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Segregation Analysis · Tutorial 202 of 989

Penetrance and Phenocopy Models

Define penetrance and phenocopy probabilities, then calculate disease and unaffected probabilities for specified carrier and noncarrier genotypes.

Intermediate 9 min read

What You'll Learn

  • Define penetrance as a conditional disease probability for a specified genotype and assessment horizon.
  • Distinguish incomplete penetrance in carriers from phenocopies in noncarriers.
  • Calculate disease and unaffected probabilities from genotype-specific penetrance values.
  • Explain when carrier status alone does not determine one unique disease probability.
  • Check the assumptions behind a simple penetrance and phenocopy model.

Why Genotype-Specific Disease Probabilities Matter

A disease-associated genotype does not guarantee that a person will develop the disease, and a person without that genotype may still develop it. A segregation model must allow for both possibilities. Penetrance describes the disease probability associated with a genotype; a phenocopy is a disease phenotype in someone who lacks the genotype defined as disease-associated in the model.

These probabilities describe disease status conditional on genotype. They are not probabilities that a genotype will be inherited, and they do not by themselves describe how a genotype is transmitted through a family. The genotype configurations established for a family and the disease probabilities assigned to those genotypes are separate parts of the model.

Definition and Basic Model

Let \(G\) denote an individual's genotype at the locus being modeled, and let \(D=1\) indicate that the individual has the disease by a specified assessment time. Let \(D=0\) indicate that the individual does not have the disease by that time. In a simple model, the penetrance for genotype \(g\) is the conditional probability of disease given that genotype.

Definition: The genotype-specific penetrance is \(f_g=P(D=1\mid G=g)\). The corresponding probability of being unaffected is \(P(D=0\mid G=g)=1-f_g\). If \(g_0\) is defined as a noncarrier genotype, then \(f_{g_0}\) is the model's phenocopy probability: the probability of disease in a noncarrier.

A carrier is an individual with at least one copy of the susceptibility allele specified by the model. In a simple two-class model, all carriers share one penetrance \(f_C\), and all noncarriers share one phenocopy probability \(f_N\). Then \(P(D=1\mid\text{carrier})=f_C\) and \(P(D=1\mid\text{noncarrier})=f_N\). Usually, incomplete penetrance means \(f_C<1\), while a nonzero phenocopy probability means \(f_N>0\).

These two-class assumptions are simplifying choices, not universal properties of genetic disease. For a genotype with two copies of the susceptibility allele, for example, the model may assign a different probability from the one assigned to a heterozygote. In that case, “carrier” includes multiple genotypes, and carrier status alone does not specify one penetrance unless the model gives those genotypes the same probability.

A penetrance model requires a clear disease definition and assessment horizon. The probabilities apply to the disease status defined for that horizon; a probability for status by a specified age, for instance, should not be interpreted as a probability for status at every age. This tutorial uses fixed, clearly stated assessment times rather than modeling changes in risk over age.

Assumptions for a Simple Model

Before calculating with penetrance values, make explicit what the model assumes. In the examples below:

  • Disease status is binary and is defined consistently for all individuals.
  • Genotype is classified correctly, and the genotype categories in the model are specified in advance.
  • The penetrance values apply to the population and assessment horizon being considered.
  • Within a genotype category, the model assigns the same disease probability to each person. Differences due to other genes, environment, or other causes are not modeled separately.
  • The noncarrier disease probability is permitted to be greater than zero; it represents disease cases not explained by the modeled susceptibility genotype.

These assumptions define a statistical probability model. They do not establish that the modeled genotype causes disease, and they do not imply that every disease in a noncarrier has one particular known cause.

Worked Example: Incomplete Penetrance and Phenocopies

Suppose a hypothetical model assigns a disease probability of 0.70 to carriers and 0.03 to noncarriers by a stated assessment time. Calculate the disease and unaffected probabilities for each group.

Carriers. By definition, \(P(D=1\mid\text{carrier})=0.70\). The unaffected probability is the complement:

$$ P(D=0\mid\text{carrier})=1-0.70=0.30. $$

Thus, this model has incomplete penetrance: 30% of carriers are predicted to remain unaffected by the specified time.

Noncarriers. The model assigns \(P(D=1\mid\text{noncarrier})=0.03\), so:

$$ P(D=0\mid\text{noncarrier})=1-0.03=0.97. $$

The 3% disease probability is the phenocopy probability under this model. It does not mean that 3% of noncarriers must be affected in every finite group; it is a probability for each noncarrier under the stated assumptions.

Calculating Risk for More Than One Genotype

When the model assigns separate probabilities to different genotypes, calculate disease and unaffected probabilities separately for each genotype. A useful table makes it harder to accidentally treat all carriers as having the same probability.

Worked Example: Heterozygotes and Homozygotes

Consider a hypothetical susceptibility allele \(A\), with \(aa\) defined as noncarrier. Suppose a model assigns penetrance 0.02 to \(aa\), 0.60 to \(Aa\), and 0.85 to \(AA\), all by the same assessment time. Find the disease and unaffected probabilities for each genotype.

For \(aa\), the disease probability is 0.02 and the unaffected probability is \(1-0.02=0.98\). The 0.02 is the noncarrier phenocopy probability.

