Tumor Regrowth After Treatment Withdrawal: A Worked Example

Following One Tumor Clone Through Treatment and Withdrawal

A tumor burden measured at treatment withdrawal is not, by itself, a prediction of when a clinically relevant threshold will be reached again. That prediction also depends on the growth rate after drug activity ends. In a single-clone model, the calculation is direct once those rates and the withdrawal burden are specified: treatment changes the burden through the balance of cell growth and drug-induced killing, and withdrawal removes the drug-kill term.

This tutorial works through that calculation with hypothetical values. The example uses two treatment intervals with different kill rates to represent a changing treatment effect, followed by drug withdrawal. It is a deterministic illustration, not a patient-specific prediction.

Definition: Let $N(t)$ be the burden of a single tumor-cell clone at time $t$, measured in cells. Let $g$ be its per-capita growth rate and $k(t)$ its per-capita drug-kill rate, both in $\text{day}^{-1}$. While treatment is active, the model is $$ \frac{dN}{dt} = \bigl(g-k(t)\bigr)N. $$ After complete treatment withdrawal, assume drug killing is zero and the clone regrows at rate $g$. The burden at withdrawal is denoted $N_w$.

The model assumes the clone is adequately represented by one burden state; growth and kill rates are constant within each treatment interval; killing is proportional to the current burden; and, after withdrawal, the relevant drug effect stops immediately. It also assumes no immune-mediated killing, migration, new clone, or density-dependent limitation is needed to describe the time window of interest. A positive growth rate is required for upward threshold crossing after withdrawal. These assumptions make the worked calculation transparent, but they are not guaranteed to hold in a real tumor.

From Rates to Burden

During any interval of length $\Delta t$ with constant $g$ and $k$, the net per-capita rate is $g-k$. If the burden at the interval's start is $N_{\mathrm{start}}$, its burden at the end is

N_{\mathrm{end}}=N_{\mathrm{start}}\exp\!\bigl((g-k)\Delta t\bigr).

When treatment rates change between intervals, calculate the end burden of one interval and use it as the starting burden for the next. After withdrawal, if $g>0$, the burden at elapsed time $\tau$ is $N_w\exp(g\tau)$. For a threshold $H$, the time to cross it from below is therefore

\tau_H=\frac{\ln(H/N_w)}{g}, \qquad \text{provided } H>N_w,\;N_w>0,\;\text{and }g>0.

If $N_w=0$ and $H>0$, this model leaves the burden at zero, so the threshold is never reached. If $H\leq N_w$, the threshold is already met at withdrawal; it is not a future upward crossing. These conditions matter as much as the formula.

Worked Example: Burden After Two Treatment Intervals

Suppose a hypothetical tumor begins with $N_0=1.00\times10^9$ cells. Its single clone has $g=0.04500\ \text{day}^{-1}$. For the first 30 days, the drug-kill rate is $k_1=0.068925\ \text{day}^{-1}$; for the next 10 days, it is $k_2=0.07500\ \text{day}^{-1}$. Treatment is withdrawn at day 40. Find the burden at days 30 and 40.

For days 0–30, the net rate is $g-k_1=0.04500-0.068925=-0.023925\ \text{day}^{-1}$. Thus

N_{30}=(1.00\times10^9)\exp\!\bigl((-0.023925)(30)\bigr) =(1.00\times10^9)\exp(-0.71775) \approx4.8785\times10^8\ \text{cells}.

For days 30–40, the net rate is $g-k_2=0.04500-0.07500=-0.03000\ \text{day}^{-1}$. Starting from $N_{30}$,

N_{40}=N_{30}\exp\!\bigl((-0.03000)(10)\bigr) =(1.00\times10^9)\exp(-1.01775) \approx3.6141\times10^8\ \text{cells}.

Both net rates are negative, so the predicted burden falls during both treatment intervals. The burden at withdrawal is $N_w=N_{40}$, not the initial burden and not the day-30 burden. It is this withdrawal burden that starts the post-treatment regrowth calculation.

