Question

If the frequency of p = 0.8, what is the frequency of the heterozygous genotype?

Responses

0.16

0.16

0.32

0.32,

1.6
1.6

3.2

Answer 1

0.32

Explanation: According to Hardy Weinberg principal,

we know that p+q=1 for genotype calculations

if p=0.8,therefore q=0.2

hence,

p and q here are homozygous genotype

2pq is heterozygous genotype

according to question ,using formulea

p^2+q^2+2pq=1 for genotype

hence,

2pq=2*0.8*0.2

2pq=0.32

hence the frequency of heterozygous is =0.32