Chapter 19 -- Population Genetics: Hardy-Weinberg Equilibrium...

P. 548, #1. One hundred persons from a small town in Pennsylvania were tested for their MN blood types. Is the population they represent in Hardy-Weinberg proportions? The genotypic data are: MM, 41; MN, 38; and NN, 21.The first step is to calculate the allele frequencies for M and N. 2*(# of MM) + (# of MN) 2*(41) + (38) freq(M) = ---------------------------- = ----------------- 2*(total # of individuals) 2*(41 + 38 + 21) freq(M) = 120/200 = 0.6 freq(N) = 1 - freq(M) = 1 - 0.6 = 0.4 The next step is to calculate the expected frequencies of each of the genotypes given that the population is in Hardy-Weinberg proportions. freq(MM) = 0.6^2 = 0.36 freq(MN) = 2*0.6*0.4 = 0.48 freq(NN) = 0.4^2 = 0.16 The expected numbers of each genotype are determined by multiplying the expected frequency of each genotype by the number of individuals in the sample. expected number of MM = 0.36 * 100 = 36 expected number of MN = 0.48 * 100 = 48 expected number of NN = 0.16 * 100 = 16 The final step is to compare the expected and observed distributions of numbers with chi-square analysis. This will test the null hypothesis that the population is in Hardy-Weinberg proportions. MM MN NN observed numbers 41 38 21 expected numbers 36 48 16 ((obs-exp)^2)/exp 0.69 2.08 1.56 chi-square value = 0.69 + 2.08 + 1.56 = 4.33 The degrees of freedom for the critical value is 1 (number of phenotypes - number of alleles) = (3-2) = 1 Thus the critical value to reject is 3.814. As the calculated value is greater than the critical value, we reject the null hypothesis that our population is in Hardy-Weinberg proportions.P. 548, #2. In the following two sets of data, calculate allelic and genotypic frequencies and determine whether the populations are in Hardy-Weinberg proportions. Do a statistical test if one is appropriate. a. Allele A is dominant to a: A-, 91; aa, 9.In this problem we have to assume that the population is in Hardy-Weinberg to be able to estimate the frequencies of the alleles, thus we will not be able to test if the population is in Hardy-Weinberg. First, freq(aa) = freq(a) x freq(a) = q^2 = 9/100 = 0.09. As q^2 = 0.09, taking the square root of both sides yields q = 0.30. As q = 0.30, then p = 1 - q = 1 - 0.3 = 0.7. Thus freq(A) = p = 0.7. Second, given that p = 0.7 and q = 0.3, then freq(AA) = p^2 = 0.7^2 = 0.49 freq(Aa) = 2pq = 2 x 0.7 x 0.3 = 0.42 freq(aa) = 9/100 = q^2 = 0.09 As the number of phenotypes is 2 and the number of alleles is 2, the degrees of freedom is (2-2) 0. Thus, we can not do a test for Hardy-Weinberg proportions because we need at least 1 degree of freedom to do a chi-square test.b. Electrophoretic alleles F and S are codominant at the alcohol the dehydrogenase locus in Drosophila: FF, 137; FS, 196; SS, 87.As we have three phenotypes and two alleles, we can do a chi-square test for Hardy-Weinberg proportions, thus this problem is just like problem #1. The first step is to calculate the allele frequencies for F and S. 2*(# of FF) + (# of FS) 2*(137) + (196) freq(F) = ---------------------------- = ----------------- 2*(total # of individuals) 2*(137 + 196+ 87) freq(F) = 470/840 = 0.56 freq(S) = 1 - freq(S) = 1 - 0.56 = 0.44 The next step is to calculate the expected frequencies of each of the genotypes given that the populations is in Hardy-Weinberg proportions. freq(MM) = 0.56^2 = 0.3136 freq(MN) = 2*0.56*0.44 = 0.4928 freq(NN) = 0.44^2 = 0.1936 The expected numbers of each genotype is determined by multiplying the expected frequency of each genotype by the number of individuals in the sample. expected number of FF = 0.3136 * 420 = 131.7 expected number of FS = 0.4928 * 420 = 207.0 expected number of NN = 0.1936 * 420 = 81.3 The final step is to compare the expected and observed distributions of numbers with chi-square analysis. This will test the null hypothesis that the population is in Hardy-Weinberg proportions. MM MN NN observed numbers 137 196 87 expected numbers 131.7 207.0 81.3 ((obs-exp)^2)/exp 0.213 0.585 0.400 chi-square value = 0.213 + 0.585 + 0.400 = 1.198 The degrees of freedom for the critical value is 1 (number of phenotypes - number of alleles) = (3-2) = 1 Thus the critical value to reject is 3.814. As the calculated value is less than the critical value, we fail to reject