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Genome-Wide Association Study Design · Tutorial 350 of 989

Genome-Wide Thresholds in Sample-Size Planning

Translate a chosen genome-wide threshold and target power into a required sample size, while keeping prospective planning distinct from correcting observed results.

Intermediate 11 min read

What You'll Learn

  • Convert a two-sided genome-wide significance threshold into its critical normal value.
  • Relate target power to the required noncentrality parameter.
  • Calculate quantitative-trait sample size using allele frequency and per-allele effect.
  • Calculate case-control effective sample size and translate it into recruitment numbers.
  • Identify assumptions and distinguish design calculations from adjustment of observed p-values.

Why the Threshold Matters Before Recruitment

A genome-wide association study can have many opportunities to detect an association, so its planned significance threshold is often much smaller than the conventional 0.05 used for a single test. A smaller threshold makes the evidence required for a discovery more demanding. If the study is planned around a particular effect size and power, that stricter threshold generally means recruiting more participants.

This tutorial treats the genome-wide threshold as a chosen design input. It shows how to combine that threshold with a target power, allele frequency, and effect size to calculate a sample size. It does not derive a threshold from the number or dependence of tests, nor does it adjust an observed result. Those are distinct tasks: the calculation here begins with a threshold already selected for the planned analysis.

Definition: A prospective sample-size calculation specifies a target type I error threshold \(\alpha_{\mathrm{GW}}\), a target power \(1-\beta\), and a scientifically relevant alternative effect. It then estimates the sample size required for the planned test to reach that power at that threshold, subject to the stated model and approximation.

From a Threshold to a Required Signal

Suppose the planned association test is two-sided and its null statistic is approximately standard normal. Write \(\Phi\) for the cumulative distribution function of a standard normal random variable. The critical value corresponding to the chosen threshold is

$$ z_{\mathrm{crit}}=\Phi^{-1}(1-\alpha_{\mathrm{GW}}/2). $$

A test rejects when the absolute value of its statistic exceeds \(z_{\mathrm{crit}}\). For example, \(\alpha_{\mathrm{GW}}=5\times10^{-8}\) gives \(z_{\mathrm{crit}}\approx5.4513\). Under a specified alternative, suppose the test statistic is approximately normal with variance 1 and mean \(\lambda>0\), where \(\lambda\) is the noncentrality parameter. Its power is

$$ P(Z<-z_{\mathrm{crit}})+P(Z>z_{\mathrm{crit}}) = \Phi(-z_{\mathrm{crit}}-\lambda)+1-\Phi(z_{\mathrm{crit}}-\lambda). $$

For a positive effect and a stringent two-sided threshold, the lower-tail term is usually negligible. A convenient planning approximation for target power \(1-\beta\) is therefore

$$ \lambda_{\mathrm{req}}\approx z_{\mathrm{crit}}+z_{1-\beta}, \qquad z_{1-\beta}=\Phi^{-1}(1-\beta). $$

This approximation chooses the alternative mean so that the upper-tail rejection probability is approximately the target power. For more exact planning, solve the full two-sided power equation for \(\lambda\); the formulas below then use that solved value in place of \(z_{\mathrm{crit}}+z_{1-\beta}\). The simple sum is typically adequate for standard GWAS planning when the effect is in the specified direction and the threshold is very stringent.

Key relationship: The threshold determines how far into the null distribution the rejection boundary lies; the target power determines how far the alternative distribution must be shifted beyond that boundary. A more stringent threshold or higher target power increases the required noncentrality and, consequently, the required sample size.

Quantitative-Trait Sample Size

Consider an additive association model for a quantitative trait. Let \(G\) be allele dosage, taking values 0, 1, or 2, and let \(f\) be the frequency of the allele whose dosage is counted. Under Hardy–Weinberg equilibrium, the dosage variance is \(2f(1-f)\). Let \(\beta\) denote the change in the trait mean per additional allele, and let \(\sigma\) denote the residual standard deviation after the planned covariate adjustment. Under the usual large-sample normal approximation, the association statistic has noncentrality approximately

$$ \lambda \approx \frac{\beta\sqrt{N\,2f(1-f)}}{\sigma}, $$

where \(N\) is the number of participants with usable phenotype and genotype data. Equating this to the required noncentrality and solving for \(N\) gives

$$ N_{\mathrm{req}}\approx \frac{(z_{\mathrm{crit}}+z_{1-\beta_{\mathrm{type\,II}}})^2\,\sigma^2} {2f(1-f)\,\beta^2}. $$

Here \(\beta_{\mathrm{type\,II}}\) is the type II error probability, not the genetic effect \(\beta\); the distinct notation in the formula indicates their different roles. Equivalently, if the trait is standardized so that the residual standard deviation is 1, the effect is expressed in residual standard-deviation units per allele. The estimate assumes a well-calibrated association test, independent participants, the stated allele frequency and effect, and an adequate normal approximation. It also treats \(N\) as the analyzable sample size, not necessarily the number initially recruited.

