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

i believe it would cause an increase of percipitation, though i could be wrong. so i would say b is the answer.

i was thinking 4.0 v hope this