Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 40 / 6 / b

To identify the unlabelled individuals in the original pedigree, call the three children of $A,B$ from left to right $U,V,W$. Then $G$ has parents $C,U$, $D$ has parents $V,W$, and $E,F$ have parents $G,D$. The three full siblings $U,V,W$ are noninbred because $A,B$ are unrelated noninbred founders in a pedigree. Their pairwise kinship coefficients areSince $D$ is the child of the sibling pair $V,W$, its inbreeding coefficient is

The founder $C$ is unrelated to the descendants of $A,B$, while the kinship coefficient between $D$ and its parents' sibling $U$ isAs $G$ has parents $C,U$, the kinship coefficient between $D,G$ isThe parents of $E$ are $D,G$, soThe same calculation gives $F_F=1/8$.

The self-kinship coefficient of the inbred individual $D$ is ${\varphi }_{DD}=(1+F_D)/2=5/8$. Since $E$ receives a uniformly chosen parental allele from each of $D,G$,The ${\varphi }_{DD}$ term is essential: self-kinship coefficient is not zero, even for a noninbred individual.

The full siblings $E,F$ have parents $G,D$. Since $G$ is noninbred, ${\varphi }_{GG}=1/2$, whereas ${\varphi }_{DD}=5/8$ and ${\varphi }_{GD}=1/8$. Applying the parental kinship coefficient recursion to both siblings givesThe familiar outbred-sibling value $1/4$ does not apply to this inbred pedigree.