Common Errors in Interpreting Heritability Enrichment

Why Enrichment Is Easy to Misread

Heritability enrichment compresses two quantities into one ratio: the share of modeled SNP heritability attributed to an annotation and the share of SNPs carrying that annotation. That ratio is useful, but it can be misread if its numerator and denominator refer to different quantities, if categories overlap, or if a point estimate is treated as certain. Even a correctly estimated enrichment is a statistical description, not proof that an annotation causes a trait.

The key is to keep separate three objects: an annotation’s contribution to modeled heritability, the proportion of SNPs in that annotation, and the conditional coefficient used to estimate the contribution. When annotations overlap, the last two objects do not turn the annotations into mutually exclusive groups.

Definition: Let $M$ be the number of SNPs in the analysis and $H$ their total modeled SNP heritability. For a binary annotation $c$, let $M_c$ be the number of SNPs carrying it and $H_c$ its modeled heritability component. The heritability proportion is $H_c/H$, the SNP proportion is $M_c/M$, and the enrichment is $$ E_c=\frac{H_c/H}{M_c/M}. $$ Enrichment is a dimensionless ratio. An estimate above 1 means that the annotation’s modeled heritability share exceeds its SNP share; it does not, by itself, establish a causal mechanism.

Match the Numerator and Denominator

The numerator of enrichment is a proportion of modeled heritability; the denominator is the proportion of SNPs in the same annotation. Comparing a heritability proportion directly with a raw SNP count, or dividing a heritability percentage by the SNP count, does not calculate enrichment. Both quantities must refer to the same SNP universe and the same annotation definition.

A useful algebraic form shows what the ratio compares:

E_c=\frac{H_c/M_c}{H/M}.

Thus enrichment compares the annotation’s heritability component per annotated SNP with the overall modeled heritability per SNP. This is an average comparison, not a claim that every SNP in the annotation has the same effect or that every annotated SNP contributes heritability.

Worked Example: Calculating Enrichment From Proportions

Suppose a hypothetical analysis contains $M=1{,}000$ SNPs and estimates total SNP heritability $H=0.40$. An annotation covers $M_c=100$ SNPs and has an estimated component $H_c=0.12$. Assume these are the matched SNP universe and component estimates from the same fitted model.

The SNP proportion is $100/1{,}000=0.10$, and the heritability proportion is $0.12/0.40=0.30$. Therefore,

E_c=\frac{0.30}{0.10}=3.00.

The estimated component accounts for 30% of modeled SNP heritability while the annotation covers 10% of the SNPs, giving threefold enrichment. This describes a relative concentration in the fitted partition; it does not mean that each annotated SNP has three times the causal effect, or that the annotation causes the trait.

Overlapping Annotations Are Not Exclusive Categories

In functional genomic data, a SNP can carry several annotations at once. For example, it might be both near a transcription start site and inside a conserved region. Consequently, annotation counts such as $M_A$ and $M_B$ are usually counts of membership, not counts in disjoint groups. Their sum can exceed $M$.

In a joint partitioned model, an annotation’s heritability component is associated with its coefficient conditional on the other annotations in that model. With binary annotations and coefficients $\tau_c$, the modeled per-SNP variance can be written as

\operatorname{Var}(\beta_j)=\sum_c a_{cj}\tau_c, \qquad H_c=\tau_c\sum_{j=1}^{M}a_{cj}=\tau_c M_c,

where $a_{cj}=1$ if SNP $j$ has annotation $c$, and $0$ otherwise. A SNP with two annotations receives both corresponding terms in the model. The component $H_c$ is the contribution of the $c$ term, not necessarily the total heritability of every SNP that carries $c$. Summing annotation-specific SNP proportions does not give a union proportion when annotations overlap.

Worked Example: Keeping Overlapping Shares Separate

Consider a hypothetical model with $M=100$ SNPs. Annotation A covers 20 SNPs, annotation B covers 30, and 10 SNPs carry both. The fitted model estimates $H=0.40$, with components $H_A=0.08$, $H_B=0.06$, and $0.26$ from the remaining modeled terms. These components sum to the stated total: $0.08+0.06+0.26=0.40$.

For A, the SNP proportion is $20/100=0.20$ and its heritability proportion is $0.08/0.40=0.20$. Its enrichment is $0.20/0.20=1.00$.

For B, the SNP proportion is $30/100=0.30$ and its heritability proportion is $0.06/0.40=0.15$. Its enrichment is $0.15/0.30=0.50$.

The two SNP proportions sum to $0.50$, but the union of A and B contains $20+30-10=40$ SNPs, or 40% of the set. The overlap must be subtracted to calculate that union. The separate enrichment calculations remain valid as annotation-specific model components; they do not turn A and B into disjoint categories or justify adding their SNP proportions as though they were exclusive.

A Conditional Coefficient Is Not an Enrichment

An annotation coefficient and an enrichment answer different questions. In the model above, $\tau_c$ is a per-SNP variance component associated with annotation $c$, conditional on the other included annotations and the model specification. For a binary annotation, multiplying it by $M_c$ gives the corresponding component $H_c$. Enrichment then compares that component’s share of total heritability with the annotation’s share of SNPs.

The coefficient is therefore not the enrichment, nor is it an unadjusted estimate of the average heritability of SNPs in the category. Its meaning depends on the other annotations and on how the model encodes them. If annotations are correlated, a coefficient can be difficult to estimate precisely or can change when annotations or other model choices change. A negative estimated coefficient can also occur from estimation noise or model limitations; it should not be silently changed to zero and interpreted as a positive enrichment.

