7.03 Genetics: Answer Key to Problem Set 6 (2001)

7.03 PS #6 Answer Key Fall 2001 1. a) Let f(b) = q Population X: q2 = 1000 100000 q = 0.1 We can also determine the statistics for the rest of the population: f(B) = p = 0.9 f(B) = p = 0.9 f(b) = q = 0.1 BB p2 = 0.81 Bb pq = 0.09 f(b) = q = 0.1 Bb pq = 0.09 f(BB) = 0.81; f(Bb) = 0.18; f(bb) = 0.01 Population Y: q2 = 5000 50000 q = 0.316 f(B) = p = 0.684 f(b) = q = 0.316 BB p = 0.468 Bb pq = 0.216 f(B) = p = 0.684 f(b) = q = 0.316 2 Bb pq = 0.216 f(BB) = 0.468; f(Bb) = 0.432; F(bb) = 0.1 b) bb q = 0.01 2 bb q2 = 0.1 Heterozygous rabbits = 2pq x population size Population X: 2(.9)(.1)(100000) = 18,000 heterozygous rabbits Population Y: 2(.684)(.316)(50000) . 21,600 heterozygous rabbits c) In order to get a white-tailed offspring, the cross has to be Bb (father from Y) x bb (mother) p(bb offspring) = p(brown-tailed rabbit in pop Y is Bb) x p(having bb offspring) = .44 x .5 . 0.24 .9 d) The probability that two rabbits will have brown-tailed offspring is the same as 1 - p(white-tailed offspring) 1 - p(brown-tailed rabbit in pop X is Bb) x p(brown-tailed rabbit in pop Y is Bb) x p(bb offspring) 1 - .18 x .99 .44 .9 x .25 . 0.98 e) Since there is no force acting to change allele frequency (like mutation or selection), the allele frequency will remain equal to what it is when the two populations are combined. Total number of b alleles in XY = 200000(0.1) + 100000(0.316) Divide this by the total number alleles in XY (300000) . 0.17 = qXY f) Number of rabbits with brown tails = 1 - q2 = 1 - (.17)2 = 0.97 g) Since mating is random, it takes only one generation for genotype frequencies to adjust to expected Hardy-Weinberg equilibrium states. 2. a) Let q = f(CF) q2 = 1 1600 q = 1 = .025 40 b) The heterozygote advantage today is lower than it was hundreds of years ago. The reason for the advantage was resistance to an infectious disease that is no longer present today. Therefore, there is no selective force giving a heterozygote advantage today. c) Since there is no longer any (or perhaps very little) heterozygote advantage, the main force acting on CF allele frequency is selection against homozygotes. This will lower the CF allele frequency. Since CF is rare, the heterozygote frequency is approximately 2q, so the heterozygote frequency will be lowered as f(CF) decreases. d) 400 years ago, there was a case of balanced polymorphism. The CF allele was rare, and mutation was negligible, so ∆q = hq - Sq2 = 0. Then, the disease disappeared, so we can say that h became 0. Consequently, q began to change at a rate of -Sq2 per generation. S = 0.95 in this case, since CF individuals have only 5% as many offspring as normal individuals (fitness = 5%). Going from one generation to the next: q' = q - Sq2 We need to go backwards from the present, so we know q' and S and need to solve for q: .025 = q - .95q2 .95q2 - q + .025 = 0 q = 1 " % 1 - 4(.95)(.025) . .0256 2(.95) This is what f(CF) was one generation, or 20 years, ago. To get an exact answer for f(CF) 400 years ago, do this calculation 20 times - once for each generation: Years f(CF) -200 0.0331 0 (present) 0.025 -220 0.0342 -20 0.0256 -240 0.0354 -40 0.0263 -260 0.0367 -60 0.0270 -280 0.0380 -80 0.0277 -300 0.0395 -100 0.0285 -320 0.0411 -120 0.0293 -340 0.0429 -140 0.0301 -360 0.0448 -160 0.0311 -380 0.0469 -180 0.0320 -400 0.0491 Therefore, f(CF) 400 years ago was approximately 0.0491. e) f(CF/CF) = q2 . 0.0024 f) The CF allele is rare, and mutation is negligible, so we can approximate h: h = Sq = (.95)(.0491) . .047 3. a) Just like in first-cousin matings, the ?? offspring has a 1/16 probability of being heterozygous by decent at a given locus from one set of great-grandparents. However, there are two such sets of great-grandparents, so the inbreeding coefficient is 1/8. Another way of looking at this is the ?? offspring has a 1/64 chance of being homozygous for one allele from a great grandparent. There are 8 such alleles at a given locus, so the inbreeding coefficient is 8 x 1/64 = 1/8. b) Population 1 is in Hardy-Weinberg equilibrium, while population 2 has non-random mating. Population 1: Disease incidence = q2 = (.002)2 = 4 x 10-6 Population 2: Disease incidence = sum of probabilities from each type of mating p(disease from given mating) = inbreeding coefficient x q x p(that type of mating) p(disease from first-cousin mating) = (1/16)(.002)(.15) = 1.875 x 10-5 p(disease from double f-c mating) = (1/8)(.002)(.05) = 1.25 x 10-5 p(disease from random mating) = (.002)2(.80) = 3.2 x 10-6 Disease incidence = 1.875 x 10-5 + 1.25 x 10-5 + 3.2 x 10-6 = 3.445 x 10-5 c) Population 1: q2 = .0001 q = 0.01 Population 2: Same calculations as above, except we now know disease incidence instead of allele frequency. (1/16)(.15)q + (1/8)(.05)q + .8q2 = .0001 .8q2 + .015625q - .0001 = 0 q = -.015625 " %(.015625)2 - 4(.8)(.0001) . 5.08 x 10-3 2(.8) d) (Disease incidence due to first-cousin or double first-cousin mating)/(total disease incidence) Disease S: (1/16)(.15)(5.08 x 10-3) + (1/8)(.05)(5.08 x 10-3) = 7.94 x 10-5 . 0.79 .0001 .0001 Disease T: (1/16)(.15)(.002) + (1/8)(.05)(.002) = 3.125 x 10-5 . 0.91 3.445 x 10-5 3.445 x 10-5