The Steady-State Velocity Model

From RNA Kinetics to a Phase Portrait

RNA velocity uses the balance between unspliced and spliced RNA to estimate whether a cell’s spliced-RNA abundance is increasing or decreasing. The steady-state version makes a simplifying assumption: the cells being analyzed share a common kinetic relationship between these two RNA compartments. Under that assumption, cells near equilibrium lie along a line in a plot of unspliced abundance against spliced abundance, and a cell’s deviation from that line provides a directional signal.

For one gene, let \(u\) denote unspliced RNA abundance and \(s\) spliced RNA abundance. The two-compartment kinetic model describes transcription, splicing, and degradation with rates \(\alpha\), \(\beta\), and \(\gamma\), respectively. In this tutorial, \(u\) and \(s\) are abundance measurements on comparable scales; \(\alpha\) has units of abundance per time, whereas \(\beta\) and \(\gamma\) have units of inverse time.

Definition: The steady-state RNA velocity model uses the kinetic equations \(du/dt=\alpha-\beta u\) and \(ds/dt=\beta u-\gamma s\). A steady state is a point at which both derivatives equal zero. Its phase portrait plots spliced abundance \(s\) on the horizontal axis and unspliced abundance \(u\) on the vertical axis.

The equations express a flow of RNA through two compartments. Transcription adds unspliced RNA at rate \(\alpha\); splicing removes it at rate \(\beta u\) and adds spliced RNA at the same rate; degradation removes spliced RNA at rate \(\gamma s\). A steady state is not merely a cell with a particular observed count pair: it is a kinetic balance in which these inflows and outflows cancel.

The Steady-State Line and Its Interpretation

At steady state, \(du/dt=0\) implies \(\alpha=\beta u\), while \(ds/dt=0\) implies \(\beta u=\gamma s\). Combining the latter balance with the first gives the phase-portrait relationship.

\beta u=\gamma s \quad\Longrightarrow\quad u=a s,\qquad a=\frac{\gamma}{\beta}.

Thus the ideal steady-state phase portrait is a line through the origin. Its slope \(a\) is the degradation-to-splicing rate ratio, not either rate separately. Different transcription rates can move equilibrium cells to different positions along this line: for positive \(\alpha\), the equilibrium abundances are \(u^*=\alpha/\beta\) and \(s^*=\alpha/\gamma\), and their ratio remains \(u^*/s^*=\gamma/\beta\) if \(\beta\) and \(\gamma\) are shared.

In observed data, a fitted line is used in place of the ideal relationship. For example, a least-squares fit constrained through the origin has slope \(\hat a=\sum_i s_i u_i/\sum_i s_i^2\), where the cells indexed by \(i\) are selected to represent the common kinetic regime. The line \(u=\hat a s\) is the fitted steady-state phase portrait. Choosing cells and fitting that line require care; the formula does not make an inappropriate cell set representative of the model.

Key relationship: The residual \(r_i=u_i-\hat a s_i\) is the vertical displacement of cell \(i\) from the fitted phase-portrait line. Under the common-rate steady-state model, it estimates the sign of the cell’s spliced-RNA velocity: positive residuals indicate increasing \(s\), and negative residuals indicate decreasing \(s\). It is a directional estimate, not automatically a velocity in abundance-per-time units.

The interpretation follows directly from the spliced-RNA equation. If the fitted slope approximates \(a=\gamma/\beta\), then \(ds/dt=\beta(u-a s)\), so the true spliced-RNA velocity is approximately \(\beta r_i\). Since \(\beta\) is positive, multiplying by it changes the residual’s scale but not its sign. If \(\beta\) is not known, the residual alone gives a relative directional signal, not a calibrated rate of change.

Why the Residual Has a Directional Meaning

The residual argument depends on a specific kinetic comparison: for the cell’s current \(s\), the fitted line predicts the unspliced abundance that would balance splicing and degradation under the shared regime. An observed \(u\) above that prediction means the splicing inflow \(\beta u\) exceeds the degradation outflow \(\gamma s\); below it, the inflow is smaller.

u=\hat a s+r \quad\text{and, if }\hat a\approx\frac{\gamma}{\beta},\quad \frac{ds}{dt}=\beta u-\gamma s\approx\beta r.

