Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 57 / 6

Take “cousins” to mean first cousins with unrelated, outbred founders and unrelated spouses in the intervening generation. Interpret a homozygous chromosome as two copies identical by descent, and first use an idealization of independent, nonrecombining chromosome blocks. Ordinary sequence homozygosity is a different quantity: unrelated copies can have the same allele, so its probability also requires allele frequencies. Similarly, identity of an entire recombining chromosome requires a genetic map, not just a pedigree.

At a single autosomal locus, the probability of autozygosity in their child is the parents' kinship coefficient. Each of the two common grandparents contributes a path of length two from each parent, with probability $(1/2{)}^{2+2+1}$. Thus the first-cousin inbreeding coefficient isIf there are $L$ independent autosomal blocks, the probability that none is autozygous is $(1-F{)}^L$. Hence the usual 23-block idealization givesThe exponent describes the stated independent-block approximation; it is not a claim that all 23 human chromosome pairs inherit like autosomes.

Humans have 22 autosomal pairs. A son has one X and one Y and cannot have two homologous X copies. A daughter can have an autozygous X, with probability $F_X$ determined by the sex-linked cousin kinship calculation in part (b). Thus a sex-aware version iswhere $p_D$ is the probability of a daughter, or is one or zero for a specified daughter's or son's sex. The independent assortment approximation applies between the autosomes and the X. Opposite-sex parentage alone does not determine $F_X$: the parents may be cousins through their mothers, their fathers or one of each. Without these pedigree and chromosome-block assumptions, the wording does not specify a unique probability of an entire homozygous human chromosome.

Use the conventional additive pedigree relatedness $r_{UV}=2{\varphi }_{UV}$, where ${\varphi }_{UV}$ is the kinship coefficient, the probability that randomly selected gene copies from $U$ and $V$ are identical by descent. For noninbred individuals this convention gives ordinary full siblings relatedness $1/2$, as required. It is important that this is not the probability of sharing at least one copy: that probability is $3/4$ for ordinary full siblings.

Let the parents be $M,F$. They are otherwise outbred first cousins, so ${\varphi }_{MF}=1/16$ and ${\varphi }_{MM}={\varphi }_{FF}=1/2$. For two children $C,D$, independent Mendelian segregation givesTherefore the autosomal additive relatedness isThe additional $1/16$ beyond $1/2$ comes from maternal-paternal cross-origin matches. Each child has inbreeding coefficient $1/16$. The additive coefficient is a sum of descent probabilities rather than a literal bounded probability in all inbred pedigrees; a relationship normalized to self-relatedness would be $r_{CD}/(1+F)=9/17$. Likewise, identity-by-descent sharing of inbred full siblings gives a different matched-copy statistic. The exam's calibration to $1/2$ supports the additive convention, but the quoted probability wording needs this qualification.

For sex chromosomes the route of cousinship matters, not merely the fact that the parents have opposite sex. Let $P_M$ be the mother-cousin's parent in the shared sibling pair, and $P_F$ the father-cousin's parent in that pair. Let $F_X$ be the probability that the mother's randomly transmitted X matches the father's single X by descent. With no founder inbreeding, the possibilities areIn the first two rows the father-cousin is related through his father, whose X he does not inherit. In the third row his mother is a sister of the mother-cousin's father: the relevant brother and sister match their maternal X copies with probability $1/4$, and the mother-cousin transmits her paternal X with probability $1/2$, giving $1/8$. In the last row the two sisters' random X gametes match with probability $3/8$: their common father's X contributes $1/4$, and their common mother's two X copies contribute $1/8$. Transmission by the mother-cousin halves this to $3/16$. This derives the sex-linked cousin kinship values.

For an unweighted comparison by chromosome pair, two daughters have sex-pair additive relatedness $3/4+F_X$: their paternal X always matches, maternal X matches with probability $1/2$, and the two cross-origin terms each contribute $F_X$. Two sons have mean sex-pair relatedness $3/4$, from an X match of probability $1/2$ and a common paternal Y with probability one. A daughter-son pair has sex-pair value $1/4+F_X/2$, with the Y contribution zero. These calculations treat chromosomes as unrecombined blocks and weight X and Y equally; chromosome-length-weighted or locus-specific coefficients are different.

For two children with independent equally likely sexes, these three cases have weights $1/4,1/4,1/2$. Their mean sex-pair relatedness is $1/2+F_X/2$, so averaging the 22 autosomal pairs and the sex pair givesIf the cousins' route has $F_X=0$, this is $103/184$, slightly below $9/16$. If, as an additional sampling assumption, the four parent-sex routes in the table are equally likely, $\mathbb{E}{F}_X=5/64$ and the answer is $1653/2944$. These are different legitimate numerical corrections under different specified pedigrees. There is no unique sex-chromosome correction from opposite-sex cousinhood alone. The table and formula give it once the missing pedigree and offspring-sex information is supplied.

