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RNA Velocity · Tutorial 386 of 700

Steady-State RNA Velocity

Use the two-compartment model to derive the zero-velocity relationship between unspliced and spliced RNA and interpret deviations from it cautiously.

Advanced 11 min read

What You'll Learn

  • Derive the unspliced-to-spliced abundance ratio at zero spliced-RNA velocity.
  • Distinguish the spliced-RNA zero-velocity relation from the full kinetic equilibrium.
  • Calculate velocity signs and magnitudes from hypothetical RNA abundances and rates.
  • Identify assumptions and measurement issues that can undermine a steady-state interpretation.
  • Compare a steady-state sign inference with conclusions that require a fuller dynamical model.

Why a Steady-State Relationship Is Useful

Single-cell RNA sequencing records RNA abundances at sampled cells, not the time derivative of RNA abundance. RNA velocity uses a kinetic model to interpret the balance between unspliced and spliced RNA as evidence about whether spliced RNA is increasing or decreasing. The steady-state relationship gives a simple reference for that interpretation: for fixed splicing and degradation rates, it identifies combinations of unspliced and spliced abundance at which spliced RNA has zero instantaneous velocity.

This relationship is useful only with a clear account of what “steady state” means. The condition for spliced-RNA velocity to be zero is not, by itself, the condition that the entire two-compartment system is at equilibrium. That distinction matters when using a gene’s position relative to the relationship to infer a direction of change.

Model and Steady-State Relationship

Let \(u(t)\) denote unspliced RNA abundance and \(s(t)\) denote spliced RNA abundance for one gene at time \(t\). In the two-compartment kinetic model, transcription adds unspliced RNA, splicing transfers RNA from the unspliced to the spliced compartment, and degradation removes spliced RNA. With transcription rate \(\alpha\), splicing rate \(\beta\), and spliced-RNA degradation rate \(\gamma\), the model is

$$ \frac{du}{dt}=\alpha-\beta u,\qquad \frac{ds}{dt}=\beta u-\gamma s. $$

The rates \(\beta\) and \(\gamma\) are positive and have units of inverse time. The abundances \(u\) and \(s\) have the same abundance units, so each term in either derivative has units of abundance per time. For the sign interpretation below, assume the model describes the gene in the cell of interest, and that the same positive \(\beta\) and \(\gamma\) apply to the compared cells. The equations describe instantaneous rates; they do not require that transcription be constant over all time.

Definition: The spliced-RNA velocity is \(v_s=ds/dt\), the instantaneous rate of change in spliced RNA abundance. The zero-spliced-velocity relation is the set of \(u,s\) values for which \(v_s=0\). Under the two-compartment model with \(\beta,\gamma>0\), it is \(u=(\gamma/\beta)s\).

To derive the relation, set the spliced-RNA derivative in the model equal to zero:

$$ 0=\frac{ds}{dt}=\beta u-\gamma s \quad\Longrightarrow\quad u=\frac{\gamma}{\beta}s \quad\text{or, equivalently,}\quad s=\frac{\beta}{\gamma}u. $$

This is a linear relationship through the origin, with slope \(\gamma/\beta\) when \(u\) is expressed as a function of \(s\). More importantly, subtracting its right-hand side from the observed \(u\) gives a direct expression for the velocity:

$$ v_s=\beta u-\gamma s =\beta\left(u-\frac{\gamma}{\beta}s\right). $$

Because \(\beta>0\), the sign of \(v_s\) is the sign of the difference \(u-(\gamma/\beta)s\). If \(u\) exceeds the value on the zero-velocity relation for the given \(s\), then \(v_s>0\). If it is below that value, then \(v_s<0\). On the relation, \(v_s=0\). This conclusion follows from the assumed model; it is not a causal claim about what would happen if RNA abundances were experimentally changed.

