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Partitioned Genetic Heritability · Tutorial 548 of 989

Continuous Annotation Scores and Heritability

Model per-variant genetic variance as a function of a continuous annotation score, interpret the coefficient on its proper scale, and assess whether the predictions are meaningful.

Advanced 10 min read

What You'll Learn

  • Define a linear conditional-variance model for per-variant effects using a continuous annotation score
  • Interpret the intercept and score coefficient in the units implied by the score and genotype coding
  • Re-express a model after centering or standardizing its annotation score
  • Calculate predicted effect variances and their implications for genetic variance under stated assumptions
  • Check predictions across the score range and recognize limits of a linear functional form

Why Model a Continuous Annotation?

A genetic annotation need not divide variants into categories. It may assign each variant a numerical score, such as a measure of regulatory activity or sequence conservation. A continuous score retains distinctions between variants that would be grouped together by a threshold. The statistical question is how the distribution of variant effects changes as the score changes.

The target in the model below is the conditional variance of a variant’s effect, not automatically its expected squared effect. This distinction matters: the expected square equals the variance plus the square of the conditional mean. We will state the mean assumption explicitly when using expected squared effects.

Definition: Let \(c_j\) be a numerical annotation score for variant \(j\), and let \(\beta_j\) be its effect on a phenotype under a specified genotype coding. A linear continuous-annotation variance model is $$ \operatorname{Var}(\beta_j\mid c_j)=\tau_0+\tau_1c_j. $$ Here \(\tau_0\) is the predicted variance at score zero, and \(\tau_1\) is the change in predicted variance for a one-unit increase in score. The model applies over the score range being analyzed, and its predictions must be nonnegative over that range.

This model describes how effect variance varies with the annotation; it does not say that the mean effect varies with the score. A common additional assumption is \(E[\beta_j\mid c_j]=0\). Under that assumption, $$ E[\beta_j^2\mid c_j]=\operatorname{Var}(\beta_j\mid c_j)=\tau_0+\tau_1c_j. $$ Without the zero-mean assumption, the expected squared effect instead equals \(\operatorname{Var}(\beta_j\mid c_j)+E[\beta_j\mid c_j]^2\). Thus, a variance model alone does not determine the expected squared effect.

Scale and Interpretation

The coefficient depends on how both the effect and the annotation are scaled. If genotypes are standardized to variance one and the phenotype is measured in standard-deviation units, \(\beta_j\) is in phenotype standard deviations per one standard deviation of genotype. Its variance is in squared phenotype-effect units. If the annotation score is dimensionless, \(\tau_1\) is the change in effect variance per one score unit. If the score is measured in physical units, the coefficient’s units include the inverse of those units.

The intercept is not always a general “baseline effect variance.” It is the predicted variance at \(c_j=0\). If zero is outside the observed score range, \(\tau_0\) is an extrapolated intercept and may not have a useful direct biological interpretation. Centering the score can make the intercept correspond to a typical score instead.

Worked Example: Interpreting a Coefficient on the Raw Score Scale

Suppose a hypothetical analysis uses standardized genotypes and a phenotype measured in standard-deviation units. For a continuous annotation score \(c_j\), the fitted conditional-variance model is $$ \operatorname{Var}(\beta_j\mid c_j)=0.0008+0.0004c_j. $$ Assume the score ranges from 0 to 2. What variance and effect standard deviation does the model predict at score \(c_j=0.5\), and what does the slope mean?

Predicted variance. Substitute \(c_j=0.5\): $$ 0.0008+0.0004(0.5)=0.0008+0.0002=0.0010. $$ The predicted conditional variance is \(0.0010\) squared phenotype-effect units per standardized genotype unit.

Predicted standard deviation. Taking the square root gives $$ \sqrt{0.0010}=0.03162. $$ Thus the conditional standard deviation of \(\beta_j\) is about \(0.03162\) phenotype standard deviations per one standard deviation of genotype. This is a spread of effects, not a prediction that the effect itself equals \(0.03162\).

Slope interpretation. For a one-unit increase in the annotation score, the model predicts an increase of \(0.0004\) in conditional effect variance, in squared phenotype-effect units. The increase is in variance, not in effect standard deviation. Because the model is nonnegative throughout the stated score range, its predictions are admissible there: at the minimum score the variance is \(0.0008\), and at the maximum it is \(0.0008+0.0004(2)=0.0016\).

