Surface Area to Volume Ratios (CIE A Level Biology)

Calculate the surface area-to-volume ratios of the two following microorganisms:

1. A bacterial cell from the species Staphylococcus aureus; you can assume that each cell is a cube with side length of 800 nm (8 × 10^-9 m)
2. A bacterial cell from the species Bacillus subtilis; these are rod-shaped cells which you can assume to be cylindrical in shape. They are 5 µm long and 1 µm in diameter

Comment on your calculated answers. 

Solution

1. For the Staphylococcus aureus cell: side length = 800 nm

Convert this value into µm by dividing by 1000

Surface area of cube = 6(s²)

= 6 × (0.8 × 0.8)
= 3.84 µm²

Volume =  0.8  × 0.8 × 0.8

= 0.512 µm³

A ratio is one number divided by another with the larger number divided by the smaller number, so

___________

2. For the Bacillus subtilis cell: radius = 0.5 µm (half of diameter 1µm) 
Surface area of a cylinder =  2πrh + 2πr²
= (2π × 0.5 × 5) + (2π × 0.5²)
= 15.708 + 1.571 µm²
= 17.278 µm²

Volume of a cylinder: formula is πr²h
= π × 0.5² × 5
= 3.927 µm³

The cuboid cell of Staphylococcus aureus has a surface area-to-volume ratio of 7.5 : 1

The cylindrical cell of Bacillus subtilis aureus has a surface area-to-volume ratio of 4.4 : 1

The smaller cell has the largest surface area-to-volume ratio of the two shapes. This allows efficient diffusion into / out of the cell. Think of how long it might take to diffuse out of cell if a molecule starts its diffusion pathway right in the centre of the cell; the shortest distance to the edge will always be found in a cube. In a long cylinder, a molecule may travel parallel to the straight sides and so will have travel further to diffuse out (remember that molecular movement is random). So the larger the surface area to volume ratio, the better adapted to simple diffusion that organism is.