Deriving Sensitive-Resistant Clone Selection Under Therapy

Why Relative Selection Matters

A tumor can shrink during treatment while the fraction of its cells that are resistant increases. These observations are not contradictory: total abundance depends on whether cells are growing or dying in absolute terms, while clonal selection depends on how the clones’ net rates compare. A two-clone model makes that distinction precise and shows when a resistant clone already present before treatment can become the more abundant clone.

The analysis below assumes constant treatment and constant clone-specific net rates. It uses the exponential growth and net-kill framework established in the earlier material on tumor growth and treatment effects, with a separate abundance for each clone.

Definition: Let \(S(t)\) and \(R(t)\) denote the sensitive- and resistant-clone abundances at time \(t\), respectively. Under constant treatment, let \(r_S\) and \(r_R\) be their per-capita growth rates and \(k_S\) and \(k_R\) their per-capita treatment-induced kill rates. Define their net rates as \(g_S=r_S-k_S\) and \(g_R=r_R-k_R\). Each rate has units of inverse time. The resistant-to-sensitive abundance ratio is \(Q(t)=R(t)/S(t)\), and the resistant fraction of the two-clone population is \(F_R(t)=R(t)/(R(t)+S(t))\).

Model and Assumptions

The constant-rate clone equations are

\frac{dS}{dt}=g_S S,\qquad \frac{dR}{dt}=g_R R.

This model assumes that treatment exposure is constant over the interval being analyzed; each clone’s net rate is therefore constant. It also assumes that there is no conversion between the two clones, and that their per-capita rates do not depend on tumor abundance or on the abundance of the other clone. The initial abundances \(S_0=S(0)\) and \(R_0=R(0)\) are nonnegative. To use a finite ratio \(Q(t)\), both must be positive. These are deterministic equations: they do not represent random extinction when a clone is present in very small numbers.

The model calls a clone “resistant” in a relative sense: treatment affects it less, or its net rate under treatment is more favorable, than for the sensitive clone. Resistance does not require \(g_R\) to be positive. If \(g_R<0\), the resistant clone is declining in absolute abundance even while it may be gaining relative to the sensitive clone.

Solving the Clone Equations and the Ratio

Each equation has the form of exponential growth with a constant net rate. Separating variables and integrating from time zero to \(t\) gives the two trajectories:

S(t)=S_0e^{g_St},\qquad R(t)=R_0e^{g_Rt}.

When \(S_0>0\) and \(R_0>0\), divide the trajectories to obtain the ratio. The initial ratio is \(Q_0=R_0/S_0\), and the difference between the net rates is \(\Delta g=g_R-g_S\).

Q(t)=\frac{R(t)}{S(t)} =\frac{R_0e^{g_Rt}}{S_0e^{g_St}} =Q_0e^{(g_R-g_S)t} =Q_0e^{\Delta g t}.

This derivation isolates selection: the ratio changes according to the difference in net rates, not either rate by itself. The resistant fraction follows directly from the ratio, since dividing its numerator and denominator by \(S(t)\) gives \(F_R(t)=Q(t)/(1+Q(t))\).

F_R(t)=\frac{Q_0e^{\Delta g t}}{1+Q_0e^{\Delta g t}}.

The condition for the resistant clone to be more abundant at time \(t\) is \(Q(t)>1\). If it starts less abundant (\(0<Q_0<1\)) and \(\Delta g>0\), the time at which the clones have equal abundance is found by setting \(Q(t)=1\):

t_{\mathrm{equal}}=\frac{\ln(1/Q_0)}{\Delta g} =\frac{\ln(S_0/R_0)}{\Delta g}.

This is a positive, finite time under those stated conditions. If \(Q_0\geq 1\), the resistant clone is already at least as abundant at treatment start. If \(\Delta g=0\), the ratio remains constant; if \(\Delta g<0\), the resistant fraction declines. When \(R_0=0\), the equations keep \(R(t)=0\), so this model has no mechanism that can create a resistant clone.

Worked Example: Enrichment While Both Clones Decline

Suppose a hypothetical tumor begins treatment with \(S_0=9{,}900\) sensitive cells and \(R_0=100\) resistant cells. Under constant treatment, let \(g_S=-0.30\) per day and \(g_R=-0.06\) per day. Find the ratio and resistant fraction after 10 days, and determine whether the resistant clone is growing.

