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SNP Heritability Estimation · Tutorial 536 of 989

Interpreting SNP Heritability Without Overclaiming

Interpret SNP heritability as a population- and scale-specific variance ratio, not as a statement about an individual or a guarantee of causal explanation or prediction.

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What You'll Learn

  • Define SNP heritability as a variance proportion for a specified population, phenotype scale, and SNP set
  • Interpret a numerical estimate without treating it as an individual-level percentage
  • Explain why variance tagged by SNPs is not necessarily variance caused by those SNPs
  • Distinguish SNP heritability from total genetic variance and from prediction accuracy
  • Recognize how phenotype scale, population, and environmental variation affect interpretation
  • Identify common overclaims and replace them with appropriately qualified conclusions

What an SNP Heritability Estimate Describes

An estimate such as “SNP heritability is 0.40” can sound like a statement about how much of a person’s trait comes from genes. It is not. SNP heritability describes a proportion of variation among individuals in a specified population, for a specified phenotype and measurement scale, under a model that uses a specified set of genetic markers. Its interpretation depends on those choices.

The term “SNP heritability” is often written \(h^2_{\mathrm{SNP}}\). In an additive model, it refers to the fraction of phenotypic variance attributed to the component captured by the analyzed SNPs. “Captured” matters: a marker can tag genetic variation through linkage disequilibrium even when it is not itself the causal variant. The estimate is therefore not automatically a measure of the variance caused by the typed SNPs alone.

Definition: For a specified population, phenotype scale, covariate adjustment, and SNP set, \(h^2_{\mathrm{SNP}}\) is the proportion of phenotypic variance attributed by the model to the additive genetic component captured by that SNP set. In a variance-components model with uncorrelated genetic and residual components, it is the captured genetic variance divided by the sum of captured genetic and residual variance.

For example, write a covariate-adjusted phenotype as \(y_i=g_i+e_i\), where \(g_i\) is the additive component captured by the SNPs and \(e_i\) is the remaining component. If the components have mean zero and are uncorrelated, then

$$ \operatorname{Var}(y_i) =\operatorname{Var}(g_i)+\operatorname{Var}(e_i) =\sigma_g^2+\sigma_e^2, \qquad h^2_{\mathrm{SNP}} =\frac{\sigma_g^2}{\sigma_g^2+\sigma_e^2}. $$

This expression is a model-based variance ratio, not a universal property of the trait. In particular, “phenotypic variance” must be understood on the analysis scale and under the analysis’s treatment of covariates. For a binary phenotype, for instance, a proportion on the observed scale is not automatically the same as one on a liability scale; those scales require explicit distinction.

Worked Example: Interpreting a Variance Ratio

Worked Example: Interpreting a Variance Ratio

Suppose a hypothetical study estimates that, in its target population and on its stated phenotype scale, the SNP-captured additive variance is 18 squared trait units and the residual variance is 42 squared trait units. Calculate and interpret the SNP heritability.

The model gives total variance \(18+42=60\) squared trait units. Therefore,

$$ \hat h^2_{\mathrm{SNP}} =\frac{18}{18+42} =\frac{18}{60} =0.30. $$

The estimate says that 30% of the between-individual phenotypic variance in this defined population, for this phenotype scale and model, is attributed to the additive component captured by the analyzed SNPs. It does not say that 30% of any one person’s trait is genetic, that every person has the same genetic contribution, or that the SNPs caused 30% of the trait in each individual.

Population Variance Is Not Individual Determinism

Heritability is a comparison of variation across individuals. It does not partition one individual’s observed value into a genetic percentage and an environmental percentage. The model’s variance components describe how values differ across the study population, not the composition of a particular person’s phenotype.

A high value also does not mean that the trait is fixed or cannot respond to environmental change. Heritability can be high in one setting because people in that setting experience relatively similar environments, while the same trait could show lower heritability in a setting with more varied environmental exposures. The estimate concerns variation under the conditions represented in the analysis; it is not a measure of how much a trait can change.

