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Investigate the effect of the concentration of the enzyme catalase on the decomposition of hydrogen peroxide.

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Introduction

Helen Bewick 10C Investigate the effect of the concentration of the enzyme catalase on the decomposition of hydrogen peroxide. During my investigation I hope to find out if the concentration of the enzyme catalase will have an effect on the decomposition of hydrogen peroxide. I will do this by changing the surface area of a potato cube and reacting it with hydrogen peroxide. What are enzymes? Enzymes are proteins that are involved in every chemical reaction in living cells. Enzymes are biological catalysts that speed up reactions but do not get used up in the reaction. An enzyme has an active site that is specific to its substrate. The substrate fits into the active site where it is broken down. The lock and key diagram shown below illustrates this. Substrate fits into enzyme's specific active site. Enzyme breaks down substrate. In my investigation I will be using the enzyme catalase in a potato cube. Hydrogen peroxide is decomposed by the enzyme catalse into oxygen and water. The equation for the reaction is: Catalyses Hydrogen peroxide water + oxygen 2H2O2 (aq) 2H2O (I) + O2 (Equation source - Chemistry for You written by Lawrie Ryan) Prediction I predict that as the surface area increases, thereby increasing the concentration of catalase, oxygen will be produced more quickly as the rate of reaction will increase. As the concentration of catalase is increased the rate of reaction will increase because the collision theory states that in a higher concentration there is a larger number of particles in the same volume. For the particles to react with each other they need to collide. So in a higher concentration there will be more collisions, therefore there will be more successful collisions, which increases the rate of reaction. This means that in a higher concentration the number of enzymes will increase thereby increasing the number of active sites causing the hydrogen peroxide to be decomposed more quickly and produce oxygen more quickly. ...read more.

Middle

Equipment I will need to perform my investigation Equipment Reason for use Graduated test tube Enables me to measure the amount of oxygen released accurately. I could have used a measuring cylinder but it would be harder to turn it upside down because the entrance isn't thumb sized. I could have used a burette but it would be difficult to hold it upright because it is long. Delivery tube Collects all oxygen that is released. Measuring cylinder and pipette Accurately measures amount of hydrogen peroxide to the nearest cm3 Conical flask and bung Prevents any oxygen being lost during the reaction Scalpel and tile Enables me to increase the surface area of the potato without contaminating the potato. Stopwatch Allows me to accurately measure the time to a hundredth of a second. Water trough Allows me to use downward displacement of water to collect the oxygen produced during the experiment. Method When I am measuring the hydrogen peroxide I will make sure the measuring cylinder is on a flat level surface and measure to the bottom of the meniscus. The different surface areas I will be investigating will be 24cm2, 32cm2, 40cm2, 48cm2, 56cm2 and 64cm2. I have chosen to use 6 different concentrations of the enzyme catalase because I think that I will gain a range of results so that I can make a reliable conclusion. 1- I will ensure that all the apparatus is airtight and then I will set up the apparatus as shown below. 2- I will measure out 60cm3 of hydrogen peroxide into the conical flask. 3- I will fill the graduated test tube with water then turn it upside down with my thumb placed over the entrance whilst it is still emerged in water thus ensuring that no air bubbles are let into the graduated test tube. 4- I will place a 2cm3 potato cube into the conical flask. ...read more.

Conclusion

To prevent this I could use a wider conical flask so I could spread the potato out more so the pieces wouldn't stick together. * It was also difficult to cut a 2cm3 cube of potato 5 times so if I was to redo the investigation I would start with a larger cube maybe a 4cm cube. * One of the variables that affect enzymes is pH. Every enzyme has an optimum pH level at which it can bond best with its substrate. Any change in the pH level will denature the enzyme and change the shape of the enzyme's active site so it cannot bond as well with the substrate, which in this case is the hydrogen peroxide. A change in the pH could have reduced the rate of reaction and make my results unfair. I could improve the investigation by maintaining a constant pH using a pH buffer of pH 7. I would use a pH 7 buffer because this is equal to the natural environment of the enzyme in the potato tissue. I could extend my investigation by using higher concentrations of the enzyme catalase by increasing the surface area of the potato even more. If I did this I could find out at what concentration the rate of reaction is quick enough to decompose all the hydrogen peroxide. If I did this I could draw a graph more like my predicted graph. I could also extend my investigation by using a larger number of concentrations between those I have already recorded so I could draw a more accurate line of best fit. I used a potato as my source of catalase. The concentration of catalase could have changed in the different potatoes which would have made my results unfair. To make my investigation a lot more accurate I could use a 1 molar solution of the enzyme catalase instead of the potato cubes. I could have diluted the enzyme to create different concentrations. This way I could have measured the concentrations far more accurately and greatly reduce the chances of error in my investigation. ...read more.

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