# MCB142/IB163 Thomson Mendelian and Population Genetics Sep.16, Mendelian and Population Genetics Equations, Tables, Formulas, etc.

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1 MCB142/IB163 Thomson Mendelian and Population Genetics Sep.16, 2003 Practice questions lectures 6-13 Mendelian and Population Genetics Equations, Tables, Formulas, etc. confidence limits of an allele frequency estimate: denoting by p the true frequency of allele A, the variance of the allele frequency estimated by the method of gene (allele) counting is p(1-p)/(2n) where n is the sample size. An estimate of this variance is obtained using this formula and the estimated allele frequency. The standard deviation (sd) is the square root of the variance. The 95% confidence interval for the range of the true allele frequency p is twice the sd on either side of the estimate. Theoretical Chi-square values with different probabilities: Degrees Probability Of Freedom 0.05* 0.01** 0.001*** degrees of freedom (df): when the expected values are independent of any features of the observed data, then the df is given by the number of genotype classes considered (with expected value >=5) minus 1. When the observed data is used in calculations to determine the expected values, e.g., for testing HWP the allele frequencies must be estimated from the observed data to determine the expected HWP for that data set, then the df is as above minus the number of independent parameters estimated (which is k-1 for a k allele system). linkage disequilibrium (LD): Consider two genes with two alleles each (A, a and B, b), and hence four gametic types (referred to as haplotypes): AB, Ab, ab, and ab. The frequencies of the four gametes (denoted f(ab) etc.) can be described in terms of the allele frequencies p A (p a = 1 - p A ) and p B (p b = 1 - p B ) at the two loci, and the LD parameter D, as follows: f (AB) = p A p B + D, f (Ab) = p A p b - D, f (ab) = p a p B - D, f (ab) = p a p b + D, where D = f(ab) - p A p B. 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 - s s2 then the equilibrium frequency for allele A1 is p1* = s2/(s1 + s2), and that for allele A2 is p2* = s1/(s1 + s2), with p1* + p2* = 1. Page 1

2 Question 1 You are studying an X-linked trait. There are two alleles, one showing complete dominance over the other. In females, 84 percent show the dominant phenotype. What percentage of the males will show the dominant phenotype, assuming that individuals in this large population are in Hardy Weinberg proportions, and that the allele frequencies are equal in males and females? Question 2 Human albinism is an autosomal recessive trait. Suppose that you find a village in the Andes where 1/4 of the population is albino. If the population size is 1000 and the population is in Hardy-Weinberg proportions with respect to this trait, how many individuals are expected to be heterozygotes? Question 3 An RFLP probe shows four different morphs (alleles) in a population of three-spined sticklebacks (a fish Gasterosteus aculeatus). How many different heterozygous genotypes will there be in this population with regard to this RFLP? Question 4 Suppose that you are a genetic counselor, and a couple seeks your advice about BHT tasting (people who can taste BHT in processed food are recessive homozygotes for the "taster" allele and non-tasters are homozygous dominant or heterozygous for the "non-taster" allele). Both prospective parents are non-tasters, but a careful analysis of the husband's pedigree reveals that he is a carrier of the taster allele. The proportion of BHT tasters in the population is 16 percent. What is the probability that the couple's first child will be a non-taster of BHT assuming the population is in Hardy Weinberg proportions? Question 5 In a population under study, the frequency of an autosomal recessive disease is 1 percent. Assuming Hardy Weinberg proportions, what is the probability of two heterozygotes mating. Among all the cases of the disease in the population, what percentage of these cases resulted from heterozygous matings? Question 6 In a Pygmy group in central Africa, the frequencies of alleles determining the ABO blood groups were estimated as 0.2 for I A, 0.1 for I B, and 0.7 for i (allele O). Assuming Hardy Weinberg proportions, determine the expected frequencies of the blood types A, B, AB, and O. (Alleles A and B are codominant, and O is recessive.) Page 2

