An Introduction to Cellular Transport and the Surface Area:Volume Ratio

A lot goes on inside a cell – respiration in the mitochondria, the processing of lipids, turning amino acids into proteins for use in enzymes and cell signalling, and numerous other processes all keeping the cell alive and well. However, for all these things that are happening inside the cell to work, things have to come in and out of the cell, too. You have oxygen (O2) and glucose (among other things) entering through endocytosis, and carbon dioxide (CO2) and other wastes exiting the cell through exocytosis. Not only that, but you have ions like Sodium (Na+) and Potassium (K+) crossing the phospholipid bilayer membrane to regulate other processes.

Everything that happens inside a cell requires resources from outside the cell, and for wastes to leave the cell to go elsewhere. These resources and wastes move in and out of the cell through the cell’s phospholipid bilayer membrane. All this cellular transport has to happen at an adequate rate for all the processes to be carried out and keep the cell healthy.
Cells divide when the surface area-to-volume ratio of a cell becomes too small. What do we mean by this? Let’s first look at the way cells grow.


Surface Area:Volume Ratio Explained
While a cell may begin small, it increases in size over time as the functions that occur inside it produce the components necessary for it to grow. When this happens, the volume of the inside of the cell increases at a faster rate than the surface area of the outside of the cell does. This is more easily explained using simple maths and a cube.


Where l is the length of the side of a cube, the surface area will be equal to l 2 (the area of one of the cube’s faces) multiplied by 6 (the number of faces in a cube).
SA = 6l 2

Similarly, the volume of a cube is l 2 (the surface area of one of the cube’s faces) multiplied again by l.
V = l 3

So here’s a cube with a side length of 1cm.


With our equations from above, we calculate the surface area and volume of this cube:
SA = 6 x 12
     = 6cm2
V = 13
         = 1cm3
The SA:V ratio for Cube 1 is 6:1.

Now we have a cube that has sides with twice the length of the original cube.


SA = 6 x 22
     = 24cm2
V = 23
         = 8cm3
The SA:V ratio for Cube 2 is 24:8, or 3:1.
This means that in relation to its volume, the SA for Cube 2 is half the size of Cube 1.
The principle is the same in cells. The larger a cell gets, the smaller its surface area is in relation to its volume.