Question

If one in 25 carries a recessive allele what must be the frequency of the homozygous recessive genotype in a population at hardy- Weinberg equilibrium.​

Answer 1

Dear Student,

Let the frequency of dominant allele be p
Let the frequency of recessive allele be q
also, we know that
p + q = 1

The frequency of recessive allele (q) is 1 in 25 i.e 1/25 = 0.04 (given)
therefore ,
p + 0.04 = 1
hence , p = 1 - 0.04 = 0.09

For a population in Hardy - Weinberg equilibrium
p² + q² + 2pq = 1

So,
q² = (0.04)² = 0.0016 (frequency of homozygous recessive)
p² = (0.96)² = 0.9216 (frequency of homozygous dominant)
2pq = 2 x 0.04 x 0.96 = 0.0768 (frequency of heterozygous)

therefore,
Frequency of homozygous recessive genotype is 0.0016

Regards.