Zero Spliced Velocity Is Not Necessarily Full Equilibrium

Full equilibrium requires both compartments to have zero derivative. If \(\alpha\) is constant and \(\beta,\gamma>0\), the equilibrium equations give

$$ 0=\alpha-\beta u^* \quad\Longrightarrow\quad u^*=\frac{\alpha}{\beta}, \qquad 0=\beta u^*-\gamma s^* \quad\Longrightarrow\quad s^*=\frac{\alpha}{\gamma}. $$

The equilibrium point \((u^*,s^*)\) lies on the zero-spliced-velocity relation, but the relation contains other points too. At one of those other points, spliced RNA can momentarily have zero velocity while unspliced RNA is changing. Thus, “on the steady-state line” is shorthand for being on the spliced-RNA zero-velocity relation; it does not establish that the cell is at full kinetic equilibrium.

Worked Example: A Zero Spliced Velocity Without Full Equilibrium

Suppose a hypothetical cell has \(u=20\) and \(s=40\) abundance units per cell, with \(\beta=0.50\ \text{h}^{-1}\), \(\gamma=0.25\ \text{h}^{-1}\), and \(\alpha=15\) abundance units per cell per hour. Calculate both derivatives.

First calculate the spliced-RNA velocity:

$$ v_s=\beta u-\gamma s =(0.50)(20)-(0.25)(40) =10-10=0 \quad\text{abundance units per cell per hour}. $$

The cell is on the zero-spliced-velocity relation: \((\gamma/\beta)s=(0.25/0.50)(40)=20=u\). But the unspliced-RNA derivative is

$$ \frac{du}{dt}=\alpha-\beta u =15-(0.50)(20)=5 \quad\text{abundance units per cell per hour}. $$

Unspliced RNA is increasing even though spliced RNA has zero instantaneous velocity. The state is therefore not at full equilibrium. This example illustrates why the relation identifies the sign of the spliced-RNA derivative, not whether every part of the system has stopped changing.

Interpreting Deviations With Numerical Examples

The next examples use hypothetical abundances and rates. They assume that \(u\) and \(s\) are measured on a compatible abundance scale, and that the rates are positive and applicable to the cell being evaluated. A positive velocity means spliced abundance is increasing instantaneously under the model; a negative velocity means it is decreasing. It does not say how long that trend will continue.

Worked Example: Unspliced Abundance Above the Relation

Suppose a hypothetical cell has \(u=30\) and \(s=40\) abundance units per cell. Let \(\beta=0.40\ \text{h}^{-1}\) and \(\gamma=0.20\ \text{h}^{-1}\). Determine the sign and magnitude of \(v_s\).

At \(s=40\), the unspliced abundance on the zero-velocity relation is

$$ \frac{\gamma}{\beta}s =\frac{0.20}{0.40}(40) =20 \quad\text{abundance units per cell}. $$

The observed \(u=30\) exceeds 20, so the model predicts positive velocity. Direct substitution gives the magnitude:

$$ v_s=\beta u-\gamma s =(0.40)(30)-(0.20)(40) =12-8 =4 \quad\text{abundance units per cell per hour}. $$

Under the stated model, spliced RNA is increasing instantaneously at 4 abundance units per cell per hour. The sign agrees with the comparison to the relation.

Worked Example: Unspliced Abundance Below the Relation

In another hypothetical cell, let \(u=24\), \(s=20\), \(\beta=0.25\ \text{h}^{-1}\), and \(\gamma=0.50\ \text{h}^{-1}\). Find the zero-velocity value of \(u\) for this \(s\), then calculate the velocity.

The reference value is

$$ \frac{\gamma}{\beta}s =\frac{0.50}{0.25}(20) =40 \quad\text{abundance units per cell}. $$

Here \(u=24\) is below 40, suggesting a negative velocity. Substituting in the model confirms the sign:

$$ v_s=(0.25)(24)-(0.50)(20) =6-10 =-4 \quad\text{abundance units per cell per hour}. $$

The model predicts that spliced RNA is decreasing instantaneously at 4 abundance units per cell per hour. The negative sign does not mean that the spliced abundance itself is negative; \(s\) remains 20 abundance units per cell.

