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Rare-Variant Study Designs · Tutorial 467 of 989

Power Under Allelic Heterogeneity and Uncertain Effects

Quantify how allelic heterogeneity and uncertain effect directions can strengthen, dilute, or cancel gene-level association signals.

Advanced 10 min read

What You'll Learn

  • Relate rare-variant carrier frequency to genotype variance and gene-level signal.
  • Calculate approximate burden-test power under a specified effect configuration.
  • Explain why neutral variants can dilute an unweighted burden statistic.
  • Compare aligned, neutral, and protective effects using signed and squared signals.
  • Average power across uncertain effect configurations rather than relying on a single scenario.

Why Gene-Level Power Depends on the Effect Pattern

A gene may contain several rare variants that affect the same phenotype, but their effects need not be alike. Some variants may increase risk, some may be protective, and many may have no effect. Their frequencies and effect sizes can also differ. A gene-level test combines information across variants, so its power depends not just on how many variants are included, but on how their signals combine under the test being used.

This tutorial uses a simple burden-test approximation to make those dependencies explicit. It also contrasts signed burden signals with a quadratic, SKAT-like signal that does not cancel effects of opposite signs in the same way. These calculations are planning approximations, not substitutes for a design-specific analysis: finite samples, covariates, linkage disequilibrium, genotype uncertainty, and the exact test implementation can alter power.

Definition: Allelic heterogeneity means that multiple variants in a gene may contribute to a phenotype, with potentially different frequencies and effects. A variant's effect is neutral for the modeled outcome when its effect parameter is zero, risk-increasing when the parameter is positive, and protective when it is negative. The sign convention here uses the logistic-regression coefficient \(\beta_j\): positive values increase the log odds of the outcome, and negative values decrease them.

A Burden-Test Approximation

Consider a case-control study with \(N\) participants, of whom a proportion \(\pi\) are cases. For variant \(j\), let \(G_j\) be the participant's alternate-allele count, coded 0, 1, or 2, and let \(q_j\) be the allele frequency. Assume Hardy–Weinberg equilibrium, so the genotype variance is \(v_j=2q_j(1-q_j)\). For rare alleles, this variance is small; a variant carried by few participants contributes little information unless its effect is sufficiently large.

For the approximation below, assume unrelated participants, independent variants (no linkage disequilibrium), accurate genotypes, and a correctly specified logistic model with small effects. Let \(\beta_j\) be the log-odds effect per alternate allele. The model is \(\operatorname{logit}\{P(Y=1\mid G)\}=a+\sum_j\beta_jG_j\), where \(Y=1\) denotes case status. Consider an unweighted burden \(B=\sum_{j=1}^{K}G_j\). Under the null, its variance is approximately \(\sum_jv_j\).

The standardized burden score is approximately normally distributed with variance one. Under the alternative, its mean, denoted by \(\lambda\), is approximately:

$$ \lambda \approx \sqrt{N\pi(1-\pi)} \frac{\sum_{j=1}^{K}v_j\beta_j} {\sqrt{\sum_{j=1}^{K}v_j}}. $$

The factor \(N\pi(1-\pi)\) reflects the information from the case-control allocation. With a two-sided test at significance level \(\alpha\), approximate power is \(P(|Z|>c)\), where \(Z\sim N(\lambda,1)\) and \(c=z_{1-\alpha/2}\). Equivalently, power is \(\Phi(-c-\lambda)+1-\Phi(c-\lambda)\). A large absolute value of \(\lambda\) gives greater power; the sign matters for a one-sided test, and can determine whether effects cancel in a burden.

Why the Approximation Has This Form

A short score argument explains the numerator and denominator. Under the null model, the case indicator has probability \(\pi\), and the score for a burden coefficient is proportional to \(\sum_i(Y_i-\pi)B_i\). Its null variance is approximately \(N\pi(1-\pi)\operatorname{Var}(B)\). With independent variants, \(\operatorname{Var}(B)=\sum_jv_j\). Under small effects, the change in the expected case probability is approximately \(\pi(1-\pi)\sum_j\beta_jG_j\). Consequently, the score's expected shift is proportional to \(N\pi(1-\pi)\operatorname{Cov}(B,\sum_j\beta_jG_j)\), and independence gives \(\operatorname{Cov}(B,\sum_j\beta_jG_j)=\sum_jv_j\beta_j\). Dividing that shift by the null standard deviation yields the displayed expression for \(\lambda\).

This justification also identifies the approximation's limits. Larger effects can make the small-effect expansion inaccurate. Correlated variants add covariance terms, and adjusted analyses must account for covariates and the corresponding score variance. The formula is best used to understand design sensitivities and to compare simple scenarios, not as an exact power guarantee.

Worked Example: Power for One Rare Risk Variant

Suppose a hypothetical study has 2,000 cases and 2,000 controls, so \(N=4{,}000\) and \(\pi=0.5\). It tests one rare variant with genotype variance approximated as \(v=0.0099\) and log-odds effect \(\beta=0.30\). Assume the model and independence conditions above hold; with one variant, independence across variants is automatic. Use a two-sided significance level of 0.05.