The initial ratio is \(Q_0=100/9{,}900=1/99\). The relative net-rate difference is \(\Delta g=-0.06-(-0.30)=0.24\) per day. Therefore,

Q(10)=\frac{1}{99}e^{0.24(10)} =\frac{e^{2.4}}{99} =\frac{11.0232}{99} \approx 0.1113.

The resistant fraction is

F_R(10)=\frac{0.111345}{1+0.111345} \approx 0.1002,

or about 10.0%. The resistant clone’s net rate is negative, \(g_R=-0.06\) per day, so its abundance declines. The sensitive clone declines faster, at \(-0.30\) per day, so the resistant clone becomes a larger fraction of the shrinking tumor. Relative enrichment does not imply absolute resistant-cell growth.

When Does the Resistant Clone Dominate?

For a pre-existing resistant clone, the sign of \(\Delta g\) determines the direction in which the ratio moves. A positive difference means the resistant clone gains relative to the sensitive clone; a negative difference means it loses relative abundance. A zero difference means neither clone is selected over the other by this model. These statements concern the ratio, not whether either clone’s absolute abundance rises or falls.

The following hypothetical example shows how a clone can cross from minority to majority even when both net rates are negative.

Worked Example: Calculating the Time to Resistant-Clone Dominance

Suppose \(S_0=900\), \(R_0=100\), \(g_S=-0.10\) per day, and \(g_R=-0.04\) per day. Calculate the resistant-to-sensitive ratio and resistant fraction at days 20 and 40, then find the time of equal abundance.

Here \(Q_0=100/900=1/9\) and \(\Delta g=-0.04-(-0.10)=0.06\) per day. At day 20,

Q(20)=\frac{1}{9}e^{0.06(20)} =\frac{e^{1.2}}{9} =\frac{3.3201}{9} \approx 0.3689, \qquad F_R(20)=\frac{0.3689}{1+0.3689} \approx 0.2695.

The resistant fraction is about 27.0% at day 20. At day 40,

Q(40)=\frac{1}{9}e^{0.06(40)} =\frac{e^{2.4}}{9} =\frac{11.0232}{9} \approx 1.225, \qquad F_R(40)=\frac{1.2248}{1+1.2248} \approx 0.5505.

Thus the resistant clone is slightly more than half of the two-clone population at day 40. The equal-abundance time is

t_{\mathrm{equal}}=\frac{\ln(900/100)}{0.06} =\frac{\ln(9)}{0.06} =\frac{2.1972}{0.06} \approx 36.62\ \text{days}.

Both net rates are negative, so both clones decline in absolute abundance. The resistant clone becomes dominant because its net rate is less negative, not because it is growing.

Absolute Abundance and Relative Selection Are Different Questions

The separate trajectories answer whether each clone grows or declines: \(g_i>0\) means clone \(i\) grows exponentially, \(g_i=0\) means its abundance is constant, and \(g_i<0\) means it declines. The ratio answers a different question: whether one clone gains relative to the other. A resistant clone can be dominant while declining, or can grow without becoming dominant if it begins rare and the observation interval is short.

For example, if \(g_R>0\) but \(g_S>g_R\), both clones grow while the resistant fraction falls. Conversely, if \(g_R<0\) and \(g_S<g_R\), both decline while the resistant fraction rises. This distinction matters when interpreting tumor shrinkage: a reduction in total tumor burden does not, by itself, show that the resistant fraction is falling.

Worked Example: Equal Abundance With a Initially Rare Resistant Clone

Consider a hypothetical tumor with \(S_0=800\) and \(R_0=200\). Under constant treatment, let \(g_S=-0.10\) per day and \(g_R=0.05\) per day. Find the time to equal abundance and state what the rates imply.

The initial ratio is \(Q_0=200/800=0.25\), and \(\Delta g=0.05-(-0.10)=0.15\) per day. Since \(Q_0<1\) and \(\Delta g>0\), the equality time is

t_{\mathrm{equal}}=\frac{\ln(800/200)}{0.15} =\frac{\ln(4)}{0.15} =\frac{1.3863}{0.15} \approx 9.242\ \text{days}.

