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Cell Population Dynamics · Tutorial 179 of 799

Phenotype Switching Between Cell States

Learn to derive and interpret a two-state cell model with reversible phenotype switching, including its effects on total abundance and state composition.

Intermediate 12 min read

What You'll Learn

  • Write state-specific abundance equations that include division, death, and switching fluxes.
  • Distinguish switching’s direct effect on composition from its indirect effect on total growth.
  • Derive the total-count and state-fraction equations from the two-state model.
  • Calculate instantaneous changes and switching-only composition dynamics from hypothetical data.
  • Interpret how reversible switching can change a population’s long-run growth rate.
  • Identify assumptions and diagnostics relevant to applying the model.

Why Track Phenotype Switching?

A cell population can change its phenotype as well as its size. If cells move between two states, a change in the number of cells observed in one state may reflect switching, state-specific growth or death, or both. A useful model must keep these processes distinct: proliferation and death change abundance, while switching transfers cells between states.

Consider two states, \(A\) and \(B\). Let \(N_A(t)\) and \(N_B(t)\) denote their cell abundances at time \(t\). A cell in state \(A\) switches to state \(B\) at per-cell rate \(k_{AB}\); the reverse switch occurs at rate \(k_{BA}\). Each state also has its own division and death rates. All rates in this tutorial are nonnegative and have units of inverse time.

Definition: Reversible phenotype switching is a process in which cells transfer between distinct states in both directions. Under a constant per-cell switching rate \(k_{AB}\), the flux from \(A\) to \(B\) per unit time is \(k_{AB}N_A\). Switching removes cells from one state and adds the same number to the other; by itself, it does not create or remove cells from the total population.

The model treats each state as internally homogeneous: all cells in a state share its rates at a given time. It also assumes that switching is a continuous-time process with flux proportional to the abundance in the state of origin. For the constant-rate results below, rates do not vary over the time interval considered. These assumptions make the model a first-order ordinary differential equation, not a claim that individual cells switch at perfectly regular times.

Deriving the Abundance Equations

Write \(p_A\) and \(p_B\) for per-cell division rates and \(d_A\) and \(d_B\) for per-cell death rates. A division is assumed to replace one parent cell with two daughters that remain in the parent's state. Thus each division increases that state's abundance by one cell. Define the net growth rates \(r_A=p_A-d_A\) and \(r_B=p_B-d_B\). If a different division mechanism produces a different number or state of daughters, its abundance contribution must be changed accordingly.

For state \(A\), division adds \(p_A N_A\), death removes \(d_A N_A\), switching to \(B\) removes \(k_{AB}N_A\), and switching from \(B\) adds \(k_{BA}N_B\). The same flux accounting applies to state \(B\):

$$ \begin{aligned} \frac{dN_A}{dt} &= (p_A-d_A-k_{AB})N_A+k_{BA}N_B = (r_A-k_{AB})N_A+k_{BA}N_B,\\ \frac{dN_B}{dt} &= k_{AB}N_A+(p_B-d_B-k_{BA})N_B = k_{AB}N_A+(r_B-k_{BA})N_B. \end{aligned} $$

The two switching terms for each direction have equal magnitude and opposite effects across the equations. In matrix form, with \(\mathbf{N}=(N_A,N_B)^\mathsf{T}\), the system is \(d\mathbf{N}/dt=M\mathbf{N}\), where

$$ M= \begin{pmatrix} r_A-k_{AB} & k_{BA}\\ k_{AB} & r_B-k_{BA} \end{pmatrix}. $$

When the rates are constant, the solution is \(\mathbf{N}(t)=e^{Mt}\mathbf{N}(0)\). The off-diagonal entries of \(M\) are nonnegative, so nonnegative initial abundances remain nonnegative. A negative predicted abundance is not a biological consequence of this model; it indicates, for example, an implementation error or use of the model outside its appropriate assumptions.

Worked Example: Computing the Instantaneous State Changes

Suppose a hypothetical cell population has \(N_A=600\) cells and \(N_B=400\) cells. Per day, state \(A\) has \(p_A=0.12\) and \(d_A=0.04\), while state \(B\) has \(p_B=0.05\) and \(d_B=0.03\). Let \(k_{AB}=0.03\) and \(k_{BA}=0.01\) per day. Assume constant rates over the instant being evaluated.

