Deriving Saturable Expansion Models for Infused Cells

Why the Per-Cell Growth Assumption Matters

After an infused cell product has a viable starting count, a model must describe how that count changes over time. The earlier material on converting cell product dose into viable starting conditions supplies the initial count \(N_0\); the choice of expansion equation determines what happens after that time. A model with constant proliferation per cell predicts continuing exponential growth. A model in which crowding reduces the per-cell rate predicts expansion that slows toward a limiting count.

These equations are not just different curve shapes. Each encodes a distinct assumption about how cells proliferate as the population changes. The simplest derivation starts with the change in count over a short time interval and expresses it as a per-cell rate multiplied by the number of cells.

Definition: Let \(N(t)\) be the number of viable cells in the modeled population at time \(t\), and let \(N_0=N(0)>0\). The net per-cell growth rate is the instantaneous change in cell count per cell, \((1/N)(dN/dt)\), with units of inverse time. It is a net rate: proliferation and any losses represented in the same population balance contribute to it.

For the equations below, time is continuous, the count is treated as a continuous quantity, and the modeled population is considered a single well-mixed population. The derivations assume no external cell input or removal during the modeled interval. Thus, an observed change in count is attributed to the net per-cell process specified by the model, not to cells entering or leaving the measured compartment. These are modeling assumptions, not consequences of observing a cell count.

Constant Per-Cell Growth Gives Exponential Expansion

Suppose each cell has the same constant net contribution \(r\) to the population’s per-cell growth rate. Here \(r>0\) has units of inverse time. In a short interval \(\Delta t\), the expected net change is approximately \(rN(t)\Delta t\). Dividing by \(\Delta t\) and taking the limit gives the exponential growth equation.

\frac{dN}{dt}=rN,\qquad N(0)=N_0

The solution follows by separating variables and integrating from the initial count to the count at time \(t\):

\frac{dN}{N}=r\,dt \quad\Longrightarrow\quad \ln\!\left(\frac{N(t)}{N_0}\right)=rt \quad\Longrightarrow\quad N(t)=N_0e^{rt}.

The logarithm of the count therefore increases linearly with time, with slope \(r\). The absolute growth rate \(dN/dt=rN\) increases as the count increases. The model has no finite upper limit: for \(r>0\), its predicted count keeps increasing without bound. This is a useful short-term approximation when crowding or other density constraints are negligible, but it should not be mistaken for a universal long-term biological law.

Worked Example: From a Constant Per-Cell Rate to a Count

Suppose a hypothetical infused population begins with \(N_0=1.20\times10^6\) viable cells and has a constant net per-cell growth rate \(r=0.250\ \mathrm{day}^{-1}\). Under the exponential model, calculate the predicted count after 3.00 days.

The assumptions are that this net rate remains constant over the interval and that there is no external input or removal in the modeled population. The solution is \(N(t)=N_0e^{rt}\). Substituting the stated values gives

N(3.00)=1.20\times10^6 e^{(0.250\ \mathrm{day}^{-1})(3.00\ \mathrm{days})} =1.20\times10^6 e^{0.750} \approx 2.54\times10^6\ \mathrm{cells}.

The exponent is dimensionless because the rate in inverse days is multiplied by time in days. This is a model prediction, not a claim that every cell divides at exactly the same time or that a measured count must equal this value.

Density-Dependent Per-Cell Growth Gives the Logistic Model

To represent saturable expansion, assume that the net per-cell growth rate declines as the population approaches a finite capacity \(K>0\). A common first-order density constraint is a linear decline from \(r\) at very low density to zero at \(N=K\):

\frac{1}{N}\frac{dN}{dt}=r\left(1-\frac{N}{K}\right).

Here \(r>0\) is the low-density net per-cell growth rate, and \(K\) is the limiting cell count, with units of cells. Multiplying by \(N\) gives the logistic differential equation. Its density factor is an explicit assumption: the per-cell rate is reduced in proportion to \(N/K\).

\frac{dN}{dt}=rN\left(1-\frac{N}{K}\right),\qquad N(0)=N_0.

For \(0<N<K\), both \(N\) and \(1-N/K\) are positive, so the count grows. At \(N=K\), the growth rate is zero. For \(N>K\), the rate is negative, and the deterministic model predicts a decline toward \(K\). These signs are useful checks on the equation and on any interpretation of a trajectory.

