Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 45 / 2

Genetic linkage analysis follows co-segregation of genetic marker and disease alleles within pedigrees. It estimates or tests a recombination fraction, with independent assortment at $\theta =1/2$ as the ordinary null. It can locate a disease genetic locus even when different families have different genetic marker alleles on their disease chromosomes.

Genetic association analysis tests whether allele or genotype frequencies differ with disease status. It can compare unrelated cases and controls, or compare transmitted and untransmitted parental alleles. A noncausal genetic marker can be associated through population linkage disequilibrium with a causal genetic locus. Association therefore concerns population-frequency relationships, whereas linkage concerns inheritance; population stratification can confound the former without producing genuine within-family linkage.

Only heterozygous parents contribute an alternative-allele transmission contrast to the transmission disequilibrium test. Both parents are homozygous in rows 1, 13, 14 and 15, so rows 1, 13, 14 and 15 are uninformative. Their counts total 15 families. Every other row has at least one heterozygous parent. When both parents are heterozygous and the child is heterozygous, there is one transmission of each allele, regardless of the unresolved parent-specific phase; these transmissions contribute to both discordant cells.

Write $b$ for transmission of allele 1 with allele 2 untransmitted, and $c$ for the reverse. The informative transmissions give $b=20$, $c=30$. Under the fair-transmission null, conditional on $b+c=50$, $b$ is binomial with success probability $1/2$. Its standardized squared deviation isThe reference distribution is asymptotically chi-squared with one degree of freedom, giving a two-sided $p$-value about $0.157$. Allele 2 is overtransmitted and hence is the allele more associated with disease in this sample, although this result is not strong evidence against the null. The diagonal transmission cells contribute no discordant contrast.

Summing the transmission table by its rows and columns givesThus the marginal case-versus-pseudo-control odds ratio isThis is the requested approximate allelic effect. Keeping the parent-specific matching instead gives the discordant-pair odds ratio $c/b=30/20=1.5$, numerically close here. Neither calculation makes the within-family allele observations independent population samples.

For each family combine the two untransmitted parental alleles into one family pseudo-control genotype. A heterozygous child of two heterozygous parents has a heterozygous pseudo-control, whatever its parent-specific transmission phase. Combining the rows givesFor example, pseudo-control genotype $1/1$ receives contributions $1,2,5,5$ from the relevant rows; genotype $2/2$ receives $3,1,2,6$, and the remaining 25 are heterozygous. These counts also give the untransmitted allele totals 51 and 49.

The estimated allele frequencies are ${\hat{p}}_1=0.51$, ${\hat{p}}_2=0.49$. The Hardy-Weinberg equilibrium expected genotype counts areThey are almost identical to the observations, so these pseudo-control counts conform extremely closely to HWE proportions. The usual Pearson discrepancy is only about $8.0\times 10^{-6}$. This numerical agreement is not a general guarantee of HWE for family pseudo-controls, whose construction involves mating structure and case ascertainment in a genetic study.

Family comparisons condition on the parental genotypes, so cases are compared with the alleles they could have inherited. This protects a transmission test from ordinary population stratification, avoids having to recruit unrelated controls matched for ancestry, and can resolve haplotypes and identify Mendelian inconsistencies. The same pedigree may also provide linkage information.

The costs are recruiting and genotyping relatives, missing or unavailable parents, and fewer informative comparisons because homozygous parents contribute no transmission contrast. These problems are especially important for late-onset disease. Related cases or multiple siblings also require an analysis accounting for dependence, and genotyping error or transmission distortion can mimic overtransmission. Population-based case/control studies are usually easier to assemble and can offer greater power per genotyped person, but need careful control of ancestry and other sources of confounding. Family sampling does not remove every form of ascertainment in a genetic study or observation bias.