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A genetic genetic carrier has one disease-associated recessive allele and one normal allele at a genetic locus, and is unaffected under a fully penetrant recessive model. Two genetic carriers have affected-offspring probability $1/4$ under Mendelian segregation; a genetic carrier and an individual with no disease allele have affected-offspring probability zero under that same mutation-free model.

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.

Assume a single disease allele $a$, complete penetrance of $aa$, no disease in $Aa$ or $AA$, and Hardy-Weinberg equilibrium in the outbred population. If its population frequency is $q$, the disease prevalence is $q^2=10^{-6}$, so $q=10^{-3}$.

For a child with inbreeding coefficient $F$, the two inherited copies are identical by descent with probability $F$. In that component there is just one ancestral allele draw, and it is $a$ with probability $q$. In the remaining component there are two independent ancestral draws, with probability $q^2$ of $aa$. Thus the inbreeding genotype frequencies giveUsing the first-cousin inbreeding coefficient $F=1/16$, the disease-risk ratio isThe absolute risk in this model is $6.34375\times 10^{-5}$. This calculation averages over the parental genotypes in first-cousin marriages before observing an affected child. It is not the risk conditional on both parents being known genetic carriers, and assumes the specified cousin relationship is independent of disease status rather than conditioning on a selected disease pedigree.

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.

Each relevant cousin marriage is a cross between two genetic carriers of $q$. A child receives two copies of $q$ with probability $(1/2)(1/2)=1/4$, and fails to receive two with probability $3/4$. Conditional on the parental genotypes, distinct offspring transmissions are independent. There are eight specified double-copy children and eight specified remaining children, givingThere are no binomial coefficients because the subjects have been individually selected. If only the counts in the three sibships were specified, rather than their identities, the probability would instead be multiplied by $\left(\genfrac{}{}{0ex}{}{1}{1}\right)\left(\genfrac{}{}{0ex}{}{7}{4}\right)\left(\genfrac{}{}{0ex}{}{8}{3}\right)=1960$.

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.

Under the original assumptions the extra affected child is impossible. A generation-3 woman can have at most one copy of the unique introduced disease allele, while an unrelated outside spouse has none. Their children cannot inherit two defective copies. Thus both linkage and no-linkage likelihoods of the enlarged data are zero, and the original single-entry model has no defined updated LOD score. At least one assumption must be relaxed: another introduced disease allele, a new mutation, a phenocopy, or the relationship of the outside spouse.

If the model is enlarged to allow an independent outside genetic carrier, the reported extra information contains disease status but no new genetic marker genotypes. In the simple extension with fixed outside heterozygous-carrier probability $c$, the old disease observations force both generation-2 family ancestors to be genetic carriers. A further generation-3 sister inherits the disease copy with probability $1/2$, independently of the existing offspring transmissions. Given that she and her spouse are genetic carriers, exactly one of their three children is affected with probabilityThe additional phenotype likelihood is therefore $27c/128$. It is the same at $\theta =0$ and $\theta =1/2$, so it cancels and the LOD remains unchanged in this phenotype-only, independent-carrier extension. If the identity of the affected child is specified, the factor is $9c/128$, which also cancels. No such conclusion applies automatically if new genetic marker data are supplied or the revised pedigree founder model couples this branch to the old genetic marker evidence.

In general an enlarged data set changes the score byafter choosing a coherent model with positive likelihood. The additional affected child alone does not justify assigning it the autozygosity evidence of an affected child of a cousin marriage.

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

Sibling recurrence risk is the conditional probability that another child has the disease, given an affected sibling. For unaffected parents, complete penetrance and a single recessive disease allele, the affected child identifies both parents as genetic carriers. Independent Mendelian segregation then gives recurrence probability $1/4$, whether or not the parents are related. Incomplete penetrance, phenocopies, different causal loci or affected parents change this calculation.

// Target: biology.bigb