Deriving the Additive Variance Ratio for SNP Heritability

From Genotype Effects to a Variance Ratio

A SNP heritability ratio connects two quantities that must be defined in the same population: the variation in phenotype and the variation in the genetic values represented by a specified set of SNPs. The ratio is not obtained by adding effect estimates alone. Each effect is weighted by the variation in its genotype, and covariances between genotypes can also contribute.

This tutorial derives that relationship from a linear genotype-effects model. The result makes explicit what “additive variance tagged by a SNP set” means under the model, why linkage disequilibrium matters, and which allele-scaling and population assumptions permit a simpler formula.

Definition: For a defined set $S$ of SNPs, let $X_j$ be the allele count at SNP $j$, centered at its population mean, and let $\beta_j$ be its additive effect per counted allele. The genetic value represented by the set is $G_S=\sum_{j\in S}X_j\beta_j$. Its additive variance is $V_{A,S}=\operatorname{Var}(G_S)$. If $Y$ is the phenotype in the target population, the SNP-set additive variance ratio is $h^2_S=V_{A,S}/V_P$, where $V_P=\operatorname{Var}(Y)$.

The set $S$ is part of the estimand: changing which SNPs are included can change $V_{A,S}$ and therefore the ratio. The ratio describes the fraction of phenotypic variance represented by the additive genetic value from that set under the stated model. It does not, by itself, say that the included SNPs are causal or that the same ratio applies in another population.

The Linear Model and Its Assumptions

For the derivation, consider individuals drawn from one target population. Let $X_j$ denote the centered allele count for SNP $j$, and write the phenotype as

Y=\mu+\sum_{j\in S}X_j\beta_j+E =\mu+G_S+E.

Here $\mu$ is the population mean, $\beta_j$ is measured in phenotype units per additional counted allele, and $E$ collects the remaining deviation from the mean, including components not represented by this SNP-set genetic value. The basic variance decomposition requires $\operatorname{Cov}(G_S,E)=0$. This is an assumption about the model in the target population, not an automatic consequence of writing down the equation.

The allele-scaling convention used here is the uncentered allele count, which takes values $0,1,2$ for the number of copies of a specified effect allele; $X_j$ is this count minus its population mean. Under random mating and Hardy–Weinberg equilibrium, if the effect-allele frequency is $p_j$, then the uncentered count has mean $2p_j$ and variance $2p_j(1-p_j)$. Thus the centered count has the same variance. Hardy–Weinberg equilibrium is needed for this particular variance formula; without it, use the genotype-count variance in the population directly.

The model also assumes that the effects are additive on the chosen phenotype scale and that the same set of effects and genotype distribution describe the population being analyzed. It omits dominance and interaction terms. If covariates or population strata are relevant, the estimand must be defined consistently—for example, using phenotype and genotype variation conditional on the same covariates or within a specified population. An unmodeled association between $G_S$ and $E$ introduces a covariance term and prevents the simple decomposition below.

Deriving the Additive Variance

Let $\boldsymbol{X}$ be the vector of centered allele counts in $S$, let $\boldsymbol{\beta}$ be the corresponding vector of effects, and let $\boldsymbol{\Sigma}_X=\operatorname{Cov}(\boldsymbol{X})$ be the genotype covariance matrix in the target population. Since $G_S=\boldsymbol{X}^{\mathsf T}\boldsymbol{\beta}$, the variance of a linear combination gives

V_{A,S} =\operatorname{Var}(\boldsymbol{X}^{\mathsf T}\boldsymbol{\beta}) =\boldsymbol{\beta}^{\mathsf T}\boldsymbol{\Sigma}_X\boldsymbol{\beta}.

This expression includes both each SNP’s individual genotype variance and the covariance between every pair of SNPs. Expanding it shows the contributions explicitly:

V_{A,S} =\sum_{j\in S}\beta_j^2\operatorname{Var}(X_j) +2\sum_{\substack{j<k\\j,k\in S}} \beta_j\beta_k\operatorname{Cov}(X_j,X_k).

