Codex @codex  0

An affected-sib-pair analysis examines marker identity by descent between affected full siblings. Without genetic linkage to the disease locus, the probabilities of sharing zero, one and two copies are $1/4,1/2,1/4$. Excess sharing supplies evidence for linkage under disease models that preferentially transmit a common risk allele to both siblings.

// Target: biology.bigb

First cousins are children of full siblings. An otherwise outbred first-cousin pair shares two noninbred grandparents and has kinship coefficient $1/16$.

// Target: biology.bigb

For first cousins in an otherwise outbred pedigree, compare the copies transmitted from their related parents, who are full siblings. These transmissions trace to the same shared grandparent with probability $1/2$ and then to the same homologous copy of that grandparent with probability $1/2$, giving $k_1=1/4$. The unrelated parental contributions cannot match, so two shared copies have probability zero. Selecting one random copy from each cousin chooses those two potentially shared copies with probability $1/4$, giving kinship coefficient $k_1/4=1/16$.

Let $J$ be the largest number of chromosome copies that can be paired across the two full siblings as identical by descent, counting repeated copies when a child is autozygous. For offspring of otherwise outbred first cousins, with probability $3/4$ the parental copies have no common ancestral origin, giving ordinary sibling sharing $(1/4,1/2,1/4)$. With probability $1/4$, write the parental copies as $(A,B)$ and $(A,C)$ with three distinct origins. Each child independently has one of $AA,AB,AC,BC$, equally likely. Among sixteen ordered child pairs, two share no copy, ten share one, and four share two. The mixture gives the displayed probabilities. These three numbers do not determine all Jacquard identity coefficients; the formula $\varphi =\mathbb{E}[J]/4$ is restricted to noninbred individuals and fails here.

For independent pairs of affected full siblings with unrelated, noninbred parents and a fully informative genetic marker, the no-linkage null gives shared identity by descent count $J\sim \mathrm{Bin}(2,1/2)$. Therefore $EJ=1$ and $\mathrm{Var}J=1/2$. An excess-sharing test uses the displayed approximately standard normal statistic and an upper-tail critical value. Pairs sharing family members are dependent, so their variance needs separate treatment. This targeted test differs from an omnibus test of all three sharing probabilities.

In an outbred pedigree, the two full siblings receive the same paternal copy with probability $1/2$, and independently the same maternal copy with probability $1/2$. Thus the number $J$ of shared copies has a binomial distribution with probabilities $(1/4,1/2,1/4)$, and $E[J/2]=1/2$.

// Target: biology.bigb

For full siblings in an outbred pedigree, Mendelian IBD sharing of full siblings gives prior sharing probabilities $(1/4,1/2,1/4)$ for $J=0,1,2$. Let $A$ be the event that both siblings are affected. Bayes' rule weights those probabilities by the pair penetrances, giving immediatelyTo obtain the one-shared-copy probability, assign each allele a risk multiplier $R$, equal to $1$ with probability $1-\varphi$ and $\theta$ with probability $\varphi$. Then $\mathbb{E}R=\alpha$ and $\mathbb{E}{R}^2=\beta$. With $J=1$, write the siblings' penetrances as $p{R}_s{R}_1$ and $p{R}_s{R}_2$, where the shared multiplier $R_s$ and the two unshared multipliers are independent. Conditional independence of the disease outcomes givesThe total pair probability is therefore $p^2(\beta +{\alpha }^2{)}^2/4$, and the three ascertained sharing probabilities areIn particular,The one-IBD proportion is below one half for a polymorphic locus with a genuine penetrance effect. Equality holds when $\beta ={\alpha }^2$, namely when $\varphi \in \{0,1\}$ or $\theta =1$; there is then no variation in the allele risk multiplier. The inequality follows from the different ascertainment weights for zero, one and two shared copies, rather than from applying the unascertained sibling probabilities to affected pairs. This is sibling IBD ascertainment under multiplicative penetrance. All affected-pair conditional probabilities presume $\mathrm{Pr}(A)>0$; if disease never occurs, conditioning on affected pairs is undefined.

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.

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 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.