Before that time the resistant clone remains less abundant; after it, the resistant clone is more abundant under this model. Here \(g_R>0\), so the resistant abundance increases, while \(g_S<0\), so the sensitive abundance decreases. This absolute-growth pattern is one possible route to dominance, but the earlier example shows it is not required.

Checking the Model and Choosing an Approach

The closed-form solutions are useful when the assumptions of constant net rates and no exchange between clones are reasonable over the time interval. Their structure also suggests practical checks. If measurements are available at several times, calculate or estimate the resistant-to-sensitive ratio at each time. Under this model, the natural logarithm of that ratio should change linearly with time, with slope \(\Delta g\):

\ln Q(t)=\ln Q_0+\Delta g\,t.

Strong curvature, abrupt changes in slope, or ratios that cannot be reconciled with the estimated rates may indicate that constant net rates are inadequate, the clone measurements are noisy, or an assumption such as independent clone dynamics is failing. Ratios are especially unstable when the measured sensitive abundance is near zero. If no resistant cells are detected, that is not proof that the true resistant abundance is exactly zero; the deterministic model’s \(R_0=0\) case should not be used to infer that resistance can never appear.

For a constant-rate interval, solving the two equations directly is the simplest approach; the ratio formula then gives selection without repeatedly calculating both abundances. An equivalent approach is to track total abundance and resistant fraction, but it must encode the same clone-specific rates to give the same result. When rates vary over time, the constant-rate exponential formulas no longer apply as written; numerical integration or a piecewise-constant approximation can be used instead. Those alternatives do not remove the need to assess whether the model structure is plausible.

The model is deliberately limited. It omits stochastic loss or expansion at low cell counts, changing treatment exposure, changes in clone-specific rates over time, interactions between clones, and the creation of resistant cells from sensitive cells. Consequently, a calculated equality time is a model-based projection, not a guaranteed clinical time of relapse or dominance. It also assumes the two abundances can be meaningfully distinguished and measured.

Common Mistake

A common error is to conclude that a resistant clone can increase in fraction only when \(g_R>0\). That reasoning confuses absolute growth with relative selection. The ratio equation shows that the fraction increases whenever \(\Delta g=g_R-g_S>0\), even if both rates are negative. The relevant comparison is between net rates, while the sign of each rate separately determines whether that clone grows or declines in absolute abundance.

Key takeaway: Under constant treatment and independent exponential clone dynamics, \(Q(t)=Q_0e^{(g_R-g_S)t}\). Resistant-clone dominance is governed by the relative net-rate difference and starting ratio; absolute resistant growth is governed by \(g_R\) alone.

Check Your Understanding

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

If \(g_R=g_S\), what happens to the resistant-to-sensitive ratio? Can total tumor abundance still change?
A tumor starts with \(S_0=800\) and \(R_0=200\), with \(g_S=-0.10\) per day and \(g_R=0.05\) per day. Calculate the equal-abundance time.
A tumor starts with \(S_0=950\) and \(R_0=50\), and \(\Delta g=0.08\) per day. Calculate the time at which the clones have equal abundance.
If \(Q_0=0.20\) and \(\Delta g=-0.03\) per day, calculate \(Q(30)\) and the resistant fraction at day 30. Is the resistant fraction increasing or decreasing?
Can a resistant clone become dominant if both clone-specific net rates are negative? State the condition on their difference.

Solutions. (1) The ratio stays at \(Q_0\), because \(Q(t)=Q_0e^0=Q_0\). Total abundance can still grow or decline according to the clone rates. (2) The rate difference is \(0.15\) per day, so \(t_{\mathrm{equal}}=\ln(800/200)/0.15=\ln(4)/0.15=1.3863/0.15\approx 9.242\) days. (3) Here \(Q_0=50/950=1/19\), so \(t_{\mathrm{equal}}=\ln(19)/0.08=2.9444/0.08\approx 36.81\) days. (4) \(Q(30)=0.20e^{-0.03(30)}=0.20e^{-0.9}\approx0.08131\). The resistant fraction is \(0.08131/(1+0.08131)\approx0.07520\), or 7.520%; it is decreasing because \(\Delta g<0\). (5) Yes. If \(g_R<0\), \(g_S<0\), and \(g_R>g_S\), both clones decline but the resistant clone gains relative abundance and can become dominant if it starts below the sensitive clone and treatment continues long enough.