding lines allows the researcher to be sure that he/she is working with homozygous genotypes. If a parent is known to be homozygous, then all of its gametes will have the same genotype. This simplifies the definition of parental genotypes and therefore the calculation of recombination frequencies.
7.3
This would suggest that individuals with a particular earlobe phenotype may also carry one or more alleles that increased their risk of cardiovascular disease. These individuals could therefore be informed of their increased risk and have an opportunity to seek increased monitoring and reduce other risk factors.
7.4 a)
It assumes that the loci are completely unlinked.
b)
If the parental gametes were AB and ab, then the gametes produced by the dihybrids would also be AB and ab, and the offspring of a cross between the two dihybrids would all be genotype AABB:AaBb:aabb,in a 1:2:1 ratio.
If the parental gametes were Ab and aB, then the gametes produced by the dihybrids would also be Ab and aB, and the offspring of a cross between the two dihybrids would all be genotype AAbb:AaBb:aaBB, in a 1:2:1 ratio.
fur
tail
behaviour
white
short
normal
16
brown
short
agitated
0
brown
short
normal
955
white
short
agitated
36
white
long
normal
0
brown
long
agitated
14
brown
long
normal
46
white
long
agitated
933
7.5 a) Parental: CcEe and ccee; Recombinant: Ccee and ccEe.
b) Parental: Ccee and ccEe; Recombinant: CcEe and ccee.
7.6 Let WwYy be the genotype of a purple-flowered (W), green seeded (Y) dihybrid . Half of the progeny of the cross WwYy × wwyy will have yellow seeds whether the loci are linked or not. The proportion of the seeds that are also either white or purple flowered would help you to know about the linkage between the two loci only if the genotypes of the parents of the dihybrid were also known.
7.7
Let tt be the genotype of a short tassels, and rr is the genotype of pathogen resistant plants. We need to start with homozygous lines with contrasting combinations of alleles, for example:
P: RRtt (pathogen sensitive, short tassels) × rrTT (pathogen resistant, long tassels)
F1: RrTt (sensitive, long) × rrtt (resistant, short)
F2: parental Rrtt (sensitive, short) , rrTt (resistant, long)
Recombinant rrtt (resistant, short) , RrTt (sensitive, long)
7.8 Let mm be the genotype of a mutants that fail to learn, and ee is the genotype of orange eyes. We need to start with homozygous lines with contrasting combinations of alleles, for example (wt means wild-type):
P: MMEE (wt eyes, wt learning) × mmee (orange eyes, failure to learn)
F1: MmEe (wt eyes, wt learning) × mmee (orange eyes, failure to learn)
F2: parental MmEe (wt eyes, wt learning) , mmee (orange eyes, failure to learn)
Recombinant Mmee (wt eyes, failure to learn) , mmEe (orange eyes, wt learning)
7.9 Given a triple mutant aabbcc , cross this to a homozygote with contrasting genotypes, i.e. AABBCC, then testcross the trihybrid progeny, i.e.
P: AABBCC × aabbcc
F1: AaBbCc × aabbcc
Then, in the F2 progeny, find the two rarest phenotypic classes; these should have reciprocal genotypes, e.g. aaBbCc and AAbbcc. Find out which of the three possible orders of loci (i.e. A-B-C, B-A-C, or B-C-A) would, following a double crossover that flanked the middle marker, produce gametes that correspond to the two rarest phenotypic classes. For example, if the rarest phenotypic classes were produced by genotypes aaBbCc and AAbbcc, then the dihybrid’s contribution to these genotypes was aBC and Abc. Since the parental gametes were ABC and abc the only gene order that is consistent with aBC and Abc being produced by a double crossover flanking a middle marker is B-A-C (which is equivalent to C-A-B).
7.10 If the progeny of the cross aaBB x AAbb is testcrossed, and the following genotypes are observed among the progeny of the testcross, what is the frequency of recombination between these loci?
AaBb 135
Aabb 430
aaBb 390
aabb 120
(135 + 120)/(135+120+390+430)= 24%
7.11
Based on the information given, the recombinant genotypes with respect to these loci will be Aabb and aaBb. The frequency of recombination between A-B is 1cM=1%, based on the information given in the question, so each of the two recombinant genotypes should be present at a frequency of about 0.5%. Thus, the answer is 0.5%.
7.12
a) 4cM
b) Random sampling effects; the same reason that many human families do not have an equal number of boys and girls.
7.13
There would be approximately 2% of each of the recombinants: (yellow, straight) and (black, curved), and approximately 48% of each of the parentals: (yellow, curved) and (black, straight).
7.14
a) Without calculating recombination frequencies, determine the relative order of these genes.
A-C-B
b)
A-B 4.6%
A-C 2%
B-C 3%
B C A
|--------------|---------|
3cM 2cM
A-B
A-C
B-C
aBC
0
0
0
ABc
15
0
15
ABC
10
10
0
aBc
0
1
1
abC
13
0
13
Abc
0
0
0
AbC
0
1
1
abc
8
8
0
TOTAL
46
20
30
%
4.6
2
3
c) Recalculate recombination frequencies accounting for double recombinants
A-B
A-C
B-C
aBC
0
0
0
ABc
15
0
15
ABC
10
10
0
aBc
1 x 2
1
1
abC
13
0
13
Abc
0
0
0
AbC
1 x 2
1
1
abc
8
8
0
TOTAL
50
20
30
%
5
2
3
7.15
A is fur color locus
B is tail length locus
C is behaviour locus
fur (A)
tail (B)
behaviour (C)
AB
AC
BC
white
short
normal
16
aBC
R
R
P
brown
short
agitated
0
ABc
P
R
R
brown
short
normal
955
ABC
P
P
P
white
short
agitated
36
aBc
R
P
R
white
long
normal
0
abC
P
R
R
brown
long
agitated
14
Abc
R
R
P
brown
long
normal
46
AbC
R
P
R
white
long
agitated
933
abc
P
P
P
B C A
|--------------|---------|
4.1cM 1.5cM
Pairwise recombination frequencies are as follows (calculations are shown below):
A-B 5.6%
A-C 1.5%
B-C 4.1%
AB
AC
BC
16
16
0
0
0
0
0
0
0
36
0
36
0
0
0
14
14
0
46
0
46
0
0
0
112
30
82
5.6%
1.5%
4.1% | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/07.E%3A_Linkage_and_Mapping_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
7.1 Compare recombination and crossover. How are these similar? How are they different?
7.2 Explain why it usually necessary to start with pure-breeding lines when measuring genetic linkage by the methods presented in this chapter.
7.3 If you knew that a locus that affected earlobe shape was tightly linked to a locus that affected susceptibility to cardiovascular disease human, under what circumstances would this information be clinically useful?
7.4 In a previous chapter, we said a 9:3:3:1 phenotypic ratio was expected among the progeny of a dihybrid cross, in absence of gene interaction.
a) What does this ratio assume about the linkage between the two loci in the dihybrid cross?
b) What ratio would be expected if the loci were completely linked? Be sure to consider every possible configuration of alleles in the dihybrids.
7.5 Given a dihybrid with the genotype CcEe:
a) If the alleles are in coupling (cis) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?
b) If the alleles are in repulsion (trans) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?
7.6 Imagine the white flowers are recessive to purple flowers, and yellow seeds are recessive to green seeds. If a green-seeded, purple-flowered dihybrid is testcrossed, and half of the progeny have yellow seeds, what can you conclude about linkage between these loci? What do you need to know about the parents of the dihybrid in this case?
7.7 In corn (i.e. maize, a diploid species), imagine that alleles for resistance to a particular pathogen are recessive and are linked to a locus that affects tassel length (short tassels are recessive to long tassels). Design a series of crosses to determine the map distance between these two loci. You can start with any genotypes you want, but be sure to specify the phenotypes of individuals at each stage of the process. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.
7.8 In a mutant screen in Drosophila, you identified a gene related to memory, as evidenced by the inability of recessive homozygotes to learn to associate a particular scent with the availability of food. Given another line of flies with an autosomal mutation that produces orange eyes, design a series of crosses to determine the map distance between these two loci. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.
7.9 Image that methionine heterotrophy, chlorosis (loss of chlorophyll), and absence of leaf hairs (trichomes) are each caused by recessive mutations at three different loci in Arabidopsis. Given a triple mutant, and assuming the loci are on the same chromosome, explain how you would determine the order of the loci relative to each other.
7.10 If the progeny of the cross aaBB x AAbb is testcrossed, and the following genotypes are observed among the progeny of the testcross, what is the frequency of recombination between these loci?
AaBb 135
Aabb 430
aaBb 390
aabb 120
7.11 Three loci are linked in the order B-C-A. If the A-B map distance is 1cM, and the B-C map distance is 0.6cM, given the lines AaBbCc and aabbcc, what will be the frequency of Aabb genotypes among their progeny if one of the parents of the dihybrid had the genotypes AABBCC?
7.12 Genes for body color (B black dominant to b yellow) and wing shape (C straight dominant to c curved) are located on the same chromosome in flies. If single mutants for each of these traits are crossed (i.e. a yellow fly crossed to a curved-wing fly), and their progeny is testcrossed, the following phenotypic ratios are observed among their progeny.
black, straight
yellow, curved
black, curved
yellow, straight
17
12
337
364
a) Calculate the map distance between B and C.
b) Why are the frequencies of the two smallest classes not exactly the same?
7.13 Given the map distance you calculated between B-C in question 12, if you crossed a double mutant (i.e. yellow body and curved wing) with a wild-type fly, and testcrossed the progeny, what phenotypes in what proportions would you expect to observe among the F2 generation?
7.14 In a three-point cross, individuals AAbbcc and aaBBCC are crossed, and their F1 progeny is testcrossed. Answer the following questions based on these F2 frequency data.
aaBbCc
480
AaBbcc
15
AaBbCc
10
aaBbcc
1
aabbCc
13
Aabbcc
472
AabbCc
1
aabbcc
8
a) Without calculating recombination frequencies, determine the relative order of these genes.
b) Calculate pair-wise recombination frequencies (without considering double cross overs) and produce a genetic map.
c) Recalculate recombination frequencies accounting for double recombinants.
7.15 Wild-type mice have brown fur and short tails. Loss of function of a particular gene produces white fur, while loss of function of another gene produces long tails, and loss of function at a third locus produces agitated behaviour. Each of these loss of function alleles is recessive. If a wild-type mouse is crossed with a triple mutant, and their F1 progeny is test-crossed, the following recombination frequencies are observed among their progeny. Produce a genetic map for these loci.
Chapter 7 - Answers
7.1
Crossovers are defined cytologically; they are observed directly under the microscope.
Recombination is defined genetically; it is calculated from observed phenotypic proportions.
Some crossovers lead to recombination, but not all crossovers result in recombination.
Some recombinations involve crossovers, but not all recombinations result from crossovers.
Crossovers happen between sister and non-sister chromatids. If the chromatids involved the crossover have identical alleles, there will not be any recombination.
Crossovers can also happen without causing recombination when there are two crossovers between the loci being scored for recombination.
Recombination can occur without crossover when loci are on different chromosomes.
7.2
The use of pure breeding lines allows the researcher to be sure that he/she is working with homozygous genotypes. If a parent is known to be homozygous, then all of its gametes will have the same genotype. This simplifies the definition of parental genotypes and therefore the calculation of recombination frequencies.
7.3
This would suggest that individuals with a particular earlobe phenotype may also carry one or more alleles that increased their risk of cardiovascular disease. These individuals could therefore be informed of their increased risk and have an opportunity to seek increased monitoring and reduce other risk factors.
7.4 a)
It assumes that the loci are completely unlinked.
b)
If the parental gametes were AB and ab, then the gametes produced by the dihybrids would also be AB and ab, and the offspring of a cross between the two dihybrids would all be genotype AABB:AaBb:aabb,in a 1:2:1 ratio.
If the parental gametes were Ab and aB, then the gametes produced by the dihybrids would also be Ab and aB, and the offspring of a cross between the two dihybrids would all be genotype AAbb:AaBb:aaBB, in a 1:2:1 ratio.
fur
tail
behaviour
white
short
normal
16
brown
short
agitated
0
brown
short
normal
955
white
short
agitated
36
white
long
normal
0
brown
long
agitated
14
brown
long
normal
46
white
long
agitated
933
7.5 a) Parental: CcEe and ccee; Recombinant: Ccee and ccEe.
b) Parental: Ccee and ccEe; Recombinant: CcEe and ccee.
7.6 Let WwYy be the genotype of a purple-flowered (W), green seeded (Y) dihybrid . Half of the progeny of the cross WwYy × wwyy will have yellow seeds whether the loci are linked or not. The proportion of the seeds that are also either white or purple flowered would help you to know about the linkage between the two loci only if the genotypes of the parents of the dihybrid were also known.
7.7
Let tt be the genotype of a short tassels, and rr is the genotype of pathogen resistant plants. We need to start with homozygous lines with contrasting combinations of alleles, for example:
P: RRtt (pathogen sensitive, short tassels) × rrTT (pathogen resistant, long tassels)
F1: RrTt (sensitive, long) × rrtt (resistant, short)
F2: parental Rrtt (sensitive, short) , rrTt (resistant, long)
Recombinant rrtt (resistant, short) , RrTt (sensitive, long)
7.8 Let mm be the genotype of a mutants that fail to learn, and ee is the genotype of orange eyes. We need to start with homozygous lines with contrasting combinations of alleles, for example (wt means wild-type):
P: MMEE (wt eyes, wt learning) × mmee (orange eyes, failure to learn)
F1: MmEe (wt eyes, wt learning) × mmee (orange eyes, failure to learn)
F2: parental MmEe (wt eyes, wt learning) , mmee (orange eyes, failure to learn)
Recombinant Mmee (wt eyes, failure to learn) , mmEe (orange eyes, wt learning)
7.9 Given a triple mutant aabbcc , cross this to a homozygote with contrasting genotypes, i.e. AABBCC, then testcross the trihybrid progeny, i.e.
P: AABBCC × aabbcc
F1: AaBbCc × aabbcc
Then, in the F2 progeny, find the two rarest phenotypic classes; these should have reciprocal genotypes, e.g. aaBbCc and AAbbcc. Find out which of the three possible orders of loci (i.e. A-B-C, B-A-C, or B-C-A) would, following a double crossover that flanked the middle marker, produce gametes that correspond to the two rarest phenotypic classes. For example, if the rarest phenotypic classes were produced by genotypes aaBbCc and AAbbcc, then the dihybrid’s contribution to these genotypes was aBC and Abc. Since the parental gametes were ABC and abc the only gene order that is consistent with aBC and Abc being produced by a double crossover flanking a middle marker is B-A-C (which is equivalent to C-A-B).
7.10 If the progeny of the cross aaBB x AAbb is testcrossed, and the following genotypes are observed among the progeny of the testcross, what is the frequency of recombination between these loci?
AaBb 135
Aabb 430
aaBb 390
aabb 120
(135 + 120)/(135+120+390+430)= 24%
7.11
Based on the information given, the recombinant genotypes with respect to these loci will be Aabb and aaBb. The frequency of recombination between A-B is 1cM=1%, based on the information given in the question, so each of the two recombinant genotypes should be present at a frequency of about 0.5%. Thus, the answer is 0.5%.
7.12
a) 4cM
b) Random sampling effects; the same reason that many human families do not have an equal number of boys and girls.
7.13
There would be approximately 2% of each of the recombinants: (yellow, straight) and (black, curved), and approximately 48% of each of the parentals: (yellow, curved) and (black, straight).
7.14
a) Without calculating recombination frequencies, determine the relative order of these genes.
A-C-B
b)
A-B 4.6%
A-C 2%
B-C 3%
B C A
|--------------|---------|
3cM 2cM
A-B
A-C
B-C
aBC
0
0
0
ABc
15
0
15
ABC
10
10
0
aBc
0
1
1
abC
13
0
13
Abc
0
0
0
AbC
0
1
1
abc
8
8
0
TOTAL
46
20
30
%
4.6
2
3
c) Recalculate recombination frequencies accounting for double recombinants
A-B
A-C
B-C
aBC
0
0
0
ABc
15
0
15
ABC
10
10
0
aBc
1 x 2
1
1
abC
13
0
13
Abc
0
0
0
AbC
1 x 2
1
1
abc
8
8
0
TOTAL
50
20
30
%
5
2
3
7.15
A is fur color locus
B is tail length locus
C is behaviour locus
fur (A)
tail (B)
behaviour (C)
AB
AC
BC
white
short
normal
16
aBC
R
R
P
brown
short
agitated
0
ABc
P
R
R
brown
short
normal
955
ABC
P
P
P
white
short
agitated
36
aBc
R
P
R
white
long
normal
0
abC
P
R
R
brown
long
agitated
14
Abc
R
R
P
brown
long
normal
46
AbC
R
P
R
white
long
agitated
933
abc
P
P
P
B C A
|--------------|---------|
4.1cM 1.5cM
Pairwise recombination frequencies are as follows (calculations are shown below):
A-B 5.6%
A-C 1.5%
B-C 4.1%
AB
AC
BC
16
16
0
0
0
0
0
0
0
36
0
36
0
0
0
14
14
0
46
0
46
0
0
0
112
30
82
5.6%
1.5%
4.1% | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/07.E%3A_Linkage_and_Mapping_(Exercises).txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
8.1 What information, and what reagents would you need to use PCR to detect HIV in a blood sample?
8.2 A 6.0 kbp PCR fragment flanked by recognition sites for the HindIII restriction enzyme is cut with HindIII then ligated with a 3kb plasmid vector that has also been cut with HindIII. This recombinant plasmid is transformed into E. coli. From one colony a plasmid is prepared and digested with HindIII.
a) When the product of the HindIII digestion is analyzed by gel electrophoresis, what will be the size of the bands observed?
b) What bands would be observed if the recombinant plasmid was cut with EcoRI, which has only one site, directly in the middle of the PCR fragment?
c) What bands would be observed if the recombinant plasmid was cut with both EcoRI and HindIII at the same time?
