Penetrance Functions for Parametric Linkage

Why Penetrance Enters Linkage Analysis

A parametric linkage analysis needs a disease model as well as a recombination model. The recombination fraction describes how often disease and marker alleles are separated in transmission; the penetrance model describes how likely each person’s recorded phenotype is under a particular disease genotype. These are distinct ingredients. A person who carries a disease-associated genotype is not represented in the likelihood by a label alone: the model assigns a probability to that person’s phenotype, taking age into account when disease risk changes over time.

This tutorial focuses on the phenotype part of the calculation and how it combines with inheritance information. For clarity, the examples use specified disease-genotype risks and hypothetical pedigrees. They do not estimate penetrance parameters from data.

Definition: A penetrance function gives the probability of a specified disease phenotype conditional on genotype and, when relevant, age. For a binary phenotype recorded by age \(a\), write \(F_g(a)=P(Y=1\mid G=g,\text{ observed by age }a)\), where \(G=g\) is the person’s disease genotype and \(Y=1\) denotes affected. The corresponding probability of being unaffected through age \(a\) is \(1-F_g(a)\).

A penetrance value is a probability for the phenotype, not a probability that the person has a genotype. If the genotype is known, the value directly supplies the phenotype likelihood contribution. If the genotype is uncertain, the contribution must average over the possible genotypes using their probabilities under the inheritance model.

Genotype-Specific Models

The labels dominant and recessive describe which genotypes are assigned disease risk in a model. They do not, by themselves, specify the numerical risk. A model must state the penetrance for each genotype and, for age-dependent disease, the age at which the phenotype is assessed.

Definition: In a simple fully penetrant dominant model with disease allele \(D\), genotypes \(DD\) and \(Dd\) have affected probability 1, while \(dd\) has affected probability 0. In a simple fully penetrant recessive model with disease allele \(d\), genotype \(dd\) has affected probability 1, while \(Dd\) and \(DD\) have affected probability 0. These are idealized genotype-to-phenotype rules; other models specify other probabilities.

For age-dependent disease, the penetrance for a genotype is a function of age rather than a single constant. If \(F_g(a)\) is the cumulative probability of onset by age \(a\), it should not decrease with age: for \(a_2>a_1\), \(F_g(a_2)\geq F_g(a_1)\). Under a model in which disease onset is the event being recorded, an unaffected person examined at age \(a\) contributes the probability of having had no onset by that age, \(1-F_g(a)\). This makes age part of the phenotype observation, not merely a descriptive detail.

The standard calculation assumes that, conditional on a person’s genotype, age, and the specified model, that person’s phenotype has the stated probability. For a pedigree, multiplying individual phenotype probabilities additionally assumes conditional independence of the recorded phenotypes given the relevant genotypes and model parameters. The inheritance model supplies probabilities for genotypes that are not observed directly.

Worked Example: Writing Dominant and Recessive Penetrance Rules

Consider two separate hypothetical disease models, both assessed at an age when they are defined as fully penetrant. In the first, \(D\) is a dominant disease allele; in the second, \(d\) is a recessive disease allele. State the affected probabilities for all genotypes.

Dominant model. The model assigns \(F_{DD}=1\) and \(F_{Dd}=1\), because either genotype carries at least one \(D\) allele. It assigns \(F_{dd}=0\). Thus, for a known \(Dd\) person, an affected observation contributes 1 and an unaffected observation contributes \(1-1=0\).

Recessive model. The model assigns \(F_{dd}=1\), while \(F_{Dd}=0\) and \(F_{DD}=0\). A known \(Dd\) person therefore contributes 0 for an affected observation and \(1-0=1\) for an unaffected observation.

The same phenotype can thus have very different likelihood contributions under different genotype models. These rules are exact only because this example stipulates full penetrance at the assessment age.

From a Person’s Phenotype to a Pedigree Likelihood

Let person \(i\) have observed phenotype \(Y_i\), examination age \(a_i\), and disease genotype \(G_i\). If \(Y_i=1\), the phenotype contribution is \(F_{G_i}(a_i)\). If \(Y_i=0\), it is \(1-F_{G_i}(a_i)\). Define this contribution as \(q_i(G_i)\). Under the conditional independence assumption, the likelihood of the phenotype records given all disease genotypes is the product of the individual contributions.

q_i(G_i)= \begin{cases} F_{G_i}(a_i), & Y_i=1,\\ 1-F_{G_i}(a_i), & Y_i=0, \end{cases} \qquad P(\mathbf{Y}\mid\mathbf{G},\mathbf{a})=\prod_i q_i(G_i).

