Solutions For An Introduction to Genetic Analysis Chapter 18 Problem 39P

631190-18-39P AID: 2498 | 27/12/2015

a.

In the given data, the allele frequencies and relative fitness of the current generation are given.

So, genotypic frequency of

B/B = p² = (0.95)² = 0.9025

Genotypic frequency of

b/b = q² = (0.5)² = 0.0025

Genotypic frequency of

B/b = 2p² = (0.95)(0.5) = 0.095

To get the relative contributions, multiply the obtained values by the relative fitnesses.

Use the resulting genotype frequencies to calculate the allele frequencies in the next generation using the Hardy–Weinberg law.

Repeat the same to find the p”, i.e, 0.9545, and q” i.e, 0.0455

b.

If the dominant B/B individuals survived and reproduced, then there no chance for recessive b allele to get into the next generation, so the b allele would be eliminated after one generation.

c.

In part a, B is the dominant allele, so the fitness of the B/B and B/b individuals are favored and their fitness is more than the b/b individuals. Here, heterozygotes also survive and reproduce at the same rate as B/B homozygotes, thus the recessive b allele is maintained.

In case of part b, heterozygotes and true recessive individuals have a relative fitness of 0, so only B/B individuals only survive, and thus are able to contribute to the next generation.