Two Tests, Two Association Questions
At one variant, an association test depends on how genotype enters the model. An additive test treats the number of effect alleles as a numerical dosage; a genotype-factor test allows the three genotype groups to have separate effects. These choices do not merely change the calculation. They specify different alternatives to the null hypothesis.
The distinction matters when the genetic effect is not well described by a constant change per additional allele. It also matters when an analyst tries several codings and reports whichever produces the smallest p-value. That minimum is not automatically a valid p-value for a single prespecified test.
Both models adjust for the same covariates, if covariates are included. The additive model imposes a linear relationship between allele dosage and log odds: moving from \(G=0\) to \(G=1\), or from \(G=1\) to \(G=2\), changes the log odds by \(\beta\). The genotype-factor model does not impose that restriction. Its two coefficients describe the log-odds differences for \(G=1\) and \(G=2\), respectively, compared with \(G=0\).
Null Hypotheses and Degrees of Freedom
For the additive test, the null hypothesis is \(H_0:\beta=0\); the two-sided alternative is \(H_a:\beta\ne0\). This is a one-parameter test, so its test statistic is compared with a one-degree-of-freedom reference distribution. Rejecting this null is evidence against no linear per-allele log-odds effect under the additive model. It is not a test of every possible genotype-specific pattern.
For the genotype-factor test, the null hypothesis is \(H_0:\theta_1=\theta_2=0\); the alternative is that at least one of these coefficients is nonzero. This is a joint two-parameter test, with two degrees of freedom. It asks whether either nonreference genotype differs from \(G=0\), without requiring the two differences to follow a per-allele trend.
These tests are nested in a useful sense. The additive model is a restricted form of the genotype-factor model: it requires \(\theta_1=\beta\) and \(\theta_2=2\beta\). Thus the factor model contains the additive patterns but also allows other patterns. A two-degree-of-freedom test may have less power than an additive test when the true pattern is additive, because it spends an additional degree of freedom. It can nevertheless detect patterns that a one-degree-of-freedom trend test describes poorly.
How the Test Statistics Arise
A likelihood-ratio test compares a model under the null with a larger model under the alternative. If \(\ell_0\) and \(\ell_1\) are their maximized log-likelihoods, respectively, the statistic is
Under the null and standard large-sample regularity conditions, \(\Lambda\) is approximately chi-square distributed, with degrees of freedom equal to the number of additional parameters in the larger model. The additive model adds one genotype parameter to the covariate-only null model; the genotype-factor model adds two. This parameter count gives the one and two degrees of freedom, respectively.
For a Wald test, let \(\widehat{\boldsymbol{\theta}}=(\hat{\theta}_1,\hat{\theta}_2)^\mathsf{T}\) be the fitted genotype-factor coefficients and let \(\widehat V\) be their estimated covariance matrix. The joint Wald statistic is
which is compared with a chi-square distribution with two degrees of freedom under \(H_0\). For the additive Wald test, the statistic is \(Z=\hat{\beta}/SE(\hat{\beta})\); \(Z^2\) is compared with a one-degree-of-freedom chi-square distribution. The likelihood-ratio and Wald procedures are alternatives for testing the same stated restrictions, but their numerical results need not be identical in a finite sample.
Worked Example: Comparing Additive and Genotype-Factor Results
Suppose a hypothetical case-control analysis uses the same participants and covariates for both models. The fitted additive coefficient is \(\hat{\beta}=0.30\), with \(SE(\hat{\beta})=0.12\). The fitted factor-model coefficients are \(\hat{\theta}_1=0.10\) and \(\hat{\theta}_2=0.62\); for this illustration, suppose their estimated covariance matrix is diagonal, with standard errors 0.20 and 0.25.
Additive test. The Wald statistic is
Against a chi-square distribution with 1 degree of freedom, this gives \(p\approx0.0124\). The fitted per-allele odds ratio is \(\exp(0.30)\approx1.35\): under the additive model, each additional effect-allele copy is associated with about 1.35 times the odds, conditional on the included covariates. The p-value is evidence against the additive null; it is not the probability that the null is true.
Genotype-factor test. With the stated diagonal covariance matrix, the joint Wald statistic is
For 2 degrees of freedom, the chi-square upper-tail probability is approximately \(0.0408\). This test asks whether the two genotype coefficients are jointly zero; it does not assert a constant per-allele effect. Here the additive test gives the smaller p-value, but the tests still address different alternatives.
When the Genotype Pattern Is Not Additive
A dosage trend can be weak even when genotype-specific effects differ from the reference. For example, a heterozygote effect may be stronger than either homozygote effect, or the two nonreference groups may differ in opposite directions. A genotype-factor test can represent such patterns, while the additive model cannot represent them exactly.
Worked Example: A Nonlinear Pattern
Consider another hypothetical, covariate-adjusted logistic analysis. Suppose the factor-model estimates are \(\hat{\theta}_1=0.60\) and \(\hat{\theta}_2=-0.10\), each with standard error 0.25, and suppose their estimated covariance is zero for this calculation. Suppose the additive fit gives \(\hat{\beta}=0.08\) with \(SE(\hat{\beta})=0.15\).
The factor-model joint Wald statistic is
For 2 degrees of freedom, \(p=\exp(-5.92/2)\approx0.0518\), just above 0.05. The additive statistic is \(Z=0.08/0.15\approx0.533\), so \(Z^2\approx0.284\) and its one-degree-of-freedom p-value is about 0.594. Neither result crosses a 0.05 threshold here. The contrast between the estimates illustrates why the hypotheses are different: the factor coefficients do not follow the additive constraint \(\theta_2=2\theta_1\). It would be incorrect to describe the factor-model result as a significant additive per-allele effect.
