Choosing Between Steady-State and Dynamical Models

Why the Model Choice Matters

RNA velocity uses the balance between unspliced and spliced RNA to estimate the local rate of change of spliced abundance. The steady-state and dynamical approaches make different assumptions about how cells arrived at their observed expression levels. That difference affects what the fitted model can represent: a steady-state model describes deviations from an equilibrium relationship, while a dynamical model can represent trajectories through transcriptional induction and repression.

The choice is not simply between a “simple” and a “better” model. A steady-state approximation can be useful when the observed phase portrait is consistent with a stable equilibrium relationship. A dynamical model can be more appropriate when there is evidence of a transition between transcriptional regimes and enough information to estimate that transition. More flexibility alone is not evidence that a gene truly switched transcription.

What the Two Models Assume

Let \(u(t)\) denote unspliced RNA abundance and \(s(t)\) denote spliced RNA abundance for a gene at time \(t\). In the two-compartment kinetic model, transcription produces unspliced RNA at rate \(\alpha(t)\), splicing moves it into the spliced compartment at rate \(\beta u(t)\), and degradation removes spliced RNA at rate \(\gamma s(t)\):

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

Here \(\beta\) and \(\gamma\) are positive kinetic rates. The steady-state approach uses the equilibrium relationship as a reference across observed cells. The dynamical approach instead fits a trajectory under a specified transcriptional history, often with transcription switching between regimes. Both approaches rely on measurements that are comparable across cells and on a kinetic model that is a reasonable approximation for the gene and population being analyzed.

Definition: A steady-state velocity model treats the equilibrium relationship between \(u\) and \(s\) as a reference for the observed cells. A dynamical velocity model fits the time-dependent kinetic equations, allowing transcription to change between regimes and estimating the resulting trajectory. The first primarily uses deviation from an equilibrium relation; the second represents departures from equilibrium as part of a fitted temporal process.

The steady-state assumption is not that every measured cell has exactly zero velocity. Instead, it assumes that a shared equilibrium relation is a useful reference for the cells being compared. Its interpretation becomes questionable if cells occupy distinct phases of a transient trajectory that cannot be represented as deviations around that relation.

Deriving the Equilibrium Reference

At equilibrium, both compartments have zero rate of change. Setting the two derivatives to zero gives \(\alpha=\beta u\) and \(\beta u=\gamma s\). Therefore, the equilibrium phase-portrait relationship is a line through the origin:

u=\frac{\gamma}{\beta}s.

This relation also explains the steady-state directional signal. The spliced-RNA velocity is \(ds/dt=\beta u-\gamma s\). If the equilibrium ratio is \(m=\gamma/\beta\), then \(ds/dt=\beta(u-ms)\). A point above the line has positive modeled spliced velocity; a point below it has negative modeled spliced velocity, provided the rate and abundance scales are meaningful and consistent with the model.

The dynamical model does not assume that cells lie near one equilibrium line. For example, when transcription switches off, unspliced RNA can fall before spliced RNA does. The resulting path can pass through regions of the phase portrait that a single equilibrium reference cannot describe as a stable relationship.

Worked Example: Checking a Steady-State Phase-Portrait Fit

Suppose four hypothetical cells have spliced abundances \(s=2,4,6,8\) and corresponding unspliced abundances \(u=1.0,2.1,2.9,4.1\), in normalized abundance units. Assume a steady-state line through the origin is appropriate for this initial check. Its least-squares slope is \(\sum s_i u_i/\sum s_i^2\).

\hat{m} = \frac{(2)(1.0)+(4)(2.1)+(6)(2.9)+(8)(4.1)} {2^2+4^2+6^2+8^2} = \frac{60.6}{120} = 0.505.

The fitted unspliced abundance is \(\hat{u}=0.505s\). The residuals, calculated as observed \(u\) minus fitted \(u\), are \(-0.01,\ 0.08,\ -0.13,\ 0.06\) abundance units. For example, at \(s=6\), the fitted value is \(0.505(6)=3.03\), so the residual is \(2.9-3.03=-0.13\).

