Why power for a locus pair depends on its allele frequencies
For a prespecified pair of loci, an interaction test asks whether the joint genotype contribution to a trait is consistent with a model containing only the intercept and the two locus main effects. Even if the interaction effect is scientifically important, the test can have low power when the interaction predictor varies little across participants. Allele frequencies help determine that variation: if one allele is rare, relatively few participants carry its dosage, and the product of the two dosages may be especially variable across the genotype distribution.
This tutorial develops an approximate power calculation for a quantitative trait and a one-degree-of-freedom additive-by-additive interaction test. The approximation needs a specified interaction effect, residual variance, sample size, allele frequencies, and significance threshold. Its central ingredient is the variance of the interaction predictor after accounting for the intercept and locus main effects.
The model and its assumptions
For individual \(i\), consider the linear model
Here \(Y_i\) is a quantitative trait and \(\varepsilon_i\) is the residual, with mean zero and variance \(\sigma^2\). The null hypothesis for the interaction test is \(H_0:\beta_I=0\), and the two-sided alternative is \(H_a:\beta_I\ne 0\). This is a one-degree-of-freedom test because the alternative adds one interaction coefficient. The main effects remain in both models; omitting them would change the test and generally would not be an appropriate hierarchical interaction test.
The simple allele-frequency formula below assumes Hardy–Weinberg equilibrium at each locus, linkage equilibrium between loci, and independent sampling of participants. It also assumes that the linear model is correctly specified, the residual variance is approximately constant, and the interaction predictor is not confounded with omitted covariates. For an exact finite-sample \(t\) test, errors must additionally be normally distributed conditional on the genotypes and the design matrix must have full rank. The normal-approximation calculation is most useful as a planning approximation, not as a replacement for checking the study design.
From allele frequencies to interaction information
Centering each dosage gives \(A=G_1-2p_1\) and \(B=G_2-2p_2\). The centered interaction predictor is \(U=AB\). Expanding it shows why it represents the same interaction coefficient as the product \(G_1G_2\), while shifting only the intercept and main-effect terms:
Thus, when both main effects and an intercept are included, using the centered product rather than the raw product leaves the coefficient of the interaction unchanged. Under linkage equilibrium, \(A\) and \(B\) are independent, each has mean zero, and their variances are \(2p_1q_1\) and \(2p_2q_2\). Consequently, the centered product is uncorrelated with the intercept and with either main-effect predictor. Its variance is
The equality \(E(A^2B^2)=E(A^2)E(B^2)\) uses independence between loci. It is not guaranteed when the loci are in linkage disequilibrium. With \(n\) independent participants, the expected sum of squared centered interaction predictors is approximately \(n\operatorname{Var}(U)\). This quantity determines how precisely the interaction coefficient can be estimated.
The absolute value is used because power for a two-sided test depends on the magnitude, not the direction, of the specified effect. A larger interaction effect or sample size increases \(\lambda\); a larger residual standard deviation decreases it. At a given effect size and sample size, the frequency factor is largest when both allele frequencies are near one half and becomes small when either allele is rare.
Turning noncentrality into power
For a two-sided test at significance level \(\alpha\), the large-sample normal approximation rejects when the absolute standardized test statistic exceeds \(z_{1-\alpha/2}\), where \(z_r\) is the \(r\)-quantile of the standard normal distribution. Under the alternative, that statistic is approximately normal with mean \(\lambda\) (taking the specified effect as positive) and variance one. The power is therefore the probability of falling in either rejection tail:
where \(\Phi\) is the standard normal cumulative distribution function. Equivalently, the squared test statistic is approximately a noncentral chi-square random variable with one degree of freedom and noncentrality parameter \(\lambda^2\). The chi-square form gives power as the probability of exceeding the central \(\chi^2_1\) critical value for the chosen \(\alpha\).
Worked Example: Power for a Prespecified Pair
Suppose a hypothetical quantitative-trait study has \(n=1{,}000\) participants. The allele frequencies are \(p_1=0.30\) and \(p_2=0.20\), the hypothesized interaction coefficient is \(\beta_I=0.50\) trait units per dosage-product unit, and the residual standard deviation is \(\sigma=2.00\) trait units. Assume linkage equilibrium, Hardy–Weinberg equilibrium, and the model conditions above. Find approximate power for a two-sided test at \(\alpha=0.05\).
First, \(q_1=0.70\), \(q_2=0.80\), and the interaction-predictor variance is
The noncentrality parameter is
At \(\alpha=0.05\), \(z_{1-\alpha/2}=z_{0.975}\approx1.960\). Thus
The approximate power is 82.6%. Under the assumed model and effect, repeated studies of this design would reject the no-interaction null in about 82.6% of repetitions. This is not the probability that the interaction exists or that the specified effect is true.
Planning a sample size
A sample size can be approximated by first finding the noncentrality required for the desired power. For a two-sided test, a convenient approximation is \(\lambda_{\mathrm{req}}\approx z_{1-\alpha/2}+z_{\text{power}}\), where \(z_{\text{power}}\) is the standard normal quantile at the target power. This follows by making the upper-tail rejection probability approximately equal to the target: \(1-\Phi(z_{1-\alpha/2}-\lambda_{\mathrm{req}})=\text{power}\). Rearranging gives the stated sum. The neglected opposite-tail probability is small for usual positive effect sizes and target powers; an exact normal or noncentral-\(t\) calculation can refine the result.
