# 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.

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

**is 1 in 25 i.e 1/25 =**

*(q)***0.04**(given)

therefore ,

p + 0.04 = 1

hence ,

*= 1 - 0.04 =*

**p****0.09**

For a population in Hardy - Weinberg equilibrium

p

^{2 }+ q

^{2}+ 2pq = 1

So,

q

^{2 }= (0.04)

^{2}=

**0.0016**(frequency of homozygous recessive)

p

^{2 }= (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.

**
**