An Autosomal Recessive Diagnosis: Conditional Probability in Genetic Counselling
A couple, both unaffected, have a son diagnosed with an autosomal recessive metabolic disorder; enzyme testing confirms that each parent carries one pathogenic allele. The mother is pregnant again, and the couple also have a healthy daughter aged four. The population carrier frequency for this allele is about 1/50. A newborn screening test for the disorder detects 95% of affected infants and gives a false positive result in 2% of unaffected infants. The parents ask the clinical geneticist three things: whether the new baby is certain to be affected because a sibling is; how to interpret a positive screening result; and what the chance is that their healthy daughter will eventually have an affected child with a partner drawn from the general population.
1. State the probability that the next child is affected, and calculate the probability that the healthy four-year-old daughter is a carrier. Explain why the carrier probability is not 1/2.
Read the worked answer
Both parents are heterozygotes (Aa × Aa), since each carries one pathogenic allele and is unaffected. Each child therefore has independent probabilities: 1/4 aa (affected), 1/2 Aa (carrier), 1/4 AA (unaffected, non-carrier).
Next child: P(affected) = 1/4 (25%) - the previous affected child does not change this, because each conception is an independent event. The parents' belief that a sibling makes it 'certain' confuses independence with tendency.
Healthy daughter: we are told she is unaffected, so the aa outcome is excluded. The remaining possibilities conditional on being unaffected are AA (1/4) and Aa (1/2), total 3/4. Hence P(carrier | unaffected) = (1/2)/(3/4) = 2/3 ≈ 67%. The answer is not 1/2 because the question is asked after conditioning on her healthy phenotype; among unaffected children, carriers outnumber non-carriers two to one. This 2/3 applies to any unaffected sibling, not only to this daughter.
2. The newborn's screening test returns positive. Calculate the probability that the infant is actually affected, and explain what the 95% figure does and does not mean.
Read the worked answer
Use Bayes' theorem with the prior from the pedigree, P(affected) = 0.25. P(+ | affected) = 0.95 (sensitivity); P(+ | unaffected) = 0.02 (false-positive rate).
P(+) = 0.25 × 0.95 + 0.75 × 0.02 = 0.2375 + 0.015 = 0.2525. P(affected | +) = 0.2375/0.2525 = 0.941, about 94%.
Interpretation. The 95% is the sensitivity - the chance of a positive result given disease - not the chance of disease given a positive result. The posterior here is high mainly because the prior is high (1 in 4 from the pedigree). Still, roughly 6% of positives in this family would be false positives, so a confirmatory enzyme or DNA test is required before any clinical decision. Contrast: for a random newborn with prior 1/10 000, P(affected | +) = (0.0001 × 0.95)/((0.0001 × 0.95) + (0.9999 × 0.02)) ≈ 0.0047, under 0.5%. The same test, the same sensitivity, a completely different meaning - the prior drives the conclusion.
3. Estimate the probability that the healthy daughter will have an affected child with a partner from the general population, stating your assumptions.
Read the worked answer
Multiply the independent probabilities along the pathway:
- She is a carrier: 2/3 (from the conditional calculation above).
- Her partner is a carrier: ≈ 1/50, the population carrier frequency, assuming he has no family history and is unrelated.
- If both are carriers, a child is affected with probability 1/4.
P = (2/3) × (1/50) × (1/4) = 2/600 = 1/300 ≈ 0.33% - about 1 in 300, roughly 3 per 1000 pregnancies.
Assumptions and cautions: the 1/50 carrier frequency applies to the partner's population background; the partners are unrelated (consanguinity would raise the carrier probability); new mutations are negligible; and no prenatal testing is assumed. The estimate is a prior risk - testing the daughter for the familial allele would replace the 2/3 with 1 or 0 and change the counselling entirely, which is why carrier testing is usually offered before pregnancy rather than during it.