the null hypothesis that our population is in Hardy-Weinberg proportions.P. 548, #3. The dominant ability to taste PTC is due to the allele T. Among a sample of 215 individuals from a population, 150 could detect the taste of PTC and 65 could not. Calculate the frequencies of T and t. Is the population in Hardy-Weinberg proportions?This problem is very similar to Problem 2a because we have only two phenotypes and two alleles and thus we have to assume that the population is in Hardy-Weinberg to be able to estimate the frequencies of the alleles, thus we will not be able to test if the population is in Hardy-Weinberg. First, freq(tt) = freq(t) x freq(t) = q^2 = 65/215 = 0.302. As q^2 = 0.302, taking the square root of both sides yields q = 0.55. As q = 0.55, then p = 1 - q = 1 - 0.55 = 0.45. Thus freq(T) = p = 0.45. Second, given that p = 0.45 and q = 0.55, then freq(TT) = p^2 = 0.45^2 = 0.2025 freq(Tt) = 2pq = 2 x 0.45 x 0.55 = 0.495 freq(tt) = 65/215 = q^2 = 0.302 As the number of phenotypes is 2 and the number of alleles is 2, the degrees of freedom is (2-2) 0. Thus, we can not do a test for Hardy-Weinberg proportions because we need at least 1 degree of freedom to do a chi-square test.P. 548, #4. The frequency of children homozygous for the recessive allele for cystic fibrosis is about one in 2,500. What is the percentage of the heterozygotes in the population?This problem is very similar to Problem 2a because we have only two phenotypes (normal and with cystic-fibrosis) and two alleles (the cystic fibrosis allele (c) and the normal allele (C)) and thus we have to assume that the population is in Hardy-Weinberg to be able to estimate the frequencies of the alleles. Thus we will not be able to test if the population is in Hardy-Weinberg. First, freq(cc) = freq(c) x freq(c) = q^2 = 1/2500 = 0.0004. As q^2 = 0.0004, taking the square root of both sides yields q = 0.02. As q = 0.02, then p = 1 - q = 1 - 0.02 = 0.98. Thus freq(C) = p = 0.98. Second, given that p = 0.98 and q = 0.02, then freq(CC) = p^2 = 0.98^2 = 0.9604 freq(Cc) = 2pq = 2 x 0.98 x 0.02 = 0.0392 freq(cc) = 1/2500 = q^2 = 0.0004P. 548, #8. PTC tasting is dominant in human beings. a. Should most human populations be heading toward a 3:1 ratio of tasters to nontasters? Explain.You can assume that most human populations are in Hardy-Weinberg equilibrium at this locus and thus that allele or genotype frequencies are not changing and thus not heading towards any given ratio. In addition, we would expect a 3:1 ratio only if the alleles were in equal frequency (p = q = 0.5). There is no reason to suspect that this might be true.b. Confronted with a population sample of human beings of unknown origin, would you expect more or less than half the sample would be tasters.Given that we do not know what the frequency of the T allele is and that T is dominant, we would expect more individuals to be tasters than nontasters because the frequency of tasters would be p^2 + 2pq and this is less than 0.5 only when p is less than 0.30. If we assume a uniform distribution of allele frequencies across all human populations, then we could assume that populations with more tasters would be more likely to occur than populations with more nontasters.P. 549, #16. What allelic frequency will generate twice as many recessive homozygotes as heterozygotes?Algebraically, we need to solve the equation freq(aa) = 2 * freq(Aa) Given that the population is in Hardy-Weinberg proportions then, q^2 = 2*(2*p*q) which we then solve for q q^2 = 4*p*q, substitute 1-q for p q^2 = 4*(1-q)*q q^2 = 4*q - 4*q^2, dividing through by q q = 4 - 4*q, subtracting 4 from both sides and q from both sides gives -4 = -5*q, dividing through by -5 -4/-5 = q 0.8 = q.P. 549, #20. On a small island, 235 mating individuals are all true-breeding for brown eyes. An epidemic eliminates all the population except 10 young women, two young men, and four older (postmenopausal)women. A boatload of foreigners arrives: the foreign population consists of six heterozygous brown-eyed females, four homozygous brown-eyed males, and 10 blue-eyed males. Assuming that eye color is controlled by one locus, that mating is random with respect to eye color, and that each male and female capable of breeding does so, calculate the allelic frequencies of their offspring.