Genotypes in a pedigree may be latent. Let \(z\) denote a possible inheritance state: the set of founder alleles and transmissions that determines the disease genotypes, and, when markers are included, the marker alleles transmitted as well. A state has probability \(P(z\mid\theta)\), where \(\theta\) is the recombination fraction, together with any specified founder-genotype probabilities. Let \(M\) denote the observed marker data. Given a state \(z\), the state determines the genotypes relevant to the phenotype model and the marker alleles relevant to the marker observations.

For a fixed state, the likelihood contribution combines the probability of that inheritance state, the probability of the marker data under it, and the phenotype probability. The likelihood for the observed pedigree is obtained by summing over the possible states. This is the ordinary law of total probability applied to unobserved inheritance states.

L(\theta)=\sum_z P(z\mid\theta)\,P(M\mid z)\, \prod_i q_i\!\left(G_i(z)\right).

If the founder distribution is not already included in \(P(z\mid\theta)\), its probability must also appear in each state’s weight. Likewise, any other modeled observations require their own conditional likelihood contributions. The equation makes the role of penetrance precise: it weights each possible inheritance state according to how well that state’s implied disease genotypes explain the observed phenotypes. It does not replace the transmission probabilities or marker likelihood.

Key result: With latent inheritance states, the pedigree phenotype likelihood is not generally the phenotype probability for one guessed genotype assignment. It is a probability-weighted sum across the possible assignments, with each assignment’s phenotype contribution computed from the relevant genotype-specific penetrances.

Worked Example: An Age-Dependent Phenotype With an Uncertain Genotype

Suppose a hypothetical dominant model assigns a disease genotype \(D d\) a cumulative onset probability \(F_{Dd}(50)=0.70\) by age 50 and assigns \(F_{dd}(50)=0\). A person examined at age 50 has probability 0.60 of genotype \(Dd\) and probability 0.40 of genotype \(dd\), based on the inheritance information. Calculate the phenotype contribution for an affected record and for an unaffected record.

Affected at age 50. Conditional on \(Dd\), the contribution is 0.70; conditional on \(dd\), it is 0. The probability averaged across genotype uncertainty is

P(Y=1\mid\text{inheritance information}) =0.60(0.70)+0.40(0)=0.42.

Unaffected at age 50. The genotype-specific probabilities are \(1-0.70=0.30\) for \(Dd\) and \(1-0=1\) for \(dd\). Therefore,

P(Y=0\mid\text{inheritance information}) =0.60(0.30)+0.40(1)=0.18+0.40=0.58.

The two probabilities sum to \(0.42+0.58=1\), as they must for the two possible binary outcomes under the same model and information. The calculation averages phenotype probabilities over genotypes; it does not average the genotype labels themselves.

Deriving the Sum Over Inheritance States

The likelihood formula follows by partitioning the possible data-generating outcomes according to the unobserved inheritance state. For each state \(z\), the joint probability of marker and phenotype data is the marker probability times the phenotype probability, conditional on that state, under the model’s conditional-independence assumptions. The joint probability of the data and state is then multiplied by the state’s probability. Summing these mutually exclusive state-specific probabilities gives the probability of the observed data without conditioning on a particular hidden state.

Condition on one state.
For a particular \(z\), calculate its weight \(P(z\mid\theta)\), its marker contribution \(P(M\mid z)\), and the phenotype product \(\prod_i q_i(G_i(z))\).

Form the joint contribution.
Under the stated conditional-independence model, the contribution for this state is \(P(z\mid\theta)P(M\mid z)\prod_i q_i(G_i(z))\).

Sum over states.
Different inheritance states are mutually exclusive and cover the possible latent transmissions, so adding their joint contributions yields the likelihood of the observed pedigree data.

