Age-Specific Penetrance for Late-Onset Disease

Why Age Matters in Penetrance

For a late-onset disease, an unaffected person in their thirties and an unaffected person in their seventies do not necessarily provide the same evidence about a genetic model. The older relative has had more time to develop disease. A model that assigns one fixed probability of disease to every carrier, regardless of age, can therefore misrepresent the family data.

Age-specific penetrance describes how the probability of disease changes with age for each genotype class. It lets us calculate, for example, the probability that a carrier has developed disease by age 60, or the probability that a relative with an uncertain genotype has developed disease by their observed age. The probabilities are conditional on the genetic model; they do not by themselves show that a genotype causes disease.

Definition: Let $T$ denote age at disease onset and let $G=g$ denote a specified genotype class. The cumulative age-specific penetrance for genotype $g$ at age $a$ is $$ F_g(a)=P(T\leq a\mid G=g). $$ It is the probability that a person in genotype class $g$ has developed the disease by age $a$, under the specified model. The probability of being disease-free at age $a$ is $1-F_g(a)$.

A penetrance model may use genotype classes such as carriers and noncarriers, or more detailed classes such as the three genotypes at a locus. The class definitions must be part of the model. In a model that allows phenocopies, for example, noncarriers can also have positive penetrance: their disease risk need not be zero.

For any fixed genotype class, cumulative penetrance must be between 0 and 1 and cannot decrease with age. Someone who has developed the disease by age $a$ has also developed it by any later age $b>a$. Thus a valid model requires $F_g(a)\leq F_g(b)$. Penetrance is not the probability of becoming ill during a particular interval; that is a different quantity.

Reading an Age-Specific Penetrance Model

A model can specify $F_g(a)$ with a formula or with values at selected ages. The table below gives a hypothetical model for a late-onset disease. “Carrier” and “noncarrier” are the two genotype classes in this example; the noncarrier values allow for disease arising through other causes.

Age $a$, in years	Carrier $F_{\mathrm{car}}(a)$	Noncarrier $F_{\mathrm{non}}(a)$
40	0.12	0.01
50	0.30	0.025
60	0.48	0.04
70	0.68	0.065
80	0.78	0.09

For a carrier, the model assigns probability 0.48 to having developed disease by age 60 and probability $1-0.48=0.52$ to being disease-free at age 60. These are cumulative probabilities from the model's time origin, not probabilities for the decade from age 50 to 60. The increase from 0.30 at age 50 to 0.48 at age 60 is 0.18, the probability of being a carrier and developing disease during that interval without conditioning on being disease-free at age 50.

Worked Example: Carrier Status at Two Ages

Suppose a hypothetical relative is known to be a carrier. Using the table, find (1) the probability of disease by age 60, (2) the probability of remaining disease-free through age 60, and (3) the probability of developing disease between ages 60 and 80, given that the relative is disease-free at age 60.

By age 60. The cumulative penetrance is $F_{\mathrm{car}}(60)=0.48$. Thus the probability of disease by age 60 is 0.48, or 48%.

Disease-free through age 60. This is the complement: $$ 1-F_{\mathrm{car}}(60)=1-0.48=0.52. $$ The model assigns probability 0.52, or 52%, to remaining disease-free through age 60.

Onset from 60 through 80, conditional on being disease-free at 60. The probability of onset after age 60 and by age 80, before conditioning, is $F_{\mathrm{car}}(80)-F_{\mathrm{car}}(60)=0.78-0.48=0.30$. Divide by the probability of being disease-free at 60: $$ P(60<T\leq 80\mid T>60,G=\mathrm{car}) =\frac{F_{\mathrm{car}}(80)-F_{\mathrm{car}}(60)}{1-F_{\mathrm{car}}(60)} =\frac{0.78-0.48}{1-0.48} =\frac{0.30}{0.52} \approx 0.5769. $$ Among carriers who are disease-free at 60, the model gives about a 57.69% probability of developing disease by age 80. This conditional interval risk is not the same as the 30% increase in cumulative penetrance.

