Question

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?

a. 0.7 b. 0.49 c. 0.42 d. 0.21

Answer 1

Option C

Step-by-step explanation:

Given ,

A1 allele is carried by
`70` % people

Let us assume A1 s dominant genotype

This means
`p= (70)/(100) = 0.7`

Thus, frequency of allele in the given population is
`0.7`

It is also given that the population is in Hardy-Weinberg equilibrium thus

`p+q=1\\0.7+q=1\\q = 1-0.7\\q= 0.3`

Frequency of fruit fly with genotype A2A2 will be

`q^2\\= (0.3)^2\\= 0.09`

As per Hardy-Weinberg's second equation of equilibrium

`p^(2) + q^(2) + 2pq = 1\\0.49+0.09+2pq = 1\\2pq = 1-0.49-0.09\\2pq= 0.42`

Hence, option C is correct