8.3 If you started with 10 molecules of double stranded DNA template, what is the maximum number of molecules you would you have after 10 PCR cycles?
8.4 What is present in a PCR tube at the end of a successful amplification reaction? With this in mind, why do you usually only see a single, sharp band on a gel when it is analyzed by electrophoresis?
8.5 A coat protein from a particular virus can be used to immunize children against further infection. However, inoculation of children with proteins extracted from natural viruses sometimes causes fatal disease, due to contamination with live viruses. How could you use molecular biology to produce an optimal vaccine?
8.6 How would cloning be different if there were no selectable markers?
8.7 Research shows that a particular form of cancer is caused by a 200bp deletion in a particular human gene that is normally 2kb long. Only one mutant copy is needed to cause the disease.
a) Explain how you would use Southern blotting to diagnose the disease.
b) How would any of the blots appear if you hybridized and washed at very low temperature?
8.8 Refer to question 8.7.
a) Explain how you would detect the presence of the same deletion using PCR, rather than a Southern blot.
b) How would PCR products appear if you annealed at very low temperature?
8.9 You have a PCR fragment for a human olfactory receptor gene (perception of smells). You want to know what genes a dog might have that are related to this human gene. How can you use your PCR fragment and genomic DNA from a dog to find this out? Do you think dogs have more or less of these genes?
8.10 You add ligase to a reaction containing a sticky-ended plasmid and sticky-ended insert fragment, which both have compatible ends. Unbeknownst to you, someone in the lab left the stock of ligase enzyme out of the freezer overnight and it degraded (no longer works). Explain in detail what will happen in your ligation experiment in this situation should you try and transform with it. | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/08E%3A_Techniques_of_Molecular_Genetics_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
8.1 What information, and what reagents would you need to use PCR to detect HIV in a blood sample?
8.2 A 6.0 kbp PCR fragment flanked by recognition sites for the HindIII restriction enzyme is cut with HindIII then ligated with a 3kb plasmid vector that has also been cut with HindIII. This recombinant plasmid is transformed into E. coli. From one colony a plasmid is prepared and digested with HindIII.
a) When the product of the HindIII digestion is analyzed by gel electrophoresis, what will be the size of the bands observed?
b) What bands would be observed if the recombinant plasmid was cut with EcoRI, which has only one site, directly in the middle of the PCR fragment?
c) What bands would be observed if the recombinant plasmid was cut with both EcoRI and HindIII at the same time?
8.3 If you started with 10 molecules of double stranded DNA template, what is the maximum number of molecules you would you have after 10 PCR cycles?
8.4 What is present in a PCR tube at the end of a successful amplification reaction? With this in mind, why do you usually only see a single, sharp band on a gel when it is analyzed by electrophoresis?
8.5 A coat protein from a particular virus can be used to immunize children against further infection. However, inoculation of children with proteins extracted from natural viruses sometimes causes fatal disease, due to contamination with live viruses. How could you use molecular biology to produce an optimal vaccine?
8.6 How would cloning be different if there were no selectable markers?
8.7 Research shows that a particular form of cancer is caused by a 200bp deletion in a particular human gene that is normally 2kb long. Only one mutant copy is needed to cause the disease.
a) Explain how you would use Southern blotting to diagnose the disease.
b) How would any of the blots appear if you hybridized and washed at very low temperature?
8.8 Refer to question 8.7.
a) Explain how you would detect the presence of the same deletion using PCR, rather than a Southern blot.
b) How would PCR products appear if you annealed at very low temperature?
8.9 You have a PCR fragment for a human olfactory receptor gene (perception of smells). You want to know what genes a dog might have that are related to this human gene. How can you use your PCR fragment and genomic DNA from a dog to find this out? Do you think dogs have more or less of these genes?
8.10 You add ligase to a reaction containing a sticky-ended plasmid and sticky-ended insert fragment, which both have compatible ends. Unbeknownst to you, someone in the lab left the stock of ligase enzyme out of the freezer overnight and it degraded (no longer works). Explain in detail what will happen in your ligation experiment in this situation should you try and transform with it.
09.E: Changes in Chromosome Number and Structure (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
9.1
Make diagrams showing how an improper crossover event during meiosis can lead to: (a) an inversion or (b) a translocation.
9.2
Make a diagram showing how a nondisjunction event can lead to a child with a 47,XYY karyotype.
9.3
How many Barr bodies would you expect to see in cells from people who are: (a) 46, XY, (b) 46,XX, (c) 47, XYY, (d) 47,XXX, (e) 45,X, and (f) 47,XXY ?
9.4
Why can people survive with trisomy-21 (47,sex,+21) but not monosomy-21 (45,sex,-21)?
9.5
If Drosophila geneticists want to generate mutant strains with deletions they expose flies to gamma rays. What does this imply about gamma rays?
9.6
What would happen if there was a nondisjunction event involving chromosome 21 in a 46,XY zygote?
9.7
Design a FISH based experiment to find out if your lab partner is a 47,XXX female or a 47,XYY male.
9.8
What would Figure 9.18 look like if it also showed metaphase chromosomes from another cell?
09.E: Changes in Chromosome Number and Structure (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
9.1
Make diagrams showing how an improper crossover event during meiosis can lead to: (a) an inversion or (b) a translocation.
9.2
Make a diagram showing how a nondisjunction event can lead to a child with a 47,XYY karyotype.
9.3
How many Barr bodies would you expect to see in cells from people who are: (a) 46, XY, (b) 46,XX, (c) 47, XYY, (d) 47,XXX, (e) 45,X, and (f) 47,XXY ?
9.4
Why can people survive with trisomy-21 (47,sex,+21) but not monosomy-21 (45,sex,-21)?
9.5
If Drosophila geneticists want to generate mutant strains with deletions they expose flies to gamma rays. What does this imply about gamma rays?
9.6
What would happen if there was a nondisjunction event involving chromosome 21 in a 46,XY zygote?
9.7
Design a FISH based experiment to find out if your lab partner is a 47,XXX female or a 47,XYY male.
9.8
What would Figure 9.18 look like if it also showed metaphase chromosomes from another cell? | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/08E%3A_Techniques_of_Molecular_Genetics_(Exercises).txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
10.1 Three different polymorphisms have been identified at a particular molecular marker locus. A single pair of PCR primers will amplify either a 50bp fragment (B2), a 60bp fragment (B3), or a 100bp fragment (B4).
Draw the PCR bands that would be expected if these primers were used to amplify DNA from individuals with each of the following genotypes:
a) B2B2
b) B4B4
c) B2B3
d) B2B4
10.2 In addition to the primers used to genotype locus B (described above), a separate pair of primers can amplify another polymorphic SSR locus E, with either a 60bp product (E1) or a 90bp (E2) product. DNA was extracted from six individuals (#1- #6), and DNA from each individual was used as a template in separate PCR reactions with primers for either locus B or primers for locus E, and the PCR products were visualized on electrophoretic gels as shown below.
Based on the following PCR banding patterns, what is the full genotype of each of the six individuals?
10.3 Based on the genotypes you recorded in Question 10.2, can you determine which of the individuals could be a parent of individual #1?
10.4 Here is part of the DNA sequence of a chromosome:
TAAAGGAATCAATTACTTCTGTGTGTGTGTGTGTGTGTGTGTGTTCTTAGTTGTTTAAGTTTTAAGTTGTGA
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
ATTTCCTTAGTTAATGAAGACACACACACACACACACACACACAAGAATCAACAAATTCAAAATTCAACACT
Identify the following features on the sequence:
a) the region of the fragment that is most likely to be polymorphic
b) any simple sequence repeats
c) the best target sites for PCR primers that could be used to detect polymorphisms in the length of the simple sequence repeat region in different individuals
10.5 In a particular diploid plant, seed color is a polygenic trait. If true-breeding plants that produce red seeds are crossed with true breeding plants that produce white seeds, the F1 produces seeds that are intermediate in color (i.e. pink). When an F1 plant self-fertilizes, white seeds are observed in the next generation. How many genes are involved in seed color for each of the following frequencies of white seeds in the F2 generation?
a) 1/4 white seeds
b) 1/16 white seeds
c) 1/64 white seeds
d) 1/256 white seeds
10.6 If height in humans is a polygenic trait, explain why it occasionally happens that two tall parents have a child who grows up to be much shorter than either of them.
10.7 In quantitative trait (QTL) mapping, researchers cross two parents that differ in expression of some quantitative trait, then allow chromosomes from these parents to recombine randomly, and after several generations of inbreeding, produce a large number of offspring (“recombinant inbred lines”). Because the position of crossovers is random, each of the offspring contain a different combination of chromosomal regions from each of the two parents. The researchers then use molecular markers to determine which chromosomal regions have the greatest influence on the quantitative trait, e.g. in tall offspring, which chromosomal regions always come from the tall parent?
Imagine that two mice strains have been identified that differ in the time required to complete a maze, which may be an indication of intelligence. The time for maze completion is heritable and these parental strains “breed true” for the same completion time in each generation. Imagine also that their chromosomes are different colors and we can track the inheritance of chromosomal regions from each parent based on this color.
Based on the following diagrams of one chromosome from each individual in a pedigree, identify a chromosomal region that may contain a gene that affects time to complete a maze. The time for each individual is shown below each chromosome. Assume that all individuals are homozygous for all loci.
Parents:
Selected individuals from among F8 progeny of the above parents:
10.8 In a more realistic situation (as compared to question 7), where you could not distinguish the parental origin of different chromosomal regions just by appearance of chromosomes, explain how you could identify which parent was the source of a particular region of a chromosome in recombinant offspring. | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/10.E%3A_Molecular_Markers_and_Quantitative_Traits_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
10.1 Three different polymorphisms have been identified at a particular molecular marker locus. A single pair of PCR primers will amplify either a 50bp fragment (B2), a 60bp fragment (B3), or a 100bp fragment (B4).
Draw the PCR bands that would be expected if these primers were used to amplify DNA from individuals with each of the following genotypes:
a) B2B2
b) B4B4
c) B2B3
d) B2B4
10.2 In addition to the primers used to genotype locus B (described above), a separate pair of primers can amplify another polymorphic SSR locus E, with either a 60bp product (E1) or a 90bp (E2) product. DNA was extracted from six individuals (#1- #6), and DNA from each individual was used as a template in separate PCR reactions with primers for either locus B or primers for locus E, and the PCR products were visualized on electrophoretic gels as shown below.
Based on the following PCR banding patterns, what is the full genotype of each of the six individuals?
10.3 Based on the genotypes you recorded in Question 10.2, can you determine which of the individuals could be a parent of individual #1?
10.4 Here is part of the DNA sequence of a chromosome:
TAAAGGAATCAATTACTTCTGTGTGTGTGTGTGTGTGTGTGTGTTCTTAGTTGTTTAAGTTTTAAGTTGTGA
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
ATTTCCTTAGTTAATGAAGACACACACACACACACACACACACAAGAATCAACAAATTCAAAATTCAACACT
Identify the following features on the sequence:
a) the region of the fragment that is most likely to be polymorphic
b) any simple sequence repeats
c) the best target sites for PCR primers that could be used to detect polymorphisms in the length of the simple sequence repeat region in different individuals
10.5 In a particular diploid plant, seed color is a polygenic trait. If true-breeding plants that produce red seeds are crossed with true breeding plants that produce white seeds, the F1 produces seeds that are intermediate in color (i.e. pink). When an F1 plant self-fertilizes, white seeds are observed in the next generation. How many genes are involved in seed color for each of the following frequencies of white seeds in the F2 generation?
a) 1/4 white seeds
b) 1/16 white seeds
c) 1/64 white seeds
d) 1/256 white seeds
10.6 If height in humans is a polygenic trait, explain why it occasionally happens that two tall parents have a child who grows up to be much shorter than either of them.
10.7 In quantitative trait (QTL) mapping, researchers cross two parents that differ in expression of some quantitative trait, then allow chromosomes from these parents to recombine randomly, and after several generations of inbreeding, produce a large number of offspring (“recombinant inbred lines”). Because the position of crossovers is random, each of the offspring contain a different combination of chromosomal regions from each of the two parents. The researchers then use molecular markers to determine which chromosomal regions have the greatest influence on the quantitative trait, e.g. in tall offspring, which chromosomal regions always come from the tall parent?
Imagine that two mice strains have been identified that differ in the time required to complete a maze, which may be an indication of intelligence. The time for maze completion is heritable and these parental strains “breed true” for the same completion time in each generation. Imagine also that their chromosomes are different colors and we can track the inheritance of chromosomal regions from each parent based on this color.
Based on the following diagrams of one chromosome from each individual in a pedigree, identify a chromosomal region that may contain a gene that affects time to complete a maze. The time for each individual is shown below each chromosome. Assume that all individuals are homozygous for all loci.
Parents:
Selected individuals from among F8 progeny of the above parents:
10.8 In a more realistic situation (as compared to question 7), where you could not distinguish the parental origin of different chromosomal regions just by appearance of chromosomes, explain how you could identify which parent was the source of a particular region of a chromosome in recombinant offspring. | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/10.E%3A_Molecular_Markers_and_Quantitative_Traits_(Exercises).txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
11.1 What are the advantages of high-throughput –omics techniques compared to studying a single gene or protein at a time? What are the disadvantages
11.2 What would the chromatogram from a capillary sequencer look like if you accidentally added only template, primers, polymerase, and fluorescent terminators to the sequencing reaction?
11.3 What are the advantages and disadvantages of clone-by-clone vs. whole genome shotgun sequencing?
11.4 How could you use DNA sequencing to identify new species of marine microorganisms?
11.5 Explain how you could use a microarray to identify wheat genes that have altered expression during drought?
11.6 A microarray identified 100 genes whose transcripts are abundant in tumors, but absent in normal tissues. Do any or all of these transcripts cause cancer? Explain your answer.
11.7 How could you ensure that each spot printed on a microarray contains DNA for only one gene?
11.8 What would the spots look like on a microarray after hybridization, if each spot contained a random mixture of genes?
11.9 What would the spots look like if the hybridization of green and red labeled DNA was done at low stringency?
11.E: Genomics and Systems Biology (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions
11.1 What are the advantages of high-throughput –omics techniques compared to studying a single gene or protein at a time? What are the disadvantages
11.2 What would the chromatogram from a capillary sequencer look like if you accidentally added only template, primers, polymerase, and fluorescent terminators to the sequencing reaction?
11.3 What are the advantages and disadvantages of clone-by-clone vs. whole genome shotgun sequencing?
11.4 How could you use DNA sequencing to identify new species of marine microorganisms?
11.5 Explain how you could use a microarray to identify wheat genes that have altered expression during drought?
11.6 A microarray identified 100 genes whose transcripts are abundant in tumors, but absent in normal tissues. Do any or all of these transcripts cause cancer? Explain your answer.
11.7 How could you ensure that each spot printed on a microarray contains DNA for only one gene?
11.8 What would the spots look like on a microarray after hybridization, if each spot contained a random mixture of genes?