Centering and Standardizing the Score

Changing the score scale changes the numerical coefficient, even when it leaves every predicted variance unchanged. Suppose the raw score is \(c_j\), its chosen centering value is \(m\), and its scaling constant is \(s_c>0\). Define \(z_j=(c_j-m)/s_c\). If the raw-scale model is \(\tau_0+\tau_1c_j\), substitution of \(c_j=m+s_cz_j\) gives $$ \tau_0+\tau_1c_j=(\tau_0+\tau_1m)+(\tau_1s_c)z_j. $$ The centered-scale intercept is therefore the predicted variance at \(c_j=m\), and the coefficient on \(z_j\) is the change in variance per one unit of \(z_j\). When \(s_c\) is the score’s standard deviation, that unit is one standard deviation of the raw score.

Worked Example: Converting Between Raw and Standardized Scores

In a hypothetical data set, the annotation score has mean 10 and standard deviation 2. A fitted model written using the standardized score \(z_j=(c_j-10)/2\) is $$ \operatorname{Var}(\beta_j\mid z_j)=0.0012+0.0003z_j. $$ Find the predicted variance at raw score \(c_j=14\), and express the same model using the raw score.

At \(c_j=14\), the standardized score is $$ z_j=\frac{14-10}{2}=2. $$ The predicted variance is $$ 0.0012+0.0003(2)=0.0012+0.0006=0.0018. $$

To convert to the raw scale, replace \(z_j\) by \((c_j-10)/2\): $$ 0.0012+0.0003\left(\frac{c_j-10}{2}\right) =0.0012+0.00015(c_j-10) =-0.0003+0.00015c_j. $$ The raw-scale slope is \(0.00015\) variance units per raw score unit, while the standardized-scale slope \(0.0003\) is the variance change per one standard deviation of score. At \(c_j=14\), either expression gives \(0.0018\). The negative raw-scale intercept is the extrapolated prediction at score zero; it is not itself evidence of a negative predicted variance within the observed score range. In this example, the centered intercept \(0.0012\) is the predicted variance at the mean score and has a direct interpretation.

From Per-Variant Variance to Genetic Variance

A per-variant variance prediction is not by itself a heritability estimate. To see how it relates to aggregate genetic variance, consider a phenotype model with a vector of genotypes \(X\), effects \(\beta\), and genetic component \(g=X\beta\). Let \(\Sigma\) be the genotype covariance matrix. Conditional on the genotypes, the genetic variance across individuals is \(\beta^\mathsf{T}\Sigma\beta\). If the effects have conditional mean vector \(m\) and covariance matrix \(V\), then averaging over the effect distribution gives $$ E[\beta^\mathsf{T}\Sigma\beta\mid c] =\operatorname{tr}(\Sigma V)+m^\mathsf{T}\Sigma m. $$ This follows by writing \(\beta=m+(\beta-m)\), expanding the quadratic form, and taking expectations: the terms linear in \(\beta-m\) have expectation zero, and the centered quadratic term has expectation \(\operatorname{tr}(\Sigma V)\).

In the special case of zero conditional means and uncorrelated effects across variants, \(m=0\) and \(V\) is diagonal. The expression reduces to $$ E[\beta^\mathsf{T}\Sigma\beta\mid c]=\sum_j\Sigma_{jj}\operatorname{Var}(\beta_j\mid c_j). $$ If every genotype is standardized to variance one, \(\Sigma_{jj}=1\), so this becomes the sum of the per-variant variances. This expectation does not require variants to be in linkage equilibrium: with uncorrelated, zero-mean effects, the off-diagonal terms average to zero. However, the genetic variance for a particular fixed effect vector can include LD cross-terms. If effects are correlated or have nonzero conditional means, the simplified sum need not apply.

Worked Example: Summing Predicted Variances Under Stated Assumptions

Consider three hypothetical variants with standardized genotypes and annotation scores \(c_1=0\), \(c_2=1\), and \(c_3=2\). Suppose $$ \operatorname{Var}(\beta_j\mid c_j)=0.001+0.0005c_j. $$ Assume the effects have conditional mean zero and are uncorrelated across variants. Calculate the expected genetic variance contribution from these three variants.

The predicted effect variances are $$ v_1=0.001+0.0005(0)=0.001,\qquad v_2=0.001+0.0005(1)=0.0015,\qquad v_3=0.001+0.0005(2)=0.002. $$ Each standardized genotype has variance one. Therefore, the expected genetic variance is $$ 1(0.001)+1(0.0015)+1(0.002)=0.0045. $$ This is an expected genetic variance in squared phenotype units under the stated model and assumptions. If the phenotype has variance one, it is also the expected fraction of phenotypic variance contributed by these variants under this model. It is not automatically the total SNP heritability for a larger set of variants, nor does it imply that each realized effect has squared value equal to its predicted variance.

