 Solution

The disease-ascertained transmission probability follows directly from Bayes' theorem. Assume that, given the child's genotype $g$, its disease probability is a common baseline penetrance times the genotype relative risk $R_g$, with no additional dependence on the parental genotypes. Write $q_g=P(g\mid g_f,g_m)$ for the Mendelian segregation probability. ThenThere are four equally likely labelled parental transmission outcomes. For one specified outcome $t$, this becomes $P(t\mid g_f,g_m,D)=R_{g(t)}/{\sum }_{t^{\prime }}{R}_{g(t^{\prime })}$, which is the four-outcome interpretation of the displayed formula.

For an unordered child genotype, let $m_g$ be the number of those outcomes producing it. Its probability is insteadFor example, two heterozygous parents have multiplicities $(1,2,1)$ for $11,12,22$. A heterozygous child therefore has probability $2R_{12}/(R_{11}+2R_{12}+R_{22})$. The four possible outcomes must be counted with multiplicity; treating their unordered genotypes as four distinct equally likely values would be incorrect.