11.9 What would the spots look like if the hybridization of green and red labeled DNA was done at low stringency? | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/11.E%3A_Genomics_and_Systems_Biology_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
12.1 List all the mechanisms that can be used to regulate gene expression in eukaryotes.
12.2 With respect to the expression of β-galactosidase, what would be the phenotype of each of the following strains of E. coli?
a) I+, O+, Z+, Y+ (no glucose, no lactose)
b) I+, O+, Z+, Y+ (no glucose, high lactose)
c) I+, O+, Z+, Y+ (high glucose, no lactose)
d) I+, O+, Z+, Y+ (high glucose, high lactose)
e) I+, O+, Z-, Y+ (no glucose, no lactose)
f) I+, O+, Z-, Y+ (high glucose, high lactose)
g) I+, O+, Z+, Y- (high glucose, high lactose)
h) I+, Oc, Z+, Y+ (no glucose, no lactose)
i) I+, Oc,Z+, Y+ (no glucose, high lactose)
j) I+, Oc, Z+, Y+ (high glucose, no lactose)
k) I+, Oc, Z+, Y+ (high glucose, high lactose)
l) I-, O+, Z+, Y+ (no glucose, no lactose)
m) I-, O+, Z+, Y+ (no glucose, high lactose)
n) I-, O+, Z+, Y+ (high glucose, no lactose)
o) I-, O+, Z+, Y+ (high glucose, high lactose)
p) Is, O+, Z+, Y+ (no glucose, no lactose)
q) Is, O+, Z+, Y+ (no glucose, high lactose)
r) Is, O+, Z+, Y+ (high glucose, no lactose)
s) Is, O+, Z+, Y+ (high glucose, high lactose)
12.3 In the E. coli strains listed below, some genes are present on both the chromosome, and the extrachromosomal F-factor episome. The genotypes of the chromosome and episome are separated by a slash. What will be the β-galactosidase phenotype of these strains? All of the strains are grown in media that lacks glucose.
a) I+, O+, Z+, Y+ / O-, Z-, Y- (high lactose)
b) I+, O+, Z+, Y+ / O-, Z-, Y- (no lactose)
c) I+, O+, Z-, Y+ / O-, Z+, Y+ (high lactose)
d) I+, O+, Z-, Y+ / O-, Z+, Y+ (no lactose)
e) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (high lactose)
f) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (no lactose)
g) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
h) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
i) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
j) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
k) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (high lactose)
l) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (no lactose)
m) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (high lactose)
n) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (no lactose)
o) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
p) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
12.1 Transcriptional: initiation, processing & splicing, degradation
Translational: initiation, processing, degradation
Post-translational: modifications (e.g. phosphorylation), localization
Others: histone modification, other chromatin remodeling, DNA methylation
12.2 Legend:
+++ Lots of β-galactosidase activity
+ Moderate β-galactosidase activity
-- No β-galactosidase activity
-- a) I+, O+, Z+, Y+ (no glucose, no lactose)
+++ b) I+, O+, Z+, Y+ (no glucose, high lactose)
-- c) I+, O+, Z+, Y+ (high glucose, no lactose)
+ d) I+, O+, Z+, Y+ (high glucose, high lactose)
-- e) I+, O+, Z-, Y+ (no glucose, no lactose)
-- f) I+, O+, Z-, Y+ (high glucose, high lactose)
+ g) I+, O+, Z+, Y- (high glucose, high lactose)
+++ h) I+, Oc, Z+, Y+ (no glucose, no lactose)
+++ i) I+, Oc,Z+, Y+ (no glucose, high lactose)
+ j) I+, Oc, Z+, Y+ (high glucose, no lactose)
+ k) I+, Oc, Z+, Y+ (high glucose, high lactose)
+++ l) I-, O+, Z+, Y+ (no glucose, no lactose)
+++ m) I-, O+, Z+, Y+ (no glucose, high lactose)
+ n) I-, O+, Z+, Y+ (high glucose, no lactose)
+ o) I-, O+, Z+, Y+ (high glucose, high lactose)
-- p) Is, O+, Z+, Y+ (no glucose, no lactose)
-- q) Is, O+, Z+, Y+ (no glucose, high lactose)
-- r) Is, O+, Z+, Y+ (high glucose, no lactose)
-- s) Is, O+, Z+, Y+ (high glucose, high lactose)
12.3 Legend:
+++ Lots of β-galactosidase activity
+ Moderate β-galactosidase activity
-- No β-galactosidase activity
+++ a) I+, O+, Z+, Y+ / O-, Z-, Y- (high lactose)
-- b) I+, O+, Z+, Y+ / O-, Z-, Y- (no lactose)
+++ c) I+, O+, Z-, Y+ / O-, Z+, Y+ (high lactose)
+ d) I+, O+, Z-, Y+ / O-, Z+, Y+ (no lactose)
+++ e) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (high lactose)
-- f) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (no lactose)
+++ g) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
-- h) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
+++ i) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
+++ j) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
+++ k) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (high lactose)
+++ l) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (no lactose)
-- m) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (high lactose)
-- n) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (no lactose)
-- o) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
-- p) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
12.4 You could demonstrate this with just I+OcZ-/I+O+Z+. The fact that this does not have constitutive lactose expression shows that the operator only acts on the same piece of DNA on which it is located. There are also other possible answers.
12.5 You could also demonstrate this with just I+O+Z-/I-O+Z+. The fact that this has the same lactose-inducible phenotype as wild-type hows that a functional lacI gene can act on operators on both the same piece of DNA from which it is transcribed, or on a different piece of DNA. There are also other possible answers.
12.6 For all of these, the answer is the same: The lacoperon would be inducible by lactose, but only moderate expression of the lac operon would be possible, even in the absence of glucose
a) loss-of-function of adenylate cyclase
b) loss of DNA binding ability of CAP
c) loss of cAMP binding ability of CAP
d) mutation of CAP binding site (CBS) cis-element so that CAP could not bind
12.7 Both involve trans-factors binding to corresponding cis-elements to regulate the initiation of transcription by recruiting or stabilizing the binding of RNApol and related transcriptional proteins at the promoter. In prokaryotes, genes may be regulated as a single operon. In eukaryotes, enhancers may be located much further from the promoter than in prokaryotes.
12.8 These fish would all have spiny tales like the deep-water population.
12.9 These could have arisen from loss-of-function mutation in FLC, or in the cis-element to which FLC normally binds.
12.10 If there was no deacetylation of FLC by HDAC, transcription of FLC might continue constantly, leading to constant suppression of flowering, even after winter. | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/12E%3A_Regulation_of_Gene_Expression_%28Exercises%29.txt |
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
12.1 List all the mechanisms that can be used to regulate gene expression in eukaryotes.
12.2 With respect to the expression of β-galactosidase, what would be the phenotype of each of the following strains of E. coli?
a) I+, O+, Z+, Y+ (no glucose, no lactose)
b) I+, O+, Z+, Y+ (no glucose, high lactose)
c) I+, O+, Z+, Y+ (high glucose, no lactose)
d) I+, O+, Z+, Y+ (high glucose, high lactose)
e) I+, O+, Z-, Y+ (no glucose, no lactose)
f) I+, O+, Z-, Y+ (high glucose, high lactose)
g) I+, O+, Z+, Y- (high glucose, high lactose)
h) I+, Oc, Z+, Y+ (no glucose, no lactose)
i) I+, Oc,Z+, Y+ (no glucose, high lactose)
j) I+, Oc, Z+, Y+ (high glucose, no lactose)
k) I+, Oc, Z+, Y+ (high glucose, high lactose)
l) I-, O+, Z+, Y+ (no glucose, no lactose)
m) I-, O+, Z+, Y+ (no glucose, high lactose)
n) I-, O+, Z+, Y+ (high glucose, no lactose)
o) I-, O+, Z+, Y+ (high glucose, high lactose)
p) Is, O+, Z+, Y+ (no glucose, no lactose)
q) Is, O+, Z+, Y+ (no glucose, high lactose)
r) Is, O+, Z+, Y+ (high glucose, no lactose)
s) Is, O+, Z+, Y+ (high glucose, high lactose)
12.3 In the E. coli strains listed below, some genes are present on both the chromosome, and the extrachromosomal F-factor episome. The genotypes of the chromosome and episome are separated by a slash. What will be the β-galactosidase phenotype of these strains? All of the strains are grown in media that lacks glucose.
a) I+, O+, Z+, Y+ / O-, Z-, Y- (high lactose)
b) I+, O+, Z+, Y+ / O-, Z-, Y- (no lactose)
c) I+, O+, Z-, Y+ / O-, Z+, Y+ (high lactose)
d) I+, O+, Z-, Y+ / O-, Z+, Y+ (no lactose)
e) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (high lactose)
f) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (no lactose)
g) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
h) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
i) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
j) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
k) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (high lactose)
l) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (no lactose)
m) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (high lactose)
n) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (no lactose)
o) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
p) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
12.1 Transcriptional: initiation, processing & splicing, degradation
Translational: initiation, processing, degradation
Post-translational: modifications (e.g. phosphorylation), localization
Others: histone modification, other chromatin remodeling, DNA methylation
12.2 Legend:
+++ Lots of β-galactosidase activity
+ Moderate β-galactosidase activity
-- No β-galactosidase activity
-- a) I+, O+, Z+, Y+ (no glucose, no lactose)
+++ b) I+, O+, Z+, Y+ (no glucose, high lactose)
-- c) I+, O+, Z+, Y+ (high glucose, no lactose)
+ d) I+, O+, Z+, Y+ (high glucose, high lactose)
-- e) I+, O+, Z-, Y+ (no glucose, no lactose)
-- f) I+, O+, Z-, Y+ (high glucose, high lactose)
+ g) I+, O+, Z+, Y- (high glucose, high lactose)
+++ h) I+, Oc, Z+, Y+ (no glucose, no lactose)
+++ i) I+, Oc,Z+, Y+ (no glucose, high lactose)
+ j) I+, Oc, Z+, Y+ (high glucose, no lactose)
+ k) I+, Oc, Z+, Y+ (high glucose, high lactose)
+++ l) I-, O+, Z+, Y+ (no glucose, no lactose)
+++ m) I-, O+, Z+, Y+ (no glucose, high lactose)
+ n) I-, O+, Z+, Y+ (high glucose, no lactose)
+ o) I-, O+, Z+, Y+ (high glucose, high lactose)
-- p) Is, O+, Z+, Y+ (no glucose, no lactose)
-- q) Is, O+, Z+, Y+ (no glucose, high lactose)
-- r) Is, O+, Z+, Y+ (high glucose, no lactose)
-- s) Is, O+, Z+, Y+ (high glucose, high lactose)
12.3 Legend:
+++ Lots of β-galactosidase activity
+ Moderate β-galactosidase activity
-- No β-galactosidase activity
+++ a) I+, O+, Z+, Y+ / O-, Z-, Y- (high lactose)
-- b) I+, O+, Z+, Y+ / O-, Z-, Y- (no lactose)
+++ c) I+, O+, Z-, Y+ / O-, Z+, Y+ (high lactose)
+ d) I+, O+, Z-, Y+ / O-, Z+, Y+ (no lactose)
+++ e) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (high lactose)
-- f) I+, O+, Z-, Y+ / I-, O+, Z+, Y+ (no lactose)
+++ g) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
-- h) I-, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
+++ i) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
+++ j) I+, Oc, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
+++ k) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (high lactose)
+++ l) I+, O+, Z-, Y+ / I+, Oc, Z+, Y+ (no lactose)
-- m) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (high lactose)
-- n) I+, O+, Z-, Y+ / Is, O+, Z+, Y+ (no lactose)
-- o) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (high lactose)
-- p) Is, O+, Z+, Y+ / I+, O+, Z-, Y+ (no lactose)
12.4 You could demonstrate this with just I+OcZ-/I+O+Z+. The fact that this does not have constitutive lactose expression shows that the operator only acts on the same piece of DNA on which it is located. There are also other possible answers.
12.5 You could also demonstrate this with just I+O+Z-/I-O+Z+. The fact that this has the same lactose-inducible phenotype as wild-type hows that a functional lacI gene can act on operators on both the same piece of DNA from which it is transcribed, or on a different piece of DNA. There are also other possible answers.
12.6 For all of these, the answer is the same: The lacoperon would be inducible by lactose, but only moderate expression of the lac operon would be possible, even in the absence of glucose
a) loss-of-function of adenylate cyclase
b) loss of DNA binding ability of CAP
c) loss of cAMP binding ability of CAP
d) mutation of CAP binding site (CBS) cis-element so that CAP could not bind
12.7 Both involve trans-factors binding to corresponding cis-elements to regulate the initiation of transcription by recruiting or stabilizing the binding of RNApol and related transcriptional proteins at the promoter. In prokaryotes, genes may be regulated as a single operon. In eukaryotes, enhancers may be located much further from the promoter than in prokaryotes.
12.8 These fish would all have spiny tales like the deep-water population.
12.9 These could have arisen from loss-of-function mutation in FLC, or in the cis-element to which FLC normally binds.
12.10 If there was no deacetylation of FLC by HDAC, transcription of FLC might continue constantly, leading to constant suppression of flowering, even after winter.
13.E: Cancer Genetics (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
13.1 Why do oncogenes tend to be dominant, but mutations in tumor suppressors tend to be recessive?
13.2 What tumor suppressing functions are controlled by p53? How can a single gene affect so many different biological pathways?
13.3 Are all carcinogens mutagens? Are all mutagens carcinogens? Explain why or why not.
13.4 Imagine that a laboratory reports that feeding a chocolate to laboratory rats increases the incidence of cancer. What other details would you want to know before you stopped eating chocolate?
13.5 Do all women with HPV get cancer? Why or why not? Do all women with mutations in BRCA1 get cancer? Why or why not?
13.E: Cancer Genetics (Exercises)
These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.
Study Questions:
13.1 Why do oncogenes tend to be dominant, but mutations in tumor suppressors tend to be recessive?
13.2 What tumor suppressing functions are controlled by p53? How can a single gene affect so many different biological pathways?
13.3 Are all carcinogens mutagens? Are all mutagens carcinogens? Explain why or why not.
13.4 Imagine that a laboratory reports that feeding a chocolate to laboratory rats increases the incidence of cancer. What other details would you want to know before you stopped eating chocolate?
13.5 Do all women with HPV get cancer? Why or why not? Do all women with mutations in BRCA1 get cancer? Why or why not? | textbooks/bio/Genetics/Exercise_-_Genetics/Exercises%3A__Online_Open_Genetics_(Nickle_and_Barrette-Ng)/12E%3A_Regulation_of_Gene_Expression_(Exercises).txt |
• 1.1: Mitosis and Meiosis
• 1.2: DNA- The Genetic Material
• 1.3: DNA Mutations
• 1.4: PCR and Gel Electrophoresis
The polymerase chain reaction laboratory technique is used in a variety of applications to make copies of a specific DNA sequence. This lesson describes how a PCR reaction works, what it accomplishes, and its basic requirements for success. Examples of interpreting results are given. PCR’s strengths, weaknesses, and applications to plant biotechnology are explained.
• 1.5: Gene Expression- Transcription
Organisms such as plants and animals have tens of thousands of genes. The impact that a single gene’s information can have on an organism, however, is tremendous. Furthermore, organisms have all their genes in each of their cells, but they only need to use the information from a subset of these genes, depending on the type of cell and the cell’s stage of development. Therefore, the key to gene function is controlling its expression.
• 1.6: Gene Expression- Translation
• 1.7: Gene Expression- Applied Example (Part 1)
• 1.8: Gene Expression- Applied Example (Part 2)
This lesson describes how changes in the DNA sequence of a gene can alter the synthesis of a protein and thus influence traits such as herbicide resistance. In this lesson we will describe how changes in the gene can alter the gene expression process and influence traits in an organism. The specific example of ALS-inhibitor herbicide resistance is used to demonstrate the impact of genetic change on trait expression in a plant.
• 1.9: Regulation of Gene Expression
Every cell in a plant contains the same genetic information, the same set of genes. Yet Therefore different sets of genes are required for the various functions of different cells or tissues, as well as for plant responses to environmental stimuli or stresses. This is achieved by regulating the activity of genes according to the physiological demands of a particular cell type, developmental stage, or environmental condition. This regulation of activity is known as gene expression.
• 1.10: Genetic Pathways
• 1.11: Recombinant DNA Technology
Recombinant DNA (rDNA) technology has resulted in breakthroughs in crop and animal biotechnology. The power of rDNA technology comes from our ability to study and modify gene function by manipulating genes and transform them into cells of plant and animals. To arrive at this several tools of molecular biology are used including, DNA isolation and analysis, molecular cloning, quantification of gene expression, and many others.
• 1.12: Genetic Engineering
• 1.13: Introduction to Mendelian Genetics
In plant and animal genetics research, the decisions a scientist will make are based on a high level of confidence in the predictable inheritance of the genes that control the trait being studied. This confidence comes from a past discovery by a biologist named Gregor Mendel, who explained the inheritance of trait variation using the idea of monogenic traits.
• 1.14: Deviations from Mendelian Genetics- Linkage (Part 1)
• 1.15: Deviations from Mendelian Genetics- Linkage (Part 2)
In this lesson you will learn to make predictions about inheritance using map unit distances and genetic markers, assemble maps from multiple-point linkage data, define the relationship between linkage maps, linkage groups and genome maps, and describe how DNA or molecular markers are observed and used in gene mapping.
01: Chapters
Learning Objectives
• Describe the cell cycle.
• Explain the events of all stages of mitosis.
• Track chromosome and chromatid number through all stages of mitosis.
• Explain the events of all stages of meiosis.
• Track chromosome and chromatid number through all stages of meiosis.
• Describe the role of meiosis in gamete formation and how it relates to inheritance.
Introduction
Multicellular organisms such as plants and animals are composed of millions to trillions (1,000,000,000) of cells that work together. The cells that make up different tissues have different shapes and perform different functions for the plant or animal. Even though they have diverse functions, each somatic cell in the organism normally has the same chromosomes and therefore the same genetic makeup. Furthermore, the millions of cells that makeup a mature organism originated from a single cell formed when the male and female gametes from the parents of the organism fused. This single cell established the life of the organism. Understanding multicellular organisms requires an understanding of the lifecycle of the cells that make up the organism.
