Tutorials › Statistical Genetics › Planning a Two-Stage GWAS Discovery Study

Genome-Wide Association Study Design · Tutorial 355 of 989

Planning a Two-Stage GWAS Discovery Study

Plan a two-stage GWAS by balancing the chance of selecting a true association in discovery against the chance of confirming it in an independent follow-up sample.

Intermediate 10 min read

What You'll Learn

  • Distinguish stage-one screening from independent stage-two confirmation
  • Calculate overall power from screening and confirmation probabilities
  • Compare participant allocations using stage-specific noncentrality
  • Account for top-variant selection and follow-up capacity
  • Identify assumptions that make confirmation evidence independent
  • Recognize how stage-two testing thresholds affect design power

Why Divide a GWAS Into Two Stages?

A genome-wide association study can genotype every participant at every variant, or it can use a staged design: screen many variants in a discovery sample, then genotype a selected set in a separate follow-up sample. The staged approach can reduce genotyping costs when follow-up assays are targeted, but it creates a design trade-off. Too little information in discovery can cause true associations to be missed; too little information in follow-up can leave selected signals unconfirmed.

The design question is therefore not simply how many people to recruit. It is how to allocate participants, genotyping, and testing effort so that a variant with a scientifically relevant effect has a good chance of passing both stages. The calculations below concern power for a specified true variant and effect; they do not by themselves determine the probability that a finding is true.

Definition: In a two-stage GWAS, stage one (discovery) tests a broad set of variants and selects candidates according to a prespecified screening rule. Stage two (follow-up or confirmation) tests selected variants in participants whose association data are independent of stage one. A variant is a confirmed discovery only if it meets the prespecified stage-two rule.

Stage-Specific Power and Overall Power

Suppose a particular variant has a true association in a specified direction. Let \(Z_j\) be its stage-specific association statistic in stage \(j\), for \(j=1,2\). Under a large-sample normal approximation, write \(Z_j\sim N(\delta_j,1)\), where \(\delta_j\) is the stage-specific noncentrality parameter: the expected standardized association statistic under the true effect. Its value depends on the effect, allele frequency, phenotype variability, sample size, and analysis. For a fixed variant and effect, more informative participants generally increase \(|\delta_j|\).

Let \(c_1\) be the stage-one screening cutoff and \(c_2\) the stage-two confirmation cutoff. For a two-sided test, the probability that a true variant passes stage \(j\) is

$$ \pi_j =P(|Z_j|>c_j) =\Phi(-c_j-\delta_j)+1-\Phi(c_j-\delta_j), $$

where \(\Phi\) is the standard normal cumulative distribution function. If effects are expected to be positive, the first term is usually very small for a stringent cutoff, but it is part of the two-sided power. A one-sided rule is appropriate only if the direction was specified in advance and the study's confirmation criterion permits it.

When the participant samples in the two stages are independent, the stage-two statistic is independent of the stage-one statistic under the model. Thus, for a variant that must pass both rules, the probability of passing the complete design is

$$ \pi_{\mathrm{overall}} =P(\text{pass stage 1 and stage 2}) =\pi_1\pi_2. $$

The product is not a universal shortcut: it depends on independent stage data and on stage-two power being calculated for the same variant and effect. Overlapping participants can induce dependence between the statistics, in which case the joint probability need not equal the product.

Worked Example: Combining the Two Stage Probabilities

Consider a hypothetical variant with anticipated noncentrality \(\delta_1=6.133\) and screening cutoff \(c_1=5.450\). Suppose its independent follow-up has \(\delta_2=2.123\) and a two-sided cutoff \(c_2=1.960\). These are illustrative stage-specific inputs, not estimates from a real study.

In stage one, the upper-tail contribution is \(1-\Phi(5.450-6.133)=1-\Phi(-0.683)=\Phi(0.683)\approx0.753\). The lower-tail contribution, \(\Phi(-5.450-6.133)=\Phi(-11.583)\), is negligible at the displayed precision. Thus \(\pi_1\approx0.753\).

In stage two, the upper-tail contribution is \(1-\Phi(1.960-2.123)=1-\Phi(-0.163)=\Phi(0.163)\approx0.565\). The lower-tail contribution is \(\Phi(-1.960-2.123)=\Phi(-4.083)\), approximately \(0.00002\). Thus \(\pi_2\approx0.565\).

