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, or the total amount of space inside the organism

  • As the surface area and volume of an organism increase (as the size of an 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:

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

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

    • acquire or dissipate thermal energy (heat)

    • exchange chemicals with the surroundings, e.g. secreting or binding to hormones at the cell surface

Calculating surface area-to-volume ratios

  • The SA:V ratios of different biological shapes can be calculated using mathematical formulae

 

Sphere

Cube

Rectangular solid

Cylinder

 

Cross section of a sphere showing a radius labeled "r" with a dashed line; a dotted line marks the circle's surface intersection.
Diagram of a cube with sides labelled "s". Each side is equal in length, illustrating a three-dimensional shape with lines and shading for depth.
Diagram of a rectangular box showing dimensions: width and length are on the base, height is vertical. Arrows indicate measurement directions.
Diagram of a cylinder showing its height (h) and radius (r) with dashed lines representing the circular bases.

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

Worked Example

Calculate the surface area-to-volume ratios of the 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 for Staphylococcus aureus:

Step 1: determine the radius of a cell

  • The diameter of a cell = 800 nm

radius = 800 ÷ 2

= 400 nm

Step 2: convert the units to µm

  • There are 1000 nm in a µm, so we divide by 1000

= 400 ÷ 1000

= 0.4 µm

Step 3: use the formula to calculate surface area

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

Step 4: use the formula to calculate volume

= 4 over 3 pi space r cubed

= 4/3 x 3.14 x 0.43

= 0.268 µm3

Step 5: convert to a ratio

  • The standard way of stating a ratio is X : 1, so we begin by converting the second number to 1

0.268 ÷ 0.268 = 1

  • Then we divide the first number by the same factor

2.01 ÷ 0.268 = 7.5

  • We can then express this as a ratio:

= 7.5:1

Answer for Bacillus subtilis

Step 1: determine the radius of a cell

  • The diameter of a cell = 1 µm

1 ÷ 2 = 0.5 µm

Step 2: use the formula to determine surface area

  • The length of a cell = 5 µm

= 2 π r h + 2 π r2

= (2π × 0.5 × 5) + (2π × 0.52)

= 15.708 + 1.571 µm2

= 17.278 µm2

Step 3: use the formula to determine volume:

= π r2 h

= π × 0.52 × 5

= 3.927 µm3

Step 4: convert to a ratio

  • Begin by converting the second number to 1

3.927 ÷ 3.927 = 1

  • Then we divide the first number by the same factor

17.278 ÷ 3.927 = 4.4

  • We can then express this as a ratio:

4.4:1

___________

  1. Staphylococcus aureus has a surface area-to-volume ratio of 7.5:1

  2. Bacillus subtilis aureus has a surface area-to-volume ratio of 4.4:1

Examiner Tips and Tricks

In an exam you are expected to be able to:

  • calculate the SA:V ratio of a sphere, cube, rectangular solid or cylinder, using formulae from a formula sheet

  • explain how the increasing size of an organism affects the SA:V ratio.

Maximizing surface area-to-volume ratio

  • As cells increase in size, their surface area–to–volume (SA:V) ratio decreases and their metabolic requirements increase

    • This means that larger cells have proportionally less surface available for exchange of substances, relative to their greater internal demand

    • The maximum size to which single cells can grow is restricted by this relationship

  • The surface area of the plasma membrane must be large enough to adequately exchange materials, so larger cells may have complex cellular structures, such as membrane folds, to enable adequate exchange of materials with the environment

Cell feature

Role

Adaptation for adequate exchange with environment

Root hairs

Diagram of a root hair cell, labelled with ribosomes, mitochondrion, nucleus, cytoplasm, cell membrane, cell wall, vacuole, and root hair.

Root hairs are single-celled extensions found on epidermis cells in plant roots

They aid absorption of water and minerals from the soil

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

Opening and closing guard cells

Diagram showing open and closed stoma with labelled parts: epidermal cells, guard cells, vacuoles, chloroplasts, nucleus, and cellulose microfibrils.

Guard cells surround stomata on the lower surface of a leaf, where their role is to open and close the stomata

This allows plants to carry out gas exchange while also controlling water loss

Guard cells can become turgid, expanding and curving outwards to allow stomata to open

This allows gases to diffuse into the leaf, where they come into contact with the large surface area of the spongy mesophyll

Microvilli on gut epithelial cells

Diagram of a cell showing microvilli, smooth endoplasmic reticulum, Golgi apparatus, nucleus, and mitochondrion, each labelled with pointers.

Gut epithelial cells are located on the inside surface of the small intestine

Their role is to absorb nutrients for distribution around the rest of the body

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

Cilia

Illustration of ciliated epithelial cells with nuclei, cilia, goblet cell, and labelled basement membrane on a purple base.

Cilia are hair-like projections on cell surfaces that can beat to move fluid over the surface of cells, e.g. unicellular eukaryotes living in a watery environment may have many cilia

Cilia keep the fluid next to the surface of cells moving, bringing in fresh fluid; this maintains a steep concentration gradient and increases the rate of diffusion

Examiner Tips and Tricks

Be careful not to confuse microvilli and cilia:

  • microvilli are formed from folded regions of the plasma membrane, and function to increase membrane surface area

  • cilia are hair-like structures located on the plasma membrane, and they beat to move substances past the cell surface

Surface area and heat exchange

  • The rate of heat exchange with the environment is also affected by cell size

    • As cell mass increases, SA:V decreases, and the rate of heat exchange with the ambient environment also decreases

  • Because of the relationship between SA:V and heat exchange, there is a also a relationship between metabolic rate per unit body mass and the size of multicellular organisms; this can be explained as follows:

    • small animals, with a higher SA:V ratio, will lose more heat to their surroundings, meaning that they need a relatively high metabolic rate to maintain body temperature

    • large animals, with a lower SA:V ratio, will lose less heat, meaning that they can maintain body temperature at a relatively low metabolic rate

Graph showing basal metabolic rate vs. mass for animals including shrew, mouse, sheep, and elephant. Rate decreases as mass increases
There is a relationship between metabolic rate per unit body mass and the size of multicellular organisms; typically, the smaller the organism, the higher the metabolic rate per unit body mass.

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Cara Head

Author: Cara Head

Expertise: Biology & Psychology 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

Reviewer: Lára Marie McIvor

Expertise: Biology, Psychology & Sociology Subject 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.