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Parametric Linkage Analysis · Tutorial 236 of 989

Sensitivity to Disease Allele Frequency and Penetrance Assumptions

Learn how disease-allele frequency and penetrance assumptions affect pedigree likelihoods and how to compare the resulting LOD scores transparently.

Intermediate 10 min read

What You'll Learn

  • Write a pedigree likelihood that includes disease-genotype priors and penetrance.
  • Recalculate linkage evidence under alternative disease-allele frequencies.
  • Compare LOD scores after changing penetrance assumptions.
  • Distinguish disease-allele frequency from marker-allele frequency.
  • Report sensitivity analyses without treating assumptions as observed facts.

Why Disease-Model Assumptions Matter

Parametric linkage analysis evaluates how well observed family data fit linkage at a specified recombination fraction, relative to no linkage. The calculation also depends on a disease model: in particular, the assumed frequency of the disease allele and the probability that each disease genotype produces the phenotype. These are inputs to the likelihood, not quantities established by a linkage result.

A linkage result can therefore change when the disease model changes, even if the observed pedigree and marker data stay fixed. Sensitivity analysis makes that dependence visible: calculate the likelihood and LOD score under a primary set of assumptions, repeat the calculation under defensible alternatives, and compare the results. This is a comparison of model-based evidence, not a way to choose assumptions simply because they give the largest LOD score.

Definition: A disease-allele frequency \(q\) is the assumed frequency of the disease allele among population alleles. A penetrance is the probability of the defined phenotype for a specified disease genotype. For a simple dominant model, let \(g=0\) denote no disease allele and \(g=1\) denote at least one disease allele; write the penetrances as \(f_0=P(D\mid g=0)\) and \(f_1=P(D\mid g=1)\), where \(D\) means affected under the study's phenotype definition.

Where the Assumptions Enter

For a pedigree, the disease genotypes and the transmissions that produced them may be unknown. Let \(z\) represent a possible set of these latent states, \(M\) the observed marker data, and \(Y\) the observed disease phenotypes. A schematic likelihood at recombination fraction \(\theta\) is

$$ L(\theta;q,f_0,f_1) = \sum_z P(z\mid q)\,P(M\mid z,\theta) \prod_i f_{g_i}^{y_i}(1-f_{g_i})^{1-y_i}, $$

where \(y_i=1\) if person \(i\) is affected and \(y_i=0\) otherwise, and \(g_i\) is that person's disease-genotype class in state \(z\). The product is the phenotype likelihood conditional on the genotypes. The summation adds the contributions from all possible latent states, weighted by their probabilities. A full pedigree likelihood also includes the appropriate transmission probabilities and any required conditioning for the way families were ascertained.

The frequency \(q\) affects the prior probabilities of disease genotypes. For example, under Hardy–Weinberg equilibrium, the probability that a founder carries at least one copy of a dominant disease allele is \(1-(1-q)^2\). Penetrance affects the phenotype likelihood terms: an affected person with \(g=1\) contributes \(f_1\), while an unaffected person with \(g=1\) contributes \(1-f_1\). A disease-allele frequency is not a marker-allele frequency; the latter belongs to the marker model.

For each assumed disease model, the LOD score at a specified \(\theta\) compares its likelihood with the likelihood at no linkage, \(\theta=0.5\):

$$ \operatorname{LOD}(\theta;q,f_0,f_1) = \log_{10} \left[ \frac{L(\theta;q,f_0,f_1)} {L(0.5;q,f_0,f_1)} \right]. $$

The numerator and denominator must use the same disease-allele frequency, penetrances, phenotype definition, and other model inputs. Comparing a numerator from one disease model with a denominator from another does not give the LOD score for either model.

Worked Example: Changing the Disease-Allele Frequency

Worked Example: Changing the Disease-Allele Frequency

Setup. Consider a hypothetical dominant-disease pedigree with one affected phenotype configuration and marker data. Assume founders are in Hardy–Weinberg equilibrium, the penetrance and marker assumptions are held fixed, and ascertainment is either independent of phenotype or properly accounted for. For illustration, the latent-state calculation gives the following conditional likelihood contributions: if a founder carries the disease allele, the contribution is \(0.120\) at \(\theta=0.10\) and \(0.040\) at \(\theta=0.5\); if not, the contribution is \(0.002\) at either value. These are hypothetical likelihood components for this example, not observed frequencies.

First take \(q=0.01\). The probability that a founder carries at least one disease allele is

$$ 1-(1-q)^2=1-(0.99)^2=0.0199. $$

Thus the noncarrier probability is \(1-0.0199=0.9801\). Weight each conditional contribution by its genotype prior probability and add:

$$ \begin{aligned} L(0.10;0.01) &=(0.0199)(0.120)+(0.9801)(0.002)\\ &=0.002388+0.0019602=0.0043482,\\ L(0.5;0.01) &=(0.0199)(0.040)+(0.9801)(0.002)\\ &=0.000796+0.0019602=0.0027562. \end{aligned} $$

The LOD score is therefore

$$ \operatorname{LOD}(0.10;0.01) = \log_{10}\left(\frac{0.0043482}{0.0027562}\right) = \log_{10}(1.578)\approx 0.198. $$

Now change only the disease-allele frequency to \(q=0.05\). The carrier probability becomes \(1-(0.95)^2=0.0975\), and the noncarrier probability is \(0.9025\). The same conditional contributions give

$$ \begin{aligned} L(0.10;0.05) &=(0.0975)(0.120)+(0.9025)(0.002)\\ &=0.011700+0.001805=0.013505,\\ L(0.5;0.05) &=(0.0975)(0.040)+(0.9025)(0.002)\\ &=0.003900+0.001805=0.005705. \end{aligned} $$

So \(\operatorname{LOD}(0.10;0.05)=\log_{10}(0.013505/0.005705)\approx 0.3743\). In this particular example, the higher assumed frequency raises the LOD score. That direction is not guaranteed: changing \(q\) changes weights across possible genotype states, and those states need not favor linkage by the same amount.

