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Practice Problems: Population Genetics 1. The gamma globulin of human blood serum exists in two forms, Gm(a+) and Gm(a-), spedified respectively by an autosomal dominant gene Gm(a+) and its recessive allele Gm(a-). Broman et al. (1963) recorded the tabulated phenotypic frequencies in three Swedish populations. Assuming the populations were at Hardy-Weinberg equilibrium, calculate the frequency of heterozygotes in each population. Region No. Tested Phenotype % Norbotten County Stockholm city and rural district Malmohus and Kristianstad counties 139 509 Gm(a+) 55.40 57.76 Gm(a-) 44.60 42.24 293 54.95 45.05 Since the populations ARE in equilibrium, this problem can be solved by defining the proportion of individuals displaying the recessive phenotype (GM (A-) as q2. You can then calculate q, and using 1-q = p, you can calculate p. Finally, the frequency of heterozygotes in a population in equilibrium is 2pq. Norbotten County: Stockholm Malmohus q2 = 0.446 q = 0.668 p = 0.332 2pq = 0.444 2pq = 0.455 2pq = 0.442 2. Among 2,820 Shorthorn cattle, 260 are white, 1,430 are red, and 1,130 are roan. Is this consistent with the assumption that the traits are controlled by a single pair of autosomal alleles and that mating has been at random for this allele pair? To determine if a population is in equilibrium, you must calculate the allele frequencies, use those allele frequencies to calculate the expected number of white, red and roan cattle, and then use chi-square analysis to compare the expected numbers with the observed numbers. First: R1 = red allele R2 = white allele Allele frequencies: p = f(R1) = (1430 + ½(1130)) / 2820 = 0.71 q = 0.29 Expected # Red = p2 x total cattle =( 0.712 )( 2820) = 1421.562 Expected # Roan = 2pq x total cattle = 2(0.71)(0.29)(2820) = 1161.276 Expected # White = q2 x total cattle = (0.292)(2820) = 237.162 c =å 2 (O - E ) = 2 E (1430 -1421.562) 1421.562 2 + (1130 -1161.276) 1161.276 2 + (260 - 237.162) 2 237.162 = 0.05 + 0.84 + 2.199 = 3.089 df = 3-1 = 2 critical value = 5.991 Accept the hypothesis; The traits appear to be controlled by a single pair of autosomal genes under random mating. 3. On the basis of allele-frequency analysis of data from a randomly mating population Snyder (1934) concluded that the ability vs. inability to taste phenylthiocarbamide (PtC) is determined by a single pair of autosomal alleles, of which T for taster is dominant to T for nontaster. Of the 3,643 individuals tested in this population, 70% were tasters and 30% were nontasters. Assume the population satisfies the conditions of HardyWeinberg equilibrium. a) Calculate the frequencies of the alleles T and T and the frequencies of the genotypes TT, TT and TT. q2 = 30% = 0.30 q = f(T) = 0.55 p = f(T) = 0.45 p2 = f (TT) = 0.20 2pq = f (TT) = 0.50 b) Determine the probability of a nontaster child from a taster x taster mating. Probability (tt) child = P(Tt parent) x P(Tt parent) x P (tt child) However, we know the parents are not tt, so we must calculate the frequency of Tt parents within the taster population (not the population at large) as follows: = frequency Tt/ total frequency of tasters (Tt + TT) = 0.50 / 0.50 + 0.20 = 0.71 So P(tt child from taster parents) = 0.71 x 0.71 x 0.25 = 0.126 4. In Drosophila melanogaster, Cncn (red vs. cinnabar eyes), Bb( gray vs. black body, (and Byby (normal vs. blistery wing) are autosomal pairs of alleles. Samples of three large natural adult populations, each classified for a different pair of traits, are found to have the following genotypes: Population A Population B Population C 31 cncn 182 BB 100 ByBy 171 Cncn 391 Bb 372 Byby 60 CnCn 152 bb 40 byby. Total 262 Total 725 Total 512 Compare these distributions with those expected for a population at HardyWeinberg equilibrium. Propose a reasonable explanation to account for any differences. Population A: p = (60 + 85.5)/262 = 0.56; q = 0.44 Expected CnCn = p2 = 0.31 x 262 = 82 Expected Cncn = 2pq = 0.49 x 262 = 129 Expected cncn = q2 = 0.19 x 262 = 51 Population B: p = 0.52; q = 0.48 Expected BB = 196 Expected Bb = 362 Expected bb = 167 Population C: p = 0.56 ; q = 0.44 Expected ByBy = 161 Expected Byby = 252 Expected byby = 99 Observed 60 Observed 171 Observed 31 Observed BB = 182 Observed = 391 Observed = 152 Observed = 100 Observed = 372 Observed = 40 Without doing Chi-square analysis, all three populations show increased heterozygosity compared to the expected, as well as decreased homozygosity of both genotypes. There may be selection for the heterozygote, or outbreeding (you’d need to see the same pattern of increased heterozygotes at all genes) or negative assortative mating (if this pattern only existed for one gene and closely linked genes). All three of these situations would give an increase in heterozygotes. As a side note, whatever is going on is much more pronounced in population C. 5. A yak population is in Hardy-Weinberg equilibrium with allele frequencies p(A) = 0.5 and q(a) = 0.5 for a gene governing color differences. If a new type of predator appears in the area, calculate the new values of q if: a) sa/a = 1.0 b) sa/a = 0.70 c) sa/a = 0.10 To solve this problem, you must first calculate the fitness values for each recessive genotype (1-S = W). Since AA and Aa are not mentioned, they have a fitness of 1.0. Then, calculate the mean population fitness using: W = p2WAA + 2 pqWAa + q 2Waa Then calculate the new values for q using the formula: qa = q2Waa + pqWAa W W a)Waa = 1-1=0 W .52 (1) 2(.5)(.5)(1) .52 (o) 0.75 qa = .52(0) + (.5)(.5)(1) 0.75 0.75 = 0.33 b) Waa = 1-.7=0.3 W .52 (1) 2(.5)(.5)(1) .52 (.3) 0.825 qa = .52(0.3) + (.5)(.5)(1) 0.825 0.825 = 0.39 c) Waa = 1-.1 = 0.9 W .52 (1) 2(.5)(.5)(1) .52 (.9) 0.975 qa = .52(0.9) + (.5)(.5)(1) 0.975 0.975 = 0.49 6. If the frequency of an allele d is 0.25 in a migrant population and 0.5 in The recipient population, and if the migration rate is 0.1, what is the frequency of d in the recipient population after one generation of migration? = -0.025 new p = 0.5-0.025 = 0.475