Deriving Power for Quantitative-Trait GWAS

Why Quantitative-Trait Power Depends on Allele Frequency

A quantitative-trait genome-wide association study (GWAS) asks whether a genetic variant is associated with variation in a measured trait, such as blood pressure or a laboratory measurement. Before collecting data, researchers need to know whether a study of a proposed size could detect an effect of scientific interest at the chosen significance threshold. For a single additive variant, the approximate answer depends on the sample size, allele frequency, per-allele effect, residual variability, and test threshold.

The key link is genotype variability. If nearly everyone has the same genotype, the study has little information about a per-allele effect, even if the sample is large. Under an additive model and Hardy–Weinberg equilibrium, the variance of genotype dosage is \(2f(1-f)\), where \(f\) is the frequency of the allele whose dosage is counted. That variance determines the standard error of the estimated effect.

Definition: Let \(G\) be an individual’s additive genotype dosage, taking values 0, 1, or 2 for the counted allele, and let \(Y\) be a quantitative trait. In the single-variant linear model, \(Y=\mu+\beta G+\varepsilon\), where \(\mu\) is an intercept, \(\beta\) is the change in the trait’s conditional mean per additional counted allele, and \(\varepsilon\) is residual variation with mean 0 and variance \(\sigma^2\). The allele frequency is \(f\), and under Hardy–Weinberg equilibrium, \(\operatorname{Var}(G)=2f(1-f)\).

The effect \(\beta\) is measured in trait units per allele. For example, if \(Y\) is measured in millimeters of mercury, \(\beta=0.5\) means that the model’s expected trait value differs by 0.5 millimeters of mercury per additional counted allele. It describes an association in the specified model; by itself, it does not establish a causal effect.

Model and Approximation

For planning, assume that the \(n\) analyzed participants are independent, their genotype dosages are measured without error, and the additive effect is the same across participants. Assume the stated allele frequency describes the study population, genotype dosage has variance \(2f(1-f)\), and residual variance conditional on genotype and any modeled covariates is \(\sigma^2\). The calculation also assumes the analysis uses a two-sided test of \(H_0:\beta=0\), with a prespecified per-variant significance threshold \(\alpha\). A normal approximation to the test statistic is used; it is generally appropriate for a large-sample association test, but it is not an exact guarantee for every study design.

In the simple model with an intercept, the estimated slope has approximate variance \(\sigma^2/\sum_{i=1}^n(G_i-\bar{G})^2\). Since the sum of squared centered dosages is approximately \(n\operatorname{Var}(G)\), the standard error is approximately \(\sigma/\sqrt{n\,2f(1-f)}\). Thus the expected signal in standard-error units, called the noncentrality parameter here, is \(\delta=\beta\sqrt{n\,2f(1-f)}/\sigma\). Its magnitude increases with the sample size and effect size, and decreases with residual variability.

SE(\hat{\beta})\approx\frac{\sigma}{\sqrt{n\,2f(1-f)}}, \qquad \delta=\frac{\beta}{SE(\hat{\beta})} \approx\frac{\beta\sqrt{n\,2f(1-f)}}{\sigma}.

Let \(c=z_{1-\alpha/2}\), the standard normal quantile with probability \(1-\alpha/2\) to its left. Under the alternative, the standardized test statistic is approximately normal with mean \(\delta\) and variance 1. The two-sided test rejects when the absolute value of that statistic exceeds \(c\). Consequently, power is the probability that a normal random variable with mean \(\delta\) and variance 1 falls below \(-c\) or above \(c\). For a negative \(\beta\), the same power results as for a positive effect of equal magnitude.

\operatorname{Power} \approx P(Z<-c)+P(Z>c) =\Phi(-c-\delta)+1-\Phi(c-\delta), \qquad Z\sim N(\delta,1).

