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Single-Cell and Cell-Type Genetics · Tutorial 963 of 989

Power and Sample Size for Cell-Type QTLs

Learn to translate donor and cell sampling into a power calculation, and to distinguish the gains from recruiting donors from the smaller gains of sequencing more cells per donor.

Advanced 12 min read

What You'll Learn

  • Define a donor-level QTL effect and the assumptions behind its variance.
  • Derive how the number of cells per donor and intradonor correlation affect precision.
  • Calculate approximate power for a single cell-type QTL test.
  • Quantify how allele frequency changes additive genotype information.
  • Explain how expression level can affect power through measurement noise.
  • Compare the gains from additional donors with diminishing returns from additional cells.

Why Power Has Two Sampling Scales

A cell-type QTL study samples donors, then measures many cells from each donor. These are not interchangeable sources of information. A donor contributes a genotype observation, while additional cells can improve the precision of that donor’s expression summary. Because cells from the same donor tend to resemble one another, measuring more cells does not generally provide the same gain as recruiting more donors.

Power is the probability that a specified test rejects \(H_0\) when a stated alternative is true, under the model and design used for planning. A useful power calculation therefore makes the target effect, analysis unit, genotype distribution, expression variability, and testing threshold explicit. The calculations here concern one prespecified cell type and one prespecified QTL test at a chosen significance level \(\alpha\). They do not account for a later adjustment across many genes or cell types.

A Donor-Level Model and Its Assumptions

Let \(Y_{ij}\) be a continuous expression measure for cell \(j\) from donor \(i\), on a scale appropriate for the analysis. Let \(G_i\) be the donor’s additive genotype dosage, coded 0, 1, or 2 copies of the effect allele. A simple random-intercept model is

$$ Y_{ij}=\mu+\beta G_i+b_i+\varepsilon_{ij}. $$

Here \(\beta\) is the expected expression difference per additional effect-allele copy, in the units of \(Y_{ij}\). The donor effect \(b_i\) represents shared donor-level variation, and \(\varepsilon_{ij}\) represents cell-level variation around that donor’s expected expression. For the basic calculation, assume donors are independent; \(b_i\) and all \(\varepsilon_{ij}\) are mutually independent and independent of genotype; \(E(b_i)=E(\varepsilon_{ij})=0\); \(\operatorname{Var}(b_i)=\tau^2\); and \(\operatorname{Var}(\varepsilon_{ij})=\sigma_e^2\). Assume the cell-level errors have the same variance and are independent across cells. These independence and mean assumptions are what permit the variance decomposition below; conditional independence of errors alone would not suffice.

Suppose each of \(N\) donors contributes \(m\) cells to the cell-type analysis. The donor-level pseudobulk mean is \(\bar{Y}_i=m^{-1}\sum_{j=1}^{m}Y_{ij}\). Averaging the model over cells gives

$$ \bar{Y}_i=\mu+\beta G_i+b_i+\bar{\varepsilon}_i, \qquad \operatorname{Var}(\bar{Y}_i\mid G_i)=\tau^2+\frac{\sigma_e^2}{m}. $$

The term \(\tau^2\) remains even if many cells are measured: it is variation between donors, not cell-level sampling noise. The second term decreases as more cells are averaged. The intradonor correlation (ICC) between two cells from the same donor under this model is

$$ \rho=\frac{\tau^2}{\tau^2+\sigma_e^2}. $$

Writing the marginal variance of a cell as \(\sigma^2=\tau^2+\sigma_e^2\), the variance of the donor mean is equivalently

$$ v(m)=\operatorname{Var}(\bar{Y}_i\mid G_i) =\sigma^2\left(\rho+\frac{1-\rho}{m}\right). $$

This formula shows the cell-count trade-off directly. When \(\rho=0\), all variation is cell-level and the donor mean variance falls as \(1/m\). When \(\rho\) is positive, the variance approaches the floor \(\sigma^2\rho=\tau^2\) as \(m\) grows.

From Donor-Mean Variance to Power

Regress the \(N\) donor-level means on genotype, with an intercept. If genotype is independent of other covariates and the model has no additional covariates, the approximate variance of the estimated additive effect is \(v(m)/(N V_G)\), where \(V_G=\operatorname{Var}(G_i)\). Under Hardy–Weinberg equilibrium with effect-allele frequency \(q\), genotype dosage has variance \(V_G=2q(1-q)\). Thus an approximate standard error is

$$ SE(\hat{\beta})\approx \sqrt{\frac{\sigma^2\left(\rho+(1-\rho)/m\right)} {N\,2q(1-q)}}. $$

This approximation uses the expected genotype variance in the sample. If the realized genotype dosages are available, a more direct calculation replaces \(N V_G\) with \(S_{GG}=\sum_i(G_i-\bar{G})^2\). If covariates are included, genotype information is the variation in genotype remaining after accounting for those covariates; the simple expression above is not automatically valid when genotype and covariates are associated.