The factor of two appears because the matrix product contains both the $j,k$ and $k,j$ terms, which are equal for a covariance matrix. The pairwise contribution can be positive or negative: its sign depends on both the sign of the effects and the sign of the genotype covariance. The total variance cannot be negative, because it is the variance of $G_S$.

Theorem: Under the linear model $Y=\mu+G_S+E$ with $\operatorname{Cov}(G_S,E)=0$, phenotypic variance decomposes as $V_P=V_{A,S}+\operatorname{Var}(E)$. Therefore, $$ h^2_S=\frac{\boldsymbol{\beta}^{\mathsf T}\boldsymbol{\Sigma}_X\boldsymbol{\beta}}{V_P}. $$ If the SNPs are in linkage equilibrium, their genotype covariances are zero, so $V_{A,S}=\sum_{j\in S}\beta_j^2\operatorname{Var}(X_j)$. If, in addition, each SNP is in Hardy–Weinberg equilibrium under allele-count scaling, then $V_{A,S}=\sum_{j\in S}2p_j(1-p_j)\beta_j^2$.

The variance decomposition follows by expanding $\operatorname{Var}(Y)$. Since $\mu$ is constant, it contributes no variance. Hence

\begin{aligned} V_P &=\operatorname{Var}(G_S+E)\\ &=\operatorname{Var}(G_S)+\operatorname{Var}(E) +2\operatorname{Cov}(G_S,E)\\ &=V_{A,S}+\operatorname{Var}(E), \end{aligned}

where the last equality uses the assumed zero covariance. Dividing $V_{A,S}$ by $V_P$ gives the ratio. If the SNPs are uncorrelated, all off-diagonal entries of $\boldsymbol{\Sigma}_X$ are zero, leaving only the diagonal terms. Hardy–Weinberg equilibrium then supplies $\operatorname{Var}(X_j)=2p_j(1-p_j)$. This establishes the simplified formula and shows exactly which assumptions it uses.

Worked Calculations

Worked Example: One SNP Under Allele-Count Scaling

Suppose a hypothetical population has one SNP in the defined set. The effect-allele frequency is $p=0.30$, the phenotype effect is $\beta=0.40$ phenotype units per additional effect allele, and the remaining component has variance $4.00$ phenotype-units squared. Assume Hardy–Weinberg equilibrium and zero covariance between the SNP genetic value and the remainder.

The centered allele count has variance $2p(1-p)$, so the SNP-set additive variance is

V_{A,S}=2p(1-p)\beta^2 =2(0.30)(0.70)(0.40)^2 =0.0672.

The units are phenotype-units squared: the allele-count variance is dimensionless, and $\beta^2$ has squared phenotype units. By the assumed decomposition, the phenotypic variance is $V_P=0.0672+4.00=4.0672$ phenotype-units squared. Therefore,

h^2_S=\frac{0.0672}{4.0672}=0.0165.

In this hypothetical population, the additive variance represented by this SNP is about 1.65% of the phenotypic variance under the model. This is a population variance ratio, not the percentage of an individual’s phenotype caused by the SNP.

Worked Example: Two SNPs With Linkage Disequilibrium

Consider a hypothetical two-SNP set with centered allele-count variances $\operatorname{Var}(X_1)=0.375$ and $\operatorname{Var}(X_2)=0.480$, and covariance $\operatorname{Cov}(X_1,X_2)=0.120$. Let the per-allele effects be $\beta_1=0.50$ and $\beta_2=-0.20$ phenotype units. Assume the remainder is uncorrelated with the set’s genetic value and has variance $1.50$ phenotype-units squared.

