Question

About one child in 2,500 is born with phenylketonuria (an inability to metabolize the amino acid phenylalanine). This is known to be a recessive autosomal trait.

A. If the population is in Hardy-Weinberg equilibrium for this trait, what is the frequency of the phenylketonuria allele?

B. What proportion of the population are carriers of the phenylketonuria allele (that is what proportion are heterozygous)?

Answer 1

A. The frequency of the PKU allele (q) is 0.02 or 2%.

B. The proportion of the population that are carriers (heterozygous for PKU) is approximately 3.92%.

Step-by-step explanation:

The question is asking about phenylketonuria (PKU), an autosomal recessive disorder. We want to calculate the allele frequency of the PKU allele and determine the proportion of heterozygous carriers in the population.

Frequency of the PKU allele

Given that about one child in 2,500 is born with PKU and following the Hardy-Weinberg principle, we can denote the frequency of the homozygous recessive genotype (q²) as 1/2,500 or 0.0004. We need to calculate the square root of this number to find the frequency of the allele 'q', which would be the frequency of the PKU allele.

q = √0.0004 = 0.02

Proportion of carriers for PKU (heterozygotes)

Now, we want to calculate the proportion of heterozygous carriers (2pq). Given that 'q' is 0.02, we can deduce 'p' as well (p = 1 - q = 0.98). Using the Hardy-Weinberg equation for heterozygotes:

2pq = 2(0.98)(0.02) = 0.0392

So, approximately 3.92% of the population are carriers of the PKU allele.