How Cell Size Affects Exchange of Materials | College Board AP Biology Revision Notes 2020

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

Surface Area	$4 {πr}^{2}$	6s²	2lh + 2lw + 2wh	$1 r e c t a n g l e = 2 πrh 2 c i r c u l a r e n d s = 2 {πr}^{2} S A = 2 πrh + 2 {πr}^{2}$
Volume	$\frac{4}{3} {πr}^{3}$	s³	l × w × h	${πr}^{2} h$
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

$\frac{400 nm}{1000} = 0.4 μm$

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

Volume = {formula from the formula sheet} = $\frac{4}{3} {πr}^{3}$

^{$= \frac{4}{3} \times π \times 0 . 4^{3}$}

= 0.268 µm³

A ratio is one number divided by another, so

$SA : V ratio = \frac{2.01}{0.268} = \frac{7.5}{1} = 7.5 : 1 (or simply 7.5)$

___________

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

$SA : V ratio = \frac{17.278}{3.927} = \frac{4.4}{1} = 4.4 : 1 (or simply 4.4)$

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.

Exam Tip

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.

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

Revision Note

Surface Area-to-Volume Ratios

Importance of a High Surface Area-to-Volume Ratio

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

Calculating Surface Area-to-Volume Ratios Table

Worked example

Exam Tip

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Author: Phil