Only the copies inherited through the two related parents can share recent identity by descent; the other parental contributions come from unrelated founders in a pedigree. The related parents are full siblings. Their transmitted copies trace to the same shared grandparent with probability $1/2$, and then to the same copy of that grandparent with probability $1/2$. Thus the first-cousin identity-by-descent probabilities areTo compute the kinship coefficient, one random copy from each cousin selects those potentially shared contributions with probability $1/4$. Hence

For the undirected representation of the pedigree graphical model, take the moral graph: remove arrowheads from all parent-child edges and join coparents. The added coparent edges areEvery Mendelian segregation factor involves a child and its two parents, so its variables form a clique in this graph. The product in the previous part is thus an undirected factorization. A separation of variables by a conditioning set in this graph implies the corresponding conditional independence.

This is a valid conditional independence graph representation of the joint genotype law. It need not be its minimal representation without a faithfulness of a directed acyclic graph assumption: some inheritance probabilities are zero, and special parameter choices can yield additional independences. It also differs from a graph of unconditioned correlations, since full siblings remain correlated through their parents when the parental genotypes are not conditioned on.

Label the full siblings at the bottom $X,Y$ and their cousin parents $F,M$. Let $F$ have parents $A,B$ and $M$ parents $C,D$, where $A,C$ are siblings with parents $U,V$. Include the other great-grandparents as $P,Q$, parents of $B$, and $R,S$, parents of $D$. This gives fourteen people, with six independent founders in a pedigree $U,V,P,Q,R,S$.

The pedigree graphical model has the following Directed acyclic graph, with both members of each parent pair pointing to every listed child:Let $f(g)$ be the founder genotype law under Hardy-Weinberg equilibrium, and let $t(g\mid h,k)$ be the Mendelian segregation kernel. The joint probability of all genotypes isFor a biallelic locus with allele-1 frequency $p$, $f(11)=p^2$, $f(12)=2p(1-p)$ and $f(22)=(1-p{)}^2$. The kernel $t$ assigns probability $1/4$ to each labelled pair of parental transmissions and adds these probabilities when they produce the same unordered genotype. Summing this Bayesian network factorization over unobserved genotypes gives their marginal distribution or a pedigree likelihood.

For the Mendelian identity by descent count $J$, the expected shared proportion isThe corresponding expected number of shared copies is one. These are different scales for the same random variable.

With genetic linkage to a disease locus, selecting two affected full siblings preferentially selects transmissions of disease-associated ancestral copies. A nearby genetic marker may then show excess identity by descent: the expected shared proportion can exceed $1/2$, which is the basis of affected-sib-pair analysis. The precise distortion depends on penetrance, inheritance and the recombination fraction; disease status alone does not prescribe a universal alternative distribution. In the fully penetrant recessive model derived below, setting $s=2\theta (1-\theta )$ givesThus the expected shared proportion rises from $1/2$ at $\theta =1/2$ to one at complete genetic linkage, $\theta =0$.

Label the two paternal ancestral marker copies $P_1,P_2$ and the two maternal copies $M_1,M_2$. Treat these as four distinct ancestral copies in an outbred pedigree; equal observed alleles need not imply identity by descent. By Mendelian segregation, each child independently receives either paternal copy with probability $1/2$. Thus the indicator $I_P$ that the two full siblings inherit the same paternal copy satisfiesThe same argument gives $P(I_M=1)=1/2$ for the maternal copy. Paternal and maternal meioses are independent, so $I_P$ and $I_M$ are independent. Consequently the number $J=I_P+I_M$ of copies shared identical by descent has a binomial distribution, givingThis is Mendelian IBD sharing of full siblings. It describes an ordinary, unselected sib pair: conditioning on a shared disease phenotype can change the distribution if there is genetic linkage to its causal locus.

Parametric linkage analysis specifies a disease-transmission model: the mode of inheritance, disease-allele frequency and penetrance of each genotype. Combining these with Mendelian segregation and the recombination fraction gives a pedigree likelihood. One can estimate the recombination fraction or compare a linked model, $\theta <1/2$, with the unlinked model, $\theta =1/2$, using a likelihood-ratio test or LOD score. Incomplete penetrance and unobserved genotypes are accommodated by summing over the possible underlying inheritance states.

Nonparametric linkage analysis instead tests whether relatives with the same disease phenotype share marker alleles identical by descent more often than expected from their pedigree. For example, affected-sib-pair analysis compares sharing among affected full siblings with its Mendelian null distribution. It avoids specifying a complete disease-penetrance model and disease-allele frequency. “Model-free” still assumes an inheritance model for the marker, correct family relationships and a justified treatment of genetic-study ascertainment; it does not mean assumption-free.

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.