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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.

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Full siblings share both parents. With unrelated, noninbred founders in a pedigree, the two parental contributions give independent opportunities for identity by descent.

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Homozygosity means that the two allele states at a diploid genetic locus are the same; an individual with that property is a homozygote. Observed homozygosity does not by itself establish identity by descent.

A homozygote carries two identical alleles at a locus; a heterozygote carries different alleles. Homozygosity of an observed haplotype is identity by state, and need not be identity by descent.

Two allele copies are identical by state when their observed states agree, regardless of whether they derive from one recent ancestral copy. Equal marker haplotypes can therefore reflect either identity by descent or matching founder haplotypes.

For two individuals carrying labelled copies $(a,b)$ and $(c,d)$, condensed identity coefficients give the probabilities of nine identity by descent patterns, identifying states that differ only by exchanging the two homologues within either individual. They retain within-individual autozygosity, which three ordinary outbred sharing coefficients omit. One convention isEach block consists of copies identical by descent, and different blocks have distinct ancestral origins. For siblings whose parents are otherwise outbred first cousins, these coefficients are $(1,0,2,1,2,1,15,30,12)/64$. This follows by conditioning on whether the parents share one ancestral copy and enumerating the four independent Mendelian segregation choices to their two children. The kinship coefficient is ${\Delta }_1+({\Delta }_3+{\Delta }_5+{\Delta }_7)/2+{\Delta }_8/4$, here $9/32$. Weir's discussion of identity states uses this convention.

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.

Nonparametric linkage analysis tests excess identity by descent among relatives with a shared phenotype, without specifying its complete penetrance model or disease-allele frequency. Its null still uses Mendelian segregation and the specified pedigree.

// Target: biology.bigb

For calculations of identity by descent, an outbred pedigree has unrelated, noninbred founders in a pedigree, and no additional ancestral relationships beyond those explicitly specified. Founder gene copies are treated as distinct ancestral copies even if they have identity by state.

// Target: biology.bigb

Use the usual genetic model in which disease outcomes are conditionally independent given the two individuals' genotypes, and unshared founder alleles are independently drawn from the same outbred population. Let $J$ denote their number of allele copies shared by identity by descent. Conditional independence then givesWhen $J=2$, both individuals have the same genotype $G$. Its distribution remains the population genotype distribution under Hardy-Weinberg equilibrium, so their individual trait means are $K=p{\alpha }^2$ andThereforeWhen $J=0$, all four unshared copies are independent, making $G_1,G_2$ independent population genotypes. Thus $\mathbb{E}(q_{G_1}{q}_{G_2})=(p{\alpha }^2{)}^2$ andThe residual-independence assumption is necessary. Marginal penetrances alone do not determine joint disease probabilities. For example, with $\theta =1$, setting both traits equal to one shared Bernoulli variable of probability $p$ preserves every marginal penetrance, but gives covariance $p(1-p)$ even with no shared alleles. Shared environmental disease determinants or population structure therefore require an expanded model. The displayed answers are the intended locus-only covariance results, not a consequence of the penetrance table without that usual assumption.

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.

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 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.

Write $D$ for the disease observations and $M$ for the genetic marker observations. The LOD score for complete linkage isThe disease marginal cancels because the single-locus disease segregation law does not depend on the genetic marker–disease recombination fraction. The ten genetic markers form one nonrecombining haplotype; they do not supply ten independent copies of the pedigree's segregation evidence.

There is a distinction between identity by state and identity by descent. The reported observation that affected subjects all have $h/h$ establishes the former. An exact genetic marker likelihood requires pedigree founder haplotype frequencies, or the actual pedigree founder genotypes if conditioning on them. Here is an explicit expression using independent pedigree founder haplotypes, with frequency $p$ for the observed haplotype $h$, and linkage equilibrium between pedigree founder genetic marker and disease states. There are eight genetic marker copies in the four pedigree founders: the original couple and the two generation-2 outside spouses. For genetic marker inheritance configuration $v$, let $K(v)$ be the number of distinct pedigree founder copies ancestral to the sixteen chromosome copies in the eight selected subjects. Thenis the unlinked probability that all those copies have state $h$. At complete linkage the affected subjects all inherit the disease pedigree founder copy twice, and the probability that this copy has genetic marker state $h$ is $p$. Therefore

The denominator can be evaluated without an unspecified inheritance sum. Let $B_a=\left(\genfrac{}{}{0ex}{}{2}{a}\right)p^a(1-p{)}^{2-a}$ for $a=0,1,2$. If parental genetic marker genotypes contain $a,b$ copies of $h$, the child count distribution isFor a generation-2 couple defineFor a particular generation-3 parent with $j$ genetic marker copies, the chance it transmits $h$ to all its $r$ selected offspring is $(j/2{)}^r$, which explains each factor. Conditional on the original founding couple, its two generation-2 siblings and their independent outside spouses give independent left and right contributions. ThusExpanding givesIt satisfies $R(1)=1$, as it must for a monomorphic haplotype. This expression supplies a numerical LOD once $p$ is specified; the problem does not specify that frequency.

In the idealized limit where a matching rare haplotype identifies one ancestral copy, the leading coefficient is particularly simple. Dropping restrictions on the eight unaffected subjects, the probability that all selected subjects are autozygous for one common copy isAccordingly $R(p)/p\to 2^{-22}$ as $p\to 0$, and the affected-only IBD calculation gives $Z(0)=22{\log }_{10}2\simeq 6.623$.

If the genetic marker observations additionally establish the exact pattern $E$ of part (d), including that the eight unaffected subjects are not autozygous, its idealized IBD likelihood ratio is insteadThis latter value requires the additional negative genetic marker observations. It cannot be inferred merely from homozygosity in the affected subjects, and neither simplified IBD score is an exact identity-by-state likelihood for an unspecified common haplotype.