The given question is incorrect. The correct question is as follows:
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 frequency of B after one generation with selection? Please give your answer to two decimal places.
Answer:
0.63
Step-by-step explanation:
According to hardy Weinberg equilibrium the frequencies of the population remains stable from one generation to the next generation unless no selection or mutation is experienced by the population.
The frequency of B after one generation with selection can be calculated as follows:
Given, BB = 500, Bb = 250, and bb = 250.
Frequency (B) = 2 × BB + Bb / 2 × total number of individual.
Frequency (B) = 2 × 500 + 250 / 2 × 1000.
Frequency (B) = 1250 / 2000 = 0.625 = 0.63.
Thus, the frequency of B in population is 0.63.