Cells & Surface Area (College Board AP® Biology): Study Guide

Cara Head

Written by: Cara Head

Reviewed by: Lára Marie McIvor

Updated on

Surface area-to-volume ratio

  • Surface area and volume are both very important factors in the exchange of materials in organisms

    • Surface area refers to the total area of the organism that is exposed to the external environment

    • 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

surface-area-to-volume-ratio-changes-with-size
As size increases, the surface area : volume ratio decreases

Importance of surface area-to-volume ratios

  • Having a high surface area-to-volume ratio increases the ability of a biological system to perform the following important functions

    • obtaining necessary resources e.g. oxygen, glucose, amino acids

    • eliminating waste products e.g. carbon dioxide, urea

    • acquiring or dissipating thermal energy (heat)

    • exchanging chemicals and energy with the surroundings e.g. absorbing hormones at the cell surface in the hormone's target organ

Calculating surface area-to-volume ratios

 

Sphere

Cube

Rectangular Solid

Cylinder

 

eAYKvTNd_sphere-with-radius-showing-r
cube-all-sides-length-s
cuboid-with-length-width-height-showing
cylinder-with-radius-height-showing

Surface Area

4 space pi space r squared

6s2

2lh + 2lw + 2wh

2 pi space r h space plus space 2 pi space r squared

Volume

4 over 3 pi space r cubed

s3

l × w × h

pi space r to the power of 2 space end exponent h

Example

If r = 1 cm
SA = 4π(1)2
= 4π cm2

V = 4/3 πr3
= 4/3 π13
4/3 π

 ∴ SA:V ratio = 4:4/3
= 3:1

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; these are spherical cells with a diameter of 800 nm (8 × 10-7 m)

  2. A bacterial cell from the species Bacillus subtilis; these are rod-shaped cells which you can assume to be cylindrical. They are 5 µm long and 1 µm in diameter

Comment on your calculated answers. 

Answer:

  1. Spherical cell: diameter = 800 nm, therefore radius = 400 nm

In order to compare cell 1 with cell 2, convert the radius value into µm by dividing by 1000

Radius = 400 over 1000 equals space 0.4 space mu m

Surface area = {formula from the formula sheet}

= 4 space pi space r squared
= 4 × 3.14 × 0.42
= 2.01 µm2

Volume =  {formula from the formula sheet}

= 4 over 3 pi space r cubed

= 4/3 x 3.14 x 0.43

= 0.268 µm3

A ratio is one number divided by another, therefore SA:Vol ratio

= fraction numerator 2.01 over denominator 0.268 end fraction

= 7.5: 1

___________

  1. Cylindrical cell: radius= 0.5 µm (half of diameter 1µm) 

Surface area = {formula from the formula sheet}

= 2 space pi space r h space plus space 2 space pi space r squared
= (2π × 0.5 × 5) + (2π × 0.52)
= 15.708 + 1.571 µm2
= 17.278 µm2

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

___________

The spherical 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 spherical cell has the largest surface area-to-volume ratio of all shapes of cells for a given radius/diameter. The larger the surface area-to-volume ratio, the better adapted to simple diffusion that organism is. 

Examiner Tips and Tricks

You will be able to use a formula sheet in your AP Exam, although you will be expected to be able to calculate the SA:V ratio for a sphere, cube, rectangular solid or cylinder from these formulae, and explain how the increasing size of an organism affects the SA:V ratio.

Maximizing surface area-to-volume ratio

  • As organisms increase in size, their surface area to volume ratio decreases

    • Small cells have a high surface area-to-volume ratio and more efficient exchange of materials with the environment

    • Larger cells have a lower surface area-to-volume ratio and are less efficient at exchanging materials with their environment

    • Materials exchanged include

      • oxygen

      • carbon dioxide

      • water

      • heat exchange

  • Some single-celled organisms can rely on simple diffusion alone to exchange materials with their environment

  • More complex cellular structures (e.g. membrane folds) are necessary to exchange materials with the environment adequately

  • As organisms evolved from single-celled to multicellular, more sophisticated exchange systems (e.g. lungs) evolved to maintain diffusion rates that were high enough to sustain life

Examples of exchange structures

  • Exchange structures may evolve to be specially adapted to maximize exchange

Gut epithelial cells

  • A cell on the inside surface of the small intestine is in contact with the flow of food passing through the lumen of the intestine

  • The cell's principal role is to absorb important nutrient molecules for distribution around the rest of the organism's body

  • Only one surface of the cell is in contact with the food flow; this is the upper surface as shown

    • The remaining surfaces of the cell are in direct contact with other cells in the tissue, so cannot absorb food molecules directly from the lumen

  • So the surface area of the food-contacting surface is maximized by folded structures called microvilli

  • These increase the internal surface area of the small intestine by a factor of around 100

Diagram of membrane folds to increase surface area
Folds in the upper surface of the cell's membrane increase the surface area over which materials can be exchanged by the cell

Root hair cells

  • Root hairs are single-celled extensions of epidermis cells in the root

    • They grow between soil particles and absorb water and minerals from the soil

  • Root hair cells are adapted for the efficient uptake of water (by osmosis) and mineral ions (by active transport)

    • Root hairs increase the surface area of plant roots, increasing the rate at which water and minerals can be absorbed

Root Hair Cell, IGCSE & GCSE Biology revision notes
A root hair cell

Guard cells

  • Guard cells surround stomata on the lower surface of a leaf; their role is to expand and contract which opens and closes the stomata allowing for gas exchange

  • The surface area-to-volume ratio of guard cells changes as they expand and contract

  • As guard cells become turgid, they curve outward which creates a larger stomatal opening

    • This increases the effective surface area for gas exchange

  • The small size of guard cells relative to the open stomata ensures efficient diffusion of gases, as the surface area-to-volume ratio is high for the pore area

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Cara Head

Author: Cara Head

Expertise: Biology Content Creator

Cara graduated from the University of Exeter in 2005 with a degree in Biological Sciences. She has fifteen years of experience teaching the Sciences at KS3 to KS5, and Psychology at A-Level. Cara has taught in a range of secondary schools across the South West of England before joining the team at SME. Cara is passionate about Biology and creating resources that bring the subject alive and deepen students' understanding

Lára Marie McIvor

Author: Lára Marie McIvor

Expertise: Biology Lead

Lára graduated from Oxford University in Biological Sciences and has now been a science tutor working in the UK for several years. Lára has a particular interest in the area of infectious disease and epidemiology, and enjoys creating original educational materials that develop confidence and facilitate learning.