Phase Portraits and Gene Kinetics

Reading a Phase Portrait

A phase portrait places a gene’s unspliced abundance against its spliced abundance, so its path shows how the two RNA compartments change relative to one another. It can help distinguish a transition toward higher transcription from one toward lower transcription, but a point’s position alone is not enough: interpretation depends on the direction of motion and on the kinetic model.

Use \(s\) for spliced abundance on the horizontal axis and \(u\) for unspliced abundance on the vertical axis. Under the two-compartment model, transcription produces unspliced RNA at rate \(\alpha\), splicing transfers it to the spliced compartment at rate \(\beta u\), and degradation removes spliced RNA at rate \(\gamma s\). For a time interval in which the rates are constant, the equations are:

Here \(u\) and \(s\) are abundances, and \(\alpha\), \(\beta\), and \(\gamma\) are nonnegative kinetic rates. If abundance is measured in counts and time in days, \(\alpha\) has units of counts per day, while \(\beta\) and \(\gamma\) have units of inverse days. The model treats abundance as a continuous quantity and assumes the rates are constant within each interval considered. It does not include other sources or sinks, such as alternative processing or changes in cell volume.

This vector interpretation is important because abundance and abundance change are different. In particular, increasing transcription raises \(du/dt\) immediately by the amount of the increase, but does not guarantee that \(u\) itself initially rises: its initial change is determined by the new value of \(\alpha-\beta u\).

Steady State and the Geometry of the Portrait

A steady state is a point at which both abundances have zero instantaneous change. If \(\beta>0\) and \(\gamma>0\), solving the two rate equations gives:

The star marks the steady-state abundance. At this point, transcription balances splicing, and splicing balances degradation. For fixed positive \(\beta\) and \(\gamma\), steady-state points at different values of \(\alpha\) lie on the line \(u=(\gamma/\beta)s\). This relation is not a claim that every point along a gene’s time-course path is in steady state. A trajectory after a switch generally approaches its new steady-state point without lying on that line.

The steady-state result requires positive splicing and degradation rates. If a rate is zero, the corresponding steady-state formula may be undefined or a finite steady state may not exist; the formulas should not be used by division without checking those conditions.

Why Switching Transcription Changes the Trajectory

Induction and repression refer here to an increase or decrease in transcription rate. To see their phase-portrait signatures, consider a gene initially at steady state under transcription rate \(\alpha_0\). At time zero, transcription switches to a constant \(\alpha_1\), while \(\beta\) and \(\gamma\) remain unchanged. The abundances cannot jump at the switch. Immediately afterward, the new derivatives are:

Thus, if \(\alpha_1>\alpha_0\), the trajectory initially moves upward with no instantaneous horizontal motion. If \(\alpha_1<\alpha_0\), it initially moves downward, also with no instantaneous horizontal motion. Spliced abundance begins changing just after the switch: differentiating its rate equation gives \(d^2s/dt^2=\beta\,du/dt-\gamma\,ds/dt\). At the switch, this equals \(\beta(\alpha_1-\alpha_0)\), so for \(\beta>0\), spliced abundance begins to accelerate upward after induction and downward after repression.

These immediate directions rely on starting at the old steady state. From an arbitrary starting point, the signs must instead be calculated from \(\alpha_1-\beta u\) and \(\beta u-\gamma s\). A transcription increase raises \(du/dt\) by \(\alpha_1-\alpha_0\) compared with the old derivative, but the new derivative can still be negative if \(\beta u\) exceeds \(\alpha_1\).

More generally, the integral expression is valid for nonnegative rates. If \(\beta=0\), the unspliced equation gives \(u(t)=u(0)+\alpha t\); if \(\gamma=0\), the spliced equation gives \(s(t)=s(0)+\beta\int_0^t u(r)\,dr\). If \(\beta=\gamma>0\), the closed-form integral term is obtained by its limit as \(\gamma\) approaches \(\beta\). These cases do not justify using a formula whose denominator is zero.

How Kinetic Rates Shape the Curves

The phase portrait reflects all three rates, not transcription alone. Changing \(\alpha\) changes the input to the unspliced compartment and, when \(\beta\) and \(\gamma\) are positive, changes both steady-state abundances proportionally. Increasing \(\beta\) increases the splicing flux at a given \(u\), while also changing the steady-state unspliced abundance \(\alpha/\beta\). Increasing \(\gamma\) increases spliced-RNA removal at a given \(s\), and lowers the steady-state spliced abundance \(\alpha/\gamma\). The speed and curvature of a trajectory depend on the rates jointly and on the starting abundances.

One useful local check is the phase-plane slope when \(ds/dt\ne0\): \(du/ds=(\alpha-\beta u)/(\beta u-\gamma s)\). When the denominator is zero, the horizontal component vanishes and the direction is vertical, so this slope is undefined; the vector components themselves remain interpretable. A curve’s visual steepness alone should therefore not be assigned to a single rate without considering both equations.

Diagnostics, Comparisons, and Limitations

A phase portrait and a time-course plot answer related but different questions. The phase portrait displays the relationship between \(u\) and \(s\) and the direction of change in that plane. A time-course plot shows how each abundance changes with time and makes timing more explicit. The steady-state line summarizes the equilibrium relationship for fixed \(\beta\) and \(\gamma\); it is not a replacement for the full trajectory after a switch. Using these views together can reveal a mistaken direction reading or an apparent pattern that is actually caused by a scale or axis convention.

In single-cell data, each cell is usually observed once, rather than followed continuously. A cloud of cells in a phase portrait is not automatically a time-course path. Interpreting its shape as a trajectory requires that the cells represent the same relevant biological process, that their ordering is meaningful, and that the kinetic model is a reasonable approximation. Branching states, cell-to-cell rate variation, technical variation in counts, and changes in normalization can all distort the apparent curve.

• Check the axes. This tutorial uses spliced abundance horizontally and unspliced abundance vertically. Reversing the axes reverses the visual interpretation of up, down, left, and right.
• Check the state, not only the location. High \(u\) is not by itself proof of induction. The relevant derivative is \(\alpha-\beta u\), and direction requires a temporal or model-based interpretation.
• Check the assumptions. A single constant-rate interval may not describe a gene with gradual or repeated transcriptional changes. A visibly curved cloud need not be caused by a transcription switch alone.
• Check whether equilibrium is plausible. A point satisfying the steady-state equations is stationary under the model; proximity to the steady-state line alone does not establish that a cell is at steady state.