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GWAS Replication and Meta-Analysis · Tutorial 383 of 989

Handling Variants Missing From Some Cohorts

See how to pool only valid cohort estimates for each variant, and how to interpret the resulting estimate when availability and precision differ across cohorts.

Advanced 10 min read

What You'll Learn

  • Define the contributing-cohort set for each variant
  • Distinguish an unavailable estimate from a valid estimate of zero
  • Calculate inverse-variance fixed-effect estimates using available cohorts
  • Explain how allele frequency can affect precision and meta-analysis weights
  • Assess how variant availability limits the populations represented by a pooled result

Why Variant Availability Matters

A GWAS meta-analysis may combine cohorts that did not all analyze every variant. A variant might be absent from a genotyping array, fail a cohort’s quality-control rules, or have too little variation to estimate an association. The analysis therefore has to distinguish a missing estimate from a measured estimate near zero. That distinction determines which data contribute to the pooled result.

The fixed-effect method introduced in earlier material combines cohort estimates of a shared effect, weighting more precise estimates more heavily. When a variant is unavailable in some cohorts, the method is applied to the valid estimates for that variant only. The resulting calculation is straightforward; interpreting what population it represents requires more care.

Definition: For variant \(k\), let \(\hat{\beta}_{ik}\) be the estimated effect in cohort \(i\), and let \(SE_{ik}\) be its standard error. The contributing-cohort set \(S_k\) contains the cohorts with a valid, usable estimate and standard error for variant \(k\). A cohort with no usable estimate is excluded from \(S_k\); its effect is not set to zero.

Fixed-Effect Pooling With Available Estimates

Suppose each contributing cohort estimates the same effect \(\beta_k\), on the same allele and phenotype scale. Assume the estimates are unbiased, their standard errors are valid, and estimates from different cohorts are independent. Under the fixed-effect model, cohort \(i\)'s estimate has variance \(SE_{ik}^2\), so its inverse-variance weight is \(w_{ik}=1/SE_{ik}^2\). The pooled estimate and its standard error are

$$ \hat{\beta}_{k,\mathrm{meta}} = \frac{\sum_{i\in S_k} w_{ik}\hat{\beta}_{ik}} {\sum_{i\in S_k} w_{ik}}, \qquad SE_{k,\mathrm{meta}} = \sqrt{\frac{1}{\sum_{i\in S_k} w_{ik}}}. $$

The sums include only cohorts in \(S_k\). A cohort with no estimate contributes neither an effect nor a weight. A valid estimate of exactly zero is different: it remains in the sums and receives a weight determined by its standard error.

Assumptions for this calculation: The contributing estimates must refer to the same effect allele, phenotype definition, and effect scale; be unbiased estimates of a common effect under the fixed-effect model; have valid standard errors; and be independent across cohorts. Harmonizing alleles and aligning phenotype scales are essential before pooling. If cohorts overlap, the independence assumption may fail, so this formula cannot be used as if their estimates were independent.

Worked Example: Pooling Two Estimates When a Third Is Unavailable

Suppose three hypothetical cohorts estimate the change in a continuous phenotype, in standard-deviation units per effect allele. Cohort A reports \(\hat{\beta}_{Ak}=0.080\) with \(SE_{Ak}=0.020\); cohort B reports \(\hat{\beta}_{Bk}=0.040\) with \(SE_{Bk}=0.040\). Cohort C did not analyze the variant and reports no estimate. Assume the two reported estimates are aligned, independent, and valid estimates of a common effect.

Cohort C is not in \(S_k\), so the weights are calculated for A and B only:

$$ w_{Ak}=\frac{1}{0.020^2}=2500, \qquad w_{Bk}=\frac{1}{0.040^2}=625. $$

The pooled estimate is

$$ \hat{\beta}_{k,\mathrm{meta}} = \frac{(2500)(0.080)+(625)(0.040)}{2500+625} = \frac{200+25}{3125} = 0.0720. $$

Its standard error is

$$ SE_{k,\mathrm{meta}} = \sqrt{\frac{1}{3125}} = 0.0179. $$

The estimate is \(0.0720\) standard deviations per effect allele among the contributing cohorts under the common-effect model. A normal-approximation 95% confidence interval is \(0.0720\pm1.96(0.0179)\), or approximately \(0.0369\) to \(0.107\) standard deviations per allele. Cohort C has not supplied evidence for a zero effect; it supplied no estimate at all.

