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 <em>p+q= 1 p=1-q</em>
q= 0.1 p= 1-0.1= 0.9
p²=0.9²= 0.81
It should be translation.
Answer:
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Explanation:
