Carrier Screening After a Negative Result
Elena and Marco are both unaffected and have a son, Luca, who has been diagnosed with a condition inherited in an autosomal recessive pattern. Elena's brother is also affected, and her parents are unaffected. Marco has no family history of the condition. The carrier frequency of pathogenic variants in the relevant gene in the general population is 1 in 25. A carrier-screening test is available: it detects 90% of true carriers and returns a false positive in 1% of non-carriers. The counsellor points out that the test does not detect every pathogenic variant - some large deletions are missed - and that 'no family history' is weak information for a recessive condition, since carriers are healthy. Elena asks: 'If I test negative, can we stop worrying?'
1. Calculate the probability that a second child would be affected, before any carrier testing.
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Step 1 - Elena's carrier probability. Her parents are unaffected but have an affected son, so both must be carriers. An unaffected child of two carriers has a 1/4 chance of being homozygous unaffected and a 2/4 chance of being a carrier; conditional on being unaffected, P(Elena is a carrier) = (1/2)/(3/4) = 2/3. This is the key step people skip: the 1/4 affected outcome is excluded by the information that she is unaffected.
Step 2 - Marco's carrier probability. His prior is the population figure, 1/25.
Step 3 - risk to a child. Both must be carriers and both must pass on the variant:
P = (2/3) × (1/25) × (1/4) = 2/300 = 1/150 ≈ 0.67%, about 1 in 150.
Note. This is roughly four times the risk in the general population for a recessive condition with this carrier frequency (1/25 × 1/25 × 1/4 = 1/2500), which is why the family history matters.
2. Elena takes the test and the result is negative. Calculate her revised probability of being a carrier and the revised risk to a second child. Comment on the phrase 'stop worrying'.
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Bayes' theorem. Let C = carrier.
- Prior P(C) = 2/3 = 0.667; prior P(not C) = 1/3 = 0.333.
- P(negative | C) = 1 − 0.90 = 0.10 (the test misses 10% of carriers).
- P(negative | not C) = 0.99 (the false-positive rate of 1% concerns positives).
Numerator: 0.667 × 0.10 = 0.0667. Denominator: 0.0667 + (0.333 × 0.99) = 0.0667 + 0.330 = 0.3967.
P(C | negative) = 0.0667 / 0.3967 ≈ 0.168, about 17% - not 10%, and not zero.
Revised child risk: 0.168 × (1/25) × (1/4) ≈ 0.00168, i.e. about 1 in 600 (0.17%).
Comment. A negative result cuts the risk roughly fourfold, from 1 in 150 to about 1 in 600 - clinically useful, but still more than twice the background risk for this condition. 'Stop worrying' is unjustified because a negative test is evidence, not proof: a residual chance remains that she carries a variant the assay cannot see.
3. Marco is also tested and is also negative. Estimate the couple's remaining risk and explain why it cannot be reduced to zero by testing alone.
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Marco's posterior. Prior 0.04; P(negative | carrier) = 0.10; P(negative | non-carrier) = 0.99.
P(C | negative) = (0.04 × 0.10) / [(0.04 × 0.10) + (0.96 × 0.99)] = 0.004 / (0.004 + 0.950) ≈ 0.0042, about 1 in 240.
Combined risk. Assuming the two results are independent:
0.168 × 0.0042 × 0.25 ≈ 0.00018, i.e. about 1 in 5,700.
Why not zero.
- Each test has 10% sensitivity loss; both being negative still leaves both partners simultaneously undetected carriers with a small but non-zero probability.
- The quoted 90% sensitivity is an average over known variants; sensitivity is lower for variants the assay does not target, such as the large deletions mentioned.
- A de novo variant in either partner's germline, though rare, is not excluded.
Practical conclusion. The couple can be told the risk has fallen by more than an order of magnitude and is now well below the population figure, while being told plainly that only a definitive test (e.g. sequencing) or prenatal testing in a pregnancy can give a much tighter answer.