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

Answer:

Option C

Explanation:

Given ,

A1 allele is carried by $70$ % people

Let us assume A1 s dominant genotype

This means $p= \frac{70}{100} = 0.7$

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