The closed-form solution can be derived by separating variables. Since \(1/[N(1-N/K)]=1/N+1/(K-N)\), integration gives a linear expression in the log of the odds of being below capacity.

\frac{dN}{N(1-N/K)}=r\,dt

\ln\!\left(\frac{N(t)}{K-N(t)}\right) -\ln\!\left(\frac{N_0}{K-N_0}\right)=rt \qquad (0<N_0<K).

Exponentiating and rearranging yields the logistic trajectory:

N(t)=\frac{K}{1+\left(\frac{K}{N_0}-1\right)e^{-rt}}.

This form assumes \(0<N_0<K\), the usual case for expansion toward capacity. If \(N_0=K\), the solution is constant at \(K\). If \(N_0>K\), the differential equation still applies, but the displayed rearranged form with the same initial-value interpretation should be handled with care; the solution approaches \(K\) from above as the count declines.

Interpretation: In the logistic model, \(r\) is the low-density per-cell rate, not the initial absolute slope unless density is negligible. The initial slope is \(rN_0(1-N_0/K)\). The parameter \(K\) is the model’s limiting count under the assumed conditions; it is not automatically a measured biological capacity.

Comparing Predicted Trajectories

Both models share the same initial count and use a rate parameter \(r\), but their rates differ as soon as the population is appreciable relative to \(K\). The logistic model has an initial slope reduced by the factor \(1-N_0/K\). At \(N=K/2\), its absolute growth rate is maximal: substituting gives \(r(K/2)(1/2)=rK/4\). Below \(K/2\), the absolute logistic growth rate increases with \(N\); between \(K/2\) and \(K\), it remains positive but decreases toward zero as the count approaches \(K\). The per-cell rate, by contrast, decreases throughout the interval \(0<N<K\).

Worked Example: Comparing Exponential and Logistic Predictions

Consider hypothetical parameters \(N_0=2.00\times10^6\) cells, \(r=0.400\ \mathrm{day}^{-1}\), and \(K=5.00\times10^7\) cells. Assume each model’s stated rate and population assumptions hold for 10.0 days. Calculate both predictions and compare their initial slopes.

For exponential expansion,

N_{\mathrm{exp}}(10) =2.00\times10^6 e^{(0.400)(10.0)} =2.00\times10^6 e^4 \approx 1.092\times10^8\ \mathrm{cells}.

For logistic expansion, \(K/N_0-1=5.00\times10^7/(2.00\times10^6)-1=24.0\), so

N_{\mathrm{log}}(10) =\frac{5.00\times10^7}{1+24.0e^{-4}} =\frac{5.00\times10^7}{1+24.0(0.0183156)} \approx 3.473\times10^7\ \mathrm{cells}.

The exponential prediction exceeds the logistic prediction by approximately \(1.092\times10^8-3.473\times10^7=7.45\times10^7\) cells. The predictions differ because exponential growth retains the full per-cell rate, while the logistic rate is reduced by density. At the start, the exponential slope is

rN_0=0.400(2.00\times10^6) =8.00\times10^5\ \mathrm{cells/day},

whereas the logistic initial slope is

rN_0\left(1-\frac{N_0}{K}\right) =0.400(2.00\times10^6)\left(1-\frac{2.00\times10^6}{5.00\times10^7}\right) =7.68\times10^5\ \mathrm{cells/day}.

The two slopes are already different at time zero, although the difference is small because \(N_0/K=0.0400\). A small early difference does not imply similar long-term predictions: the exponential model has no capacity, while the logistic prediction bends toward \(K\).

Worked Example: Locating the Maximum Logistic Growth Rate

For a hypothetical population with \(r=0.300\ \mathrm{day}^{-1}\) and \(K=4.00\times10^7\) cells, find the count at which absolute growth is maximal and calculate that maximum rate. Assume \(0<N<K\).

The absolute rate is \(g(N)=rN(1-N/K)=rN-rN^2/K\). Differentiating with respect to \(N\) gives \(g'(N)=r-2rN/K\). Setting this derivative to zero gives \(N=K/2\); the second derivative is \(-2r/K<0\), confirming a maximum.

N=\frac{K}{2}=2.00\times10^7\ \mathrm{cells}, \qquad g_{\max}=\frac{rK}{4} =\frac{(0.300)(4.00\times10^7)}{4} =3.00\times10^6\ \mathrm{cells/day}.

