###### Answer: The expected genotype frequencies are 0.64 0.32 and 0.04 for A1A1 A1A2 and A2A2 respectively. (Why: the expected frequency of the A1A1 genotype is p^2=(0.8)(0.8)=0.64; the expected frequency of the A1A2 genotype is 2pq=2(0.8)(0.2)=0.32; the expected frequency of the A2A2 genotype is q^2=(0.2)(0.2)=0.4. To verify your calculations confirm that the three frequencies add up to one.

What genotype frequencies are expected under Hardy-Weinberg equilibrium for a po…