Problem box answers

Initial allele frequencies set 1: p = 0.75, q = 0.20, r = 0.05 T = 0.679(0.75)2 + 0.153(0.2)2 + 1(0.05)2 + (0.763)2(0.75)(0.2) + (0.679)2(0.75)(0.05) + (0.534)2(0.2)(0.05) T = 0.382 + 0.00612 + 0.0025 + 0.2289

+ 0.051 + 0.0107 = 0.6812 TC = 1(0.05) + (0.679)(0.75) + (0.534)(0.2)

The C allele goes to loss because its marginal fitness is lower than the mean fitness.

Initial allele frequencies set 2: p = 0.7, q = 0.20, r = 0.1 T = 0.679(0.7)2 + 0.153(0.2)2 + 1(0.1)2 + (0.763)2(0.7)(0.2) + (0.679)2(0.7)(0.1) + (0.534)2(0.2)(0.1)

P 0.6792

The C allele goes to fixation because its marginal fitness is greater than the mean fitness, a condition that will hold until the C allele is fixed in the population.

Problem box 7.2 answer

At selection events #1 and #2, the incoming branch displaces the continuing branch. The incoming and continuing branches have identical states at selection event #3. The result is that all lineages have the state A in the present. In this case the height of the genealogy is shorter because of the outcome of selection event #1. See Figure 7.13.

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