Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 40 / 5

Genetic linkage analysis asks whether a genetic marker and a disease genetic locus cosegregate within families more often than would be expected under independent assortment. It uses pedigrees and Mendelian segregation to estimate the recombination fraction $\theta$, with evidence for linkage arising when the data favor $\theta <1/2$. Even when founders are in linkage equilibrium, linked loci can show familial cosegregation.

Genetic association analysis instead asks whether observed allele or genotype frequencies differ with disease status in a sampled population, often comparing cases with controls. It can detect a causal variant or a correlated genetic marker through linkage disequilibrium; population structure and selection can also create association. Linkage is about familial transmission; association is about population dependence. Neither an association alone nor a low family recombination estimate establishes that the marker itself causes disease.

Distinguish the disease alleles $D$ (normal) and $d$ (recessive disease) from the marker alleles $A,a$. The affected mother and two affected children are $dd$. The unaffected father must be $Dd$, since he has affected children; the unaffected child is $Dd$, since the mother always transmits $d$. The mother is $AA$ at the marker, so the marker genotypes identify the father's transmissions to the three children as $a,A,A$. Their disease transmissions from the father are $d,d,D$.

The father's unphased double heterozygosity has two possible pairs of haplotypes:Under $s_1$, the three paternal transmissions $(ad,Ad,AD)$ are respectively nonrecombinant, recombinant and nonrecombinant. Under $s_2$, they are recombinant, nonrecombinant and recombinant. Each particular gamete has probability $(1-\theta )/2$ or $\theta /2$, giving the conditional pedigree likelihoodsAssuming linkage equilibrium among founder haplotypes, the two phases have equal prior weights conditional on the observed paternal genotype. Phase averaging in a linkage likelihood therefore givesAny factors for the parental genotypes omitted by this conditioning are constant in $\theta$. The derivative is $(1-2\theta )/16$, so the maximum likelihood estimate is $\hat{\theta }=1/2$ and there is no evidence for linkage: the maximum LOD score is zero. Choosing the best phase after observing the offspring would give a different and inappropriate calculation; the unknown phase must be averaged. The linkage-equilibrium assumption is modified explicitly in part (v).

Arrange rows by marker genotype, with cases and controls as the two columns. Both column totals are 100, and the three row totals are 100, 60 and 40. Under the null of no genetic association, estimate each common row probability from its pooled proportion. The expected count is row total times column total divided by the grand total. Thus the expected contingency table isThe Pearson chi-squared test of independence usesUnder independent sampled individuals and the no-association null, its reference chi-squared distribution has $(3-1)(2-1)=2$ degrees of freedom. All expected cells are well above five. The null is overwhelmingly rejected: the approximate p-value is $e^{-93.44/2}\simeq 5.13\times 10^{-21}$. This establishes association in the sampling model, not by itself a causal role for the marker.

Each diploid individual contributes two marker alleles. Count $A$ copies as twice the $AA$ count plus the $Aa$ count, and count $a$ copies analogously. The observed chromosome-level contingency table, followed by its null expected table, isThere are 200 chromosomes in each group; pooled totals are 140 $A$ and 260 $a$. The usual allelic Pearson chi-squared test of independence givesWith the ordinary independent-chromosome null model, use one degree of freedom; its p-value is about $1.04\times 10^{-25}$, providing very strong evidence of genetic association.

The two alleles from the same person are not automatically two independent observations. The usual one-degree-of-freedom calibration is justified, for example, for unrelated individuals with Hardy-Weinberg equilibrium under the null. If that assumption is not suitable, retain individuals as the sampling units and test their allele dosages using an appropriate variance or permutation of case-control labels, or use the genotype-level test from part (iii). No Hardy-Weinberg equilibrium assumption was needed for that genotype-level test.

Using cases as disease haplotypes and controls as control haplotypes in the model requested here, the chromosome counts estimateThese are conditional marker allele frequencies, not the reverse conditional probabilities $P(d\mid A)$. Under independent founder chromosomes, conditional on the father's $Dd,Aa$ genotype, the phase $AD/ad$ has weight proportional to $0.60\cdot 0.90=0.54$, and $Ad/aD$ has weight proportional to $0.10\cdot 0.40=0.04$. Their normalized weights are $27/29$ and $2/29$. This conditions on the model's observed parental genotypes; the omitted disease-allele frequencies cancel in the phase-weight ratio.

Using the two phase-specific pedigree likelihoods already derived,The normalizing factor is independent of $\theta$, so maximize $g(\theta )=0.54\theta -1.04{\theta }^2+0.50{\theta }^3$. Its derivative gives exactlyOnly the smaller root lies in $[0,1/2]$:On this interval $g^{\prime \prime }(\theta )=-2.08+3\theta <0$, so this stationary point is the unique maximum. The maximum LOD score, retaining the same linkage disequilibrium frequencies in numerator and denominator, isThis is only a likelihood ratio of about $1.145$ against $\theta =1/2$, hence very weak evidence. The population frequencies are treated as fixed plug-in estimates, as requested; uncertainty in those estimates and possible disease-model differences between controls and normal-allele chromosomes would matter in a fuller analysis.