Worked Example: Why the Same Genetic Variance Can Give Different Ratios

Consider two hypothetical populations measured on the same trait scale. Suppose, solely to isolate the effect of residual variance, each has SNP-captured genetic variance of 20 squared units. In Population A, residual variance is 30; in Population B, it is 60. Calculate both ratios.

For Population A,

$$ h^2_{\mathrm{SNP},A} =\frac{20}{20+30} =\frac{20}{50} =0.40. $$

For Population B,

$$ h^2_{\mathrm{SNP},B} =\frac{20}{20+60} =\frac{20}{80} =0.25. $$

Under the stipulated values, the captured genetic variance is the same, but the proportion is smaller in Population B because its total variance is larger. This calculation does not establish that real populations have identical genetic effects: allele frequencies, linkage disequilibrium, environments, measurement, and other features may differ. It shows why a difference between heritability estimates is not, by itself, evidence that the underlying biology differs.

Captured Variance Is Not the Same as Causal Attribution

An SNP can be statistically associated with a trait because it is correlated with a causal variant. In that case, it can help tag genetic variation without being the biological cause of the trait difference. More generally, an SNP heritability estimate summarizes a model’s captured variance; the estimate alone does not identify which variants are causal, which biological mechanism is involved, or whether an association is free of confounding.

The distinction can be seen in a simple hypothetical example. Let \(C\) be a standardized causal genotype and \(S\) a different standardized marker, with correlation 0.8. Suppose the phenotype follows \(y=2C+e\), where \(e\) is uncorrelated with both genotypes. The marginal regression coefficient of \(y\) on \(S\) is

$$ \beta_S =\frac{\operatorname{Cov}(S,y)}{\operatorname{Var}(S)} =\frac{\operatorname{Cov}(S,2C+e)}{1} =2(0.8) =1.6. $$

Thus \(S\) has a nonzero association in this example even though the phenotype model assigns the direct effect to \(C\), not \(S\). The association illustrates tagging through correlation; it does not demonstrate that changing \(S\) would change the phenotype. Real analyses require appropriate design and assumptions to address confounding, and a heritability estimate by itself cannot supply a causal interpretation.

Worked Example: An Associated Marker Need Not Be the Cause

Using the hypothetical setup above, calculate the expected phenotype difference associated with a one-unit increase in the standardized marker \(S\), and state what the result does and does not imply.

Because \(C\) and \(S\) each have variance 1 and correlation 0.8, their covariance is 0.8. With \(y=2C+e\) and \(\operatorname{Cov}(S,e)=0\),

$$ \operatorname{Cov}(S,y) =2\operatorname{Cov}(S,C)+\operatorname{Cov}(S,e) =2(0.8)+0 =1.6. $$

Since \(\operatorname{Var}(S)=1\), the marginal slope is \(1.6/1=1.6\) phenotype units per one standardized unit of \(S\). This is an association under the stated model. It does not show that \(S\) is causal: in this setup its association arises because it is correlated with \(C\). Nor does this small calculation establish the cause of any real-world association.

SNP Heritability Is Not Total Genetic Variance

SNP heritability is limited by what the analyzed markers and model capture. If the SNP set does not tag some genetic variation well, that variation need not contribute fully to the estimate. Depending on the study and model, relevant differences can arise from marker coverage, allele frequencies, linkage disequilibrium, the variants included, and the assumed effect structure. Genetic contributions not represented by the fitted additive component are not automatically included.

For that reason, \(h^2_{\mathrm{SNP}}\) should not be read as “all genetic influence.” It is also not automatically equal to narrow-sense heritability, which refers to the additive genetic contribution as defined for a population and model. The two quantities may differ because the SNP-based estimate reflects the genetic variation tagged by the analyzed markers under its estimation assumptions. Neither quantity is a fixed percentage that applies to every population or environment.

SNP Heritability Is Not Prediction Accuracy

A heritability estimate describes a variance ratio. Prediction accuracy describes how well a predictor, such as a polygenic score, predicts phenotypes for individuals. They are related in principle but are not interchangeable. Prediction depends not only on the amount of genetic variance available to capture, but also on the quality of the estimated effects, the sample size and design used to estimate them, the target population, and the phenotype measurement.

