Repeated-Measures QTL Models Across Tissues

Why Repeated Tissue Measurements Need Their Own Model

A molecular QTL study may measure the same expression phenotype in several tissues from each donor. These measurements are not independent: they share the donor’s genotype, and they may also share biological and technical influences specific to that donor. Treating every tissue measurement as a separate, independent observation can therefore exaggerate the amount of information in the study and produce standard errors that are too small.

The central modelling task is to represent two features at once: tissue measurements can have different average levels, and measurements from one donor can be correlated. This tutorial focuses on estimating and testing a common genotype effect across the observed tissues, while allowing for that repeated-measures correlation. It does not assume that the correlation is negligible simply because the tissues are distinct.

Definition: A repeated-measures QTL model describes molecular measurements \(Y_{ij}\) from donor \(i\) in tissue \(j\), with a covariance model that allows measurements from the same donor to be dependent. A common genotype effect is one coefficient for the change in the expected phenotype per allele copy, shared across the tissues included in that model.

A Common-Effect Model With a Random Intercept

Let \(Y_{ij}\) be the normalized molecular phenotype for donor \(i\) in tissue \(j\), and let \(g_i\in\{0,1,2\}\) be the donor’s allele dosage at the variant. A basic model is

Y_{ij}=\mu+\alpha_j+\beta g_i+\boldsymbol{\gamma}^{\mathsf T}\mathbf{x}_i+b_i+\varepsilon_{ij}.

Here \(\mu\) is an intercept, \(\alpha_j\) is the fixed baseline difference for tissue \(j\), and \(\beta\) is the common additive genotype effect in phenotype units per allele copy. The vector \(\mathbf{x}_i\) contains any included donor-level covariates, with coefficients \(\boldsymbol{\gamma}\). The donor-specific random intercept \(b_i\) captures a shared shift in that donor’s measurements. A standard random-intercept specification assumes \(b_i\sim N(0,\tau^2)\), \(\varepsilon_{ij}\sim N(0,\sigma^2)\), and independence between all random intercepts and residual errors. Conditional on the random intercepts and covariates, the residuals are independent across observations. These distributional and independence assumptions are modelling assumptions, not guaranteed properties of the data.

The genotype is constant across a donor’s tissues. Thus, having four tissue measurements for one donor does not make that donor equivalent to four independent genotype observations. The model uses the repeated measurements to characterize the donor’s molecular phenotype while accounting for their dependence.

Random-intercept covariance: Under the model above, a measurement has variance \(\tau^2+\sigma^2\). Two distinct measurements from the same donor have covariance \(\tau^2\), and hence correlation \(\tau^2/(\tau^2+\sigma^2)\). Measurements from different donors have covariance zero under this model, conditional on the included covariates and absent any additional donor-relatedness covariance.

The random intercept is one way to specify the repeated-measures covariance. If donor \(i\) has \(m_i\) tissue measurements, the random-intercept model gives

\operatorname{Var}(\mathbf{Y}_i)=\tau^2\mathbf{1}\mathbf{1}^{\mathsf T}+\sigma^2 I_{m_i},

where \(\mathbf{Y}_i\) stacks that donor’s measurements, \(\mathbf{1}\) is a vector of ones, and \(I_{m_i}\) is the identity matrix. This structure gives every pair of tissues from a donor the same covariance. An alternative is to specify a covariance matrix directly—for example, an unstructured matrix that estimates a separate variance for each tissue and a separate covariance for each tissue pair. Such a matrix is more flexible, but it uses more parameters and must be estimated from enough donors to be stable. The covariance choice should reflect the design and the amount of data; it should not be selected solely because it yields the smallest p-value.

Testing the Genotype Effect

For a common-effect analysis, the null and two-sided alternative hypotheses are \(H_0:\beta=0\) and \(H_a:\beta\ne0\). The coefficient is estimated while accounting for the chosen covariance structure. In a linear mixed model, this is commonly done by maximum likelihood or restricted maximum likelihood (REML) for the variance components, together with estimation of the fixed effects. A Wald statistic is

Z=\frac{\hat{\beta}}{SE(\hat{\beta})}.

With a large enough number of independent donors, this statistic is often compared with a standard normal distribution. In smaller samples, an appropriate \(t\)-reference distribution with a finite-sample degrees-of-freedom approximation may be preferable. A likelihood-ratio test comparing models with and without the genotype term is another option; its reference distribution and variance-component treatment must match the fitted models. In every case, the uncertainty calculation must account for within-donor dependence.