3 Question 7 In a Northern European population, the frequencies of alleles at the HLA A gene are: Alleles A1 A2 A3 A11 A29 Frequencies Assume Hardy Weinberg proportions (HWP) to answer all the following questions. In a population of size 1,000 individuals: (a) how many are expected to be heterozygotes A2A29? (b) how many are expected to be heterozygotes A1AX, where AX denotes all alleles except A1? (c) How many total homozygotes are expected, i.e., counting all homozygous classes? Question 8 In a test of Hardy Weinberg proportions for a gene with 3 alleles, and hence 6 genotype classes, determine whether the chi-square value given below for a number of different cases is compatible with the null hypothesis, or rejects the null hypothesis of Hardy Weinberg proportions, in which case give the P value. You will need to determine the degrees of freedom (df) in each case. (a) The chi-square value is 12.5 and all expected values for the 6 genotype classes are >5. (b) The chi-square value is 4.3, one genotype class with expected value <5 was combined with another, for a total of 5 genotype classes. (c) The chi-square value is 20.5, two genotype classes with expected values <5 were combined together (the expected value was then >5). (d) Describe what the P value means in each of the cases where the null hypothesis was rejected. Question 9 Perform a chi-square goodness of fit test of Hardy Weinberg proportions (HWP) for the genotype data given below for a single gene with 2 codominant alleles. Determine the following quantities (a.-c.) and describe (d.) (show all work): (a) The total chi-square value (b) The degrees of freedom (df) (c) Whether the chi-square value indicates that the observed genotype values are compatible with the null hypothesis of HWP, or if they are significantly different from HWP, in which case give the P value. (d) Briefly describe what a P value represents when a test is significant, i.e., the null hypothesis is rejected. Observed data Genotypes AA AB BB Observed (O) Total = 100 Page 3

4 Question 10 In samples from three different populations, the following genotypes were observed for the codominant alleles A 1 and A 2 : Population A 1 A 1 A 1 A 2 A 2 A (a). Calculate the frequency of alleles A 1 and A 2 in each population. (b). What are the Hardy-Weinberg frequencies and expected genotypic numbers for the three populations. (c). Calculate the chi-square value for test to fit to Hardy-Weinberg proportions for these populations, and indicate if the populations are statistically different from Hardy-Weinberg proportions. Question 11 A plant of genotype C/C ; d/d is crossed to c/c ; D/D and an F 1 testcrossed to c/c ; d/d. (a) If the genes are on separate chromosomes, what percentage of c/c d/d offspring will be seen? (b) If the genes are linked, and 20 map units apart, what percentage of c/c d/d offspring will be seen? Question 12 In Drosophila, the two genes w and sn are X-linked and 25 map units apart. A female fly of genotype w + sn + /w sn is crossed to a male from a wild-type line. What percent of male progeny will be w + sn? Question 13 A female fruit fly of genotype A/a ; B/b is crossed to a male fruit fly of genotype a/a ; b/b. The progeny of this cross were Genotype Number of individuals A/a ; B/b 38 a/a ; b/b 37 A/a ; b/b 12 a/a ; B/b 13 (a) What gametes were produced by parent 1 and in what proportions? (b) Do these proportions represent independent assortment of the two genes? (c) What can be deduced from these proportions? (d) Draw a diagram or explain the origin of the two rarest progeny genotypes. Page 4

5 Question 14 In mice the allele for color expression is C (c = albino). Another gene determines color (B = black and b = brown). Yet another gene modifies the amount of color so that D = normal amount of color and d = dilute (milky) color. Two mice that are C/c ; B/b ; D/d are mated. What proportion of progeny will be dilute brown (assume complete dominance at each locus and independent assortment of the genes)? Question 15 In pigs, when a pure-breeding red is crossed to a pure-breeding white the F 1 are all red. However, the F 2 shows 9/16 red, 1/16 white, and 6/16 are a new color, sandy. (a) Provide a genetic hypothesis for this observation (b) How would you test this hypothesis? Question 16 The genes Dm/dm, Lu/lu, and Sc/sc are inherited autosomally in humans. In a large population, you find women who are phenotypically dominant for all three traits whose parents consisted of one phenotypically dominant individual (for all three traits) and one phenotypically recessive individual (for all three traits). You observe the progeny of all such women whose husbands have a recessive phenotype for these three traits, and obtain the following data Phenotypes Number of individuals Dm Lu Sc 407 Dm Lu sc 4 Dm lu sc 22 Dm lu Sc 72 dm Lu Sc 18 dm lu Sc 3 dm Lu sc 70 dm lu sc 404 1,000 (a) What is the map order of the three loci? (b) What are the map distances between the three markers? Page 5

