Question

PKU (Phenylketonuria) is an autosomal recessive disease, in which the synthesis of amino acid Tyrosine from Phenylalanine is blocked. As a result, an excess of Phenylalanine gets converted into phenylketones, which appear in the urine. In severe conditions it may also result in damage to the brain. The gene responsible for this is p, whereas the gene P is responsible for normal synthesis of Tyrosine. In a small population of Brazilian natives, the frequency of gene p, responsible for this disease, is 0.3. What must be the frequency of people who are heterozygous for this disease? ( p + q = 1, p2 + 2pq + q2 = 1) (A) 0.56
(B) 0.35
(C) 0.42

Answer 1

To calculate the frequency of the heterozygote genotype (Pq) for this gene we must use the Hardy-Weinberg equation ( p2 + 2pq + q2 = 1 ). This equation relies on the Hardy-Weinberg principle, a model in population genetics that states that the frequency of the alleles in a population is never changing, only the combinations (the genotypes) are changing.

If there are only two alleles (variations) of this gene in a population, then their frequencies should add up to 1 (100%). From this, we can calculate the frequency of the q allele.

p +q=1

0,3 +q=1

q= 1-0,3

q= 0,7

Now hat we have the frequency of the q allele we can use the HW equation to calculate the frequency of the heterozygotes.

$p^{2} + 2pq + q^{2} = 1$

$(0,3)^{2} + 2pq +(0.7)^{2} = 1$

0,09 + 2pq +0.49= 1

2pq +0,58= 1

2pq= 1-0.58

2pq=0,42

The freqency of the heterozygotes in this population is 0.42