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Modern Genetic Analysis Chapter 4. The Inheritance of Genes
Solved Problem

2. | Phenylketonuria (PKU) is a rare autosomal recessive disease. A couple wants to have children but consults a genetic counselor because the man has a sister with PKU and the woman has a brother with PKU. There are no other known cases in their families. They ask the genetic counselor to determine the probability that their first child will have PKU. What is this probability? See answer

Solution

What can we deduce? If we let the allele causing the PKU phenotype be p and the respective normal allele be P, then the man's sister and woman's brother must have been p/p. In order to produce these affected individuals, all four grandparents must have been heterozygous normal (P/p). The pedigree can be summarized as follows:

The only way the couple can have a child with PKU is if both of them are heterozygotes (it is obvious that they themselves do not have the disease). Both the grandparental matings are simple monohybrid crosses (P/p × P/p) expected to produce progeny in the following proportions:

We know that the man and the woman are normal, so for each of them the probability of being a heterozygote is 2/3, because within the normal P/ class, 2/3 are P/p and 1/3 are P/P. Therefore, using the product rule, the probability of both the man and the woman being heterozygotes is 2/3 × 2/3 = 4/9. If they are both heterozygous, then one-quarter of their children would have PKU, so the probability that their first child will have PKU is 1/4, and the probability of their being heterozygous and of their first child having PKU is 4/9 × 1/4 = 4/36 = 1/9, which is the answer to the question.

See question

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