What a SNP Coefficient Measures
In an additive quantitative-trait GWAS, a SNP coefficient describes how the model’s expected phenotype changes with one additional copy of the designated effect allele, holding the included covariates fixed. The phrase per effect allele is essential: the coefficient refers to a one-unit increase in allele dosage, not to a comparison between any two arbitrary genotype groups.
Let \(Y_i\) be the quantitative phenotype for individual \(i\), \(G_i\) the dosage of the effect allele, and \(C_i\) the included covariates. A dosage is typically between 0 and 2; with imputation it can be fractional. The additive linear model is
Here \(\beta_G\) is the SNP coefficient, \(\beta_0\) is the intercept, \(\gamma\) is the vector of covariate coefficients, and \(\varepsilon_i\) is the residual. For the usual conditional interpretation, the model assumes that the conditional mean of \(Y_i\) is linear in dosage and the included covariates, so \(E(\varepsilon_i\mid G_i,C_i)=0\). The additive specification says that each additional effect-allele copy has the same modeled change in expected phenotype. Conventional standard errors and tests also require appropriate assumptions about the residual variation and sampling or study design; the coefficient’s units, however, come from the variables in the model.
This is a model-based association, not by itself evidence that changing the genotype would cause the phenotype to change. Covariate adjustment defines the comparison represented by the coefficient; it does not automatically remove every source of confounding or guarantee that the model is correctly specified.
Reading the Coefficient in Original Units
Suppose a phenotype is measured in milligrams per deciliter and the estimated dosage coefficient is \(0.8\). The fitted conditional mean is higher by \(0.8\) milligrams per deciliter for each additional effect-allele copy. Under the additive model, the fitted difference between dosage 0 and dosage 2 is \(2(0.8)=1.6\) milligrams per deciliter, at the same covariate values. This is an expected difference, not a claim that every individual with dosage 2 has a phenotype exactly 1.6 units higher.
Worked Example: Interpreting a Coefficient in Physical Units
Consider hypothetical data in which the outcome is fasting glucose in milligrams per deciliter. A fitted additive model, adjusted for age and sex, estimates \(\hat{\beta}_G=1.25\) milligrams per deciliter per effect-allele copy. Interpret the estimate for dosage increases of one and two copies.
One-copy increase. At the same modeled age and sex, the fitted mean difference for a one-unit dosage increase is
Two-copy increase. The fitted mean difference between dosage 0 and dosage 2 is
Thus, the estimated conditional mean is 1.25 mg/dL higher per additional effect-allele copy, or 2.50 mg/dL higher at dosage 2 than at dosage 0. Both comparisons follow from the additive model. They do not describe an individual-level deterministic change.
The same interpretation applies if a dosage is fractional. For example, a change from 0.4 to 1.4 is a one-unit increase and corresponds to the same modeled mean difference as a change from 0 to 1. Dosages are estimates of allele count, so fractional values do not indicate that an individual carries a fractional physical allele.
Interpreting Effects After Phenotype Standardization
A quantitative phenotype may be standardized by subtracting a mean and dividing by a standard deviation. If \(Y_i^*=(Y_i-\bar{x})/s_Y\), where \(\bar{x}\) and \(s_Y\) are the sample mean and sample standard deviation of the phenotype, then \(Y_i^*\) is measured in sample standard-deviation units. Fitting the same model to \(Y_i^*\) changes the scale of the outcome and therefore the scale of its regression coefficients.
With the same individuals, dosage, covariates, and linear model, standardizing only the phenotype divides the SNP coefficient by \(s_Y\). This follows directly: subtracting a constant from every outcome changes the intercept, while dividing every outcome by \(s_Y\) divides every fitted coefficient by \(s_Y\).
The standardized coefficient is interpreted as the expected difference in phenotype standard deviations per additional effect-allele copy. It is not a percentage change, and it is not a standard deviation of genotype. If the phenotype was standardized using a different reference mean and standard deviation, the coefficient uses that reference scale; comparisons across analyses require attention to how standardization was done.
Worked Example: Converting to Standard-Deviation Units
Suppose a hypothetical study measures a lipid trait in milligrams per deciliter. Its sample standard deviation is \(s_Y=24\) mg/dL, and the additive-model estimate is \(\hat{\beta}_G=0.72\) mg/dL per effect-allele copy. Find and interpret the coefficient after standardizing the phenotype by this sample standard deviation.
The conversion is
The units of mg/dL cancel. In this model, one additional effect-allele copy is associated with an estimated \(0.0300\)-sample-standard-deviation higher mean lipid trait, conditional on the included covariates. For a two-copy dosage difference, the modeled difference is \(2(0.0300)=0.0600\) standard deviations. The original-scale estimate remains \(0.72\) mg/dL per copy; standardization changes the reporting scale, not the underlying fitted comparison.
This conversion assumes the outcome alone was rescaled and that the model, observations, and covariates are otherwise unchanged. If genotype dosage is also standardized, the coefficient instead describes a change per standard deviation of genotype, not per effect-allele copy. Always check which variables were transformed before attaching units to a reported coefficient.
Effect Size Is Not Statistical Significance
The coefficient estimates the direction and magnitude of the modeled association. Its standard error \(SE\) describes sampling uncertainty under the model and analysis assumptions. In a large-sample Wald test of \(H_0:\beta_G=0\), the test statistic is commonly calculated as \(\hat{\beta}_G/SE\), and the p-value measures how incompatible the observed test statistic is with the null model. Neither the standard error nor the p-value is the effect size.