6 Question 17 In corn, the alleles c + and c result in colored (c + ) versus colorless (c) seeds, wx + and wx in nonwaxy (wx + ) versus waxy (wx) endosperm, and sh + and sh in plump (sh + ) versus shrunken (sh) endosperm. Plants grown from seeds heterozygous for each of these pairs of alleles (from crosses of a pure breeding colorless waxy parent and pure breeding shrunken parent) were testcrossed with plants homozygous for colorless, waxy, shrunken seeds. The progeny seeds were as follows: colorless, waxy, plump 334 colored, nonwaxy, shrunken 339 colorless, nonwaxy, plump 105 colored, waxy, shrunken 98 colorless, nonwaxy, shrunken 55 colored, waxy, plump 61 colorless, waxy, shrunken 5 colored, nonwaxy, plump 3 Total 1,000 (a) Determine the map order of the three genes, and the recombination fraction (RF) between each of the three pairs of genes. (b) If the three genes were on three separate chromosomes, what numbers (or ratios) would be expected for each of the 8 classes of progeny seeds? Question 18 For the following 3 examples (I) (III) of mapping disease genes in humans, indicate whether they are an example of: (A) positional cloning (B) functional cloning (C) candidate gene approach (The same answer may apply to more than one example.) (I) Sickle-cell anemia (an autosomal recessive disease) was demonstrated to be due to a mutation in the β chain of the hemoglobin gene (Hb-β) and the Hb-β gene was later mapped (II) In a screen of genetic markers across the genome, Huntington disease (an autosomal dominant disease with variable age of onset) was shown to be linked to an RFLP marker on chromosome 4. (III) Genes of the immune response HLA complex are studied for linkage and association with autoimmune and infectious diseases and cancers. Page 6

7 Question 19 Most genetic variation for many human loci lies within local populations rather than between populations or races. What does this observation tell you about human genetic evolution? Question 20 Most genetic variation for many human loci lies within local populations rather than between populations or races. There are some contradictions to this general trend, such as hair and skin color, and sickle cell anemia. What do these exceptions tell you about the evolution of those traits? Question 21 Compare and contrast the neutral, balance, and evolutionary lag schools. Question 22 What are the consequences on the frequency of the heterozygote class of complete assortative mating among three genotypes, A 1 A 1, A 1 A 2, and A 2 A 2? Question 23 For each of the following sets of genotype relative fitness values, predict whether a stable polymorphism will be maintained (in which case give the equilibrium frequency of the two alleles) or whether one or the other of the alternate alleles will go to fixation (in which case state which allele will become fixed in the population). Assume that the population size is very large and that selection is the only force acting to alter the frequencies of the alleles t this locus. AA AB BB (a) (b) (c) Question 24 At a locus with two alleles A and B, the homozygous AA individual has a fitness of 0.5 relative to the heterozygote AB, while the BB genotype is lethal. Under this selection scheme (a) what is the equilibrium allele frequency of A? (b) what are the three genotype frequencies expected at zygote formation with this equilibrium allele frequency, assuming random mating? Question 25 What will determine the rate of a molecular clock for a gene that codes for a functional protein? Page 7

8 Question 26 A molecular biologist determines the DNA sequence for a particular gene in 4 different species (labeled a, b, c, and d) and finds the following nucleotide differences between each pair of species a b c d a b c - 30 d - From the fossil record she knows that species a and c diverged 20 Myr. ago. If the rate of nucleotide substitutions is constant over years and among lineages, how long ago did species c and d diverge? Question 27 To what extent is evolution a chance process and to what extent a directed process, i.e., how does random change contribute to evolution? How, if at all, is evolution a directed process? Question 28 In the thousands of years that dogs have been domesticated, breeders have developed numerous breeds with many desirable traits. In many breeds of dogs, however, undesirable characteristics have appeared as well. The undesirable features include single gene traits such as retinal degeneration or deafness. Why do these undesirable traits persist? Page 8