Here \(N\pi(1-\pi)=4{,}000(0.5)(0.5)=1{,}000\). For one variant, the burden noncentrality is:

$$ \lambda=\sqrt{1{,}000}\, \frac{(0.0099)(0.30)}{\sqrt{0.0099}} =\sqrt{1{,}000(0.0099)}(0.30) =\sqrt{9.9}(0.30) \approx 0.9439. $$

At \(\alpha=0.05\), \(c\approx1.96\), so the approximate power is \(\Phi(-1.96-0.9439)+1-\Phi(1.96-0.9439)\), or \(\Phi(-2.9039)+1-\Phi(1.0161)\approx0.0018+0.1548=0.1566\). Thus, under these assumptions, the study has about 15.7% power to detect this effect with this test. The result illustrates that even a nonzero effect may be difficult to detect when the variant is rare.

Variant Count, Direction, and Effect-Size Variation

When variants have equal genotype variance \(v\), the burden mean simplifies to \(\lambda=\sqrt{N\pi(1-\pi)v}\,(\sum_j\beta_j)/\sqrt{K}\). If all \(K\) effects equal a common positive value \(\beta\), this becomes \(\sqrt{N\pi(1-\pi)vK}\,\beta\): adding aligned causal variants increases the burden signal. But the sum is signed. Equal positive and negative effects can cancel, while adding neutral variants increases the denominator without adding to the numerator. Both mechanisms can lower burden power.

A quadratic statistic behaves differently. To see the basic contrast, suppose variant-specific score statistics have been standardized and are approximately independent. If \(Z_j\) is the standardized score for variant \(j\), then the simple quadratic statistic \(Q=\sum_jZ_j^2\) has a central chi-square distribution with \(K\) degrees of freedom under the null. Under the alternative it is approximately noncentral chi-square, with noncentrality parameter \(\Lambda=\sum_j\lambda_j^2\), where \(\lambda_j=\sqrt{N\pi(1-\pi)v_j}\,\beta_j\). The squares prevent opposite signs from cancelling. Actual SKAT statistics generally use weights and have a weighted mixture of chi-square distributions; their power must be evaluated for that specified statistic. The simple \(Q\) calculation is an illustration of the directional contrast, not a universal SKAT power formula.

Thus, more variants do not automatically mean more power. The consequence of increasing the variant set depends on whether added variants carry effects, whether effects point in a common direction, and how the test weights information across variants. A burden test can be efficient when included effects are mostly aligned, while a variance-component test can be more robust to mixed directions. Neither is guaranteed to dominate across all configurations.

Worked Example: Aligned, Protective, and Opposing Effects

Consider four independent hypothetical variants with the same variance \(v=0.0099\), in a case-control design with \(N\pi(1-\pi)=1{,}000\). Assume the small-effect approximation is adequate. First suppose all four effects are \(+0.20\). The burden noncentrality is:

$$ \lambda=\sqrt{1{,}000(0.0099)} \frac{4(0.20)}{\sqrt{4}} =\sqrt{9.9}(0.40) \approx1.2586. $$

Now suppose two effects are \(+0.20\) and two are \(-0.20\). Their signed sum is \(2(0.20)+2(-0.20)=0\), so the burden approximation gives \(\lambda=0\). For a two-sided test, that corresponds to null power, approximately 0.05, under the stated approximation. The result does not mean the variants have no effects; it means this unweighted burden has no expected directional shift under that configuration.

For the simple quadratic illustration, each variant has \(\lambda_j=\sqrt{9.9}(0.20)\approx0.6293\), regardless of its sign. Therefore \(\Lambda=\sum_j\lambda_j^2=4(0.6293^2)\approx1.584\) in both the aligned and opposing configurations. The quadratic signal retains information in the opposing scenario, though this calculation alone does not give the power of a particular weighted SKAT implementation. It shows why effect direction is central to burden power but does not create the same cancellation in a squared-score approach.

Neutral Variants and Unequal Carrier Frequencies

An unweighted burden gives each included allele count the same coefficient, even though variants can differ substantially in frequency and effect. Under the approximation, a variant contributes \(v_j\beta_j\) to the signed numerator and \(v_j\) to the denominator's variance. A neutral variant has \(\beta_j=0\): it adds no expected signal but can increase burden variance. A rarer variant has a smaller \(v_j\), so it contributes less information at a fixed effect size. Weighting schemes can change these contributions, but their performance depends on whether the weights reflect the true frequencies and effects.

Worked Example: Adding Neutral Variants to a Burden

Return to the hypothetical design and causal variant from the first example: \(N\pi(1-\pi)=1{,}000\), \(v=0.0099\), and \(\beta=0.30\). Suppose the burden now includes that variant plus nine independent neutral variants, each with the same genotype variance. This is a simplifying scenario; all ten variants are assumed measured accurately and the same unweighted burden is tested.