Worked Example: From Conditional Coefficients to Enrichment

Suppose a hypothetical joint model has 100 SNPs. It includes a baseline annotation with coefficient $\tau_0=0.002$ for every SNP, annotation A on 20 SNPs with $\tau_A=0.003$, and annotation B on 30 SNPs with $\tau_B=0.001$. Ten SNPs carry both A and B. Assume the coefficients are estimated on the same variance scale and all other components are absent.

The baseline component is $100(0.002)=0.20$, the A component is $20(0.003)=0.06$, and the B component is $30(0.001)=0.03$. Thus total modeled heritability is $H=0.20+0.06+0.03=0.29$. This sum is also consistent with the per-SNP model: SNPs in neither annotation have variance $0.002$, those in A only have $0.005$, those in B only have $0.003$, and those in both have $0.006$.

For A, the heritability proportion is $0.06/0.29\approx0.2069$, and the SNP proportion is $20/100=0.20$. Its enrichment is $0.2069/0.20\approx1.034$. For B, the heritability proportion is $0.03/0.29\approx0.1034$, and the SNP proportion is $0.30$, giving enrichment $0.1034/0.30\approx0.3448$.

The coefficients $0.003$ and $0.001$ are not these enrichment estimates. The enrichment calculations use each coefficient’s resulting component, the total modeled heritability, and the relevant SNP proportion. Because the coefficients come from the joint model, their interpretation is conditional on the baseline term and the other annotation.

Uncertainty Applies to the Enrichment Estimate

An enrichment is estimated from data, so its point estimate should not be reported as though it were exact. Its uncertainty depends on the estimated annotation component and total heritability, which are themselves related. In partitioned heritability analyses, block jackknife procedures can estimate uncertainty while accounting for the dependence induced by genomic regions. Uncertainty for the ratio should be obtained from the fitted analysis, not by treating its numerator and denominator as independent without justification.

A confidence interval that includes 1 is compatible with no enrichment relative to the SNP proportion at the corresponding confidence level. A point estimate above 1 does not override that uncertainty. In addition, when many annotations are examined, nominal evidence for one annotation may not remain compelling after accounting for multiple comparisons.

Worked Example: An Enrichment Estimate With Uncertainty

Suppose a hypothetical block-jackknife analysis estimates enrichment as 1.30 with standard error 0.20. For a simple normal approximation, an approximate 95% confidence interval is

1.30\pm1.96(0.20)=1.30\pm0.392=(0.908,\ 1.692).

The point estimate is above 1, but the approximate interval includes 1. The data therefore do not rule out an enrichment of 1 at this approximate 95% confidence level. This does not establish that there is no enrichment; it indicates that the estimate is not precise enough here to distinguish it from 1 using this interval. The interval describes the uncertainty procedure’s repeated-sampling coverage, not a probability that the fixed true enrichment lies in this particular interval.

Enrichment Is Not Evidence of Biological Causation

An annotation may be enriched because its SNPs are associated with larger modeled heritability components under the analysis, conditional on the included annotations and assumptions. That result alone does not show that the annotated biological feature causes the trait, that the annotation is the mechanism through which variants act, or that changing the feature would change the trait. Correlated annotations, imperfect annotation definitions, LD modeling choices, and other sources of model misspecification can affect the estimate.

The appropriate conclusion is bounded: for the specified SNP set, trait data, annotations, and fitted model, the estimated heritability component is concentrated in the annotation relative to its SNP share, with the reported uncertainty. Biological explanations require additional evidence beyond an enrichment ratio.

Common Mistakes

Adding overlapping SNP proportions. The wrong reasoning is that two annotations covering 20% and 30% of SNPs must cover 50% together. The correction is to subtract their overlap when calculating a union; separate annotation proportions are not mutually exclusive shares.
Using mismatched denominators. Dividing a heritability proportion by a raw SNP count, or using a different SNP universe for the two proportions, does not yield the defined enrichment.
Calling a conditional coefficient an enrichment. A coefficient is a model term on a variance scale; enrichment is a ratio of heritability and SNP proportions. Calculate the component and both proportions before interpreting enrichment.
Ignoring uncertainty. A point estimate above 1 is not conclusive evidence of enrichment if its uncertainty includes 1. Report the uncertainty estimate and, where relevant, account for testing many annotations.
Making a causal claim. “The annotation is enriched in this fitted model” is not equivalent to “the biological feature causes the trait.” Keep the conclusion at the level supported by the analysis.

Key takeaway: Interpret enrichment only after matching the heritability component and SNP denominator, recognizing that annotations can overlap, distinguishing conditional coefficients from ratios, and considering uncertainty. Enrichment describes a model-based concentration of heritability; it is not a causal conclusion.

Check Your Understanding

Answer each question using the distinctions made in this tutorial.

An annotation covers 15% of SNPs and has 30% of modeled heritability. What is its enrichment, and what does that value mean?
Two annotations each cover some of the same SNPs. Why can their SNP proportions not automatically be added to obtain the proportion in their union?
In a joint model, what is the difference between an annotation coefficient and the annotation’s enrichment?
A point estimate of enrichment is 1.25, but its confidence interval includes 1. What should be said about the evidence for enrichment?
Why does a statistically supported annotation enrichment not, by itself, show that the annotated biological feature causes the trait?