Cystic fibrosis: a negative carrier test in a family with an affected child
A couple, both aged 32, have a 4-year-old daughter with cystic fibrosis. Neither parent has symptoms. The mother is now 10 weeks pregnant. Cystic fibrosis is caused by recessive mutations in the CFTR gene on chromosome 7; the most common mutation in Ireland is F508del, but more than 1,000 different mutations have been described. A genetic counsellor explains that a carrier test can look for a panel of common mutations but cannot detect every possible mutation. The couple's daughter has two identical F508del mutations. The father's brother also has cystic fibrosis; the mother has no family history. The mother takes the carrier test and the result is negative for all mutations on the panel. The couple ask: 'What is the chance this baby will have CF, and can you guarantee that the test will tell us?'
1. Explain why both parents must be carriers, and calculate the probability that the new baby will have cystic fibrosis.
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CF is autosomal recessive, so an affected child has two mutant alleles, one from each parent. The parents have no symptoms, so each must have one mutant allele and one normal allele: they are heterozygous carriers. The father's brother being affected is consistent with a family mutation, but the decisive evidence is the affected daughter. Use a Punnett square: Aa × Aa gives AA, Aa, Aa, aa. Only aa is affected, so the probability is 1 in 4 (25%) for each pregnancy, independent of the previous child. The 3 in 4 unaffected includes 1 AA and 2 Aa.
2. The mother's carrier test is negative. Explain how this can happen and what it means for the risk estimate.
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The test screens for a panel of common mutations, not the whole gene. The mother already has an affected child, so she must carry a CF-causing mutation unless the child's second mutation arose de novo, which is rare. A negative panel result is therefore a false negative caused by allelic heterogeneity: her mutation is not on the panel. The pregnancy risk remains about 1 in 4 because the family genotype, not the panel result, is the stronger evidence. The counsellor cannot guarantee a test result. The useful next step is to test the affected daughter and parents for the family mutations, then offer prenatal testing for those specific variants.
3. A healthy sibling is tested and does not have CF. What is the probability that this sibling is a carrier?
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This is a conditional probability. The possible genotypes from Aa × Aa are AA (1/4), Aa (1/2) and aa (1/4). We know the sibling is not aa, so we condition on the 3/4 that are unaffected. Among these, AA is 1 part and Aa is 2 parts, so P(carrier | unaffected) = 2/3. Answer: 2 in 3, not 1 in 2. The calculation matters for counselling because a healthy sibling is more likely than a random person to carry the family mutation.