Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 2

The kinship coefficient ${\varphi }_{XY}$ is the probability that a randomly selected allele copy from individual $X$ and an independently selected copy from individual $Y$ at the same genetic locus are identical by descent. The sampling is of gene copies, not of whole genotypes. For a noninbred individual $X$, sampling its two copies with replacement gives ${\varphi }_{XX}=1/2$; in general ${\varphi }_{XX}=(1+F_X)/2$, where $F_X$ is its inbreeding coefficient.

For two full siblings in an outbred pedigree, the two sampled copies are both paternal with probability $1/4$, both maternal with probability $1/4$, and of different parental origin with probability $1/2$. Given that both are paternal, their independent Mendelian segregation transmissions select the same paternal ancestral copy with probability $1/2$; likewise for both maternal. The different-parent case cannot be identical by descent when the parents are unrelated. ThereforeEquivalently, conditional on an IBD-sharing count $J$, the chance that the two random copies match is $J/4$, so ${\varphi }_{\mathrm{sib}}=E[J]/4=1/4$. The expected shared proportion $E[J/2]=1/2$ and the usual additive relationship coefficient $2\varphi =1/2$ are twice this kinship coefficient; confusing these conventions would double the subsequent inbreeding coefficient.

Let the first cousins be $X,Y$, with parents $(P,U)$ and $(Q,V)$ respectively, where $P,Q$ are full siblings and $U,V$ have no ancestral relationship to one another or to that sibling family. To compute ${\varphi }_{XY}$, choose one allele copy from each cousin. They originate from $P$ and $Q$ with probability $1/4$. All other parental combinations have zero kinship coefficient under the otherwise outbred pedigree assumption. HenceThe two gene copies within a child of $X,Y$ are random transmissions from these parents. They are identical by descent precisely in the event defining parental kinship, so the child's inbreeding coefficient isFor a direct ancestral-path check, there are two shared noninbred grandparents. Through either grandparent, the two cousin-to-grandparent paths have two meioses each, and matching the ancestral grandparental copy supplies a further factor $1/2$. Each contributes $2^{-2-2-1}=1/32$, giving the same first-cousin inbreeding coefficient. Additional ancestral relationships or inbred grandparents would change the result.

Assume a single disease allele $a$, complete penetrance of $aa$, no disease in $Aa$ or $AA$, and Hardy-Weinberg equilibrium in the outbred population. If its population frequency is $q$, the disease prevalence is $q^2=10^{-6}$, so $q=10^{-3}$.

For a child with inbreeding coefficient $F$, the two inherited copies are identical by descent with probability $F$. In that component there is just one ancestral allele draw, and it is $a$ with probability $q$. In the remaining component there are two independent ancestral draws, with probability $q^2$ of $aa$. Thus the inbreeding genotype frequencies giveUsing the first-cousin inbreeding coefficient $F=1/16$, the disease-risk ratio isThe absolute risk in this model is $6.34375\times 10^{-5}$. This calculation averages over the parental genotypes in first-cousin marriages before observing an affected child. It is not the risk conditional on both parents being known genetic carriers, and assumes the specified cousin relationship is independent of disease status rather than conditioning on a selected disease pedigree.

Assume the parents are unaffected, the disease has complete penetrance with a single recessive causal allele, and there are no phenocopies, new disease mutations or alternative causal loci. An affected first child has genotype $aa$, so each unaffected parent must be a heterozygote $Aa$. Conditional on these parental genotypes, Mendelian segregation in the second child's paternal and maternal meioses is independent of the transmissions to the first child. ThereforeThis is the same sibling recurrence risk as for unrelated, unaffected parents under the same recessive model. Parental kinship increases the prior probability that both parents carry the same rare disease allele; after both are identified as carriers, it does not alter their Mendelian transmission probabilities. The factor $63.4$ from the preceding part is therefore not an additional multiplier on the sibling recurrence risk.

If unaffected parents are not explicitly required, the rare-disease assumption is an approximation supporting the usual carrier-by-carrier calculation, rather than a logically exact consequence of an affected child alone. Under complete penetrance, the compatible parental matings are $Aa\times Aa$, $Aa\times aa$ and $aa\times aa$, with second-child risks $1/4,1/2,1$ respectively. Let their prior probabilities be $w_{11},w_{12},w_{22}$, including both orders in $w_{12}$. Conditioning on the affected first child givesWithout negligible or excluded affected-parent matings, this mixture need not equal $1/4$ and may depend on consanguinity. Likewise incomplete penetrance or disease heterogeneity invalidates the assertion that an affected child of unaffected parents necessarily identifies an $Aa\times Aa$ mating. The stated $1/4$ answer uses the unaffected-parent, fully penetrant single-locus assumptions.