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

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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

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