Here \(\Phi\) is the standard normal cumulative distribution function. For the usual planning calculation, the lower-tail term is negligible when the effect points toward one tail and the signal is not extremely small. To achieve target power \(1-\gamma\), where \(\gamma\) is the target probability of failing to reject under the specified alternative, a convenient approximation is \(|\delta|\approx c+z_{1-\gamma}\). Solving this relation for \(n\) gives the sample-size formula below. It is an approximation because it neglects the small opposite-tail contribution and because the inputs and normal approximation may themselves be uncertain.

n\approx \frac{(z_{1-\alpha/2}+z_{1-\gamma})^2\sigma^2} {2f(1-f)\beta^2}.

Derivation: To obtain target power \(1-\gamma\) approximately, require \(|\beta|\sqrt{n\,2f(1-f)}/\sigma\approx z_{1-\alpha/2}+z_{1-\gamma}\). Squaring both sides gives \(\beta^2 n\,2f(1-f)/\sigma^2\approx(z_{1-\alpha/2}+z_{1-\gamma})^2\). Rearranging yields the sample-size formula. Since participants are counted in whole numbers, round the resulting requirement up.

Worked Power and Sample-Size Calculations

Worked Example: Required Sample Size at a Stringent Threshold

Suppose a hypothetical study aims for 80% power to detect a per-allele effect of \(\beta=0.04\) trait units, with residual standard deviation \(\sigma=1\) trait unit and allele frequency \(f=0.20\). The prespecified two-sided threshold is \(\alpha=5\times10^{-8}\). Use \(z_{1-\alpha/2}\approx5.451\) and \(z_{0.80}\approx0.842\), so their sum is 6.293.

The dosage variance is \(2f(1-f)=2(0.20)(0.80)=0.32\). Substituting into the approximate sample-size formula gives

n\approx \frac{(5.451+0.842)^2(1)^2}{(0.32)(0.04)^2} =\frac{(6.293)^2}{(0.32)(0.0016)} =\frac{39.601849}{0.000512} \approx77{,}347.36.

The study therefore needs at least 77,348 analyzable participants under these assumptions. The result is an approximate planning requirement, not a promise that this sample size will yield 80% power in practice. The calculation assumes the effect, allele frequency, residual standard deviation, and test threshold are as specified.

Worked Example: Power for a Proposed Study

Consider a hypothetical study with \(n=100{,}000\) analyzable participants. Suppose \(f=0.30\), \(\beta=0.02\) trait units per allele, \(\sigma=0.80\) trait units, and \(\alpha=5\times10^{-8}\) for a two-sided test. Use \(c=5.451\).

First, \(2f(1-f)=2(0.30)(0.70)=0.42\). The noncentrality parameter is

\delta\approx \frac{0.02\sqrt{100{,}000(0.42)}}{0.80} =\frac{0.02\sqrt{42{,}000}}{0.80} \approx\frac{0.02(204.939)}{0.80} \approx5.123.

The approximate power is \(\Phi(-5.451-5.123)+1-\Phi(5.451-5.123)\), or \(\Phi(-10.574)+1-\Phi(0.328)\). The first term is negligible, and \(1-\Phi(0.328)\approx0.372\). Thus the proposed study has about 37.2% power for this specified effect under the model. This is not the probability that the association is real; it is the probability of rejecting the null at this threshold if the specified effect and assumptions hold.

Worked Example: A Lower-Frequency Variant

Suppose another hypothetical design targets 80% power for a per-allele effect of \(\beta=0.10\) trait units when \(\sigma=1.2\) trait units and \(f=0.05\). Retain the two-sided threshold \(\alpha=5\times10^{-8}\), with \(z_{1-\alpha/2}\approx5.451\) and \(z_{0.80}\approx0.842\).

The genotype variance is \(2(0.05)(0.95)=0.095\), and the squared residual standard deviation is \((1.2)^2=1.44\). The approximate sample size is

n\approx \frac{(5.451+0.842)^2(1.44)}{(0.095)(0.10)^2} =\frac{(6.293)^2(1.44)}{0.095(0.01)} =\frac{57.02666256}{0.00095} \approx60{,}028.07.

