11.4: Deriving a differential equation for the growth of cell mass

In Section 1.2, we asked how the size of a living cell influences the balance between the rates of nutrient absorption (called \(A\) ) and consumption (denoted C). But what if the two processes do not balance? What happens to the cell if the rates are unequal?

If a cell absorb nutrients faster than nutrients are consumed \((A>C)\), some of the excess nutrients accumulate, and this buildup of nutrient mass can be converted into cell mass. This can result in growth (increase of cell mass). Conversely, if the consumption rate exceeds the rate of absorption of nutrients \((C>A)\), the cell has a shortage of metabolic "fuel", and needs to convert some of its own mass into energy reserves that can power its metabolism this would lead to loss in cell mass.

Derivation of a differential equation that describes cell growth resulting from absorption and consumption of nutrients.

We can keep track of such changes in cell mass by using a simple "balance equation". The balance equation states that "the rate of change of cell mass is the difference between the rate of nutrient (mass) coming in \((A)\) and the rate of nutrient (mass) being consumed \((C)\), i.e.

\[\frac{d m}{d t}=A-C \]

Each term in this equation must have the same units, mass of nutrient per unit time. A contributes positively to mass increase, whereas \(C\) is a rate of depletion that makes a negative contribution (hence the signs associated with terms in the equation). It also makes sense to adopt the assumptions previously made in Section \(1.2\) (and Featured Problem 9.1) that

\[A=k_{1} S, \quad C=k_{2} V, \quad m=\rho V, \nonumber \]

where \(S, V, \rho\) are the surface area, volume, and density of the cell, and \(k_{1}, k_{2}, \rho\) are positive constants. Then Equation (11.13) becomes

\[\frac{d m}{d t}=A-C \quad \Rightarrow \quad \frac{d(\rho V)}{d t}=k_{1} S-k_{2} V \]

The above equation is rather general, and does not depend on cell shape.

Master Material Check

What are the units of \(k_{1}, k_{2}, \rho\) ?

Now consider the special case of a spherical cell for which \(V=(4 / 3) \pi r^{3}\), \(S=4 \pi r^{2}\). This simplification will permit us to convert the balance equation into a differential equation that describes changes in cell radius over time. Now Equation (11.4.2) can be rewritten as

\[\frac{d\left[\rho \cdot(4 / 3) \pi r^{3}\right]}{d t}=k_{1}\left(4 \pi r^{2}\right)-k_{2}(4 / 3) \pi r^{3} \]

We can simplify the derivative on the right hand side using the chain rule, as done in Featured Problem 9.1, obtaining

\[\rho \frac{4 \pi}{3} \pi\left(3 r^{2}\right) \frac{d r}{d t}=k_{1}\left(4 \pi r^{2}\right)-k_{2}(4 / 3) \pi r^{3} \]

What does this tell us about cell radius?

One way to satisfy Equation (11.4.4) is to set \(r=0\) in each term. While this is a "solution" to the equation, it is not biologically interesting. (It merely describes a "cell" of zero radius that never changes.) Suppose \(r \neq 0\). In that case, we can cancel out a factor of \(r^{2}\) from both sides of the equation. (We can also cancel out \(4 \pi\).) After some simplification, we arrive at

\[\rho \frac{d r}{d t}=k_{1}-\frac{k_{2}}{3} r, \quad \Rightarrow \quad \frac{d r}{d t}=\frac{1}{\rho}\left(k_{1}-\frac{k_{2}}{3} r\right) \nonumber \]

With appropriate units and taking into account typical cell size and density, this equation might look something like

\[\frac{d r}{d t}=(1-0.1 r) \]

Hint:

If we use units of \(\mu \mathrm{m}\left(=10^{-6} \mathrm{~m}\right)\) for cell radius, \(\mathrm{pg}\left(=10^{-12} \mathrm{gm}\right)\) for mass, and measure time in hours, then approximate values of the constants are

\(\rho=1 \mathrm{pg} \mu \mathrm{m}^{-3}\)

\(k_1=1 \mathrm{pg} \mu \mathrm{m}^{-2} \mathrm{hr}^{-1}\)

\(k_2=0.3 \mathrm{pg} \mu \mathrm{m}^{-3} \mathrm{hr}^{-1}\)

In that case, the equation for cell radius is \(d r / d t=(1-0.1 \cdot r)\).

From a statement about how cell mass changes, we have arrived at a resultant prediction about the rate of change of the cell radius. The equation we obtained is a differential equation that tells us something about a growing cell. In an upcoming chapter, we will build tools to be able to understand what this equation says, how to solve it for the cell radius \(r(t)\) as a function of time \(t\), and what such analysis predicts about the dynamics of cells with different initial sizes.