Assume a closed population with even size $N$, unrelated founders, monogamous mating, nonoverlapping generations and exactly two offspring per couple. Sex labels are exchangeable for the autosomal calculation, with enough individuals of each sex to form the stipulated pairs. A randomly chosen distinct offspring pair are siblings with probability $1/(N-1)$: among the $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ pairs, exactly $N/2$ are within-family pairs. This fixed-family-size fact, rather than independent sampling of all parents, determines the recurrence.

First give the elementary chromosome-sharing idealization consistent with taking self-relatedness as one and ordinary sibling relatedness as $1/2$. In this model each individual carries one ancestral label, passed through one randomly chosen parent. Let $r_G$ be mean sharing between distinct individuals, and let $m_G$ be the mean sharing between mates. Siblings share their parental label with probability $(1+m_G)/2$. The average relatedness between parents from different couples isTo see the latter count, the ordered pairs from different couples are all ordered distinct pairs minus the $N$ ordered mating pairs. Each pair of parents from different couples contributes equally to the four possible cross-family offspring relationships. Combining sibling and nonsibling offspring givesWith random mating, $m_G=r_G$ in expectation, so $r_0=0$ givesThe mean sharing tends to one, on a generation scale of order $2(N-1)$. It is not linear indefinitely: repeated ancestry saturates the probability. This is an exact result for the stated one-label approximation, not an exact diploid formula after substantial inbreeding has developed.

Now prohibit full-sibling matings, with $N\ge 4$ and representative sampling of allowed nonsibling mating pairs. The following closed recurrences assume that opposite-sex mating pairs have the same mean autosomal sharing as all nonsibling pairs; a sex-specific or assortative scheme requires separate mating means. In the exchangeable one-label model write $u_G$ for mean relatedness of a nonsibling pair, so $m_G=u_G$. The two required recurrences areThe second relation includes sharing through grandparents and more distant ancestors; setting $u_G=0$ forever would be wrong. For $v_G=1-r_G$, $w_G=1-u_G$, the linear recurrence matrix isIts eigenvalues areboth of magnitude less than one for $N\ge 4$. Thus $r_G\to 1$ even when siblings never mate. Explicitly, $1-r_G=A{\lambda }_{+}^G+(1-A){\lambda }_{-}^G$, where $A=[1-1/(2(N-1))-{\lambda }_{-}]/({\lambda }_{+}-{\lambda }_{-})$. Sibling avoidance prevents immediate close-relative matings, but it does not make the remaining population unrelated. In a fixed-two-offspring model it need not reduce average pairwise sharing at every generation: a smaller mating mean actually increases the displayed increment in $r_G$.

For a fully diploid formulation, self-kinship also changes with inbreeding and cannot remain artificially fixed. Let ${\Phi }_G$ be mean kinship coefficient between distinct individuals, $F_G$ mean inbreeding coefficient, and ${\Phi }_{m,G}$ mean kinship of mates. A sibling pair has kinship $(1+F_G)/4+{\Phi }_{m,G}/2$, because parental self-kinship is $(1+F_G)/2$. Repeating the pair count yields finite-population chromosomal coancestry:Initially ${\Phi }_0=F_0=0$. For random mating take ${\Phi }_{m,G}={\Phi }_G$. An explicit solution is the affine matrix recurrenceIts characteristic polynomial is ${\lambda }^2-[1-1/(2(N-1))]\lambda -1/[4(N-1)]$; its two roots have magnitude less than one. Hence ${\Phi }_G,F_G\to 1$. The normalized relationship $2{\Phi }_G/(1+F_G)$ tends to one, while unnormalized additive relationship $2{\Phi }_G$ tends to two, demonstrating why that coefficient cannot always be read literally as a probability.

With sibling avoidance replace ${\Phi }_{m,G}$ by the nonsibling mean $U_G$ and addThese equations compute every generation's kinship and inbreeding, including delayed consanguinity through cousins. Under random allowed mating in a connected closed population, backward ancestral gene-copy lineages have a positive chance to meet in each sufficiently long block of generations; repeated blocks force eventual coalescence. Thus sibling avoidance alone does not prevent long-term identity by descent and genetic drift. Maintaining distinct isolated breeding groups, introducing immigrants, or changing offspring-number variance gives a different model. The elementary exponential formula answers the fixed-self-sharing approximation; the diploid recurrences give the consistent inbreeding-aware answer.