Two Families, One Recessive Disorder
A couple, Priya and Daniel, both unaffected, have a son with an autosomal recessive condition. Genetic testing shows Priya is heterozygous for a known pathogenic variant (allele a) and Daniel carries a different pathogenic variant in the same gene (allele a′). Their son is a/a′ - a compound heterozygote, affected because neither allele produces functional protein. Priya is pregnant again. Daniel's brother, who is unaffected, is planning children with a woman from a population where carrier frequency for any pathogenic variant in this gene is 1 in 25. A relative says, 'Since Daniel is a carrier, his brother must be too, and the risk is one in four.'
1. Calculate the probability that Priya and Daniel's next child is affected, and explain why the 'one in four' rule applies here.
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Calculation. Priya's genotype is A/a; Daniel's is A/a′.
Punnett square of gametes (A or a) × (A or a′):
- A/A - unaffected, non-carrier (1/4)
- A/a - unaffected carrier (1/4)
- A/a′ - unaffected carrier (1/4)
- a/a′ - affected (1/4)
Probability of an affected child = 1/4 (25%).
Why the rule applies. Both parents are confirmed carriers, and the alleles are at the same locus, so each child independently receives one allele from each parent. The 1/4 is a per-pregnancy probability; it does not mean one in four children, and previous outcomes do not change the next one. The unusual feature here is that the two pathogenic variants are different (compound heterozygosity), but the segregation logic is identical to the standard case.
2. Evaluate the relative's statement about Daniel's brother and calculate the risk for his planned child.
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Evaluating the statement. The relative is wrong on both counts.
Carrier probability. Daniel's parents must each have contributed one pathogenic allele to Daniel. Their genotypes are therefore carrier × carrier (assuming both variants are rare and neither parent is affected). For an unaffected sibling such as Daniel's brother, the possible genotypes are A/A, A/a or A/a′ in ratio 1:2:1. Given he is unaffected, he is not a/a′. Of the three remaining equally likely outcomes, two are carrier states, so P(brother is a carrier) = 2/3, not 1.
Risk to his child. If he is a carrier, the partner's probability of carrying any pathogenic variant is 1/25. Risk = 2/3 × 1/25 × 1/4 = 2/300 ≈ 1 in 150 (about 0.67%).
Key point. The relative conflated 'Daniel is a carrier' with 'his brother is certainly a carrier', ignoring that unaffected siblings of an affected child have a 2/3 carrier probability, and ignored the partner's population risk entirely.