A fruit 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?

0.42

Explanation:

According to Hardy-Weinberg equilibrium,

p + q = 1

p² + 2pq + q² = 1

where,

p = frequency of dominant allele

q = frequency of recessive allele

p² = frequency of homozygous dominant genotype

2pq = frequency of heterozygous genotype

q² = frequency of homozygous recessive genotype

Here,

p = A1 = 70% = 0.7

So, q =  A2

= 1 - p

= 1 - 0.7 = 0.3

Flies carrying both A1 and A2 are heterozygous flies so their frequency

= 2pq

= 2 * 0.7 * 0.3 = 0.42

Hence, the correct answer is 0.42.