The net rates are \(r_A=0.12-0.04=0.08\) per day and \(r_B=0.05-0.03=0.02\) per day. Substituting into the abundance equations gives

$$ \begin{aligned} \frac{dN_A}{dt} &=(0.08-0.03)(600)+(0.01)(400) =30+4=34\ \text{cells/day},\\ \frac{dN_B}{dt} &=(0.03)(600)+(0.02-0.01)(400) =18+4=22\ \text{cells/day}. \end{aligned} $$

The total abundance is \(N=N_A+N_B=1000\) cells, and its instantaneous rate of change is \(34+22=56\) cells per day. The switching flux from \(A\) to \(B\) is \(18\) cells per day; the reverse flux is \(4\) cells per day. Their net effect shifts \(14\) cells per day from \(A\) to \(B\), but neither flux directly adds cells to or removes cells from the total.

Total Abundance and State Composition

Adding the two state equations cancels both switching fluxes. This gives the total-count equation:

$$ \frac{dN}{dt}=r_A N_A+r_B N_B,\qquad N=N_A+N_B. $$

Thus switching does not appear as a direct source or sink in the total-count equation. It can still affect total abundance over time by changing how many cells occupy each state, particularly when \(r_A\ne r_B\). To see that distinction explicitly, define the state-\(A\) fraction \(f=N_A/N\), for \(N>0\). The quotient rule gives \(df/dt=(N\,dN_A/dt-N_A\,dN/dt)/N^2\). Substituting the abundance and total-count equations and simplifying yields

$$ \frac{df}{dt}=k_{BA}(1-f)-k_{AB}f+f(1-f)(r_A-r_B). $$

The first two terms are the direct switching contribution: cells enter the \(A\) fraction from \(B\) and leave it from \(A\). The last term reflects the different net growth rates. If \(r_A>r_B\), growth tends to increase \(f\); if \(r_A<r_B\), growth tends to decrease it. These are effects on the state composition, not proof that switching itself creates cells.

A Short Derivation: Switching Alone

If there is no net growth or death in either state (\(r_A=r_B=0\)), the fraction equation reduces to \(df/dt=k_{BA}-(k_{AB}+k_{BA})f\). For constant switching rates, its equilibrium fraction is the value \(f^*\) that makes the derivative zero, namely \(f^*=k_{BA}/(k_{AB}+k_{BA})\), provided the sum of switching rates is positive. Separating the deviation from equilibrium gives

$$ \frac{d(f-f^*)}{dt}=-(k_{AB}+k_{BA})(f-f^*), \qquad f(t)=f^*+[f(0)-f^*]e^{-(k_{AB}+k_{BA})t}. $$

This shows why the equilibrium fraction is not generally one-half: it depends on the relative rates of switching into and out of \(A\). In this switching-only case, \(dN/dt=0\), so the total abundance stays constant while the composition approaches its equilibrium.

Worked Example: Composition Under Switching With No Net Growth

Suppose a hypothetical population contains 800 cells in state \(A\) and 200 in state \(B\). Set both net growth rates to zero, with \(k_{AB}=0.20\) and \(k_{BA}=0.10\) per week. The initial fraction is \(f(0)=800/1000=0.80\), and the switching-only equilibrium is

$$ f^*=\frac{0.10}{0.20+0.10}=\frac{1}{3}. $$

After five weeks, the fraction in \(A\) is

$$ f(5)=\frac{1}{3}+\left(0.80-\frac{1}{3}\right)e^{-(0.30)(5)} \approx 0.437. $$

The total remains 1000 cells under the stated assumptions, so the predicted state abundances are approximately \(437\) cells in \(A\) and \(563\) in \(B\). This is a redistribution, not an increase in total cell number. The calculation assumes that switching rates remain constant and that there is no net growth or death.

How Switching Can Change Long-Run Growth

Although switching cancels from the instantaneous equation for total abundance, it changes the mixture of states and therefore can change the subsequent total growth rate. For a constant-rate two-state model, the long-run behavior is governed by the eigenvalue of \(M\) with the largest real part, assuming the initial population has a component in its associated mode. A positive dominant eigenvalue corresponds to eventual exponential growth; a negative one corresponds to eventual decay.

