Deriving Equilibrium Fractions Under Reversible Resistance

Why Equilibrium Fractions Matter

A tumor cell population can contain sensitive cells that switch into a resistant state and resistant cells that switch back. When both directions occur under stable conditions, neither state must disappear. Instead, their fractions can approach a stable composition. Deriving that composition clarifies what the switching rates imply—and what an observed resistant fraction alone cannot tell us.

Here, “equilibrium” refers to a stable composition, not necessarily a stable tumor size. The calculation isolates switching: it does not assume that tumor burden itself is constant because of growth and death. The earlier material on acquired resistance through phenotypic switching motivates the two-state representation; this tutorial derives its equilibrium fractions and convergence properties.

Definition: Let \(S(t)\) and \(R(t)\) denote the numbers of sensitive and resistant cells at time \(t\), and let \(N(t)=S(t)+R(t)\). Let \(k_{SR}\) be the rate constant for switching from sensitive to resistant, and \(k_{RS}\) the rate constant for switching from resistant to sensitive. Both rates have units of inverse time. The sensitive and resistant fractions are \(f_S(t)=S(t)/N(t)\) and \(f_R(t)=R(t)/N(t)\), respectively.

Model and Assumptions

Assume the population has only these two states, and cells switch between them with constant, nonnegative rate constants. A cell in the sensitive state switches at rate \(k_{SR}\); a cell in the resistant state switches at rate \(k_{RS}\). For the fraction equations below, assume also that switching is the only process changing the state counts: neither state has differential proliferation or death, and there is no influx or removal of cells. Thus \(N(t)\) is constant. These assumptions define a closed two-state switching model, not a general tumor-growth model.

Under constant-rate switching, the expected flow out of a state per unit time is its cell count multiplied by its outgoing rate. The count equations are therefore:

\frac{dS}{dt}=-k_{SR}S+k_{RS}R,\qquad \frac{dR}{dt}=k_{SR}S-k_{RS}R.

The two equations conserve total cell number: adding them gives \(dN/dt=0\). With \(N\) constant, dividing the resistant-cell equation by \(N\) gives a closed equation for the resistant fraction. Since \(f_S=1-f_R\), either fraction then determines the other.

\frac{df_R}{dt}=k_{SR}(1-f_R)-k_{RS}f_R =k_{SR}-(k_{SR}+k_{RS})f_R,\qquad f_S=1-f_R.

Deriving the Equilibrium Fractions

At equilibrium, the fraction no longer changes, so \(df_R/dt=0\). Setting the net rate of change to zero gives:

0=k_{SR}-(k_{SR}+k_{RS})f_R^*.

If \(k_{SR}+k_{RS}>0\), solving for \(f_R^*\) and then using \(f_S^*=1-f_R^*\) yields the equilibrium composition:

f_R^*=\frac{k_{SR}}{k_{SR}+k_{RS}},\qquad f_S^*=\frac{k_{RS}}{k_{SR}+k_{RS}}.

The same result follows by balancing the two switching flows. At equilibrium, sensitive-to-resistant switching replaces resistant cells at the same rate that resistant-to-sensitive switching replaces sensitive cells: \(k_{SR}f_S^*=k_{RS}f_R^*\). Together with \(f_S^*+f_R^*=1\), this balance gives the fractions above. When both rates are positive, each equilibrium fraction is between zero and one, and the fractions sum to one.

Key result: With constant positive switching rates and no state-specific growth or loss, the resistant equilibrium fraction is the sensitive-to-resistant rate divided by the sum of the two rates. The reverse rate contributes to the denominator because it removes cells from the resistant state.

Worked Example: Computing the Equilibrium Composition

Suppose a hypothetical tumor-cell population has \(k_{SR}=0.06\) per day and \(k_{RS}=0.14\) per day. Assume switching is the only process changing state counts, and both rates remain constant. Find the equilibrium fractions.

The total switching rate is \(0.06+0.14=0.20\) per day, which is positive, so the equilibrium formulas apply:

f_R^*=\frac{0.06}{0.20}=0.300,\qquad f_S^*=\frac{0.14}{0.20}=0.700.

The model predicts an equilibrium composition of 30.0% resistant cells and 70.0% sensitive cells. The switching flows agree: \(0.06(0.700)=0.042\) per day per cell in the population’s fraction scale, and \(0.14(0.300)=0.042\) on the same scale. This is a balance of flows, not an absence of switching.

How the Population Approaches Equilibrium

The equilibrium calculation gives a target composition, but not by itself how quickly the population approaches that target. Define \(\lambda=k_{SR}+k_{RS}\). Subtract the equilibrium condition \(k_{SR}=\lambda f_R^*\) from the fraction equation. The deviation \(f_R-f_R^*\) then obeys a simple decay equation:

\frac{d}{dt}(f_R-f_R^*)=-\lambda(f_R-f_R^*).

For constant rates, separating variables and integrating from time zero to time \(t\) gives the solution below. The sensitive fraction follows from \(f_S=1-f_R\).

f_R(t)=f_R^*+\bigl(f_R(0)-f_R^*\bigr)e^{-\lambda t},\qquad f_S(t)=1-f_R(t).

Because \(\lambda>0\), the exponential term tends to zero as time increases, so the fraction converges to \(f_R^*\). The difference from equilibrium decays with time constant \(\tau=1/\lambda\), measured in the same time units as the rate constants. A time constant is not the time at which equilibrium is reached exactly: the model approaches its limit asymptotically.

Worked Example: Calculating a Fraction Before Equilibrium

Use the rates from the previous hypothetical example, and suppose the initial resistant fraction is \(f_R(0)=0.100\). Find the resistant fraction after 10 days. The switching-only assumptions and constant rates are unchanged.

