Question

A population of 200 individuals is in Hardy Weinberg equilibrium for a gene with only two alleles - B and b. If the gene frequency of an allele B is 0.8, the genotype frequency of Bb is

• A. 0.64
• B. 0.4
• C. 0.04
• D. 0.32

Answer 1

The correct option is D 0.32

Frequnce of allele B = p = 0.8
Frequence of allele b = q =?
We know that, p + q = 1
$\therefore$ q = 1 - 0.8 = 0.2

According to Hardy Weinberg principle,
$p^2+2pq+q^2=1$, where
$p^2\to$ Frequency of BB
$q^2\to$ Frequency of bb
2pq $\to$ Frequency of Bb
$2pq=2\times 0.8\times 0.2$
= 0.32