Calculating Regrowth and Comparing Withdrawal Times

The post-withdrawal calculation uses the untreated growth rate $g$, not the on-treatment net rate $g-k$. In the example, complete withdrawal sets $k=0$, so the burden grows at $0.04500\ \text{day}^{-1}$. For threshold $H=1.00\times10^9$ cells, the burden at withdrawal is positive and below threshold, and the growth rate is positive. All conditions for the crossing-time formula are satisfied.

Worked Example: Time to Return to the Initial Burden

Using the day-40 withdrawal burden from the previous calculation, find the time after withdrawal to reach $H=1.00\times10^9$ cells.

Substitute $N_w=3.6141\times10^8$ cells and $g=0.04500\ \text{day}^{-1}$:

\tau_H =\frac{\ln\!\left((1.00\times10^9)/(3.6141\times10^8)\right)} {0.04500\ \text{day}^{-1}} =\frac{1.01775}{0.04500\ \text{day}^{-1}} \approx22.62\ \text{days}.

The model predicts that this hypothetical clone reaches $1.00\times10^9$ cells about 22.62 days after withdrawal. The logarithm's argument is a ratio of burdens, so it is dimensionless; division by a rate in $\text{day}^{-1}$ gives time in days. This is a model-based threshold prediction, not a probability or a guarantee of an observed clinical event.

The impact of treatment duration can be compared using the same rates and threshold. After 30 days of treatment, the burden is $4.8785\times10^8$ cells. If treatment is then withdrawn and the same $g$ applies, the time to $1.00\times10^9$ cells is

\tau_{H,30} =\frac{\ln\!\left((1.00\times10^9)/(4.8785\times10^8)\right)} {0.04500\ \text{day}^{-1}} =\frac{0.71775}{0.04500\ \text{day}^{-1}} =15.95\ \text{days}.

With withdrawal at day 40, the prediction is 22.62 days after withdrawal. The difference is $22.62-15.95=6.67$ days. This comparison holds the post-withdrawal growth rate and threshold fixed; the later withdrawal leaves a smaller burden, so more regrowth time is needed to reach the same threshold. The comparison does not say that the later treatment schedule is clinically better: it accounts only for the specified burden model and rates.

Worked Example: A Different Threshold and the Zero-Burden Boundary

Suppose withdrawal occurs at day 40 as above, but the threshold of interest is $H=6.00\times10^8$ cells. The threshold is above $N_w$, and $N_w>0$ and $g>0$, so the formula applies:

\tau_H =\frac{\ln\!\left((6.00\times10^8)/(3.6141\times10^8)\right)} {0.04500\ \text{day}^{-1}} \approx\frac{0.5069}{0.04500\ \text{day}^{-1}} \approx11.26\ \text{days}.

The lower threshold is reached sooner than $1.00\times10^9$ cells, as expected for monotonically increasing burden. Now consider the boundary case $N_w=0$ with the same positive threshold. The post-withdrawal differential equation is $dN/dt=gN$, and $N(0)=0$ has the solution $N(t)=0$. Thus the threshold is never reached in this model; taking a logarithm of $H/N_w$ would be undefined and would not be a valid calculation.

Justification of the Threshold Formula

After withdrawal, the assumptions give $dN/dt=gN$, with $N(0)=N_w$. For $N_w>0$, divide both sides by $N$ and integrate from withdrawal to elapsed time $\tau$:

\int_{N_w}^{N(\tau)}\frac{dN}{N} =\int_0^\tau g\,dt \quad\Longrightarrow\quad \ln\!\left(\frac{N(\tau)}{N_w}\right)=g\tau \quad\Longrightarrow\quad N(\tau)=N_w e^{g\tau}.

At an upward crossing, $N(\tau_H)=H$. Substituting $H$ and solving gives $\tau_H=\ln(H/N_w)/g$. If $H>N_w$ and $g>0$, both the logarithm and the resulting time are positive. If $H=N_w$, the crossing is at withdrawal, $\tau_H=0$; if $H<N_w$, the burden is already above the threshold at withdrawal. The derivation also shows why the zero-burden case must be handled separately: dividing by $N$ requires $N>0$.