From Cumulative Penetrance to Interval Risk

The interval formula follows directly from conditional probability. If a person is disease-free at age $a$, then $T>a$. Among those people, the event of developing disease by age $b$, where $b>a$, is $a<T\leq b$. Its probability before conditioning is $F_g(b)-F_g(a)$, while the probability of the conditioning event is $1-F_g(a)$.

\begin{aligned} P(a<T\leq b\mid T>a,G=g) &=\frac{P(a<T\leq b\mid G=g)}{P(T>a\mid G=g)}\\ &=\frac{F_g(b)-F_g(a)}{1-F_g(a)}. \end{aligned}

This expression is defined when $1-F_g(a)>0$, meaning that the model assigns a positive probability of being disease-free at age $a$. If a model gives $F_g(a)=1$, no one in that genotype class remains disease-free at age $a$, so the conditional interval probability is not defined.

When ages fall between tabulated ages, a model must say how to obtain $F_g(a)$. A piecewise-constant, linear, or parametric rule gives different values between table entries. Do not silently interpolate unless the specified model allows it. In all cases, the resulting values should remain between 0 and 1 and be nondecreasing with age.

Key distinction: $F_g(a)$ is cumulative disease probability by age $a$. The difference $F_g(b)-F_g(a)$ is the unconditional probability, within genotype class $g$, of onset in the interval $(a,b]$. The ratio $\{F_g(b)-F_g(a)\}/\{1-F_g(a)\}$ is the interval probability conditional on being disease-free at age $a$.

Relatives With Uncertain Genotypes

A relative's genotype may not be known with certainty. If the pedigree information and transmission model give genotype probabilities $P(G=g\mid I)$, where $I$ denotes the available family information, the relative's disease probability by age $a$ is the weighted average of the penetrances for the possible genotypes:

P(T\leq a\mid I)=\sum_g P(G=g\mid I)F_g(a).

Each genotype-specific penetrance is weighted by the probability of that genotype given the information. This is the law of total probability. The genotype probabilities must refer to the relative's actual pedigree information, not to an assumed 50:50 split unless the transmission model and known parental genotypes justify that split.

Worked Example: A Child With an Uncertain Genotype

In a hypothetical family, one parent is a known heterozygous carrier and the other parent is a known noncarrier. Assume Mendelian transmission and no other relevant genotype uncertainty. A child therefore has probability $1/2$ of being a carrier and probability $1/2$ of being a noncarrier. Using the table, calculate the child's probability of disease by age 60 and the probability of being disease-free through age 60.

The carrier penetrance at age 60 is 0.48 and the noncarrier penetrance is 0.04. Averaging over the two possible genotypes gives $$ \begin{aligned} P(T\leq 60) &=P(G=\mathrm{car})F_{\mathrm{car}}(60) +P(G=\mathrm{non})F_{\mathrm{non}}(60)\\ &=(0.50)(0.48)+(0.50)(0.04)\\ &=0.24+0.02=0.26. \end{aligned} $$ The child's model-based probability of disease by age 60 is 0.26, or 26%.

The probability of being disease-free by age 60 is the complement, $1-0.26=0.74$. It can also be calculated by averaging genotype-specific disease-free probabilities: $$ (0.50)(1-0.48)+(0.50)(1-0.04) =(0.50)(0.52)+(0.50)(0.96) =0.26+0.48=0.74. $$ The two calculations agree, as they must. This result is not the penetrance of either genotype alone; it averages across the child's uncertainty about genotype.

Combining Disease Statuses at Different Ages

When several relatives are included, each person's disease-status probability must use that person's age and genotype class. If the relatives' disease outcomes are conditionally independent given their genotypes and the model, the probability of a particular set of statuses is the product of their individual probabilities. That conditional-independence assumption is a modeling choice; shared exposures or other dependence not represented in the model could make it inappropriate.

