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Structural Variants and Copy Number · Tutorial 483 of 989

Carrier Tests For Rare Structural Variants

Learn to define the carrier and reference groups, fit carrier-indicator models for quantitative and binary outcomes, and interpret the resulting effects and uncertainty in context.

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What You'll Learn

  • Define a carrier indicator for one prespecified structural variant and name its reference group.
  • Interpret a linear-regression coefficient as a mean outcome difference between carriers and noncarriers.
  • Connect the unadjusted carrier coefficient to the difference in group means.
  • Interpret a logistic-regression coefficient as a log odds ratio, and its exponential as an odds ratio.
  • Distinguish unadjusted from covariate-adjusted carrier associations.
  • Check model assumptions and avoid interpreting an association as a causal effect.

Why Use a Carrier Indicator?

For a specified rare structural variant, the primary comparison may be between people who carry the event and people who do not. A carrier indicator encodes that comparison directly. It is useful when the question concerns the presence of a particular event, rather than a linear change across copy-number states. The estimate always depends on how the event and the reference group are defined.

The event must be specified before analysis: for example, a deletion at a defined locus that passes the study's variant- and sample-level quality criteria. Here, \(C_i=1\) means participant \(i\) is classified as carrying that event, and \(C_i=0\) means the participant is classified as a noncarrier. This coding does not distinguish different carrier copy counts or dosages; it asks only whether the specified event is present. The conclusions are associations, not evidence that the event caused an outcome.

Definition: A carrier indicator for a prespecified structural variant is \(C_i=1\) for a participant classified as carrying that event and \(C_i=0\) for a participant classified as a noncarrier. In a model with \(C_i\) as a predictor, noncarriers are the reference group unless the coding or model explicitly says otherwise.

Quantitative Outcomes: A Difference in Means

Let \(Y_i\) be a quantitative outcome, such as a blood biomarker measured in mg/L. The simplest carrier-association model is

$$ Y_i=\beta_0+\beta_1 C_i+\varepsilon_i. $$

In this model, \(\beta_0\) is the mean outcome among noncarriers, and \(\beta_1\) is the carrier-minus-noncarrier mean difference. Thus, a positive \(\beta_1\) indicates a higher mean outcome among carriers; a negative \(\beta_1\) indicates a lower mean. The units of \(\beta_1\) are the units of \(Y\).

For the usual small-sample \(t\)-based inference in this simple model, assume independent participants, normally distributed errors within each carrier group, and a common error variance in the two groups. The model also assumes the outcome is adequately summarized by a group mean in each category. If covariates are needed, they can be added to the regression; their inclusion changes the interpretation to a conditional, adjusted difference.

Derivation: With an intercept and a binary predictor, the fitted value for every noncarrier is \(\hat{\beta}_0\), and the fitted value for every carrier is \(\hat{\beta}_0+\hat{\beta}_1\). Least squares minimizes the sum of squared deviations within each group, so each group's fitted value is its sample mean. Therefore, \(\hat{\beta}_0=\bar{Y}_0\) and \(\hat{\beta}_0+\hat{\beta}_1=\bar{Y}_1\), giving \(\hat{\beta}_1=\bar{Y}_1-\bar{Y}_0\), where subscripts 0 and 1 denote noncarriers and carriers.

Worked Example: Unadjusted Biomarker Difference

Suppose a hypothetical study measures a biomarker in mg/L for five carriers and five noncarriers of one specified structural variant. The carrier values are 10, 11, 12, 13, and 14 mg/L; the noncarrier values are 7, 8, 9, 10, and 11 mg/L. Assume independent participants and, for this example, normally distributed outcomes with a common variance in the two groups.

The group means are \(\bar{Y}_1=(10+11+12+13+14)/5=12\) mg/L and \(\bar{Y}_0=(7+8+9+10+11)/5=9\) mg/L. The estimated carrier effect, with noncarriers as the reference group, is

$$ \hat{\beta}_1=\bar{Y}_1-\bar{Y}_0=12-9=3\text{ mg/L}. $$

For each group, the squared deviations from its mean sum to \(4+1+0+1+4=10\), so both sample variances are \(10/(5-1)=2.5\;(\text{mg/L})^2\). The pooled variance is

$$ s_p^2=\frac{(5-1)(2.5)+(5-1)(2.5)}{5+5-2} =\frac{20}{8}=2.5\;(\text{mg/L})^2. $$