The numerator remains \(0.0099(0.30)=0.00297\), because the nine neutral effects are zero. The denominator becomes \(\sqrt{10(0.0099)}\). Thus:

$$ \lambda=\sqrt{1{,}000} \frac{0.00297}{\sqrt{0.099}} =\frac{\sqrt{1{,}000(0.0099)}(0.30)}{\sqrt{10}} =\frac{0.9439}{\sqrt{10}} \approx0.2985. $$

At the same two-sided 0.05 threshold, power is approximately \(\Phi(-1.96-0.2985)+1-\Phi(1.96-0.2985)=\Phi(-2.2585)+1-\Phi(1.6615)\approx0.0120+0.0483=0.0603\). The burden's power falls from about 15.7% to about 6.0% in this approximation. The added neutral variants have not reduced the causal variant's effect; they have made the unweighted aggregate less precise for that effect pattern.

Uncertain Effects and Average Power

Power is conditional on a particular set of effect parameters. Before data are collected, that set is usually uncertain. A design can therefore be assessed across plausible configurations: for example, all effects aligned, a mixture of neutral and risk-increasing effects, or a mixture of risk-increasing and protective effects. If configuration \(s\) has probability \(r_s\) and conditional power \(P_s\), the average power under that planning distribution is \(\sum_s r_sP_s\), with \(\sum_s r_s=1\). This is an average over assumed alternatives, not the probability that the null hypothesis is true.

Averaging the effects first and then calculating power is generally not equivalent to averaging conditional powers. Because the burden depends on signed effects and power is a nonlinear function of \(\lambda\), the average effect can hide configurations that yield substantial signals in opposite directions. Planning should report the assumed effect scenarios and their probabilities rather than presenting an average as if it were guaranteed.

Worked Example: Averaging Across Two Effect Configurations

For the four-variant design in the second example, suppose the planning model assigns equal probability to two configurations: all four effects are \(+0.20\), or two are \(+0.20\) and two are \(-0.20\). Assume the two-sided test, significance level, and approximations used there. The aligned scenario has \(\lambda\approx1.2586\), giving power \(\Phi(-3.2186)+1-\Phi(0.7014)\approx0.0006+0.2416=0.2422\). The opposing scenario has \(\lambda=0\), giving power 0.05.

The average power across these equally weighted scenarios is \((0.5)(0.2422)+(0.5)(0.05)=0.1461\), or about 14.6%. The average signed effect across scenarios is nonzero: the averaged effect vector has mean effect 0.10 per variant and total signed effect 0.40. Substituting this vector into the burden formula gives \(\lambda\approx0.6293\), and power \(\Phi(-2.5893)+1-\Phi(1.3307)\approx0.0048+0.0916=0.0964\), not null power. That calculation would discard the fact that one plausible configuration is aligned and detectable. The scenario-average power instead answers the stated planning question, conditional on the assumed 50:50 distribution of configurations.

Design Implications and a Common Pitfall

Power planning under allelic heterogeneity should make the effect configuration explicit. At a minimum, vary the number of causal variants, their allele frequencies or genotype variances, the range of effect sizes, and the proportions of neutral and protective effects. Also state the test and its weights: the same variant set can yield quite different power for a directional burden and a quadratic test. If the gene-level significance threshold is more stringent than 0.05 because of a broader testing plan, use that actual threshold; a higher threshold lowers power.

The assumptions behind a calculation matter. Unequal ancestry composition can change allele frequencies and induce confounding; correlated variants change score covariance; and differential callability or genotype error can reduce or distort the signal. The formulas above assume these complications are absent. In an actual design, they require appropriate modeling or a more realistic power analysis.

  • Pitfall: counting variants as if each added variant must improve power. More aligned causal variants can strengthen a burden signal, but neutral variants can inflate its variance, and protective variants can cancel risk-increasing effects. The correction is to specify the effect pattern and recompute signal and variance for the chosen test.
  • Pitfall: using the average signed effect as the whole uncertainty analysis. A zero average can arise from uncertainty between two nonzero, oppositely directed scenarios. The correction is to calculate conditional power under each scenario and average those powers only when a planning distribution over scenarios is justified.
  • Pitfall: interpreting a quadratic noncentrality as exact SKAT power. A real weighted kernel statistic may have a mixture-of-chi-square null distribution. The correction is to use the actual statistic and its calibration when making a quantitative power claim.
Key takeaway: Gene-level power is determined by how a test combines variant-specific information. For a burden, signed effects can reinforce or cancel, and neutral variants can dilute the signal. Quadratic tests avoid the same directional cancellation, but their power still depends on frequencies, effect magnitudes, weights, and the exact null distribution.

Check Your Understanding

Use the assumptions and definitions in this tutorial to answer the following.

  1. For an unweighted burden with independent variants, what happens to the numerator and denominator when a neutral variant is added?
  2. Why can two risk-increasing effects and two protective effects yield a burden noncentrality of zero in the equal-variance example?
  3. How does a lower allele frequency affect genotype variance under Hardy–Weinberg equilibrium, and why does that matter for power?
  4. In what sense does a quadratic score avoid the cancellation that can occur in a burden test?
  5. Why is average power across uncertain configurations not generally obtained by plugging the average effect into a power formula?
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