This is a local model-based interpretation. A positive residual does not mean that the cell has more spliced RNA than another cell, nor does it establish that the cell will move to a particular future state. It means that, conditional on the model and fitted line, the current balance predicts net production of spliced RNA. The estimate is gene-specific; it does not by itself describe an entire cell-state trajectory.

Worked Example: Converting a Residual into a Direction

Suppose a hypothetical reference set from one gene gives the phase-portrait points \((s,u)=(2,1),(4,2),(6,3)\), in normalized abundance units. These points are consistent with a through-origin line. For a separate focal cell, let \(s=4\) and \(u=2.6\). Estimate its residual-based direction.

For the reference set, the through-origin slope is \(\hat a=[(2)(1)+(4)(2)+(6)(3)]/[2^2+4^2+6^2]=28/56=0.5\). The fitted line predicts \(u=0.5(4)=2\) for the focal cell. Its residual is \(r=2.6-2=0.6\) normalized abundance units.

The residual is positive, so the model indicates that spliced abundance is increasing. If, hypothetically, the splicing rate were \(\beta=0.4\) per unit time and the fitted slope equaled the true rate ratio, the corresponding model velocity would be \(ds/dt\approx\beta r=(0.4)(0.6)=0.24\) normalized abundance units per unit time. Without that rate calibration, the justified conclusion is the positive direction, not the numerical speed.

The model assumptions are plausible for this calculation only insofar as the reference cells represent the same kinetic regime as the focal cell, the abundance scales are comparable, and the reference relationship is adequately described by the line. The exact alignment of the reference points supports the line form in this small illustration but cannot establish those assumptions for biological data.

Assumptions About the Cells Being Compared

A fitted steady-state line is informative only when the cells used to estimate and interpret it can reasonably be treated as sampling a common kinetic regime. This does not require identical transcription rates or identical abundance values. It does require that the kinetic relationship linking \(u\) and \(s\), especially the rate ratio \(\gamma/\beta\), be sufficiently shared for one line to serve as a reference.

Comparable kinetic rates: Splicing and degradation rates, or at least their ratio, should not differ so much across included cells that a single slope ceases to describe them.
Comparable measurement scale: The unspliced and spliced measurements must be processed consistently across cells. Gene-specific offsets, library effects, or incompatible normalization can displace observations independently of their kinetic state.
Meaningful reference cells: Cells used to define the line should provide a reasonable representation of the steady-state relationship. A set dominated by cells in a transient or distinct regime can define the wrong reference.
A suitable relationship: A straight line through the origin is implied by the ideal model without additive offsets. Strong systematic curvature or a nonzero baseline is evidence that this simple reference may not describe the data.

These assumptions are not guaranteed just because a phase portrait looks roughly linear. Measurement noise can obscure structure, and a visually plausible pooled line can conceal groups with different slopes. The steady-state estimate is therefore conditional on the kinetic regime and data scale, rather than a direct observation of motion.

Worked Example: Pooling Two Kinetic Regimes

Consider four hypothetical cells from two regimes. Regime A has steady-state points \((s,u)=(10,5)\) and \((20,10)\); regime B has \((10,10)\) and \((20,20)\). Each regime has its own exact steady-state line, but the slopes differ. Calculate a single through-origin fit to all four cells and examine its residuals.

The pooled slope is \(\hat a=[(10)(5)+(20)(10)+(10)(10)+(20)(20)]/[10^2+20^2+10^2+20^2]\). The numerator is \(50+200+100+400=750\), and the denominator is \(100+400+100+400=1000\). Therefore \(\hat a=750/1000=0.75\).

For the regime A cell with \(s=20,u=10\), the pooled residual is \(10-0.75(20)=-5\). For the regime B cell with \(s=20,u=20\), it is \(20-0.75(20)=5\). The other two residuals are \(5-0.75(10)=-2.5\) for regime A and \(10-0.75(10)=2.5\) for regime B.

All four cells are exactly at steady state within their own regimes, yet the pooled line gives nonzero residuals and opposite apparent directions for the two regimes. Here the arithmetic is consistent with the data, but the pooled residuals do not represent genuine departures from a shared steady-state balance. The calculation illustrates why the common-regime assumption is substantive, not a technical detail.

