The Dynamical RNA Velocity Model

From a Steady-State Relation to a Time-Dependent Trajectory

A steady-state phase portrait gives a useful reference for a gene’s spliced and unspliced RNA abundances. But cells need not be at steady state: transcription may turn on or off, and the two RNA compartments respond at different rates. After a change in transcription, unspliced abundance can respond before spliced abundance does. The resulting path through the unspliced–spliced plane is a dynamical trajectory, not simply a collection of points on one steady-state line.

The model below describes a gene over an interval during which its transcriptional state is fixed. It treats abundances as continuous quantities and describes their rates of change. In a single-cell experiment, the cells are typically snapshots rather than repeated measurements of one cell; interpreting a cloud of cells as positions along a shared trajectory therefore also requires a biological assumption that those cells follow sufficiently similar kinetics.

Definition: Let $u(t)$ denote unspliced RNA abundance and $s(t)$ denote spliced RNA abundance at time $t$. The dynamical two-compartment model is $$ \frac{du}{dt}=\alpha(t)-\beta u(t), \qquad \frac{ds}{dt}=\beta u(t)-\gamma s(t). $$ Here $\alpha(t)$ is the transcription rate, $\beta$ is the splicing rate, and $\gamma$ is the spliced-RNA degradation rate. In a transcriptional on phase, $\alpha(t)=\alpha_{\mathrm{on}}>0$; in an off phase, $\alpha(t)=0$. The rates are nonnegative, with $\beta>0$ and $\gamma>0$, and $\beta$ and $\gamma$ are treated as constant within the phases considered.

The first equation balances transcription entering the unspliced compartment against splicing out of it. The second balances that splicing input against degradation of spliced RNA. Thus the velocity at a particular state $(u,s)$ depends on the state and on whether transcription is on or off. The two components of velocity are $du/dt$ and $ds/dt$, measured in abundance units per unit time.

Solutions Within an On or Off Phase

Suppose transcription is on at a constant rate $\alpha$ from time $0$, with initial abundances $u(0)=u_0$ and $s(0)=s_0$. Solving the first differential equation gives

u(t)=\frac{\alpha}{\beta}+ \left(u_0-\frac{\alpha}{\beta}\right)e^{-\beta t}.

This means that unspliced abundance approaches $\alpha/\beta$ during a sustained on phase. The spliced compartment responds to the changing unspliced abundance. Solving its linear differential equation using an integrating factor gives the following expression, which also works when $\beta=\gamma$:

s(t)=s_0e^{-\gamma t} +\beta\int_0^t e^{-\gamma(t-r)}u(r)\,dr.

The integral adds up the spliced RNA contributed by unspliced RNA at earlier times, with each contribution discounted by degradation since it entered the spliced compartment. If $\beta\ne\gamma$, evaluating the integral gives

s(t)=s_0e^{-\gamma t} +\frac{\alpha(1-e^{-\gamma t})}{\gamma} +\beta\left(u_0-\frac{\alpha}{\beta}\right) \frac{e^{-\beta t}-e^{-\gamma t}}{\gamma-\beta}.

If transcription switches off at a particular state $(u_0,s_0)$, let $r$ denote elapsed time since that switch. Now $\alpha=0$, so

u(r)=u_0e^{-\beta r}, \qquad s(r)=s_0e^{-\gamma r} +\beta u_0\frac{e^{-\beta r}-e^{-\gamma r}}{\gamma-\beta} \quad (\beta\ne\gamma).

The off-phase formula does not say that spliced abundance immediately decreases. Splicing of the unspliced RNA already present continues after transcription stops, and can temporarily exceed the loss due to degradation. When the two rates are equal, the off-phase expression for $s$ is $e^{-\beta r}(s_0+\beta u_0r)$, obtained by integrating the same differential equation rather than dividing by $\gamma-\beta$.

Worked Example: Following an On Phase Away From Steady State

Consider hypothetical abundances measured in arbitrary abundance units, with $\alpha=4$ units per hour, $\beta=2$ per hour, and $\gamma=1$ per hour. Suppose both compartments start at zero and transcription remains on for one hour. These rates are positive and constant over the stated interval, as required by the on-phase model.

