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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.

A case chromosome is transmitted and a pseudo-control chromosome is untransmitted. Summing the paired table by columns for cases and by rows for controls givesFor the ordinary unpaired chi-squared test of independence, the expected cells are 30,30,70,70. Its uncorrected Pearson statistic isFor the paired McNemar's test, only the discordant transmissions matter, with counts 24 and 18. ThusBoth uncorrected statistics happen to coincide and have the same approximate one-degree-of-freedom chi-squared distribution reference, giving $p\approx 0.355$. The data do not show persuasive transmission imbalance at the conventional 5% level.

The equality is a numerical coincidence, not a justification for ignoring pairing. To see the equality criterion for paired and unpaired allele tests, write the original matched cells as $a,b,c,d$ and $n=a+b+c+d$. The unpaired marginal-table statistic iswhereas McNemar's is $(b-c{)}^2/(b+c)$. Here $n=100$, $b+c=42$, and $(2a+b+c)(2d+b+c)=60\times 140=2n(b+c)$, giving equality. Changing concordant counts can change the unpaired statistic without changing McNemar's. The transmission disequilibrium test should retain its conditional paired interpretation.

If a continuity correction is used, state it explicitly: McNemar's corrected statistic is $(|24-18|-1{)}^2/42=25/42\approx 0.5952$; the corresponding Yates-corrected marginal statistic also coincides here. An exact conditional test uses $\mathrm{Binomial}(42,1/2)$ and gives the two-sided probability $2\mathbb{P}(K\le 18)\approx 0.4408$.