Phenocopies and Incomplete Penetrance in Disease Pedigrees

Why Disease Status Is Probabilistic Evidence

A disease pedigree is not always a perfect record of who carries a disease allele. A carrier may remain unaffected, and a person without the disease-associated genotype may nevertheless develop the disease. These observations do not automatically contradict a genetic model. They can be represented by incomplete penetrance and a phenocopy probability, respectively.

In linkage analysis, this distinction matters because the observed disease statuses help evaluate possible inheritance patterns. If disease status were treated as a direct label for genotype, an unaffected carrier or affected noncarrier could incorrectly receive probability zero under the model. Instead, each person’s status contributes a probability conditional on their latent genotype. When that genotype is uncertain, the likelihood averages over possible genotypes using their inheritance probabilities.

The values \(f_1\) and \(f_0\) are model probabilities, not diagnoses of why a particular person is affected or unaffected. An affected noncarrier is compatible with the model when \(f_0>0\); the model does not establish that this individual’s disease was caused by a non-genetic factor. Similarly, an unaffected carrier is compatible with the model when \(1-f_1>0\).

This tutorial uses a binary disease status for clarity. With an age-specific penetrance model, the relevant probability is the one for the person’s observed age or follow-up status. The probabilities must refer to the same disease definition and observation conditions as the pedigree data.

Likelihood Contribution of an Observed Status

Let \(Y_i=1\) if person \(i\) is affected and \(Y_i=0\) if unaffected. Given that person’s genotype \(g_i\), their disease-status likelihood contribution is

Thus an affected person contributes \(f(g_i)\), while an unaffected person contributes \(1-f(g_i)\). Under the simple carrier/noncarrier model, an affected carrier contributes \(f_1\), an unaffected carrier contributes \(1-f_1\), an affected noncarrier contributes \(f_0\), and an unaffected noncarrier contributes \(1-f_0\).

For multiple people whose disease statuses are conditionally independent given their genotypes and the model parameters, their phenotype likelihood is the product of their individual contributions:

This conditional-independence assumption concerns disease statuses after conditioning on the modeled genotypes and any relevant covariates. If the model includes additional shared causes or dependence between relatives’ disease statuses, the product form may need to be extended. The inheritance model supplies probabilities for the genotypes; penetrance supplies probabilities for the observed statuses given those genotypes.

When Genotype Is Uncertain: Average Over Possibilities

A pedigree analysis often does not know every person’s disease genotype with certainty. Let \(Z\) represent a complete assignment of the relevant latent inheritance states and resulting genotypes in a pedigree. For specified model parameters, the observed disease-status likelihood is obtained by summing over all possible assignments, weighting each by its probability under the inheritance model.

Here \(g_i(z)\) is the genotype assigned to person \(i\) under inheritance state \(z\). A state that is unlikely under the transmission model receives little weight; a state that makes the observed disease statuses unlikely receives a small phenotype contribution. The sum combines these two sources of evidence without declaring any observed status to be a certain genotype label.

In a linkage likelihood, the latent states also encode marker inheritance and disease-allele transmission. The same principle applies: sum the probability of the observed data over compatible latent states, including transmission probabilities, marker-data probabilities where relevant, and genotype-specific disease-status probabilities. The exact factors depend on the specified model and observed data. The phenotype term must not be replaced by a rule that forces all affected people to be carriers or all unaffected people to be noncarriers.

A Small Pedigree Likelihood

Consider a hypothetical family with one known heterozygous carrier parent and one known noncarrier parent under a dominant disease model. Assume the carrier parent transmits the disease allele to each child with probability 0.50, and transmissions to the two children are independent. Use \(f_1=0.80\) and \(f_0=0.02\). Suppose the first child is affected and the second is unaffected.

For the affected child, the probability of the observed status is 0.80 if the child inherited the disease allele and 0.02 otherwise. Since each genotype has probability 0.50, the affected child’s marginal contribution is

For the unaffected child, the corresponding probabilities are \(1-f_1=0.20\) for a carrier and \(1-f_0=0.98\) for a noncarrier. Therefore,

The children’s transmissions and disease statuses are independent conditional on the specified parental genotypes and model. Summing over each child’s possible genotype therefore gives the joint likelihood contribution

This value is the likelihood of the two observed child statuses under the stated model and known parental genotypes. It does not mean that 24.19% of families with these parents will have exactly these statuses under every possible model; it is specific to the hypothetical parameters and assumptions above. If parental disease statuses were also part of the observed data, their genotype-specific contributions would also be included.

Justification: Why the Sum Over Genotypes Is Required

The marginalization formula follows from the law of total probability. The possible assignments \(z\) partition the latent inheritance states, so the event of observing the disease statuses can be separated according to which assignment occurred. For each assignment, multiply its probability by the probability of the statuses conditional on that assignment, then add across assignments:

The last equality uses the conditional-independence assumption for disease statuses given the genotypes and model. This calculation explains why one cannot select only the genotype assignment that best matches the observed statuses and treat it as certain. Every assignment with positive probability contributes, weighted by its inheritance probability and its phenotype likelihood.

Practical Checks and Common Pitfalls

Before calculating a pedigree likelihood, state the disease-status probabilities and the genotype probabilities being used. Check that each penetrance lies between 0 and 1, that an unaffected contribution is \(1-f(g)\), and that inheritance weights for mutually exclusive states sum to 1. When an age-specific model is used, make sure the penetrance matches the observed age or follow-up information rather than applying a lifetime probability indiscriminately.

• Wrong reasoning: “The person is unaffected, so they cannot be a carrier.” Correction: An unaffected carrier has likelihood \(1-f_1\), which is positive whenever penetrance is below 1.
• Wrong reasoning: “The affected person is a noncarrier, so the disease model is impossible.” Correction: If \(f_0>0\), an affected noncarrier has a nonzero status contribution \(f_0\).
• Wrong reasoning: “An affected noncarrier proves a phenocopy occurred.” Correction: The model assigns a probability to the observation; disease status alone does not establish its cause.
• Wrong reasoning: “Use the most plausible genotype and ignore the alternatives.” Correction: When genotype is uncertain, sum over the possible states with their inheritance probabilities, as required by the law of total probability.
• Wrong reasoning: “The phenotype likelihood is the same as the genotype probability.” Correction: The inheritance model gives genotype probabilities; penetrance gives the probability of observed disease status conditional on genotype. Both enter the pedigree likelihood.