Assuming independent participants between stages, overall power is approximately \(0.753\times0.565=0.425\). The design therefore has about 42.5% power to select and then confirm this particular association under the stated model and cutoffs. Its stage-one power alone, about 75.3%, would overstate the chance of a completed discovery.

How the Allocation Changes Each Stage's Power

With equal information per participant and a fixed total number of participants, assigning more participants to discovery tends to increase \(\delta_1\) and \(\pi_1\), while leaving fewer for follow-up and reducing \(\delta_2\) and \(\pi_2\). The reverse allocation can make confirmation strong but leave many true variants undiscovered. Because overall power is a product, the best allocation need not give either stage the largest possible power in isolation.

A useful planning approximation for a quantitative trait illustrates the dependence on sample size. If genotype is coded as the number of effect alleles, the effect per allele is \(\beta\), residual trait standard deviation is \(\sigma\), and allele frequency is \(f\), then under a simple additive model the stage-specific noncentrality is approximately

$$ \delta_j \approx \frac{\beta}{\sigma}\sqrt{n_j\,2f(1-f)}. $$

This approximation assumes unrelated participants, an additive genetic effect, appropriate phenotype and genotype measurement, and no unmodelled structure or confounding that changes the association statistic's distribution. Real designs may require a more specific information calculation, including imputation quality, covariates, case-control allocation, or phenotype harmonization. The central design point remains: the two stage-specific noncentralities depend on their own samples.

For a cost-constrained study, participant counts alone are not enough. A planning objective can be to maximize \(\pi_1\pi_2\) for a prespecified effect and variant, or to maximize the expected number of confirmed true associations across plausible effects. The constraint should account for the cost of broad discovery genotyping, the number of variants followed up, follow-up assay costs, and any recruitment or phenotyping costs. If all participants receive genome-wide genotyping at stage one, reducing the stage-one sample may save less than expected; if follow-up assays are targeted, the stage-two cost may depend strongly on how many variants are selected.

Worked Example: Comparing Two Participant Allocations

A hypothetical study has 1,000 distinct participants and compares allocations of 700/300 and 500/500 to discovery and follow-up. Assume equal per-participant information across stages and that \(\delta=2.000\) for a sample of 500. Use the deliberately simplified two-sided screening cutoff \(c_1=2.500\) and confirmation cutoff \(c_2=1.960\), for one selected variant. The example isolates the allocation trade-off; a real genome-wide study must use cutoffs appropriate to its testing plan.

Under the 700/300 allocation, the noncentralities scale with the square root of sample size: \(\delta_1=2\sqrt{700/500}=2.366\) and \(\delta_2=2\sqrt{300/500}=1.549\). The discovery power is \(\Phi(-2.500-2.366)+1-\Phi(2.500-2.366)\approx0.447\). The follow-up power is \(\Phi(-1.960-1.549)+1-\Phi(1.960-1.549)\approx0.341\). So overall power is approximately \(0.447\times0.341=0.152\), or 15.2%.

Under the 500/500 allocation, both noncentralities equal 2.000. Discovery power is \(\Phi(-4.500)+1-\Phi(0.500)\approx0.309\); follow-up power is \(\Phi(-3.960)+1-\Phi(-0.040)\approx0.516\). Overall power is approximately \(0.309\times0.516=0.159\), or 15.9%.

For these illustrative cutoffs and assumptions, the balanced allocation gives slightly greater overall power, even though its discovery power is lower. The result is specific to the chosen effect, cutoffs, and information assumptions; it does not establish that equal allocation is generally optimal. A practical design should compare candidate allocations across a range of plausible effect sizes and allele frequencies, and include costs and follow-up capacity.

Variant Selection and Follow-Up Capacity

Stage-one selection may be defined by a fixed threshold, by ranking and following the top \(K\) variants, or by a combination of rules. These choices affect the probability that any particular true variant reaches stage two. For a threshold rule, the stage-one selection probability for that variant is its stage-one power at the threshold. For a top-\(K\) rule, selection also depends on the other variants' statistics and their correlations; the probability of ranking among the top \(K\) is not generally the same as the probability of crossing a fixed cutoff.

The stage-two cutoff must be set for the actual follow-up plan. If several selected variants are tested, the confirmation rule must address the number of follow-up tests and the intended error control. A threshold that would be suitable for one prespecified variant may not be suitable for a large selected set. This affects stage-two power: a more stringent cutoff reduces the chance of confirming any given true effect.

