Carrier testing in a consanguineous family with spinal muscular atrophy
A couple from a town in Tamil Nadu seeks genetic counselling. The woman, 27, is healthy, but her 19-year-old brother has spinal muscular atrophy (SMA) with homozygous loss of SMN1 exon 7; both her parents are unaffected. Her husband, 29, is her first cousin - the woman's mother and the husband's father are siblings - and he has no family history of the disease. The husband's parents are alive and unaffected and have three other children. The couple already have one healthy child. The regional carrier frequency for SMA is estimated at 1 in 40. The clinic uses an MLPA kit that reports SMN1 copy number, and the couple can pay for only one carrier test this month. The counsellor must decide whom to test first, how to interpret a two-copy MLPA result, and what recurrence risk to quote for the next pregnancy.
1. Estimate the probability that the couple's next child will be affected, showing every step and stating your assumptions. Is the 1-in-40 population figure relevant here?
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Step 1 - the woman. An affected child proves both her parents are carriers. An unaffected sibling of an affected child has carrier probability 2/3 (of the three possible unaffected genotypes, AA, Aa, aA, two are carriers).
Step 2 - the husband. The woman's mother is an obligate carrier. Her brother (the husband's father) therefore has a 1/2 chance of carrying the same allele; if he does, the husband has a 1/2 chance. Husband's carrier probability = 1/2 × 1/2 = 1/4.
Step 3 - the child. 2/3 × 1/4 × 1/4 = 2/48 = 1/24 ≈ 4.2% (about 1 in 24), roughly 17 times the population-based figure.
Is 1/40 relevant? Only as a small add-on. Combining independent routes: 1 − (1 − 0.25)(1 − 0.025) ≈ 0.27, which barely changes the answer. Using the population frequency alone for the husband would be a serious error because consanguinity makes the same ancestral allele far more likely than a random allele.
Assumptions: full penetrance, no de novo event, the same familial variant in both branches, no additional consanguineous loops, and a deletion-type variant (as indicated by absent exon 7).
2. The husband's MLPA reports two SMN1 copies. Recalculate the risk and explain why the result is reassuring but not a guarantee of zero.
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Two copies means he is not a conventional deletion carrier, so his carrier probability falls from 25% to a residual figure.
Sources of residual risk
- Point mutations or small intragenic changes in SMN1 are not detected by a dosage assay; such variants account for a few per cent of carrier chromosomes.
- The "2+0" configuration - both copies on one homologue, none on the other - is invisible to MLPA, which counts copies but does not assign them to chromosomes.
- Technical failure: partial digestion, poor DNA quality, borderline dosage ratios.
Recalculation. Taking a residual carrier probability of about 3–5%: 0.25 × 0.04 ≈ 1%. Child risk ≈ 2/3 × 0.01 × 1/4 ≈ 0.17% (about 1 in 600). That is a 25-fold reduction from 4.2% but still above the background risk for the population.
Key point: a normal MLPA changes the number, it does not license the statement "there is no risk". The residual is dominated by the possibility of a rare non-deletion allele, which could be resolved by testing the affected brother for a point variant if the family wishes, or by testing the woman herself.
3. With money for only one test, whom would you test and what would a two-copy result in that person let you say?
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Test the woman first. She has the higher prior probability (2/3 versus 1/4), so the expected information gain per rupee is greater, and her result is decisive in a way the husband's is not.
Why her result is cleaner. Her affected brother is homozygous for an SMN1 deletion, so both parents' carrier chromosomes lack exon 7. Each parent therefore carries a single copy on the other homologue. The woman has two copies only if she inherited the intact homologue from both parents - in that case she is not a carrier of the family variant. The 2+0 ambiguity that clouds the husband's result does not apply to her, because a 2+0 arrangement cannot arise from two parents who each carry a deletion.
Interpretation
- One copy → she is a carrier; the couple's risk is essentially the husband's residual risk, around 1/4 × 1/4 ≈ 6% if he is untested, and testing him becomes the next step.
- Two copies → she is not a carrier; the child's risk drops to the population-level route through the husband, roughly 1/40 × 1/4 ≈ 0.6%, falling further if he is later tested.
Either way the couple leaves with an actionable number rather than a range.