Question

What are some examples of surface area to volume ratio?

Answer 1

The surface-area-to-volume ratio or SA:V, is the amount of surface area of an organism divided by its volume.

Assume that you are a spherical cell.

Your SA:V is important because you depend on diffusion through your cell wall to obtain oxygen, water, and food and get rid of carbon dioxide and waste materials.

Let's calculate SA:V for three cell sizes.

`"SA" = 4πr^2` and `V = 4/3πr^3`

r = 1 mm: `SA = 4π " mm"^2; V= 4/3π " mm"^3; "SA:V" = 3.0`

r = 2 mm: `SA = 16π " mm"^2; V= 32/3π " mm"^3; "SA:V" = 1.5`

r = 3 mm: `SA = 36π " mm"^2; V= 108/3π " mm"^3; "SA:V" = 1.0`

Your surface area to volume ratio decreases as you get bigger.

Now let's assume that nutrients can diffuse into your cell at the rate of 0.05 mm/min. In 10 min they would reach 0.5 mm to the centre. What fraction of your cell would still be unfed after 10 min?

r = 1 mm

`V_"tot" = 4/3π " mm"^3`

`r_"unfed" = "0.5 mm"`

`V_"unfed" = 4/3πr^3 = 4/3π×(0.50" mm")^3 = 0.50/3π" mm"^3`

`% "unfed" = V_"unfed"/V_"tot" × 100 % = (0.50/cancel(3) cancel("π mm³"))/(4/cancel(3) cancel("π mm³")) × 100 % = 12 %`

r = 2 mm

`V_"tot" = 32/3π" mm"^3`

`r_"unfed" = "1.5 mm"`

`V_"unfed" = 4/3πr^3 = 4/3π × ("1.5 mm")^3 = 13.5/3π" mm"^3`

`% "unfed" = V_"unfed"/V_"tot" × 100 % = (13.5/cancel(3) cancel("π mm³"))/(32/cancel(3) cancel("π mm³")) × 100 % = 42 %`

r = 3 mm

`V_"tot" = 108/3π" mm"^3`

`r_"unfed" = "1.5 mm"`

`V_"unfed" = 4/3πr^3 = 4/3π×("2.5 mm")^3 = 62.5/3π" mm"^3`

`% "unfed" = V_"unfed"/V_"tot" × 100 %(62.5/cancel(3) cancel("π mm³"))/(108/cancel(3) cancel("π mm³")) × 100 % = 58 %`

The bigger you get, the longer it takes for the nutrients to reach your interior.

Beyond a certain limit, not enough nutrients will be able to cross the membrane fast enough to accommodate your increased volume.

You will have to stop growing if you want to survive.