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To investigate how long it takes catalase (an enzyme) to react with the hydrogen peroxide and if the molarity of the substrate affects the rate of reaction of the enzyme.

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Catalase Investigation. Aim: To investigate how long it takes catalase (an enzyme) to react with the hydrogen peroxide and if the molarity of the substrate affects the rate of reaction of the enzyme. Introduction: Enzymes are used to speed up the rate of a chemical reaction, known as a catabolic or 'breaking down' reaction, and are made of protein. Each enzyme is specific to the material or substrate it is used with due to the shape of its active site. During the reaction, the substrate molecule fits into the enzyme molecule's active site like a 'lock and key'. The enzyme molecule then breaks the substrate molecule down and is then left free to be re-used with another molecule. In the experiment I am doing, I am using potato juice as the enzyme and hydrogen peroxide as the substrate. Catalase is a very efficient enzyme found in all living cells. All cells build up a chemical called hydrogen peroxide as a waste product. The hydrogen peroxide is toxic and dangerous and so catalase is produced to break it down into the harmless products, water and oxygen. This is done by two types of reactions called oxidation (losing electrons) and reduction (gaining electrons). When the substrate concentration is changed, it can alter the rate of the reaction. When more substrate molecules are added, more active sites can be occupied meaning that more molecules can be broken and more products formed. ...read more.


Molarity Table: Molarity Hydrogen peroxide (ml) Water (ml) 2 50 0 1.6 40 10 1.2 30 20 0.8 20 30 0.4 10 40 Prediction: I think that if you increase the substrate concentration, the product (oxygen) will double too. I predict that as the molarity of the substrate goes up, the length of time it takes the disc of filter paper to sink and return to the surface will get shorter and shorter. This is because the higher the concentration of substrate, the quicker the reaction will be. This means that the oxygen from the reaction is produced quicker. The oxygen causes the filter paper disc to rise to the surface of the tube and therefore the filter disc will return to the surface quicker. The graph below shows that if the substrate is doubled, the product doubles too. It also shows that when the substrate concentration reaches a certain point, some of the substrate is left unused and it causes the reaction to remain at its best, shown on the graph by the plateau. Fair test: To ensure a fair test, I will make sure I only change one variable at a time. This means I will have to be careful of things such as temperature of the water or hydrogen peroxide and the quantities of water, hydrogen peroxide and potato I am using. ...read more.


The time should have been faster. This could have been because of a number of reasons. The disc may not have been thoroughly soaked in the potato juice or there could have been an error in measuring out the concentration of the hydrogen peroxide. As it was only one result, I feel my results are of a good quality. Improvements: To improve the method I could have timed how long I soaked the filter paper disc in the potato juice for to ensure that all the discs soaked up the same amount of enzyme. To make the test fairer, I could also have changed the hydrogen peroxide after each repeat to ensure the hydrogen peroxide concentration was pure and clean. Further Work: Another experiment I could do to measure the rate that different concentrations of catalase break down hydrogen peroxide, would be in an inverted measuring cylinder, to measure the oxygen produced by the displacement of water. I would mix the enzyme and substrate together in a test tube and then use a tube to collect the gas given off and send it to a measuring cylinder and it would be measured at regular times for a total duration of a minute. For this experiment, I would expect the amount of oxygen produced in the minute to increase as the concentration of the catalase increased, this is because as the catalase concentration increases, the rate of reaction also increases. ...read more.

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