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In this investigation I will investigate the effect of the enzyme Catalase on Hydrogen Peroxide by measuring the volume of oxygen gas produced in a certain time.

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Introduction

The Effect of Catalase on Hydrogen Peroxide Aim In this investigation I will investigate the effect of the enzyme Catalase on Hydrogen Peroxide by measuring the volume of oxygen gas produced in a certain time. All living cells contain Catalase in them and so instead of using a Catalase solution I will use a yeast solution. The reaction of this experiment will be as follows: Hydrogen Peroxide Water + Oxygen Hypothesis My hypothesis for this experiment is that I think as I increase the concentration of Catalase, the rate of oxygen produced will increase as long as there is always excess substrate (H2O2). I am going to now explain what an enzyme is and this will help me to justify my hypothesis. An enzyme is a protein which acts like a biological catalyst, and the definition of a catalyst is a substance that speeds up a reaction, but is not chemically changed at the end of the reaction and therefore is never used up in a reaction. From preliminary work I have done on enzymes I know this to be true because I have done an experiment in which I measured the mass of Manganese (IV) oxide (MnO2) (also a catalyst of H2O2) before adding it to Hydrogen Peroxide. Then after the reaction had finished I measured the mass again. And as I suspected the mass remained exactly the same and so this proves the fact that the catalyst is unchanged. ...read more.

Middle

After having done this I began to find the best possible time range. I did this by reacting 5 cm3 H2O2 with 5 cm3 of Catalase and measuring the volume of gas given off at different times. I found that 15 seconds after the start of the reaction was a good time range because it gave the reaction plenty of time and the volume of gas given off was about 75 cm3 which means that for my own reactions if I have anything that gives off more gas within this time will not go off the scale of the gas syringe. In the reactions there will permanently be 5 cm3 of H2O2 and then I will use 5 cm3 (100% yeast), 4 cm3 (80% yeast), 3 cm3 (60% yeast), 2 cm3 (40% yeast) and 1 cm3 (20% yeast) of the yeast solution, but since I have to have a total volume of 10 cm3 I will then add the corresponding amount of water to the total volume in order to get 10 cm3. (For example 5 cm3 of H2O2 + 3 cm3 of Yeast + 2 cm3 of water = 10 cm3 total volume). Also, as a control I will have 0 cm3 (0% yeast) of yeast with 5 cm3 of water and 5 cm3 of H2O2. This control experiment will show me that pure water will not react or catalyse the hydrogen peroxide. ...read more.

Conclusion

a higher concentration of a substance in a solution it will have more collision which means more reactions and therefore this will make my results quicker than if I used a lower concentration of it. I can solve this problem by using the same volume of H2O2 in all my reactions. The second way that substrate concentration can affect my results is that it can level off the rate of oxygen made, if all the substrate is catalysed within my time range. I can solve this problem by simply using excess substrate but at the same time I will still have to keep the volume the same because of the problem I discussed earlier. Total volume must also be kept the same because otherwise the experiment would be unfair due to collision theory. This is because if the numbers of particles are changed, the number of collisions would be changed and therefore this means the number of reactions would change leading to a different amount of gas produced in my experiment. I can solve this by adding water when I use a smaller amount of enzyme. Time is another one of the variable that must be kept the same because if not the reaction would have more or less time to react and so producing different volumes of gas. I can solve this by always letting the reaction go on for the same amount of time; this will be 15 seconds in my experiments. Shayaan Nackvi ...read more.

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