Question

Phenylketonuria occurs in the population with a frequency of 1 in 2000. Calculate the frequency of the carrier genotype.

Answer 1

The carrier genotype is heterozygous. The Hardy-Weinberg equation is used to calculate genotype frequencies.

p² + 2pq + q² = 1 

where p² = frequency of homozygous dominant genotype, 2pq = frequency of heterozygous genotype, q² = frequency of homozygous recessive genotype.

The incidence of phenylketonuria in the population appears in individuals with the homozygous recessive genotype, hence q² is in 2,000 or 1/2000 = 0.0005

Therefore `q = sqrt0.0005`

= 0.0224

Since, p + q = l

⇒  p = 1 − q

= 1 − 0.0224

= 0.9776

The frequency of the heterozygous genotype (2pq) is therefore,

= 2 × (0.9776) × (0.0224)

= 0.044 = 1 in 23

= 5%

Approx. 5% of the population are carriers of the recessive gene for phenylketonuria.