Why Include Phenotypic Switching?
A tumor can become enriched in treatment-resistant cells because sensitive cells are killed more quickly, leaving resistant cells to make up a larger share of the remaining population. That is selection among phenotypes already present. A different mechanism is phenotypic switching: a cell changes from a treatment-sensitive state to a resistant state, or back again. The mechanisms can act together, but they represent different flows between cell populations.
A switching model is useful when the cell state can change without requiring a new, heritable genetic clone. It tracks cell numbers in two phenotypic compartments and makes the transition assumptions explicit. The model does not, by itself, establish that switching occurs in a particular tumor; its rates and biological interpretation require evidence.
A Two-Phenotype Model
Let \(r_S(t)\) and \(r_R(t)\) denote the net per-cell growth rates of the sensitive and resistant populations. Each is a net rate: proliferation contributes positively and cell death, including drug-induced killing, contributes negatively. Thus, a negative \(r_S\) means the sensitive population would decline in the absence of switching. Rates can vary with time if treatment exposure changes.
Assume that each cell switches independently at its compartment's specified rate; switching transfers a cell between compartments but does not itself create or destroy a cell. Also assume that the two compartments adequately represent the relevant phenotypes, and that all cells within a compartment share its net growth and switching rates at a given time. Under these assumptions, the model is
The transition terms have opposite signs in the two equations. Forward switching removes cells from \(S\) and adds the same number to \(R\); reverse switching does the opposite. For example, over a short interval \(\Delta t\), the expected number switching from \(S\) to \(R\) is approximately \(k_{SR}S\Delta t\), when rates and cell numbers change little over that interval. The corresponding flux in the differential equation is \(k_{SR}S\), with units of cells per unit time.
How Switching and Growth Change Composition
When \(N>0\), define the resistant fraction \(f=R/N\), so the sensitive fraction is \(1-f=S/N\). Differentiating \(f=R/N\) gives \(df/dt=(N\,dR/dt-R\,dN/dt)/N^2\). Substituting the two-compartment model yields
The first two terms are the direct contribution of phenotype switching: forward switching increases \(f\), while reverse switching decreases it. The last term is selection due to different net growth rates. If resistant cells have a greater net growth rate than sensitive cells, then \(r_R-r_S>0\), and this term increases the resistant fraction when both phenotypes are present. If resistant cells have the lower net growth rate, it decreases that fraction.
This equation describes an instantaneous rate of change. It does not say that the fraction changes linearly over time: \(f\), the net growth rates, and possibly the switching rates may change. It also separates a change in composition from a change in absolute cell number. A resistant fraction can fall even while the resistant cell count rises, if the total tumor grows faster than the resistant compartment.
Worked Example: Forward Switching During Treatment
Consider a hypothetical treated tumor with \(S=990\) million cells and \(R=10\) million cells. Suppose \(r_S=-0.30\) day\(^{-1}\), \(r_R=0.05\) day\(^{-1}\), \(k_{SR}=0.002\) day\(^{-1}\), and \(k_{RS}=0.0005\) day\(^{-1}\). Treat these rates as constant for the instantaneous calculation.
The initial resistant fraction is \(f=10/1000=0.0100\), or 1.00%. The sensitive-cell rate of change is
where the cell counts in this calculation are in millions. Thus, \(S\) initially decreases by about 299 million cells per day. For the resistant compartment,
so \(R\) initially increases by 2.475 million cells per day. The total changes at \(-298.975+2.475=-296.5\) million cells per day, which also equals \((-0.30)(990)+(0.05)(10)\); switching cancels from the total.
The composition equation gives
The resistant fraction is initially rising at 0.544 percentage points per day. This is a local rate, not a prediction that the fraction will rise by 0.544 points every day. Both forward switching and the resistant cells’ relative growth advantage contribute to the increase.
Isolating Switching from Differential Growth
Setting \(r_S=r_R=0\) isolates the switching process: the total cell number remains constant, but cells can move between phenotypes. This is a useful model check as well as a way to see what the transition terms mean. It does not imply that growth and treatment effects are generally negligible.
Worked Example: Switching Fluxes With No Net Growth
Suppose a hypothetical tumor has \(S=800\) million cells and \(R=200\) million cells. Let \(k_{SR}=0.010\) day\(^{-1}\), \(k_{RS}=0.004\) day\(^{-1}\), and \(r_S=r_R=0\). The forward flux is \(0.010(800)=8.0\) million cells per day, while the reverse flux is \(0.004(200)=0.8\) million cells per day.
Therefore,
with counts in millions. The total rate is zero. Since \(f=0.20\), the composition equation also gives
The resistant fraction is initially increasing by 0.72 percentage points per day, even though the total number of cells does not change. This calculation describes the initial direction and rate; it should not be extrapolated as a constant daily increment.
