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

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.