The common-effect model includes tissue baseline terms but no genotype-by-tissue interaction. Consequently, \(\beta\) summarizes the genotype association under the constraint that the per-allele effect is the same across the included tissues. If that constraint is not scientifically appropriate, the common-effect estimate may not describe every tissue well. Assessing whether effects differ across tissues is a separate question.

Worked Examples

Worked Example: Quantifying Correlation Within a Donor

Suppose a hypothetical study measures one phenotype in two tissues per donor. For illustration, assume a random-intercept model with \(\tau^2=0.36\) and \(\sigma^2=0.64\), both in squared normalized-expression units. Check the implied correlation and the uncertainty in the average of the two measurements.

The marginal variance of one measurement is \(0.36+0.64=1.00\). The covariance between the two measurements from the same donor is \(0.36\), so their correlation is

\frac{0.36}{0.36+0.64}=0.36.

For the donor’s average across two tissues, the shared intercept remains in the average, while the independent residual errors are averaged. Its variance is therefore

\operatorname{Var}\left(\frac{Y_{i1}+Y_{i2}}{2}\right) =\tau^2+\frac{\sigma^2}{2} =0.36+\frac{0.64}{2} =0.68.

The standard deviation of that average is \(\sqrt{0.68}=0.8246\) normalized-expression units. If the two measurements were incorrectly treated as independent with marginal variance \(1.00\), the calculated variance of their average would be \(1.00/2=0.50\), giving a standard deviation of \(0.7071\). That calculation misses the donor-level component shared across tissues and understates the uncertainty.

Worked Example: Testing a Common Per-Allele Effect

Suppose a hypothetical study measures normalized expression in three tissues for each of 120 donors. A model with tissue fixed effects and a donor random intercept estimates a common genotype coefficient of \(\hat{\beta}=0.31\) normalized-expression units per allele copy, with \(SE(\hat{\beta})=0.12\) units per allele copy. Assume the dosage model, covariates, and random-intercept covariance are adequate, and that a large-sample normal approximation is suitable.

For \(H_0:\beta=0\) against \(H_a:\beta\ne0\), the Wald statistic is

Z=\frac{0.31}{0.12}=2.583.

The two-sided normal-reference p-value is approximately \(0.0098\). Under the model and null hypothesis, this is the probability of obtaining a test statistic at least as extreme as \(2.583\) in absolute value; it is not the probability that the null hypothesis is true.

A 95% Wald confidence interval is

0.31\pm1.96(0.12) =0.31\pm0.2352 =[0.0748,\,0.5452]\ \text{normalized-expression units per allele copy}.

The interval excludes zero and agrees with the two-sided test at the 5% significance level. Under this model, the estimated common association is an increase of 0.31 normalized-expression units per additional allele copy, with a 95% confidence interval from 0.0748 to 0.5452 units per allele copy. This is an association estimate; the model alone does not establish that the variant causes the expression change.

Worked Example: Why Repeated Rows Do Not Mean Independent Genotypes

Consider a hypothetical balanced study with six donors, two tissues per donor, and dosages \(g_i=0,0,1,1,2,2\). The tissue measurements, in normalized-expression units, are: for dosage 0, \((0.5,0.7)\) and \((1.5,1.7)\); for dosage 1, \((1.0,1.2)\) and \((2.0,2.2)\); and for dosage 2, \((1.5,1.7)\) and \((2.5,2.7)\). Each pair lists tissue 1 and tissue 2. Assume for this calculation that the random-intercept variance is \(\tau^2=0.25\) and the residual variance is \(\sigma^2=0.25\), in squared normalized-expression units, and that these variance components are known.

The donor averages are \(0.6,1.6,1.1,2.1,1.6,2.6\), in dosage order. The mean donor averages at dosages 0, 1, and 2 are \(1.1,1.6,\) and \(2.1\). Their increase is \(0.5\) units for each one-copy increase in dosage, so the fitted common slope is \(\hat{\beta}=0.5\) normalized-expression units per allele copy. In this balanced design, including a tissue fixed effect does not change this slope: tissue 2 is consistently 0.2 units above tissue 1.

The sample mean dosage is 1, and the sum of squared dosage deviations across the six donors is

\sum_{i=1}^{6}(g_i-\bar{g})^2 =2(0-1)^2+2(1-1)^2+2(2-1)^2=4.