The Cell Cycle
Let us think about the cell cycle from a personal point of view. Your age plus about nine months ago you were a zygote, a single cell formed when the sperm and egg from your biological parents fused in a fallopian tube (or possibly in a test tube if in vitro fertilization factored into your birth). You have come a long way since then, progressing one cell cycle at a time. The cell cycle is the life cycle of a single cell. We depict the cell cycle in a circular diagram although the cells in your body do not actually go around in circles. The main idea is that when new cells are made from existing cells, the new cells start their lifecycle, and the old cells end theirs.
The first part of the cell cycle is the G1 phase. Cells can go through growth and development in this phase. Some cells differentiate into specialized cells and then never leave this phase. However, a zygote does not grow in size, but instead continues the cell cycle so it can quickly give rise to more cells.
The second part of the cell cycle is the S phase (synthesis phase). Here, the cell replicates its chromosomes so that it has a copy of each chromosome to pass on to daughter cells. The third part of the cell cycle is the G2 phase where the cell prepares for division. The G1, S and G2 phases together are called interphase. The M phase completes the cell cycle. ’M’ could be mitosis or meiosis depending on the type of cell. For the zygote, the goal is to make more somatic cells. Therefore, it goes through mitosis and gives rise to two daughter cells. This completes the life cycle of the zygote and starts the lifecycle of the new cells. Rounds of the cell cycle continued over and over to form the body you have today. As long as you live, some of your cells must be able to complete the cell cycle.
From a cytogenetics point of view, you have two types of cells in your body. You have cells with 46 chromosomes (somatic or body cells) and cells with 23 chromosomes (gamete or sex cells). Since you started as a single cell with 46 chromosomes, there must be two types of cell division taking place in your body to accommodate both somatic and gamete cells. Mitosis and meiosis are the two types of cell division.
Mitosis: Somatic cell division
The objective of mitosis is to make two genetically identical cells from a single cell. In the cells of our body, we start with 46 chromosomes in a single cell and end up with 46 chromosomes in two cells. Obviously, replicating the chromosomes is a prerequisite to mitosis. Remember, replication takes place during interphase when the chromosomes are dispersed structures in the nucleus. Mitosis is an organized procession of activity in the cell that allows the replicated chromosomes to be properly divided into two identical cells. Chromosomes are important because they contain genes. Therefore, we will include genes on our chromosome diagrams and slide show. These pictures depict the four stages of mitosis. We will describe the events at each stage that are important in understanding the distribution of genes during cell division.
Mitosis: Prophase
Prophase is the beginning of mitosis (Figure 3).
During interphase, the chromosomes look like a plate of spaghetti in the nucleus. It is difficult to pick out an individual chromosome because they are each so spread out. The chromosomes in the nucleus change from being loosely dispersed to becoming more condensed. This change in chromosome structure makes them easier to move around the cell, an important issue for what is about to happen. As the chromosomes condense, they get shorter and thicker and can be seen through the microscope as individual structures (Figure 4). The chromosomes at prophase will consist of two identical parts called sister chromatids that stay connected at the centromere. It is now clear that the chromosomes have been replicated. Chromosome replication occurs during the S-phase of interphase (Figure 1). Next, the nucleus is dissolved. At the end of prophase the replicated chromosomes are moved by the spindle apparatus to the center of the cell.
Mitosis: Metaphase
When the chromosomes are maneuvered to the center of the cell, metaphase begins (Figure 5). The spindle fiber network connects the centromere of the replicated chromosome to the outer part of each cell. The chromosomes now resemble the line of scrimmage of a football team, poised and waiting for some cellular signal to begin the next phase.
Mitosis: Anaphase
At anaphase (Figure 6) the chromatids making up each chromosome are pulled apart and begin to move away from each other under the control of the spindle fibers. Once they are pulled apart, the chromatids are now considered individual chromosomes. As anaphase progresses, the chromosomes on each end of the cell are pulled into a bundle. When they stop moving, telophase begins.
Mitosis: Telophase
The final stage of mitosis is telophase (Figure 7). A nuclear envelope will form around each bundle of chromosomes. The cell will undergo cytokinesis and the cytoplasm is split between the two identical daughter cells. Cell membranes (and cell walls in plants) form between the two cells. The chromosomes begin to decondense. They loosen up and spread around the nucleus because they no longer will need to move around. The new cells have now begun their lifecycle.
Meiosis: Gamete formation
The objective of meiosis is to make four cells from a single somatic cell. The four cells each have half the chromosome number found in the somatic cell. In our human bodies, the four gametes will each have 23 chromosomes which means the 46 chromosomes in the somatic cell must replicate during interphase prior to meiosis just as they would before mitosis. Meiosis occurs in specialized cells of the body called germline cells.
To appreciate meiosis and gamete formation it is important to first understand two ideas, chromosome sets and homologous chromosomes.
Chromosome sets: The 46 chromosomes you have consist of two sets. You are a diploid organism (‘di’ means two and ‘ploid’ means sets). One set of chromosomes came from each parent when their gametes fused. Therefore, human gametes are haploid (one set).
Homologous chromosomes: The 46 chromosomes in a somatic cell can be arranged into 23 homologous or similar pairs. One chromosome from each pair came from the male parent, the other from the female. Homologous chromosomes have the same genes although in heterozygous people the genes would be different alleles (A,a). The exception to this would be the sex (X and Y) chromosomes. Passing on a complete set of human genes requires one chromosome from each pair to end up in each gamete.
There are several key differences between meiosis and mitosis that are summarized in the following table:
Table 1. The key events that happen in each of the stages of meiosis are summarized.
Mitosis
Meiosis
Chromosome number stays the same
Chromosome number is halved
One division occurs to make two cells. Four stages of this division.
Two divisions occur to make four cells. Eight stages in these divisions.
Similar or homologous chromosomes do not pair.
Homologous chromosomes pair during prophase l. Pairing is called synapsis.
Crossover exchanges between homologous chromosomes is rare.
Synapsis allows crossing over between homologous chromosomes.
Two cells made are genetically identical.
Four cells made are genetically different.
Meiosis: Prophase I
This stage starts meiosis and is the same as prophase of mitosis with one important change. As the chromosomes condense, they form groups of four chromatids called tetrads or bivalents. Close inspection reveals that each chromosome is replicated and consists of two sister chromatids. The two chromosomes in each cell that are homologous and have the same genes (but perhaps different alleles if the organism is heterozygous) will pair closely. This close association, or synapsis, allows the homologous chromosomes to crossover and exchange identical parts. The impact of crossing over is that genes that are on the same chromosome (Figure 8) can be recombined so that they are not always inherited together. The tetrad or bivalent formed during synapsis remains assembled as prophase progresses. The tetrad therefore moves as a unit to the center of the cell.
Meiosis: Metaphase I
Metaphase I starts when the tetrads are at the center of the cell (Figure 9). The tetrads have stayed together which insures that during the first division, each cell will get one chromosome from each homologous pair. The chromosomes remain at the center of the cell until the homologous pairs are ready to move away from each other.
Meiosis: Anaphase I
The chromosomes that make up each tetrad separate during anaphase I (Figure 10). However, the sister chromatids will stay connected at the centromere. Anaphase I proceeds until the chromosomes are pulled into a bundle at opposite ends of the cell.
Meiosis: Telophase l
The cell divides into two cells during telophase I (Figure 11). The bundle of chromosomes may have a nuclear envelope develop around them. The germline cells in some organisms such as human females, go through the first four stages of meiosis prior to birth. The germline cells remain at telophase I for some time. The second round of division occurs when the gamete is needed for reproduction. In other situations, telophase I is an abbreviated stage, and the second round of division proceeds without delay.
Meiosis: Prophase II
If the chromosomes became dispersed in telophase I, they will condense again at prophase II. The spindle apparatus moves the chromosomes to the middle of the cell. Look! The centromeres are still holding the sister chromatids together (Figure 12).
Meiosis: Metaphase II
In metaphase II the chromosomes are aligned at the center of the cell (Figure 13). This time there are not homologous chromosomes to be paired with. This metaphase looks similar to metaphase of mitosis but there is a key difference. What is the difference? (Compare Figures 5 and 13).
Meiosis: Anaphase II
During anaphase II, the chromatids are pulled apart by the spindle fibers. Now they are classified as chromosomes, not chromatids. The chromosomes move apart to opposite ends of the cell (Figure 14).
Meiosis: Telophase II
In the final stage of meiosis, telophase II, the nucleus forms around the bundle of chromosomes (Figure 15). The cell divides. Now four cells exist that originated from one germline cell. Each cell is a gamete with half the number of chromosomes and genes as a somatic cell.
Parallel Behavior and the Chromosome Theory
The successful completion of mitosis or meiosis requires the cell to move large objects with precision and control many detailed events. The process surpasses any engineering accomplishments of NASA. The importance of mitosis and meiosis to an organism is obvious when we consider that genes are a part of the chromosome and the genes must be copied and distributed properly to produce viable daughter cells. The mechanisms of these events are far from being completely understood. From our current understanding, we can appreciate how the principles of segregation and independent assortment are controlled by the mechanics of meiosis.
When cell division was first observed and described by cytogeneticists, biologists were just beginning to accept the idea that genes were tiny objects that controlled traits and existed in the cells of living things. Two biologists, a German named Boveri and an American graduate student named Sutton, recognized that chromosome behavior during meiosis matched Mendel’s principles of gene behavior. Both scientists proposed the idea that while genes had not yet been directly observed, they must be a part of the chromosome. Sutton and Boveri are both given credit for proposing this chromosome theory; genes are a part of chromosomes.
Segregation predicts gene behavior that matches the chromosome behavior observed by cytogeneticists. Genes are in pairs because chromosomes are in pairs. The gene pairs associate during gamete formation when the homologous chromosomes pair in prophase I. The gene pairs separate when the homologous pairs separate in anaphase I followed by chromatid separation in anaphase II. Thus, chromosome behavior dictates the gene’s segregation behavior.
Independent assortment of gene pairs also correlates with chromosome behavior. Let us consider a dihybrid individual with the genotype BbEe (Figure 16). How many kinds of gametes can it make? The four possible combinations (BE, bE, Be and be) are made at equal frequencies. We can understand why this occurs if we think about what happens to chromosomes at metaphase I of meiosis. The chromosome pair that carries the ‘E’ and ‘e’ genes will be moved independently from the chromosome pair with the ‘B’ and ‘b’ chromosomes. When the chromosomes arrive at the center of the cell at metaphase I, the two tetrads may be aligned so that the ‘E’ and ‘B’ genes move to one cell and the ‘e’ and ‘b’ genes move to the other during the first division. In the other half of the cells that go through meiosis, the tetrads will line up so that the ‘E’ genes will be passed on with the ‘b’ genes and the ‘e’ genes will go with the ‘B’ genes. Keep in mind that organisms that make gametes make thousands of them. The chance alignment of the tetrads at metaphase I therefore dictates the overall frequency of gametes with different combinations of genes when the genes are on separate chromosomes.
When more gene pairs are considered, the same scenarios described above will be true as long as the genes are on separate chromosomes. When genes are on the same chromosome, the role of crossing over on gene inheritance needs to be considered in more detail. We will cover the inheritance of genes on the same chromosome in a later lesson.
Learning Activities
1. Watch this video about Mitosis from Daily Med Ed
2. Watch this video about Meiosis from Daily Med Ed | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.01%3A_Mitosis_and_Meiosis.txt |
Learning Objectives
• Detail the molecular nature of a gene.
• Recognize when and where replication of the DNA that is packaged in an organism’s chromosome is happening for growth, maintenance, and reproduction.
• Describe how chromosome replication speed can be increased with multiple origins of replication and bidirectional replication.
• Explain the role of each of the primary DNA replicating enzymes in conducting replication of a chromosome. Include the role of these enzyme in ensuring the accuracy of the replication process.
• Predict how the semiconservative replication of a double stranded DNA will happen given the structure of the molecule and the specific function of the DNA replicating enzymes.
Introduction
The genetic (hereditary) material for all living things is composed of DNA (deoxyribonucleic acid). The structure of DNA must enable this substance to store coded information that control the biological function of cells. The genetic material transmits this hereditary information in a stable form for the cell and organism through accurate replication of DNA. Although DNA is capable of change (we will discuss how this happens when we talk about DNA mutations), the replication process ensures high accuracy in copying the genetic information so that all progeny cells receive the same information. The DNA from all chromosomes in a human cell would be more than 6.5 feet long! Therefore, to fit inside the nucleus of a cell, DNA is packaged into chromosomes.
Read this article about Watson and Crick for more insight on the experimental facts discovered by chemists and biologist which contributed to the determination of the structure of DNA.
Chemical Structure of DNA Subunits
DNA is a polymer made of nucleotide subunits. A nucleotide consists of 3 chemical groups; a sugar, a phosphate and a nitrogenous base (Figure 1). In the case of DNA, the sugar is deoxyribose.
There are 4 different nucleotides in DNA that contain 4 different bases (Figure 2). These bases are named adenosine (A), cytosine (C), guanine (G) and thymine (T).
Complementary, antiparallel DNA strands
DNA codes information by ordering the sequence of the A,T,G and C nucleotides in a long polymer. The nucleotides are connected by phosphate bonds to form a strand. Two strands form a double stranded molecule (Figure 3). Note that in Figure 1, the carbon atoms on the deoxyribose sugar are numbered. The phosphate bond connects carbon #3 (the 3 prime or 3′ carbon) of one sugar to the 5′ carbon of the next. This gives a DNA strand directionality because the 5′ carbon faces one way and the 3′ carbon faces the other.
The bases on one strand form hydrogen bonds with the bases on the other. This is called “base pairing” and it is very specific, A only pairs with T and C only pairs with G. Thus, the sequences on the two strands of DNA are “complementary”; whenever there is an “A” on one strand, there is a “T” in the corresponding position of the complementary strand.
In addition to containing complementary nucleotide sequences, the strands are in opposite orientations. The 5′ end of one strand is oriented toward the 3′ end of the other. For this reason, the strands are referred to as “antiparallel”.
Double helix
The final feature of the molecular structure is that DNA assumes a helical conformation (Figure 4). This is the most stable configuration that accommodates all the molecular structures and chemical bonds that make up DNA.
Activity
Given the following sequence of a DNA strand, predict the sequence and orientation of the complementary strand.
5′-G-A-C-C-G-T-A-A-T-C-G-C-3′
Show Answer
Answer: 3′-C-T-G-G-C-A-T-T-A-G-C-G-5′
Packaging of DNA into chromatin and chromosomes
Chromosomes are long double stranded DNA molecules. Over 150,000,000 nucleotide pairs make up the human X chromosome. The complete replication and orderly transfer of something this large requires the chromosome to be packaged for stability and organization which creates genetic stability in the cells of a species. Every cell in a diploid organism contains two copies (2n) of every chromosome present in that organism (Figure 5). For example, humans have 46 chromosomes in their body, 23 were inherited from the father and 23 from the mother. Gametes, the reproductive cells of an organism, (egg or sperm), have only one set (1n) of chromosomes. When the two gametes unite, they form a living embryo with two sets of genetic information. Therefore, we actually have two copies of the genetic information for each trait. Sometimes one copy controls trait expression, and other times both copies influence a trait. As a result, the offspring will have characteristics of both the mother and the father.
Inside cell nuclei, DNA is assembled, or packaged, into material called chromatin, which is composed of both DNA and proteins (Figure 6). Chromatin functions to protect and regulate DNA, as well as to efficiently store the very long DNA molecules that make up chromosomes within the limited space of the nucleus.
DNA from higher eukaryotic organisms is about 109 to 1010 bp. Human and maize chromosomal DNA is about 109 bp, which is equivalent to 1.8 m if all chromosomal DNA were stretched end to end in a linear manner. The diameter of the nucleus is only about 4-6 μm making it a challenge to fit chromosomal DNA inside the cell nucleus. This is accomplished by the ability of DNA to assume a very condensed structure to fit inside the nucleus. Therefore, the organization of eukaryotic DNA into chromatin is an important aspect of DNA packaging.
Nucleosomes and histones
The ordered coiling of chromosomal DNA around a histone protein core forms the chromatin. Chromatin is made up of nucleosomes (Figure 6) which represent the association of chromosomal DNA with histone proteins. A nucleosome is made up of about 145-147 base pairs of DNA coiled around each histone octamer, for about two complete turns. A histone octamer consists of two copies of each core histone. The total mass of the histones in the nucleus approaches that of the DNA – 2 molecules of each core histone to approximately 200 bp of DNA.
Higher order structures
If the higher order chromatin structure is disrupted, electron microscopy reveals the appearance of “beads on a string” with a diameter of about 10 nm. The “beads” represent the DNA wrapped around histones. Nucleosome formation results in a DNA fiber that is about 10 nm and a packaging ratio of about 7. The higher order chromatin structure results when the 10 nm fiber is coiled into a solenoid. The result of nucleosome coiling is a chromatin fiber of 30 nm that is observed by electron microscopy.
