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Oncology: Tumor Growth and Therapy Response · Tutorial 267 of 799

Predicting Relapse From Two-Clone Tumor Dynamics

Model sensitive and resistant clone burdens separately, then determine when their combined burden first reaches a specified threshold.

Advanced 11 min read

What You'll Learn

  • Write the two-clone exponential model from initial burdens and constant net growth rates.
  • Define relapse time as the first threshold crossing after a below-threshold starting burden.
  • Identify when total burden is monotonic and when a mixed-rate model may first decline.
  • Bracket and numerically calculate a threshold-crossing time.
  • Check growth-rate units, threshold units, crossing direction, and model limitations.

Why Predict Relapse From Both Clones?

A tumor can contain drug-sensitive cells that decline while treatment is active and resistant cells that persist or grow. A threshold-based relapse prediction must therefore account for the sum of the clone burdens, not just the clone that is growing fastest or the clone that is most abundant at one time. The prediction depends on each clone’s burden at the start of the forecast and its net growth rate during the forecast.

The calculation below is a deterministic model over a specified time window. Its result is the first modeled time that total burden reaches a chosen threshold; it is not, by itself, a clinical definition of relapse or a guarantee that a patient will relapse at that time.

Two-Clone Model and Relapse Time

Let \(S(t)\) and \(R(t)\) denote the sensitive- and resistant-clone burdens at time \(t\). Use the same burden unit for both clones, such as millions of cells. At \(t=0\), their explicit initial burdens are \(S_0\) and \(R_0\). Assume that each clone has a constant net growth rate over the prediction window: \(g_S\) for the sensitive clone and \(g_R\) for the resistant clone. Rates have units of inverse time, such as day\(^{-1}\); a negative rate represents net decline.

Definition: Under constant net growth rates, the two-clone model is \(S(t)=S_0e^{g_St}\) and \(R(t)=R_0e^{g_Rt}\). Total burden is \(B(t)=S(t)+R(t)\). For a specified relapse threshold \(B_{\mathrm{rel}}\), the modeled relapse time is the first time \(t\geq 0\) for which \(B(t)\geq B_{\mathrm{rel}}\), provided the burden starts below the threshold. If \(B(0)\geq B_{\mathrm{rel}}\), the threshold has already been reached at the start of the forecast.

This model follows from the clone-specific differential equations \(dS/dt=g_SS\) and \(dR/dt=g_RR\), with initial conditions \(S(0)=S_0\) and \(R(0)=R_0\). It assumes constant rates and no exchange between clones during the forecast. The rates are net rates: if proliferation and drug-mediated killing are represented separately in a more detailed model, their net effect must be reflected in \(g_S\) and \(g_R\) for this calculation.

The threshold and clone burdens must use the same measurement scale. The two initial burdens must refer to the same time and tumor, and the rates must apply to the same forecast conditions. For example, a rate estimated under one treatment condition cannot be used without justification for a forecast under a different condition.

When Does the Total Burden Rise?

The clone trajectories are exponential, but their sum is generally not a single exponential. Differentiating the total burden gives the instantaneous direction of change:

$$ \frac{dB(t)}{dt} =g_SS_0e^{g_St}+g_RR_0e^{g_Rt} =g_SS(t)+g_RR(t). $$

If both rates are nonnegative and at least one clone with positive burden has a positive rate, total burden increases. If both rates are nonpositive, total burden does not increase, so a threshold above the initial total cannot be reached. With one negative and one positive rate, the total may first fall and later rise, or first rise and later fall. The signs of the two rates alone do not establish the direction of the total at every time.

For the common mixed case \(g_S<0<g_R\), a turning point occurs when the declining sensitive contribution to the derivative balances the growing resistant contribution. Setting the derivative to zero and solving gives:

$$ t_{\min} =\frac{\ln\!\left(\dfrac{-g_SS_0}{g_RR_0}\right)} {g_R-g_S}. $$

This time is relevant only if it is nonnegative. The derivative changes from negative to positive there, because the ratio of the positive derivative contribution to the magnitude of the negative one grows as \(e^{(g_R-g_S)t}\). Thus, when \(t_{\min}>0\), the total has a minimum. A threshold search should not assume the total rises from \(t=0\): the first upward crossing may occur only after that minimum.

