Worked Ascertainment-Corrected Pedigree Analysis

Why Ascertainment Changes a Pedigree Likelihood

A family recruited because one member is affected is not a random sample of families with respect to that member’s phenotype. A likelihood that treats the affected proband as though the family had been sampled without selection can therefore give the wrong contribution for an ascertained family. The correction is to condition the phenotype likelihood on the event that caused the family to enter the study.

This tutorial works through that correction for a two-child pedigree. The parents’ genotypes are unobserved, and two possible parental mating classes are assigned specified prior probabilities. The proband is affected by the recruitment rule; the sibling is observed to be unaffected. We calculate the likelihood of that sibling observation conditional on the proband being affected, under two different penetrance models.

Pedigree, Genotype Summation, and Assumptions

Let \(A\) denote a disease-associated allele and \(a\) the other allele. Write \(g\in\{0,1,2\}\) for the number of \(A\) alleles in a child’s genotype. The pedigree has two children: child 1 is the designated proband, selected because affected, and child 2 is an observed sibling. Neither child’s genotype nor either parent’s genotype is observed.

Represent uncertainty about the parental genotypes with a latent mating class \(C\). Before observing phenotypes, the two classes each have probability \(1/2\):

• \(C_1\): parents \(Aa\times aa\), so each child has probabilities \(P(g=0\mid C_1)=1/2\), \(P(g=1\mid C_1)=1/2\), and \(P(g=2\mid C_1)=0\).
• \(C_2\): parents \(Aa\times Aa\), so each child has probabilities \(P(g=0\mid C_2)=1/4\), \(P(g=1\mid C_2)=1/2\), and \(P(g=2\mid C_2)=1/4\).

These are stipulated prior probabilities for this hypothetical example; they are not estimates from a population allele frequency. Given a mating class, children’s genotypes are independent draws from the class-specific Mendelian probabilities. Given their genotypes, the children’s phenotypes are independent. Both children have the same penetrance function within a model, and there are no additional age or environmental covariates in this example.

For a given class \(C_c\), the probability that a child is affected is found by summing over that child’s possible genotypes:

This genotype sum is necessary because the genotypes are unobserved. The two children’s phenotypes are not generally independent after the mating class is averaged out: they share the same uncertain parental class. That shared uncertainty is why conditioning on the proband’s phenotype can change the sibling’s likelihood.

Deriving the Ascertainment-Corrected Likelihood

Let \(S\) be the selection event that child 1, the designated proband, is affected. The observed phenotype pattern is \(D=(\text{child 1 affected},\text{child 2 unaffected})\). For one mating class, conditional independence of the children given that class gives probability \(q_c(1-q_c)\) for this pattern. Averaging over the prior mating classes gives the joint probability. The selection probability is the marginal probability that child 1 is affected:

where \(w_c=P(C_c)\). Dividing the joint probability by the selection probability yields the conditional likelihood:

The division has a useful interpretation. Observing an affected proband changes the relative weights of the mating classes. In fact, \(P(C_c\mid S)=w_cq_c/\sum_jw_jq_j\), so equivalently \(L=\sum_c P(C_c\mid S)(1-q_c)\). Thus the correction does not merely remove an affected-proband factor; it also accounts for the information that the proband’s affected phenotype provides about the unobserved parental class.

Comparing the Two Penetrance Models

For Model 2, the genotype sums give \(q_1=0.275\) and \(q_2=0.500\). The probability of selecting a family through an affected proband is

The joint probability of an affected proband and unaffected sibling is

The corrected likelihood is therefore

For the observed pattern in this one ascertained family, the likelihood ratio comparing Model 2 with Model 1 is

Under these specified models and priors, the observed unaffected sibling is about 1.249 times as likely under Model 2 as under Model 1, conditional on selection through an affected proband. This is a likelihood comparison for the observed data, not a probability that either model is true. It also uses the conditional likelihoods: comparing the uncorrected joint probabilities would answer a different question because it would include the probability of entering the study.

A General Calculation Workflow

For a small pedigree with an ascertainment rule and unobserved genotypes, keep the genotype summation and selection conditioning explicit. A reliable sequence is:

Common Mistakes and Why the Correction Matters

A frequent error is to calculate \(P(\text{proband affected},\text{sibling unaffected})\) and call it the likelihood for a family recruited through an affected proband. That joint probability describes a family sampled before applying the recruitment condition. Once inclusion requires an affected proband, the relevant probability for the sibling data is conditional on that event, so the joint probability must be divided by \(P(\text{proband affected})\).

A second error is to average sibling unaffected probabilities using the original \(1/2,1/2\) class weights after observing an affected proband. Those are the prior weights. The proband’s phenotype changes the class weights to \(w_cq_c/\sum_jw_jq_j\). The quotient formula automatically makes this update.

Finally, the denominator must reflect the actual ascertainment rule. If selection required at least one affected member among several relatives, conditioning only on a designated proband being affected would not represent that design. The correction is tied to the sampling mechanism, not chosen for algebraic convenience.