I read in a book that the smaller the cell the greater its surface-to-volume ratio (the surface area of a cell compared to its volume).

And larger cells have limited surface area compared to its volume.

How is this possible? I don't understand.


closed as off-topic by canadianer, David, theforestecologist, Daniel Standage, Tyto alba Jun 7 '17 at 16:20

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  • $\begingroup$ also consider the width of the surface and the number of atoms it contains as a lipid, it's not strictly geometry. $\endgroup$ – com.prehensible Jun 6 '17 at 10:02
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    $\begingroup$ @comprehensible It is just geometry: brooklyn.cuny.edu/bc/ahp/LAD/C5/C5_ProbSize.html $\endgroup$ – kmm Jun 6 '17 at 13:46
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    $\begingroup$ This is more about understanding geometry than biology. $\endgroup$ – Daniel Standage Jun 6 '17 at 20:44
  • $\begingroup$ This is a very basic question that requires you to think in terms of geometry. Just think. $\endgroup$ – Adam Radek Martinez Jun 7 '17 at 16:01
  • $\begingroup$ Did you try finding an answer yourself? Please share it and see our homework policy. $\endgroup$ – Tyto alba Jun 7 '17 at 16:20

This is a basic property of geometry: for any given shape, area will be generally proportional to the square of the length, and volume will be proportional to its cube. (this isn't true for all shapes by any means; the "generally" refers to the principle that length, area and volume are 1, 2 and 3-dimensional properties respectively, not that all or even most shapes behave this way... although a lot of the shapes that happen to be common in the Universe do. See end of the answer).

For example, if a sphere has a radius of R, its surface will have an area of 4*Pi*R², and its interior will have a volume of 4/3*Pi*R³. This means its surface-to-volume ratio will be 3/R, which is inversely proportional to R, meaning when R increases it will decrease and vice-versa.

Like so:

\begin{array} {|l|l|} \hline R~(\sim length) & R^2~(\sim surface~area) & R^3~(\sim volume) & 1/R~(\sim surface~area / volume) \\ \hline 1 & 1 & 1 & 1~(= 1/1) \\ \hline 2 & 4 & 8 & 1/2~(= 4/8)\\ \hline 3 & 9 & 27 & 1/3~(=9/27)\\ \hline 4 & 16 & 64 & 1/4~(=16/64)\\ \hline 5 & 25 & 125 & 1/5~(=25/125) \\ \hline \end{array}

Put another way, if your shape is growing in 3D then whenever you add a bit of length you'll tend to add much more area (because area is increasing along two directions of length) but you'll be adding even more volume (because volume is increasing along three directions of length), and more volume than area means the area-to-volume ratio becomes smaller.

It helps that cells are convex and can often be approximated as spheres, so the rule absolutely applies to them. It's different with squiggly shapes, or cases where you aren't growing along every axis (for example this principle is true if you increase a cylinder's radius and length, but not if you just increase its length; at that point length, surface area and volume are all increasing along a single dimension and thus all at the same rate; check the equations to see this). You often do find such squiggly, long or thin shapes in nature, and this is often an adaptation to get around the surface-to-volume decrease issue. Examples include your intestines, or the interface between the placenta and the uterus.

  • $\begingroup$ Its still not clear but thank you a lot for the answer $\endgroup$ – roxaite Jun 6 '17 at 14:10
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    $\begingroup$ If the above answer is not clear to you, you might want to review your basics in geometry and algebra. You can as well comment specifically on what you don't understand from the current answer to let @Rozenn (and other users) what specific point needs to be clarified. $\endgroup$ – Remi.b Jun 6 '17 at 15:37
  • $\begingroup$ Got it now,............. $\endgroup$ – roxaite Jun 7 '17 at 16:45

The answer of Rozenn Keribin and Gerardo Furtado already explains the basic principle very well. I'd like to add another aspect, though. The effect of the increased surface to volume ratio also comes into play in the case off nanotechnology where small particles are compared to bulk material.

Consider the following example, where a cube with edge length 1 m is devided into smaller cubes with edge length s:

\begin{array} {c c c} \hline edge length ~s & number~of~particles & surface~area ~A (m²) & total~volume\\ \hline 1~m & 10^0 & 6*10^0 & 1~m^3 \\ 1~cm & 10^6 & 6*10^3 & 1~m^3 \\ 1~mm & 10^9 & 6*10^5 & 1~m^3 \\ 1~µm & 10^{18} & 6*10^{11} & 1~m^3 \\ 1~nm & 10^{27} & 6*10^{17} & 1~m^3 \\ \hline \end{array}

As you can see, the total volume of all particles is always 1 m³ since you only cut the original cube of 1x1x1 m into smaller pieces. However, this example highlights that you have a much larger total surface area for all particles if they are smaller when compared to larger particles.
This means that the effective surface area of the obtained particles increases with the inverse of the square of the size while the total mass (and volume) stays constant. The surface of the particles is proportional to their (bio-)activity.

  • $\begingroup$ The other answer is not mine, I just edited it to add a Mathjax table (as you did here) :-). $\endgroup$ – user24284 Jun 6 '17 at 11:07
  • $\begingroup$ Anyway, you contributed ;-) $\endgroup$ – Christoph Engwer Jun 6 '17 at 11:12
  • $\begingroup$ Still not clear but thank you a lot $\endgroup$ – roxaite Jun 6 '17 at 14:10
  • $\begingroup$ @roxaite I modified my answer - please check, if it is more clear to you now. $\endgroup$ – Christoph Engwer Jun 6 '17 at 15:31
  • $\begingroup$ Got it chris..thank you for simplifying your answer... $\endgroup$ – roxaite Jun 7 '17 at 16:44

Others have mentioned cubes, and the principle is the same, but I prefer to think in terms of spheres.

The volume of a sphere is $\frac43 \pi r^3$

The surface area of a sphere is $4 \pi r^2$

The ratio is thus $\frac3{r}$.

You can plot the function $\frac3{r}$ (see here), or just know that this is an inverse relationship -- as $r$ goes up, the ratio goes down; as $r$ goes down, the ratio goes up.

This is a different way of saying -- a smaller object (which has smaller $r$) has a greater surface-volume ratio.


Let's approximate a cell by the simplest possible shape, a cube. Let's say the sides of the cube are 1mm (that'd be a large cell...). It's volume is obviously 1 mm3, and since it has 6 sides, a surface area of 6 mm2. That gives us a surface to volume ratio of 6. Now let's cut this cube in half. Now every part has a volume of 0.5 mm3. But what happens to the surface area? Four of the six sides are cut in half, and we keep one of the original side. But we also get a new side from where the cut took place. So our new area is 1+1+4*0.5 = 4 mm2. This gives us a surface to volume ratio of 4/0.5 = 8, which is greater than the original 6.

  • $\begingroup$ Tthnx bro.. your answer was also helpful $\endgroup$ – roxaite Jun 7 '17 at 16:44

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