The incidence of phenylketonuria (PKU), an autosomal recessive genetic disorder, is approximately 1 in 4,500 in Ireland. Assuming Hardy- Weinberg equilibrium, what is the probability that a phenotypically normal Irish female, with no family history, and a phenotypically normal Irish male, with an affected sister, will have a child with PKU?
A) 0.028%
B) 0.49%
C) 1.96%
D) 0.55%
E) 2.2%

Question

Quizplus · Accepted Answer

To solve this, first, we calculate the frequency of the recessive allele (q) using the incidence of PKU (1 in 4,500 or q^2 = 1/4500). Solving for q gives us the square root of 1/4500, which is approximately 0.015. The frequency of the dominant allele (p) is 1 - q, which is approximately 0.985. A phenotypically normal female with no family history has a 2pq chance of being a carrier, which is 2 * 0.985 * 0.015 ≈ 0.0295 or 2.95%. The male, with an affected sister, has a 2/3 chance of being a carrier because we know he is not affected (thus, he must either be a carrier or not, and given his sister is affected, it's 2 chances out of 3 that he's a carrier if we exclude the possibility of him being affected). The probability that both carriers mate and have an affected child is 1/4. Therefore, the probability that their child will have PKU is 2.95% * 2/3 * 1/4 ≈ 0.49%.

The Incidence of Phenylketonuria (PKU), an Autosomal Recessive Genetic Disorder