For \(Aa\), the disease probability is 0.60 and the unaffected probability is \(1-0.60=0.40\). For \(AA\), the disease probability is 0.85 and the unaffected probability is \(1-0.85=0.15\).

$$ \begin{aligned} P(D=1\mid G=aa)&=0.02, & P(D=0\mid G=aa)&=0.98,\\ P(D=1\mid G=Aa)&=0.60, & P(D=0\mid G=Aa)&=0.40,\\ P(D=1\mid G=AA)&=0.85, & P(D=0\mid G=AA)&=0.15. \end{aligned} $$

Here, both \(Aa\) and \(AA\) are carrier genotypes, but they do not have the same penetrance. It would therefore be incorrect to report one carrier disease probability without stating whether it refers to \(Aa\), \(AA\), or a defined mixture of these genotypes.

Why the Complement Formula Holds

For any fixed genotype \(g\), the two disease-status events \(D=1\) and \(D=0\) are mutually exclusive and together include every possible status under the binary definition. Conditional on \(G=g\), their probabilities must therefore sum to one:

$$ P(D=1\mid G=g)+P(D=0\mid G=g)=1. $$

Rearranging gives \(P(D=0\mid G=g)=1-P(D=1\mid G=g)=1-f_g\). This complement is a conditional probability for the same genotype and the same assessment horizon. It should not be replaced by an unconditional proportion of unaffected people, which would also depend on how common each genotype is.

Worked Example: Expected Affected Counts in Genotyped Groups

Suppose a hypothetical sample contains 20 people with genotype \(aa\), 10 with \(Aa\), and 4 with \(AA\). Assume the penetrances from the previous example apply, and that all genotypes are known. Calculate the expected number of affected people in each genotype group.

For \(aa\), the expected affected count is group size multiplied by the genotype-specific probability:

$$ 20(0.02)=0.4. $$

For \(Aa\):

$$ 10(0.60)=6. $$

For \(AA\):

$$ 4(0.85)=3.4. $$

The total expected affected count is \(0.4+6+3.4=9.8\) among 34 people. The expected count need not be an integer: it is an average over repeated samples under the model, not a prediction that exactly 0.4 people in the \(aa\) group will be affected. This calculation uses known genotype-group sizes; it does not determine an individual’s genotype from disease status.

Interpreting Carrier and Noncarrier Probabilities

A penetrance value is a conditional probability, not a certainty. If \(P(D=1\mid G=g)=0.60\), the interpretation is: among individuals with genotype \(g\), the model assigns a 60% probability of disease by the stated assessment time. It does not say that 60% of all affected people have genotype \(g\). That reverse conditional probability would require information about genotype frequencies as well as disease probabilities.

Likewise, a phenocopy probability is not the probability that a particular affected person is a noncarrier. It is \(P(D=1\mid\text{noncarrier})\). The distinction between \(P(D=1\mid\text{noncarrier})\) and \(P(\text{noncarrier}\mid D=1)\) is important: they condition on different events and generally have different values.

When all carriers share one probability and all noncarriers share another, the model’s carrier and noncarrier risks can be compared directly. For example, with carrier risk 0.70 and noncarrier risk 0.03, the difference is \(0.70-0.03=0.67\), or 67 percentage points. This is a contrast between modeled conditional probabilities, not evidence by itself that carrier status causes the difference.

Common Mistakes and Checks

  • Treating a susceptibility genotype as deterministic. The incorrect reasoning is “the person is a carrier, so the person must have the disease.” The correction is to use the assigned penetrance; if it is below 1, unaffected carriers have positive probability.
  • Setting noncarrier risk to zero automatically. A model may allow noncarriers to develop the phenotype. Use the specified phenocopy probability rather than assuming every disease case is due to the modeled genotype.
  • Using one carrier probability when genotypes differ. If \(Aa\) and \(AA\) have different penetrances, state which genotype is being discussed or provide a defined carrier-group calculation. Do not silently treat their probabilities as equal.
  • Reversing the conditional probability. \(P(D=1\mid G=g)\) is not \(P(G=g\mid D=1)\). The latter cannot be obtained from penetrance alone.
  • Confusing penetrance with transmission. Penetrance concerns disease conditional on genotype. The probability that a parent transmits an allele is a separate probability.
Key takeaway: Specify disease status, genotype categories, and assessment time first. Then use \(P(D=1\mid G=g)=f_g\) for disease and \(P(D=0\mid G=g)=1-f_g\) for unaffected status. A nonzero noncarrier value allows phenocopies, while a carrier value below one allows incomplete penetrance.

Check Your Understanding

Use the stated genotype and assessment time in each answer.

  1. A model assigns penetrance 0.72 to a carrier genotype. What is the probability that a person with that genotype is unaffected?
  2. If the disease probability for noncarriers is 0.04, what is the model’s phenocopy probability, and what is the noncarrier unaffected probability?
  3. Why is a penetrance of 0.60 not the probability that an affected person is a carrier?
  4. A model assigns different penetrances to \(Aa\) and \(AA\). What information is needed before reporting one probability for “carriers” as a group?
  5. In a group of 25 people with the same genotype and penetrance 0.08, what is the expected number affected? Explain why the answer need not be a whole number.
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