A small p-value can accompany a small estimated effect when the estimate is precise, and a comparatively large estimate can have weak statistical evidence when it is imprecise. Sample size, allele frequency, residual variation, and study design can affect precision. Therefore, statistical significance alone does not establish that an effect is large, clinically important, or practically consequential.
Worked Example: A Smaller Effect Can Be More Significant
Compare two hypothetical, independent large-sample analyses of quantitative traits. In the first, \(\hat{\beta}_G=0.12\) trait units per copy and \(SE=0.04\) trait units per copy. In the second, \(\hat{\beta}_G=0.04\) trait units per copy and \(SE=0.01\) trait units per copy. Use the two-sided normal-approximation Wald test for \(H_0:\beta_G=0\).
For the first analysis,
giving a two-sided normal-approximation p-value of about \(0.00270\). The estimated effect is \(0.12\) trait units per copy.
For the second analysis,
giving a two-sided normal-approximation p-value of about \(0.0000633\). Its estimated effect is only \(0.04\) trait units per copy, but it has the smaller p-value because its standard error is smaller relative to its estimate.
These calculations assume that the stated standard errors are appropriate and that the large-sample normal approximation is reasonable. The second result provides stronger evidence against a zero coefficient by this test, but its estimated effect is smaller. The traits’ units and scales also matter: coefficients from different phenotypes are not automatically comparable just because both have numerical values.
Sign, Allele Orientation, and Comparability
The sign of \(\hat{\beta}_G\) depends on which allele is counted. If the dosage is recoded from effect-allele count \(G_i\) to the other allele count \(2-G_i\), the fitted association has the opposite sign, while the underlying dosage contrast is the same with its direction reversed. A positive coefficient therefore means “higher phenotype per additional counted allele,” not that the allele is intrinsically beneficial.
Before comparing reported estimates, check the effect allele, phenotype units, phenotype standardization, covariate model, and whether the estimate is per allele copy or per another genotype scale. Allele harmonization is necessary for comparing directions across results. A coefficient in milligrams per deciliter cannot be compared numerically with a coefficient in standardized units without a valid scale conversion.
Assumptions, Diagnostics, and Common Pitfalls
The per-copy interpretation relies on the model’s additive linear mean: the expected phenotype must change linearly with dosage over the values being compared, conditional on covariates. If the pattern of genotype-specific means is not approximately linear in dosage, one coefficient may be an inadequate summary. Model residual behavior and the validity of standard errors matter for inferential claims, even though they do not change the basic units of the coefficient.
Useful checks include confirming allele and phenotype coding, examining whether the fitted mean pattern across observed dosages is plausibly linear, and checking that the reported units match the data transformation. Strongly unequal residual spread, influential observations, or dependence not handled by the analysis can make conventional standard errors and p-values unreliable. These are reasons to investigate the model and inference, not reasons to reinterpret a coefficient as a different quantity.
- Pitfall: treating the coefficient as a genotype-group label. A coefficient of 0.5 is not automatically the difference between heterozygotes and homozygotes. It is the model’s change for one dosage unit; the comparison depends on the dosages being contrasted.
- Pitfall: calling a standardized coefficient a percent change. A coefficient of 0.03 means 0.03 phenotype standard deviations per copy, not a 3% increase.
- Pitfall: equating a small p-value with a large or important effect. A p-value depends on both the estimate and its uncertainty. Report the effect and its units alongside inferential results.
- Pitfall: interpreting the sign without identifying the effect allele. Reversing the counted allele reverses the sign. State the allele and direction explicitly.
The additive coefficient is also a summary constrained by the chosen model. It does not show that each genotype group differs by exactly the same amount in the population, nor does it prove a causal mechanism. Other genotype parameterizations may represent different patterns, but the per-copy interpretation belongs specifically to the additive dosage model.
Check Your Understanding
Answer each question, then compare with the worked solutions below.
- A hypothetical additive model estimates \(0.6\) millimeters of trait difference per effect-allele copy. What modeled difference corresponds to a dosage contrast from 0 to 2, with covariates held fixed?
- A phenotype has sample standard deviation \(15\) units and an estimated coefficient of \(-0.45\) units per copy. What is the coefficient when the phenotype is standardized, and what does its sign mean?
- Two results have estimates \(0.20\) with \(SE=0.10\), and \(0.08\) with \(SE=0.02\). Calculate their Wald statistics and explain why the smaller estimate can have stronger evidence against a zero coefficient.
- What happens to the coefficient’s sign if allele dosage is recoded from \(G_i\) to \(2-G_i\)? What should be checked when comparing the two reports?
Solutions. (1) The dosage difference is \(2-0=2\), so the fitted phenotype difference is \(0.6(2)=1.2\) mm. (2) The standardized coefficient is \(-0.45/15=-0.0300\) standard deviations per copy. The negative sign means the modeled mean is lower with each additional counted effect-allele copy, conditional on the covariates. (3) The statistics are \(0.20/0.10=2.00\) and \(0.08/0.02=4.00\). The second estimate is smaller but is four standard errors from zero rather than two, so its Wald test gives stronger evidence against the null under the stated approximation. (4) The sign reverses because the counted dosage direction reverses. Check which allele is counted, as well as the phenotype scale, units, and model, before comparing estimates.