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 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? 0.7 0.09 0.21 0.42 0.49

Hardy Weinberg principle in genetics asserts that in the lack of evolutionary influence the alleles or the recurrence of genotypes will be unchanged.

It is calculated by using the formula:

p²+2pq+q²=1

Where,

The correct answer is:

option D. 0.42

The explanation for this is:

p+q=1

Where,

70% of the given population contains a1 allele so,

p= 70% = 0.7

If p=0.7 and p+q=1

Then,

q = 1 - p  

= 1 - 0.7  

= 0.3

Hence the population of the flies that carries a1a2 genotype will be:

= 2pq

= 2 × 0.7 × 0.3  

= 0.42

Therefore, the population of flies containing both a1 and a2 alleles is 0.42.

To learn more about Hardy-Weinberg equilibrium follow the given link:

link

this is a good question! i'll do some research and come back when i have the answer!