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

The kinship coefficient ${\varphi }_{XY}$ is the probability that one allele copy sampled uniformly from each of individuals $X,Y$, independently conditional on their copies, is identical by descent. The inbreeding coefficient $F_X$ is the probability that the two homologous allele copies within $X$ are identical by descent. For distinct offspring and older individual, Mendelian segregation givesFor the last identity, two independent selections choose the same copy with probability $1/2$, and choose distinct copies with probability $1/2$, in which case identity has probability $F_X$. An unrelated, noninbred founder has $F=0$ and self-kinship $1/2$, while distinct such founders have zero mutual kinship coefficient. The source's statement of zero founder kinship must be read as zero kinship between distinct founders; zero self-kinship would contradict the definition.

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.

For the usual no-linkage reference model with unrelated, noninbred parents, define $J$ as the number of homologous allele copies shared identical by descent by two full siblings at the marker. The two siblings receive the same paternal copy with probability $1/2$, and independently receive the same maternal copy with probability $1/2$. Therefore Mendelian IBD sharing of full siblings givesHenceFor affected siblings, absence of genetic linkage makes the marker transmissions independent of selection by disease, so these are the relevant null moments. This is a fresh outbred reference calculation, not an application of these sharing probabilities to the inbred siblings $E,F$ in part (b). Without the outbred-parent assumption the three ordinary sharing probabilities require modification.

Use the prespecified excess-sharing alternative: affected full siblings linked to a disease genetic locus are expected to share more marker copies identical by descent. The total sharing count is $0(42)+1(98)+2(60)=218$, so $\bar{J}=218/200=1.09$. Under the no-linkage null, independent sibling pairs have $E[J]=1$ and $\mathrm{Var}(J)=1/2$, giving the mean allele-sharing test for affected siblingsThe upper 5% standard normal critical value is approximately $1.64$, so there is evidence of excess sharing, and hence linkage, at the one-sided 5% level. The normal approximation gives upper-tail p-value $1-\Phi (1.8)\simeq 0.0359$.

The conclusion depends on the intended directional test. A two-sided test of the sharing mean uses $1.96$, so it would not reject at 5%. An omnibus Pearson chi-squared goodness-of-fit test of the three counts against $(50,100,50)$ gives $64/50+4/100+100/50=3.32$ with two degrees of freedom and also would not reject. The directional mean-sharing test targets the scientifically specified excess-sharing alternative; these different tests should not be selected after seeing which rejects. Independence of the 200 pairs is assumed: if pairs overlap within families, the standard error needs adjustment.