Enzyme Lab Questions

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Enzyme lab questions 1. How do you account for the differences in rates? 1. In the first experiment, 2 mL of H2O2 was mixed with 0.1g sand. In this mixture, there are no enzymes present and therefore no products are formed. The reaction rate was rated 0, just as we anticipated. 2. The 2nd experiment was a mixture of 2 mL H2O2 and 0.1g Manganese dioxide. We can use the same principal we used for the first experiment, Manganese dioxide is not a protein, not an enzyme, it is a catalyst. Due to this, no products are formed, and just as we anticipated, the reaction rate was given a 0. 3. The 3rd experiment is a mixture of 2 mL H2O2 and liver. In this situation, both the substrate (H2O2) and the enzyme (catalase) are present, and…show more content…
Experiment 8 is composed of potato, sand and 2 mL of H2O2. The contents of the test tube were grinded as in experiment 7, and therefore we may use the same principal as we used in analyzing experiment 7. The increase in surface are causes a greater concentration of enzymes to be present to react and therefore the reaction rate goes up. Additionally, because of the increase in surface area, the reaction rate should be higher than that of experiment 4, which it was but also lower than that of experiment 7. This is because there is a greater concentration of enzymes in liver than potatoes. Based on observations, the experiment received a 3 for reaction…show more content…
Experiment 8 is composed of liver at 370C and 2 mL H2O2. 370C is the approximate body temperature of a calf and therefore the enzymes should function well because they are exposed to optimum conditions. Just as we anticipated the reaction rate was high and based on observations we gave the reaction rate a 4. 11. Experiment 11 is composed of cold liver and 2 mL H2O2. We anticipated that the reaction rate should be very low because a low temperature means a low reaction rate. At low temperatures, particles move slower and therefore there is a lower amount of collisions between the enzymes and the substrates. Based on observations, the reaction rate was fairly high which came as a surprise for us, this is not what we had anticipated. The reaction rate was given a 3. 2. Hydrogen peroxide can be broken down by catalysts other than those found in living systems. Hydrogen peroxide will break down in the presence of any substance that can oxidize it. Manganese dioxide is an inorganic molecule that can achieve this. The catalyst lowers the activation energy of the decomposition which is already noticeable without the addition of a catalyst. In our lab we observed that the Manganse dioxide had no effect on the reaction, however it should have a significant

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