Cells, growing old

As each cell reproduces once every doubling time, we can write the number of cells at time $t$ as $N(t) = N_i2^{t/T_d}\equiv N_ie^{\mu t}$ where $\mu = \frac{\ln 2}{T_d}$

Suppose at time $t$ in a bacterial culture with growth rate $\mu$ , the fraction $\phi_a(t)$ cells are of the age $a$ . If we let a small time $\Delta t$ pass, the culture will grow by the factor $e^{\mu \Delta t}$ . However, all the same cells which were in the fraction of cells age $a$ at $t$ will be in the fraction of cells age $a+\Delta t$ at time $t+\Delta t$ . Therefore, we have:

$\phi_{a+\Delta t}(t+\Delta t) = e^{-\mu\Delta t}\phi_a(t)$

However, the fraction of cells of a given age does not change over time (as we are in the steady state), so the distribution is time independent, and we have:

$\phi_{a+\Delta t} = e^{-\mu\Delta t}\phi_a$

We could solve this relation with calculus, or we can simply observe that this equation is satisfied if $\phi_m = e^{-\mu m}$ . Therefore, the fraction of cells of age $a$ is proportional to $e^{-\mu a}$

By definition, all cells are aged between 0 and $T_d$ . Therefore, we can normalize the age distribution over the interval $\left[0,T_d\right]$ to obtain the fully quantitative age distribution of the cells.

$\displaystyle 1 = C \int \limits_0^{T_d} e^{-\mu a} da = C \left( \frac{ 1 - e^{-\mu T_d} }{\mu} \right) = C \cdot \frac{T_d}{2\ln 2}$

which shows that $C \equiv \frac{2\ln 2}{T_d}$ . Hence, the age distribution is given by $\displaystyle \phi_a = \frac{2 \ln 2}{T_d} e^{-\mu a}$ or $\displaystyle 2\mu e^{-\mu a}$

The average cell age $\langle a\rangle$ is given by

$\displaystyle\int\limits_0^{T_d}a\phi_a da$

which we can calculate directly

Rewriting the expression slightly, we have

$\displaystyle 2\mu \left[\int a e^{-\mu a} da \right]_0^{T_d}$

The bracketed indefinite integral can be solved by differentiating under the integral sign. We recognize that $ae^{-\mu a}$ is the derivative of $-e^{-\mu a}$ with respect to $\mu$ . Thus

$\displaystyle \int a e^{-\mu a} da = -\int \frac{d}{d\mu}e^{-\mu a} da = -\frac{d}{d\mu} \int e^{-\mu a} da$

The derivative can be pulled in front of the integral as it is on $\mu$ while the integral is over $a$ .

The integral $\displaystyle \int e^{-\mu a} da$ is easy and is equal to $\displaystyle \frac{-e^{-\mu a}}{\mu}$

Now, we evaluate the expression and recall $\displaystyle \mu = \frac{\ln 2}{T_d}$ .

$\begin{aligned} \langle a \rangle =& 2\mu \left[\frac{1}{\mu^2} - T_d\frac{e^{-\mu T_d}}{\mu} - \frac{e^{-\mu T_d}}{\mu^2} \right] \\ =& 2 \left[ \frac{1}{\mu} - T_d e^{-\mu T_d} - \frac{e^{-\mu T_d}}{\mu} \right] \\ =& 2 \left[ \frac{T_d}{\ln 2} - \frac{T_d}{2} - \frac{T_d}{2\ln2}\right] \\ =& \boxed{T_d\left(\frac{1}{\ln 2} - 1\right)} \\ \end{aligned}$

Therefore, the average cell age is roughly equal to $\displaystyle\frac{44}{100} T_d$ , or $44\%$ of one doubling time.

Interestingly, the average age is independent of how quickly the cell culture is growing and is slightly less than halfway between birth and division.

The probability distribution is an exponential curve from x = 0 to 1 where P(0) = 2 * P(1). using calculus we can show that mean x is 1/ln2 - 1. The mean age is 60(1/ln2 -1) = 26.56 or roughly 27 minutes