Tutorials › Statistical Genetics › Heritability Proportions and Enrichment Ratios

Partitioned Genetic Heritability · Tutorial 540 of 989

Heritability Proportions and Enrichment Ratios

Calculate how much SNP heritability falls in an annotation, compare that share with the annotation’s share of variants, and interpret the resulting enrichment carefully.

Advanced 10 min read

What You'll Learn

  • Define annotation-specific heritability and variant proportions on a consistent SNP set and scale
  • Calculate enrichment as the heritability proportion divided by the variant proportion
  • Interpret enrichment as relative heritability per variant, not as a causal effect
  • Distinguish exhaustive disjoint partitions from overlapping annotation summaries
  • Check why disjoint category enrichments have a variant-share-weighted average of one

Why Compare Heritability Share With Variant Share?

A functional annotation may contain a small fraction of the variants in a genetic analysis but account for a larger fraction of its estimated SNP heritability. Reporting only the heritability assigned to that annotation does not show whether its share is large relative to its size. The comparison requires two quantities: the annotation’s share of total SNP heritability and its share of the variants being analyzed.

The ratio of those shares is called enrichment. It provides a compact descriptive measure: an annotation is enriched when its heritability share is greater than its variant share. The calculation is straightforward, but its interpretation depends on using the same variant set and heritability scale throughout. It also matters whether the annotations form disjoint groups or overlap.

Definitions and Calculation

Let the analysis include \(M\) variants, with total SNP heritability \(h_g^2\) on a specified phenotype scale. For variant \(j\), let \(v_j\) denote its contribution to that total under the chosen additive model, so that \(h_g^2=\sum_{j=1}^{M}v_j\). The contribution \(v_j\) must be defined on the same scale as the total. For example, with standardized genotypes it can be expressed using the variance of the per-variant effect; with unstandardized genotype dosage, the genotype variance must also be included. In either case, the contributions being summed must match the heritability convention used in the analysis.

For an annotation \(C\), let \(M_C\) be the number of variants in that annotation. Define its annotation-specific heritability as the sum of the contributions of the variants it contains, \(h_g^2(C)=\sum_{j\in C}v_j\). The annotation’s heritability proportion is \(h_g^2(C)/h_g^2\), while its variant proportion is \(M_C/M\).

Definition: The heritability proportion of annotation \(C\) is \(P_C=h_g^2(C)/h_g^2\). Its variant proportion is \(Q_C=M_C/M\). The enrichment is \(E_C=P_C/Q_C\), provided the total heritability and the annotation’s variant proportion are positive. Equivalently, \(E_C=[h_g^2(C)/M_C]/[h_g^2/M]\): the average heritability contribution per variant in \(C\), relative to the average across all analyzed variants.

An enrichment of 1 means the annotation’s heritability share matches its variant share. An enrichment greater than 1 means it accounts for a larger share of heritability than of variants; an enrichment below 1 means its heritability share is smaller than its variant share. Enrichment is a ratio, not a heritability estimate, a probability, or evidence that annotation membership causes a variant to have a particular effect.

Worked Example: Enrichment in a Disjoint Partition

Suppose a hypothetical analysis includes 1,000 variants, with total SNP heritability \(h_g^2=0.20\). A disjoint annotation A contains 200 variants whose contributions sum to \(0.06\). The remaining 800 variants form group B and contribute \(0.14\). Calculate the proportions and enrichments.

For A, the heritability proportion is \(P_A=0.06/0.20=0.30\), or 30%. Its variant proportion is \(Q_A=200/1{,}000=0.20\), or 20%. Therefore,

$$ E_A=\frac{P_A}{Q_A}=\frac{0.30}{0.20}=1.50. $$

A accounts for 1.5 times its share of variants in heritability. For B, \(P_B=0.14/0.20=0.70\) and \(Q_B=800/1{,}000=0.80\), so \(E_B=0.70/0.80=0.875\). B accounts for 87.5% of its share of variants in heritability. The two heritability proportions sum to 1, as do the two variant proportions, because these groups are disjoint and exhaustive.

Disjoint Groups: A Partition of the Total

Annotations are a disjoint partition when every analyzed variant belongs to exactly one group. If the groups are also exhaustive, each variant and each variant contribution is counted once across the categories. Thus, for categories \(C_1,\ldots,C_K\), their heritability proportions sum to 1 and their variant proportions sum to 1.

This gives a useful check on calculations. For an exhaustive disjoint partition, the variant-share-weighted average of the category enrichments is 1. The result follows directly from the definitions:

$$ \sum_{c=1}^{K}Q_{C_c}E_{C_c} = \sum_{c=1}^{K}Q_{C_c}\frac{P_{C_c}}{Q_{C_c}} = \sum_{c=1}^{K}P_{C_c} = 1. $$

The equality does not say that the unweighted average of enrichments is 1. Categories can have very different numbers of variants, so their variant proportions provide the appropriate weights. Nor does the result apply when the categories leave some variants out or count some variants more than once.

Worked Example: Checking a Three-Group Partition

In a hypothetical analysis, three disjoint and exhaustive groups contain 50%, 30%, and 20% of the variants. Their respective heritability proportions are 40%, 30%, and 30%. Calculate each enrichment and verify the weighted-average check.

The enrichments are \(E_1=0.40/0.50=0.80\), \(E_2=0.30/0.30=1.00\), and \(E_3=0.30/0.20=1.50\). Group 1 has a smaller heritability share than variant share, group 2 matches its share, and group 3 has 1.5 times its share.

Weighting by variant proportions gives \(0.50(0.80)+0.30(1.00)+0.20(1.50)=0.40+0.30+0.30=1.00\). This agrees with the theorem. The unweighted mean, \((0.80+1.00+1.50)/3=1.10\), is not the relevant check.

