Cystic Fibrosis Risk After a Negative Carrier Panel
Mele and Sera are both unaffected by cystic fibrosis (CF), an autosomal recessive condition, but each has a brother with CF. Neither has any other family history, and neither has been tested. The population carrier frequency is about 1 in 25. Their clinic offers a panel that detects the three most common CF variants, which together account for about 90% of CF alleles in the region. Both take the test; both results are negative. The counsellor tells them, 'Your baby's risk is now the same as anyone else's, about 1 in 2500.' Sera is 10 weeks pregnant. The couple asks whether they should test the fetus. Chorionic villus sampling (CVS) would give a definite answer at 11–13 weeks but carries a 0.5% risk of miscarriage. Both say they would consider ending the pregnancy if the fetus were affected, but they are anxious about the procedure. The counsellor's notes record only the test result and the population frequency, not the family history.
1. Show, using the family history and the test's detection rate, why the counsellor's '1 in 2500' is wrong. Calculate the couple's residual risk after both negative tests.
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Each parent is an unaffected sibling of an affected child. For an autosomal recessive condition, an unaffected sibling of an affected child has a 2/3 chance of being a carrier, because the possible genotypes are AA, Aa and aA; aa is excluded. A negative panel with 90% detection does not remove that. For each parent, P(carrier | negative) = (2/3 × 0.10) / [(2/3 × 0.10) + (1/3 × 1.0)] = 0.0667 / 0.400 = 1/6. The chance that both are carriers is (1/6)² = 1/36. If both are carriers, the chance of an affected child is 1/4. So the fetal risk is 1/36 × 1/4 = 1/144 ≈ 0.7%, or about 1 in 144. The counsellor's 1 in 2500 applies to a couple from the general population with no family history and is about 17 times too low for this couple.
2. Explain what the 90% detection rate means and why a negative result does not prove that a person is not a carrier.
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A 90% detection rate is the panel's sensitivity: it finds 9 of every 10 carrier chromosomes in this population, but 1 in 10 carriers has a variant the panel does not test. A negative result therefore lowers the probability of being a carrier; it does not make it zero. The higher the prior probability, the more carrier risk remains after a negative test. Here the prior is 2/3 because of the affected brothers, so the residual is 1/6. If the same negative test were applied to a person with prior 1/25, the residual would be about (1/25 × 0.10) / [(1/25 × 0.10) + (24/25)] ≈ 1/241, a very different result. Interpretation must start from the family history, not from the population frequency. The test also cannot detect variants outside its three targets, which is why full sequencing may be offered.
3. Using the numbers, advise the couple on whether CVS is justified. Refer to the risk of the procedure and the risk of CF, and note what else they should be told.
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With a fetal risk of about 0.7% (1 in 144) and a procedure-related miscarriage risk of 0.5% (1 in 200), CVS does not offer a large net safety gain: the chance of losing an unaffected pregnancy is of similar magnitude to the chance of finding an affected fetus. That does not make CVS wrong, but it means the couple's values matter. If they would definitely terminate an affected pregnancy, CVS gives information they would act on, and 0.7% is greater than 0.5%. Before CVS, the clinic should offer full sequencing of both parents' CF genes; if both disease-causing variants are found, the couple's risk can be defined much more precisely and the fetus can be tested with near-complete certainty for those variants. They should also be told that a negative CVS is highly reassuring but not perfect, and that the 0.5% figure is an average.