For these initial conditions, the on-phase solutions simplify to $u(t)=2(1-e^{-2t})$ and $s(t)=4(1-e^{-t})^2$. At $t=1$ hour,

u(1)=2(1-e^{-2})=2(1-0.135335)=1.729

and

s(1)=4(1-e^{-1})^2 =4(1-0.367879)^2 =1.598.

The instantaneous derivatives at that time are

\frac{du}{dt}=4-2(1.72933)\approx0.5413, \qquad \frac{ds}{dt}=2(1.72933)-1.59830\approx1.860.

Both abundances are still increasing. Under sustained transcription, the equilibrium is $u=\alpha/\beta=2$ and $s=\alpha/\gamma=4$, but the one-hour state has not reached it. In particular, it does not lie on the steady-state relation $s=(\beta/\gamma)u=2u$: at $u=1.729$, that line gives $s=3.459$, whereas the trajectory has $s=1.598$. The difference reflects transient accumulation, not an arithmetic contradiction.

Why the Trajectory Differs From a Steady-State Phase Portrait

At steady state, both derivatives equal zero. During an on phase, solving those two conditions gives

u_{\mathrm{eq}}=\frac{\alpha}{\beta}, \qquad s_{\mathrm{eq}}=\frac{\alpha}{\gamma}.

The steady-state phase-portrait relation follows by setting only $ds/dt=0$, which gives $s=(\beta/\gamma)u$. It describes states where spliced abundance is momentarily not changing. It does not describe every state visited during a transient. Along a dynamical trajectory, both coordinates generally change, so there is no reason for the trajectory to remain on that relation.

For a smooth trajectory with $du/dt\ne0$, its local slope in the phase plane is obtained by dividing the two derivatives:

\frac{ds}{du} =\frac{ds/dt}{du/dt} =\frac{\beta u-\gamma s}{\alpha-\beta u}.

This slope depends on the current state as well as the transcriptional phase. If $du/dt=0$ but $ds/dt\ne0$, the trajectory has a vertical tangent: it is moving in the $s$ direction while $u$ is instantaneously unchanged. If both derivatives are zero, the state is an equilibrium for that constant-transcription phase; there is no instantaneous direction of motion at that point. In particular, the ratio above must not be used to call an equilibrium a vertical tangent.

Worked Example: Reading Motion at and Near a Vertical Tangent

Suppose transcription is on with $\alpha=2$ units per hour, $\beta=1$ per hour, and $\gamma=1$ per hour. At $u=2$ and $s=1$, these rates are assumed constant and nonnegative, so the model applies. The derivatives are

\frac{du}{dt}=2-1(2)=0, \qquad \frac{ds}{dt}=1(2)-1(1)=1.

Thus $u$ is momentarily unchanged while $s$ increases at 1 abundance unit per hour: the phase-plane trajectory has a vertical tangent there. The slope formula has a zero denominator and nonzero numerator, consistent with that interpretation.

At $u=2$, $s=2$, the same rates instead give $du/dt=0$ and $ds/dt=0$. This is the on-phase equilibrium. The slope ratio is again undefined, but the system has no instantaneous motion there; calling it a vertical tangent would be wrong.

Worked Example: Spliced RNA Can Rise After Transcription Turns Off

Continue the hypothetical example with $\alpha=4$, $\beta=2$, and $\gamma=1$ per hour. After one hour on, the state is $u_0=1.72933$, $s_0=1.59830$. Now set transcription to zero and follow the system for one more hour. The off-phase assumptions are that transcription stops at the switch and the two kinetic rates stay fixed.

Using the off-phase solution, with $e^{-2}=0.135335$ and $e^{-1}=0.367879$,

u(1)=1.72933e^{-2}=0.2340.

