A fruit fly population has a gene with two alleles, A1 and A2. Molecular genetic tests show that 40% of the gametes produced in the population contain the A1 allele. If the population is in Hardy-Weinberg equilibrium, what proportion of the flies are homozygous for the A1 allele?

= 0.48

Explanation:

the genetic frequency can be calculated using hardy - Weinberg equation

p + q = 1

here p represent the dominant trait and q represent the recessive trait.

the proportion A1 or p = 40/100

                                    = 0.4

now, the proportion of A2 or q will be

p + q = 1

0.4 + q = 1

q = 1- 0.4

q = 0.6

proportion of the flies are homozygous for the A1 allele

2pq

2 x 0.4 x 0.6

= 0.48

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