Worked Example: A Shrinking Sensitive Clone and a Growing Resistant Clone

Suppose the initial sensitive burden is 15.0 million cells and its net rate is \(-0.0200\) day\(^{-1}\). The resistant burden is 0.800 million cells with net rate \(0.0600\) day\(^{-1}\). Find when their total first reaches 20.0 million cells.

The model is \(B(t)=15.0e^{-0.0200t}+0.800e^{0.0600t}\), in millions of cells. Initially \(B(0)=15.8\) million, below the threshold. Because the sensitive clone declines and the resistant clone grows, check the total rather than assuming its direction from either clone alone.

At 48.00 days, the contributions are approximately \(15.0e^{-0.9600}=5.7434\) million and \(0.800e^{2.8800}=14.2514\) million. Thus \(B(48.00)\approx19.9948\) million, still below threshold. At 48.01 days, \(e^{-0.9602}\approx0.382816\) and \(e^{2.8806}\approx17.82496\). The contributions are approximately \(15.0(0.382816)=5.7422\) million and \(0.800(17.82496)=14.2600\) million, totaling \(20.0022\) million.

The crossing is between 48.00 and 48.01 days. Numerical interpolation or root finding gives approximately 48.007 days. The total initially falls because its derivative at time zero is \((-0.0200)(15.0)+(0.0600)(0.800)=-0.252\) million cells per day. It later turns upward as the resistant contribution becomes dominant. The reported time is the first threshold crossing after the starting burden, not the time at which the resistant clone alone reaches 20.0 million cells.

Finding a Crossing Numerically

Except in special cases, the equation \(S_0e^{g_St}+R_0e^{g_Rt}=B_{\mathrm{rel}}\) has no simple closed-form solution for \(t\). Define \(f(t)=B(t)-B_{\mathrm{rel}}\). A numerical root is appropriate once a bracket is found: times \(t_L\) and \(t_U\) for which \(f(t_L)<0\) and \(f(t_U)\geq0\) locate an upward crossing in that interval. Bisection repeatedly halves the bracket and preserves the sign change; it is reliable for a continuous function on the bracket. A numerical solver should still be used with care if there may be more than one crossing, since a bracket for a later crossing does not establish that it is the first one.

A practical approach is to evaluate the model at \(t=0\), inspect the rate signs and derivative, and then increase time in sensible increments until the total crosses the threshold. If mixed rates give an initial decline, evaluate around the possible minimum before looking for the upward crossing. Once a bracket is obtained, refine it with bisection. Newton’s method can converge faster when started near a root, but it uses the derivative and may leave the relevant interval or converge to a different crossing.

Worked Example: Both Clones Grow

Suppose \(S_0=4.00\) million cells, \(g_S=0.0250\) day\(^{-1}\), \(R_0=0.500\) million cells, and \(g_R=0.0800\) day\(^{-1}\). The threshold is 10.0 million cells. Both rates are positive, so the total increases and there can be at most one threshold crossing after \(t=0\).

Here \(B(t)=4.00e^{0.0250t}+0.500e^{0.0800t}\). At 22.30 days, \(B(22.30)=4.00e^{0.5575}+0.500e^{1.784}\approx6.985+2.977=9.962\) million cells. At 22.40 days, \(B(22.40)=4.00e^{0.5600}+0.500e^{1.792}\approx7.0027+3.0007\approx10.003\) million cells. The threshold is bracketed between these times.

Refining the bracket numerically gives \(t\approx22.392\) days. At that time, the sensitive contribution is about 7.001 million cells and the resistant contribution about 2.999 million cells, totaling approximately 10.000 million cells. Under the stated constant-rate assumptions, the predicted threshold time is therefore about 22.39 days.

Worked Example: A Decline Before the Threshold Is Reached

Let \(S_0=10.0\) million cells with \(g_S=-0.0500\) day\(^{-1}\), and \(R_0=0.500\) million cells with \(g_R=0.0400\) day\(^{-1}\). The initial total is 10.5 million cells, and the specified threshold is 12.0 million cells.

