Exchange Surfaces and Exchanging Substances Noadswood Science, 2016

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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