Centromeres
The centromere is a chromosomal region that controls chromosome segregation at mitosis and meiosis (Figure 7). Centromeres connect to microtubules of the spindle apparatus, which directs their movement to opposite poles (daughter cell nuclei) during cell division. Scientists were successful in isolating these centromeric sequences from yeast and engineered brewer’s yeast (Saccharomyces cerevisiae) plasmids that were able to replicate like the chromosomes. Through these efforts scientists were able to point centromeric function to a DNA stretch of about 120 bp that was resistant to DNAse and bound to a single microtubule.
Telomeres
The telomere lies at the end of the chromosome and confers stability by “sealing” the end of a chromosome (Figure 7). Telomeres consist of long series of short repeated DNA sequence that occurs in tandem arrays and are added to the end of the chromosome during DNA replication by an enzyme called telomerase. In most plant species the sequence TTTAGGG constitutes a conserved telomere motif.
The knowledge about the structure and function of centromeres and telomeres has had a profound impact in molecular genetics. For example, yeast artificial chromosomes (YACs) and bacterial artificial chromosomes (BACs) have proven useful in the physical mapping of plant genomes requiring the cloning and multiplying large (>100 kb) DNA fragments in yeast or bacterial cells.
DNA Replication
In 1953, J.D. Watson and F.H.C. Crick published a note in NATURE; one of the world’s most widely read research journals. The note cited just six references.
Watson and Crick’s writing had limited chemistry details given the title of their note, “A Structure for Deoxyribonucleic Acid”. A later article would share more of the chemistry behind their proposed double helix structure for Deoxyribonucleic Acid (DNA). This note in Nature was written for a broader audience, including biologists who recognized that the structure of DNA was a prerequisite for understanding the function of the genetic material. They included a diagram to allow readers to visualize this unique double helix structure. To emphasize the structure determines function importance, Watson and Crick included this statement in their note…
“It has not escaped our notice that the specific pairing we have postulate immediately suggests a possible copying mechanism for the genetic material.”
Accurate replication prior to cell division is one of the three key functions of the genetic material. While later experiments would reveal the details of how cells replicate DNA, Watson and Crick were obviously impressed that the proposed double helix structure would fit with the concept that living cells are needed to produce more living cells and the transmission of a complete set of biological instructions. The genetic material must have a structure that lends itself to being copied. Using old to help build new leads to a proposed semiconservative model for replicating the double stranded DNA molecule (Figure 8).
Prior to both mitosis and meiosis cell division in multicellular organisms, and cell division in prokaryotes, the DNA inside the cell must be replicated. (see lesson Mitosis and Meiosis). This replication process generates the genetic information needed for either two cells, genetically identical with the original cell or in sexually reproducing organisms, four gamete cells with half of the original cell’s genetic information.
When and Where:
DNA replication will happen before cells divide. In hours old embryos that are developing into multicellular organisms, all the cells are replicating their DNA and dividing into new cells. Hours or days later, multicellular organisms have developed specific meristem regions where new cell production occurs. A visual example of meristem cells where DNA replication is needed is the tip of a plant root (Figure 9). New cells made at the root tip allow the root to grow. The root must also replace the cells that are damaged as the growing root moves through soil. In Figure 8, the meristem region (1) contains meristem cells that are replicating their chromosomes, then dividing into two identical cells by mitosis. Cells on the tip part of the meristem specialize in root cap cells (2,3) with the job of protecting the meristem. The fate of these cells is to die (4), and they need to be replaced by new cells made in the meristem. Cells made on the other half of the meristem (5) can elongate and eventually specialize. These root cells will then function for the plant, all season for annual plants, years for perennial plants. Replication of DNA will only happen in the meristem cells.
Speed of chromosome replication
Fifty-two years after Watson and Crick’s Nature note, a group of about 50 molecular geneticists published the complete DNA sequence of the human X chromosome (Figure 5). This discovery revealed that X chromosome is one double-stranded DNA molecule that is about 155 million nucleotide pairs long. Every time a human cell divides, one or two of these 155,000,000 nucleotides must be replicated. This must be done with speed and accuracy.
The main DNA replication enzyme (DNA polII, see below) works fast. Estimates of its replicating speed are around 750 nucleotides per second.
Activity
Let us do the math.
155,000,000 nucleotides / chromosome X 1 second / 750 nucleotides
= 206,666 seconds per X chromosome
Let us convert these seconds to hours.
1 hour / 360 seconds X 206,666 seconds/chromosome = 547 hours per chromosome
That means one replication enzyme would take 547 hours or about 24 days to replicate an X chromosome. What are the biological ramifications of this 24 days? If you cut yourself, it would take 24 days to replicate the chromosomes before the skin cells could divide. The formation of new cells needed for healing cuts in our skin would keep us in a wounded state.
To speed up the chromosome replication process, two tactics are used by living cells. Both tactics are shown in Figure 10. First, replication will not start on just one end of the chromosome. Instead, there are hundreds of origins of replication (ori) along the chromosome. In addition, the replication process moves bidirectionally from the ori. This bidirectional replication creates forks of replication that move toward each other as the chromosome is replicated.
As a result of these two tactics, a large chromosome can be replicated in hours rather than weeks and cell division can happen quickly when it is needed for growth or cell replacement.
The Chromosome Replicating Team
Like all processes in living cells, a collection of specific proteins works together to perform functions which control and complete chromosome replication. All the proteins involved in DNA replication make and break bonds, so they are enzymes. The function of the main DNA replicating enzymes is described below.
Five of the main DNA replicating enzymes
• Helicase: This is one of the enzymes that unwinds the double stranded DNA molecule. The unwound single strands no longer hydrogen bond to their complementary strand but the sugar-phosphate bonds remain intact.
• Topoisomerase: The unwinding work of Helicase creates tension in the double stranded DNA ahead of the replication fork. Topoisomerase relieves this tension by catalyzing a series of sugar phosphate bond breaking and making steps. Without this tension relief, the Double stranded DNA could break and the intact chromosome cannot finish replicating.
• DNA Polymerase III (DNA pol III): This is the main DNA synthesizing enzyme. The enzyme reads the single strand as a template and places in the complementary deoxyribonucleotide. DNA pol III reads the template in the 3′ to 5′ direction and builds the new strand in the 5′ to 3′ direction, adding the next nucleotide to the 3′ end by catalyzing sugar-phosphate bonding. DNA pol III illustrates the specificity of enzymes several ways; it reads and proofreads the placement of new nucleotides to insure accurate replication. It can only read and build in one direction. DNA pol III can only add nucleotides to a free 3′ end. This last specification means that DNA pol III cannot start the replication process on the single stranded template. Another enzyme needs to be part of the in vivo replication team to prime the work of DNA pol III.
• Primase: Biochemists named this enzyme to describe its role in starting or priming the replication process. DNA (or RNA) primase is a special RNA polymerase. The enzyme reads the single stranded DNA template 3′ to 5′ and adds ribonucleic acid (RNA) nucleotides in the 5′ to 3′ direction. Once a few hundred RNA nucleotides are added, primase falls off the template strand and leaves the 3′ end that DNA Pol III needs to continue the process.
• DNA Polymerase I (DNA pol I): The DNA polymerase specializes in the removal of the RNA primers and replacing them with DNA nucleotides. The enzyme works with the same 3′ to 5′ reading and 5′ to 3′ building.
• DNA Ligase: After the five enzymes described above have completed their work, some sugar phosphate bonds need to be made to complete the double stranded molecule. DNA ligase has this backbone bond sealing assignment.
Now that we have introduced the replication enzymes, we can describe the step-by-step action of this team. We will focus on the action that happens at one origin of replication.
Step 1: Initiating replication at the Ori
Helicase enzymes bind to the ori sequences and start to unwind the double stranded DNA. This establishes two replication forks (one shown in Figure 10a) and a helicase enzyme will be working to unwind the DNA at each fork as the forks move away from each other. The unwinding exposes single strands of DNA that are the template for building a new strand. Topoisomerase can be seen working to relieve tension ahead of the replication fork in Figure 10a.
Step 2: Priming DNA with RNA
The process of priming with primase and then replicating with DNA pol III happens at both replication forks (Figure 10a and b). DNA pol III enzymes can replicate new strands as fast as the DNA is unwound at each replication fork. The unwinding creates two template strands at each fork so primase enzyme must work to prime both old strands.
Step 3: Synthesis of leading and lagging strands
The double strands of DNA are anti-parallel and ‘run’ in opposite directions and DNA pol III can only work in one direction (Figure 10c). Thus, there is continuous and discontinuous replication happening at each replication fork. The strand that can be replicated as helicase unwinds is the leading strand. The DNA pol III must replicate the other strand in the opposite direction so this strand will lag behind. (Figure 10c).
There is more priming needed to replicate the lagging strand and the priming leaves RNA-DNA hybrid stretches in the new double stranded molecule. This means there is work for the enzyme DNA pol I. This enzyme will be part of the removal of the RNA nucleotide primers, and replacement with DNA nucleotides (Figure 10c and d).
Both replication forks have a leading and a lagging strand of new DNA. The DNA nucleotide stretches between the red RNA primers would be the Okazaki fragments.
One of the first scientists to provide evidence to support this model of DNA replication was the team of Reiji and Tsuneko Okazaki in the 1960s. They used radioactive nucleotides to track the newly replicating fragments and found that many of the fragments were short. This fits with the discontinuous model.
As the four enzymes just described attend to their specific roles in replication, two double stranded DNA molecules with an old strand and a new strand are built. If the DNA pol III makes a mistake and adds the wrong, non-complementing nucleotide, the enzyme can proof read its work and replace these replication mistakes. When replication is perfect, the two double stranded DNA molecules will have identical sequences.
Step 4: Ligation of Okazaki fragments
The fifth enzyme that has a final role in completing replication is DNA ligase (not shown in Figure 10d and e). This enzyme will seal the sugar phosphate bond between the last replacement nucleotide added by DNA pol I and the first nucleotide that had been added after priming by DNA pol III. With this bond sealing work, the replication process is complete.
The two double stranded DNA molecules that get made will have identical sequences unless rare mistakes happen as the enzymes conduct their work.
Lesson Summary:
DNA is composed of nucleotides. Nucleotides are connected together by phosphate bonds to form a strand. The bases on one strand form hydrogen bonds with the bases on the other to form a double stranded molecule. The final feature of the molecular structure is that DNA assumes a helical conformation. To fit inside the nucleus, DNA assumes a very condensed structure. Therefore, chromosomal DNA is coiled around a histone protein core to form chromatin. The tight packaging of DNA in chromatin must be modified to allow DNA replication and transcription. In the process of replication, the two DNA strands separate and act as templates for the synthesis of complementary daughter strands.
Learning Activities
Place the single strands, replication enzymes, and 3′ / 5′ labels in the appropriate places (#1-#7) | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.02%3A_DNA-_The_Genetic_Material.txt |
Learning Objectives
• List the types of mutation.
• Predict the effects of specific mutations in gene coding regions on the protein encoded by the gene.
• Describe ways in which mutations arise, naturally and experimentally.
Introduction
Crop improvement relies on genetic variation, which is synonymous with DNA variation. Mutations are the source of genetic variation that is the driver for trait improvement. In general, mutations are rare events, and most are deleterious and get selected against during the course of evolution. However, occasionally mutations create genetic variation that is beneficial. As such, it is often of interest for plant improvement or for genetic studies to induce genetic mutations by exposing seed or plant tissue to certain agents, called mutagens.
By definition, a mutation is a heritable change in DNA sequence. This can happen in several ways: substitution of a DNA base, insertion or deletion of one or more DNA bases, or by large-scale chromosomal rearrangements, the latter of which will not be considered here. It has also become clear in recent years that chromatin carries additional information to the DNA sequence in the form of base modifications such as cytosine methylation. This is known as epigenetic modification and can impact gene function and plant traits.
A. DNA base substitutions
The simplest type of mutation is a substitution of one base for another in the DNA sequence. Substitutions most often arise as errors during DNA replication or repair. The most common type is the transition, where one pyrimidine may be substituted by the other, or one purine by the other. The less common type is the transversion, in which a purine may be substituted by a pyrimidine or vice versa. As described in the next section, these can have various effects on gene function depending on where they occur in a gene.
B. Insertions & deletions
Insertions or deletions of DNA are other types of mutations that occur quite frequently. They can vary in size from one to thousands or more nucleotide bases. Their effects on gene function depend on the size of the insertion or deletion, and the location relative to the gene. As described below, insertions and deletions that are in multiples of 3 are referred to as in-frame insertions and deletions and result in the addition or deletion of amino acids from the protein sequence. When not in multiples of three, insertions and deletions lead to frameshift mutations, altering the amino acid sequence of the protein encoded by the mutated gene.
A common source of inserted DNA is from transposons, sequences of DNA that are able to move from one genomic site and insert into another. Transposons were discovered in maize by Barbara McClintock. For this discovery, she was awarded the Nobel Prize in Physiology and Medicine in 1983.
Some transposons copy themselves and multiply in number within a genome. Transposons can cause wide scale chromosome rearrangements and gene mutations. A feature of mutations caused by many plant transposons is that they are unstable. Excision of a transposon from the mutated gene can often restore it to wild type or create a new stable mutant allele. Transposons are found in every living organism studied and are a major driving force in genome evolution.
Effects of mutations on gene function
A. Coding region mutations
Silent mutation
A silent mutation is a mutation that results in the change of a codon without a change in the amino acid represented by the codon. Because of the redundancy of the genetic code, DNA base substitutions may not lead to the incorporation of an incorrect amino acid in the protein.
Missense mutation (amino acid substitution)
A missense mutation is a single nucleotide base substitution that alters a codon such that it codes for a different amino acid that is incorporated into the encoded protein. The amino acid substitution may affect the function of the protein if it occurs in a critical portion of the protein.
Frameshift mutation
Due to the triplet nature of the genetic code, an insertion or deletion can change the reading frame for the entire subsequent sequence. For example, if a particular sequence is read sequentially (Figure 4).
Because the new sequence of codons is different from the original, the entire amino acid sequence is changed from the point of insertion. A similar effect is seen for deletions. This effect is seen whenever the number of nucleotide bases inserted or deleted is not a multiple of 3 and it usually will result in the loss of the function of the protein.
Amino acid additions or deletions
On the other hand, if insertions or deletions are in multiples of 3, amino acid additions or deletions can arise. Consider the following 3-base insertion (XXX):
UUU CCC XXX AAA GGG (codons)
aa1 aa2 aa5 aa3 aa4 (amino acids)
If such insertions or deletions occur in important regions of a protein, they might impair function.
Nonsense mutation
A nonsense mutation arises when a functional codon is changed to a stop codon. Nonsense mutations may cause premature termination of translation.
B. Regulatory region mutations
Promoters and enhancers control temporal, spatial and quantitative aspects of gene expression. As such, mutations in these regulatory regions of a gene can alter the regulation of the gene’s expression. Genes can become expressed at inappropriate times, places or levels, which can affect a plant’s phenotype.
Teosinte, the progenitor of modern maize, produces multiple branches (Figure 6A) due to reduced apical dominance. In teosinte and modern maize, apical dominance is under the control of a gene called teosinte branched1, tb1 (Doebley et al. 1997). An insertion of a transposon in the regulatory region of the tb1 gene in modern maize causes the gene to be strongly expressed (Doebley et al. 1997; Studer et al. 2011), resulting in enhanced apical dominance, and suppression of excessive branching (Figure 6B).
C. Recessive vs. dominant mutations
Recessive mutations
Recessive mutations most commonly result from a loss of function. A loss of function mutation causes a decrease in the function of a gene. A complete loss of function is known as a null mutation while partial loss of function mutants are often referred to as hypomorphs or “leaky”. Loss of function mutations can affect gene expression or the function of the resultant protein. Most loss of function mutations are recessive because in a heterozygote, the normal allele can provide the necessary function.
Dominant mutations
Dominant mutations can affect genes in several different ways. One general class is gain of function mutations. These encompass several different types.
1. Overexpression mutants. As the name implies, the affected gene is expressed at levels that are inappropriately high, causing too much of a gene product to be produced.
2. Neomorphic mutations arise when a mutation causes a qualitatively new effect not seen with normal alleles. This can occur when a mutation affects the regulation of a gene such that it becomes expressed in a new time or place in the plant. It can also happen when a mutation alters the properties of the encoded protein. For example, a transcription factor could be mutated to recognize a different promoter sequence, or an enzyme could recognize a new substrate
3. Dominant negative mutations alter a gene product such that the mutant gene product interferes with the function of the normal one. This can occur at the protein level. For example, a protein might function as a multimeric complex and the presence of one mutant subunit could compromise the function of the entire complex. Dominant negatives can also occur at the RNA level for example if an insertion, deletion or rearrangement causes the wrong strand to be transcribed, creating an antisense or RNAi suppression effect.
