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First teaching 2023

First exams 2025

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Surface Area to Volume Ratios (CIE AS Biology)

Revision Note

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Principles of SA:V

  • Surface area and volume are both very important factors in the exchange of materials in organisms
  • The surface area refers to the total area of the organism that is exposed to the external environment
  • The volume refers to the total internal volume of the organism (total amount of space inside the organism)
  • As the surface area and volume of an organism increase (and therefore the overall ‘size’ of the organism increases), the surface area : volume ratio decreases
  • This is because volume increases much more rapidly than surface area as size increases

Importance of a High Surface Area to Volume Ratio

  • Having a high surface area to volume ratio increases the ability of a biological system to perform the following important functions
    • Obtaining necessary resources eg, oxygen, glucose, amino acids
    • Eliminating waste products eg. carbon dioxide, urea
    • Acquiring or dissipating thermal energy (heat)
    • Otherwise exchanging chemicals and energy with the surroundings eg. absorbing hormones at the cell surface in the hormone's target organ

How Surface Area-to-Volume Ratio Changes With Size Diagram

surface-area-to-volume-ratio-changes-with-size

As size increases, the surface area : volume ratio decreases

Calculating Surface Area to Volume Ratios Table

  Cube Cuboid Cylinder
  cube-all-sides-length-s cuboid-with-length-width-height-showing cylinder-with-radius-height-showing
Surface Area 6s2 2lh + 2lw + 2wh 1 space r e c t a n g l e space equals space 2 straight pi rh
2 space c i r c u l a r space e n d s space equals 2 straight pi straight r squared
S A space equals space space 2 straight pi rh plus 2 straight pi straight r squared
Volume s3 l × w × h straight pi straight r squared straight h
Example

If s = 1 cm
then
SA = (1×1)×6
SA = 6cm2

V = s3 = 13 =1cm3
∴ SA:V ratio = 6:1

If l = 4cm, w = 2cm, h = 1cm, then
SA = 2((4×1)+(4×2)+(2×1))
= 28cm2

V = 4 × 2 × 1
= 8cm3

∴ SA:V ratio = 28:8
= 3.5:1

If r = 2 cm and h = 6cm, then
SA = 2πrh + 2πr2
= 8π +24π
= 32π cm2

V = π(2)2 × 6 = 24π cm2

∴ SA : V ratio
= 32 : 24
= 1.33:1

The surface area:volume ratio calculation differs for different shapes (these shapes can reflect different cells or organisms)

Worked example

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

fraction numerator 800 space nm over denominator 1000 end fraction equals space 0.8 straight mu straight m

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

SA colon straight V space ratio space equals space fraction numerator 3.84 over denominator 0.512 end fraction equals fraction numerator 7.5 over denominator 1 end fraction equals 7.5 space colon space 1 space open parentheses or space simply space 7.5 close parentheses

___________

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

SA colon straight V space ratio space equals space fraction numerator 17.278 over denominator 3.927 end fraction equals fraction numerator 4.4 over denominator 1 end fraction equals 4.4 space colon space 1 space open parentheses or space simply space 4.4 close parentheses

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. 

Examiner Tip

This worked example assumes that Staphylococcus aureus is a cuboid shaped cell; in fact it is a spherical cell. You are not required to know how to calculate the surface area and volume of a sphere for this exam. A sphere has the highest surface area to volume ratio of any shape of a given radius/size, so is the best-adapted shape for diffusion. 

However, you are expected to be able to calculate the SA:V ratio for a cube, cuboid or cylinder and explain how the increasing size of an organism affects the SA:V ratio.

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Phil

Author: Phil

Expertise: Biology

Phil has a BSc in Biochemistry from the University of Birmingham, followed by an MBA from Manchester Business School. He has 15 years of teaching and tutoring experience, teaching Biology in schools before becoming director of a growing tuition agency. He has also examined Biology for one of the leading UK exam boards. Phil has a particular passion for empowering students to overcome their fear of numbers in a scientific context.