Connection to Annotation-Based LD Regression

The same variance model can be represented in the expected association statistics used in stratified LD score regression. For standardized genotypes, let \(r_{jk}\) be the correlation between variants \(j\) and \(k\). Define the ordinary LD score \(l_j=\sum_k r_{jk}^2\), and the LD score weighted by the continuous annotation \(l_{j,c}=\sum_k r_{jk}^2c_k\). If effects have zero conditional means and are uncorrelated across variants conditional on the annotations, and the phenotype/residual-variance scaling makes the association noncentrality approximately \(N\) times the genetic variance contribution (or the corresponding scale factor is absorbed into the variance coefficients), then under the usual approximation for the regression of association chi-square statistics the model \(\operatorname{Var}(\beta_k\mid c_k)=\tau_0+\tau_1c_k\) gives $$ E[\chi_j^2]\approx a+N\tau_0l_j+N\tau_1l_{j,c}, $$ where \(N\) is the sample size and \(a\) is an intercept that can include the null contribution and modeled inflation not attributable to polygenic signal.

This equation helps explain why the coefficient on a continuous annotation is a variance coefficient: it weights the annotation score carried by variants in LD with the tested variant. The coefficient is not the expected increase in an individual variant’s estimated GWAS effect, and it is not an enrichment ratio. Its numerical value depends on the annotation’s scale and the effect coding. The regression interpretation also relies on the assumptions and approximations of the LD score model; it does not remove the need to examine whether the underlying variance model is sensible.

Check the Functional Form and the Predictions

A linear function is a convenient starting point, not a guarantee that biology or the effect distribution follows a straight line. The model makes two basic checks essential. First, its predicted variance must be nonnegative at every score value to which the model is applied. Second, the score’s scale and observed range must be understood, since a coefficient can look numerically small or large simply because the score is expressed in different units.

For example, if the score ranges from 0 to 2, then \(\tau_0+\tau_1c\) is linear in \(c\), so it is enough to check its values at both endpoints to ensure nonnegativity throughout that range. If the score range is broader, if the relation appears curved, or if predictions become negative near an endpoint, the linear form may be inadequate. Possible remedies include reconsidering the score transformation or using a different variance model, but any alternative must still define a valid, nonnegative variance over the range of use.

1
State the modeled quantity.
Specify that the annotation predicts \(\operatorname{Var}(\beta_j\mid c_j)\), and state any additional assumptions, such as zero conditional mean, before interpreting it as \(E[\beta_j^2\mid c_j]\).
2
Record the scales.
Identify genotype coding, phenotype units, and annotation units. These determine the units of the variance and the meaning of a one-unit score change.
3
Calculate predictions over the score range.
Substitute representative scores, including the range endpoints, and check that the predicted variance is nonnegative where the model is applied.
4
Interpret without overreach.
Describe how the modeled spread of effects changes with score. Do not treat the coefficient as a causal effect of changing an annotation or as a direct prediction of an individual variant’s realized effect.

Common Pitfall: Confusing Variance With Expected Squared Effect

A tempting but incomplete statement is: “The expected squared effect increases linearly with the annotation score because the variance does.” That conclusion does not follow unless the conditional mean is zero. In general, $$ E[\beta_j^2\mid c_j] =\operatorname{Var}(\beta_j\mid c_j)+E[\beta_j\mid c_j]^2. $$ If the conditional mean varies with the score, the second term can also vary, and the expected squared effect need not follow the same linear model. State the model as a conditional-variance model unless the zero-mean assumption has been made and is appropriate.

Key takeaway: A continuous annotation model specifies how conditional per-variant effect variance changes with score. The slope is measured per score unit, the intercept refers to score zero, and predictions must be nonnegative over the modeled range. Converting these predictions into aggregate genetic variance requires assumptions about genotype scaling and the joint distribution of effects.

Check Your Understanding

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

  1. For \(\operatorname{Var}(\beta_j\mid c_j)=0.002+0.0003c_j\), what is the predicted variance at \(c_j=4\), and what does the slope mean in context?
  2. Why does a linear model for \(\operatorname{Var}(\beta_j\mid c_j)\) not automatically imply a linear model for \(E[\beta_j^2\mid c_j]\)? State the additional assumption that would make them equal.
  3. If an annotation is centered at its sample mean, what score value does the model intercept describe? How does that differ from an uncentered model’s intercept?
  4. Suppose the score ranges from \(-3\) to \(2\). What prediction check should be made before interpreting the fitted linear variance model?
  5. Under what assumptions can the expected genetic variance for standardized genotypes be calculated by summing the per-variant effect variances?
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