Question

Assuming that cells are spheres, compare three cells with diameters of 2 μm, 10 μm and 20 μm. Calculate the surface area to volume ratio for all three cells.

Based on your calculations, rank these in order (lowest SA/V ratio first). Which cell is most suited for rapid exchange of material with the extracellular environment?

For a sphere; V = 4/3 πr3 and A = 4πr2

Answer 1

Cell 1. Cell with diameter of 2µm, it means radius = 1µm

Volume = 4/3πr³ = (4/3) x 3.14 x 1 x 1 x 1 = 4.187 µm³

Surface area = 4πr² = 4 x 3.14 x 1 x 1 = 12.56 µm²

SA/V = 12.56/4.187 = 2.999 = 3

Cell 2. Cell with diameter of 10µm, it means radius = 5µm

Volume = 4/3πr³ = (4/3) x 3.14 x 5 x 5 x 5 = 523.33 µm³

Surface area = 4πr² = 4 x 3.14 x 5 x 5 = 314 µm²

SA/V = 314/523.33 = 0.6

Cell 3. Cell with diameter of 20µm, it means radius = 10µm

Volume = 4/3πr³ = (4/3) x 3.14 x 10 x 10 x 10 = 4186.67 µm³

Surface area = 4πr² = 4 x 3.14 x 10 x 10 = 1256 µm²

SA/V = 1256/4186.67 = 0.2999 = 0.3

Rank from lowest to highest SA/V ratio will be Cell 3 < Cell 2 < Cell 1 as they have value of 0.3<0.6< 3.0

Cell 1 (diameter of 2µm) is most suited for rapid exchange of the material with the extracellular environment, because it has small in size and higher SA/V ratio.