= Solution
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 $\psi_i,\psi_j$, so
$$
\mathbb P(\text{transmit }i\mid\text{parent }i/j,\text{affected child})=\frac{\psi_i}{\psi_i+\psi_j}.
$$
The 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.
\b[Prefer the within-family conditional transmission test to an unconditional population case–control analysis when the population-frequency assumptions and ancestry comparability are not secure.]
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