For a two-sided test at level \(\alpha\), a convenient large-sample approximation uses \(z_{1-\alpha/2}\), the standard normal quantile. Under a true effect \(\beta\), define the noncentrality parameter \(\delta=|\beta|/SE(\hat{\beta})\). Approximate power is

$$ \operatorname{Power}\approx \Phi\left(-z_{1-\alpha/2}-\delta\right) +1-\Phi\left(z_{1-\alpha/2}-\delta\right), $$

where \(\Phi\) is the standard normal cumulative distribution function. A finite-sample analysis may instead use a noncentral \(t\) distribution, with degrees of freedom determined by the fitted regression. The normal approximation is most useful for planning when the number of donors and genotype groups are not too small.

Worked Example: Cells per Donor and Diminishing Returns

Worked Example: Cells per Donor and Diminishing Returns

Consider a hypothetical study with \(N=120\) donors, effect-allele frequency \(q=0.30\), and a target effect \(\beta=0.15\) expression units per allele. Suppose the marginal cell-level variance is \(\sigma^2=1\) and the ICC is \(\rho=0.20\). Compare \(m=20\) with \(m=80\) cells per donor, using a two-sided \(\alpha=0.05\) test.

First, the expected genotype variance is \(2q(1-q)=2(0.30)(0.70)=0.42\). With 20 cells per donor, the donor-mean variance is

$$ v(20)=1\left(0.20+\frac{0.80}{20}\right)=0.240. $$

The approximate standard error and noncentrality parameter are

$$ SE=\sqrt{\frac{0.240}{120(0.42)}}= \sqrt{0.004762}=0.0690, \qquad \delta=\frac{0.15}{0.0690}=2.17. $$

Using \(z_{0.975}=1.96\), approximate power is \(\Phi(-1.96-2.17)+1-\Phi(1.96-2.17)\), or about \(0.583\). With 80 cells per donor,

$$ v(80)=1\left(0.20+\frac{0.80}{80}\right)=0.210, \qquad SE=\sqrt{\frac{0.210}{120(0.42)}}=0.06455, \qquad \delta=\frac{0.15}{0.06455}=2.324. $$

The corresponding approximate power is about \(0.642\), using the unrounded noncentrality parameter. Quadrupling the cells per donor raises power by roughly 5.7 percentage points here, not by a factor of four. Even with infinitely many cells, \(v(m)\) cannot fall below \(0.20\) under these assumptions. The calculation assumes a common cell count, the stated ICC, and independent donors; unequal cell counts or model misspecification would require a design-specific calculation.

Why Donors, Genotype Frequency, and Expression Matter

The standard error formula separates several design contributions. Increasing \(N\) increases genotype information approximately in proportion to \(N\), so the standard error falls roughly as \(1/\sqrt{N}\). Increasing \(m\) reduces only the cell-level component of donor-mean variance. A large ICC makes that component smaller relative to the between-donor floor, and therefore reduces the benefit of collecting still more cells.

Genotype frequency matters because it determines how much dosage variation is available for estimating an additive effect. Under Hardy–Weinberg equilibrium, \(2q(1-q)\) is largest at \(q=0.5\) and becomes small when the allele is rare or nearly fixed. With few copies of a rare allele, the study has less information to estimate the per-allele effect, even if its total cell count is large.

Expression level affects power through the variability and reliability of the expression outcome, rather than through a universal multiplier. Lowly expressed genes can have greater sampling noise or more zero counts; higher expression can make a donor-level expression estimate more precise. The size of this effect depends on the measurement scale, library sizes, normalization, and count distribution. It should be represented by a gene- and cell-type-appropriate variance estimate, not assumed to be identical across genes.

Worked Example: Genotype Frequency Changes Information

Suppose a hypothetical design has \(N=300\) donors, donor-mean variance \(v(m)=0.25\), and a target additive effect \(\beta=0.12\) units per allele. Compare an effect-allele frequency of \(q=0.50\) with \(q=0.05\), using a two-sided \(\alpha=0.05\) test. Assume Hardy–Weinberg equilibrium and use the normal approximation.

At \(q=0.50\), \(V_G=2(0.50)(0.50)=0.50\). Therefore

$$ SE=\sqrt{\frac{0.25}{300(0.50)}}=0.0408, \qquad \delta=\frac{0.12}{0.0408}=2.94. $$

Power is \(\Phi(-1.96-2.94)+1-\Phi(1.96-2.94)\), approximately \(0.836\). At \(q=0.05\), \(V_G=2(0.05)(0.95)=0.095\), so

$$ SE=\sqrt{\frac{0.25}{300(0.095)}}=0.0937, \qquad \delta=\frac{0.12}{0.0937}=1.28. $$

The resulting power is approximately \(0.2488\). The effect and donor count are unchanged; the lower genotype variance at the rarer allele frequency gives substantially less information. In a real small study, the number of minor-allele homozygotes could also be very low, making a normal approximation less reliable.