Effects of mutations on traits
Example of a loss of function mutation – gene function is required for a trait
Sweetness is an important quality trait in maize (corn). One of the loss of function mutants used to improve sweetness in corn is sugary1 (su1). The Su1 gene is expressed in the endosperm (Lertrat and Pulam, 2007), and it encodes and enzyme that is required for normal starch biosynthesis (Rahman et al. 1998). In the su1 mutant, with defective starch biosynthesis, the concentration of sugar is 3 times higher than wild type making this mutant valuable for sweet corn (Lertrat and Pulam, 2007). In addition, su1 kernels accumulate a highly branched polysaccharide called phytoglycogen, which gives the kernels a creamy texture. New commercial sweet corn hybrids are developed using a combination of starch biosynthesis mutants (Nelson and Pan, 1995; Lertrat and Pulam, 2007).
Example of a gain of function mutation – gene function is sufficient for a trait
The plant growth hormone, gibberellin (GA) regulates plant height, thus, a plant that fails to produce GA, or does not respond to GA will have a dwarfed phenotype (Figure 7). At the molecular level, the cell response to GA involves several processes, including the destruction of repressor proteins, referred to as DELLAs.
The “green revolution gene” Reduced height-1 (Rht-1) from wheat encodes a DELLA repressor of GA signaling (Peng et al. 1999). A stretch of five amino acids referred to as the DELLA domain is required for the destruction of the repressor in response to GA. Mutations in the DELLA domain of Rht-1 genes result in a protein product that is resistant to degradation, leading to failure to signal a GA response. The failure to signal the presence of GA in the rht-1 mutants is responsible for their short stature. When the mutant rht-1 gene is transformed into rice, the result is also short plants, which is proof that the new function of the DELLA protein is sufficient to induce dwarfism (Peng et al. 1999).
Generation of mutations
Because mutations are the ultimate source of genetic variation for breeding and genetic studies, it is sometimes of interest to generate new mutations. New DNA mutations (Figure 8) can arise spontaneously or they can be induced by experimental methods.
Spontaneous mutations arise “naturally”
That is, no particular effort was made on the part of the scientist to increase the mutation rate. Spontaneous mutations generally occur at very low rates, on the order of 1 per gene per million gametes.
Mistakes in DNA replication
Very rarely, incorrect bases are incorporated or bases omitted from a DNA strand during synthesis. Such mistakes can lead to spontaneous substitutions, insertions or deletions. For example, strand slippage due to the formation of a loop.
Environmental mutagens
There are several chemical agents present in the environment with potential to damage DNA and create mutations. These include naturally occurring compounds as well as man-made pollutants. Some modify DNA nucleotide bases through deamination (e.g., nitrous acid), while others promote oxidative reactions that may damage DNA (e.g., ozone).
Radiation is another type of environmental mutagen. Ionizing radiation (e.g., X-rays) can shatter DNA sequences and promote chromosome rearrangements. The less powerful ultraviolet rays can penetrate the cell and promote the formation of pyrimidine dimers which may inhibit replication and transcription.
Experimental mutagenesis involves applying a mutagenic agent to a plant or plant part to generate a mutation. For the mutation to be useful, it must be heritable, which imposes limitations on what tissues can be treated. Following the mutagenic treatment, a subsequent breeding scheme must be implemented to generate pure breeding lines of appropriate genotypes to display mutant phenotypes. These will depend on the species and the design of the mutagenesis.
Chemical mutagenesis
A commonly used chemical mutagen for experimental plant biology and mutation breeding is ethyl methanesulfonate (EMS). The compound is known as an alkylating agent and induces a CG to TA transition.
Seed Mutagenesis
This is the most common method of chemical mutagenesis in most species. Seeds are soaked in EMS solution, planted and allowed to flower and set seed. During EMS mutagenesis, every cell in the embryo will be independently mutagenized and the resultant seed and subsequent plant will be chimeric. These seeds can be sown and plants screened for traits of interest. For self-pollinating species, the required breeding is very simple, whereas for dioecious species it would be much more elaborate and labor-intensive.
Pollen Mutagenesis
For a few species such as maize, pollen mutagenesis is the method of choice. Pollen are treated with EMS and applied to silks of an acceptor plant. The benefit is that the resultant seed is not chimeric but rather all the cells of the embryo carry the same mutagenized paternal genome inherited from the pollen grain.
Insertional mutagenesis
The idea behind insertional mutagenesis is that a mobile or introduced piece of DNA can sometimes insert into a gene, thereby inactivating or modifying the function of the gene. The approach is useful in gene cloning because the inserted DNA serves as a “tag” that can be used to locate the gene of interest, and subsequently isolate that gene.
Another strategy used by researchers is mutagenesis using T-DNA from the Agrobacterium tumefaciens Ti plasmid. A larger number of plants (up to 10,000 for Arabidopsis) are transformed with T-DNA to generate mutants at random.
Similar to the example of transposon mutagenesis above, a gene can be “tagged” by a T-DNA and the tag used to isolate the gene. However, a T-DNA insertions are stable and thus disrupt the gene permanently.
In recent years, insertional mutagens have been engineered in many innovative ways. For example, strong promoters have been engineered into transposons such that if the transposon inserts near a gene, it can cause strong transcriptional expression of that gene. Such “activation tagging” approaches have been successful in identifying many genes that were not amenable to other mutational approaches.
Summary
Mutations alter A-T and G-C base pairs in DNA. A mutation in a coding sequence may alter the sequence and function of the protein product. A frameshift mutation changes the reading frame through insertions or deletions to produce an entirely novel product. A point mutation on the other hand alters only the amino acid represented by the codon in which the mutation exists. Mutations in regulatory regions might alter the expression of a gene. Mutations have been powerful tools for genetic studies as well as to study biological functions, for example, growth and development. The natural occurrence of mutations can be enhanced experimentally by applying agents referred to as mutagens. Researchers have used certain mutagens to induce mutations to study gene function and apply new traits for crop improvement.
Activity 1
For this activity, it may be helpful to refer to the genetic code chart.
For the following DNA sequence:
5′ ATG TTG GAG AAG GTT GAA ACT TTC 3′
Write the coding strand
3′ TAC AAG CTC TTC CAA CTT TGA AAG 5′
Write the mRNA sequence from the given DNA sequence
AUG UUC GAG AAG GUU GAA ACU UUG
Write the resulting peptide sequence
Met-Phe-Glu-Lys-Val-Glu-Thr-Leu
If the third triplet of the original DNA sequence is mutated to GAA to result in the following sequence
5′ ATG TTG GAA AAG GTT GAA ACT TTC 3′
What will be the mRNA sequence from the mutated DNA sequence?
AUG UUC GAA AAG GUU GAA ACU UUG
What will be the peptide sequence from the mutated DNA sequence?
Met-Phe-Glu-Lys-Val-Glu-Thr-Leu
What type of coding region mutation is this? Give a reason for your answer.
1. Silent
2. Amino acid substitution
3. Frameshift
4. Nonsense
If the sixth triplet of the original DNA sequence is mutated to TAA to result in the following sequence
5′ ATG TTG GAG AAG GTT TAA ACT TTC 3′
What will be the mRNA sequence from the mutated DNA sequence?
AUG UUC GAG AAG GUU UAA ACU UUG
What will be the peptide sequence from the mutated DNA sequence?
Met-Phe-Glu-Lys-Val
What type of coding region mutation is this? Give reason for your answer.
1. Silent
2. Amino acid substitution
3. Frameshift
4. Nonsense
The original DNA sequence is mutated to result in the following sequence
5′ ATG TTG GAG AAG GTT CGA AAC TTT C 3′
What will be the mRNA sequence from the mutated DNA sequence?
AUG UUC GAG AAG GUU CGA AAC UUU C
What will be the peptide sequence from the mutated DNA sequence?
Met-Phe-Glu-Lys-Val-Arg-Asn-Phe
What type of coding region mutation is this? Give a reason for your answer.
1. Silent
2. Amino acid substitution
3. Frameshift
4. Nonsense
Activity 2
1. A hypothetical plant gene encodes a protein with the following amino acid sequence:
Met-Leu-Lys-Thr-Phe-Val-Glu
A mutation of a single nucleotide alters the amino acid sequence to:
Met-Tyr-Asp-Pro-Gly-Ala-Lys-Ile-Arg
Describe the type of the mutation and the position of the codon that is affected.
2. For the following DNA sequence
3′ ATG GCC GGC AAT CAA CTA TAT TGA 5′
1 24 (nucleotide number)
a. Write the coding strand
b. Write the mRNA strand
c. Write the protein strand
d. Give the altered amino acid sequence of the protein for each of the following mutations:
i. A transition at nucleotide 11
ii. A transition at nucleotide 13
iii. A single nucleotide deletion at nucleotide 7
iv. A T to A change at nucleotide 15
v. An addition of ACC after nucleotide 6
vi. A transition at nucleotide 9 | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.03%3A_DNA_Mutations.txt |
Learning Objectives
At the completion of this PCR lesson, learners will be able to:
• List the 5 chemical components of a PCR reaction and describe their roles.
• List the functions of the 3 temperature cycles which are repeated during a PCR reaction.
• Describe the process of observing results and interpreting results of a PCR experiment.
• List possible uses of PCR in genetic testing and in research.
Overview
The polymerase chain reaction laboratory technique is used in a variety of applications to make copies of a specific DNA sequence. This lesson describes how a PCR reaction works, what it accomplishes, and its basic requirements for success. Examples of interpreting results are given. PCR’s strengths, weaknesses, and applications to plant biotechnology are explained.
The Discovery of PCR
In 1983, Kary Mullis was driving along a Californian mountain road late one night. As a molecular biologist, Dr. Mullis was imagining a better way to study DNA. This late-night thinking led to a revolutionary way to make laboratory copies of DNA molecules (Saiki et al. 1985, Mullis 1990). In the decades since, the polymerase chain reaction or PCR, has become the standard method used for detecting specific DNA or RNA sequences. Selling the equipment and reagent kits for PCR is a multi-billion-dollar business because DNA and RNA detection is critical information in many applications.
In Vitro vs. In Vivo Replication
PCR is an In Vitro process; a series of chemical reactions that happen outside of a living cell. This laboratory technique is modeled after an In vivo process, the living cell’s natural ability to replicate DNA during normal cell cycles (see Lesson on DNA: The Genetic Material). Every living cell makes a duplicate copy of each chromosome before the cell is ready to divide. Figure 1 below illustrates the key parts of In vivo DNA replication that are the basis for PCR success.
There are other enzymes that play an important role in in vivo replication. However, PCR works as an in vitro DNA replication process by using just one of these enzymes. Mullis imagined a chemical reagent and a temperature change step in the method that could perform the work of the other two enzymes. It should be noted that because Dr. Kerry Mullis had learned about the details of in vivo DNA replication, he could create this science changing in vitro method.
Before reading the description of PCR components and processes, watching this video can help you visualize the importance of each step,
Name and Chemical Components of PCR
The name ‘Polymerase Chain Reaction’ represents the nature of the process. ‘Polymerase’ because DNA Polymerase III is required for constructing new DNA strands, just like in a living cell. ‘Chain Reaction’ describes repeating cycles of replication which target a specific segment of a chromosome and use a “copy the copies” progression each cycle that doubles the amount of DNA copies of a specific segment of DNA present each cycle. In just 20 cycles of the chain reaction, over one million (220) copies of that specific segment of DNA can be produced. This is enough DNA to see with your naked eye. The goal of PCR is to make millions of copies of a specific segment of DNA that all originate from a single DNA sample.
The five chemical components that must be added to a test tube for the PCR reaction to work, include a DNA template, DNA polymerase III enzyme, single stranded DNA primers, nucleotides, and reaction buffer.
1. The DNA template is a sample of DNA that contains the target sequence of DNA for copying.
2. DNA pol III. There are two requirements for a suitable DNA polymerase enzyme for PCR. First, the enzyme must have a good activity rate around 75°C. Second, the enzyme should be able to withstand temperatures of 95-100°C without denaturing and losing activity.
3. Two primers. Primers are short oligonucleotides of DNA, usually around 20 base pairs in length. Because the purpose of PCR is to amplify a specific section of DNA in the genome, such as a known gene, then primers of specific sequences must be used. The geneticist planning the PCR reaction will design a forward primer to bind to one strand and a reverse primer that complements and binds to the other strand. The primer design process to select forward and reverse primers is requiring appropriate genetics thinking and is describe later in this reading.
4. The four different deoxyribonucleotide triphosphates (dNTPs). Adenine (A), guanine (G), cytosine (C), and thymine (T) are needed to provide the building blocks for DNA replication. DNA polymerase will add each complementary base to the new growing DNA strand according to the original strand’s sequence following normal A-T and C-G pairings.
5. Finally, a reaction buffer. This creates a stable pH and provides the Mg2+ cofactor needed for DNA pol III activity.
Three Temperature Cycles
A key insight to the success of PCR as an in vitro DNA replication process which generates millions of specific sequence copies was a three-temperature cycle which accomplish three parts of DNA replication: denaturation of the double stranded template, annealing of the primers to the single strands and extension of new strand synthesis by DNA pol III.
In general, a single PCR run will undergo 25-35 cycles. The first step for a single cycle is the denaturation step, in which the double-stranded DNA template molecule (Figure 2) is made single-stranded(Figure 3). The temperature for this step is typically in the range of 95-100°C, near boiling. The high heat breaks the hydrogen bonds between the strands but does not break the sugar-phosphate bonds that hold the nucleotides of a single strand together (Figure 3).
Thousands of copies of the single stranded primers and the individual nucleotides were added to the test tube prior to beginning the cycles. Both the primers and nucleotides will become part of the new DNA strands. The second step in the PCR reaction is to cool the temperature in the test tube to 45-55°C. This is the primer annealing step in which the primers bind to complementary sequences in the single-stranded DNA template. The two primers are called the forward and the reverse primer and are designed because their sequences will target the desired segment of the DNA template for replication (Figure 4).
The geneticist planning the PCR analysis must “design” the forward and reverse primers and then buy them from a vendor who can synthesize single stranded DNA that has a specific sequence and length. The two most important criteria for primer design are the following.
1. One primer must have a sequence that complements one of the template strands and the other primer must be complementary to the other strand. BOTH strands need to be primed for the replication process.
2. The primers must bind so that their 3′ ends are ‘pointing’ in the direction of the other primer. This ensures that the sequence between the primers is replicated in the PCR cycles.
Extension: The final PCR step is when the DNA polymerase enzyme reads the template and connects new nucleotidesto the primer’s 3’ end, extending a new complementary strand of DNA (Figure 5). Completion of the final step and the first cycle of PCR, will make two double stranded DNA copies from the original template DNA, doubling of the amount of DNA present. The test tube is heated to around 75°C, optimizing DNA pol. III activity and the newly synthesizing DNA strand is extended as the template strand is read by DNA pol. III. The Extension step will run for a few minutes and this step completes one PCR cycle.
For cycle 2, the denaturation, annealing, and extension steps are repeated (Figure 6 a, b, c). This time, though there will be twice as many DNA template molecules compared to what there was at the beginning of cycle 1. Copies are being made by reading the original template and copies are made by reading the copies made in the previous cycle. Therefore, if everything is working correctly, the DNA replication in the test tube is a chain reaction where at the end of a cycle, there is double the amount of that DNA sequence as what was found at the beginning of the cycle.
Because thousands of copies of the forward and reverse primer are added at the start of PCR, all the single strand templates, both the original, the copies in cycle 3 and beyond, and the copies of the copies made from previous cycles will be primed for the extension step of the cycles.
Thermal cycler
When PCR was first invented, scientists used water baths set at different temperatures and the hand transfer of test tubes at timed intervals to run the PCR reaction. Once the technique became a proven technology for DNA analysis, engineers went to work to create PCR machines. The instrument which heats and cools the DNA samples is called a thermal cycler (Figure 7). Each small tube or sample well in a plate contains all the chemical components needed for a PCR reaction. Adding a specific sample to the reaction mix provides the template DNA. A thermal cycler can be programmed for specific temperatures and the amount of time spent at each temperature. The engineered design of thermal cyclers to maximize the accurate replication of the targeted DNA in a small sample volume with the minimum amount of time can be critical in many applications of PCR.
Taq DNA polymerase
When Dr. Kerry Mullis ran the first PCR experiments, he needed to add a new sample of DNA pol III after each denaturation step. This was because the high temperature needed to denature the double stranded DNA template also denatured the DNA pol III protein structure. The DNA pol III enzyme commonly available to molecular geneticists was from E. coli bacteria and this enzyme had no stability at near boiling temperatures. Fortunately, biologists had been investigating Thermus aquaticus, (Taq) a thermophilic eubacterium found in hot springs (Chien et al. 1976). The Taq version of DNA pol III does not easily denature in the hot temperatures required in PCR; plus, it has a good efficiency, able to add 60 base pairs/sec at 70°C. Like all other DNA polymerases, Taq DNA pol III cannot begin DNA replication without the addition of a starting primer. Thus, the discovery of Taq DNA pol III and the commercial availability of this enzyme made PCR a more reliable and doable technology which hastened its application to science investigation and diagnostic testing.