Diagnostics, Alternatives, and Limitations

Before interpreting a threshold time, check the calculation against the trajectory's basic properties. When $k>g$, the on-treatment burden should decline; when $k<g$, it should rise; and when $k=g$, it should remain constant. After withdrawal with $g>0$, a positive burden should increase. A computed future crossing time that is negative when $H>N_w$, or that decreases when the threshold is raised while all else is fixed, signals a likely sign, unit, or substitution error.

The exponential single-clone model is useful when a compact calculation is the goal, but its assumptions can be restrictive. Real tumors may slow as they become larger, drug concentrations and effects may change over time, and a measured burden may combine clones with distinct growth or treatment sensitivity. If evidence supports density dependence, a logistic-type growth model may be more appropriate; if clones differ, a multi-clone model can represent their separate burdens and rates. Those alternatives do not share the same simple threshold-time formula in general. Their added structure is useful only when the data and question support estimating it.

A practical limitation is that the formula treats $g$ and $N_w$ as known. In applications, both may be uncertain, and the predicted time can be sensitive to them. In particular, the denominator $g$ means a smaller positive growth rate produces a longer predicted time. A rate estimated over one observation window may not remain constant after withdrawal. Also, the model's burden is a cell count: a clinical imaging threshold may use a different measurement scale and require a justified link between that measurement and $N(t)$.

Common Pitfalls

Using the treatment net rate after withdrawal. The rate $g-k$ applies while the drug-kill term is active. Under complete withdrawal here, the rate is $g$, not $g-k$.
Using the wrong starting burden. Regrowth starts from $N_w$, the burden at withdrawal. Use the burden at the correct treatment duration, especially when kill rates change between intervals.
Dropping the formula's conditions. For a positive threshold above the withdrawal burden, require $N_w>0$ and $g>0$. If $N_w=0$, positive exponential growth from zero does not spontaneously create tumor cells in this deterministic model.
Mixing time units. Rates in $\text{day}^{-1}$ produce times in days. Rates expressed per week require durations and reported times in weeks, or a consistent unit conversion.
Over-interpreting a point prediction. A calculated crossing time follows from the assumed rates and model structure. It is not evidence that the tumor will follow that exact trajectory.

Key takeaway: Calculate the burden at withdrawal by applying the correct net growth rate in each treatment interval. Then, for $H>N_w>0$ and $g>0$, the time to reach a higher threshold after complete withdrawal is $\ln(H/N_w)/g$. Verify the assumptions, units, and boundary cases before interpreting that time.

Check Your Understanding

Answer each question using the stated single-clone model. Solutions are included after the questions.

A clone has $N_w=2.00\times10^8$ cells at withdrawal and $g=0.0500\ \text{day}^{-1}$. How long does it take to reach $4.00\times10^8$ cells?
If the burden at withdrawal is $5.00\times10^8$ cells and $H=4.00\times10^8$ cells, is the threshold a future upward crossing?
What does the model predict for $N_w=0$, $H=1.00\times10^6$ cells, and $g>0$?
During treatment, $g=0.040\ \text{day}^{-1}$ and $k=0.030\ \text{day}^{-1}$. Does the burden rise or fall, and what is the net rate?

Solutions. (1) The conditions hold because $H>N_w>0$ and $g>0$. Thus $\tau_H=\ln(4.00\times10^8/2.00\times10^8)/(0.0500\ \text{day}^{-1})=\ln(2)/0.0500\approx13.86$ days. (2) No future upward crossing is being predicted: $H<N_w$, so the threshold is already met at withdrawal. (3) The solution is $N(t)=0$ for all $t$, so the positive threshold is never reached. (4) The net rate is $g-k=0.040-0.030=0.010\ \text{day}^{-1}$, which is positive; the model predicts that the burden rises during treatment.