Question

A couple with European ancestry seeks genetic counseling before having children because of a history of cystic fibrosis (CF) in the husband's family. ASO testing for CF reveals that the husband is heterozygous for the \(\Delta 508\) mutation and that the wife is heterozygous for the \(R 117\) mutation. You are the couple's genetic counselor. When consulting with you, they express their conviction that they are not at risk for having an affected child because they each carry different mutations and cannot have a child who is homozygous for either mutation. What would you say to them?

Short answer

Answer: The probability of the couple having an affected child is 25%, as the child could inherit one mutated allele from each parent resulting in a compound heterozygote for the disease. This occurs even though the parents carry different cystic fibrosis mutations (\(\Delta 508\) and \(R 117\)).

Step-by-step solution

Understanding cystic fibrosis mutations

Cystic fibrosis is an autosomal recessive genetic disorder, meaning that a person must inherit two copies of the mutated gene (one from each parent) to be affected. In this case, the husband is heterozygous for the \(\Delta 508\) mutation, and the wife is heterozygous for the \(R 117\) mutation. Being heterozygous means that they each carry one normal allele and one mutated allele.

Examine the possible outcomes for child's genotype

To understand the possible outcomes for the couple's potential children, we need to examine all possible combinations of the two alleles from each parent. Using the Punnett square, we can see the following possible genotypes for the couple's child: 1. Normal allele from both parents (unaffected) 2. Normal allele from father and \(R 117\) mutation from mother (unaffected carrier) 3. \(\Delta 508\) mutation from father and normal allele from mother (unaffected carrier) 4. \(\Delta 508\) mutation from father and \(R 117\) mutation from mother (affected)

Calculating the probability of having an affected child

From the four possible genotypes listed above, only one of them results in an affected child (combination of both mutated alleles). The probability of this outcome is \(\frac{1}{4}\), as there is a 50% chance for each parent to pass on their mutated allele, and to have an affected child, both mutated alleles must be inherited simultaneously (0.5 x 0.5 = 0.25).

Explaining the risk to the couple

Contrary to the couple's belief, they can indeed have a child who is affected by cystic fibrosis, even if they carry different mutations. Their child would inherit one mutated allele from each parent, resulting in a compound heterozygote for the disease. They have a 25% chance of having an affected child, a 50% chance of having an unaffected child who is a carrier of the mutation, and a 25% chance of having an unaffected child with no mutated alleles. It is important to discuss these risks with the couple and help them make informed decisions regarding family planning.