For spliced abundance,

\begin{aligned} s(1) &=1.59830e^{-1} +2(1.72933)\frac{e^{-2}-e^{-1}}{1-2}\\ &\approx0.5880+0.8043\\ &\approx1.392. \end{aligned}

Spliced abundance has declined from about 1.598 to 1.392 units, but it remains far above the unspliced abundance. The off-phase derivative at the end is $ds/dt=2(0.2340)-1.392\approx-0.924$ units per hour, so it is then decreasing. The continued splicing term explains why the off-phase path need not immediately follow a steady-state relation.

Assumptions, Diagnostics, and Limits

The equations define a compact kinetic model, not a guarantee that every observed cell cloud follows one idealized path. Their interpretation relies on several assumptions: transcription can be represented by constant on and off levels within intervals; the splicing and degradation rates remain constant within those intervals; the two compartments are adequately represented by continuous abundances; and cells being compared can reasonably be viewed as sharing a kinetic program. The model also omits additional processes that might alter measured abundances, such as cell-specific rate variation or unmodeled transcript processing.

Useful checks are therefore both mathematical and biological. A proposed trajectory should have nonnegative abundances, and its directions should agree with the equations: in an off phase, $u$ decreases whenever $\beta>0$ and $u>0$ (and is constant when $\beta=0$), while $s$ increases only when $\beta u>\gamma s$. A trajectory that implies increasing unspliced RNA while transcription is off, or assigns the wrong sign to $ds/dt$ given the stated state and rates, signals an inconsistency in the model calculation or interpretation. In data, large departures from the predicted shape may instead indicate that the assumptions do not adequately describe the cells.

The main alternative in this context is a steady-state phase portrait. It is simpler and appropriate when the relevant cells are plausibly near a sustained transcriptional equilibrium. The dynamical model is more informative when cells may be in transient phases, but it requires assumptions about how transcription changes over time and about shared kinetics. Neither representation establishes a causal developmental history from a snapshot alone. A dynamical curve describes consequences of the specified equations; deciding whether real cells follow that curve requires evidence beyond the equations themselves.

Common pitfall: A point away from $s=(\beta/\gamma)u$ is not automatically a bad observation or a failed model. The line is a steady-state relation for the spliced compartment; transient states can lie away from it. Conversely, an undefined slope ratio alone does not establish vertical motion: check whether $ds/dt$ is nonzero. If both derivatives vanish, the point is an equilibrium for that phase.

Specify the phase.
Set transcription to its constant on-state rate or to zero for the off phase being considered.

Compute the velocity components.
Evaluate $du/dt=\alpha-\beta u$ and $ds/dt=\beta u-\gamma s$ at the state of interest.

Interpret the path with care.
Use the signs of the derivatives for direction; use their ratio for phase-plane slope only when $du/dt\ne0$.

Check Your Understanding

Try each question, then compare your reasoning with the solutions below.

During an off phase, $u=3$, $s=2$, $\beta=1$ per hour, and $\gamma=1$ per hour. What are the two derivatives, and which abundance is changing faster in units per hour?
For an on phase with $\alpha=6$, $\beta=2$, and $\gamma=3$, find the equilibrium abundances.
At a point where $du/dt=0$ and $ds/dt<0$, what is the phase-plane interpretation?
Why can a cell state off the steady-state line $s=(\beta/\gamma)u$ still be consistent with the dynamical model?

Solutions. (1) In the off phase, $du/dt=-\beta u=-3$ units per hour and $ds/dt=\beta u-\gamma s=3-2=1$ unit per hour. Unspliced abundance decreases, while spliced abundance increases; the unspliced derivative has the larger absolute magnitude. (2) The on-phase equilibrium is $u=\alpha/\beta=6/2=3$ and $s=\alpha/\gamma=6/3=2$ abundance units. (3) Since $du/dt=0$ but $ds/dt\ne0$, the trajectory has a vertical tangent and is moving downward in $s$. (4) The steady-state line describes states with $ds/dt=0$, whereas a transient state can have $ds/dt\ne0$ and lie away from that line.

The Dynamical RNA Velocity Model

What You'll Learn