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MCB142/IB163 Thomson Mendelian and population genetics Fall 2002 Handout 3 • Non-random mating, selection (ch. 17: 568-579) deviations from Hardy Weinberg assumptions: the strength of the Hardy Weinberg (HW) law is that one can deviate from the assumptions quite a bit and the data will still approximate Hardy Weinberg proportions (HWP). The weakness of the HW test is that the deviation from the assumptions has to be very strong in order to detect the effect of this evolutionary force, e.g., selection. Deviations from HW assumptions involve: (1) non-random mating, e.g., inbreeding, (2) effects of mutation are negligible as mutation rates are low—but mutation is an important force in creating new variation, (3) migration is important if the migration rate is high and the two populations are very distinct genetically, (4) genetic drift due to small population size (chance effects)—genetic drift effects are important in both small and large (but finite) populations in terms of long term effects of changes in allele frequencies over generations due solely to drift effects, however the finite size of a sample taken from a population is taken into account in the statistical tests for HWP and finite population size itself does not cause significantly detectable deviations from HWP, (5) selection has to be strong to cause deviations from HWP, e.g., it can be detected with sickle cell anemia, see below. non-random mating: individuals with certain genotypes sometimes mate with one another more commonly than would be expected on a random basis. When like mates more often with like we term this positive assortative mating, e.g., height, IQ. Positive assortative mating increases the proportion of homozygous individuals but does not alter the allele frequencies. With self-fertilizing plants the level of heterozygosity is reduced by 1/2 each generation (see Figure 17.7 of text.) Self-fertilizing plants have more homozygotes than expected under Hardy-Weinberg and often show significant deviations from HWP. inbreeding: mating with close relatives is another form of non-random mating. Relatives are more likely to carry the same recessive allele for a rare recessive trait—inbreeding increases the number of affected individuals with rare recessive traits. Marriages between first cousins have about twice the rate of birth defects as random matings. mutation: in and of itself does not change allele frequencies to a noticeable extent as mutation rates are low. However, mutations are the raw material of evolution, the ultimate source of genetic variation. Although the frequencies of mutants are initially rare, and most are lost from the population, nevertheless some increase in frequency due to genetic drift effects and selection (see below). migration: is the movement of individuals from one population into another, which can alter allele frequencies, and if there are large genetic differences cause a statistically significant deficiency of heterozygotes from Hardy-Weinberg expectations. relative fitness: the average number of offspring produced by individuals with a certain genotype, relative to the number produced by individuals with other genotypes. heterozygote advantage: the heterozygote has a higher relative fitness than both homozygotes, this leads to a balanced polymorphism. If we denote the relative fitness values of the 3 genotypes in a 2-allele system as follows: A1A1 A1A2 A2A2 relative fitness 1 - s1 1 1 - s2 then the equilibrium frequency for allele A1, denoted p1* = s2/(s1 + s2), and that for allele A2 is p2* = s1/(s1 + s2), with p1* + p2* = 1. The allele A1 is more frequent if the selection is stronger against the A2 homozygote than the A1 homozygote and vice versa. 1 sickle cell anemia: Individuals with sickle cell trait (AS heterozygotes) are more resistant to malaria and have a higher fitness than individuals who are homozygous AA or SS (these individuals are also more resistant to malaria than AA individuals, but they suffer the deleterious effects of sickle cell anemia). The relative fitness values in a malarial environment have been estimated as: genotype AA AS SS relative fitness 0.88 1.0 0.14 The predicted equilibrium frequencies with this selection regime (s1 = 0.12, s2 = 0.86) are f(A) = 0.88, and f(S) = 0.12 using the formula above. The observed allele frequencies are very close to these values. The selection in this case is very strong and causes detectable deviations from HWP. Genotypes AA AS SS Observed no. among adults 9,365 2,993 29 Total: 12,387 Observed freq. among adults 0.756 0.242 0.002 Expected no. among adults (HWP) 9,523 2,676 188 Total: 12,387 Expected freq. among adults (HWP) 0.769 0.216 0.015 Also see problem 3 below. frequency dependent