Choosing a Coding After Seeing the Results
Additive, dominant, recessive, and genotype-factor tests are not interchangeable labels for one hypothesis. Each coding represents a different model restriction and can produce a different p-value. If the coding is chosen before examining the association results, a p-value can be interpreted as the result of that prespecified test, subject to its model and approximation assumptions.
If several codings are tried and only the smallest p-value is reported, the reported value is selected using the data. Even if each candidate test has a 0.05 type I error rate when considered alone, the chance that at least one rejects under the null can exceed 0.05. The candidate tests are usually correlated because they use the same participants and genotypes, so the exact chance is not generally obtained by treating them as independent. A valid analysis must account for the selection or present the tests as exploratory rather than interpreting the minimum as an ordinary prespecified p-value.
Worked Example: Adjusting for Three Tried Codings
Suppose an analyst examines additive, dominant, and recessive codings at one hypothetical variant and obtains p-values 0.012, 0.031, and 0.44. The smallest is 0.012. If these three tests are treated as one family of possible tests, a Bonferroni-adjusted p-value for the selected minimum is
This adjusted value is below 0.05. Bonferroni adjustment controls the familywise error rate at no more than 0.05 for these three tests under the global null, without requiring the tests to be independent. It may be conservative when tests are correlated. The unadjusted 0.012 is not the appropriate p-value for the claim that the best of the three codings was selected after looking at all three results.
If, purely for comparison, three tests were independent and each had a 0.05 null rejection probability, the chance of at least one rejection would be \(1-(1-0.05)^3=1-0.95^3\approx0.1426\). Actual genotype-coding tests are generally dependent, so this is not their exact error rate; it shows why repeated opportunities to find a small p-value matter.
Diagnostics, Comparisons, and Limitations
Before interpreting either test, check that the genotype counts and fitted model support the approximation being used. A very small count in one genotype group can make factor-model coefficients imprecise; a logistic fit may also encounter separation. Inspect the estimates and standard errors, confirm that the same participants and covariates were used for the comparison, and verify the intended reference genotype and allele coding. If likelihood-ratio and Wald results differ substantially, the difference can be a warning of weak information or a poor large-sample approximation, not a reason to report whichever p-value is smaller.
| Approach | Genotype parameters tested | Degrees of freedom | Main interpretation |
|---|---|---|---|
| Additive test | One dosage coefficient, \(\beta\) | 1 | Evidence for a linear per-allele log-odds effect |
| Genotype-factor test | Two coefficients, \(\theta_1,\theta_2\), tested jointly | 2 | Evidence that at least one nonreference genotype differs from the reference |
| Several codings, then select the minimum | More than one candidate test | Depends on the testing procedure | Requires accounting for the selection among tests |
The additive test is more parsimonious and can be more powerful when a per-allele pattern is plausible. The factor test allows a broader genotype pattern but uses an additional degree of freedom and can be unstable with sparse genotype counts. A prespecified choice based on the scientific model is often clearer than selecting a coding by its observed p-value. If the pattern is uncertain and several tests are reported, describe all of them and account for the set of tests when making confirmatory claims.
The same distinction between a one-parameter dosage restriction and a joint test of genotype-specific effects can be formulated for other outcome models. The reference distribution and details of the statistic depend on the model and estimation method. The hypotheses must still be stated explicitly; the number of degrees of freedom comes from the number of independent genotype restrictions being tested.
Common Mistakes
- Calling both tests “the SNP association test.” The additive null tests one dosage effect; the factor null tests two genotype contrasts jointly. State which null was tested.
- Reading a two-degree-of-freedom rejection as a per-allele effect. The factor test does not require a constant change per allele. Interpret its coefficients or genotype-specific pattern separately.
- Reporting the smallest p-value from several codings as if prespecified. Selection changes the error rate. Prespecify the test or use an adjustment or other valid calibration for the selection procedure.
- Assuming a smaller p-value proves a better biological model. A smaller p-value reflects the test statistic and its reference distribution; it does not by itself establish that the coding is biologically correct.
- Ignoring sparse genotype groups. Weak information can destabilize estimates and undermine chi-square approximations. Examine counts and fit diagnostics before relying on the nominal p-value.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- In the genotype-factor logistic model, what is the null hypothesis and why does the joint test have 2 degrees of freedom?
- What genotype pattern does the additive model impose on \(\theta_1\) and \(\theta_2\) in the factor model?
- A fitted additive coefficient is 0.20 with standard error 0.10. Calculate its Wald statistic \(Z\) and chi-square statistic \(Z^2\).
- Why is a nominal p-value of 0.02 not automatically valid as a confirmatory p-value if it is the smallest of four codings tried?
- Name one reason a two-degree-of-freedom genotype-factor test may be unreliable in a particular data set.
Solutions. (1) The null is \(H_0:\theta_1=\theta_2=0\); two independent genotype coefficients are tested jointly, giving 2 degrees of freedom. (2) The additive restriction is \(\theta_1=\beta\) and \(\theta_2=2\beta\), so the log-odds change per additional effect-allele copy is constant. (3) \(Z=0.20/0.10=2.00\), and \(Z^2=4.00\), compared with a chi-square distribution with 1 degree of freedom. (4) The coding was selected after examining multiple p-values, so the chance of a small result across the set can exceed the single-test error rate; the selection needs to be accounted for. (5) A very small genotype group can lead to imprecise or unstable coefficients and poor large-sample chi-square approximation.