If the rate ratio were \(\gamma/\beta=0.5\), the fitted slope of \(0.505\) would be close to that ratio. These small residuals show no obvious separate induction and repression branches in this tiny hypothetical data set. They do not, by themselves, establish that the biological system is at equilibrium: the data are few, and a larger phase portrait and other evidence would be needed.

How a Transcriptional Switch Produces a Trajectory

A dynamical model is useful when a change in transcription can explain the observed path. Consider a hypothetical gene for which transcription switches off at time zero. Suppose \(u(0)=4\), \(s(0)=4\), \(\beta=0.5\) per hour, and \(\gamma=0.25\) per hour. After the switch, set \(\alpha(t)=0\). The unspliced equation becomes \(du/dt=-\beta u\), so \(u(t)=4e^{-0.5t}\).

To obtain \(s(t)\), substitute this expression into \(ds/dt+\gamma s=\beta u(t)\). Multiplying by the integrating factor \(e^{\gamma t}\) gives the derivative of \(e^{\gamma t}s(t)\). Integrating from zero to \(t\) and using \(s(0)=4\) yields:

\begin{aligned} \frac{d}{dt}\left(e^{\gamma t}s(t)\right) &=\beta u(0)e^{(\gamma-\beta)t},\\ e^{\gamma t}s(t)-s(0) &=\frac{\beta u(0)}{\gamma-\beta} \left(e^{(\gamma-\beta)t}-1\right),\\ s(t) &=s(0)e^{-\gamma t} +\frac{\beta u(0)}{\beta-\gamma} \left(e^{-\gamma t}-e^{-\beta t}\right), \qquad \beta\ne\gamma. \end{aligned}

This derivation shows why the two abundances need not decline together: the spliced compartment continues to receive RNA from the unspliced compartment after transcription has switched off. The delay can create a transient phase-portrait path rather than a single equilibrium relationship.

Worked Example: A Switch-Off Trajectory

Use the hypothetical rates and initial abundances above to calculate the state one hour after transcription switches off. The unspliced abundance is

u(1)=4e^{-0.5}\approx 2.426.

For spliced RNA, substitute \(\beta=0.5\), \(\gamma=0.25\), \(u(0)=4\), and \(s(0)=4\):

s(1)=4e^{-0.25} +\frac{(0.5)(4)}{0.5-0.25} \left(e^{-0.25}-e^{-0.5}\right) \approx 3.115+1.378 =4.493.

The spliced abundance has risen from 4 to about 4.493 abundance units even though transcription is off, because splicing of the remaining unspliced RNA still adds to the spliced compartment. The instantaneous spliced velocity is

\frac{ds}{dt} =(0.5)(2.426)-(0.25)(4.493) \approx 1.213-1.123 =0.090 \quad\text{abundance units per hour}.

The positive velocity is consistent with the calculated increase in \(s\) at this time. A steady-state interpretation based only on the fact that transcription is off would miss this transient behavior.

Choosing Between the Models

Model choice should use the phase-portrait structure, the biological plausibility of a switching history, and a comparison of fit that accounts for flexibility. A steady-state model is a reasonable candidate when the cells form a broadly coherent relationship around one equilibrium reference, residuals do not show a clear ordered pattern, and there is no strong evidence that the sampled population spans a transcriptional transition. A dynamical model deserves consideration when the phase portrait has distinct branches or a loop-like pattern, and a plausible induction or repression history can explain those patterns.

The following comparison is a guide, not an automatic decision rule:

Feature	Steady-state approach	Dynamical approach
Main reference	One equilibrium relation between \(u\) and \(s\)	A time-dependent trajectory under a transcriptional history
Useful phase-portrait evidence	Cells broadly follow one relation without clear ordered branches	Distinct transient branches or a path consistent with switching
Primary output	Direction relative to the fitted equilibrium reference	Fitted kinetic trajectory and switching-related parameters
Main risk	Transient structure is mistaken for equilibrium deviation	Extra flexibility fits noise or poorly identified switching patterns

A likelihood comparison is informative only when both models use the same observations and a comparable observation model. A complexity-penalized criterion such as BIC can help prevent a more flexible model from winning solely because it has more parameters. For maximized likelihood \(L\), parameter count \(k\), and sample size \(n\), BIC is \(-2\log L+k\log n\); lower values are preferred within the comparison. This criterion is a model-selection aid, not a test that proves a transcriptional switch occurred.

