Afruit fly population has a gene with two alleles, a1 and a2. tests show that 70% of the gametes produced in the population contain the a1 allele. if the population is in hardy-weinberg equilibrium, what proportion of the flies carry both a1 and a2? a) 0.49b) 0.7c) 0.42d) 0.21

The correct answer is C)0.42

When there is Hardy-Weinberg equilibrium like in this case of a single locus with two alleles denoted A1 and A2 with frequencies f(A1) = p and f(A2) = q, the expected genotype frequencies under random mating are f(A1A1) = p² for the A1A1 homozygotes, f(A2A2) = q² for the A2A2 homozygotes, and f(A1A2) = 2pq for the heterozygotes. We have:

p+q= 1      q=1-p

Frequency of allele A1 is 70%=0.7    p=0.7

Frequency of allele A2 is q=1-0.7=0.3

Flies that carry both A1 and A2 alleles are heterozygotes (2pq)

Proportion is 2*0.7*0.3=0.42

sorry, but you do not have posted the diagram.