When Reverse Switching and Growth Favor Different Outcomes
Switching and relative growth need not push composition in the same direction. In particular, resistant cells can continue to increase in number while becoming a smaller fraction of a growing tumor. Checking both the compartment equations and the fraction equation helps distinguish these statements.
Worked Example: A Falling Resistant Fraction While Resistant Cells Grow
Consider a hypothetical state with \(f=0.40\), \(r_S=0.10\) day\(^{-1}\), \(r_R=0.02\) day\(^{-1}\), \(k_{SR}=0.003\) day\(^{-1}\), and \(k_{RS}=0.020\) day\(^{-1}\). The tumor has \(S=0.60N\) and \(R=0.40N\).
The switching contribution to the resistant fraction is
The selection contribution is
Together, \(df/dt=-0.0254\) day\(^{-1}\), an initial decline of 2.54 percentage points per day. Yet the resistant-cell count itself has rate
Thus, \(R\) is initially growing by 0.18% of the current total cell count per day. The total tumor grows at \((0.10)(0.60)+(0.02)(0.40)=0.068\) day\(^{-1}\), so the faster-growing sensitive compartment increases the denominator sufficiently for the resistant fraction to fall. A decreasing resistant fraction does not, by itself, mean that resistant cells are being eliminated.
Assumptions, Diagnostics, and Alternatives
The model is a compact description, not a guarantee that the biology follows two homogeneous compartments. Its interpretation is strongest when \(S\) and \(R\) correspond to measurable states and when transition rates can reasonably be treated as constant or specified functions of time over the interval studied. The predicted cell numbers should remain nonnegative; the switching rates must be nonnegative; and switching out of one compartment must appear as an equal flux into the other if transitions conserve cells.
Useful checks include confirming units in every term (each term in a cell-count equation must have units of cells per time), verifying that switching cancels from \(dN/dt\), and checking the predicted direction of change in simple cases. With \(r_S=r_R=0\) and \(k_{SR}S>k_{RS}R\), the resistant count must initially rise. If the equations instead imply that the same forward flux removes cells from both compartments, there is a sign or bookkeeping error.
A fixed-phenotype selection model is the main alternative when the biological question is whether treatment enriches a pre-existing resistant population without cells changing state. It is obtained by setting \(k_{SR}=k_{RS}=0\). The switching model allows both selection and state transitions. The same observed rise in resistant fraction could be consistent with either mechanism or both; the fraction trend alone does not identify which process caused it. Phenotype measurements over time, or other information that distinguishes state transitions from changes in clone abundance, may be needed.
Tumor-volume observations alone often provide limited information about separate switching rates, because switching cancels from total cell number and can affect volume only indirectly through changed composition and subsequent growth. Further limitations arise if there are more than two states, if rates depend on cell density or treatment history, or if switching is associated with division or death. In those cases the model may need additional compartments or different transition terms. A good fit to total tumor size is not, by itself, evidence that the inferred switching mechanism is correct.
Check Your Understanding
Try each problem, then compare your reasoning with the solutions below.
- A hypothetical tumor has \(S=600\) million cells and \(R=400\) million cells. If \(k_{SR}=0.005\) day\(^{-1}\) and \(k_{RS}=0.002\) day\(^{-1}\), with no net growth or death, calculate the initial rates of change in \(S\), \(R\), and the resistant fraction.
- In the two-phenotype model, why does switching not appear in the equation for \(dN/dt\)?
- If \(f=0.25\), \(r_R-r_S=0.12\) day\(^{-1}\), and there is no switching, calculate the initial selection contribution to \(df/dt\).
- Can the resistant fraction fall while the resistant-cell count rises? State the condition in words that makes this possible.
Solutions. (1) The forward flux is \(0.005(600)=3.0\) million cells per day, and the reverse flux is \(0.002(400)=0.8\) million cells per day. Thus \(dS/dt=-3.0+0.8=-2.2\) million cells per day and \(dR/dt=3.0-0.8=2.2\) million cells per day. With \(f=0.40\), \(df/dt=(0.005)(0.60)-(0.002)(0.40)=0.0022\) day\(^{-1}\). (2) Each switching term that is negative in one compartment is positive by the same amount in the other, so the terms sum to zero in \(d(S+R)/dt\). (3) The contribution is \(f(1-f)(r_R-r_S)=(0.25)(0.75)(0.12)=0.0225\) day\(^{-1}\). (4) Yes. This can occur if resistant cells grow in number, but the total tumor grows proportionally faster, for example because the sensitive compartment grows faster and makes up most of the tumor.