Worked Example: Transcriptional Induction and Repression

Induction and Repression as a Piecewise Kinetic Process

A transcriptional switch creates a useful example of how RNA velocity depends on the gene’s recent history. When transcription turns on, unspliced RNA can rise before spliced RNA accumulates. When transcription turns off, the remaining unspliced molecules can still be spliced, so spliced RNA may continue increasing briefly even as unspliced RNA declines. The signs of the two velocities therefore depend on both the current expression levels and the switch history.

We use the standard two-species kinetic model for one gene. Let \(u(t)\) and \(s(t)\) denote unspliced and spliced RNA abundance at time \(t\), respectively. Abundance is in arbitrary normalized units, and time is in hours. The rates \(\alpha\), \(\beta\), and \(\gamma\) denote transcription, splicing, and spliced-RNA degradation, respectively; each rate is nonnegative and has units per hour, except that \(\alpha\) has abundance units per hour.

The calculations below assume a homogeneous population trajectory for this gene, constant splicing and degradation rates, and a transcription rate that changes instantaneously between fixed levels. They also assume that \(u\) and \(s\) are on a common abundance scale and that there is no additional source or loss term, such as cell division or transport. These assumptions define an idealized deterministic trajectory; single-cell measurements need not lie exactly on it.

Deriving the Induction Trajectory

Suppose transcription switches on at time zero with constant rate \(\alpha>0\), and the gene begins with \(u(0)=s(0)=0\). During induction, the unspliced equation is a first-order linear differential equation. Multiplying it by the integrating factor \(e^{\beta t}\), integrating, and applying the initial condition gives, for \(\beta>0\), the unspliced trajectory (if \(\beta=0\), direct integration gives \(u(t)=\alpha t\)).

To obtain the spliced trajectory, solve \(\dot s+\gamma s=\beta u(t)\) using the integrating factor \(e^{\gamma t}\). Substituting the expression for \(u(t)\) and using \(s(0)=0\) yields, when \(\gamma>0\) and \(\beta\ne\gamma\),

The integral shows the mechanism: each contribution to spliced RNA comes from unspliced RNA present at an earlier time \(r\), then is reduced by degradation between \(r\) and \(t\). If \(\gamma=0\) and \(\beta>0\), the integral instead gives \(s(t)=\alpha\left[t-(1-e^{-\beta t})/\beta\right]\). If \(\beta=\gamma>0\), the expression with denominator \(\gamma-\beta\) is replaced by its limit, \(s(t)=\frac{\alpha}{\gamma}[1-(1+\gamma t)e^{-\gamma t}]\). The velocities follow directly from the model equations, rather than from a visual estimate of the curve.

After Transcription Turns Off

Now suppose transcription turns off at time \(T\), so that \(\alpha=0\) after the switch. Write \(\tau=t-T\) for elapsed time since shutdown, and let \(u_T=u(T)\) and \(s_T=s(T)\) be the abundances at the switch. These are initial conditions for the off phase. With constant \(\beta\) and \(\gamma\), solving the unspliced equation gives \(u(T+\tau)=u_Te^{-\beta\tau}\). Substituting this into the spliced equation gives, if \(\beta\ne\gamma\),

This solution separates the spliced RNA already present at shutdown, \(s_Te^{-\gamma\tau}\), from the contribution of unspliced RNA that is spliced after shutdown. In particular, the fact that \(\alpha=0\) does not imply that \(s\) immediately declines: its direction is determined by \(\dot s=\beta u-\gamma s\). Spliced abundance is momentarily stationary when \(\beta u=\gamma s\), increasing when \(\beta u>\gamma s\), and decreasing when \(\beta u<\gamma s\).

Reading the Phase Portrait

A phase portrait plots expression levels rather than time: here, \(s\) is horizontal and \(u\) is vertical. At each point, \(\dot s\) determines horizontal motion and \(\dot u\) determines vertical motion. During induction, \(\dot u=\alpha-\beta u\) is positive whenever \(u<\alpha/\beta\), and \(\dot s=\beta u-\gamma s\) is positive whenever \(s<\beta u/\gamma\). For the example’s rates, the induction equilibrium is \((s,u)=(4,2)\), where both velocities are zero. Starting from zero, the trajectory initially moves upward with no initial horizontal change because \(\dot s(0)=0\); after unspliced RNA accumulates, the spliced component becomes positive too.

After shutdown, \(\dot u=-\beta u<0\) whenever \(u>0\), so the phase-point moves downward. The sign of \(\dot s\), however, can be positive first and negative later. In the worked example, the trajectory initially moves down and right, reaches its spliced peak where \(\dot s=0\), and then moves down and left. Eventually both species approach zero. These directions describe the ideal continuous-time model, not a guarantee that every measured cell follows that exact path.

Assumptions, Diagnostics, and Common Pitfalls

A sharp switch is an idealization. Real transcription may change gradually, and rate constants may vary among cells or over time. The derivations also treat observed unspliced and spliced abundances as comparable to the model quantities. In single-cell data, sampling noise, sparse counts, scaling, isoform effects, and cell-cycle-associated changes can alter the apparent positions and directions. A phase portrait is therefore a way to interpret model-implied dynamics, not proof of a transcriptional switch or of causal regulation.

Several checks help identify when the ideal trajectory is a poor description. Under induction from zero, \(u(t)\) and \(s(t)\) must be nonnegative, and their velocities at any proposed state must agree with \(\alpha-\beta u\) and \(\beta u-\gamma s\). Under shutdown, unspliced RNA must decline when \(u>0\), but spliced RNA can rise only while \(\beta u>\gamma s\). If a fitted or plotted path repeatedly contradicts these implications, possible explanations include measurement noise, a non-instantaneous switch, incorrect rate assumptions, or a trajectory that is not adequately represented by this two-species model. Such checks diagnose tension; they do not identify its cause on their own.

The main contrast is between a time-dependent switch trajectory and a steady-state interpretation. At steady state with constant transcription, both derivatives are zero, so \(\alpha=\beta u\) and \(\beta u=\gamma s\); when \(\beta>0\) and \(\gamma>0\) this gives \(u=\alpha/\beta\) and \(s=\alpha/\gamma\), while zero-rate cases must be handled separately. A switch trajectory is generally not at that equilibrium: its position depends on elapsed time and on the previous state. Treating every point as steady state can therefore miss the delayed splicing contribution after induction or shutdown. Conversely, a visually plausible induction–repression loop does not establish that the switch model is correct; different histories or rate changes can produce similar observed paths.

The useful interpretation is local and conditional: given the model, rates, and state, the velocity signs specify the instantaneous direction of change. They do not give a probability that a cell will reach a particular fate, and they do not establish that transcription caused an observed cell-state transition. The example is most informative when the switch time, rate assumptions, and abundance scale are plausible and when the observed data show no systematic contradiction with the implied trajectories.