After the epidemic (the elderly survivors of the epidemic are not part of the breeding populations) and with the arrival of the foreigners, we have in the resulting breeding population,survivors foreigners 10 BB women 10 bb men 2 BB men 4 BB men 6 Bb womenWe need to determine the allele frequencies in the breeding males and females. Thus2(10 for BB) + (6 for Bb) freq(B) in the women is --------------------------- = 0.8125 2(total women) freq(b) in the women is 1 - 0.8125 = 0.1875 and 2(6 for BB) + (0 for Bb) freq(B) in the men is --------------------------- = 0.3750 2(total men) freq(b) in the men is 1 - 0.3750 = 0.6250Now to calculate the expected frequency of each genotype in the offspring of these individuals, we need to assume that Hardy-Weinberg will hold. Thus the frequency of each genotype is given by the frequency of each gamete from the male gamete pool and from the female gamete pool. Thusfreq(BB) = freq(B) for males x freq(B) for females = 0.3750 * 0.8125 = 0.3047 freq(Bb) = freq(B) for males x freq(b) for females plus freq(b) for males x freq(B) for females = 0.3750 * 0.1875 + 0.6250 * 0.8125 = 0.0703 + 0.5078 = 0.5781 freq(bb) = freq(b) for males x freq(b) for females = 0.6250 * 0.1875 = 0.1171Having now determined the frequency of each genotype, we can use that information to calculate the allele frequency in the populations.freq(B) = freq(BB) + 0.5 * freq(Bb) = 0.3047 + 0.5 * 0.5781 = 0.5938 freq(b) = freq(bb) + 0.5 * freq(Bb) = 0.1171 + 0.5 * 0.5781 = 0.4062P. 549, #22. In a given population, only theAandBalleles are present in the ABO system; there are no individuals with type O blood or withOalleles. If two hundred people have type A blood, seventy-five have type AB blood, and twenty-five have type B blood, what are the alleleic frequencies of this population?

To calculate the allele frequencies forAandB, we need to remember that the individuals with type A blood are homozygousAA, individuals with type AB blood are heterozygousAB, and individuals with type B blood are homozygousBB.2*(# of AA) + (# of AB) 2*(200) + (75) freq(A) = ---------------------------- = ----------------- 2*(total # of individuals) 2*(200 + 75 + 25) freq(A) = 475/600 = 0.792 freq(B) = 1 - freq(B) = 1 - 0.792 = 0.208P. 550, #24. In a sample of 100 people, there are 14 MM, 32 MN, and 54 NN individuals. Calculate the inbreeding coefficient.

We need to use the formula for the inbreeding coefficient, which isexpected # of heterozygotes - observed # of heterozygotes F = -------------------------------------------------------- expected # of heterozygoteswhere the expected # of heterozygotes is given by Hardy-Weinberg proportions. So the first thing that we need to do is to calculate the frequency of each allele and then the expected number of heterozygotes. To calculate the allele frequencies for M and N,2*(# of MM) + (# of MN) 2*(14) + (32) freq(M) = ---------------------------- = ----------------- 2*(total # of individuals) 2*(14 + 32 + 54) freq(M) = 60/200 = 0.3 freq(N) = 1 - freq(M) = 1 - 0.4 = 0.7The next step is to calculate the expected frequencies of each of the genotypes given that the population is in Hardy-Weinberg proportions.

freq(MM) = 0.3^2 = 0.09

freq(MN) = 2*0.7*0.3 = 0.42

freq(NN) = 0.7^2 = 0.49

The expected numbers of each genotype is determined by multiplying the expected frequency of each genotype by the number of individuals in the sample.

expected number of MM = 0.09 * 100 = 9

expected number of MN = 0.42 * 100 = 42

expected number of NN = 0.49 * 100 = 49

Now we use the formula for the inbreeding coefficient that we introduced above.expected # of heterozygotes - observed # of heterozygotes F = -------------------------------------------------------- expected # of heterozygotes F = (42-32)/42 = 0.238P. 550, #25. If, in a population with two alleles at an autosomal locus, p = 0.8, q = 0.2, and the frequency of heterozygotes is 0.20, what is the inbreeding coefficient?

We need to use the formula for the inbreeding coefficient, which isexp. freq of heterozygotes - obs. freq of heterozygotes F = -------------------------------------------------------------- expected freq of heterozygotes where the expected freq of heterozygotes is given by Hardy-Weinberg proportions. Thus, F = (2pq - 0.20)/2pq = (0.32 - 0.20)/0.32 F = 0.375

Last updated on 18 November 1996.