Worked Binary-Trait Association Scan

From a Binary Trait to a SNP Association

In a case-control genome-wide association scan, each participant is classified as a case or a control, and each SNP is tested for association with that binary outcome. The practical challenge is to connect the result to a clear genetic contrast: which allele is counted, how genotype is represented, what covariates are adjusted for, and what an estimated effect means. A fitted association is evidence of a statistical relationship under the specified model; by itself, it does not establish that the SNP causes the trait.

We use $Y_i=1$ for a case and $Y_i=0$ for a control. For a biallelic SNP, let $G_i$ be the number of copies of a prespecified effect allele for participant $i$. Thus $G_i$ is 0, 1, or 2. Let $Z_i$ denote the vector of covariates included in the model, such as age, sex, or ancestry principal components. Genotype quality and allele labels must be checked before these values are analyzed: a reversed allele label reverses the sign of the coefficient and changes which allele the reported odds ratio describes.

Model: An additive single-SNP logistic regression model is $$ \operatorname{logit}\{P(Y_i=1\mid G_i,Z_i)\} =\log\left(\frac{P(Y_i=1\mid G_i,Z_i)}{1-P(Y_i=1\mid G_i,Z_i)}\right) =\beta_0+\beta_G G_i+Z_i^\mathsf{T}\gamma. $$ Here $\beta_G$ is the change in log odds of being a case for each additional copy of the effect allele, holding the included covariates fixed. The model assumes independent observations, a correctly specified logit-linear relationship for the included predictors, and a finite, identifiable coefficient estimate. It also requires valid phenotype and genotype data and enough information to estimate the effects; sparse cells or separation can make ordinary logistic estimates unstable or nonexistent.

The additive model treats the genotype dosage as a numerical predictor: the model-implied log-odds difference between $G=0$ and $G=2$, at the same covariate values, is $2\beta_G$. This is a modeling assumption, not a biological fact guaranteed by the data. The association null hypothesis is $H_0:\beta_G=0$; a two-sided alternative is $H_a:\beta_G\ne 0$. Under the model, the per-allele odds ratio is $\exp(\beta_G)$. It compares the odds of case status for two otherwise covariate-matched individuals whose dosages differ by one effect-allele copy.

Worked Genotype Coding and a Crude Comparison

Before fitting the adjusted model, inspect the allele definition, dosage coding, and genotype counts. The following counts are hypothetical and are included to illustrate the calculations, not to describe a real study.

Worked Example: Coding Dosage and Calculating a Crude Odds Ratio

Suppose a hypothetical study has 400 cases and 400 controls. The effect allele is labeled A, and genotype counts are as follows.

Effect-allele dosage	Genotype	Cases	Controls
0	GG	140	180
1	AG	190	180
2	AA	70	40

For example, an AG participant receives $G_i=1$, because that genotype contains one copy of A. To calculate a simple, unadjusted comparison, combine genotypes with at least one A copy and compare them with GG. There are $190+70=260$ carrier cases and $180+40=220$ carrier controls. There are 140 noncarrier cases and 180 noncarrier controls. The crude carrier-versus-noncarrier odds ratio is

\widehat{\mathrm{OR}}_{\mathrm{crude}} =\frac{260/220}{140/180} =\frac{260\times180}{220\times140} =\frac{46800}{30800} \approx 1.519.

In this hypothetical sample, the estimated odds of case status among A-allele carriers are about 1.52 times the odds among noncarriers. This is a dominant-model contrast, not the additive per-allele odds ratio: it combines AG and AA and makes no distinction between them. It is also unadjusted for covariates. In a case-control study, the sampled case-to-control ratio is set by the study design, so this calculation does not estimate population disease probability.

Adjusted Effect, Uncertainty, and Interpretation

For an additive scan, fit the logistic model with $G_i$ as a single dosage predictor and include the prespecified covariates in $Z_i$. The estimate $\hat{\beta}_G$ is on the log-odds scale. Exponentiating it gives the per-copy odds ratio, and exponentiating a confidence interval's endpoints gives an interval on the odds-ratio scale. A Wald interval is appropriate when the estimate is sufficiently regular and its standard error is reliable; very sparse genotype patterns or near-separation can undermine that approximation.