To make the distinction concrete, suppose a hypothetical phenotype has total variance 100 units squared and the SNP-captured component has variance 40 units squared. The resulting SNP heritability is 0.40. Now suppose an independently evaluated polygenic score has correlation 0.45 with the phenotype in the target sample. For a simple linear prediction, the squared correlation is the fraction of phenotype variance accounted for by the score’s linear predictions:

$$ R^2=(0.45)^2=0.2025. $$

Worked Example: Separating a Heritability Estimate From Score Performance

Assume the hypothetical variance components above, and assume the score was evaluated in independent participants on the same phenotype scale. Compare \(h^2_{\mathrm{SNP}}\) with the score’s \(R^2\).

The variance ratio is \(40/100=0.40\), or 40%. The score’s squared correlation is \(0.45^2=0.2025\), or 20.25%. The estimate of SNP heritability describes the model-attributed share of population variance; the score’s \(R^2\) describes the variance accounted for by this particular predictor in the validation sample. The score’s performance is not 40% simply because the estimated heritability is 0.40.

The values are illustrative, not a universal relationship between the quantities. Different score construction, validation populations, phenotype scales, and modeling assumptions can produce different prediction performance. Always report prediction metrics from the population in which the score is evaluated.

A Reliable Interpretation Checklist

Before interpreting an SNP heritability estimate, identify what the number refers to. The following sequence keeps the conclusion tied to the estimand rather than extending it beyond the analysis.

1
Name the population.
State which individuals the estimate describes. The result need not transfer unchanged to another population.
2
Name the phenotype and scale.
Specify what was measured and whether the estimate is on the observed or another stated scale.
3
Name the captured component.
Describe the SNP set and model, and qualify the result as variance captured or tagged under those assumptions.
4
Keep the claim at the population level.
Do not translate a variance proportion into an individual’s genetic percentage or a causal claim.
5
Report prediction separately.
If individual-level prediction is the question, give an appropriate validation metric rather than substituting heritability.

Common Overclaims

A frequent mistake is to write, “40% of this person’s trait is determined by their SNPs,” after observing \(h^2_{\mathrm{SNP}}=0.40\). That reasoning confuses variation across people with the makeup of one person’s phenotype. The correction is: “In this population, on this phenotype scale, the model attributes 40% of phenotypic variance to the additive component captured by the analyzed SNPs.”

Other conclusions that do not follow from the number alone include:

  • “The associated SNPs caused the trait.” Association can reflect linkage disequilibrium or other sources of association; heritability does not establish causation.
  • “This is the total genetic contribution.” The estimate concerns the component captured under the specified SNP set and model, not necessarily every genetic contribution.
  • “A score will predict 40% of the trait.” Heritability and prediction accuracy are different quantities; score performance must be assessed directly.
  • “The trait is genetically fixed.” A variance ratio does not measure whether the trait can respond to environmental change.
  • “A higher estimate proves a stronger genetic effect in one population.” The ratio can change when total variance, tagging, measurement, or other population features change.
Key takeaway: Interpret \(h^2_{\mathrm{SNP}}\) as a model-dependent proportion of between-individual variance for a defined population and phenotype scale. It is not an individual-level percentage, a causal attribution, a guarantee of total genetic variance, or a measure of prediction accuracy.

Check Your Understanding

For each question, state the interpretation that follows from the information given and avoid adding claims that the estimate does not support.

  1. A study reports \(h^2_{\mathrm{SNP}}=0.25\) for a quantitative trait in a specified population. What does the 0.25 describe?
  2. Why does an SNP heritability estimate not mean that a quarter of each individual’s trait value is genetic?
  3. A marker is associated with a trait and is in linkage disequilibrium with a causal variant. Why is the marker association not, by itself, evidence that the marker causes the trait?
  4. Two populations have different SNP heritability estimates. Give two features, other than a difference in causal effects, that could contribute to the difference.
  5. A polygenic score has a validation correlation of 0.30 with a phenotype. What is its squared correlation, and why should it not be replaced by the SNP heritability estimate?
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