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Surface area to volume ratio of cells | Cell structure and function | AP Biology | Khan Academy


4m read
·Nov 11, 2024

So let's say that this is a cell.

We know that all sorts of activity is going on inside of this cell here, and we will study that in a lot more depth as we go further in our study of biology. But it's important to realize that this cell and the activity in that cell is not operating in isolation.

In order to live, that cell needs resources from the outside world. So resources need to make their way through that outer membrane of the cell so it can be used inside that cellular machinery.

As that cell does what it does, it's also going to generate waste products, and that needs to be released somehow across that membrane. So you also have waste, and you also have energy that is going to be transferred either from the inside of the cell to the outside or from the outside to the inside.

A lot of times, we imagine that all of the activity inside of the cell is generating thermal energy that has to be dissipated somehow, and that is usually the case—not always. So you have thermal energy that has to be dissipated.

Now, you might see something interesting, or maybe you haven't seen it just yet: you have all of this activity operating in the volume of the cell, but then all of this exchange—all of the resources coming in, the waste coming out, the thermal energy going in either direction—has to be somehow diffused across this surface, across this two-dimensional surface.

So this raises an interesting question. As our volume increases, what happens to the ratio of our surface to our volume? Because you could imagine maybe at some point the volume gets large enough that you don't have enough surface area to do these three things well.

So let's think about this ratio. Let's think about the ratio of our surface area to volume. I'm going to get a little bit mathy here. You don't have to know the math for the context of a biology course, but you need to know what the conclusion is that the math is going to give us.

So if this is a sphere of radius (r), the surface area of this sphere is going to be (4 \pi r^2), and the volume of this sphere is going to be (\frac{4}{3} \pi r^3).

So this (\pi) would cancel with that (\pi). If we divide the numerator and denominator by (r^2), we get a (1) there, and then we just get an (r) right over here. If we divide both of these by (4), you get a (1) there, and this is going to be a (1/3).

So we are going to be left with (\frac{1}{1/3} r), or we could just write this as (= \frac{3}{r}).

And so we see, at least for a spherical cell like this, as (r) increases, as our cell gets larger and larger, the ratio between our surface area to volume decreases.

Let me write that as (r) goes up, then the ratio between our surface area to volume is going to go down. The bigger your denominator, the lower the value is going to be.

And so what that tells us is that as the volume of our cell increases, as our cell gets bigger and bigger and bigger, we have less surface area per unit of volume. Hence, it's going to make that exchange of the resources, the waste, and that energy harder and harder and harder.

We would get a similar result if, instead of doing a spherical cell, let's say we did a cuboidal cell. So let's do it like this: a cuboidal cell.

You might see this in some plants, something that's roughly cuboidal or rectangular in some way, or rectangular prism, I should say. But let's say it's (x) by (x) by (x). We could do the same exercise.

Our ratio of surface area to volume is going to be what? Well, our surface area—you have six faces that each have an area of (x^2), so our surface area is going to be (6x^2), and then our volume is going to be (x \times x \times x = x^3).

So this is going to be (\frac{6x^2}{x^3}). If we divide the numerator and denominator by (x^2), you get (\frac{6}{x}).

So once again, you see that as (x) increases, our ratio of surface area to volume decreases. As our denominator increases, well then, that whole expression is going to decrease.

So given this phenomenon, it makes it hard for larger and larger cells to exist because for all the activity happening in the volume, they don't have enough surface area to do all of this exchange.

Now there are things we see in biological systems that help cells get further than what we see here. If you imagine the two-dimensional cross-section of this cell, one way to increase the surface area to volume is for the membrane to look more like this.

The more folds you have, the higher surface area to volume that you are going to have. And you indeed see this in a lot of biology. Anytime you want a high surface area to volume, you tend to see things like these folds in the membranes of the cells.

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