Estimating Kinetic Rates from Cell Counts

What Cross-Cell Variation Can Tell Us

Unspliced and spliced RNA counts vary across cells, and some of that variation can reveal how quickly RNA moves through a gene’s kinetic process. The key is to distinguish variation in abundance from variation over time. A collection of cells is a cross-sectional snapshot; differences among cells do not, by themselves, show how any one cell changed.

For a gene, let \(U\) and \(S\) denote unspliced and spliced RNA abundance. The two-compartment model describes their rates of change as

\frac{dU}{dt}=\alpha-\beta U,\qquad \frac{dS}{dt}=\beta U-\gamma S.

Here \(\alpha\) is the transcription rate, \(\beta\) the splicing rate, and \(\gamma\) the spliced-RNA degradation rate. The rates are nonnegative. If abundance is measured in molecules per cell and time in hours, \(\alpha\) has units of molecules per cell per hour, while \(\beta\) and \(\gamma\) have units of inverse hours. The model treats each rate as constant over the interval being described; it does not assert that all cells have identical transcription or abundance.

Definition: A steady state is a condition in which both abundance derivatives are zero. For positive \(\beta\) and \(\gamma\), the steady-state equations imply \(U=\alpha/\beta\) and \(S=(\beta/\gamma)U=\alpha/\gamma\). Without positive rates, retain the undivided equations \(\alpha=\beta U\) and \(\beta U=\gamma S\); the quotient forms need not be defined.

The central estimation opportunity comes when cells are near steady state, share the same gene-specific \(\beta\) and \(\gamma\), and differ in transcriptional activity \(\alpha\). Their steady-state abundances then vary together: higher transcription produces higher \(U\) and \(S\), while the relationship between them has slope \(\beta/\gamma\). This is a rate ratio, not either absolute rate by itself.

Deriving the Phase-Portrait Relationship

At steady state, set the two derivatives to zero. The first equation gives \(\alpha=\beta U\), and the second gives \(\beta U=\gamma S\). When \(\gamma>0\), substitution yields \(S=(\beta/\gamma)U\). If each cell has a possibly different transcription rate \(\alpha_i\), but the same positive \(\beta\) and \(\gamma\), then the same reasoning applies cell by cell:

U_i=\frac{\alpha_i}{\beta},\qquad S_i=\frac{\alpha_i}{\gamma} \quad\Longrightarrow\quad S_i=\frac{\beta}{\gamma}U_i.

Thus a plot with unspliced abundance on the horizontal axis and spliced abundance on the vertical axis should follow a line through the origin, with slope \(\beta/\gamma\), under these conditions. If both measurements use the same abundance scale, this slope is dimensionless. If the two compartments are scaled differently, the observed slope also includes that scale difference.

In real data, cells will not lie exactly on a line. A basic estimate is the ordinary least-squares slope from regressing \(S_i\) on \(U_i\), with an intercept included to assess departures from the assumed origin relationship. For cells indexed by \(i=1,\ldots,n\), the fitted slope and intercept are

\widehat{r} =\frac{\sum_{i=1}^{n}(U_i-\bar U)(S_i-\bar S)} {\sum_{i=1}^{n}(U_i-\bar U)^2}, \qquad \widehat{a}=\bar S-\widehat{r}\bar U.

Under the steady-state assumptions, \(\widehat{r}\) estimates \(\beta/\gamma\), and the intercept should be near zero. This interpretation requires meaningful variation in \(U_i\); if all cells have nearly the same unspliced abundance, the denominator is small and the slope is unstable. The regression is a practical summary, not proof that the model assumptions hold.

Worked Example: Estimating a Rate Ratio From a Phase Portrait

Suppose a hypothetical gene has the following unspliced and spliced abundances in four cells. The values are assumed to be on a common abundance scale, and the cells are provisionally treated as near steady state with shared splicing and degradation rates.

Cell	\(U_i\)	\(S_i\)
1	2	3
2	4	6
3	6	8
4	8	12

The means are \(\bar U=(2+4+6+8)/4=5\) and \(\bar S=(3+6+8+12)/4=7.25\). The cross-product sum is \(({-3})({-4.25})+({-1})({-1.25})+(1)(0.75)+(3)(4.75)=29\), and the unspliced sum of squares is \(9+1+1+9=20\). Therefore,

\widehat{r}=\frac{29}{20}=1.45,\qquad \widehat{a}=7.25-(1.45)(5)=0.

The fitted relationship is \(\widehat S=1.45U\). Under the stated assumptions, the estimated splicing-to-degradation rate ratio is \(\beta/\gamma=1.45\). This result does not say that \(\beta=1.45\) per hour, or identify either rate separately. The zero fitted intercept is consistent with the steady-state relationship, but four cells provide little basis for judging how well the assumptions hold.

When Separate Rate Estimates Are Possible

A cross-cell slope alone supplies one ratio. To calculate both absolute rates, an additional quantity on an absolute time and abundance scale is needed. For example, if \(\alpha\) is independently known for a representative steady-state condition and \(U\) and \(S\) are calibrated abundances, the undivided steady-state equations give \(\beta=\alpha/U\) and \(\gamma=\alpha/S\), provided \(U>0\) and \(S>0\). These calculations inherit the assumptions and calibration quality of all three inputs.

Worked Example: Calculating Rates With a Known Transcription Rate

Consider hypothetical calibrated steady-state abundances \(U=20\) molecules per cell and \(S=50\) molecules per cell. Suppose an independent measurement gives \(\alpha=10\) molecules per cell per hour. The positive-abundance conditions for division are satisfied, so

\beta=\frac{\alpha}{U} =\frac{10\ \text{molecules per cell per hour}}{20\ \text{molecules per cell}} =0.50\ \text{hour}^{-1},

and

\gamma=\frac{\alpha}{S} =\frac{10\ \text{molecules per cell per hour}}{50\ \text{molecules per cell}} =0.20\ \text{hour}^{-1}.