Why an Unavailable Estimate Is Not a Null Effect

An association estimate of zero is a result of an analysis: the estimated effect is zero, and its standard error describes how uncertain that estimate is. An unavailable result has no corresponding estimated effect or standard error to use. Treating it as zero would invent data and would require inventing a weight as well. Since the pooled result depends on both the estimates and their weights, that substitution can change the answer arbitrarily.

Availability can have several causes. A cohort may lack the variant because of its genotyping design; a variant may fail quality control; an allele may be absent or too rare in that cohort to support estimation; or a result may be suppressed under reporting rules. These cases do not all have the same scientific meaning, but none automatically imply that the true effect is zero. Record why estimates are unavailable where possible, rather than encoding every reason as the same numerical value.

Worked Example: A Valid Zero Estimate Still Contributes

Consider four hypothetical cohorts, again estimating phenotype standard deviations per effect allele. Cohort A reports \(0.030\) with standard error \(0.015\); cohort B reports \(0.050\) with standard error \(0.030\); cohort C has no usable estimate; and cohort D reports a valid estimate of \(0.000\) with standard error \(0.020\). Assume the three reported results meet the fixed-effect assumptions.

The contributing set is \(S_k=\{A,B,D\}\). The weights are

$$ w_{Ak}=\frac{1}{0.015^2}=4444.44,\quad w_{Bk}=\frac{1}{0.030^2}=1111.11,\quad w_{Dk}=\frac{1}{0.020^2}=2500. $$

Cohort D contributes weight despite its zero estimate; cohort C contributes nothing. Thus

$$ \hat{\beta}_{k,\mathrm{meta}} = \frac{(4444.44)(0.030)+(1111.11)(0.050)+(2500)(0.000)} {4444.44+1111.11+2500} = \frac{188.889}{8055.56} = 0.0234. $$

The pooled standard error is \(\sqrt{1/8055.56}=0.0111\) standard deviations per allele. A normal-approximation 95% confidence interval is \(0.0234\pm1.96(0.0111)\), approximately \(0.0016\) to \(0.0453\). The zero estimate from D pulls the pooled value toward zero while contributing precision. C cannot do either, because it has no valid estimate or standard error.

Justification for the Available-Cohort Formula

The inverse-variance formula follows directly from the likelihood for independent, normally distributed cohort estimates under a common effect. For a fixed variant, assume \(\hat{\beta}_{ik}\sim N(\beta_k,SE_{ik}^2)\) for every \(i\in S_k\). Apart from constants, the log likelihood as a function of \(\beta_k\) is

$$ \ell(\beta_k) = -\frac{1}{2}\sum_{i\in S_k} \frac{(\hat{\beta}_{ik}-\beta_k)^2}{SE_{ik}^2}. $$

Differentiate with respect to \(\beta_k\), set the derivative to zero, and solve:

$$ \frac{d\ell}{d\beta_k} = \sum_{i\in S_k}\frac{\hat{\beta}_{ik}-\beta_k}{SE_{ik}^2}=0 \quad\Longrightarrow\quad \hat{\beta}_{k,\mathrm{meta}} = \frac{\sum_{i\in S_k}\hat{\beta}_{ik}/SE_{ik}^2} {\sum_{i\in S_k}1/SE_{ik}^2}. $$

The information about \(\beta_k\) in this likelihood is \(\sum_{i\in S_k}1/SE_{ik}^2\), giving the variance \(1/\sum_{i\in S_k}1/SE_{ik}^2\) and the standard error shown above. A cohort outside \(S_k\) has no likelihood term for this variant, which is why it cannot be entered as a zero-valued observation.

Availability, Allele Frequency, and the Pooled Estimate

Allele frequency can affect both whether a cohort can estimate a variant and how precise that estimate is. In a simple additive model for a continuous phenotype, with genotype coded as 0, 1, or 2 copies of the effect allele, genotype variance is approximately \(2f(1-f)\), where \(f\) is the effect-allele frequency. Under otherwise comparable conditions, a rough relationship is

$$ \operatorname{Var}(\hat{\beta}_{ik}) \ \propto\ \frac{1}{n_i\,2f_{ik}(1-f_{ik})}, $$

where \(n_i\) is the cohort sample size. This is an approximation, not a universal standard-error formula: phenotype variance, covariates, case-control balance, and genotype uncertainty also matter. It nevertheless illustrates why a less frequent allele often has a larger standard error and a smaller inverse-variance weight. If a variant is absent or has too little variation in a cohort, an estimate may not be available at all.