The standard error of the difference is \(SE=\sqrt{s_p^2(1/5+1/5)}=\sqrt{2.5(0.4)}=1.00\) mg/L. Testing \(H_0:\beta_1=0\) against a two-sided alternative gives \(t=3/1=3.00\), with \(df=5+5-2=8\). The two-sided \(p\)-value is approximately 0.017. A 95% confidence interval uses \(t_{0.975,8}=2.306\):

$$ 3.00\pm 2.306(1.00)=[0.694,\;5.306]\text{ mg/L}. $$

Under the stated model, the data estimate a carrier-group mean that is 3.00 mg/L higher than the noncarrier-group mean. The interval is a range of values compatible with the model and data for that mean difference; it is not a 95% probability statement about this particular fixed parameter.

Adjustment for Covariates

A carrier association may be estimated while accounting for measured characteristics that relate to the outcome and differ between carrier groups. For example, with age \(A_i\) measured in years, a linear model might be

$$ Y_i=\beta_0+\beta_1 C_i+\beta_2(A_i-50)+\varepsilon_i. $$

If this model is adequate and contains no carrier-by-age interaction, \(\beta_1\) is the modeled difference in mean outcome between carriers and noncarriers at the same age. It is an adjusted association, not automatically a causal effect. Adjustment relies on the model's functional form and assumptions, and it does not remove bias from unmeasured or poorly measured factors.

The linear-model assumptions for conventional \(t\)-based inference include independent errors, a correct conditional mean specification, normally distributed errors for exact small-sample inference, and constant error variance. A covariate should not be included mechanically: its role in the scientific question and study design matters.

Worked Example: Age-Adjusted Carrier Difference

Consider hypothetical biomarker data for 16 participants, with four people in each carrier-by-age group. “Younger” and “older” denote two age strata. The table gives group means in mg/L; within every group, the four individual values equal the listed mean plus \(-2,-1,1,2\) mg/L, respectively.

Age stratumNoncarrier mean (n)Carrier mean (n)
Younger20 mg/L (4)23 mg/L (4)
Older30 mg/L (4)33 mg/L (4)

Fit \(Y_i=\beta_0+\beta_1C_i+\beta_2A_i+\varepsilon_i\), where \(A_i=0\) for younger and \(A_i=1\) for older participants. The four cell means match the additive model exactly: the younger noncarrier mean is 20, the carrier difference is \(23-20=3\), and the older-group increment is \(30-20=10\). The model therefore estimates \(\hat{\beta}_1=3\) mg/L, with noncarriers as the reference group, and \(\hat{\beta}_2=10\) mg/L for the older-versus-younger contrast.

The residuals within each of the four cells are \(-2,-1,1,2\), with squared sum \(4+1+1+4=10\) per cell. Thus, the residual sum of squares is \(4(10)=40\). There are 16 observations and 3 fitted coefficients (intercept, carrier indicator, and age indicator), so \(df=16-3=13\), and \(MSE=40/13=3.077\;(\text{mg/L})^2\). With four participants in each cell, there are eight carriers and eight noncarriers; in this balanced design the standard error for the carrier contrast is

$$ SE(\hat{\beta}_1)=\sqrt{MSE\left(\frac{1}{8}+\frac{1}{8}\right)} =\sqrt{\frac{40}{13}\cdot\frac{1}{4}}=0.877\text{ mg/L}. $$

The \(t\)-statistic is \(3/0.877=3.42\), with 13 degrees of freedom. The two-sided \(p\)-value is approximately 0.0046; a 95% interval using \(t_{0.975,13}=2.160\) is \(3\pm2.160(0.877)=[1.11,4.89]\) mg/L. This adjusted estimate compares carriers and noncarriers at the same age under the additive model. The example assumes independent, normally distributed errors with equal variance and no carrier-by-age interaction; the specified residual pattern satisfies the normal-model form only as an illustrative constructed data set, not as evidence that real data meet it.

Binary Outcomes: Odds Ratios

For a binary outcome \(D_i\), where \(D_i=1\) denotes the outcome and \(D_i=0\) its absence, a common model is logistic regression:

$$ \log\left(\frac{P(D_i=1\mid C_i)}{1-P(D_i=1\mid C_i)}\right) =\beta_0+\beta_1C_i. $$

The left side is the log odds of the outcome. Here, \(\beta_0\) is the log odds among noncarriers, and \(\beta_1\) is the carrier-versus-noncarrier log odds ratio. Exponentiating gives \(\exp(\beta_1)\), the odds ratio. An odds ratio above 1 means higher outcome odds among carriers; below 1 means lower odds. It is not a risk ratio, and when outcomes are common the two can differ substantially.