For an affected relative observed at age $a$, the relevant cumulative probability is $F_g(a)$. For a relative known to be disease-free through age $a$, the corresponding probability is $1-F_g(a)$. These are the basic age-specific status probabilities. When genotype is uncertain, average over the possible genotypes for that relative before combining probabilities, while respecting any dependence among relatives' genotypes implied by the pedigree.

Worked Example: Two Relatives Observed at Different Ages

Consider two hypothetical siblings, both known to be carriers. The first is affected by age 50; the second is disease-free through age 70. Assume their disease outcomes are conditionally independent given their genotypes and the specified penetrance model. Calculate the probability of this pair of statuses.

For the first sibling, the probability of being affected by age 50 is $F_{\mathrm{car}}(50)=0.30$. For the second sibling, the probability of remaining disease-free through age 70 is $$ 1-F_{\mathrm{car}}(70)=1-0.68=0.32. $$ Under the stated conditional-independence assumption, multiply these probabilities: $$ P(\text{first affected by 50, second unaffected through 70}) =(0.30)(0.32)=0.096. $$ The model assigns probability 0.096, or 9.6%, to this pair of observations, conditional on both siblings being carriers. Using the same penetrance value for both siblings would be incorrect because their observed ages differ.

Short Justification and Model Checks

The calculations above rely on conditioning on genotype and age. The identity for interval risk can be verified by partitioning the event of being disease-free at age $a$: among those disease-free at $a$, a person either remains disease-free through $b$, or develops disease in $(a,b]$. Therefore,

1-F_g(a)=\{1-F_g(b)\}+\{F_g(b)-F_g(a)\}.

Dividing the interval term $F_g(b)-F_g(a)$ by $1-F_g(a)$ gives its conditional probability among those disease-free at $a$. This also shows why the conditional interval probability cannot be obtained by simply subtracting two cumulative penetrances: subtraction gives the interval event probability in the original genotype class, not among those who were still disease-free at the interval's start.

Before using a specified model, check that its genotype classes are clear, that its age units and disease definition are consistent, and that its penetrance values are valid probabilities and nondecreasing over age. If a relative's age is not exactly one of the model's tabulated ages, identify the stated rule for evaluating penetrance there. If genotype is uncertain, use the appropriate genotype probabilities rather than assigning the relative to a genotype without justification.

Common Mistake: Treating Penetrance as a Constant

A common mistake is to say, “The carrier penetrance is 48%,” and then use 0.48 for every carrier, regardless of age. In the example model, 0.48 is specifically the probability of disease by age 60. At age 50 the cumulative probability is 0.30, and at age 80 it is 0.78. A single number cannot represent all three age-specific probabilities.

Another error is to interpret the increase from 0.48 at age 60 to 0.78 at age 80 as a 30% conditional risk for carriers who are unaffected at age 60. The increase is the probability of onset in that interval before conditioning. The conditional probability is $0.30/0.52\approx0.5769$, because only the 52% who remain disease-free at age 60 are at risk for onset in the interval under this formulation.

Key takeaway: State the genotype class and age whenever reporting penetrance. For relatives with different ages, use the appropriate age-specific probabilities; for uncertain genotypes, average over genotype probabilities supported by the pedigree model.

Check Your Understanding

Use the hypothetical penetrance table in this tutorial where needed. State any assumptions required for your calculations.

For a noncarrier, what is the model probability of being disease-free through age 70?
For a carrier disease-free at age 50, calculate the conditional probability of developing disease by age 70.
A child has probability 0.25 of being a carrier and 0.75 of being a noncarrier. Calculate the child's probability of disease by age 80.
Explain the difference between $F_g(80)-F_g(60)$ and the conditional probability of onset by age 80 among people with genotype $g$ who are disease-free at age 60.
Two conditionally independent carrier relatives are observed disease-free through ages 40 and 80, respectively. Express the probability of both statuses using the penetrance table.