selection: is another cause leading to balanced polymorphism. The reproductive success of a phenotype depends on its frequency, with higher fitness associated with lower frequency of the phenotype in the population, e.g., ‘right-mouthed’ and ‘left-mouthed’ species of cichlid fish. balanced polymorphism: 2 or more alleles are maintained in a population due to selection. Both heterozygote advantage and frequency dependent selection are examples of balancing selection, they both lead to a stable polymorphic equilibrium state. directional selection: leads to fixation of an allele. An example would be relative fitnesses AA AB BB 1.0 0.9 0.8 which would lead to fixation of the A allele. rates of change: in the table below the number of generations required for the frequency of allele A to change from 0.01 to 0.99 is given for the fitness model AA AB BB 1 1 - hs 1-s for values of h = 0 (or 1) (A is dominant or recessive), 0.1 (or 0.9) and 0.5 (additive effect model), and s = 0.005, 0.01, 0.05, 0.1, and 0.2: h s = 0.2 0.1 0.05 0.01 0.005 0.5 83 175 359 1,830 3,700 0.1 155 317 640 3,225 6,500 0 534 1,075 2,157 10,800 21,600 Weak selection, of course, takes longer to effect an allelic substitution. Even with fairly strong selection it can take quite a long time to effect a substitution, e.g., with h = 0.5 and s = 0.1, it takes 175 generations to change the allele frequency from 0.01 to 0.99, and taking an average of 25 years per generation for humans this would be 4,375 years. Weaker selection, e.g., h = 0.5, s = 0.01, which nonetheless is strong selection in an evolutionary sense, takes 1,830 generations (45,750 years for humans). The time required is approximately proportional to 1/s, e.g., the time length with s = 0.1 is close to twice the time length with s = 0.2. The total number of generations is smallest for genic selection (h = 0.5). If either allele is recessive, or nearly so, the total length of time is greatly extended, this is attributable, for the most part, to the weak effect that selection has when the recessive allele is rare. quantitative trait: determined by a large number of genes each of small effect and environmental factors, e.g., height and weight. stabilizing selection: selection acts to eliminate both extremes from an array of phenotypes (this is the quantitative equivalent of balancing selection for a single gene trait). directional selection: selection acts to eliminate one extreme from an array of phenotypes. 2 disruptive selection: (also called diversifying selection) selection acts to eliminate the intermediate type, favoring both extremes (this is the quantitative equivalent of heterozygote disadvantage for a single gene trait). Problems 1. Do problem 15 from chapter 17 on page 585 of the textbook. Write out the numbers of males and females of each genotype in the two separate groups assuming HWP and equal numbers of males and females, and also test the assumption stated in the problem that in the mixed population the 1,000 females would be in HWP (answer is given below). 2. In a population of self fertilizing plants of size 100, the genotypes at a codominant locus are 20 AA, 60 AB, and 20 BB individuals. What will the genotype counts be in the next generation (assume a population size of 100 again)? What would they be if there was random mating? 3. In Africa in an area where malaria is prevalent, 1,000 newborns and 1,000 adults are typed for the sickle-cell polymorphism: Genotypes AA AS SS Newborns (O) 780 204 16 Total: 1,000 Adults (O) 769 229 2 Total: 1,000 For both samples, determine the allele frequencies of the A and S alleles, and test fit to HWP. Comment on any differences, or not, between the allele frequencies and the test of HWP in the two samples. 4. For the following selection schemes, where the relative fitness values for each genotype are given, state whether one or the other allele will be fixed (and if so which allele) or whether a balanced polymorphism will be maintained, in which case determine the allele frequencies at this equilibrium: AA AB BB (a) 0.8 0.9 1.0 (b) 0.2 1.0 0.6 (c) 1.2 1.1 1.0 3 Answers to Problems 1. Problem 15: In population 1 q1 = 0.10 and in population 2 q2 = 0.30 (the proportion of color-blind males in each population). After equal mixing of the two populations, the frequency of colorblind males is 0.2. The numbers in each group before and after mixing are as follows, and the test of HWP on the females: females X+X+ X+Xc XcXc males X+Y XcY population 1 405 90 5 450 50 Total: 1,000 population 2 245 210 45 350 150 Total: 1,000 mixed (O) 650 300 50 800 200 Total: 2,000 mixed (E) 640 320 40 Total: 1,000 mixed (O - E) 10 20 10 Total: 1,000 2 mixed (O - E) /E 0.156 1.25 2.50 Total chi-square = 3.90, 1df, P<0.05* There is evidence of lack of fit to HWP in the females. In fact mixing of two distinct populations with different allele frequencies is a well known cause of lack of fit to HWP, so the text should not have told you to make this assumption. 