1. Exchange Surfaces and Exchanging Substances
Noadswood Science, 2016

2. Exchange Surfaces and Exchanging Substances
Monday, April 23, 2018
Exchange Surfaces and Exchanging Substances
To identify how materials can be exchanged and how the rate can be increased

3. Precise Learning
A single-celled organism has a relatively large surface area to volume ratio. This allows sufficient transport of molecules into and out of the cell to meet the needs of the organism.
Students should be able to calculate and compare surface area to volume ratios.
Students should be able to explain the need for exchange surfaces and a transport system in multicellular organisms in terms of surface area to volume ratio.
Students should be able to explain how the small intestine and lungs in mammals, gills in fish, and the roots and leaves in plants, are adapted for exchanging materials.
In multicellular organisms, surfaces and organ systems are specialised for exchanging materials. This is to allow sufficient molecules to be transported into and out of cells for the organism’s needs. The effectiveness of an exchange surface is increased by:
having a large surface area
a membrane that is thin, to provide a short diffusion path
(in animals) having an efficient blood supply
(in animals, for gaseous exchange) being ventilated.

4. Exchange Surfaces
Solutes are dissolved substances – living organisms need to be able to exchange dissolved substances with their environment in order to survive
An organism’s size and surface area affect how quickly this is done

5. Movement
Life processes need gases or other dissolved substances (solutes) before they can happen. Dissolved substances move to where they need to be by diffusion and active transport (water via osmosis)
Cells can use diffusion to take in substances they need and get rid of waste products, e.g.
Oxygen and carbon dioxide are transferred between cells and the environment during gas exchange
In humans urea diffuses from cells into the blood plasma for removal from the body by the kidneys
Exchanging substances between the organism and the environment depends on the SA: volume

6. Packing Boxes
Look at the packing box delivery experiment to identify how SA: volume works…

7. volume = length x width x height
SA: volume
A ratio shows how big one value is compared to another – the larger an organism is the smaller its SA is compared to its volume
area = length x width
All sides need their area worked out and added together which is a total area of 52 cm2 (top, bottom, left, right, front and back)
volume = length x width x height
The block is 4 cm x 3 cm x 2 cm which is a volume of 24 cm3

8. Ratio
Imagine comparing a cube mouse (1x cube) versus a cube hippo (32x cubes) – what is their SA: volume ratios?

9. Ratio
Imagine comparing a cube mouse (1x cube) versus a cube hippo (32x cubes) – what is their SA: volume ratios?
The SA is 6 x (1 cm x 1 cm) = 6 cm2
The volume is 1 cm x 1 cm x 1cm = 1cm3
So the mouse has a SA: volume of 6:1

10. Ratio
Imagine comparing a cube mouse (1x cube) versus a cube hippo (32x cubes) – what is their SA: volume ratios?
The SA is 2 x (4 cm x 4 cm) = 32 cm2 (top and bottom surfaces) and 4 x (4 cm x 2 cm) = 32cm2 (four sides) so the total is 64cm2
The volume is 2 cm x 4 cm x 4cm = 32cm3
So the hippo has a SA: volume of 64:32 or simplified a SA: volume of 2:1

11. Ratio
So the hippo at 2:1 has a smaller SA: volume than the mouse at 6:1

12. Exchange Surfaces
In single-celled organisms, gases and dissolved substances can diffuse directly into or out of the cell across the cell membrane
Multicellular organisms have a smaller SA: volume which means not enough substances can diffuse from their outside surface to supply their entire volume – for this reason exchange surfaces are needed

13. Adaptations
Exchange surfaces have the following properties to increase their effectiveness: -
Thin membranes so substances only have a short distance to diffuse (a short diffusion pathway)
Large surface area so lots of a substance can diffuse at once
In animals lots of blood vessels and are often ventilated

14. Small intestine
The inside wall of the small intestine is thin, with a large surface area. This allows absorption to happen quickly and efficiently. To get a big surface area, the inside wall of the small intestine is lined with tiny villi.
These stick out and give a big surface area. They also contain blood capillaries to carry away the absorbed food molecules

15. Lungs
To maximise the efficiency of gas exchange, the alveoli within the lungs have several adaptations:
They are folded, providing a much greater surface area for gas exchange to occur.
The walls of the alveoli are only one cell thick making the exchange surface very thin (shortening the diffusion distance across which gases have to move)
Each alveolus is surrounded by blood capillaries which ensure a good blood supply

16. Gills
Fish carry out gas exchange using their gills. These only work when water is forced across the gill filaments, so fish are restricted to living in water.

17. Roots and Leaves
Roots and leaves have large surface areas to absorb as much water / minerals (roots) and light (leaves) as possible

18. Practice Questions
Give three ways that substances can move into and out of a cell
By which of these three ways can dissolved substances move into and out of a cell?
Give three ways in which exchange surfaces in animals may be adapted for effective exchange

19. Answers
Give three ways that substances can move into and out of a cell – diffusion, osmosis and active transport
By which of these three ways can dissolved substances move into and out of a cell – diffusion and active transport
Give three ways in which exchange surfaces in animals may be adapted for effective exchange – thin / large SA / lots of blood vessels / ventilated

20. Practice questions – application
Three different cells can be represented by the shapes below
Calculate the surface area of each cell
Calculate the volume of each cell
Calculate the SA: volume for each cell and predict which is the most efficient at absorbing substances by diffusion

21. Answers Calculate the surface area of each cell
A = 6 x (1 x 1) = 6 µm2
B = 2 x (0.5 x 0.5) + 4 x (2 x 0.5) = 4.5 µm2
C = 2 x (0.2 x 0.4) + 2 x (1 x 0.4) + 2 x (1 x 0.2) = 1.36 µm2
Calculate the volume of each cell
A = 1 x 1 x 1 = 1 µm3
B = 2 x 0.5 x 0.5 = 0.5 µm3
C = 1 x 0.4 x 0.2 = 0.08 µm3

22. Answers
Calculate the SA: volume for each cell and predict which is the most efficient at absorbing substances by diffusion
A = 6:1
B = 4.5:0.5 = 9:1
C = 1.36:0.08 = 17:1
C is the best at absorbing as it has the largest SA: volume