Worked Example: Expression-Dependent Measurement Noise

Consider a hypothetical log-expression estimate formed from aggregated counts. As a simple illustration, suppose a donor has \(m\) cells, each with equal exposure and mean count \(\lambda\). For sufficiently large aggregate counts, the delta method gives an approximate sampling variance of \(1/(m\lambda)\) for the log of the mean count. Suppose additional donor-level log-expression variation has variance \(\tau^2=0.04\). Then use \(v(m)\approx 0.04+1/(m\lambda)\).

Take \(N=100\), \(m=50\), \(V_G=0.50\), and a target log-expression effect \(\beta=0.08\) per allele. For a gene with \(\lambda=2\) expected counts per cell, the approximate donor-mean variance is \(0.04+1/(50\cdot2)=0.050\). Thus

$$ SE=\sqrt{\frac{0.050}{100(0.50)}}=0.0316, \qquad \delta=\frac{0.08}{0.0316}=2.53. $$

The two-sided power at \(\alpha=0.05\) is approximately \(0.716\). For a lower-expression gene with \(\lambda=0.5\), the variance is \(0.04+1/(50\cdot0.5)=0.080\), giving

$$ SE=\sqrt{\frac{0.080}{100(0.50)}}=0.0400, \qquad \delta=\frac{0.08}{0.0400}=2.00. $$

Approximate power is then \(0.516\). This illustration isolates a measurement-noise mechanism; it is not a universal count model. In particular, the log-scale approximation can be poor when aggregate counts are very small, and actual planning should use a variance model appropriate to the assay and analysis.

A Short Derivation of the Cell-Count Limit

The diminishing return follows directly from the donor-mean variance. Under the random-intercept assumptions, the variance of a sum of \(m\) cell outcomes for one donor is \(m\tau^2+m\sigma_e^2\), because each cell shares the same \(b_i\), while independent cell errors contribute separately. Dividing the sum by \(m^2\) gives

$$ \operatorname{Var}(\bar{Y}_i\mid G_i) =\frac{m\tau^2+m\sigma_e^2}{m^2} =\tau^2+\frac{\sigma_e^2}{m}. $$

As \(m\) tends to infinity, \(\sigma_e^2/m\) tends to zero, but \(\tau^2\) remains. Consequently, at fixed \(N\), the approximate standard error of the genotype effect approaches \(\sqrt{\tau^2/(N V_G)}\), not zero. More cells can sharpen each donor’s expression estimate; they cannot create additional independent genotype observations or remove between-donor variation.

A planning calculation can be organized as a sequence: specify a meaningful effect and scale; estimate variance components from relevant pilot data; specify donor count, expected cells per donor, and genotype frequency; calculate the approximate standard error and power; then examine sensitivity to plausible changes in ICC, expression noise, and allele frequency. If cell counts vary substantially among donors, equal-\(m\) formulas are only a guide, and power should reflect the planned analysis and cell-count distribution.

Common Pitfall and Practical Interpretation

A common mistake is to treat every sequenced cell as an independent genetic observation and substitute the total number of cells for \(N\). That reasoning is wrong because genotype is measured at the donor level and cells within a donor share donor-level variation. The correction is to model or summarize the donor structure and calculate information from independent donors, with cells per donor contributing through the reduction in measurement variance.

Power is conditional on the assumed effect and variance model. It is not the probability that a detected association is true, nor does a planned high power guarantee a significant result in a particular study. Likewise, a nonsignificant test does not establish that the effect is absent. The calculation is most useful for comparing feasible designs: recruit more donors when possible, and add cells when cell-level sampling noise remains an important component of uncertainty.

Key takeaway: For a donor-level cell-type QTL, approximate precision depends on donor count, genotype variance, and the variance of each donor’s expression summary. More cells reduce cell-level noise, but intradonor correlation creates a floor; donors supply the independent genotype information.

Check Your Understanding

Use the model assumptions and distinguish donor-level information from cell-level precision.

  1. Under the random-intercept model, what is the variance of a donor mean based on \(m\) cells?
  2. Why does the gain from additional cells become small when the ICC is high?
  3. Under Hardy–Weinberg equilibrium, compare genotype variance for allele frequencies \(q=0.5\) and \(q=0.1\). Which provides more additive genotype information?
  4. If the number of cells per donor is doubled, does the approximate standard error of the genetic effect necessarily fall by a factor of \(\sqrt{2}\)? Explain.
  5. Name two ways expression level can influence power, and explain why neither implies a universal expression-level multiplier.
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