Question

Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advantage relative to b, so that the fitnesses of the three genotypes are BB = 1, Bb = 1, and bb = 1-s. In this case, s = 0.50, so that bb homozygotes have 50% fitness of heterozygotes and BB homozygotes. If the population has the following genotypic counts prior to selection of BB = 500, Bb = 250, and bb = 250, what is the expected change in the frequency of B after one generation with selection? Please give your answer to two decimal places.

Answer 1

Answer:

0.09

Explanation:

Before selection

Total number in population = 1000

Genotype frequencies

Genotype frequency of BB = 500/1000 = 0.5

Genotype frequency of Bb = 250/1000 = 0.25

Genotype frequency of bb = 250/1000 = 0.25

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.5 + (0.25/2) = 0.625

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.25 + (0.25/2) = 0.375

After selection

We are told that after selection, the genotype frequency of bb is changed as they become 50% less fit. This means the frequency of bb individuals changes from 250 to 125 individuals (50% reduction).

Now the total number of individuals is 500 + 125 + 250 = 875.

Genotype frequencies

Genotype frequency of BB = 500/875 = 0.57

Genotype frequency of Bb = 250/875 = 0.29

Genotype frequency of bb = 125/875 = 0.14

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.57 + (0.29/2) = 0.715

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.14 + (0.29/2) = 0.285

Change in frequency of B after 1 generation

0.715 - 0.625 = 0.09