Solving the noncentrality formula for \(n\) gives
Because participants are counted in whole numbers, round the result up. This formula makes the design trade-offs explicit: doubling the hypothesized interaction effect divides the required sample size by four, whereas doubling the residual standard deviation multiplies it by four. A more stringent significance threshold increases the required sample size through the larger critical value.
Worked Example: Sample Size for 80% Power
Consider a hypothetical study with allele frequencies \(p_1=0.10\) and \(p_2=0.35\), a target interaction effect of \(\beta_I=0.40\) trait units per dosage-product unit, residual standard deviation \(\sigma=1.50\) trait units, two-sided \(\alpha=0.05\), and target power 80%. Assume the same equilibrium, independence, and model conditions.
The allele-frequency factor is
Using \(z_{0.975}\approx1.960\) and \(z_{0.80}\approx0.8416\), the required noncentrality is approximately \(1.960+0.8416=2.802\). Therefore,
The planning estimate is about 1,348 participants. In practice, a design may choose a somewhat larger target to allow for missing genotypes, uncertainty in the allele frequencies or residual variance, and departures from the assumed model. Such inflation should be justified using the expected study conditions rather than treated as part of the basic formula.
How frequency changes power
The product \(4p_1q_1p_2q_2\) shows that allele frequencies affect the interaction test through the amount of variation in the product predictor. This frequency factor is not a measure of biological importance. A rare-allele pair can have a substantial interaction effect but still provide little information because only a small fraction of participants contribute informative dosage combinations.
Worked Example: Comparing a Rare and a Common Pair
Suppose two hypothetical prespecified pairs are tested in separate designs, each with \(n=5{,}000\), \(\beta_I=0.25\) trait units per dosage-product unit, \(\sigma=1.50\) trait units, and a two-sided \(\alpha=0.05\). Assume the model conditions hold. Pair A has frequencies \(p_1=0.05\) and \(p_2=0.30\); Pair B has frequencies \(p_1=p_2=0.30\).
For Pair A, the frequency factor and noncentrality are
Its approximate power is
For Pair B, the frequency factor is \(4(0.30)(0.70)(0.30)(0.70)=0.1764\), giving
With the same effect, sample size, and residual standard deviation, the common-frequency pair has much higher approximate power because its interaction predictor varies more. The calculation does not establish that either pair has an interaction; it describes sensitivity to the specified effect under the assumed model.
When to modify or distrust the approximation
The allele-frequency formula is deliberately simple. If the loci are in linkage disequilibrium, the centered dosages need not be independent, and the variance of their product cannot generally be obtained by multiplying the two dosage variances. More generally, the relevant predictor is the interaction column after projecting out the intercept, main effects, and any covariates in the fitted model. Its residualized variance, rather than \(4p_1q_1p_2q_2\), determines the information for the interaction coefficient. Covariate adjustment can also change that information.
The calculation can be a poor guide when genotype combinations are very sparse. The expected variance formula may describe the population distribution while a particular sample contains few or no observations in informative genotype patterns. In that situation, the interaction estimate can be unstable or not identifiable, and a large-sample approximation may be unreliable. The earlier material on sparse genotype cells and interaction identifiability is directly relevant to this limitation. For binary outcomes, a linear-model residual variance formula is not appropriate; power must instead use the information and variance implied by the specified binary-outcome model.
The significance threshold must also match the planned test. A threshold for a single prespecified pair differs from a threshold adjusted for testing many pairs. The formula accepts the threshold supplied to it; it does not itself account for a search across loci. Likewise, if the tested interaction is defined on a different genotype coding or trait scale, the effect size and predictor variance must correspond to that test.
Common mistake
- Wrong reasoning: “The interaction effect is large, so the test must have high power.”
- Correction: Power depends on the effect relative to its standard error. For this approximation, that ratio is \(|\beta_I|\sqrt{n\,4p_1q_1p_2q_2}/\sigma\). A rare allele, high residual variation, small sample, or stringent threshold can leave power low even for an effect that seems substantial in trait units.
- Wrong reasoning: “The computed power is the probability that this particular pair truly interacts.”
- Correction: Power is a repeated-sampling probability of rejecting \(H_0\) when the specified alternative model and effect are true. It is not a probability that a hypothesis is true.
Check Your Understanding
Answer each question using the assumptions and formulas in this tutorial.
- Under Hardy–Weinberg equilibrium, what is the variance of an allele dosage with frequency \(p\), and how does that lead to the two-locus interaction-predictor variance under linkage equilibrium?
- Keeping the effect size, sample size, and residual standard deviation fixed, what happens to approximate power when one allele frequency moves closer to zero? Explain using the frequency factor.
- A researcher doubles the hypothesized interaction effect while keeping the other design inputs fixed. By what factor does the approximate required sample size change?
- Why is the product of the two dosage variances not generally sufficient to calculate interaction-predictor variance under linkage disequilibrium?
- Does a design with 80% power imply an 80% probability that the interaction is real? Explain the correct interpretation.