2. With selfing, the frequency of heterozygotes is reduced by 1/2 each generation, with the remaining 1/2 divided equally between the two homozygous classes. The genotype counts in the next generation will be 35 AA, 30 AB, and 35 BB. If there had been random mating, the genotype counts would be 25 AA, 50 AB, and 25 BB. 3. Among the newborns, f(A) = p = 0.8820, f(S) = q = 0.1220, and in the adults f(A) = 0.8835, f(S) = 0.1165. Newborns (O) 780 204 16 Total: 1,000 Newborns (E) 777.9 208.2 13.9 Total: 1,000 (O - E) 2.1 4.2 2.1 2 (O - E) 0.006 0.085 0.317 Total chi-square: 0.408, df =1, ns Adults (O) Adults (E) (O - E) (O - E)2 769 780.6 11.6 0.17 229 205.9 23.1 2.59 2 13.5 11.5 9.80 Total: 1,000 Total: 1,000 Total chi-square: 12.56, df =1, P<0.001*** The adult sample shows very significant deviation from HWP, while the newborn sample does not. This reflects the fact that at birth (pre-selection) the genotype frequencies are expected to be in HWP, whereas after selection has acted (malaria and sickle-cell anemia), the adult population shows significant deviation from HWP. The fact that the allele frequencies are very similar in newborns and adults may indicate that the population is close to equilibrium. 4. (a) fixation of the B allele, loss of the A allele (b) stable polymorphism, s1 = 0.8, s2 = 0.4, thus p1* = 1/3 (= f(A)), p2* = 2/3 (=f(B)) (c) fixation of the A allele, loss of the B allele 4 • Genetic drift (ch. 17: 580-582) genetic drift: (chance effects) random change in the frequency of alleles at a locus. short term genetic drift effects: cause changes in allele frequencies, both in small and large populations. The change in allele frequency due to genetic drift in a small population appears larger, statistical testing can determine whether changes are larger than expected by chance. As an example, an allele frequency change in a population of size 50 from p = 0.5 to 0.56 in 1 generation is within the range expected by drift, whereas in a population of size 5,000 such a change would be much too large to be due solely to drift effects. Think in terms of tossing a coin, if you tossed a coin 50 times you would expect 25 heads and 25 tails, but due to the finite number of tosses would not be surprised to observe 28 heads and 22 tails (the change given for the allele frequency above, 56% heads). In fact, 95% of the results of tossing a coin 50 times would fall within the range from 30 heads (20 tails) to 20 heads (30 tails). If you toss a coin 5,000 times 95% of the results would fall within the range from 2,550 heads (2,450 tails) to 2,450 heads (2,550 tails), and an observed outcome of 2,800 heads (2,200 tails) (56% heads when 50% expected) is well outside the range expected by chance (this outcome would occur by chance less than 1 in a million times). long term genetic drift effects: are loss or ‘fixation’ of an allele. It is easy to understand how genetic drift in small populations can lead to loss of alleles (and hence ‘fixation’ of the other allele), e.g., suppose we have a constant population size of 10 haploid individuals, with an initial allele frequency of p = f(A) = 0.5, q = f(B) = 0.5. We assume at the moment that no new mutations arise. After the first generation reproduces, the allele frequencies in the 2nd generation are p = 0.6, q = 0.4, and we now reweight our ‘coin’ so that the probability of an A in the next (3rd) generation is 0.6 and of a B is 0.4, the observed frequencies in the current (2nd) generation. Due to drift effects the observed frequencies in the 3rd generation may differ from these expected values. Continuing this process, the genetic drift (random) effects result in eventual loss of one of the alleles, and ‘fixation’ of the other. alleles A B 5 5 6 4 8 2 7 3 9 1 10 0 In this example, the B allele is lost, and the A allele is ‘fixed’ in the population. From the starting point of equal allele frequencies it is equally likely that B would be fixed and A lost in a subsequent random sampling of the genetic material. The exact same process goes on in large populations, it just takes longer. alleles A B 500 500 495 505 510 490 ............. 