From Per-Variant Effects to Annotation-Specific Heritability

From Variant Effects to Genetic Variance

A variant’s estimated effect describes how the phenotype changes, on average, with a one-allele increase in its genotype dosage. Its effect alone does not tell us how much phenotypic variation is associated with that variant. That also depends on how much the genotype varies in the population being studied. A common allele with a modest effect can contribute more variance than a rare allele with the same effect.

Partitioning SNP heritability begins by combining these two ingredients: the per-variant effect and the variance of the corresponding genotype. We can then sum the contributions of variants assigned to a category, such as a set of variants carrying a particular annotation. This tutorial focuses on that basic conversion and on what the resulting sums do—and do not—mean.

Definition: Let $G_j$ be the genotype dosage at variant $j$, coded as the number of effect alleles, and let $\beta_j$ be its additive effect on a quantitative phenotype. The per-variant additive genetic-variance contribution is $$ v_j=\beta_j^2\operatorname{Var}(G_j). $$ For a category $C$ of variants, its summed contribution is $V_C=\sum_{j\in C}v_j$. If the categories are disjoint and exhaustive and the modeled genotypes are uncorrelated, the total modeled additive genetic variance is $V_{\mathrm{SNP}}=\sum_j v_j$, and its SNP heritability is $h^2_{\mathrm{SNP}}=V_{\mathrm{SNP}}/\sigma_Y^2$, where $\sigma_Y^2$ is the total phenotypic variance on the scale being analyzed.

The units make the calculation interpretable. If $G_j$ is an allele count, its variance is in squared allele-count units. The effect $\beta_j$ is in phenotype units per allele, so $v_j$ is in squared phenotype units. Dividing a genetic-variance contribution by total phenotypic variance gives a dimensionless proportion.

When a variant has effect-allele frequency $f_j$, and its genotype follows Hardy–Weinberg proportions, $\operatorname{Var}(G_j)=2f_j(1-f_j)$. This shortcut uses the Hardy–Weinberg assumption for the genotype distribution in the target population. If that assumption is unsuitable, use the genotype variance appropriate to the population or sample and be clear about how it was obtained.

What a Category Contribution Measures

A category contribution $V_C$ is measured in phenotype-variance units. Its corresponding contribution to phenotypic variance, $V_C/\sigma_Y^2$, describes the proportion of total phenotypic variance represented by that category under the specified model and assumptions. It is not the same quantity as the category’s share of total SNP heritability, $V_C/V_{\mathrm{SNP}}$. The first uses total phenotypic variance as its denominator; the second uses modeled SNP genetic variance.

These quantities answer different questions. A category can account for a small fraction of total phenotypic variance while accounting for a large fraction of the modeled SNP variance. This happens, for example, when total SNP heritability itself is small. A category’s contribution is also not a causal effect of the annotation: it summarizes variance associated with the variants included under the analysis’s definition and assumptions.

Worked Example: One Variant’s Contribution

Suppose a hypothetical quantitative phenotype has total variance $\sigma_Y^2=25$ units squared. At one variant, the effect-allele frequency is $f=0.30$, and the estimated additive effect is $\beta=0.50$ phenotype units per allele. Assume Hardy–Weinberg genotype proportions and an additive effect on the stated phenotype scale.

The genotype variance is $$ \operatorname{Var}(G)=2f(1-f)=2(0.30)(0.70)=0.42. $$

Therefore, the variant’s additive genetic-variance contribution is $$ v=\beta^2\operatorname{Var}(G)=(0.50)^2(0.42)=0.25(0.42)=0.105 $$ phenotype units squared. Relative to total phenotypic variance, the contribution is $$ \frac{v}{\sigma_Y^2}=\frac{0.105}{25}=0.00420. $$

Under this model, the variant contributes 0.00420, or 0.420%, of the phenotype variance. This is a variance contribution under the stated assumptions, not evidence that changing the allele would cause a change of 0.420% in an individual’s phenotype.

Summing Contributions Across Categories

To define a partition, first decide which variants belong to each category. In the simplest case, every modeled variant belongs to exactly one category. If genotypes are uncorrelated, the variance contributions add: the variance of the sum of the genetic effects is the sum of the individual variances. When the categories cover all modeled variants, the category sums therefore recover total modeled SNP genetic variance.

The uncorrelated-genotype condition matters. In general, if the additive genetic value is $A=\sum_j\beta_j(G_j-E[G_j])$, its variance includes both per-variant terms and cross-variant covariance terms:

\operatorname{Var}(A) = \sum_j \beta_j^2\operatorname{Var}(G_j) + 2\sum_{j<k}\beta_j\beta_k\operatorname{Cov}(G_j,G_k).

The first sum contains the $v_j$ terms. The second sum is zero only when the relevant genotype covariances are zero, or when the cross-terms cancel under the particular effects and covariance structure. Thus, summing $v_j$ alone is not automatically an exact decomposition of genetic variance when genotypes are correlated. With correlated genotypes, category attribution requires a model or allocation convention that handles those shared covariance terms; a simple per-variant sum should not be presented as an exact partition without its assumptions.

