PRACTICE PROBLEMS 5Answers1. (a) H = 1 - [(0.5)2 + (0.52)] = 0.50 or 2 X 0.5 X 0.5 = 0.5 (b) H = 0.0198 (c) H = 0.80 (d) H = 0.187The largest value of expected heterozygosity is reached when there are many alleles ofsimilar frequencies. When one allele is in great excess, H is low.2. (a) f(c1) = [309 + (24/2) + (1/2) + (1/2)]/335 = 0.9612f(c2) = (1/2)/335 = 0.0015f(c3) = (1/2)/335 = 0.0015f(c4) = (24/2)/335 = 0.0358 Check: 0.9612 + 0.0358 + 0.0015 + 0.0015 = 1(b) H = 0.0748(c) Observed heterozygosity is (24+1+1)/335 = 0.078, very close to H, suggesting thatthe population is random mating.3.(a) 1 - 0.001. With no additional information about the second codon and the roleplayed by the amino acid for which it codes, our best guess is that diversity at thatcodon is representative of diversity in the gene as a whole.(b) 1, because this is part of the invariant ATG start codon.4.(a) 0, because all mutations in a pseudogene are neutral. (b) 1, or nearly so, because the mutants can grow whereas the wild type fitness is 0 ornearly 0.(c) > -1 and < 0. The answer is not 0; you can tell that the mutation is detrimentalbecause bacteria normally live in the absence of antibiotic but never become entirelyresistant as they would if resistance mutations were neutral and occasionally wouldbe fixed by drift.5. (a) = (b) <, because zoo populations of animals are small and small populationsare forced to inbreed.6. Using H = 4Neu/(1 + 4Neu):(a) 0.1667 (b) 0.9524 (c) 0.9995 (d) The last two values are much larger than areobserved in real life; the first value of Ne is more likely.7.(a) This question can be understood as asking the probability that out of 20 offspringfrom a cross of A a X A A, what is the probability that none of the offspring will getthe mutant allele? The probability that any one offspring will get it is 1/2, so theanswer is (1/2)20 = 9.54 X 10-7.(b) The probability of fixation is 1/2N = 1/100, and the probability of loss is 1 - 0.01 =0.99.8.(a) No, mutation rates are similar for all sites in a gene, because the factors that causemutations (replication errors, mutagens) are the same regardless of the function of asite.(b) No, synonymous and nonsynonymous refer only to coding sequences.(c) No, promoters aren't transcribed.(d) Yes; functionally important sequences are difficult to change without bad effect, soan especially large fraction of mutations are detrimental instead of
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