Question

A population of 200 fruit flies is in Hardy Weinberg equilibrium. The frequency of the allele (a) 0.4. Calculate the following:

The number of carrier fruit flies.

Answer 1

In Hardy Weinberg, p² + 2pq + q² = 1, where 'p²' is the frequency of the homozygous dominant genotype (AA), '2pq' is the frequency of the heterozygous genotype (Aa), and 'q²' is the frequency of the homozygous recessive genotype (aa).

Given:

q = 0.4

We know p + q = 1

p = 1 − q

= 1 − 0.4

= 0.6

Number of carrier fruit flies is

As, 2pq (Aa)

= 2 × 0.6 × 0.4

= 0.48

= 0.48 × 200

= 96