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The usual transmission disequilibrium test conditions on the observed parental genotypes, rather than modelling their population probabilities. For a heterozygous parent carrying $i,j$, the two transmission outcomes are equally likely under no association. With multiplicative penetrance, ascertainment through an affected child weights them by ${\psi }_i,{\psi }_j$, soThe other parent's penetrance factor cancels. This is a special case of disease-ascertained transmission probability. Under the null ${\psi }_i={\psi }_j$, the conditional probability is $1/2$. Count the discordant transmitted/untransmitted pairs from heterozygous parents and test their balance, using McNemar's test or its exact conditional binomial form. Homozygous parental transmissions do not distinguish the alternatives.

This conditional analysis removes the nuisance population allele frequencies and retains the family matching. It is therefore robust to population stratification, which can otherwise make allele frequencies differ between cases and controls without a within-family transmission effect. It does not require the unconditional Hardy-Weinberg equilibrium and random-mating assumptions used for the population pseudo-control factorization. Independence of suitably sampled families, valid genotypes and the Mendelian null transmission model still matter.

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.