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Cell Population Dynamics · Tutorial 159 of 799

Deriving Exponential Cell Growth

You will derive the exponential-growth solution and use it to calculate cell counts and doubling times when the net per-capita growth rate is constant.

Intermediate 10 min read

What You'll Learn

  • Derive the exponential solution from a constant net per-capita rate.
  • Relate division and loss rates to net population growth.
  • Calculate cell counts at specified times and interpret the rate units.
  • Calculate doubling times and infer a growth rate from two cell counts.
  • Identify assumptions, diagnostics, and limitations of exponential predictions.
  • Distinguish continuous exponential growth from a discrete-time doubling approximation.

From a Per-Capita Rate to a Population Trajectory

A per-capita rate describes change relative to the current population. If each cell contributes the same constant net rate of change, the number of cells added per unit time increases as the population grows. This proportionality is the defining feature of exponential growth: a population that is larger changes faster, even though the rate per cell stays fixed.

Let \(N(t)\) be the number of cells at time \(t\), and let \(r\) be the constant net per-capita growth rate, measured in inverse time. When division and loss each occur at constant per-cell rates, their contributions can be combined into \(r\). For example, if division adds one net cell per division event and loss removes one cell per loss event, then \(r=\lambda-\delta\), where \(\lambda\) and \(\delta\) are the respective per-cell rates. The growth equation concerns net cell-count change; it does not imply that division and loss occur separately at rate \(r\).

Definition: Under constant net per-capita growth, the population satisfies \(dN/dt=rN\), with initial population \(N(0)=N_0\). Here \(N_0\) is the cell count at time zero and \(r\) has units of inverse time. If \(r>0\), the population grows; if \(r=0\), it is constant; and if \(r<0\), it declines.

This model assumes a closed population over the time interval being considered, a constant net rate per cell, and no dependence of that rate on the population size or elapsed time. It treats \(N(t)\) as a continuous quantity, so its predictions are modelled cell counts and need not be integers. It is most useful when these assumptions are reasonable over the time window of interest, not as a claim that populations grow exponentially indefinitely.

Deriving the Exponential Solution

For \(N(t)>0\), divide the growth equation by \(N(t)\) and separate the time and population terms. Integrating from time zero to time \(t\) gives the solution:

$$ \begin{aligned} \frac{dN}{dt} &= rN, \qquad N(0)=N_0,\\ \frac{1}{N}\,dN &= r\,dt,\\ \int_{N_0}^{N(t)}\frac{1}{N}\,dN &= \int_0^t r\,dt,\\ \ln\!\left(\frac{N(t)}{N_0}\right) &= rt,\\ N(t) &= N_0 e^{rt}. \end{aligned} $$

The initial condition is built into the solution: at \(t=0\), \(e^{r\cdot 0}=1\), so \(N(0)=N_0\). A direct check also confirms the result. Differentiating \(N_0e^{rt}\) gives \(rN_0e^{rt}=rN(t)\), as required by the model. The exponent \(rt\) must be dimensionless: if \(r\) is in days\(^{-1}\), then \(t\) must be in days.

The same derivation shows why a plot of \(\ln N(t)\) against time is a useful diagnostic under this model. Taking logarithms gives \(\ln N(t)=\ln N_0+rt\), a straight line with intercept \(\ln N_0\) and slope \(r\). A visibly curved pattern on that scale suggests that a constant net per-capita rate may not describe the observations over the interval.

Calculating Population Size and Doubling Time

For \(r>0\), doubling time is the time required for the population to reach twice its current size. Under constant \(r\), the time to double is the same no matter when the doubling starts. Set \(N(t+T_d)=2N(t)\) in the exponential solution; the initial population cancels, leaving \(e^{rT_d}=2\). Thus:

$$ T_d=\frac{\ln 2}{r}, \qquad r>0. $$

The doubling time has units of time. It is not defined as a positive growth doubling time when \(r=0\) or \(r<0\): in those cases the population does not grow to twice its size under this model. When \(r<0\), the time to halve the population is \(\ln 2/|r|\).

Worked Example: Combining Division and Loss Rates

Suppose a hypothetical cell population begins with \(200{,}000\) cells. Each cell has a constant division rate of \(0.045\) per day and a constant loss rate of \(0.015\) per day. Assume these events add one cell and remove one cell, respectively. Find the net growth rate, the predicted population after 24 days, and the doubling time.

Net rate. The net per-capita rate is the division rate minus the loss rate: \(r=0.045-0.015=0.030\) day\(^{-1}\). It is positive, so the model predicts growth.

Population after 24 days. Substitute \(N_0=200{,}000\), \(r=0.030\) day\(^{-1}\), and \(t=24\) days:

$$ N(24)=200{,}000e^{(0.030)(24)} =200{,}000e^{0.72} \approx 410{,}887\ \text{cells}. $$

Doubling time. The time is measured in days because \(r\) is in day\(^{-1}\):

$$ T_d=\frac{\ln 2}{0.030} \approx 23.10\ \text{days}. $$

The 24-day prediction is slightly more than one doubling, consistent with the calculated doubling time.