The steady-state equations check: \(\beta U=(0.50)(20)=10\) molecules per cell per hour, and \(\gamma S=(0.20)(50)=10\) molecules per cell per hour. Their ratio is \(\beta/\gamma=0.50/0.20=2.5\), which also equals \(S/U=50/20=2.5\) at steady state. Without the independent transcription-rate information and calibrated abundances, these absolute rates do not follow from the cross-cell slope alone.

Assumptions, Diagnostics, and Failure Modes

Before interpreting a slope as a kinetic rate ratio, check the conditions that connect it to the model:

Approximate steady state: the cells used for the estimate should be close enough to steady state for both balance equations to apply. A snapshot does not establish this condition.
Shared rates within the fitted group: \(\beta\) and \(\gamma\) should be reasonably common across the cells. A mixture of states with different rates need not follow one line.
Variation that reflects abundance changes: differences in transcription can generate the predicted line. Cell-specific technical effects or mismatched compartment scales can instead create or distort a relationship.
Informative range and usable measurements: there must be enough spread in \(U\), and measurement noise, zeros, and outlying cells must not dominate the fitted slope.

Inspect the scatterplot as well as the numerical slope. Curvature, separate clusters, a substantial intercept, or a few influential cells are warnings that one straight steady-state relationship may not describe the data. Compare slopes across biologically coherent subsets as a diagnostic, while recognizing that selecting subsets after looking at the plot can produce misleadingly tidy fits. Replicate information and uncertainty assessment matter: cells from the same sample are not automatically independent biological replicates.

A frequent error is to regress \(S\) on \(U\), obtain a slope of 1.45, and report a splicing rate of 1.45 per hour. The correction is that, under the model and measurement assumptions, the slope estimates \(\beta/\gamma\); its units are the ratio of the abundance scales, not inverse time. Absolute rates require additional information. Another error is to treat cell order along a trajectory as elapsed time. Pseudotime is an ordering, not a calibrated clock, so it does not by itself supply the time scale needed for rates.

Worked Example: A Pooled Slope Can Hide Different Regimes

Suppose six hypothetical cells form two apparent groups: \((U,S)=(2,2),(4,4),(6,6)\) and \((8,16),(10,20),(12,24)\). Each group lies on a line through the origin, but the first has slope 1 and the second slope 2. A pooled regression gives \(\bar U=7\), \(\bar S=12\), a cross-product sum of \(168\), and a sum of squared deviations in \(U\) of \(70\). Its slope is \(168/70=2.40\), and its intercept is \(12-(2.40)(7)=-4.80\).

The pooled line, \(\widehat S=-4.80+2.40U\), is not the relationship in either group. Its nonzero intercept and the separated patterns signal that a single shared-rate steady-state fit is questionable. Reporting 2.40 as one gene-wide rate ratio would conceal the heterogeneity rather than resolve it. The appropriate conclusion from this example is diagnostic: investigate whether the cells represent different regimes before assigning one ratio.

Choosing and Interpreting an Estimation Strategy

The cross-cell phase-portrait slope is useful when the target is a relative rate and steady-state conditions are plausible. It is simple to inspect and connects directly to the model. Ordinary least squares, however, treats \(U\) as a predictor measured without error; substantial noise in \(U\) can pull the slope toward zero. If both compartments are noisy, an errors-in-variables method may be more suitable, but it does not repair violations of steady state or mixed rates. A ratio of group means, \(\bar S/\bar U\), is another simple summary, but it is sensitive to how cells are grouped and weighted and should not be confused with a fitted regression slope when there is scatter or an intercept.

For estimating absolute rates, independently calibrated transcription or other rate information can add what an uncalibrated cross-sectional slope lacks. Such estimates are only as reliable as the external measurement, the abundance calibration, and the steady-state approximation. If those inputs are unavailable, report the estimated ratio and its assumptions rather than assigning unsupported absolute rates. A later treatment of parameter identifiability develops the broader question of which kinetic quantities the observations can distinguish.

Key takeaway: Under a shared-rate, near-steady-state model with comparable abundance scales, cross-cell variation estimates the ratio \(\beta/\gamma\) through the slope of spliced abundance on unspliced abundance. It does not, by itself, determine absolute splicing and degradation rates.

Check Your Understanding

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

At steady state, what equations hold without dividing by any kinetic rate?
A fitted steady-state phase-portrait slope is 0.80 on a common abundance scale. What quantity does it estimate, and what does it not establish?
If \(\alpha=12\) molecules per cell per hour and \(U=30\) molecules per cell at a calibrated steady state, calculate \(\beta\) and state its units.
Why can pooling cells from two groups with different phase-portrait slopes produce a misleading gene-level estimate?

Solutions. (1) The balance equations are \(\alpha=\beta U\) and \(\beta U=\gamma S\); these remain meaningful even when a rate is zero. (2) It estimates the ratio \(\beta/\gamma=0.80\), subject to the model assumptions. It does not give either absolute rate or establish that the cells are at steady state. (3) Since \(U>0\), \(\beta=\alpha/U=12/30=0.40\) hour\(^{-1}\); the abundance units cancel. (4) Each group may follow a different rate relationship, so one pooled line can have an intercept and slope that describe neither group. The pooled estimate would mask heterogeneity rather than represent either group’s ratio.