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AP®BiologyCollege BoardStudy GuidesUnit 2: Cell Structure & FunctionCell StructureHow Cell Size Affects Exchange of Materials

How Cell Size Affects Exchange of Materials (College Board AP® Biology)

Study Guide

Phil

Written by: Phil

Reviewed by: Lára Marie McIvor

Surface Area-to-Volume Ratios

  • 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

 

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 straight pi straight r squared

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

4 over 3 straight pi straight r cubed

s3

l × w × h

straight pi straight r squared straight 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:

  • A bacterial cell from the species Staphylococcus aureus; these are spherical cells with a diameter of 800 nm (8 × 10-7 m)

  • 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

For the spherical cell: diameter = 800 nm, therefore radius = 400 nm

For comparison, convert this value into µm by dividing by 1000

fraction numerator 400 space nm over denominator 1000 end fraction equals space 0.4 straight mu straight m

Surface area = {formula from the formula sheet} = 4πr2
= 4 × π × 0.42
= 2.01 µm2

Volume =  {formula from the formula sheet} = 4 over 3 straight pi straight r cubed

equals space 4 over 3 cross times straight pi cross times 0.4 cubed

= 0.268 µm3

A ratio is one number divided by another, so

SA colon straight V space ratio space equals space fraction numerator 2.01 over denominator 0.268 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

___________

For the cylindrical cell: diameter = 0.5 µm (half of diameter 1µm) 
Surface area = {formula from the formula sheet} = 2πrh + 2πr2
= (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

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 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. 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 center of the cell; the shortest distance to the edge will always be found in a sphere. In a long cylinder, a molecule may travel parallel to the straight sides and so will have travel further to diffuse out. So 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.

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    Author
    Reviewer
    Phil

    Author: Phil

    Expertise: Biology Content Creator

    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.

    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.