Visualizing the Results with Electrophoresis
Once a PCR reaction has been completed, we need to be able to see the results. To do this, a sample of the PCR mixture is loaded into an agarose gel for electrophoresis. The agarose gel contains a matrix of pores which enables it to separate DNA fragments based on their sizes. For details about setting up and running an electrophoresis gel, see Electrophoresis: How Scientist observe fragments of DNA
Figure 8 shows a picture of a gel electrophoresis gel that is running. The box on the right contains DNA loaded in the agarose gel. The gel placed in an aqueous solution of electrolytes. Depending on the type of dye used, color bands are a dye that was added to the PCR sample before it was loaded into the sample well. This allows for the tracking of the DNA’s progression through the gel. Hooked up to this gel unit is an electrical power source which provides the force to move the DNA through the gel. Since DNA molecules are negatively charged, they will migrate towards the red, positive electrode. Shorter DNA fragments move faster than longer fragments through the pores in the gel.
After the gel is run, the DNA is stained with a chemical that binds specifically to DNA molecules and then will either reflect a specific color of visible light or fluoresce a specific color when viewed with ultraviolet light. A single ‘band’ contains 1000s of individual DNA fragments, all of the same length. Figure 10 illustrates the visual information that can be obtained from an electrophoresis gel after it has run. The electric current uniformly moves all the DNA fragments through a gel in the same direction. The sample wells at the top of the gel image thus establish lanes for the DNA samples to move.
Below is a description of what information is revealed from each lane.
• Lane L: This was loaded with the DNA size ladder that contains copies of seven different lengths of DNA fragments. Commercial vendors of the DNA size ladder provide information on the lengths of each fragment in base pairs (bp). Running this lane provides an estimate of the DNA fragment lengths in the sample lanes (1-5).
• Lane 1: The PCR sample loaded in this lane has copies of a single length of DNA. The length is slightly more than 400 base pairs.
• Lane 2: The PCR sample loaded in this lane has copies of two lengths of DNA. One fragment is the same length as the fragments in lane 1 and the second fragment is slightly less than 400 bp.
• Lane 3: The PCR sample loaded in this lane has copies of one fragment that is the same length as the shorter fragments in lane 2.
• Lane 4: The positive control worked as predicted. The sample was set up with all the same reagents as the other PCR samples plus a DNA template that was known to contain the sequence targeted by the PCR primers that would generate a 410 base pair fragment.
• Lane 5: The negative control worked as predicted. The sample was set up with all the same reagents as the other PCR samples except no template DNA was added. Therefore, we would not expect the PCR reaction to work and the absence of a band of DNA fragments is as expected.
Because the positive and negative controls worked as expected, the biologist can be confident that the bands of DNA observed in lanes 1,2 and 3 reveal genetic information about the individual providing that DNA sample.
Advantages of PCR
PCR quickly became the method of choice for many types of DNA analysis because of several advantages over other DNA detection methods. The first; it is a simple procedure to set up and run. Fewer steps save time in getting a DNA analysis result. The second is the sensitivity. A very small amount of template DNA in the sample can be detected. Even just a few skin cells from one human hair contain enough DNA, making PCR useful in forensics. The third is that it can be designed to accurately differentiate genetic samples that are different by as little as single nucleotide in the targeted sequence. Finally, in application where large numbers of samples need to be subjected to the same PCR analysis, automation and robotic assistance allow the processing of many samples in a very short time. For example, an automated PCR would be critical for testing large numbers of people in hours during a pandemic.
Limitations of PCR
There are some drawbacks of using PCR that one should be aware of as well. First, the sequence of the gene or chromosome region being targeted is required. This limitation is rapidly diminishing as gene and genome sequencing technology and sharing of this sequence through Internet data bases has emerged as the norm in genetic analysis. Because of the size of the genomes of living things and a high conservation of gene sequences in many organisms, primers designed for a PCR test must be empirically tested with the proper controls. Biologists can easily generate false positive DNA from PCR that is caused by contamination or lack of specificity in primer design. There may also be a need to optimize concentrations of each chemical component. For example, changing the amount of DNA template, MgCl2 and Taq polymerase can affect both the quantity and quality of bands produced. Some studies have shown that even the brand of Taq polymerase can affect results (Holden et al. 2003). Likewise, the temperature cycles may need to be fine-tuned for a specific PCR test. Finally, as an in vitro DNA replication method, PCR cannot replicate entire chromosomes.
Summary
To briefly highlight the topics of this lesson, remember that PCR is a relatively easy lab technique which amplifies the amount of DNA present, much like living cells do in the beginning stages of a cell cycle. There are 5 chemical components of a PCR reaction: a DNA template, a DNA polymerase, primers, nucleotides, and a buffer. The 3 temperature steps for one cycle are the denaturation, primer annealing and extension steps. There are now many variations and uses of PCR ranging from forensics to genomic studies to identifying transgenic crops.
Lesson Activities
Watch these videos to learn more about PCR: | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.04%3A_PCR_and_Gel_Electrophoresis.txt |
Learning Objectives
• Describe the roles that the promoter, coding region and untranslated regions of a gene play in gene expression.
• Describe mRNA processing steps.
• Draw the process of transcription and include the following in your drawing. DNA template and non-template strands, RNA polymerase, new RNA strand, and direction of RNA synthesis.
• Draw the process of mRNA processing and include the following in your diagram, Gene (DNA), promoter, coding region, introns, exons, pre-mRNA, mature mRNA, poly A tail, cap.
Introduction
Genes are DNA sequences that control traits in an organism by coding for proteins (Figure 1). Organisms such as plants and animals have tens of thousands of genes. The impact that a single gene’s information can have on an organism, however, is tremendous. Furthermore, organisms have all their genes in each of their cells, but they only need to use the information from a subset of these genes, depending on the type of cell and the cell’s stage of development. Therefore, the key to gene function is controlling its expression.
Gene structure and transcription
The DNA sequence contains the information to control all biological functions, including the manifestation of traits important to agriculture (yield, drought tolerance, disease resistance, etc., etc.) How is the information contained in DNA sequences converted into the cellular activities necessary for plants and other organisms to function? DNA sequences are used to direct the synthesis of other molecules that actually perform these cellular functions.
Most typical genes encode proteins. The production of a protein from a gene involves several different processes (Figure 1). Transcription involves the copying of the DNA nucleotide sequence into an intermediate nucleotide molecule called RNA (ribonucleic acid). The primary RNA molecule is processed into a mature messenger RNA (mRNA) which then provides the information for the synthesis of a protein through the process of translation. Proteins are composed of amino acids connected by peptide bonds. The sequence of amino acids is determined by the sequence of nucleotide bases in the mRNA. The 20 amino acids have different chemical and physical properties and the sequence of amino acids determine the structure and function of the protein.
Gene structure
Only particular regions of chromosomal DNA are transcribed. A gene can be considered as the region of transcribed DNA, along with associated regions of DNA important for the regulation of transcription (Figure 1). A gene has several parts that are each important to the function of the gene.
The regulatory region (also known as the promoter) contains DNA sequence involved in the control of where and when the genes will be turned on to produce mRNA. The coding region is the part of the gene that is used as template to produce RNA molecules in a process called transcription. Some RNA molecules perform cellular functions directly while many others (messenger RNAs) are used to direct the synthesis of proteins in a process called translation.
a. Gene Promoter
The signals for starting and stopping transcription are located within DNA sequences. Specific nucleotide segments called promoters are recognized by RNA polymerase to start RNA synthesis. After the transcription of full-length RNA strand is completed, a second segment of DNA called terminator invokes termination of RNA synthesis and the detachment of RNA polymerases from the DNA template.
b. Protein-coding region
The protein-coding region of a gene is composed of the sequence of nucleotides that codes for amino acids. As described further in the section on translation, the coding region begins with an ATG start codon (AUG in RNA) and then ends with one of three stop codons. These sequences include only exons, but not all exonic sequences are protein coding as they may include untranslated regions.
c. Untranslated regions (UTRs)
Mature transcripts contain some sequences that do not code for amino acid sequences in proteins. These are referred to as untranslated regions or UTRs. Most mRNA transcripts contain a 5′ and a 3′ UTR. The 5′ UTR contains sequences toward the 5′ end of the mRNA sequence, before the start codon. These sequences can often be important for translational regulation, and sometimes other functions. The sequences following the stop codon are the 3′ UTR. The 3′ UTR may also have important functions regulating transcript stability or directing transcript localization within cells, or sometimes even transport (trafficking) between cells.
UTRs and introns are often useful in genetic studies. Protein coding regions are under strong selective pressure to produce functional proteins and so sequence variation is relatively rare. UTRs and introns on the other hand are under less stringent selection and are therefore sources of sequence variants that can be used to develop genetic markers.
Transcription
The genetic information of DNA is transferred to an intermediate molecule called RNA that is often translated to amino acid sequences used to build proteins. RNA is a nucleic acid, like DNA, but with some important differences. RNA contains a ribose sugar group instead of the deoxyribose found in DNA. RNA molecules are single stranded, instead of being double stranded. RNA contains a uridine (U) base and does not contain a thymidine base. The other bases (A, C, G) are contained in both RNA and DNA. U has the property of base-pairing with A.
RNA synthesis is directed by a DNA template in a process called transcription. A protein complex containing the enzyme RNA polymerase synthesizes an RNA molecule by adding nucleosides to the 3′ end of a growing chain. The principle of base pairing is used again and each nucleotide base added is complementary to the corresponding base on the DNA template. Thus, the RNA is complementary in sequence to the template strand of DNA, which is also referred to as “antisense” or “negative” strand (Figure 4). The RNA is identical in sequence (except U replaces T) to the other strand, which is called the “sense” or “positive” strand. Because RNA molecules are produced by the process of transcription, they are often referred to as transcripts.
RNA processing
Coding (transcribed) region
This is the region that is transcribed by RNA polymerase, also known as the RNA coding region. As described below, it may include introns, sequences that are removed from the mature RNA molecule during RNA processing. The transcribed region is demarcated by promoter and terminator sequences.
Introns and exons
As mentioned, and described in detail below, introns are sequences that are removed from transcripts during RNA processing. Sequences that are retained in mature transcripts are called exons. The corresponding stretches of DNA are typically referred to with the same terms. Introns are commonly found in genes of eukaryotes but are rare in prokaryotic organisms.
Intron splicing
The process of transcription produces pre-mRNA that contains both introns and exons. The process of splicing involves removal of introns from pre-mRNA and joining together the exons. A complex group of proteins that form a spliceosome perform the splicing reaction.
Introns sometimes serve as boundaries for sequences encoding functional protein domains, leading to possibility for new and variant proteins by exon shuffling. Also, introns can provide possibility for productions of variant RNA forms through alternate splicing allowing more than one gene product from a single gene. Some introns result from the insertion of transposable elements and may be spliced perfectly of imperfectly, offering more possibility for new genetic diversity.
5′ Capping
The 5’ capping is the addition of a 7-methyl guanidine to the first nucleotide of mRNA molecule, usually and adenine or guanidine. The phosphodiester linkage between 7-methyl guanidine and the target nucleotide is 5′-5′ instead of 5′-3′, and 3 phosphates rather than 1 are retained in the linkage. The cap stabilizes the 5′ end of the mRNA and plays a role in translation initiation.
Poly adenylation
The transcription of a gene may proceed beyond what ends up as 3′ end of mature mRNA. Thus the 3′ end of mRNA is formed after transcription. The enzyme poly(A) polymerase adds numerous adenosines to the 3′ end to result in what is called the poly(A) tail. The poly(A) tail is necessary for proper processing and transport of mRNA to the cytoplasm. The poly(A) tail is also important for the stability of mRNA, and initiation of translation in eukaryotic organisms.
Summary
The genetic information of DNA is transferred through transcription to an intermediate molecule called RNA. The signals for starting and stopping transcription are located within the DNA sequence and referred to as promoter and terminator sequences. The coding region of a gene is composed of a sequence of nucleotides that are transcribed into RNA. These sequences include exons and introns. Exons are the sequences that code for proteins. The coding region of a gene contains exons and introns. Also, pre-mRNA contains both introns and exons. The introns in pre-mRNA are removed through a process called intron splicing. The mRNA is processed by 5′ capping and addition of a poly(A) tail.
Learning Activities
Learning Activity 1
Given the following sequence of double-stranded DNA, predict the sequence of the RNA strand.
Learning Activity 2
Watch This Video: Transcription Detail
A video element has been excluded from this version of the text. You can watch it online here: https://iastate.pressbooks.pub/genagbiotech/?p=24 | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.05%3A_Gene_Expression-_Transcription.txt |
Learning Objectives
• Determine the amino acid sequence of a protein given the nucleotide sequence of a gene.
• Recognize the biological connection among mRNA, tRNA, amino acids, and proteins.
• Understand the concept of protein structure and its relation to protein function.
Introduction
The importance of gene expression is evident when you observe the changes plants go through during their lifecycle or during a season. Trees and bushes, for example have dormant buds through the winter. Environmental signals affiliated with the coming of spring induce genes in the buds to turn on and drive the dramatic changes of leaf development and flowering. The genes were always present in those bud cells but were controlled to turn on at the proper time. Understanding gene expression thus requires an examination of two processes, the activation of gene expression to make a “message” and the reading of this message to build a specific protein.
Amino acids are used to make proteins
Amino acids are the building blocks of proteins. Except for the amino acid proline, they all consist of a central carbon atom covalently bonded to an amino group, a hydrogen atom, a carboxyl group, and a variable R group (Figure 1). In the physiological range, both the carboxylic acid and amino groups are completely ionized allowing amino acids to act as either an acid or a base. Thus, amino acids do not assume a neutral form in an aqueous environment.
The chemical and physical properties of the side chains provide the functionally important properties to amino acids. Some of the important properties of side chains include acidic, basic, or neutral, hydrophobic vs. hydrophilic, as well as the size of side chains. They are categorized according to their side groups as nonpolar, polar, positively, or negatively charged or aromatic. All these properties and others affect how each amino acid contributes to the structure and function of a mature protein.
There are 20 different amino acids making up the subunits of proteins (Table 1). All proteins are made from differing combinations of the amino acids (Figure 2). Therefore, if you are going to make proteins you need to either make amino acids first (Figure 3) or consume amino acids in your diet. Animals consume some of their amino acids (the essential amino acids), but plants and bacteria are able make all their own amino acids.
Table 1. Amino acids and their abbreviations[1]
Full Name Abbreviation (3 Letter) Abbreviation (1 Letter)
Alanine Ala A
Arginine Arg R
Asparagine Asn N
Aspartate Asp D
Cysteine Cys C
Glutamate Glu E
Glutamine Gln Q
Glycine Gly G
Histidine His H
Isoleucine Ile I
Leucine Leu L
Lysine Lys K
Methionine Met M
Phenylalanine Phe F
Proline Pro P
Serine Ser S
Threonine Thr T
Tryptophan Trp W
Tyrosine Tyr Y
Valine Val V
Codons and the Genetic Code
The search for the genetic code (Table 2) revealed that genetic information is stored in nucleotide triplets referred to as codons. The genetic code is degenerate because many amino acids are specified by more than one codon. Sixty one of the 64 possible combinations of the three bases in a codon are used to code for specific amino acids. Three stop codons, UAA, UAG, and UGA do not code for any amino acids but specify the termination of peptide chain synthesis during translation. The AUG start codon is used to initiate polypeptide synthesis and codes for the amino acid methionine.
Table 2. The genetic code. The 20 common amino acids are listed in their three and one-letter formats Stop codons are UAA, AUG and UGA.
First Position
Second Position
Third Position
U
C
A
G
U
Phe (F)
Ser (S)
Tyr (Y)
Cys (C)
U
C
Leu (L)
Stop
Stop
A
Stop
Trp (W)
G
C
Leu (L)
Pro (P)
His (H)
Arg (R)
U
C
Gln (Q)
A
G
A
Ile (I)
Thr (T)
Asn (N)
Ser (S)
U
C
Lys (K)
Arg (R)
A
Met (M)
G
G
Val (V)
Ala (A)
Asp (D)
Gly (G)
U
C
Glu (E)
A
G
Why a Triplet Code?
Prior to understanding the details of transcription and translation, geneticists predicted that DNA could encode amino acids only if a code of at least three nucleotides was used. The logic is that the nucleotide code must be able to specify the placement of 20 amino acids. Since there are only four nucleotides, a code of single nucleotides would only represent four amino acids, such that A, C, G and U could be translated to encode amino acids. A doublet code could code for 16 amino acids (4 x 4). A triplet code could make a genetic code for 64 different combinations (4 X 4 X 4) genetic code and provide plenty of information in the DNA molecule to specify the placement of all 20 amino acids. When experiments were performed to crack the genetic code, it was found to be a code that was triplet. These three letter codes of nucleotides (AUG, AAA, etc.) are called codons.
The genetic code only needed to be cracked once because it is universal (with some rare exceptions). That means all organisms use the same codons to specify the placement of each of the 20 amino acids in protein formation. A codon table can therefore be constructed and any coding region of nucleotides read to determine the amino acid sequence of the protein encoded. A look at the genetic code in the codon table below reveals that the code is redundant meaning many of the amino acids can be coded by four or six possible codons. The amino acid sequence of proteins from all types of organisms is usually determined by sequencing the gene that encodes the protein and then reading the genetic code from the DNA sequence
Ribosomes, tRNA and Anti-codons
The ribosomes are a large molecular complexes containing an assembly of several ribosomal RNAs (rRNA) and many proteins. Ribosomes are the machinery of protein synthesis, facilitating the ordered addition of amino acids to a nascent polypeptide chain under the guidance of an mRNA template (Figure 5A). Prior to the initiation of protein synthesis the ribosome occurs in two separate subunits of size 60s and 40s (“s” stands for Svedberg units and is a measure of a particles sedimentation rate during centrifugation).