Allele frequency can differ across populations, so the same variant may be common and precisely estimated in one cohort but rare or unavailable in another. Consequently, the contributing cohorts and the distribution of weights can vary from variant to variant. A pooled result can then be driven mainly by cohorts in which that variant is more common and more precisely estimated. Frequency does not directly enter the inverse-variance formula once valid estimates and standard errors are supplied; it affects the result through availability and precision.

Worked Example: Different Frequencies Produce Different Weights

Suppose three hypothetical cohorts study the same continuous phenotype and aligned effect allele. Cohort A has effect-allele frequency \(0.30\), estimate \(0.050\), and standard error \(0.010\). Cohort B has frequency \(0.020\), estimate \(0.070\), and standard error \(0.040\). In cohort C the variant is not available for analysis. Assume A and B are independent and that the common-effect assumptions are appropriate for this calculation.

For the available cohorts,

$$ w_{Ak}=\frac{1}{0.010^2}=10000, \qquad w_{Bk}=\frac{1}{0.040^2}=625. $$

The pooled estimate and standard error are

$$ \hat{\beta}_{k,\mathrm{meta}} = \frac{(10000)(0.050)+(625)(0.070)}{10000+625} = \frac{543.75}{10625} = 0.0512, \qquad SE_{k,\mathrm{meta}} = \sqrt{\frac{1}{10625}} = 0.00970. $$

Although B’s estimate is larger, A supplies \(10000/10625=94.1\%\) of the total weight, so the pooled estimate lies much closer to A’s estimate. In this hypothetical example, A’s lower standard error is consistent with the higher allele frequency, though frequency alone does not determine precision. The result is based on A and B, not C, and its interpretation should reflect that coverage.

What the Variant-Specific Result Represents

Under a genuinely common effect, removing a cohort without a usable estimate reduces the information available but does not, by itself, change the effect being estimated. In practice, however, effects may differ across cohorts or populations. Then a fixed-effect estimate is a precision-weighted summary of the estimates that are available for that variant. If the available set differs across variants, those summaries need not represent the same mix of cohorts.

A result based on a subset should not automatically be described as an estimate for every cohort originally considered. Report the number or identities of contributing cohorts, the total sample size where available, and relevant reasons for non-availability. When allele frequencies or cohort characteristics differ, also consider whether the available cohorts adequately represent the populations to which the conclusion is intended to apply. The fixed-effect calculation alone cannot recover effects for cohorts with no usable data.

Common Mistakes

  • Replacing a missing estimate with zero. “No result means no effect” is incorrect. A missing estimate has no effect estimate or standard error; exclude it from that variant’s sums.
  • Dropping a valid zero estimate. A reported estimate of zero with a valid standard error is informative and contributes inverse-variance weight.
  • Assuming every variant has the same contributing cohorts. Availability is variant-specific. Define \(S_k\) separately for each variant and inspect its coverage.
  • Assuming allele frequency changes the effect scale. Frequency can affect precision and availability, and therefore the weights and cohort mix; it does not itself convert the effect estimate to a different scale.
  • Interpreting a subset result as universal. A pooled estimate from the available cohorts does not establish the effect in cohorts that could not estimate the variant.
Key takeaway: For each variant, pool only valid, harmonized estimates with valid standard errors. An unavailable cohort is not a null result, and allele-frequency differences can change which cohorts contribute and how strongly they influence the pooled estimate.

Check Your Understanding

Use the definitions and assumptions above to answer the following questions.

  1. A cohort has no result for a variant because it was not genotyped. What does that imply about the cohort’s effect estimate and weight in the variant-specific meta-analysis?
  2. Why should a valid estimate of zero with a standard error be treated differently from an unavailable estimate?
  3. Two cohorts have the same effect estimate, but one has a larger standard error. Which cohort receives more inverse-variance weight, and why?
  4. How could allele-frequency differences make the contributing cohorts differ across variants?
  5. If a pooled estimate is based on only a subset of the planned cohorts, what information should be reported to support a careful interpretation?
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