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Investigate the action of the Enzyme Catalyse

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Linsey Belford L5V Biology Coursework To Investigate the action of the Enzyme Catalyse Aim: The this experiment is to investigate the action of the enzyme catalase using potato and hydrogen peroxide. I am trying to find out the productive rates of the hydrogen peroxide and the potato. I am changing the surface area of the potato but I am still using the same mass to make this fair. Prediction: I predict that the larger the surface area the faster the productive rate and the more gas produced because the catalase has more active sites for the hydrogen peroxide to fit into. Catalysts are used to speed up specific reactions in cells. They are all very unique as each enzyme only performs one particular reaction. Catalase is the enzyme which I am using. It is in the cells of living organisms. It is found in potato and liver so I am using potato in my investigation. The catalase speeds up the decomposition of Hydrogen Peroxide (H2O2) into water and oxygen. catalase Hydrogen Peroxide Water + Oxygen 2H2O2 2H2O + O2 It is necessary for Hydrogen Peroxide to be broken down as it is produced by many chemical reactions in the cells and it is very dangerous. As it is such a dangerous substance it must be broken down immediately. Enzymes are proteins. The molecules have a three dimensional shape. This shape contains a small space, dent or area of which it is exactly the correct size and shape for a molecule, of the enzymes substrate, to fit into. In my case for this experiment the catalase molecules have a space for the hydrogen peroxide to fit into. ...read more.


Hydrogen Peroxide, Burette, 2 Large Potatoes, Stop Clock, Cork Borer (1cm diameter), Chopping Knife, Chopping Tile, 8 weighing boats Heatproof Mat, Weighing Scales, Ruler. This is how the apparatus should be set up: This will be the actual method for my experiment: * Cut out the potato using the cork borer and cut into the required sections and surface areas. Weigh the cylinders of potato separately in the weighing boats and measure the lengths that they are cut. Make sure the surface area has been recorded accurately. Keep the potato in the weighing boats to avoid confusion. Use a ruler with millimetres for cutting. * Measure 10cm� of hydrogen peroxide (15 vol) and suck this up into the syringe. Always wear safety glasses, as hydrogen peroxide is very irritant to the skin. (To measure against the measurements on the side, use the bottom line of the base on the syringe) * Be careful with the sharp knife when cutting. Fill the burette with water making sure the tap is closed. Let the gas in to the level 50ml (it is a reading on the side) this is actually only a volume of approx 10 cm�. Let the air in to the top reading on the burette. The measurements should be taken using the bottom of the menisci. * Connect up the delivery tube to the burette in the tub of water ready for the displacement of air. * Place the potato into the conical flask and place the bung on. * Place the syringe onto the bung and add the hydrogen peroxide to the potato in the conical flask. ...read more.


A line of best for graph 8 (rate of reaction against surface area) may be appropriate to show how the graph is going up generally. Conclusion Using my evidence I have concluded that the as the surface area increases the rate of reaction increases also. This proves that my prediction is correct. This is because if there is a larger surface area there is likely to be more area for the active sites for the enzyme substrate to fit into and be broken down. The larger the surface area the more active site there are so they can work faster. This means that more Hydrogen Peroxide will be broken down into water and oxygen. The H2O2 molecule fits into the active site like 'lock and key' and the catalase then distorts the H2O2 and releases it. If there are more active sites doing this then more oxygen is going to be released and so the rate of reaction will speed up. My results have 'agreed' with my prediction and the scientific knowledge which is earlier in my plan. On my graph the reason that the smaller surface areas have a slower rate of reaction is because the concentration is the same the active site number is different. This is because although here are the same number of H2O2 molecules some may have to wait to be broken down whereas in the larger surface areas there are enough active sites to satisfy the number of H2O2 so fewer have to wait to be broken down. The maximum rate of reaction is when all the sites are being used but in reality this is never reached due to the fact that not all the active sites are being used at the same time. 1 ...read more.

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