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

Assume a single disease allele $a$, complete penetrance of $aa$, no disease in $Aa$ or $AA$, and Hardy-Weinberg equilibrium in the outbred population. If its population frequency is $q$, the disease prevalence is $q^2=10^{-6}$, so $q=10^{-3}$.

For a child with inbreeding coefficient $F$, the two inherited copies are identical by descent with probability $F$. In that component there is just one ancestral allele draw, and it is $a$ with probability $q$. In the remaining component there are two independent ancestral draws, with probability $q^2$ of $aa$. Thus the inbreeding genotype frequencies giveUsing the first-cousin inbreeding coefficient $F=1/16$, the disease-risk ratio isThe absolute risk in this model is $6.34375\times 10^{-5}$. This calculation averages over the parental genotypes in first-cousin marriages before observing an affected child. It is not the risk conditional on both parents being known genetic carriers, and assumes the specified cousin relationship is independent of disease status rather than conditioning on a selected disease pedigree.