Tutorials › Statistical Genetics › Power Gains From Quantitative-Trait Enrichment

Genome-Wide Association Study Design · Tutorial 353 of 989

Power Gains From Quantitative-Trait Enrichment

Estimate power gains from informative or enriched quantitative-trait samples while distinguishing useful genetic signal from variance that merely adds noise.

Intermediate 10 min read

What You'll Learn

  • Use the quantitative-trait GWAS noncentrality parameter to compare designs.
  • Distinguish phenotype reliability from total phenotype variance.
  • Calculate how reduced residual variation changes approximate power.
  • Identify when higher trait variance does and does not improve power.
  • Assess how enrichment can change the effect estimated and limit generalizability.

Why Enrichment Can Change Power

A quantitative-trait GWAS can gain power without increasing its sample size if recruited participants provide more informative measurements. For example, a study may prioritize people whose phenotype is measured reliably, or recruit from a stratum in which the genetic effect is stronger. But “higher trait variance” is not, by itself, a guarantee of greater power: extra variation can be genetic signal, or it can be noise that makes the signal harder to detect.

The design question is therefore not simply whether the selected sample has a wider range of trait values. It is whether selection increases the association information relevant to the target genetic effect. The same selection can also change which population the result describes, especially when genetic effects differ between the selected participants and the broader target population.

Power in a Quantitative-Trait Association

Definition: In a simple additive association model, let \(Y_i\) be a quantitative trait for participant \(i\), and let \(G_i\) be that participant’s genotype dosage. The model is \(Y_i=\alpha+\beta G_i+\varepsilon_i\), where \(\beta\) is the change in mean trait per one-unit increase in dosage and \(\varepsilon_i\) is residual variation. For a standardized genotype with variance 1, independent participants, a common effect \(\beta\), and residual standard deviation \(\sigma_e\), the approximate noncentrality parameter is \(\delta=\beta\sqrt{n}/\sigma_e\). For an unstandardized genotype, replace \(n\) in this expression with \(n\operatorname{Var}(G)\).

This approximation assumes the linear model is appropriate, residuals have finite variance, and the association test is well calibrated. The simple expression also assumes the genotype is independent of the residual and that any covariates have been handled appropriately. With covariate adjustment, the relevant genotype variation is the variation remaining after adjustment, not necessarily the raw variance of genotype. If effects or residual variances differ across strata, a single common-effect expression may not describe the whole design.

For a two-sided test using a fixed critical value \(c\), the normal approximation gives power as the probability that a normal test statistic with mean \(\delta\) and variance 1 falls beyond either rejection boundary:

$$ \operatorname{Power}\approx \Phi(\delta-c)+\Phi(-\delta-c). $$

Here \(\Phi\) is the standard normal cumulative distribution function. For a very stringent GWAS threshold and positive \(\delta\), the second term is often negligible, but it is part of the two-sided expression. In the examples below, \(c=5.45\) is a hypothetical fixed genome-wide critical value used solely to compare designs.

What Enrichment Changes in the Formula

The noncentrality parameter shows three distinct routes to more power: a larger sample size \(n\), more genotype variation among participants, or a larger genetic effect relative to residual variation. Enrichment at fixed sample size is useful only if it changes one or more of the latter two in a favorable way. If genotype is not standardized, for example, recruiting a population with greater genotype variance can increase information; but this does not follow from greater phenotype variance.

Measurement error illustrates a direct route. If the measured trait is \(Y=Y^*+e\), where \(Y^*\) is the underlying trait and the measurement error \(e\) is independent of genotype and of the underlying residual, then the genetic slope is unchanged while the error contributes to residual variance. In this setting, more reliable measurements can lower \(\sigma_e\) and increase power. By contrast, raising the total variance of \(Y\) by adding independent noise increases the denominator and decreases power.

There is also a scale issue. Multiplying every trait value, including the genetic effect and residuals, by the same constant changes the units but not the test statistic’s noncentrality parameter. Thus, a trait measured in a scale with a larger numerical standard deviation is not inherently more informative. What matters is the effect in relation to residual variation and the genotype information in the sample.

Key distinction: A higher phenotypic variance improves power only when it represents more association signal relative to residual noise, or accompanies more informative genotype variation. Higher variance caused by unrelated noise can reduce power; a simple change of measurement units changes neither power nor information.