Worked Example: Changing Penetrance

Worked Example: Changing Penetrance

Setup. For a second hypothetical pedigree, hold \(q\), the marker model, and the observed data fixed. Assume the same dominant genotype classes as above and that ascertainment is handled consistently. Compare a baseline penetrance model, \(f_0=0.02,\ f_1=0.80\), with an alternative model, \(f_0=0.01,\ f_1=0.95\). The higher \(f_1\) in the alternative model assigns a smaller probability to an unaffected disease-allele carrier; the lower \(f_0\) assigns a smaller probability to an affected noncarrier.

Suppose summing over all latent genotypes and transmissions under the baseline penetrances gives \(L(0.10)=0.012772\) and \(L(0.5)=0.011188\). These are hypothetical likelihoods from the pedigree calculation. Substitution gives

$$ \operatorname{LOD}_{\text{baseline}}(0.10) = \log_{10}\left(\frac{0.012772}{0.011188}\right) = \log_{10}(1.14158) \approx 0.05751. $$

Under the alternative penetrances, suppose the corresponding summed likelihoods are \(L(0.10)=0.014000\) and \(L(0.5)=0.012000\). Then

$$ \operatorname{LOD}_{\text{alternative}}(0.10) = \log_{10}\left(\frac{0.014000}{0.012000}\right) = \log_{10}(1.16667) \approx 0.06695. $$

The LOD score is higher under the alternative model in this example, but the change is modest. It does not establish that the alternative penetrances are correct. The likelihoods must be recomputed because changing penetrance changes the relative support of latent genotype states that can explain the observed affected and unaffected relatives.

Why Sensitivity Changes a LOD Score

The LOD score is a log of a likelihood ratio, not a log of the numerator alone. If a change in assumptions multiplied both \(L(\theta)\) and \(L(0.5)\) by the same positive constant, their ratio—and hence the LOD score—would be unchanged. Sensitivity arises when the assumptions affect the two likelihoods differently. This happens when plausible genotype and transmission states have different relative support under linkage and no linkage.

The calculation can be organized as a controlled comparison:

1
Specify a primary disease model.
State the disease-allele frequency, penetrances, phenotype definition, and any relevant genotype assumptions.
2
Hold the observed data fixed.
Use the same pedigree phenotypes and marker data in each calculation, and apply the same ascertainment treatment.
3
Vary assumptions deliberately.
Change one input at a time when identifying its effect, or define a small set of joint alternatives when evaluating combined uncertainty.
4
Recompute both likelihoods.
For every disease model, calculate \(L(\theta)\) and \(L(0.5)\) under that same model before taking the logarithm of their ratio.
5
Report the comparison in context.
Give the assumptions and resulting LOD scores together, and describe whether the linkage evidence is stable across the alternatives considered.

A useful sensitivity table can list each combination of \(q\), \(f_0\), and \(f_1\), with the likelihoods and LOD score. If the conclusion changes substantially across scientifically defensible values, the result depends on the disease model and should be reported as such. If it remains similar, that is evidence of robustness to the particular assumptions examined—not proof that every possible model gives the same result.

Common Pitfall: Treating One LOD Score as Assumption-Free

A common mistake is to say, “The LOD score is 2, so the data provide that amount of evidence regardless of the disease model.” The correction is that a parametric LOD score is conditional on the specified model. Disease-allele frequency sets prior genotype weights, while penetrance determines phenotype probabilities given genotypes; either can change the likelihood ratio. Another mistake is to change penetrance and update only the linked likelihood. The unlinked likelihood must also be recomputed under the new penetrance model.

Sensitivity analysis does not mean selecting the frequency or penetrance that maximizes the score and presenting that score as if the assumptions were known. Instead, choose alternatives justified by relevant biological or population information, show how the evidence changes, and keep assumptions distinct from observations. The goal is to make model dependence interpretable.

Key takeaway: Disease-allele frequency and penetrance enter different parts of the pedigree likelihood. Recalculate the linked and unlinked likelihoods under each consistent disease model, then compare the resulting LOD scores to assess how sensitive the linkage evidence is to those assumptions.

Check Your Understanding

Use the definitions and calculations above to answer the following questions.

  1. Under Hardy–Weinberg equilibrium, what is the probability that a founder carries at least one copy of a dominant disease allele when \(q=0.02\)?
  2. In the likelihood expression, which factor is affected directly by penetrance: the prior probability of a latent genotype state or the phenotype likelihood conditional on that state?
  3. Why must \(L(0.5)\) be recalculated when penetrance assumptions change?
  4. For the first worked example, explain why increasing \(q\) does not guarantee that the LOD score will increase in every pedigree.
  5. What does a stable LOD score across a set of plausible penetrance assumptions support, and what does it not prove?

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