A short two-state version illustrates the same reasoning. If only \(z_1\) and \(z_2\) are possible, then \(P(M,\mathbf{Y})=P(M,\mathbf{Y},z_1)+P(M,\mathbf{Y},z_2)\). For each state, the joint term is \(P(z_j)P(M,\mathbf{Y}\mid z_j)\); factorizing the conditional data term gives the state-specific expression above. Extending this identity to every possible state gives the full sum. This derivation also explains why choosing only the most likely state generally does not give the likelihood: it discards the probabilities of the other possible states.

Worked Example: A Marker Transmission and an Unaffected Child

Consider a hypothetical parent with phased disease-marker haplotypes \(D\!-\!M\) and \(d\!-\!m\). The other parent is \(d\!-\!m/d\!-\!m\). A child is observed to have marker genotype \(M/m\) and to be unaffected at age 40. Assume the child’s disease genotype is \(Dd\) if the first parent transmits \(D\), and \(dd\) if that parent transmits \(d\). The disease model gives \(F_{Dd}(40)=0.80\) and \(F_{dd}(40)=0\). Ignore other pedigree phenotypes in this illustrative likelihood contribution.

The observed marker genotype requires the first parent to transmit \(M\). With recombination fraction \(\theta\), the probability of transmitting the \(D\!-\!M\) haplotype is \((1-\theta)/2\), and the probability of transmitting the recombinant \(d\!-\!M\) combination is \(\theta/2\). Given the first event, the child is \(Dd\), so the probability of being unaffected at 40 is \(1-0.80=0.20\). Given the second, the child is \(dd\), so the probability of being unaffected is \(1-0=1\).

Summing the two mutually exclusive transmission paths gives the joint contribution for the marker observation and phenotype:

L_{\text{child}}(\theta) =\frac{1-\theta}{2}(0.20)+\frac{\theta}{2}(1) =0.10(1-\theta)+0.50\theta =0.10+0.40\theta.

At \(\theta=0.10\), this contribution is \(0.10+0.40(0.10)=0.14\). At \(\theta=0.50\), it is \(0.10+0.40(0.50)=0.30\). For these specified observations and assumptions, the joint contribution is larger at \(\theta=0.50\) than at \(\theta=0.10\). This comparison alone is not a complete linkage analysis: a full pedigree likelihood includes all modeled individuals and observations, and a likelihood ratio compares the relevant likelihoods under its stated hypotheses.

Common Mistakes and Interpretation

Using affected penetrance for an unaffected person. An unaffected person’s contribution by age \(a\) is \(1-F_g(a)\), not \(F_g(a)\). The two probabilities represent distinct outcomes.
Ignoring age in an age-dependent model. An unaffected observation at age 25 and one at age 70 can have different probabilities under the same genotype. Use the penetrance at the person’s observed age.
Substituting a guessed genotype for an uncertain one. When genotype is latent, sum the state-specific contributions with their inheritance probabilities. Selecting one plausible state does not account for the others.
Confusing phenotype probability with transmission probability. Penetrance describes the phenotype conditional on genotype; \(\theta\) governs recombination in transmission. Both enter the likelihood, but they are not interchangeable.
Interpreting a likelihood as a probability that a model is true. A likelihood is the probability of the observed data viewed as a function of model parameters. A larger likelihood at one parameter value does not by itself give the probability that the corresponding model is true.

Penetrance functions matter because phenotype records can support or weigh against particular inheritance paths even when marker data do not identify those paths uniquely. Age-specific probabilities also prevent an unaffected young person from being treated as equivalent to an unaffected older person when the modeled disease risk accumulates over time. In a parametric linkage calculation, transparent genotype-specific penetrances make clear exactly how the phenotype evidence affects each state’s weight.

Check Your Understanding

Work through these questions using the definitions and calculations above.

In a fully penetrant recessive model, what phenotype probability does an unaffected \(Dd\) person contribute?
If \(F_g(60)=0.65\), what is the probability that a person with genotype \(g\) is unaffected through age 60 under the binary onset model?
A person has probabilities 0.30 and 0.70 of genotypes \(G_1\) and \(G_2\). Their affected penetrances at the examination age are 0.80 and 0.10. What is the affected phenotype contribution after averaging over genotype uncertainty?
Why does a pedigree likelihood sum over possible inheritance states rather than use only the most likely state?
In a linkage likelihood, what role does the recombination fraction play, and what distinct role does the penetrance function play?