Diagnostics, Alternatives, and Limits

A useful first diagnostic is to inspect the phase portrait for systematic departures from the fitted line rather than treating every residual as a biological signal. Patterns among residuals can indicate that the straight-line reference is missing structure. For instance, residuals that are predominantly negative in one region and positive in another may reflect a curved relationship, a mixture of regimes, or a poorly chosen reference set. A broad scatter around the line may instead reflect noisy abundance measurements. These patterns do not identify a unique cause, but they warn against interpreting residual signs mechanically.

The most direct alternative to one pooled steady-state line is to analyze cells within more homogeneous kinetic groups, if such groups can be justified. This can reduce the regime-mixing problem, but dividing cells into groups does not guarantee that the shared-rate assumption holds within each group. A model that explicitly represents a time-varying transcriptional process is another conceptual alternative when the steady-state assumption is inappropriate. It requires different assumptions and should not be treated as interchangeable with the steady-state residual.

The residual method is limited in what it estimates. Its sign is meaningful only under the kinetic and measurement assumptions; its magnitude depends on the scale of the abundance data and does not, by itself, give an absolute speed. Even when the assumptions are reasonable, a positive residual is evidence for a model-implied direction in spliced abundance, not proof of a future change in cell identity or a causal explanation of cell-state progression.

Worked Example: Recognizing a Background Offset

Suppose a hypothetical gene has observed points \((s,u)=(2,3),(4,4),(6,5)\), which follow \(u=2+0.5s\). Calculate the through-origin slope and assess what the resulting residuals imply.

The through-origin slope is \(\hat a=[(2)(3)+(4)(4)+(6)(5)]/[2^2+4^2+6^2]=(6+16+30)/56=52/56\approx0.9286\). The residuals are \(3-0.9286(2)\approx1.1429\), \(4-0.9286(4)\approx0.2857\), and \(5-0.9286(6)\approx-0.5714\) abundance units.

The data have a shared additive offset of 2, so they do not follow the ideal through-origin relationship. The fitted through-origin residuals make the first two points look directionally positive and the third negative, even though the points lie exactly on one offset line. This is a warning to investigate background or measurement effects before assigning a kinetic interpretation. An intercept fit would describe the observed straight line, but the intercept would not, by itself, prove a biological mechanism or validate a residual-based velocity model.

Common Mistake

A common error is to calculate residuals from a pooled line and interpret every nonzero value as evidence that individual cells are moving away from steady state. That reasoning assumes the pooled line is the correct kinetic reference. In the two-regime example, the pooled slope is \(0.75\), even though regime A has slope \(0.5\) and regime B has slope \(1\); cells at equilibrium within their own regimes consequently receive residuals of opposite signs. The correction is to check whether a common kinetic relationship is defensible before interpreting residuals. A residual is a model-based directional estimate, not an unconditional measurement of motion.

Key takeaway: In the steady-state model, the phase-portrait slope estimates \(\gamma/\beta\), and \(u-\hat a s\) estimates the sign of spliced-RNA velocity when a common kinetic regime and comparable measurement scale are plausible. Regime mixing, offsets, and model misfit can create residual directions that are not genuine cell-level departures from steady state.

Check Your Understanding

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

In the steady-state phase portrait, which abundance is on each axis, and what rate ratio does the slope represent?
A cell has \(s=8\), \(u=6\), and the fitted slope is \(0.5\). Calculate its residual and state the model-implied direction of spliced-RNA change.
Why can two groups of cells that are each at steady state show nonzero residuals relative to one pooled line?
Does a residual of \(0.6\) abundance units imply a spliced-RNA velocity of \(0.6\) abundance units per time? Explain.
What concern is raised if points follow a line with a substantial nonzero intercept while the fitted steady-state model forces the line through the origin?

Solutions. (1) Spliced abundance \(s\) is on the horizontal axis and unspliced abundance \(u\) on the vertical axis; the slope is \(\gamma/\beta\). (2) \(r=6-0.5(8)=2\) abundance units, so the model implies increasing spliced abundance, conditional on its assumptions. (3) The groups may have different rate ratios, so the pooled line need not represent either group’s equilibrium relationship. (4) No. The residual gives a directional signal; the model velocity is approximately \(\beta r\), and requires a calibrated splicing rate to have abundance-per-time units. (5) An additive background or another model mismatch may be present, making through-origin residuals misleading as directional evidence.