9 Answers to Practice questions lectures 5-12 Question 1 Let allele A represent the dominant allele, with frequency p in both males and females, and a the recessive allele, with frequency q, with p + q =1. We are given the information that in females 84% have the dominant phenotype. This means that 16% have the recessive phenotype aa, with expected frequency q 2, so using the square root formula (we can t use allele counting since we cannot distinguish the heterozygotes from the dominant homozygote class) under the assumption that the population is in Hardy Weinberg proportions, the estimate of q = 0.4, hence p = 0.6. In males, the frequency of the dominant phenotype is equal to the allele frequency, hence the answer is 0.6. Question 2 We represent the autosomal recessive allele for albinism by a, with frequency q, and the dominant allele is represented by A, with frequency p, with p + q =1. We are told to assume Hardy Weinberg proportions, thus we can use the square root formula to estimate q. Since q 2 = 0.25, the estimate of q = 0.5. The expected proportion of heterozygous individuals is 2pq = 0.5, thus in the population of size 1,000 we expect 500 individuals to be heterozygotes, and hence carriers of the disease predisposing allele. Question 3 Denoting the four alleles as A, B, C, and D, then there are 6 heterozygous genotypes, AB, AC, AD, BC, BD, and CD. (There are 4 homozygous genotypes possible: AA, BB, CC, and DD.) Question 4 Denote the recessive taster allele by t (frequency q), so that tasters are tt (frequency q 2 ). We are given that the husband is heterozygous, hence Tt. We are given that the mother is a non-taster, so she is either TT or Tt. Assuming Hardy Weinberg proportions, we estimate q using the square root formula: given that q 2 = 0.16, then the estimate is q = 0.4, hence p = 0.6. The frequency of homozygous TT individuals in the population is p 2 = 0.36, and the frequency of heterozygotes Tt is 2pq = 2(0.6)(0.4) = Given that the woman is a non-taster, i.e., she is either TT or Tt, the conditional probability that she is TT is 0.36/0.84 = , and the probability she is Tt, is 0.48/ 0.84 = We have to consider both possible genotypes of the mother, since a non-taster child is possible both if she is TT or Tt. Thus the required probability is 1 (the probability the man is Tt) x (the probability the woman is TT given she is a non-taster) x 1 (the probability that their first child will be a non-taster (TT or Tt), given these genotypes) (= ) + 1 (the probability the man is Tt) x (the probability the woman is Tt given she is a nontaster) x 3/4 (the probability that their first child will be a non-taster (TT or Tt) given these genotypes) (= ), thus the required probability is (= , note that it is a coincidence of the given allele frequencies that the probabilities from both possible genotypes of the woman are the same, usually these would be different). A quicker way to get the answer, is to calculate the probability their first child will be a taster = 1 (the probability the man is Tt) x (the probability the woman is Tt given she is a nontaster) x 1/4 (the probability that their first child will be a taster (tt), given these genotypes = , and thus the required probability is = Page 9