1000 0, or A could be lost, and B ‘fixed’ in the population. The important point with regard to genetic drift is that because all populations are of finite size, evolution (in the sense of amino acid or nucleotide (allelic) changes) will always occur. One allelic type will eventually replace another. common ancestor: the consequence of finite population sizes and hence genetic drift effects is that we have a common ancestor for all our genes. It is a different common ancestor for each gene due to independent assortment of genes on different chromosomes (Mendel’s 2nd law), and recombination between genes on the same chromosome. Very closely linked genes could share a common ancestor. For a neutral allele at an autosomal locus in a diploid organism the time to trace back to the common ancestor is on average 4Ne generations where Ne is the effective population size. 5 The female ancestor of our mitochondrial DNA, which is inherited maternally and without recombination, has been traced to Africa about 200,000 years ago. Note that this woman is only the common ancestor for our mitochondrial DNA, and further this observation does not tell us what the population size was at that time, it certainly does not mean it was just this woman and one man. Our nuclear genes trace back to many other common ancestors, some presumably from this time period, some more recent, and others which are older. Because of genetic recombination of nuclear genes, it is much more difficult than with mitochondrial DNA to trace back to the common ancestor. The Y chromosome will trace back to a male common ancestor and studies on Y chromosome variation are in progress in a number of laboratories. founder effect: the change in allele frequencies when a new colony is formed by a very small number of founding individuals from a larger population. Alleles found in the general population may be absent from the founder population, e.g., of an island population, or a religious isolate, such as the Amish and Hutterite populations in the United States. On the other hand, by chance rare alleles from the general population may be more frequent in the founder population, e.g., a form of dwarfism in the Amish. bottleneck: drastic reduction in population size due, e.g., to over fishing or a natural disaster such as a hurricane, will lead to changes in allele frequencies. Problems 1. Consider five different genes in humans, one of which is mitochondrial (inherited maternally and independently of the nuclear chromosomes); the other four genes are on four separate nuclear chromosomes (autosomes) and have evolved independently. These five genes trace back to how many different ancestors? 2. Explain how it is possible for humans and chimps to share some polymorphisms, e.g., the ABO blood group alleles. 6 Answers to Problems: Genetic drift 1. One female common ancestor for the mitochondrial gene, four different common ancestors of either sex for the four nuclear genes. 2. We need to distinguish between gene trees and species trees. Closely related species may share polymorphisms, assuming both groups were polymorphic for the same variation when they split from their common ancestor, and have maintained the polymorphism since then. In the case of ABO blood groups, almost many primate species have the 3 alleles present, and it is most likely that there is some form of selection maintaining this polymorphism. When a new species forms, the population does not necessarily go through a bottleneck, and variation can be maintained. • Evolutionary genetics (ch. 19: 619-648) evolutionary trees from DNA data: A series of evolutionary changes involves accumulation of genetic change in the DNA. Even if there were no selection operating, i.e., mutations which arose were selectively neutral, because of the finite size of populations and consequent chance events, alleles always eventually become lost from a population, so there is eventual replacement of allelic types by another. The more distant two species the more genetic differences (amino acid changes or nucleotide changes) that will have accumulated between them. So, the longer the time since the organisms diverged, the greater the number of differences in the nucleotide sequence of the gene for, e.g., cytochrome c. Evolutionary trees drawn from DNA data agree well with those drawn from the fossil record, and can be important where convergent evolution of similar characteristics can cause confusion in drawing evolutionary trees based on the characteristics of organisms, and/or when the fossil record is poor. Human Gorilla Pig Rabbit Human 0 Gorilla 1 0 Pig 19 20 0 Rabbit 26 27 27 0 substitution times: the time required for substitution of different amino acids is long, in the millions of years. For example, with hemoglobin we expect on average 1 amino acid substitution every 5 million years. molecular clock: there is a regularity with time in the rate of accumulation of genetic changes at the nucleotide and amino acid levels. Different proteins show different rates of accumulation of changes. Some evolve very slowly, such as histones, others at a medium rate such as hemoglobin, and others faster such as fibrinopeptide. But for each protein, the molecular clock generally applies. 7