Worked Example: A Quantitative Trait at the Standard GWAS Threshold

Suppose a hypothetical study plans a two-sided threshold of \(5\times10^{-8}\) and 90% power to detect a per-allele increase of \(0.10\) residual standard deviations. Assume the allele frequency is \(f=0.30\), Hardy–Weinberg equilibrium is a reasonable planning model, and all \(N\) participants have usable measurements. Find the approximate sample size.

The dosage variance is \(2f(1-f)=2(0.30)(0.70)=0.42\). The threshold gives \(z_{\mathrm{crit}}\approx5.4513\), and 90% power gives \(z_{0.90}\approx1.2816\). Since the trait effect is expressed in residual standard-deviation units, \(\sigma=1\). Thus,

$$ N_{\mathrm{req}}\approx \frac{(5.4513+1.2816)^2}{0.42(0.10)^2} = \frac{45.3319}{0.0042} \approx 10{,}793.3. $$

The calculation uses rounded normal quantiles, so the decimal is approximate. A sample size must be an integer and must meet or exceed the target; rounding up gives 10,794 analyzable participants. This result is conditional on the assumed effect, frequency, residual variation, and model. It is not a guarantee that the study will attain exactly 90% power in practice.

Case-Control Sample Size

For a case-control study, let \(n_{\mathrm{case}}\) and \(n_{\mathrm{control}}\) be the numbers of cases and controls, and let \(\beta_{\mathrm{logOR}}\) be the log odds ratio per additional allele. For a modest effect and a well-calibrated large-sample test, the approximate information for the genetic effect is \(2f(1-f)n_{\mathrm{eff}}\beta_{\mathrm{logOR}}^2/4\), where the effective sample size is

$$ n_{\mathrm{eff}}= \frac{4n_{\mathrm{case}}n_{\mathrm{control}}} {n_{\mathrm{case}}+n_{\mathrm{control}}}. $$

This definition makes \(n_{\mathrm{eff}}\) equal to the total sample size when cases and controls are equal in number. With a case-control ratio other than 1:1, it is smaller than the total. The corresponding approximate noncentrality is

$$ \lambda\approx \frac{\beta_{\mathrm{logOR}}\sqrt{2f(1-f)n_{\mathrm{eff}}}}{2}. $$

Solving for the effective sample size gives

$$ n_{\mathrm{eff,req}}\approx \frac{4(z_{\mathrm{crit}}+z_{1-\beta_{\mathrm{type\,II}}})^2} {2f(1-f)\beta_{\mathrm{logOR}}^2}. $$

This approximation treats the allele frequency as the same in cases and controls under the null and uses the planned log odds ratio as the alternative. It is most useful for large samples and modest genetic effects. It does not replace a more detailed calculation when the model, ascertainment, covariates, or sampling design makes these approximations unsuitable.

Worked Example: Planning an Imbalanced Case-Control Study

Consider a hypothetical study with a two-sided threshold of \(5\times10^{-8}\), target power of 80%, allele frequency \(f=0.20\), and per-allele odds ratio 1.22. Plan for two controls per case. Assume the large-sample approximation is suitable and that the stated effect and frequency apply to the intended analysis.

The log odds ratio is \(\beta_{\mathrm{logOR}}=\log(1.22)\approx0.1989\), and \(2f(1-f)=2(0.20)(0.80)=0.32\). With \(z_{\mathrm{crit}}\approx5.4513\) and \(z_{0.80}\approx0.8416\), the effective sample size is approximately

$$ n_{\mathrm{eff,req}}\approx \frac{4(5.4513+0.8416)^2}{0.32(0.1989)^2} \approx \frac{158.4024}{0.01266} \approx 12{,}512. $$

The displayed denominator is rounded; using the unrounded logarithm and quantiles gives approximately \(12{,}519\). The difference is a rounding effect, so use the unrounded inputs for the final planning value. For a 1:2 case-control ratio, \(n_{\mathrm{eff}}=(8/3)n_{\mathrm{case}}\). Therefore the unrounded calculation requires about \(4{,}695\) cases and \(9{,}390\) controls, or approximately \(14{,}085\) participants in total. Rounding recruitment upward to whole people and checking the ratio preserves at least the target effective sample size.