10 Question 5 Assuming Hardy Weinberg proportions, we use the square root formula to calculate the allele frequency q of the recessive allele, denoted by a: given q 2 =0.01, q = 0.1, hence p = 0.9. The number of heterozygous carriers (Aa) is thus 2pq = 0.18, and the frequency of heterozygous x heterozygous matings (assuming mating is at random) is (2pq) x (2pq) = Since ¼ of the children from such a mating will be affected (frequency = /4 = ), this represents 81% of the total cases of all affected children which overall have a frequency of 1%. Question 6 f(a) = (0.2) 2 + 2(0.2)(0.7) = 0.32 f(b) = = (0.1) 2 + 2(0.1)(0.7) = 0.15 f(ab) = 2(0.2)(0.1) = 0.04 f(oo) = (0.7) 2 = 0.49 (check that the sum of the genotype frequencies adds to 1) Question 7 (a) The expected HWP of the heterozygote A1A29 is 2(0.3)(0.1) = 0.06, In a population of size 1,000 there would be 60 A2A29 heterozygotes expected under HWP. (b) The frequency of AX is 0.8 (= 1 0.2), so the expected HWP is 2(0.2)(0.8) = 0.32 In a population of 1,000 there would be 320 A1AX heterozygotes expected under HWP. (c) The frequency of all homozygotes expected under HWP is Σpi 2 where pi is the frequency of the ith allele and summation is over all alleles, i = 1-5 in this case, = (0.2) 2 + (0.3) 2 + (0.1) 2 + (0.3) 2 + (0.1) 2 = 0.24, In a population of 1,000 there would be 240 total homozygotes expected under HWP. Question 8 (a) df = 3 (=6 1 2, 6 classes minus 1 since we know the total count, minus 2 allele frequency estimates, the third allele frequency is known once we know 2 of the allele frequencies), the chi-square value of 12.5 is between the P<0.01 and P<0.001 values for a chi-square distribution with 3 df (see table from handout), so we reject the null hypothesis with P<0.01. (b) df = 2 (5 1 2), the chi-square value of 4.3 with 2 df is compatible with the null hypothesis, so these data are not significant. (c) df = 2 (5 1 2), the chi-square of 20.5 with 2 df is very significantly different from the null hypothesis (fit to Hardy Weinberg proportions) with P< (d) The P value represents the type 1 error, which is the probabaility that you have rejected the null hypothesis when it was true, i.e., the probability you have made a mistake in rejecting the null hypothesis. So in (a) the type 1 error is 0.01, while in (c) it is smaller at Page 10

11 Question 9 Allele frequency estimates by allele (gene) counting p(a) = [29 + (62)/2]/100, OR = /2, OR = [2(29) + 62]/200 = 0.6, q(b) = 0.4 Expected (E) Total = 100 (O E) Sum = 0s (O E) 2 /E (a) Chi-square = 8.51 (b) df = 1 (= 3 1 1), since 1 allele frequency is estimated (c) significant, P<0.01 (d) The P value is the type 1 error, the probability you have rejected the null hypothesis (fit to HWP in this case) when in fact it is true, i.e., it is the probability of getting a value this extreme under the null hypothesis (the probability you are wrong in rejecting the null hypothesis). Question 10 (a) and (b) Population p q p 2 2pq q (c) p 2 N 2pqN q 2 N X n.s p< p<0.05 Question 11 (a) All F1 individuals will be C/c D/d and will produce equal numbers (since the genes are on different chromosomes) of the 4 gametic types CD, Cd, cd, and cd. Since the testcross individual only produces gametes of type cd, the proportion of c/c d/d offspring will be 25%. (b) All F1 individuals will again be C/c D/d. However, since the genes are now linked (r = 20 cm = 20% recombination) apart, and we know the parental gametic types are Cd and cd, the 4 gametic types CD, Cd, cd, and cd are expected in proportions r/2, (1-r)/2, (1-r)/2, and r/2 respectively, where r = 0.2. Since the testcross individual only produces gametes of type cd, the proportion of c/c d/d offspring will be 10%. Page 11

12 Question 12 The female will produce four gametes: the 2 parental w + sn + and w sn will occur with equal frequencies of (1-r)/2 where r = 0.25, i.e., each with 37.5% frequency, while the 2 non-parental (recombinant) gametes w + sn and w s + will each occur with a frequency of 12.5% (check total number of gametes = 100%). Since males have only one X chromosome which they receive from their mothers, the percent of male progeny that will be w + sn is 12.5%. Question 13 (a) The proportions were 38% AB, 37% ab, 12% Ab, and 13% ab. (b) No, because we expect 25% of each under independent assortment, and these numbers are very different from that. (One could do the chi-square test to show lack of independent assortment, but it is not asked for in this question.) (c) The two genes must be linked; the %R (RF) = ( ) / 100 = 25%, i.e., the two loci are 25 map units (25 cm) apart on the same chromosome. (d) The female parent received gametes AB and ab from her two parents (she is AB/ab), the male parent received ab gametes from both his parents (he is ab/ab), and the two rarest progeny classes represent recombinant classes from the female parent. Question 14 We need to determine the probability of a C/- ; bb ; dd offspring from this cross = ¾ x ¼ x ¼ = 3/64. Question 15 (a) The presence of a dominant allele at either of two loci (A/- ; b/b or a/a ; B/-) gives sandy, where red is determined by the dominant alleles of both loci (A/- ; B/-), and white by the recessive genotype at both loci (a/a ; b/b). (b) Do testcrosses of a large number of the sandy offspring with the white offspring. Under this hypothesis, the sandy offspring are A/A ; b/b (1/16) (gametes Ab) or A/a ; b/b (2/16) (gametes ½ Ab and ½ ab) or a/a ; B/B (1/16) (gametes ab) or a/a ; B/b (2/16) (gametes ½ ab and ½ ab), i.e., three types of gametes with equal frequencies: Ab, ab, and ab, so the offspring from the testcrosses with (a/a ; b/b) should be equal proportions of A/a ; b/b (sandy) a/a ; B/b (sandy) a/a ; b/b (white), i.e., 2/3 sandy and 1/3 white. (One should also perform other crosses to confirm the model, including testcrosses and crosses of the putative genotypes of specific sandy offspring.) Page 12