A case-control sampling design selects participants according to case status. Under the usual logistic model and sampling within outcome groups, this changes the intercept relative to the population model but preserves the disease odds-ratio slopes. Thus the SNP odds ratio can be interpreted as an association measure under the stated model, but the fitted intercept and predicted case probabilities are generally not population risks without additional sampling information.

Worked Example: Adjusted Per-Allele Odds Ratio and Confidence Interval

Suppose a hypothetical logistic regression of the 800 participants reports $\hat{\beta}_G=0.230$ and $SE(\hat{\beta}_G)=0.075$, adjusted for age, sex, and three prespecified ancestry principal components. Assume the fit converged, the coefficient is identifiable, and the Wald approximation is adequate. The per-allele odds ratio is

\widehat{\mathrm{OR}}=\exp(\hat{\beta}_G)=\exp(0.230)\approx1.259.

The approximate 95% confidence interval on the log-odds scale is

0.230\pm1.96(0.075) =0.230\pm0.147 =(0.083,\ 0.377).

Exponentiating both endpoints gives

\left(\exp(0.083),\exp(0.377)\right) \approx(1.087,\ 1.458).

The estimated odds of being a case are 1.259 times as large per additional A allele copy, for participants with the same included covariate values. The approximate 95% confidence interval for that adjusted per-copy odds ratio is 1.087 to 1.458. Under repeated sampling and the assumptions of the procedure, intervals constructed this way would cover the model's odds ratio about 95% of the time; this is not a statement that there is a 95% probability that this particular fixed parameter lies in this interval.

For the two-sided Wald test, $z=0.230/0.075=3.067$, giving $p\approx0.002165$. This is evidence against $H_0:\beta_G=0$ at a nominal 0.05 level. It is not, by itself, evidence that the SNP meets a genome-wide significance threshold: a scan tests many variants, so interpretation must account for the prespecified multiple-testing procedure and threshold.

Why the Odds-Ratio Transformation Works

The interpretation follows directly from the logistic model. Hold $Z_i$ fixed and compare two dosages, $G_i=g$ and $G_i=g+1$. Their linear predictors differ by $\beta_G$. Since the odds corresponding to a logit value $\eta$ are $\exp(\eta)$, the ratio of their odds is $\exp(\beta_G)$. The argument also explains why a two-copy contrast has odds ratio $\exp(2\beta_G)$.

Justification: Write the linear predictor at dosage $g$ as $\eta_g=\beta_0+\beta_Gg+Z_i^\mathsf{T}\gamma$. At dosage $g+1$, it is $\eta_{g+1}=\eta_g+\beta_G$. Therefore, $$ \frac{\operatorname{odds}(Y_i=1\mid G_i=g+1,Z_i)} {\operatorname{odds}(Y_i=1\mid G_i=g,Z_i)} =\frac{\exp(\eta_g+\beta_G)}{\exp(\eta_g)} =\exp(\beta_G). $$ For dosages 0 and 2, the ratio is $\exp(2\beta_G)$. This derivation depends on the assumed linear dosage effect on the log-odds scale.

Model Checks and Failure Modes

A numerical result is only useful if the inputs and fit are credible. Check that case and control labels are correct, that the effect allele and dosage coding are consistent across samples, and that genotype missingness or quality problems do not differ materially by outcome group. Inspect genotype counts and the number of participants in each covariate pattern. Very small or empty genotype-by-outcome cells can signal sparse-data instability; perfect or near separation can produce extreme estimates, large standard errors, or convergence warnings. A converged algorithm alone does not prove that the estimate is reliable.

Assess whether the additive dosage assumption is plausible rather than treating it as automatic. Compare observed patterns across the three genotype groups, and consider a prespecified alternative model if there is a scientific reason to expect a non-additive effect. Avoid selecting whichever genotype coding gives the smallest p-value and reporting it as if that choice had been planned: data-driven selection makes the reported evidence too optimistic unless the selection is accounted for.