The total initially declines: \(B'(0)=(-0.0500)(10.0)+(0.0400)(0.500)=-0.480\) million cells per day. Nevertheless, the resistant clone grows indefinitely under this idealized exponential model, so the total eventually rises past the threshold. At 79.00 days, \(B(79.00)=10.0e^{-3.950}+0.500e^{3.160}\approx0.1925+11.785\approx11.978\) million cells. At 79.05 days, \(B(79.05)=10.0e^{-3.9525}+0.500e^{3.162}\approx0.1921+11.809\approx12.001\) million cells.

The upward crossing is bracketed by 79.00 and 79.05 days. Numerical refinement gives approximately 79.05 days. It would be incorrect to infer “no crossing” from the negative initial derivative: the derivative describes the initial direction, while the growing resistant clone eventually changes the total’s trajectory. It would also be incorrect to report the time of the total’s minimum as the relapse time.

Assumptions, Diagnostics, and Limits

Before interpreting a calculated time, verify that the inputs and model match the intended prediction:

  • Units and scale: \(S_0\), \(R_0\), and \(B_{\mathrm{rel}}\) use the same burden units, and each growth rate is in inverse units of the time axis.
  • Starting condition: if \(B(0)\geq B_{\mathrm{rel}}\), the threshold is already met. A positive-time relapse prediction from below threshold is not appropriate without redefining the forecast start.
  • Crossing direction: check that the reported root is an upward crossing when relapse is defined as reaching the threshold from below. A root that merely touches the threshold at a turning point may not represent an upward crossing.
  • Numerical bracket: confirm that the computed burden is below threshold at the lower endpoint and at or above it at the upper endpoint. Recalculate the total from both clone contributions at the reported time.
  • Rate applicability: check that the rates are plausible and apply over the entire forecast interval. A constant rate can be a poor approximation if treatment, tumor environment, or clone behavior changes.

A common mistake is to calculate the resistant clone’s threshold time and call it the tumor relapse time. That ignores the sensitive burden, which may still contribute substantially to total burden. Another error is to average the two clone-specific rates and apply that average to the initial total. In general, \(S_0e^{g_St}+R_0e^{g_Rt}\) is not equal to \((S_0+R_0)e^{\bar g t}\) for a fixed average rate \(\bar g\). Such a one-exponential approximation can misstate both the threshold time and whether total burden initially rises or falls.

The two-clone calculation is transparent and useful when initial burdens and rates are specified, but it treats those inputs as fixed and the trajectories as deterministic. It does not represent uncertainty in measurements or rates, stochastic loss of a small clone, switching between phenotypes, new mutation, interactions between clones, or capacity-limited tumor growth. A threshold time should therefore be interpreted as a conditional model prediction: it follows if the specified initial burdens, rates, threshold, and assumptions hold over the forecast window.

Key takeaway: Compute each clone’s burden separately, add the contributions, and solve for the first threshold crossing. The total trajectory—not the behavior of either clone considered alone—determines the predicted relapse time.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. A sensitive clone starts at 3.0 million cells and a resistant clone at 1.0 million cells. Both have a constant rate of 0.020 day\(^{-1}\). Write the total-burden model and calculate when it reaches 8.0 million cells.
  2. If both clone-specific rates are negative and the starting total is below the threshold, can the threshold be reached under this model? Explain.
  3. A model begins below threshold with one declining clone and one growing clone. Why is a negative initial derivative not enough to conclude that there will be no later upward crossing?
  4. At a proposed root, the clone burdens are 4.0 and 5.5 million cells, and the threshold is 10.0 million cells. Is that time a verified threshold crossing? What should be checked?

Solutions. (1) The total is \(B(t)=4.0e^{0.020t}\) million cells because both clones have the same rate. Solve \(4.0e^{0.020t}=8.0\): \(e^{0.020t}=2\), so \(t=\ln(2)/0.020=34.66\) days. (2) No. Each burden is nonincreasing, so their sum cannot rise from a starting value below the threshold. (3) The derivative at time zero gives only the initial direction. The growing clone can become large enough to outweigh the declining clone later, so the total must be evaluated over time or analyzed using its derivative. (4) No: the stated contributions sum to \(4.0+5.5=9.5\) million cells, below the 10.0 million-cell threshold. Recheck the time, exponential factors, arithmetic, units, and whether the proposed value is actually at or above the threshold.

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