13 Question 16 (a) The non-recombinant gametes from the mother are Dm Lu Sc and dm lu sc, while the double recombinant gametes (the least frequent class) are Dm Lu sc and dm lu Sc, thus the order of the three loci is Dm Sc Lu or Lu Sc Dm. (b) The Dm Sc recombination distance is ( ) / 1,000 = 4.7% (= 4.7 mu = 4.7 cm). The Sc Lu recombination distance is ( ) / 1,000 = 14.9%. The Dm Sc recombination distance is ( x x 2) / 1,000 = 19.6%. Correct answer must include double recombinants for Dm Sc, and check that the recombination distances are consistent, i.e., = Question 17 c + colored and c colorless wx + nonwaxy and wx waxy sh + plump and sh shrunken colorless, waxy, plump 334 c wx sh + c sh + wx colored, nonwaxy, shrunken 339 c + wx + sh c + sh wx + colorless, nonwaxy, plump 105 c wx + sh + c sh + wx + colored, waxy, shrunken 98 c + wx sh c + sh wx colorless, nonwaxy, shrunken 55 c wx + sh c sh wx + colored, waxy, plump 61 c + wx sh + c + sh + wx colorless, waxy, shrunken 5 c wx sh c sh wx colored, nonwaxy, plump 3 c + wx + sh + c + sh + wx + Total 1,000 (a) (i) Map order The parental gametes (most common) and also from given parental information are colorless, waxy, plump (c wx sh + ) and colored, nonwaxy, shrunken (c + wx + sh The double recombinant class (two rarest classes) are colorless, waxy, shrunken (c wx sh) and colored, nonwaxy, shrunken (c + wx + sh + ) Thus the middle gene is shrunken, and gene order is colorless (c), shrunken (sh), waxy (wx), or vice versa. Correct answer re gene order can also be obtained by working out the 3 pairwise recombination fractions without correcting for double recombinants, which gives the largest distance for the pair c wx, if this approach is used, the corrected recombination distance including twice the double recombinants should also then be worked out for next part of question. Page 13

14 (ii) Recombination distances c sh pairwise RF fraction: colorless (c), plump (sh + ) = 439 colored (c + ), shrunken (sh) = 437 colorless (c), shrunken (sh) = 60 colored (c + ), plump (sh + ) = 64 Total 1,000 RF = ( )/1,000 = 124/1,000 = = 12.4% (= 12.4 mu = 12.4 cm) wx sh pairwise RF fraction: nonwaxy (wx + ), shrunken (sh) = 394 nonwaxy (wx + ), plump (sh + ) = 108 waxy (wx), shrunken (sh) = 103 waxy, plump = 395 Total 1,000 RF = ( )/1,000 = 211/1,000 = = 21.1% (= 21.1 mu = 21.1 cm) c wx pairwise RF fraction: colorless (c), nonwaxy (wx + ) = 160 colorless (c), waxy (wx) = 339 colored (c + ), waxy (wx) = 159 colored (c + ), nonwaxy (wx + ) = 342 Total 1,000 To determine RF need to include double recombinants with the single recombinant class, RF = ( (5) (3))/1,000 = 335/1,000 = (= 33.5% = 33.5 mu = 33.5 cm), and check that = 33.5 cm. (b) 3 genes unlinked If all 3 genes were unlinked, we expect equal numbers in each of the 8 classes, i.e., 125 in each class (1:1:1:1:1:1:1:1 ratio). Question 18 (I) B (II) A (III) C Functional cloning (disease function gene- map) Positional cloning (disease map gene function) Candidate gene genes with a known or proposed function with the potential to influence the disease phenotype are investigated for a direct role in disease. Question 19 It suggests that humans have had persistent migration between populations that has prevented localized differentiation for most genes Page 14

