Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 29 / 5

Let $I,J$ denote the two ordered alleles at the genetic locus. Under Hardy-Weinberg equilibrium they are independent with probabilities ${\pi }_i,{\pi }_j$. Write the multiplicative penetrance as $\mathbb{P}(D\mid I=i,J=j)=k{\psi }_i{\psi }_j$, with parameters chosen so these are valid probabilities, and set $Z={\sum }_u{\pi }_u{\psi }_u$. ThenThe Bayes' theorem givesThus multiplicative penetrance preserves Hardy-Weinberg equilibrium: the affected subjects still have two independent allele draws, with tilted allele frequencies. The displayed product uses ordered allele slots. For the usual unordered genotypes, the corresponding probabilities areThe factor of two is required for heterozygotes and must not be dropped when translating the ordered notation into genotype counts.

For genetic association analysis, the case genotype probabilities factor into allele frequencies, so under this model an allelic comparison with representative population controls is appropriate; separate dominance departures are not required by the penetrance model. The ratio of case allele frequency to population allele frequency is proportional to ${\psi }_i$, and relative risk multipliers satisfy ${\psi }_i/{\psi }_j=({\pi }_i^{\ast }/{\pi }_i)/({\pi }_j^{\ast }/{\pi }_j)$. Population controls are important to this exact statement. Among specifically unaffected controls,which generally does not factor. For a rare disease it is close to the population genotype law, but no rare-disease assumption was needed for the case factorization itself. Population stratification and sampling dependence must also be addressed before using an ordinary allelic chi-squared test.

Label each parent's two allele copies by its realized transmission: let $(A,B)$ be the transmitted and untransmitted alleles of the first parent and $(C,D^{\prime })$ those of the second. Under random mating and Hardy-Weinberg equilibrium, the four parental allele draws are independent with population frequencies $\pi$. Random Mendelian segregation merely swaps the two independent copies within each parent, so the transmitted/untransmitted labelled pair still has probability ${\pi }_a{\pi }_b$; the second pair has probability ${\pi }_c{\pi }_d$.

The child receives $A,C$. If $\mathcal{D}$ is its disease event, its penetrance is $k{\psi }_A{\psi }_C$ and its overall disease probability is $k{Z}^2$. ThereforeThis proves the transmitted and untransmitted alleles under multiplicative penetrance factorization. Under these assumptions, the transmitted pair $a/c$ has the affected-case distribution, while the complementary pair $b/d$ has the population distribution and is independent of the transmitted pair. It is the family pseudo-control genotype.

As in part (a), the formula records labelled transmissions. When observed genotypes are unordered, sum over every compatible parental-origin transmission. This includes multiplicities for overlapping parental alleles or two heterozygous parents producing a heterozygous child; it avoids interpreting one product as the total probability of all observationally identical configurations. The homogeneous random-mating model is also essential to the unconditional independence: mixing ancestry strata can correlate the case and pseudo-control through their shared family background.

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 $2\mathbb{P}(K\le 18)\approx 0.4408$.