Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 45 / 3

Fix one particular ancestral allele copy $q$ among the four copies in the founding couple. The pedigree founder carrying it transmits it independently to each of the two generation-2 siblings with probability $1/2$. ThereforeHere “one copy” means a specified copy, as required to follow “this ancestral copy” in the subsequent parts. If instead one asks whether the siblings share at least one of the four copies without specifying which, the probability is $1-(1/2{)}^2=3/4$: they fail to share either the paternal or maternal copy with probability $1/4$. These are different events.

Conditional on both generation-2 siblings carrying the specified copy $q$, each passes it to each of their three children with probability $1/2$. Their outside partners do not carry that ancestral copy. There are six independent meioses, one for each generation-3 individual forming the three cousin marriages. ConsequentlyThis calculation ignores phenotype ascertainment in a genetic study, as requested; conditioning on the later disease observations would force these transmissions rather than assign them their prior probabilities.

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 $E_q$ be the event that all eight selected subjects carry two copies of the same specified ancestral copy $q$, and that no other generation-4 subject is autozygous at this genetic locus. The selected subjects occur in every sibship, so this event forces both generation-2 transmissions and all six relevant generation-3 transmissions. Applying the preceding conditional probabilities,Once all six parents carry $q$, their other copies come from the two unrelated outside pedigree founders, one on each side of the pedigree. A child not receiving two $q$ copies therefore cannot become autozygous for a different founding copy. This verifies the “only” condition, rather than merely excluding homozygosity for $q$.

For the event $E$ without specifying which of the original four copies is shared, the four alternatives $E_q$ are disjoint: a selected child cannot simultaneously have both its copies descended from two different ancestral copies. HenceAll selected individuals are then autozygous for the same copy and mutually 2-IBD. The distinction from the $3/4$ probability in the alternative interpretation of part (a) is precisely this disjointness.

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.

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.

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.