A key benefit of independent follow-up is that selection in stage one does not, by itself, make the independent stage-two statistic unusually large under the true effect model. Stage-one effect estimates for selected variants can be inflated by selection, so the confirmation sample should not be treated as merely a second look at the discovery participants. Independence does not remove all design concerns: phenotype definitions, variant coding, ancestry, and analysis choices still need to be sufficiently aligned for the target association to be meaningfully tested.

Worked Example: Expected Confirmations Under a Top-Variant Plan

Suppose a hypothetical design can follow four candidate variants. For four true associated variants, assume the probabilities of surviving the discovery selection rule, including the ranking limit, are \(0.80\), \(0.60\), \(0.30\), and \(0.10\). Suppose each selected variant has independent stage-two power \(0.70\). These selection probabilities and powers are stipulated planning inputs.

For the first variant, the probability of both selection and confirmation is \(0.80\times0.70=0.56\). The corresponding probabilities for the other three are \(0.60\times0.70=0.42\), \(0.30\times0.70=0.21\), and \(0.10\times0.70=0.07\).

By adding the probabilities for the four variants, the expected number of confirmed true associations is \(0.56+0.42+0.21+0.07=1.26\). This expectation does not mean that a fractional number of variants will be confirmed in any single study. It is an average over repetitions of the design. If fewer variants can be followed, the selection probabilities must reflect the stricter ranking limit, or the design must specify which candidates will be excluded.

A Practical Planning Sequence

A defensible two-stage plan specifies the target population and association model, the discovery selection rule, the independent confirmation criterion, and the resource constraint before comparing allocations. The following sequence makes the dependencies explicit.

1
Set the target effect and variant range.
Choose plausible effect sizes, allele frequencies, and phenotype or case-control assumptions relevant to the intended population.
2
Define stage-one selection.
Specify a screening threshold, a ranking limit, or both; calculate or simulate the chance that a true variant enters follow-up.
3
Define stage-two confirmation.
Specify an independent sample, the test direction, and a threshold that reflects the number of variants to be tested.
4
Compare feasible allocations.
For each allocation, calculate stage-specific power and combine the probabilities when the independence assumptions hold; include genotyping and follow-up costs.
5
Check sensitivity and implementation.
Repeat calculations across plausible effects and frequencies, then verify that recruitment, assay coverage, ancestry representation, and phenotype definitions support the assumptions.

Derivation and Common Pitfalls

The product formula follows directly from conditional probability. Let \(S_1\) denote passing the stage-one selection rule and \(S_2\) denote passing the stage-two confirmation rule. Then

$$ P(S_1\cap S_2)=P(S_1)P(S_2\mid S_1). $$

If the stage-two participants and their test data are independent of stage one, and the variant's true effect and stage-two rule are fixed, then conditioning on stage-one selection does not alter the distribution of the stage-two statistic. Therefore \(P(S_2\mid S_1)=P(S_2)=\pi_2\), while \(P(S_1)=\pi_1\), giving \(P(S_1\cap S_2)=\pi_1\pi_2\). With participant overlap, adaptive changes to the follow-up rule, or dependence created by reusing stage-one data, that equality requires further justification and may fail.

Common mistake: “Discovery power is 80%, so the study has 80% power to report the association.” This ignores the requirement that the variant also pass independent confirmation. If confirmation power is 60% and the stages are independent, overall power is \(0.80\times0.60=0.48\), or 48%, not 80%. A related mistake is to use a permissive stage-two cutoff without accounting for the number of variants followed up. The correction is to define both selection and confirmation rules in advance, calculate each stage's probability under the same effect assumptions, and combine them only when the independence conditions are satisfied.

Key takeaway: A two-stage GWAS is powered by the chance that a true variant both survives discovery selection and meets independent confirmation. Allocate resources to balance those probabilities, while making selection limits, follow-up thresholds, and independence assumptions explicit.

Check Your Understanding

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

  1. A design has stage-one power \(0.70\) and independent stage-two power \(0.50\) for a specified true variant. What is its overall power to pass both stages?
  2. Why can a top-\(K\) selection rule have a different selection probability from a fixed stage-one threshold?
  3. What assumption allows the overall probability of passing both stages to be written as the product of the two stage-specific probabilities?
  4. If a study increases the discovery sample while keeping the total number of participants fixed, what happens to the follow-up sample, and why might overall power not increase?
  5. Why must a stage-two threshold reflect the number of selected variants being tested?
Clinical Biostats