The meaning of a codon for a specific amino acid is determined by the tRNA (transfer RNA). Each tRNA (Figure 5B) contains a triplet of nucleotides referred to as an anticodon, which is complementary to a specific codon. For each of the 20 amino acids, a specific enzyme (aminoacyl-tRNA synthetase) catalyzes its linkage to the 3′ end of its specific tRNA adapter. The function of aminoacyl-tRNA synthetase is critical as it provides a check for accuracy in protein synthesis by adding a specific amino acid to a specific tRNA molecule. In this way, one particular amino acid is targeted to each codon triplet of mRNAs.
Peptide chain formation
Protein synthesis is initiated by the 40s subunit through attachment it’s to the mRNA, and recognition of the AUG codon. This is followed by the attachment of the 60s subunit to the 40s-mRNA complex to provide the structure necessary to align each successive aminoacyl-charged tRNA for transfer of its amino acid to the growing polypeptide. Successive amino acids are attached to the growing polypeptide by the formation of peptide bonds that form between the carboxyl group of one amino acid and the amino group of the next.
Like nucleic acids, polypeptides also have molecular directionality. One end of a polypeptide chain will have a free amino group and the opposite end will have a free carboxy group. As such, these ends are referred to as the amino-terminus or carboxy-terminus, respectively. During translation, mRNA templates are read from the 5’ end toward the 3’ end. Proteins are synthesized beginning at the amino terminal end going toward the carboxy terminus.
Protein structure and function
Proteins assume complex 3-dimensional structures that are essential for their various functions as enzymes, regulatory factors, or structural proteins. Overall structure is determined by complex physico-chemical interactions among amino acid side groups.
Several levels of structure are considered (Figure 6). The primary structure of a protein is the amino acid sequence of its polypeptide chains. The secondary structure is derived through the interactions between neighboring amino acids to form local structural elements such as beta sheets or alpha helices. The tertiary structure refers to the three-dimensional structure of an entire polypeptide and is determined by the diverse properties of amino acid side groups. For example, in an aqueous environment, hydrophobic amino acids will often interact at the core of a protein structure while hydrophilic groups may be exposed on the protein surface.
Many proteins are multimeric, composed of several polypeptide chains called “subunits”, which associate non-covalently or in some cases via disulphide bonds. Such higher order spatial associations among polypeptides are also the result of interactions among amino acids and result in a quaternary structure. Some multimeric proteins consist of multiple copies of the same polypeptide.
Summary
The genetic information of DNA is transferred to through transcription to an intermediate molecule called RNA. The signals for starting and stopping transcription are located within the DNA sequence, and referred to as promoter and terminator sequences. The coding region of a gene is composed of a sequence of nucleotides that are transcribed into RNA. These sequences include exons and introns. Exons are the sequences that code for proteins. The coding region of a gene contains exons and introns. Also, pre-mRNA contains both introns and exons. The introns in pre-mRNA are removed through a process called intron splicing. The mRNA is processed by 5’ capping and addition of a poly(A) tail. Mature RNA is then translated to amino acids used to build proteins.
1. Table Source: Donald Lee, University of Nebraska-Lincoln | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.06%3A_Gene_Expression-_Translation.txt |
Learning Objectives
At the completion of this lesson you should be able to:
1. Define the roles of DNA and proteins in cell development and metabolism
2. Determine the amino acid sequence of a protein given the nucleotide sequence of a gene.
3. Describe the roles that the promoter, coding region and termination sequence of a gene play in gene expression.
4. Recognize the differences between the structure of proteins, amino acids, genes, and nucleotides.
5. Draw the process of gene expression and include the following in your drawing. Gene, RNA polymerase, promoter, coding region, termination sequence, cell, nucleus, cytoplasm, mRNA, tRNA, ribosome, anticodon, codon, amino acid, protein, peptide bond.
You have learned this fundamental biology fact: genes provide the information to instruct cells how to build specific proteins and these proteins control cell development and metabolism. Some biologists refer to this fact as the “Central Dogma”. We prefer to refer to this process as gene expression. This lesson will integrate this biology fundamental with some gene expression examples. These examples emphasize an additional fundamental biology idea that life is a continuous cycle. In order for living cells to live, they require the genes, RNAs, and proteins that came from prior existing living cells.
Example #1: Gene Expression of the Plant Acetolactate Synthase Enzyme Gene
We have selected this gene because the protein the gene encodes is the target site for several types of herbicides and thus is an important example in agriculture. The gene also illustrates an important circle of life concept. All living things need amino acids to build proteins and some living things need proteins to build amino acids.
The ALS gene is not found in animals but is found in plants and bacteria. Plants and bacteria are able make all their own amino acids. Animals must consume a subset of the 20 amino acids (see Lesson on Translation) in their diet. One gene all plants and bacteria must have to live encodes the enzyme acetolactate synthase (ALS). ALS enzyme allows plants to make their own supply of three of the 20 amino acids; leucine, isoleucine, and valine. The ALS protein is an enzyme and is needed to catalyze a reaction in the chemical pathway that makes these three amino acids. Let us detail the information transfer steps needed for gene expression: use the genetic code of DNA sequences in the gene to encode the amino acid sequence of the ALS enzyme.
Three parts of the Gene and a Two-Step Process
The ALS gene consists of several thousand nucleotides of DNA and like all genes has three main parts, the promoter, coding region, and termination sequences (Figure 1, gene sequence abbreviated for demonstration).
Each part of the ALS gene plays a role in controlling gene expression and this is a two-part process. The promoter is the on/off switch of the gene, the coding region determines the amino acids sequence of the protein that will get made and the termination sequence signals where the gene information ends. The DNA sequence that makes up a gene is a part of a much larger chromosome of continuous DNA. The gene sequence is organized to make an RNA (ribonucleic acid) copy of the gene’s nucleotide sequence to serve as a mobile message for protein building. RNA building is called transcription because DNA nucleotides encode RNA nucleotides.
Step 1: Transcription: control from the promoter and termination sequence
The promoter’s role is to selectively turn on the ALS gene in cells that will need the ALS enzyme. The promoter accomplishes this by its specific nucleotide sequences that react with regulatory molecules in the cell. We will discuss the regulation of gene expression in more detail later. What cells in a plant will need the ALS enzyme? All cells that are making proteins. This will be all cells in the plant at some stage of development. Therefore, the ALS gene has a promoter sequence that allows the gene to be turned on in all types of cells. When the ALS gene is turned on, transcription can take place. Like most chemistry in the cell, enzymes will be needed for gene expression. RNA polymerase is the transcription enzyme (Figure 2) which will bind to the DNA sequences in the ALS promoter and interact with one strand of the gene (the coding strand).
RNA polymerase works by moving along the gene promoter until it encounters a series of A and T nucleotides called the TATA box (Figure 3). The TATA box signals RNA polymerase that it has reached the end of the promoter. Now the enzyme has the “green light” to read the coding strand DNA and make RNA. The procedure works much like DNA replication. RNA polymerase will read the DNA in a 3’ to 5’ direction and build the RNA 5’ to 3’. Unlike DNA replication though, only one DNA strand is read and the transcription process must end once the RNA polymerase reaches the end of the ALS gene. If RNA polymerase kept going along that strand of the DNA making up the chromosome, other genes that are on the same chromosome might be expressed in the wrong cells or at the wrong times.
How is the RNA polymerase signaled that it has reached the end of the gene? That is the role of a termination sequence in the gene (Figure 4). The termination sequences signal the end of the gene. One transcription termination strategy will be described here. The nucleotides that make up the termination sequence of the ALS gene could be ordered to form a palindrome. This means that the nucleotides being placed in the RNA can fold back on themselves and form a hairpin loop (Figure 5).
While no geneticist has actually viewed this hairpin loop formation in action, it is believed that the hairpin formation snaps the RNA polymerase off the DNA coding strand and releases the RNA message. Transcription of an RNA that has the coding region information is now completed. Each time an RNA polymerase goes through the process, one copy of the RNA is made by reading the DNA template. The type of RNA made from transcription of the ALS gene is called a messenger RNA or mRNA for short. Some genes can encode transfer RNA (tRNA) or ribosomal RNA (rRNA). These RNA molecules have special roles in the next part of gene expression called translation. Once an mRNA is transcribed, some modifications will occur in the nucleus. We will describe some of the post-transcriptional processes later. For now, let’s follow the ALS mRNA onto the translation step of gene expression.
Video Examples
Watch this video from UNL for more information about the Transcription process
Step 2: Translation: reading the code to make proteins
The translation process requires a conversion of information from a chain of nucleotides in RNA to a chain of amino acids in protein. The translation term was coined based on the idea that we are converting to a new molecular language at this stage of gene expression. The conversion of information in translation is a little more complex than transcription and requires several biomolecules to interact and work together (Figure 7). We will describe the ribosomes and tRNAs molecules first and then describe how they work together to translate a mRNA.
• tRNA: The ‘t’ prefix means transfer. The tRNA is a small RNA (70 to 80 nucleotides) and has a sequence that allows hairpins to form which gives a secondary structure to the molecule. Their small size makes them mobile to provide the cells means to transfer amino acids to the site of protein building. Their structure also makes them specific in how they deliver their cargo of a single, specific amino acid. A tRNA is about 50 times larger than an amino acid.
• Ribosomes: Ribosomes are macromolecules that are made of both RNAs (rRNAs) and proteins. These components assemble into a workbench for the translation process to occur. The ribosome is a big macromolecule: it can hold three tRNAs in position for the translation work and can catalyze bond breaking and bond formation.
We will use a simple representation of the mRNA, tRNA, ribosome, and amino acid to show how they work together (Figure 6).
The tRNAs shown in Figure 6 are not ready for translation. The structure of the tRNA is recognized by special enzymes in the cell that attach the proper amino acid to the tRNAs. The tRNAs that are bound to their designated amino acid are called charged tRNAs (Figure 7). The near the middle of the 70-80 nucleotides in a tRNA are three nucleotides called the anticodon. When the tRNA folds, the anticodon becomes one end of the molecule and the other end will be attached to a specific amino acid. In the figures for this lesson, the tRNA is shown with its secondary structure and only the anticodon sequence is shown (Figure 7).
Let’s describe how the key players get protein building started, how they keep it going, and how they end it. Active ribosomes are assembled from small and large subunits when the ribosome “workbench” is ready to translate a mRNA (Figure 7). When mRNAs are present in the cytoplasm, the small and large ribosome subunits assemble at the 5’ end of these messages and are ready to read mRNAs like blueprints to build specific proteins.
The ribosome reads the mRNA quickly and accurately, but it must determine where to start. As the ribosome moves in the 5’ to 3’ direction along the mRNA it “looks” for the sequence AUG (Figure 7). This sequence is called the start codon and signals the ribosome that this is where to begin building the protein sequence.
Video Examples
Watch this UNL video on Translation for more information about these processes.
The Reading Frame, Codons and Anticodons
The mRNAs, tRNAs, ribosomes, and amino acids are now in position to combine their specific functions into the building of a protein. The AUG start codon establishes the beginning of what is called a reading frame on a mRNA. The nucleotides in the mRNA must have the specificity to code for 20 different amino acids to build the protein. The genetic code is read as a three-letter code which provides this specificity. The ribosome will move along and read the sequence in the mRNA one codon (three nucleotides) at a time to follow the reading frame and build the correct protein (Figure 8).
Getting Translation Started with the Start Codon
The AUG start codon signals the ribosome to place in the amino acid methionine. The A, P and E sites on the ribosome site works somewhat like a vice on the workbench, holding stuff in place so it can be worked on. In Figure 9, the ribosome has moved in the 5’ to 3’ direction on the mRNA .mRNA. This allowed a tRNA with the anticodon GGG to move into the A site and placed two amino acids very close to each other on the ribosome. The ribosome will now move one more codon in the 5’ to 3’ direction which moves the two tRNAs now to the E and P sites (Figure 9). Now the ribosome is ready to start binding together amino acids with peptide bonds to start to form a protein. A third tRNA (with the CCC anticodon) can now arrive at the open A site, bringing the amino acid designated by the GGG codon.
Peptide Bond Formation and Protein Building
As you can see, a ribosome “workbench” has catalytic tools to go along with its ‘P’ and ‘A’ “vice sites”. The ribosome has enzymatic functions that allow it to break and form bonds. The ribosome will break the bond that binds the amino acid to the tRNA at the ‘E’ site. Simultaneously the ribosome forms a peptide bond between the second and the third amino acids that were brought in by the tRNAs. The close proximity of these amino acids at the ribosome sites provides the opportunity for peptide bond formation and then release of the tRNA in a smooth continuous process (Figure 10 and 11). Peptide bonds always bind the acid end of one amino acid with the amino end of the next amino acid. Peptide bonds are strong covalent bonds which keep the amino acids connected during and after translation. The tRNA at the ‘E’ site has now accomplished its task by bringing in the first amino acid. This tRNA will then move off the ribosome (Figure 11). The tRNA released from a ribosome is recycled by the cell. It can be charged again by binding another amino acid in the cytoplasm and contribute to the synthesis of another protein. The protein. The first three amino acids in the protein have now been put together.
The synthesis of our protein is far from over. The ribosome continues to shift one codon in the 5’ to 3’ direction as a tRNA leaves the E site .site. A tRNA can now come into the A site if it has the anticodon that is complementary to the next codon on the mRNA. Proteins are hundreds of amino acids long. In our example, the steps described continue until the complete mRNA information is read to build the 667 amino acid long ALS enzyme.
The End of Translation: stop codons looking for something they cannot find
Start codons start translation so it is logical that stop codons stop translation (Figure 11) How do stop codons do this? There are 61 tRNAs with different anticodons (see Appendix 1: Codon Table). That means there are three codons that do not have corresponding tRNAs with complementary anticodons. These three codons serve as stop codons. When a ribosome encounters a stop codon on a mRNA it will wait for a tRNA with the right anticodon to come over. It will not skip the codon or shift over one nucleotide to form a new reading frame. The ribosome waits for the right tRNA, but it does not wait for long. A stalled ribosome will quickly cleave off the bound tRNA with the newly built protein chain and then move on to translate another mRNA. No tRNAs in the cell have anticodons that complement any of the three possible stop codons. Therefore, stop codons are able to end the translation process when the completed protein is made.
Sequence, structure, and function
In the case of our ALS protein, the assembly of all 667 amino acids into the polypeptide or protein chain in the correct order is important because the sequence of the amino acids determines how the protein folds. The folded structure of the protein determines its function. If some amino acids are changed and the structure changes, the ALS will not function properly as an enzyme in plant or microbe cells.
Summary
Organisms use the process of transcription to make mRNA as a temporary molecule to transport protein coding information from a gene. The translation process reads the mRNA and makes the intended protein, one amino acid at a time. This process is controlled by the promoter of the gene to ensure that proteins are made in the right cells at the right time to drive growth, development, metabolism, stress response and other processes needed to complete the organism’s life cycle.
Video Examples
Watch this video from UNL for an overview of Transcription and Translation.
Circle of Life and Gene Expression
Let us revisit the idea that gene expression works because living cells give rise to more living cells. Imagine yourself when you were just a single cell zygote with a copy of all the genes in the human genome from your biological mother and a copy from your biological father. What else did you need to get from your parents before you could express your first gene and build your first protein? You needed to already have an RNA polymerase in the nucleus of the cell to transcribe your first gene and you needed to have ribosomes, tRNAs and amino acids ready for translating that mRNA in the cytoplasm. Luckily, the egg cell from your mother had the 23 human chromosomes plus all of these gene expression biomolecules. Later, you could turn on your own genes for RNA polymerase, tRNAs and ribosomes but your mother’s egg cell was a living environment that allowed you to start your life.
A plant treated with a herbicide which binds to and blocks the action of the ALS enzyme likewise demonstrates that gene expression has a cycle of dependency on living cells. Small plants in an environment for rapid growth will have their growth halted by the ALS binding herbicides. The plants do not die from the herbicide, they just stop growing. The reason for this is that plants need to build new proteins to grow and to build new proteins they need the ALS enzyme to be working and building new leucine, isoleucine, and valine amino acids. ALS herbicide treated plants stop making these three amino acids and translation stops for every protein in the cell that requires at least one of these amino acids. Living things need amino acids to build proteins and plants and microbes need the ALS enzyme to build three of the amino acids. The next lesson (Gene expression part 2) will apply the gene expression principles from this lesson to the observation that some plants can be resistant to ALS inhibitor herbicides. | textbooks/bio/Genetics/Genetics_Agriculture_and_Biotechnology_(Suza_and_Lee)/1.07%3A_Gene_Expression-_Applied_Example_%28Part_1%29.txt |
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