1.2: How big can a cell be? A model for nutrient balance

Learning Objectives

1. Describe the derivation of a mathematical model for cell nutrient absorption and consumption.
2. Use parameters \(\left(k_{1}, k_{2}\right)\) rather than specific numbers in mathematical expressions.
3. Demonstrate the link between power functions in Section \(1.1\) and cell nutrient balance in the model.
4. Interpret the results of the model.

Consider the following biologically motivated questions:

• What physical and biological constraints determine the size of a cell?
• Why do some size limitations exist?
• Why should animals be made of millions of tiny cells, instead of a just a few large ones?

Video: A summary of the cell size model. We discuss what cell size is consistent with a balance between nutrient absorption and consumption in a cell.

We already have enough mathematical prowess to address these questions particularly if we assume a cell is spherical. Of course, this is often not the case. The shapes of living cells uniquely suit their functions. Many have long appendages, cylindrical parts, or branch-like structures. But here, we neglect all these beautiful complexities and look at a simple spherical cell because it suffices to answer our questions. Such mathematical simplifications can be very illuminating: they allow us to form a mathematical model. A mathematical model is just a representation of a real situation which simplifies things by representing the most important aspects, and neglecting or idealizing complicating details.

In this section, we follow a reasonable set of assumptions and mathematical facts to explore how nutrient balance can affect and limit cell size.

Building the model

In order to build the model we make some simplifying assumptions and then restate them mathematically. We base the model on the following assumptions:

1. The cell is roughly spherical (See Figure 1.3).
2. The cell absorbs oxygen and nutrients through its surface. The larger the surface area, \(S\), the faster the total rate of absorption. We assume that the rate at which nutrients (or oxygen) are absorbed is proportional to the surface area of the cell.
3. The rate at which nutrients are consumed (i.e., used up) in metabolism is proportional the volume, \(V\), of the cell. The bigger the volume, the more nutrients are needed to keep the cell alive.

Figure 1.3: An assumed spherical cell absorbs nutrients at a rate proportional to its surface area \(S\), but consumes nutrients at a rate proportional to its volume \(V\).

We define the following quantities for our model of a single cell:

\[\begin{aligned} & A=\text { net rate of absorption of nutrients per unit time, } \\ & C=\text { net rate of consumption of nutrients per unit time, } \\ & V=\text { cell volume, } \\ & S=\text { cell surface area, } \\ & r=\text { radius of the cell. } \end{aligned} \nonumber \]

We now rephrase the assumptions mathematically. By Assumption 2, the absorption rate, \(A\), is proportional to \(S\) : this means that

\[A=k_{1} S, \nonumber \]

where \(k_{1}\) is a constant of proportionality. Since absorption and surface area are positive quantities, only positive values of the proportionality constant make sense, so \(k_{1}\) must be positive. The value of this constant depends on properties of the cell membrane such as its permeability or how many pores it contains to permit passage of nutrients. By using a generic constant - called a parameter - to represent this proportionality constant, we keep the model general enough to apply to many different cell types.

By Assumption 3, the rate of nutrient consumption, \(C\) is proportional to \(V\), so that

\[C=k_{2} V, \nonumber \]

where \(k_{2}>0\) is a second (positive) proportionality constant. The value of \(k_{2}\) depends on the cell metabolism, i.e. how quickly it consumes nutrients in carrying out its activities.

By Assumption 1, the cell is spherical, thus its surface area, \(S\), and volume, \(V\), are:

\[S=4 \pi r^{2}, \quad V=\frac{4}{3} \pi r^{3} . \nonumber \]

Mastered Material Check

6. What does "A is proportional to B" mean?
7. What might the units be for quantities \(A, C, V, S\) and \(r\) ?
8. Given your choices for 7., what are the units associated with \(k_{1}, k_{2}\) ? Putting these facts together leads to the following relationships between nutrient absorption \(A\), consumption \(C\), and cell radius \(r\) :

\[\begin{aligned} & A=k_{1}\left(4 \pi r^{2}\right)=\left(4 \pi k_{1}\right) r^{2}, \\ & C=k_{2}\left(\frac{4}{3} \pi r^{3}\right)=\left(\frac{4}{3} \pi k_{2}\right) r^{3} . \end{aligned} \nonumber \]

Putting these facts together leads to the following relationships between nutrient absorption \(A\), consumption \(C\), and cell radius \(r\):

\[A=k_1\left(4 \pi r^2\right)=\left(4 \pi k_1\right) r^2 \nonumber \]

\[C=k_2\left(\frac{4}{3} \pi r^3\right)=\left(\frac{4}{3} \pi k_2\right) r^3 \nonumber \]

Rewriting this relationship as

\[A(r)=\left(4 \pi k_{1}\right) r^{2}, \quad \text { and } \quad C(r)=\left(\frac{4}{3} \pi k_{2}\right) r^{3} \]

we observe that \(A, C\) are simply power functions of the cell radius, \(r\), that is

\[A(r)=a r^{2}, \quad C(r)=c r^{3} . \nonumber \]

Note: the powers are \(n=3\) for consumption and \(n=2\) for absorption.

Mastered Material Check

9. What are constants \(a\) and \(c\) in terms of \(k_{1}\) and \(k_{2}\) ?
10. Why are we considering different values of \(r\) in Example 1.3?

The discussion of power functions in Section 1.1 now contributes to our analysis of how nutrient balance depends on cell size.

Nutrient balance depends on cell size

In our discussion of cell size, we found two power functions that depend on the cell radius \(r\), namely the nutrient absorption \(A(r)\) and consumption \(C(r)\) given in Equations (\(\PageIndex{1}\)). We first ask whether absorption or consumption of nutrients dominates for small, medium, or large cells.

Example 1.3

For what cell size is the consumption rate exactly balanced by the absorption rate? Which rate (consumption or absorption) dominates for small cells? For large cells?

Solution

The two rates "balance"’ (and their graphs intersect) when

\[A(r)=C(r) \Rightarrow\left(\frac{4}{3} \pi k_{2}\right) r^{3}=\left(4 \pi k_{1}\right) r^{2} \nonumber \]

A trivial solution to this equation is \(r=0\).

Note: this solution is not interesting biologically, but we should not forget it in mathematical analysis of such problems.

If \(r \neq 0\), then, canceling a factor of \(r^{2}\) from both sides gives:

\[r=3 \frac{k_{1}}{k_{2}} . \nonumber \]

This means absorption and consumption rates are equal for cells of this size. For small \(r\), the power function with the smaller power of \(r\) (namely \(A(r))\) dominates, but for very large values of \(r\), the power function with the higher power of \(r\) (namely \(C(r)\) ) dominates. It follows that for smaller cells, absorption \(A \approx r^{2}\) is the dominant process, while for larger cells, consumption rate \(C \approx r^{3}\) dominates. We conclude that cells larger than the critical size \(r=3 k_{1} / k_{2}\) are unable to keep up with the nutrient demand, and cannot survive since consumption overtakes absorption of nutrients.

Using the above simple geometric argument, we deduced that cell size has strong implications on its ability to absorb nutrients or oxygen quickly enough to feed itself. For these reasons, cells larger than some maximal size (roughly \(1 \mathrm{~mm}\) in diameter) rarely occur.

A similar strategy also allows us to consider the energy balance and sustainability of life on Earth - as seen next, in Section 1.3.