Calibration checks ask whether predicted case proportions agree with observed proportions across groups of similar predictions. Such checks are descriptive and can have limited power, especially in small groups. Residual or influence diagnostics can also flag observations or covariate patterns that have an unusually large effect on the fit. A poor fit, implausible coefficient, or sensitivity to a few observations calls for investigation, not merely a different interpretation of the same p-value.

Worked Example: Reading a Simple Calibration Check

In a hypothetical validation subset, suppose participants are divided into four groups of 100 by predicted case probability. The average predictions are 0.10, 0.20, 0.40, and 0.70; the observed numbers of cases are 12, 18, 43, and 66. The corresponding observed case fractions are $12/100=0.12$, $18/100=0.18$, $43/100=0.43$, and $66/100=0.66$. The observed fractions are fairly close to the group predictions, but the small differences do not prove that the model is correctly specified. Examine the pattern and uncertainty, and check the model in the context of the sampling design. For a case-control sample, these predicted probabilities describe case status under the analyzed sampling scheme, not necessarily population disease risks.

A common mistake is to read $\exp(\hat{\beta}_G)$ as the increase in disease probability for each allele copy. It is an odds ratio, not a risk difference or probability ratio. For example, an odds ratio of 1.259 does not mean a 25.9 percentage-point increase in disease probability. The probability change depends on the starting probability, and in a case-control sample that probability is generally shaped by the sampling scheme. State the result on the odds scale and qualify the sampling context.

Choosing and Reporting the Analysis

The additive model is compact and tests a per-allele trend. A categorical genotype model estimates separate effects for heterozygotes and effect-allele homozygotes relative to the reference genotype, allowing departures from additivity but using more parameters. A dominant or recessive model combines genotypes in a different way and answers a different contrast. These approaches are not interchangeable; the genetic contrast should be scientifically motivated or prespecified. Wald, score, and likelihood-ratio procedures test model effects in different ways and can behave differently in finite or sparse samples. The effect estimate and its uncertainty remain important even when the test statistic is the primary scan output.

A concise result statement should name the SNP or variant, the counted allele, the model, the adjusted covariates, the odds ratio and interval, and the test evidence. It should also identify whether the reported p-value is nominal or meets the study's multiple-testing criterion. Avoid causal language unless the study design and assumptions justify a causal conclusion; an association scan alone generally supports an association claim.

For example, a suitably qualified statement for the hypothetical adjusted fit is: “In this hypothetical case-control analysis, each additional A allele was associated with higher odds of case status after adjustment for age, sex, and three ancestry principal components (per-allele OR 1.259, 95% CI 1.087–1.458; Wald $p=0.002165$). This is nominal evidence and does not meet a genome-wide criterion unless it also passes the study's prespecified multiple-testing threshold.” The statement reports the model-based association without claiming a change in absolute disease risk or a causal effect.

Key takeaway: A binary-trait SNP result is interpretable only with its allele coding, genetic model, covariate adjustment, uncertainty, and sampling context. Check the genotype data and model fit, then distinguish nominal evidence from evidence that satisfies the scan's multiple-testing standard.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If an individual has genotype AG and A is the effect allele, what is the additive dosage?
A fitted additive model has $\hat{\beta}_G=-0.20$. What is the per-copy odds ratio, and how should its direction be described?
Why can a case-control logistic model estimate an odds ratio without directly estimating population disease risk?
What does a very small genotype-by-outcome cell suggest, and name one possible consequence for the fit.
A variant has a nominal p-value below 0.05 in a genome-wide scan. Is that alone sufficient to claim genome-wide significance?

Solutions. 1. The dosage is 1, because AG contains one A copy. 2. $\exp(-0.20)\approx0.819$; each additional effect-allele copy is associated with about 0.819 times the odds, or lower odds, conditional on the included covariates under the model. 3. Sampling by case status changes the sampled case fraction and generally the intercept, while under the usual case-control logistic assumptions the odds-ratio slope is preserved. 4. It suggests sparse data; the estimate may be unstable, have a large standard error, or be affected by separation. 5. No. The p-value must be considered against the scan's prespecified multiple-testing procedure and threshold.