Assumptions, Diagnostics, and Limitations

The sign rule is exact for the stated kinetic equation, but its application to single-cell measurements depends on whether that equation and its inputs are credible. In particular, the mathematical sign is determined by the true abundances and applicable rates. Sequencing measurements are noisy observations of RNA abundance, so a small apparent deviation from a reference relation does not guarantee that the true velocity has the same sign.

  • Check the abundance scale. The \(u\) and \(s\) values must be expressed on compatible scales. If one compartment is transformed differently from the other, the numerical relation \(u=(\gamma/\beta)s\) cannot be applied without accounting for that transformation.
  • Check the rate assumptions. The sign result uses positive \(\beta\) and \(\gamma\) and assumes the rates are appropriate for the cell being interpreted. If rates vary across cells or the model omits relevant processes, one shared reference relation may not describe all cells.
  • Consider measurement noise and sparse counts. When counts are low, sampling variability can place a measured point on either side of the relation. A very small difference should not be treated as strong directional evidence without considering uncertainty.
  • Do not infer duration from an instantaneous derivative. A positive \(v_s\) describes the local rate at the evaluated state. Changes in transcription, splicing, degradation, or abundance can alter that rate later.
  • Do not equate a zero velocity with no system change. The worked example showed that \(ds/dt=0\) can coexist with \(du/dt\ne 0\).

The steady-state sign rule is more specific than simply ordering cells by an inferred trajectory: it uses the model’s balance of splicing input and degradation to interpret a gene’s spliced-RNA derivative. Its simplicity is also its limitation. A fuller dynamical approach can represent changes over time in the kinetic process, but a sign inferred from this relation alone does not establish the history that produced the observed abundances. The appropriate interpretation is therefore conditional: if the two-compartment model and its rates are adequate for the cell, the deviation gives the sign of instantaneous spliced-RNA velocity.

Common Mistake

Common mistake: “The cell lies on the steady-state relation, so the entire gene-expression system is at equilibrium.” This reasoning confuses \(ds/dt=0\) with the stronger requirement that both \(du/dt=0\) and \(ds/dt=0\). The correction is to call the line the zero-spliced-velocity relation and check the unspliced derivative separately before claiming full equilibrium.

A second error is to read a positive velocity as evidence that spliced RNA abundance is large. Velocity is a rate of change, not an abundance: a cell can have low \(s\) with positive \(v_s\), or high \(s\) with negative \(v_s\). Always report the sign and, when calculated, the abundance-per-time units of the derivative.

Check Your Understanding

Answer each question using the two-compartment model. Solutions follow the questions.

  1. For \(\beta=0.30\ \text{h}^{-1}\), \(\gamma=0.15\ \text{h}^{-1}\), and \(s=50\) abundance units per cell, what value of \(u\) lies on the zero-spliced-velocity relation?
  2. Using the rates in Question 1, calculate \(v_s\) when \(u=20\) and \(s=50\) abundance units per cell. Give its sign and units.
  3. Why does a point on the zero-spliced-velocity relation not necessarily represent full equilibrium?
  4. Name two assumptions or measurement issues that could make the sign inferred from observed abundances unreliable.

Solutions. (1) \(u=(\gamma/\beta)s=(0.15/0.30)(50)=25\) abundance units per cell. (2) \(v_s=(0.30)(20)-(0.15)(50)=6-7.5=-1.5\) abundance units per cell per hour, so the model predicts decreasing spliced RNA instantaneously. (3) The relation sets only \(ds/dt=0\); full equilibrium also requires \(du/dt=0\). (4) For example, noisy sparse counts can obscure which side of the relation a cell occupies, and incompatible scaling of \(u\) and \(s\) makes the comparison invalid. A shared rate relation may also fail if kinetic rates differ across the cells being compared.

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