Worked Example: More Reliable Trait Measurements

Suppose a hypothetical study tests a standardized genotype in \(n=24{,}708\) participants. The per-unit genetic effect is \(\beta=0.027\) trait units. Compare two measurement strategies: one has residual standard deviation \(\sigma_e=0.80\), and the more reliable strategy has \(\sigma_e=0.70\). Assume the effect and genotype distribution are otherwise unchanged, and use \(c=5.45\).

For the first strategy, the noncentrality parameter is

$$ \delta_{0.80} =\frac{0.027\sqrt{24{,}708}}{0.80} \approx 5.3051. $$

The approximate power is

$$ \operatorname{Power} \approx \Phi(5.3051-5.45)+\Phi(-5.3051-5.45) \approx \Phi(-0.1449) \approx 0.442. $$

The second tail is negligible at this threshold. Under the more reliable measurement strategy,

$$ \delta_{0.70} =\frac{0.027\sqrt{24{,}708}}{0.70} \approx 6.0630, $$

so

$$ \operatorname{Power} \approx \Phi(6.0630-5.45) =\Phi(0.6130) \approx 0.730. $$

In this hypothetical comparison, reducing residual standard deviation from 0.80 to 0.70 increases approximate power from 0.442 to 0.730 at the same sample size. This conclusion depends on the stated model: the measurement improvement reduces residual variation without changing the genetic effect, genotype distribution, or participant independence.

Higher Trait Variance: Signal or Noise?

It is important to distinguish total trait variance from residual variance. In the simple model, the outcome variance includes variation associated with genotype and variation left in the residual. A larger total variance may reflect a larger effect of the tested genotype or more informative variation in that genotype, but it may instead reflect genetic variation from other sources, environmental heterogeneity, or measurement error. Only the first two possibilities necessarily supply more association information for the association under consideration.

Worked Example: Greater Residual Variation Can Lower Power

Consider another hypothetical design with \(n=41{,}180\), standardized genotype, genetic effect \(\beta=0.027\), and critical value \(c=5.45\). Compare residual standard deviations of 0.80 and 1.00. The latter might arise in a sample with more variable environmental conditions or less precise measurements.

With \(\sigma_e=0.80\),

$$ \delta_{0.80} =\frac{0.027\sqrt{41{,}180}}{0.80} \approx 6.8488, \qquad \operatorname{Power} \approx \Phi(6.8488-5.45) =\Phi(1.3988) \approx 0.919. $$

With \(\sigma_e=1.00\),

$$ \delta_{1.00} =\frac{0.027\sqrt{41{,}180}}{1.00} \approx 5.4791, \qquad \operatorname{Power} \approx \Phi(5.4791-5.45) =\Phi(0.0291) \approx 0.512. $$

The sample with residual standard deviation 1.00 has more residual variation, not more useful association information. Its approximate power is about 0.512 rather than 0.919. If the larger standard deviation instead came from multiplying the whole phenotype, effect included, by a constant, the effect and residual standard deviation would scale together and this power comparison would not apply.

A different situation occurs when a known stratum has a stronger genetic effect. Suppose a recruitment characteristic identifies a subgroup in which the same genotype has a larger effect on the trait. At a fixed residual standard deviation, that stratum can have substantially greater power for its own effect. This is not a free improvement to the estimate for everyone: it may reflect genuine effect heterogeneity, so a selected-sample estimate need not equal the population-average effect.

Worked Example: Stronger Effect in an Enriched Stratum

Suppose a hypothetical study enrolls \(n=24{,}708\) participants from a prespecified stratum where a genotype’s effect is \(\beta=0.040\), compared with \(\beta=0.027\) in the broader population. Assume standardized genotype, residual standard deviation \(\sigma_e=0.80\) in both groups, independent observations, and \(c=5.45\). This example assumes effect modification by the stratum-defining characteristic; it does not assume that merely having a wider trait distribution causes the larger slope.

