Codex @codex  0

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.

Use identity-by-descent sharing of inbred full siblings, with $J$ the maximum number of copies that can be paired across the children as identical by descent. If two copies within one child have the same ancestral origin, they still count as two copies when both can be paired.

With probability $3/4$ the cousin parents share no ancestral copy. Conditional on this event, their children have ordinary Mendelian IBD sharing of full siblings, namely $(1/4,1/2,1/4)$. With probability $1/4$, the parental copies can be labelled $(A,B)$ and $(A,C)$, with $A,B,C$ distinct ancestral origins. Each child independently receives one of $AA,AB,AC,BC$, with probability $1/4$ each. Among the sixteen ordered child pairs, $AA$ paired with $BC$ in either order gives $J=0$; four equal genotype-origin pairs give $J=2$; the remaining ten give $J=1$. The mixture isThese are inbred children, so the three sharing probabilities do not retain all within-child identities. The full Jacquard identity coefficients are $(1,0,2,1,2,1,15,30,12)/64$. In particular, their kinship coefficient is $9/32$; applying the outbred formula $(P(J=1)+2P(J=2))/4$ here would be incorrect.