Worked Example: Inferring Growth Rate and Doubling Time

In a hypothetical experiment, a culture contains \(120{,}000\) cells at time zero and \(300{,}000\) cells 10 hours later. If exponential growth with a constant net per-capita rate is a suitable model, estimate that rate and its implied doubling time.

From \(N(t)=N_0e^{rt}\), divide by \(N_0\) and take logarithms to obtain \(r=\ln(N(t)/N_0)/t\). Here the observed ratio is \(300{,}000/120{,}000=2.5\), so:

$$ r=\frac{\ln(2.5)}{10\ \text{h}} =\frac{0.9163}{10\ \text{h}} \approx 0.09163\ \text{h}^{-1}. $$

The implied doubling time is:

$$ T_d=\frac{\ln 2}{0.09163\ \text{h}^{-1}} \approx 7.565\ \text{h}. $$

This rate and doubling time are estimates conditional on the constant-rate model. Two cell counts determine a rate mathematically, but they cannot establish that the rate stayed constant between measurements.

Worked Example: Predicting Growth From a Doubling Time

A hypothetical cell line has an estimated doubling time of 18 days under a specified condition. Starting from \(75{,}000\) cells, predict the population after 30 days, assuming the constant-rate model holds.

First convert doubling time to a net growth rate: \(r=\ln 2/(18\ \text{days})\approx0.03851\ \text{day}^{-1}\). Then use that rate in the population solution:

$$ N(30)=75{,}000e^{(0.03851)(30)} \approx 75{,}000e^{1.155} \approx 238{,}100\ \text{cells}. $$

As a check, 30 days is \(30/18\approx1.667\) doubling times. The same prediction can be written \(75{,}000\times2^{30/18}\), which also gives approximately \(238{,}100\) cells. This agreement follows from \(e^{rt}=2^{t/T_d}\) when \(r=\ln 2/T_d\).

Checking the Model and Choosing an Approach

A calculated trajectory is only as credible as its assumptions. Useful checks include plotting cell counts over time and, where counts are positive, plotting their logarithms against time. Approximately linear log-counts support a constant-rate description over the measured interval; systematic curvature indicates that the net rate may be changing. Measurement error, sparse sampling, or variability between replicate cultures can also make a true pattern hard to assess, so a straight-looking plot is not proof that the assumptions hold.

The exponential equation is the continuous-time counterpart of repeated proportional multiplication. A discrete approximation that doubles the population every \(T_d\) time units uses \(N(t)=N_0\,2^{t/T_d}\). When its doubling time represents the same constant growth rate, this expression is algebraically equivalent to \(N_0e^{rt}\), not a different growth mechanism. A simple step-by-step doubling rule is often convenient for mental estimates, but the exponential form directly handles any elapsed time.

If measurements suggest that the rate varies with time, one constant \(r\) may not summarize the whole interval. If the rate depends on population size, a density-dependent model may be more appropriate; such a model requires a different equation and its own assumptions. For complex rates or model structures without a simple closed-form solution, numerical integration can approximate the trajectory, but it does not fix an inappropriate model.

Common pitfall: Treating a constant per-capita rate as a constant number of cells added per day. With exponential growth, the total change per unit time is \(dN/dt=rN\), so it increases as \(N\) increases. A constant number added per day instead describes \(dN/dt=c\), which has the linear solution \(N(t)=N_0+ct\), not \(N_0e^{rt}\).

A second limitation is extrapolation. The equation predicts continued exponential growth for every future time if \(r>0\); a real cell population may not maintain that rate indefinitely. Predictions far beyond the observations should therefore be labelled as model-based projections, not as guaranteed outcomes. In addition, an estimated rate from two measurements is sensitive to counting errors, while additional time points can reveal whether a constant-rate fit is plausible.

Check Your Understanding

Answer each question using the constant net per-capita growth model where its assumptions are stated.

  1. A population starts at \(50{,}000\) cells and has \(r=0.020\) day\(^{-1}\). What population does the model predict after 10 days?
  2. A culture grows from \(80{,}000\) to \(160{,}000\) cells in 12 hours. Under a constant-rate model, what are the estimated net rate and doubling time?
  3. What does the exponential model predict if \(r=0\)? Does it have a finite positive doubling time?
  4. Why is a straight-line fit to cell count versus time not the diagnostic pattern expected for exponential growth?

Solutions. (1) \(N(10)=50{,}000e^{(0.020)(10)}=50{,}000e^{0.2}\approx61{,}070\) cells. (2) The ratio is \(160{,}000/80{,}000=2\), so \(r=\ln(2)/(12\ \text{h})\approx0.05776\ \text{h}^{-1}\), and \(T_d=\ln(2)/r=12.00\) hours. (3) If \(r=0\), then \(N(t)=N_0e^0=N_0\): the population is constant and has no finite positive doubling time. (4) Exponential growth has \(N(t)=N_0e^{rt}\), which is generally curved on the count scale. Its logarithm, \(\ln N(t)=\ln N_0+rt\), is linear in time under the model.

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