Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 39 / 4

Count one transmitted/untransmitted allele pair per observable parent. A homozygous parent contributes to $a$ or $d$; a heterozygous parent contributes to $b$ or $c$. When two heterozygotes have a heterozygous child, the parent-specific origin is ambiguous, but the combined contribution is unambiguously one to $b$ and one to $c$.

The derived contributions, ordered as $(a,b,c,d)$, areIn family 7, the observed father's transmission cannot be determined from the heterozygous child without the mother's genotype. In family 10, the homozygous child reveals the father's allele-1 transmission, but not the missing mother's complete transmitted/untransmitted pair.

Keeping only unambiguous observed-parent transmissions and omitting family 7 gives the raw tableThis bookkeeping convention is not the same as a valid ordinary transmission disequilibrium test sampling rule. Incomplete-trio transmission-selection bias arises if only homozygous children make a missing-parent family usable. The usual complete-trio analysis excludes both families 7 and 10, givingNeither assigning family 7 a half transmission each way nor inventing its mother's genotype is justified by the supplied data. A method using all twelve families must model or validly condition on the missing information.

The transmission disequilibrium test is the paired McNemar test statistic $(b-c{)}^2/(b+c)$. From the raw unambiguous-transmission table,The formal one-degree-of-freedom chi-squared distribution approximation gives $p\simeq 0.071$, so the calculation does not reject at 5%. Its ordinary null calibration would require a valid missing-parent inclusion rule; the raw arithmetic alone does not establish one.

Using the complete-trio table gives ${\chi }_{\mathrm{TDT}}^2=36/14=18/7\simeq 2.571$, with $p\simeq 0.109$. It also does not reject. Homozygous-parent transmissions in $a,d$ supply no discordant contrast for this test.

The unmatched haplotype-relative-risk counts satisfy $w=a+b$, $x=c+d$, $y=a+c$ and $z=b+d$. For the raw table this givesThe Pearson chi-squared test of independence in this two-by-two table has statisticThus the unmatched calculation gives nominal evidence at 5%, with $p\simeq 0.030$. It differs from the transmission disequilibrium test because it uses an unmatched table, and it also inherits the missing-parent selection issue from the raw counts.

For the ten complete trios, $(w,x,y,z)=(12,8,6,14)$, giving odds ratio $7/2$ and the unmatched Pearson chi-squared test of independence statistic $40/11\simeq 3.636$, with $p\simeq 0.057$. That calculation does not reject at 5%. The nominal significance from the raw table therefore should not be presented as a robust association conclusion without a justified model for incomplete families and the unmatched-table assumptions.

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.

Let $A=R_{11}+2R_{12}+R_{22}$. Among the ten complete trios, six have two heterozygous parents: three have child $11$, two have child $12$ and one has child $22$. Their conditional likelihood contribution isTwo more trios have a heterozygous parent and a $22$ parent, with child $12$ in each; they contribute $[R_{12}/(R_{12}+R_{22}){]}^2$. The remaining two parent pairs can produce only child $12$, so their conditional probabilities are one. Multiplying givesThe factor $4$ does not depend on the genotype relative risks, so the conventional proportional likelihood function isThis is the expression in the original PDF, up to the parameter-independent constant. Its last denominator is $R_{12}+R_{22}$; the converted TeX's $R_{11}+R_{22}$ is a transcription error. The factor $4$ in the exact probability comes from the two heterozygous children of two heterozygous parents, each of which has two labelled transmission routes.