For the broader-population effect, the noncentrality parameter and power are

$$ \delta_{\text{broad}} =\frac{0.027\sqrt{24{,}708}}{0.80} \approx 5.3051, \qquad \operatorname{Power}_{\text{broad}} \approx \Phi(5.3051-5.45) \approx 0.442. $$

For the enriched stratum’s effect,

$$ \delta_{\text{stratum}} =\frac{0.040\sqrt{24{,}708}}{0.80} \approx 7.8594, \qquad \operatorname{Power}_{\text{stratum}} \approx \Phi(7.8594-5.45) =\Phi(2.4094) \approx 0.992. $$

The stratum has much greater power to detect its larger effect under the assumptions given. Its result estimates an association in that stratum. It should not automatically be reported as the effect for the broader population, where the effect is assumed to be smaller.

Justifying the Information Comparison

The central comparison follows from the uncertainty of the estimated slope. With a standardized genotype of variance 1, independent observations, and residual variance \(\sigma_e^2\), the approximate variance of the slope estimate is \(\sigma_e^2/n\). Therefore its standard error is \(\sigma_e/\sqrt{n}\), and the mean of the test statistic under a true effect \(\beta\) is the effect divided by that standard error:

$$ SE(\hat{\beta})\approx \frac{\sigma_e}{\sqrt{n}}, \qquad \delta\approx \frac{\beta}{SE(\hat{\beta})} =\frac{\beta\sqrt{n}}{\sigma_e}. $$

This derivation explains why a reliability improvement can be valuable even without recruiting more participants. It also explains the limit: a design change that increases \(\sigma_e\), with \(n\) and \(\beta\) held fixed, reduces \(\delta\). More generally, if the genotype is not standardized, the slope’s information is proportional to \(n\operatorname{Var}(G)/\sigma_e^2\), under the simple model assumptions. Any claim of a power gain should specify which part of that information has changed.

For a sample containing strata with different residual variances, genotype distributions, or effects, information should be assessed with those stratum-specific quantities rather than by substituting a pooled trait standard deviation into a single-stratum formula. A common-effect analysis can be appropriate when the effect is genuinely common and the model is correctly specified; otherwise, a selected sample may be testing a different association from the one intended for the target population.

Generalizability and Practical Design Checks

Enrichment changes the recruited sample’s composition. If effects differ by age, environment, clinical status, or another characteristic used in selection, the association estimate from the enriched sample describes that composition unless the analysis is designed to recover a target-population estimand. If effects are common and selection does not induce bias, the estimate may still apply more broadly, but the selected sample alone provides less direct evidence about underrepresented groups.

Before planning an enrichment strategy, specify the target population and effect of interest, then ask:

  • Does selection improve phenotype reliability or reduce residual variation, or does it mainly select a group with a different effect?
  • Could selection change genotype variation, residual variance, or the relationship between genotype and phenotype?
  • Does the planned analysis estimate a stratum-specific effect or a target-population effect?
  • Are participants from the target population represented well enough to support the intended interpretation?

When selection probabilities and relevant covariates are known, weighting or stratified analyses may help align estimates with a target population. Such approaches require adequate representation across the relevant strata and suitable assumptions about selection and effect variation; they cannot recover information about groups that are absent from the study.

Common Mistake

A common mistake is to reason, “The enriched group has a larger trait standard deviation, so it must give a more powerful GWAS.” That conclusion confuses total phenotype spread with association information. If the increase is independent noise, residual variation rises and power can fall. If the whole trait is merely expressed in larger units, the effect and residual standard deviation scale together and power is unchanged. A gain is justified when the genetic effect becomes larger relative to residual variation, measurement becomes more informative, or genotype information increases—and the resulting estimand must still match the scientific target.

Key takeaway: Quantitative-trait enrichment improves power through the signal-to-noise ratio and genotype information, not through trait variance alone. Always report what selection changes and whether the enriched sample supports inference for the intended population.

Check Your Understanding

Use the model assumptions stated in this tutorial when answering.

  1. At fixed sample size and genetic effect, what happens to approximate power if independent measurement error increases the residual standard deviation?
  2. Why does multiplying every trait value and genetic effect by the same constant leave the noncentrality parameter unchanged?
  3. A selected stratum has greater trait variance than the target population. What additional information is needed before concluding that GWAS power will improve?
  4. If the genetic effect differs between the enriched stratum and the broader population, which effect does a simple analysis of the enriched sample estimate?
  5. For an unstandardized genotype, what quantity replaces \(n\) in the simple noncentrality formula?
Clinical Biostats