Worked Example: Switching Changes the Long-Run Growth Rate

Consider a hypothetical population with net growth rates \(r_A=0.08\) per day and \(r_B=-0.04\) per day. Let \(k_{AB}=0.03\) and \(k_{BA}=0.01\) per day. The constant-rate matrix is

$$ M= \begin{pmatrix} 0.05 & 0.01\\ 0.03 & -0.05 \end{pmatrix}\ \text{per day}. $$

Its characteristic equation is \(\det(M-\lambda I)=0\), or \(\lambda^2-0.0028=0\). Therefore the eigenvalues are \(\lambda=\pm\sqrt{0.0028}\), approximately \(+0.05292\) and \(-0.05292\) per day. Since the dominant eigenvalue is positive, a population with cells in the dominant mode eventually grows at an exponential rate of about \(0.05292\) per day.

For comparison, without switching, the state-\(A\) population would grow at \(0.08\) per day, while state \(B\) would decline at \(0.04\) per day. Here, switching out of the faster-growing state reduces the long-run growth rate relative to growth in state \(A\) alone. At a particular instant, however, the total growth rate is still \(r_A N_A+r_B N_B\); switching changes that rate later by changing the state mixture. This conclusion assumes constant rates and the stated division and switching rules.

Assumptions, Diagnostics, and Limitations

Before interpreting estimates or predictions, check whether the model’s structure fits the question and data. The two states must be defined so that a cell can be assigned to one state at a time; the modeled abundance should be measured consistently; and the first-order flux assumption should be plausible over the time interval. If rates vary with time or treatment, they can be represented as time-dependent inputs, but the constant-rate solution and equilibrium calculations above no longer apply unchanged.

State-resolved measurements are important. Total counts alone give information about the weighted net growth \(r_A N_A+r_B N_B\), but usually cannot distinguish switching from state-specific growth and death. Even with measurements of both states, different combinations of rates may produce similar trajectories. Independent measurements or informative perturbations may be needed to identify the individual rates.

  • Check flux signs and units. A switching term must leave one state and enter the other with equal magnitude. Rates in inverse days must be paired with time measured in days.
  • Compare predicted and observed state fractions as well as total counts. A good fit to the total can conceal incorrect state-specific dynamics.
  • Inspect residual patterns and implausible rates. Systematic mismatch, negative fitted rates, or rates that change sharply with the observation interval can indicate model misspecification, measurement problems, or insufficient information.
  • Remember the homogeneous-state assumption. If cells within a nominal state have substantially different growth or switching behavior, one compartment per state may be inadequate.

This model also assumes that switching itself does not cause cell loss, division, or a delay spent outside the two modeled states. Those processes require additional terms or states. It has no density dependence or capacity constraint, so it should not be treated as a model of growth regulation when such effects are central.

Common Mistake

A frequent error is to add switching flux \(k_{AB}N_A\) to the total population because it appears as a positive term in the equation for \(N_B\). That counts a transfer as if it created a cell. The correction is to include the matching loss \(-k_{AB}N_A\) in the equation for \(N_A\) and then add the state equations: the two terms cancel. Switching can change future total growth through state composition, but it is not itself a direct source or sink under this model.

Key takeaway: State-specific division and death determine direct changes in total cell number; reversible switching redistributes cells. When states have different net growth rates, that redistribution can alter the population’s later total growth as well as its composition.

Check Your Understanding

Use the two-state model to answer each question. The solutions are included below for checking your reasoning.

  1. If both net growth rates are zero, what happens to total abundance when switching occurs?
  2. With \(k_{AB}=0.06\) per day and \(k_{BA}=0.02\) per day, what is the switching-only equilibrium fraction in state \(A\)?
  3. At a given instant, \(N_A=300\), \(N_B=700\), \(r_A=0.04\) per day, and \(r_B=-0.01\) per day. What is the total-count derivative?
  4. Explain why observing a changing state-\(A\) fraction does not by itself establish that cells are switching.

Solutions. (1) Total abundance stays constant because \(dN/dt=r_A N_A+r_B N_B=0\); switching fluxes cancel. (2) \(f^*=k_{BA}/(k_{AB}+k_{BA})=0.02/(0.06+0.02)=0.25\), or 25%. (3) \(dN/dt=(0.04)(300)+(-0.01)(700)=12-7=5\) cells per day. (4) The fraction can change through switching, but also through differential net growth: if the states grow or die at different rates, their relative abundances change even without switching. State-resolved trajectories and additional information are needed to distinguish these mechanisms.

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