The covariance-matrix formula gives

\begin{aligned} V_{A,S} &=\beta_1^2\operatorname{Var}(X_1) +\beta_2^2\operatorname{Var}(X_2) +2\beta_1\beta_2\operatorname{Cov}(X_1,X_2)\\ &=(0.50)^2(0.375)+(-0.20)^2(0.480) +2(0.50)(-0.20)(0.120)\\ &=0.09375+0.01920-0.02400\\ &=0.08895. \end{aligned}

The LD covariance term is negative because the effects have opposite signs while the genotype covariance is positive. The phenotypic variance is $V_P=0.08895+1.50=1.58895$ phenotype-units squared, and the ratio is

h^2_S=\frac{0.08895}{1.58895}=0.0560.

Thus the two-SNP additive variance is about 5.60% of the phenotypic variance in this hypothetical population. Dropping the covariance term would give $0.11295$, not $0.08895$, and would therefore produce a different ratio. The supplied covariance matrix is the relevant population quantity; the simplified independent-SNP formula is not justified here.

Worked Example: Changing the Effect-Allele Label

For a hypothetical SNP, suppose the counted allele has frequency $p=0.20$, its per-allele effect is $\beta=0.30$ phenotype units, and the phenotypic variance is $2.00$ phenotype-units squared. Under Hardy–Weinberg equilibrium, the additive variance is

V_{A,S}=2(0.20)(0.80)(0.30)^2=0.0288, \qquad h^2_S=\frac{0.0288}{2.00}=0.0144.

Now count the other allele instead. Its frequency is $1-p=0.80$. The new centered count is the negative of the original centered count, so its effect must also change sign: $\beta'=-0.30$. Then

$$ 2(0.80)(0.20)(-0.30)^2=0.0288. $$

The variance ratio remains $0.0144$. Relabeling the counted allele changes the sign of the centered genotype and the effect together, but not their product or its variance. This invariance depends on changing both the allele frequency and the effect consistently.

Interpretation and Scope

The ratio is defined relative to a phenotype scale, a population, and a SNP set. Its numerator depends on the genotype covariance matrix and the additive effects in that population; its denominator is the total phenotypic variance in the same population and on the same scale. Frequencies, LD patterns, and phenotypic variation can differ across populations, so a ratio derived in one population need not transfer unchanged to another.

The calculation is also distinct from estimating the ratio. The equations define the target under a model; estimating its components from data requires an appropriate statistical procedure and its own assumptions. Variance-component methods such as REML provide one route for estimating genetic and residual variance components, but the derivation here does not prescribe an estimator.

Finally, the term “tagged” should not be read as “caused.” A SNP set may represent additive genetic variation through correlations among its genotypes, including LD. The variance ratio describes the model-based variation represented by the set, not necessarily a sum of independently causal effects at the listed SNPs.

Common Mistake

Wrong reasoning: “The SNP-set variance is the sum of the squared effect estimates, so LD can be ignored.” Squared effects do not account for how often genotypes vary, and correlated genotypes contribute covariance terms.
Correction: Use $V_{A,S}=\boldsymbol{\beta}^{\mathsf T}\boldsymbol{\Sigma}_X\boldsymbol{\beta}$. The simpler sum $\sum_j2p_j(1-p_j)\beta_j^2$ requires allele-count scaling, Hardy–Weinberg equilibrium at each SNP, and zero covariance between SNP genotypes in the target population.

Key takeaway: The additive variance represented by a defined SNP set is the variance of its linear genetic value. Its general form is $\boldsymbol{\beta}^{\mathsf T}\boldsymbol{\Sigma}_X\boldsymbol{\beta}$, and dividing by phenotypic variance gives the SNP-set ratio when the numerator and denominator refer to the same population and phenotype scale.

Check Your Understanding

Use the model assumptions and variance formulas above to answer these questions.

Why does the general additive-variance formula include covariance terms between SNP genotypes?
Under allele-count scaling and Hardy–Weinberg equilibrium, what is the genotype variance at a SNP with effect-allele frequency $p$?
What additional assumption is needed to write $V_P=V_{A,S}+\operatorname{Var}(E)$?
If two SNPs are in positive LD and their effects have opposite signs, what is the sign of their covariance contribution to additive variance?
Why does changing the counted allele leave the variance ratio unchanged when the effect and frequency are recoded consistently?