Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 2 / d / Solution

Assume the parents are unaffected, the disease has complete penetrance with a single recessive causal allele, and there are no phenocopies, new disease mutations or alternative causal loci. An affected first child has genotype $aa$, so each unaffected parent must be a heterozygote $Aa$. Conditional on these parental genotypes, Mendelian segregation in the second child's paternal and maternal meioses is independent of the transmissions to the first child. ThereforeThis is the same sibling recurrence risk as for unrelated, unaffected parents under the same recessive model. Parental kinship increases the prior probability that both parents carry the same rare disease allele; after both are identified as carriers, it does not alter their Mendelian transmission probabilities. The factor $63.4$ from the preceding part is therefore not an additional multiplier on the sibling recurrence risk.

If unaffected parents are not explicitly required, the rare-disease assumption is an approximation supporting the usual carrier-by-carrier calculation, rather than a logically exact consequence of an affected child alone. Under complete penetrance, the compatible parental matings are $Aa\times Aa$, $Aa\times aa$ and $aa\times aa$, with second-child risks $1/4,1/2,1$ respectively. Let their prior probabilities be $w_{11},w_{12},w_{22}$, including both orders in $w_{12}$. Conditioning on the affected first child givesWithout negligible or excluded affected-parent matings, this mixture need not equal $1/4$ and may depend on consanguinity. Likewise incomplete penetrance or disease heterogeneity invalidates the assertion that an affected child of unaffected parents necessarily identifies an $Aa\times Aa$ mating. The stated $1/4$ answer uses the unaffected-parent, fully penetrant single-locus assumptions.