Rounding up gives a requirement of 60,029 analyzable participants. Although the specified effect is larger than in the first example, the lower genotype variance means relatively few participants contribute dosage variation. The required sample size reflects all three features: effect size, residual variability, and allele frequency.

Interpreting the Inputs and Results

The formula is most useful when each input is tied to the intended analysis and population. The sample size \(n\) is the number of participants whose phenotype and genotype information contribute to the test, not automatically the number enrolled. The residual standard deviation describes variability left after accounting for the modeled predictors, in the trait’s measurement units. The effect \(\beta\) must use those same units per allele. A change of trait scale changes both \(\beta\) and \(\sigma\); their ratio, which enters the signal, is unchanged if both are rescaled consistently.

The factor \(2f(1-f)\) is largest at \(f=0.5\), where it equals 0.5, and becomes smaller as the allele frequency approaches 0 or 1. Under the stated model, rarer alleles therefore generally require more participants to detect the same per-allele effect with the same residual variability and threshold. The relationship is not a statement that rare variants necessarily have smaller or larger biological effects; it concerns the statistical information available under the specified additive model.

If the analysis adjusts for covariates, the simple formula is appropriate when adjustment leaves the relevant genotype variation approximately represented by \(2f(1-f)\) and the residual variance is the conditional variance after adjustment. More generally, the information comes from genotype variation remaining after accounting for the covariates. Strong correlation between genotype and covariates can reduce that information, so plugging the marginal allele frequency into the simple formula may overstate power. In a more elaborate design, use a calculation that reflects the adjusted genotype information and the intended analysis.

Power is conditional on a specified alternative, not a general property of a study independent of the effect being considered. A study can have high power for a relatively large effect and low power for a smaller one. Likewise, a nominal significance threshold must be supplied as part of the design; these calculations do not derive a genome-wide threshold. If the threshold, allele frequency, effect, or residual variability is uncertain, it is sensible to examine a range of plausible values rather than report a single sample-size number as definitive.

Common Mistakes

Using the wrong genotype variance. For additive dosage under Hardy–Weinberg equilibrium the variance is \(2f(1-f)\), not \(f(1-f)\). Omitting the factor of 2 changes the calculated information and power.
Confusing variance with standard deviation. The formula uses \(\sigma^2\) in the sample-size numerator, while the noncentrality parameter divides by \(\sigma\). Squaring the standard deviation twice or using it without squaring in the sample-size formula gives an incorrect result.
Treating power as a probability about the null hypothesis. Power is the probability of rejecting \(H_0\) under a specified alternative and model. It is not the probability that \(H_0\) is false after observing data.
Rounding a required sample size down. A calculated requirement such as 77,347.36 must be rounded up to 77,348 participants, not down to 77,347.
Ignoring units or model assumptions. The effect and residual standard deviation must be expressed on the same trait scale, and the allele frequency and dosage variance should correspond to the study population and analysis.

Key takeaway: For an additive quantitative-trait association, approximate power is governed by the signal \(|\beta|\sqrt{n\,2f(1-f)}/\sigma\) relative to the two-sided critical value. State the model, units, allele frequency, and threshold before interpreting a power or sample-size calculation.

Check Your Understanding

Try these questions using the definitions and approximations above.

Under Hardy–Weinberg equilibrium, what is the genotype-dosage variance when the counted allele frequency is \(f=0.10\)?
For fixed allele frequency, effect size, and residual standard deviation, what happens to the noncentrality parameter when the sample size is multiplied by four?
Why must the effect \(\beta\) and residual standard deviation \(\sigma\) be expressed using compatible trait units?
What does 80% power mean for a specified effect and significance threshold, and what does it not mean?
Why might using the marginal allele frequency in the simple formula overstate power when genotype is strongly related to adjusted covariates?