The model assumes independent observations and a correctly specified logit relationship; covariates can be added to estimate a conditional odds ratio. Standard Wald tests and intervals also rely on a large-sample approximation. With very few carriers or outcome events, that approximation may be unreliable, so the apparent precision from a routine fit should not be taken for granted.

Worked Example: Carrier Association With a Binary Outcome

In a hypothetical cohort, 30 participants carry a specified variant: 12 have the outcome and 18 do not. Among 240 noncarriers, 24 have the outcome and 216 do not. Noncarriers are the explicit reference group. Assume independent participants, reliable carrier classification, and a logistic model with no covariates.

The carrier outcome odds are \(12/18=0.667\); the noncarrier outcome odds are \(24/216=0.111\). Therefore, the estimated odds ratio is

$$ \widehat{OR}=\frac{12/18}{24/216}=6.00, \qquad \hat{\beta}_1=\log(6.00)=1.792. $$

The usual large-sample standard error for the log odds ratio from a two-by-two table is the square root of the sum of the reciprocals of its four cell counts:

$$ SE(\hat{\beta}_1)=\sqrt{\frac{1}{12}+\frac{1}{18}+\frac{1}{24}+\frac{1}{216}} =\sqrt{0.1852}=0.430. $$

The Wald statistic is \(z=1.792/0.430=4.16\), giving a two-sided \(p\)-value of approximately 0.000031. A 95% interval on the log odds-ratio scale is \(1.792\pm1.96(0.430)=[0.948,2.635]\). Exponentiating both limits gives an odds-ratio interval of approximately \([2.58,13.94]\). In this hypothetical cohort, estimated outcome odds are six times as high among carriers as among noncarriers. This is an odds comparison, not a statement that carriers are six times as likely to have the outcome, and it does not establish causation.

Choosing and Reporting the Contrast

The carrier coefficient has meaning only in relation to its coding and reference group. A clear report names the event, identifies the noncarrier reference group, gives the effect estimate and its units or scale, and states whether the estimate is adjusted. For a quantitative outcome, report a mean difference in the outcome's units. For a binary outcome in logistic regression, report an odds ratio and make clear that it concerns odds.

For example, “The estimated mean biomarker level was 3.00 mg/L higher in carriers than noncarriers” states the direction and reference group. “The estimated odds ratio was 6.00 for carriers versus noncarriers” states the binary-outcome contrast. Avoid saying simply “the variant increased the outcome”: that phrasing can imply causation and obscures the comparison being estimated.

Common Mistakes

  • Leaving the reference group unstated. A coefficient's sign depends on which group is coded 1. Name carriers and noncarriers in the contrast.
  • Calling a logistic odds ratio a risk ratio. The model estimates odds; do not translate an odds ratio into a proportional change in risk.
  • Treating a carrier indicator as dosage. \(C_i\) distinguishes presence from absence of the specified event. It does not estimate a per-copy effect or distinguish carrier dosage states.
  • Interpreting an adjusted coefficient as causal. Covariate adjustment does not by itself establish causality or eliminate unmeasured confounding.
  • Ignoring uncertainty or model assumptions. A point estimate alone does not show precision, and routine \(p\)-values depend on the model and its assumptions.
Key takeaway: A carrier test estimates a defined contrast between carriers of one prespecified structural variant and noncarriers. For a quantitative outcome, the basic coefficient is a mean difference; for a binary outcome in logistic regression, it is a log odds ratio. State the reference group, model, units or effect scale, adjustment set, and uncertainty.

Check Your Understanding

Answer each question by naming the comparison and interpreting the effect on its proper scale.

  1. If \(C_i=1\) for carriers and \(C_i=0\) for noncarriers, what does the intercept represent in a linear model with no covariates?
  2. A fitted quantitative-outcome coefficient is \(-2.4\) units. Interpret its sign and reference group when carriers are coded 1.
  3. In logistic regression, what does \(\exp(\beta_1)=0.70\) mean when noncarriers are the reference group?
  4. Why is an odds ratio not generally interchangeable with a risk ratio?
  5. What additional interpretation is needed when an age covariate is added to a carrier-indicator model?
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