15 Question 20 They most likely reflect adaptations to local conditions. Dark skin is more adaptive in warm sunny climates, whereas light skin may be more adaptive when exposure to sun is minimal. The sickle cell anemia allele provides protection against malaria (in heterozygotes and homozygotes), and so is at high frequency in malarial environments, despite the deleterious effects of the allele in homozygous individuals with sickle cell anemia. Other genes which do not contribute to local adaptation would not show differentiation (the majority of genes). Question 21 The neutral school maintains that much of the genetic variation in populations represents allelic variants that are selectively equivalent to each other and polymorphisms seen in a population represent a balance of new mutations entering the population, some of which will replace older alleles by drift (chance) effects. In this way genetic changes accumulate between distinct species. The balance school maintains that much of the genetic variation in populations has adaptive significance. The evolutionary lag school argues that much of the variation seen in populations may represent transient variation, as selectively advantageous alleles replace older alleles in a race to keep up with changes in the environment. It is difficult, if not impossible, to know what fraction of the genetic variation in populations falls into each of these three categories, and all we can say is that each contributes to genetic variation. Question 22 This mating is equivalent to self-fertilization. The heterozygous class would decrease by half each generation. Question 23 (a) A will be fixed (b) Balanced polymorphism, f(a) = p1 = s2/(s1 + s2) = 0.4/0.6 = 2/3 (s1 = 0.2, s2 = 0.4), f(b) = p2 = s1/(s1 + s2) = 0.2/0.6 = 1/3. (c) A will be fixed. Question 24 (a) Using the formula to determine equilibrium allele frequencies with heterozygote advantage, s1 = 0.5, and s2 = 1.0, so the equilibrium frequency of A = p1 = s2/(s1 + s2) = 1/1.5 = 2/3 (and the equilibrium frequency of B = 0.5/1.5 = 1/3). (b) At zygote formation, the expected genotype frequencies are the Hardy Weinberg proportions 4/9 AA, 4/9 AB, and 1/9 BB. Page 15

16 Question 25 The degree to which changes in the protein will decrease the fitness of the organism. If the protein is very sensitive to change (cannot function optimally with even small changes in its amino acid composition) then the molecular clock will be extremely slow. If the protein can function adequately even with many changes, perhaps to sites that are not immediately associated with the active site on the protein, then the molecular clock will be relatively fast. Question Myr ( = (30/10) x 20 Myr Question 27 The ultimate source of genetic variation is mutation, so the building blocks of evolution come from chance events. Also, mutations which are selectively equivalent to existing alleles in the population can become established in the population by genetic drift (chance) effects. Thus, even without direct selection, evolution will occur in populations, since they are always of finite size and subject to drift effects. Evolution is not a directed process in terms of having a purpose or striving for improvement. Since there are many conceivable ways to succeed at surviving and reproducing, natural selection lacks direction in a specific sense. But there are more ways of failing to survive and reproduce than there are to succeed at any point in time, and therefore natural selection provides some direction in moving populations away from traits that fail and towards the traits that succeed Question 28 Development of a new breed of any plant or animal frequently includes inbreeding, and this may expose rare recessive alleles in the gene pool. This is a special problem when breeders start with only a few animals and create entire stocks from their descendants. Correlated responses to selection also contribute to the development of undesirable characteristics when selecting for desirable characteristics. Selection for a nicely shaped head or a particular leg length in a dog may have undesirable effects on other skeletal features. Genetic linkage and association (linkage disequilibrium) of desirable alleles at some loci with undesirable alleles at other loci contribute greatly to the correlated selection of two or more different traits. Page 16

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