Answer:
Option C
Explanation:
Given ,
A1 allele is carried by [tex]70[/tex] % people
Let us assume A1 s dominant genotype
This means [tex]p= \frac{70}{100} = 0.7[/tex]
Thus, frequency of allele in the given population is [tex]0.7[/tex]
It is also given that the population is in Hardy-Weinberg equilibrium thus
[tex]p+q=1\\0.7+q=1\\q = 1-0.7\\q= 0.3[/tex]
Frequency of fruit fly with genotype A2A2 will be
[tex]q^2\\= (0.3)^2\\= 0.09[/tex]
As per Hardy-Weinberg's second equation of equilibrium
[tex]p^{2} + q^{2} + 2pq = 1\\0.49+0.09+2pq = 1\\2pq = 1-0.49-0.09\\2pq= 0.42[/tex]
Hence, option C is correct