Reading the Calculation as a Design Tool

The formulas make several design trade-offs explicit. Holding other inputs fixed, required sample size is proportional to the square of the required noncentrality. It is inversely proportional to the squared effect size and to genotype variance. Thus, halving the effect size requires roughly four times the sample size under this approximation. Alleles with frequencies near 0.5 have greater dosage variance than very rare alleles, so they provide more information per participant for the same per-allele effect.

The chosen threshold is also consequential. A smaller \(\alpha_{\mathrm{GW}}\) increases \(z_{\mathrm{crit}}\), increasing the required noncentrality and sample size. In a planning document, state the threshold itself and its rationale rather than presenting the sample-size result as if the threshold were an intrinsic property of the association statistic. The threshold used here is an input; the calculation does not determine how a testing family should be defined or how an observed p-value should be adjusted.

The result is usually an analyzable sample size. If some recruited participants are expected to lack usable genotype or phenotype data, recruitment must be larger. For example, under a simple assumption that a proportion \(r\) of recruited participants will be usable, a target of \(N_{\mathrm{req}}\) usable participants calls for approximately \(N_{\mathrm{req}}/r\) recruits. This adjustment is only as good as the retention assumption, and it does not account for other departures from the model.

Worked Example: Comparing Two Prespecified Thresholds

Suppose a hypothetical quantitative-trait design has a standardized per-allele effect of \(0.12\), allele frequency \(f=0.25\), and target power 80%. Compare a prespecified threshold of \(10^{-6}\) with \(5\times10^{-8}\). Assume the same independent-sample, additive-model approximation as above and residual standard deviation 1.

For \(10^{-6}\), the two-sided critical value is about \(4.8916\). The required sample size is

$$ N_{\mathrm{req}}\approx \frac{(4.8916+0.8416)^2}{2(0.25)(0.75)(0.12)^2} = \frac{32.8696}{0.0054} \approx 6{,}087. $$

For \(5\times10^{-8}\), use \(z_{\mathrm{crit}}\approx5.4513\):

$$ N_{\mathrm{req}}\approx \frac{(5.4513+0.8416)^2}{0.0054} = \frac{39.6006}{0.0054} \approx 7{,}333. $$

The stricter threshold requires about 1,246 additional analyzable participants in this example. These are separate prospective calculations under two different threshold choices. The comparison neither recommends one threshold nor describes a procedure for correcting observed results.

Assumptions, Limitations, and a Common Pitfall

These calculations are useful when the normal approximation is adequate and the planning inputs are defensible. They can be misleading when the assumed allele frequency or effect does not represent the target population, when the phenotype variance differs from the planning value, or when the analysis has substantially less information than the simple model assumes. The test must also be appropriately calibrated under the null; the earlier material on association calibration and regression diagnostics is relevant to evaluating that condition.

A further limitation is that a calculation for one variant does not automatically describe the power of an entire study. Different variants may have different frequencies, effects, and genotype quality. The calculated sample size is specific to the stated alternative and test. It is not the probability of finding any association in the study, and it does not guarantee discovery of variants whose effects are smaller or frequencies rarer than assumed.

  • Wrong reasoning: “The study has a genome-wide threshold, so I can use 0.05 in the power formula and adjust the result afterward.”
  • Correction: Use the actual planned threshold in the prospective power calculation. A threshold change alters the critical value and therefore the required sample size; a later adjustment of an observed result is not a substitute for planning at the intended threshold.
  • Wrong reasoning: “The required total case-control sample size is the effective sample size.”
  • Correction: Convert the required effective sample size into cases and controls using the planned allocation. Imbalance means the total recruited number can be appreciably larger than \(n_{\mathrm{eff}}\).
Key takeaway: A genome-wide sample-size calculation links a specified two-sided threshold and target power to the noncentrality needed under a stated effect model. Report the threshold, power, allele frequency, effect size, model assumptions, and whether the result is analyzable or recruited sample size.

Check Your Understanding

Check whether you can explain the planning logic and identify what each calculation assumes.

  1. Why does a smaller two-sided significance threshold generally increase the sample size needed for a fixed effect and target power?
  2. For a quantitative trait with residual standard deviation 2, how should the sample-size formula account for a per-allele effect reported in the trait’s original units?
  3. What is the effective sample size for 2,000 cases and 4,000 controls, and why is it less than the total sample size?
  4. In the normal approximation, what role does \(z_{1-\beta}\) play, and why is the lower-tail term often negligible for a positive effect at a stringent threshold?
  5. Why is a prospective power calculation at a chosen threshold different from correcting an observed p-value?
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