Overlapping Annotations: Summaries Need Not Add to One

Annotations overlap when a variant can belong to more than one category. For each category, calculate its heritability proportion by summing the contributions of the variants in that category, and calculate its variant proportion by counting those variants. Then calculate enrichment using the same ratio as for a disjoint annotation.

A variant in two annotations contributes to each annotation’s separate summary. Therefore, adding the category-specific heritability proportions may count that variant’s heritability more than once; the same is true of variant proportions. The summaries for overlapping annotations are not necessarily parts of one total, and their proportions need not sum to 1. Each enrichment still compares one annotation’s heritability share with its own variant share, relative to the full analysis set.

Worked Example: Two Overlapping Annotations

Consider a hypothetical analysis of 100 variants with total SNP heritability \(h_g^2=0.10\). Annotation A contains 30 variants and has total contribution \(0.04\). Annotation B contains 40 variants and has total contribution \(0.03\). Ten variants are in both annotations; their contributions sum to \(0.01\). Calculate both enrichments and explain why the proportions do not form a partition.

For A, \(P_A=0.04/0.10=0.40\) and \(Q_A=30/100=0.30\). Thus \(E_A=0.40/0.30=1.33\): A’s heritability share is about 1.33 times its variant share. For B, \(P_B=0.03/0.10=0.30\) and \(Q_B=40/100=0.40\), so \(E_B=0.30/0.40=0.75\).

The two heritability proportions sum to \(0.40+0.30=0.70\), and the two variant proportions sum to \(0.30+0.40=0.70\). Neither sum is required to be 1. The overlap contributes \(0.01/0.10=0.10\) to each heritability proportion and accounts for 10 of the 100 variants in each category count. The union of A and B contains 60 variants and has contribution \(0.04+0.03-0.01=0.06\); the subtraction removes the overlap counted twice. The remaining 40 variants contribute \(0.10-0.06=0.04\).

A Short Derivation: Why Enrichment Is a Relative Per-Variant Quantity

The alternative form of the enrichment definition makes its scale easier to interpret. Starting with the ratio of proportions and substituting \(P_C=h_g^2(C)/h_g^2\) and \(Q_C=M_C/M\) gives:

$$ E_C = \frac{h_g^2(C)/h_g^2}{M_C/M} = \frac{h_g^2(C)/M_C}{h_g^2/M}. $$

The final expression is the mean contribution per variant in the annotation divided by the mean contribution per variant overall. Thus an enrichment of 2 means that the annotation’s average contribution per variant is twice the overall average under the stated definitions. It does not mean each variant in that annotation has twice the contribution, and it does not identify a mechanism. Individual contributions can vary substantially within an annotation.

For overlapping categories this interpretation remains valid one annotation at a time. The overall mean in the denominator is still calculated using the full analysis set, whereas the numerator uses the selected annotation. Since a variant can enter several annotation numerators, those numerators should not be added as though they were disjoint components.

Assumptions and Interpretation in Practice

The arithmetic is meaningful only when the quantities refer to a consistent analysis. Use the same modeled variant set to define \(M\), \(M_C\), and total heritability; use the same phenotype scale for \(h_g^2\) and \(h_g^2(C)\); and use a consistent convention for each variant’s contribution. For instance, do not compare a category’s heritability on a liability scale with a total on an observed binary-trait scale. If the annotation has no variants, its enrichment is undefined. If total heritability is zero, the heritability proportion and enrichment are also undefined.

These summaries describe how estimated heritability is distributed across the variants selected by an annotation. They do not by themselves test whether enrichment differs from 1, quantify uncertainty, or establish that the annotation causes larger genetic effects. Estimates of heritability and annotation membership can both be uncertain or imperfect. An enrichment ratio should therefore be reported with its definitions, the scale, and the analysis set, and interpreted as a descriptive comparison unless uncertainty and a suitable inferential procedure are also provided.

Key takeaway: Enrichment is the annotation’s share of heritability divided by its share of variants. For disjoint exhaustive groups, the category shares form a partition and enrichments average to 1 when weighted by variant share. For overlapping groups, calculate each category’s ratio separately; do not treat the category summaries as additive pieces of the total.

Common Mistakes

  • Calling a large heritability proportion “enrichment” by itself. The correction is to divide the heritability proportion by the annotation’s variant proportion. A large annotation can account for much heritability simply because it contains many variants.
  • Assuming overlapping category proportions must sum to 1. The correction is to recognize that variants in multiple annotations appear in each relevant summary. The proportions are category-specific, not mutually exclusive shares.
  • Interpreting enrichment as a statement about every variant. The ratio compares average contributions per variant; individual contributions may differ, and enrichment does not imply a causal effect of annotation membership.
  • Comparing values on inconsistent scales or variant sets. The correction is to define the total and category-specific quantities using the same phenotype scale, modeled variants, and contribution convention.
  • Using an unweighted mean as the partition check. For disjoint exhaustive groups, the check is \(\sum_c Q_{C_c}E_{C_c}=1\), weighted by each group’s variant proportion.

Check Your Understanding

Answer these questions using the definitions and distinctions above.

  1. An annotation contains 15% of analyzed variants and 24% of total SNP heritability. What is its enrichment, and what does that value mean?
  2. Why do heritability proportions for overlapping annotations not necessarily sum to 1?
  3. For disjoint exhaustive groups, which weighted average of category enrichments equals 1, and what are the weights?
  4. A category’s heritability share is 0.30 and its variant share is 0.60. Calculate its enrichment and interpret it in context.
  5. What must be kept consistent when comparing an annotation-specific heritability proportion with the total?
Clinical Biostats