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Autozygosity is homozygosity by descent: the maternal and paternal copies in one individual derive from the same ancestral copy. Unlike ordinary homozygosity, it concerns the ancestry of chromosome copies rather than only their observed states. In a pedigree with one ancestral disease-allele copy and fully penetrant recessive inheritance, every affected individual is autozygous for that copy.

Parametric linkage analysis forms a pedigree likelihood from a specified disease model, including penetrance, allele frequencies and recessive inheritance or another transmission model. It estimates or tests the recombination fraction between a disease locus and a genetic marker.

// Target: biology.bigb

Let $a$ denote the disease allele. Complete penetrance with recessive inheritance makes each affected child $aa$. The unaffected parents must therefore both be $Aa$ genetic carriers. Condition on these parental genotypes and the fully informative marker; the distinct paternal and maternal marker copies identify which parent's copies are shared. Average over the two equally likely marker-disease haplotype phases in each parent. This is the phase-symmetric model implicit in the requested pedigree likelihood identities.

For one parent, suppose first that both children receive the same marker copy. If that marker copy is on the parental chromosome carrying $a$, both children receive $a$ with probability $(1-\theta {)}^2$; in the opposite phase the probability is ${\theta }^2$. The phase-averaged probability isIf the children receive different marker copies, one disease-allele transmission is recombinant and the other is nonrecombinant in either phase. The probability that both receive $a$ is thereforePaternal and maternal transmissions are independent. Sharing zero marker copies means the different-copy case for both parents; sharing two means the same-copy case for both; sharing one means one of each. Thus the recessive affected-sib-pair likelihood isHere $L_j$ is the probability that both children are affected conditional on the marker sharing state $J=j$, not the probability of $J=j$ conditional on their being affected. Since $u,v\ge 0$,At $\theta =1/2$, all three conditional disease probabilities equal $1/16$, as expected from two independent children of an $Aa\times Aa$ mating. At $\theta =0$, only the two-copy-sharing state can produce two affected children. Phase weights other than $1/2$ would require a different likelihood function and need not obey these identities.

For a binary disease phenotype, put $p_i(g)=P(X_i=\text{affected}\mid G_i=g)$. The penetrance function depends on the inheritance model, the disease alleles and genotype, and potentially age, sex, environment and other genetic modifiers. Phenocopies allow disease without the putative disease genotype, while incomplete penetrance allows a genetic carrier of that genotype to remain unaffected. Thus the phenotype factor is $p_i(g)$ for an affected individual and $1-p_i(g)$ for an unaffected one.

Under a fully penetrant model with no phenocopies the genotype determines disease status, so the factor is 1 for a compatible genotype and 0 for an incompatible genotype. For example, in fully penetrant recessive inheritance, $aa$ is affected and $AA,Aa$ are unaffected. Missing phenotype observations contribute the constant factor 1 rather than excluding genotypes.

Let $D$ denote the specified disease pattern. With one introduced disease-allele copy, no new mutations or phenocopies, and fully penetrant recessive inheritance, every affected subject must carry two descendants of that one copy. Since every sibship has an affected child, all the upstream genetic carrier transmissions described above are forced. Conversely, unaffected siblings cannot carry two disease copies. Thus $D$ implies the inheritance event $E$ at the causal genetic locus, andThe rarity of the unconditional pattern in part (d) does not make its posterior probability small after observing the pattern that forces it. If the location of the one disease copy is initially uniform among the four founding copies, the likelihood of $D$ is the same for each, so its posterior location remains uniform; $P(E_q\mid D)=1/4$ for any named copy. If $q$ is already designated as the disease copy, its corresponding event has posterior probability 1.

For unaffected genetic carriers of a disease with complete penetrance and recessive inheritance, suppose parental marker-disease haplotype phases are equally likely and the marker is fully informative. Write $u=\theta (1-\theta )$ and $v=1/2-u$, where $\theta$ is the recombination fraction. Conditional on sharing a parental marker copy, the probability that both siblings inherit that parent's disease allele is $[(1-\theta {)}^2+{\theta }^2]/2=v$; conditional on inheriting different marker copies it is $\theta (1-\theta )=u$. Independent paternal and maternal transmissions therefore give $L_0=u^2,L_1=uv,L_2=v^2$. Conditional instead on both siblings being affected, the IBD probabilities are $s^2,2s(1-s),(1-s{)}^2$, where $s=2u\in [0,1/2]$.

// Target: biology.bigb