Worked Example: Fit Improvement Versus Model Complexity

Suppose the same hypothetical set of 120 cell-level observations is fit by both approaches using the same likelihood model. The steady-state fit has maximized log-likelihood \(-180\) and \(k=2\) fitted parameters. The dynamical fit has maximized log-likelihood \(-150\) and \(k=5\). For this comparison, assume the likelihoods are computed on the same data and the parameter counts include all fitted model parameters.

The steady-state BIC is

-2(-180)+2\log(120) =360+9.575 \approx369.6.

The dynamical BIC is

-2(-150)+5\log(120) =300+23.94 \approx323.9.

The dynamical model has the lower BIC by about \(369.6-323.9=45.7\), so this criterion favors it for these hypothetical observations despite its three additional parameters. The result supports choosing the dynamical fit as a statistical description, conditional on the likelihood and comparison being appropriate. It does not show that a switch is biologically certain; phase-portrait evidence and model adequacy still matter.

Diagnostics, Limitations, and Common Pitfalls

Before selecting a model, inspect the phase portrait and the fitted patterns rather than relying on a single summary. For a steady-state fit, look for systematic residual structure: for example, cells at similar spliced abundance that form clearly separated unspliced-abundance branches may not be well described by one equilibrium relation. For a dynamical fit, check whether the fitted trajectory actually covers the observed structure and whether the inferred switching history is supported by observations across the relevant phases. A flexible curve that follows a few extreme points is weak evidence if most cells do not support the path.

Possible steady-state failure: distinct branches or ordered residual patterns suggest that transient phases may be mixed together. A single reference line can then give misleading directional evidence.
Possible dynamical failure: switching times or rates may be weakly identified when observations do not cover the relevant phases. A good apparent fit can be driven by flexibility rather than strong information about the history.
Shared concerns: measurement noise, sparse observations, or an unsuitable kinetic or observation model can distort either fit. A model comparison cannot repair data or assumptions that are inadequate.

A common mistake is to choose the dynamical model whenever it has the higher likelihood. The correction is to compare fit and complexity on the same data, then ask whether the phase portrait and switching interpretation are credible. Another mistake is to treat a fitted steady-state slope as proof that all cells are at equilibrium. It is only an estimated reference relationship under the model assumptions. Neither model establishes causation or guarantees what a particular cell will do next.

Key takeaway: Prefer the steady-state approximation when one equilibrium reference adequately describes the phase portrait. Prefer a dynamical description when there is credible transient structure and a plausible switching history, with enough information to fit it. In either case, model choice is conditional on assumptions and evidence, not a biological verdict.

Check Your Understanding

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

At equilibrium, what relationship links \(u\) and \(s\), and which kinetic rates determine its slope?
After transcription switches off, why can \(s\) continue to increase for a time?
A phase portrait shows two separated branches at similar spliced abundances. Why might a single steady-state reference be inadequate?
For \(n=100\), a hypothetical steady-state model has \(\log L=-90\) and \(k=2\); a dynamical model has \(\log L=-75\) and \(k=4\). Calculate both BIC values and state which is lower.
Does a lower BIC for the dynamical fit prove that transcription switched in the biological system? Explain.

Solutions. (1) Setting both derivatives to zero gives \(\beta u=\gamma s\), so \(u=(\gamma/\beta)s\); the slope is the degradation rate divided by the splicing rate. (2) Unspliced RNA already present can still be spliced into the spliced compartment, so \(\beta u\) can exceed \(\gamma s\) temporarily. (3) A single equilibrium relationship may not capture multiple transient phases; the branches could reflect a trajectory with switching, though other explanations and model adequacy must also be considered. (4) The steady-state BIC is \(-2(-90)+2\log(100)=180+9.210=189.2\). The dynamical BIC is \(-2(-75)+4\log(100)=150+18.42=168.4\). The dynamical model has the lower BIC by about 20.8. (5) No. It favors the dynamical model under this criterion and the assumed likelihood, but it does not prove the biological mechanism; the phase portrait, assumptions, and fit diagnostics remain important.