Here \(f_R^*=0.300\) and \(\lambda=0.20\) per day. Substitute \(t=10\) days into the solution:

f_R(10)=0.300+(0.100-0.300)e^{-(0.20)(10)} =0.300-0.200e^{-2} \approx 0.300-0.02707 =0.2729.

The predicted resistant fraction after 10 days is about 27.3%; the sensitive fraction is \(1-0.2729=0.7271\), or about 72.7%. The resistant fraction has moved upward from 10.0% toward 30.0%, but has not yet reached the equilibrium value. The time constant is \(1/0.20=5\) days.

Composition and Convergence Are Different Questions

The equilibrium fraction depends on the ratio of the two rates, whereas the convergence time depends on their sum. Multiplying both rates by the same positive factor leaves the equilibrium fractions unchanged but increases the speed of switching and shortens the time constant. Thus two populations can share an equilibrium composition yet approach it at different speeds.

Worked Example: Same Equilibrium, Different Approach Time

Compare two hypothetical populations, each with an initial resistant fraction of 0.600. Population A has \(k_{SR}=0.02\) per day and \(k_{RS}=0.08\) per day. Population B has \(k_{SR}=0.10\) per day and \(k_{RS}=0.40\) per day. Assume the same switching-only model for both, and compare their fractions after 4 days.

For A, \(f_R^*=0.02/(0.02+0.08)=0.200\), \(\lambda=0.10\) per day, and \(\tau=10\) days. Therefore:

f_{R,A}(4)=0.200+(0.600-0.200)e^{-(0.10)(4)} =0.200+0.400e^{-0.4}\approx 0.4681.

For B, \(f_R^*=0.10/(0.10+0.40)=0.200\), \(\lambda=0.50\) per day, and \(\tau=2\) days. Therefore:

f_{R,B}(4)=0.200+(0.600-0.200)e^{-(0.50)(4)} =0.200+0.400e^{-2}\approx 0.2541.

Both populations have the same equilibrium resistant fraction, 20.0%, because their rate ratios are the same. After four days, however, B is much closer to equilibrium because its rates have a larger sum. Equilibrium composition alone would not reveal this difference in switching speed.

Boundary Cases, Diagnostics, and Limitations

The equilibrium formula needs careful interpretation at the boundaries. If \(k_{SR}=0\) and \(k_{RS}>0\), cells can switch only from resistant to sensitive, so the stable composition is \(f_R^*=0\). If \(k_{RS}=0\) and \(k_{SR}>0\), the stable composition is \(f_R^*=1\). If both rates are zero, no switching occurs: every initial composition stays fixed, and there is no unique equilibrium fraction. In that case the ratio formula has a zero denominator and must not be used.

Several checks help identify a calculation or model that is going wrong:

Check the fractions. Valid fractions must lie between zero and one and satisfy \(f_S+f_R=1\).
Check the flow balance. At the calculated equilibrium, \(k_{SR}f_S^*=k_{RS}f_R^*\).
Check the direction of movement. If the initial resistant fraction is below equilibrium, the model predicts it will rise toward equilibrium; if it is above, it will fall toward equilibrium.
Check units. Both rate constants must use the same time unit, and the product of a rate and elapsed time must be dimensionless in the exponential.

A common error is to interpret equilibrium as a state in which switching has stopped. That is incorrect: with positive rates, switching continues in both directions; the net change in each fraction is zero because the opposing flows balance. Another error is to treat the equilibrium resistant fraction as \(k_{SR}/k_{RS}\). That ratio is not itself a fraction and may exceed one. The normalized formula is \(k_{SR}/(k_{SR}+k_{RS})\).

The formula also has limits. If sensitive and resistant cells have different net growth or death rates, total cell number is no longer constant under switching and the fraction equation acquires additional terms. If rates vary with time, the constant-rate equilibrium and exponential solution need not describe the trajectory. A snapshot of the equilibrium fraction identifies a rate ratio under this model, not the two individual rates: additional information about switching speed is needed to separate them. These are reasons to check the assumptions, not reasons to infer switching rates from a fraction alone.

A deterministic fraction model is convenient for describing the expected composition of a sufficiently large population. A count-based model is useful when tumor burden as well as composition is needed; under the switching-only assumptions, it gives the same fractions after division by total cell number. For small populations, random switching can make realized counts fluctuate around the deterministic prediction. A measured time course can also be compared with the exponential approach derived above, whereas a single endpoint cannot assess that approach speed.

Check Your Understanding

Use the two-state switching model and state any assumptions needed. Solutions follow each question.

If \(k_{SR}=0.12\) per day and \(k_{RS}=0.18\) per day, what are the equilibrium fractions? Solution: \(f_R^*=0.12/(0.12+0.18)=0.400\), and \(f_S^*=0.18/0.30=0.600\).
For those rates, what is the time constant? Solution: \(\tau=1/(0.12+0.18)=3.33\) days.
If the resistant fraction starts below its equilibrium value, what direction of change does the model predict? Solution: It increases toward \(f_R^*\), because \(df_R/dt=(k_{SR}+k_{RS})(f_R^*-f_R)>0\) when \(f_R<f_R^*\).
Can an equilibrium resistant fraction of 0.400 determine both switching rates? Solution: No. It determines the ratio \(k_{SR}/(k_{SR}+k_{RS})=0.400\), but infinitely many positive rate pairs have that composition. A measure of the approach speed or other information is needed to determine the rate scale.
What is the implication if both switching rates are zero? Solution: Every initial composition remains unchanged. There is no unique equilibrium fraction, and the formula with denominator \(k_{SR}+k_{RS}\) is undefined.