In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele a is 0.1. What is the frequency of individuals with AA genotype? A) 0.42B) 0.20C) 0.32D) 0.81

The correct answer is D) 0,81

When there is Hardy-Weinberg equilibrium like in this case of a single locus with two alleles denoted A and a with frequencies f (A) = p and f (a) = q, the expected genotype frequencies under random mating are f (AA) = p² for the AA homozygotes, f (aa) = q² for the aa homozygotes, and f (Aa) = 2pq for the heterozygotes. And we have:

p²+2*p*q+q² = 1 p+q = 1 p=1-q

q = 0.1 p = 1-0.1 = 0.9

p²=0.9² = 0.81