This maximum concerns the total count gained per day, not the per-cell rate. At \(K/2\), the per-cell rate is \(r(1-1/2)=0.150\ \mathrm{day}^{-1}\). If a trajectory starts above \(K/2\) but below \(K\), its absolute rate is positive and decreases toward zero as it approaches \(K\); it does not first rise to the maximum.

Assumptions, Diagnostics, and Alternatives

A useful model choice depends on both the mechanism being represented and the time range of interest. Exponential growth is simpler and can be appropriate when the population is far below any effective constraint. Logistic growth adds a specific saturation mechanism: the per-cell rate declines linearly with the fraction of capacity occupied. It is not merely an exponential model with a cap imposed after the calculation.

Diagnostics should focus on whether the assumed rate structure is plausible, rather than on curve shape alone. Under exponential growth, \(\ln N(t)=\ln N_0+rt\); substantial systematic curvature in a plot of log count against time is evidence against a constant per-cell rate. Under logistic growth, the per-cell rate is \(r(1-N/K)\), so it should decline as \(N\) rises, and the trajectory should flatten as it nears \(K\). A fitted logistic curve that predicts a negative growth rate while counts are well below the proposed capacity, or that places \(K\) below observed counts without representing a decline, merits checking for a parameterization or data problem.

Sparse observations confined to low counts may not distinguish the models: when \(N/K\) is small, \(1-N/K\) is near one, so logistic growth resembles exponential growth. In that region, the data may provide little information about \(K\). A plateau can support a saturable model, but it does not by itself establish that the cause is density-dependent proliferation. Changes in the measured compartment, time-varying rates, or other unmodeled processes can also alter observed counts. If those processes matter, the one-population assumptions need revision rather than an automatic interpretation of the fitted \(K\).

Other saturation functions can encode different relationships between cell density and per-cell rate, and a model with an explicitly represented limiting resource can be more mechanistic. Such alternatives require additional assumptions and often additional data. For a first comparison, the exponential model is the constant-rate reference and the logistic model is a parsimonious saturable alternative. Neither should be selected solely because it gives a visually appealing fit.

Common pitfall: It is incorrect to say that the logistic rate is negative whenever the population is above \(K/2\). For \(K/2<N<K\), \(rN(1-N/K)>0\): the count is still increasing, but its positive absolute growth rate is decreasing toward zero. The rate becomes negative only when \(N>K\).

Practical Model-Checking Sequence

Define the modeled count.
Specify which viable cells \(N(t)\) represents, the time origin, and the units. Use the viable starting count as \(N_0\).

State the per-cell assumption.
Choose a constant net rate \(r\) for exponential growth, or a density-reduced rate \(r(1-N/K)\) for logistic growth. State the interval and assumptions under which it is being used.

Check rate signs and units.
Confirm that \(r\) has units of inverse time, \(K\) has units of cells, and the predicted derivative has units of cells per time. For logistic growth, check whether \(N\) is below, at, or above \(K\).

Compare predictions over the observed range.
Examine whether a constant per-cell rate is plausible and whether the data show evidence of slowing toward a limit. Do not infer a well-determined \(K\) from observations that remain far below it.

Check Your Understanding

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

A hypothetical population has \(N_0=5.00\times10^5\) cells and constant \(r=0.200\ \mathrm{day}^{-1}\). What does the exponential model predict at 2.00 days?
For logistic growth with \(r=0.500\ \mathrm{day}^{-1}\), \(N=0.20K\), and \(N>0\), what is the per-cell growth rate and is the count increasing or decreasing?
For \(r=0.250\ \mathrm{day}^{-1}\), \(K=8.00\times10^7\) cells, and \(N_0=2.00\times10^7\) cells, calculate the logistic initial slope.
Why can observations collected only at counts much smaller than \(K\) be insufficient to estimate \(K\) reliably?

Solutions. (1) \(N(2)=5.00\times10^5e^{0.200(2.00)}=5.00\times10^5e^{0.400}\approx7.46\times10^5\) cells. (2) The per-cell rate is \(r(1-N/K)=0.500(1-0.20)=0.400\ \mathrm{day}^{-1}\), which is positive, so the count is increasing. (3) The slope is \(rN_0(1-N_0/K)=0.250(2.00\times10^7)(1-0.250)=3.75\times10^6\ \mathrm{cells/day}\). (4) When \(N/K\) is small, \(1-N/K\) is close to one, making logistic growth resemble exponential growth. Without observations that reveal appreciable slowing, many values of \(K\) can yield similar predictions.