Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 45 / 1

Treat $G_i$ as the entire phase-known genotype of individual $i$ at the genetic loci being studied, and let $\overline{F}$ denote the nonfounders. Assume unrelated pedigree founders have independent genotype priors, each nonfounder's genotype depends only on its parents' genotypes, and the observed phenotypes are conditionally independent given the genotypes. These are the standard assumptions behind the pedigree likelihood; shared environmental effects or related pedigree founders would require additional factors.

The directed ancestry graph is acyclic, even when there is inbreeding. Order individuals with parents preceding offspring and apply the chain rule for probabilities. The parental conditional independence assumption givesConditional phenotype independence gives $P(X_1,\dots ,X_n\mid G_1,\dots ,G_n)={\prod }_{i}P(X_i\mid G_i)$. Multiplying these two expressions and summing over every possible latent genotype configuration provesThis is the law of total probability, not a claim that relatives have independent unconditional genotypes.

For a binary disease phenotype, put $p_i(g)=P(X_i=\text{affected}\mid G_i=g)$. The penetrance function depends on the inheritance model, the disease alleles and genotype, and potentially age, sex, environment and other genetic modifiers. Phenocopies allow disease without the putative disease genotype, while incomplete penetrance allows a genetic carrier of that genotype to remain unaffected. Thus the phenotype factor is $p_i(g)$ for an affected individual and $1-p_i(g)$ for an unaffected one.

Under a fully penetrant model with no phenocopies the genotype determines disease status, so the factor is 1 for a compatible genotype and 0 for an incompatible genotype. For example, in fully penetrant recessive inheritance, $aa$ is affected and $AA,Aa$ are unaffected. Missing phenotype observations contribute the constant factor 1 rather than excluding genotypes.

Without genotyping error, the observed genetic marker genotype is a deterministic function $u(g)$ of the underlying phase-known genotype. In particular $u$ can forget chromosome phase and parental origin. ThereforeSeveral different phased haplotype pairs can receive factor 1 when the observed multilocus genotype is unphased. With laboratory error, missing alleles or uncertain calls, replace the indicator by an error or observation model. A wholly unobserved genetic marker contributes factor 1. Thus a 0-or-1 factor reflects perfect observation of a deterministic genetic marker state, not a general property of every genetic measurement.

The pedigree founder factor $P(G_i)$ depends on population allele and haplotype frequencies, ancestry, mating assumptions and any population linkage disequilibrium. At one autosomal genetic locus, Hardy-Weinberg equilibrium with allele frequencies $p,q$ gives genotype probabilities $p^2,2pq,q^2$. For a specified ordered pair of independent pedigree founder haplotypes $h_1,h_2$, the probability is $p_{h_1}{p}_{h_2}$; an unordered distinct pair has twice that probability. Pedigree founders assumed to be related require a joint prior rather than a product of unrelated-founder factors.

The nonfounder factor depends on the parental genotypes and phases, Mendelian segregation, and the recombination fractions between genetic loci. In a two-locus heterozygote of known phase, the two parental haplotypes have transmission probabilities $(1-\theta )/2$ each, and the two recombinant haplotypes $\theta /2$ each. Multiply the maternal and paternal gamete probabilities and sum any gamete combinations producing the child's genotype. Sex-specific genetic recombination, mutation or segregation distortion can be incorporated by changing these gamete probabilities. At one ordinary autosomal genetic locus, each parental copy is transmitted with probability $1/2$.

Call the parent genotype $g$ and child genotype $h$, each in $\{0,1\}$. DefineThere is one pedigree founder and one nonfounder, soExplicitly this is $a_0{p}_0{b}_0{t}_{0\mid 0}+a_0{p}_0{b}_1{t}_{1\mid 0}+a_1{p}_1{b}_0{t}_{0\mid 1}+a_1{p}_1{b}_1{t}_{1\mid 1}$. The genotype-transmission model can be arbitrary for the single-parent species; no two-parent Mendelian assumption is being introduced.

In the unrearranged likelihood there are four summands. Each is a product of four supplied probabilities and takes three multiplications. Summing the four results takes three additions. ThusThis count concerns direct evaluation without reusing repeated factors; zero-or-one phenotype values could of course allow extra special-case savings.

The Elston-Stewart algorithm first sums over the child:For each of the two parent states, the inner message requires two multiplications and one addition, totaling four multiplications and two additions. Multiplying each message by its two parent factors takes two further multiplications per parent state, totaling four. The final sum uses one addition. ThereforeMoving the sum inward saves four multiplications because the factors independent of the child genotype are evaluated only once per parent state.