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For latent genotypes $G_i$ and observed phenotypes $X_i$, assume conditional phenotype independence, independent founder genotypes, and conditionally independent Mendelian transmissions. The likelihood function isThis follows by the chain rule applied to the acyclic ancestry graph, then the law of total probability over all latent genotypes. Penetrance supplies the phenotype factors and genetic recombination supplies multilocus transmission factors.

A phenotype is an observable characteristic of an organism, arising from its genotype, environment and other influences. Seed shape is a phenotype; $Aa$ is a genotype. The distinction prevents appearance from being mistaken for a complete inherited description. See NHGRI: phenotype.

For a specified binary phenotype $T$, let its penetrance in genotype $g$ be $r_g=P(T\mid g)$. The law of total probability gives $P(T)={\sum }_g{r}_{g}P(g)$. Thus the ideal $Aa\times Aa$ example has $P(T)=r_{AA}/4+r_{Aa}/2+r_{aa}/4$: the familiar $3/4$ follows only when these penetrances are $1,1,0$. A pedigree likelihood keeps inheritance and the conditional phenotype model as separate factors.

For unaffected genetic carriers of a disease with complete penetrance and recessive inheritance, suppose parental marker-disease haplotype phases are equally likely and the marker is fully informative. Write $u=\theta (1-\theta )$ and $v=1/2-u$, where $\theta$ is the recombination fraction. Conditional on sharing a parental marker copy, the probability that both siblings inherit that parent's disease allele is $[(1-\theta {)}^2+{\theta }^2]/2=v$; conditional on inheriting different marker copies it is $\theta (1-\theta )=u$. Independent paternal and maternal transmissions therefore give $L_0=u^2,L_1=uv,L_2=v^2$. Conditional instead on both siblings being affected, the IBD probabilities are $s^2,2s(1-s),(1-s{)}^2$, where $s=2u\in [0,1/2]$.

// Target: biology.bigb

Sibling recurrence risk is the conditional probability that another child has the disease, given an affected sibling. For unaffected parents, complete penetrance and a single recessive disease allele, the affected child identifies both parents as genetic carriers. Independent Mendelian segregation then gives recurrence probability $1/4$, whether or not the parents are related. Incomplete penetrance, phenocopies, different causal loci or affected parents change this calculation.

// Target: biology.bigb