Worked Example: Two Disjoint Categories

Consider a hypothetical phenotype with total variance 100 units squared and three modeled variants. Assume all three genotypes have variance 0.50, the additive effects are measured in phenotype units per allele, and the genotypes are mutually uncorrelated. Category A contains variants 1 and 3; category B contains variant 2. The categories are disjoint and together include all three variants.

For variant 1, let $\beta_1=0.40$. Its contribution is $$ v_1=(0.40)^2(0.50)=0.16(0.50)=0.0800. $$ For variant 2, let $\beta_2=0.20$. Its contribution is $$ v_2=(0.20)^2(0.50)=0.04(0.50)=0.0200. $$ For variant 3, let $\beta_3=0.60$. Its contribution is $$ v_3=(0.60)^2(0.50)=0.36(0.50)=0.180. $$ Each contribution is in phenotype units squared.

Summing within categories gives $$ V_A=v_1+v_3=0.0800+0.180=0.260, \qquad V_B=v_2=0.0200. $$ The total modeled genetic variance is $$ V_{\mathrm{SNP}}=V_A+V_B=0.260+0.0200=0.280. $$

On the phenotype-variance scale, category A contributes $0.260/100=0.00260$, and category B contributes $0.0200/100=0.000200$. Total SNP heritability is $$ h^2_{\mathrm{SNP}}=\frac{0.280}{100}=0.00280. $$ The category contributions sum to the total because the categories partition all three variants and the genotype covariance terms are zero. Category A’s share of total SNP genetic variance is $0.260/0.280=0.9286$, or 92.86%; this is distinct from its 0.260% contribution to total phenotypic variance.

Why the Denominator Changes the Interpretation

A category’s contribution to phenotypic variance and its share of total SNP heritability should not be substituted for one another. The former places a category’s genetic variance over all phenotypic variance; the latter places it over genetic variance attributed to the modeled SNPs. The distinction is especially important when total SNP heritability is modest.

Worked Example: Phenotype Contribution Versus SNP Share

Suppose a hypothetical analysis has total phenotype variance of 40 units squared. Under a model with disjoint, exhaustive categories and uncorrelated genotypes, category C contains two variants with contributions 0.0400 and 0.0800 units squared. All other modeled variants together contribute 0.0300 and 0.0500 units squared.

Category C contributes $$ V_C=0.0400+0.0800=0.120 $$ units squared. The other categories contribute $0.0300+0.0500=0.0800$ units squared, so total modeled SNP genetic variance is $$ V_{\mathrm{SNP}}=0.120+0.0800=0.200 $$ units squared.

Category C’s contribution to total phenotypic variance is $0.120/40=0.00300$, or 0.300%. Total SNP heritability is $0.200/40=0.00500$, or 0.500%. But category C’s share of total SNP variance is $0.120/0.200=0.600$, or 60.0%. These percentages differ because their denominators differ. The 60.0% figure does not mean category C explains 60.0% of all phenotypic variance.

Assumptions and Scope of Interpretation

The calculations above use an additive genetic value: each variant contributes an effect proportional to its dosage, and the modeled genetic value is the sum of those effects. Effects and genotype variances must refer to a compatible population and phenotype scale. For example, using an effect estimated on one phenotype scale with a variance measured on another would not produce a coherent variance proportion.

For the simple category sums to equal total modeled SNP heritability, the categories must be disjoint and exhaustive over the variants being counted, and covariance terms must either be absent or handled explicitly. If categories overlap, a variant may be counted more than once by naïvely summing category totals. If some modeled variants are in no category, the listed categories do not exhaust the total. These are accounting issues, separate from whether the annotation has biological meaning.

Finally, category membership is not itself proof of function or causation. A variance contribution describes how the modeled genetic effects and genotype variation combine for a defined set of variants. It does not establish that the annotation caused the phenotype, that every variant in the category is functional, or that the same contribution will hold in a different population.

Common Mistake

A frequent error is to report a category’s share of total SNP heritability as though it were that category’s fraction of total phenotypic variance. For example, saying “category C explains 60% of the phenotype variance” when $V_C/V_{\mathrm{SNP}}=0.60$ is incorrect. The correction is to name the denominator: $V_C/\sigma_Y^2$ is the category’s phenotype-variance contribution, while $V_C/V_{\mathrm{SNP}}$ is its share of modeled SNP genetic variance. Another error is to sum category contributions as an exact total while ignoring genotype covariance or overlapping categories; check those conditions before making the claim.

Key takeaway: A per-variant contribution is effect squared times genotype variance. Summing those terms within a category gives its variance contribution under the stated model; dividing by total phenotypic variance gives its heritability-scale contribution. Category sums recover total SNP heritability only when the partition and covariance assumptions justify that sum.

Check Your Understanding

Answer each question by identifying the relevant variance quantity and its denominator.

A variant has an additive effect of 0.30 phenotype units per allele and genotype variance 0.40. What is its per-variant genetic-variance contribution, in phenotype units squared?
Under Hardy–Weinberg genotype proportions, what is the genotype variance when the effect-allele frequency is 0.10?
Explain the difference between $V_C/\sigma_Y^2$ and $V_C/V_{\mathrm{SNP}}$.
Why do disjoint, exhaustive categories not by themselves guarantee that their per-variant contributions sum to exact genetic variance?
What can and cannot be concluded when a category has a large share of total modeled SNP variance?