The variance of a donor average across two tissues is \(0.25+0.25/2=0.375\). Therefore, using the donor averages, the standard error of the slope under the assumed known covariance is

SE(\hat{\beta})=\sqrt{\frac{0.375}{4}}=0.3062\ \text{normalized-expression units per allele copy}.

The corresponding normal statistic is \(0.5/0.3062=1.633\), with a two-sided p-value of approximately \(0.1025\). For comparison, treating all 12 rows as independent would use marginal variance \(0.25+0.25=0.50\). The sum of squared dosage deviations across those rows is \(2(4)=8\), giving the naive standard error \(\sqrt{0.50/8}=0.25\) and statistic \(0.5/0.25=2.00\), with a two-sided p-value of about \(0.0455\). That apparent significance results from an independence assumption contradicted by the specified shared donor effect. In real analyses, variance components are generally estimated, and inference must reflect that estimation rather than treating them as known.

When Averaging Helps—and What It Cannot Do

For a balanced design in which every donor has the same tissues measured, and the model has a common genotype effect, donor averaging can clarify the analysis. With a common random-intercept structure, the average retains the donor-level genotype information while averaging down independent residual noise. A tissue fixed effect can account for baseline differences, provided the design allows those differences to be estimated.

Averaging is not a universal replacement for a repeated-measures model. It can discard information when donors have different numbers of observed tissues, when tissue-specific residual variances differ, or when the covariance structure is more complex. It also cannot estimate a separate genotype association for each tissue after the tissue measurements have been collapsed into one number. A model fitted to the repeated observations can use the observed covariance structure directly and handle an incomplete tissue panel under appropriate assumptions about the missingness and the likelihood.

The random intercept describes correlation among measurements from the same donor. It does not, by itself, account for correlation between different donors who are genetically related or for population structure. Those are distinct sources of dependence and may require additional terms or covariates. In a study with both repeated tissues and related donors, the covariance model must represent the relevant sources without treating one as a substitute for the other.

Assumptions and Practical Checks

Before interpreting a genotype test, check that the model corresponds to the design and that its assumptions are defensible. In particular:

Outcome scale and mean model: the chosen linear model should be reasonable for the processed phenotype, and tissue baseline effects and relevant covariates should be represented.
Genotype coding: coding dosage as 0, 1, or 2 imposes an additive linear effect per allele copy. The coefficient is not automatically a separate estimate for each genotype class.
Covariance structure: the random-intercept model imposes equal within-donor covariance across tissue pairs. Residual plots and covariance summaries can help identify substantial departures, although diagnostics cannot prove the model correct.
Independent donor units: the basic model assumes different donors are independent after adjustment. If donors are related or population structure remains, additional modelling or adjustment is needed.
Inference method: the reference distribution and standard error should reflect the number of independent donors and whether variance components were estimated. A large number of tissue rows does not by itself justify a large-sample approximation if the donor count is small.

A covariance model is part of the scientific specification, not a technical afterthought. If the true within-donor dependence is stronger than the fitted model allows, the reported standard errors and p-values can be misleading even when the estimated genotype coefficient looks plausible.

Common Mistake

Common mistake: “There are three tissues for each of 100 donors, so the genotype analysis has 300 independent observations.” The genotype is measured once per donor, and the three outcomes share donor-level influences. Counting those 300 rows as independent ignores within-donor correlation. Fit a repeated-measures model with an appropriate covariance structure—or use a justified donor-level summary—and base the uncertainty on the dependence and the number of independent donors.

Key takeaway: A repeated-measures QTL analysis must model both the mean structure, including tissue baselines and genotype dosage, and the covariance among measurements from the same donor. A common genotype coefficient is interpretable as a shared per-allele association only under the model’s common-effect assumption, and its standard error must not treat repeated tissue measurements as independent genotype observations.

Check Your Understanding

Use the model assumptions and calculations in this tutorial to answer the following questions.

In the random-intercept model, what are the covariance and correlation between two different tissue measurements from the same donor?
Why does a donor measured in four tissues not count as four independent genotype observations?
A fitted common-effect model estimates \(\hat{\beta}=-0.20\) with \(SE(\hat{\beta})=0.08\). What is the Wald statistic, and what does its sign indicate?
What assumption about genotype effects across tissues is made by a model with tissue fixed effects but no genotype-by-tissue interaction?
Give one design feature that can make a simple average across tissues an inadequate replacement for a repeated-measures model.