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

The answer is; C

Use the Hardy Weinberg equiation for population in assumed equilibrium. ( p+q = 1)

If p = 0.3 then q=1-p = 1-0.3 = 0.7

People who are heterozygous for the disease are represented by 2pq = 2 * 0.3 * 0.7

0.42