Abstract
The aim of this study was to test the rate of reactivity of the enzyme catalase on hydrogen peroxide while subject to different concentrations of an inhibitor. The hypothesis was that hydrogen peroxide will be broken down by catalase into hydrogen and oxygen, where a higher concentration of inhibitor will yield less oxygen, resultant of a lower rate of reaction. Crushed potato samples of equal weight were placed in hydrogen peroxide solutions of various temperatures. The results showed that less gas was produced as the concentration of the inhibitor rose. This Is because more enzymes were inhibited, and so less active sites were available for reaction.
Reasearch and rationale
Hypothesis:
Catalase will break down hydrogen …show more content…

This may change the structure of the active site, stopping any reaction between enzyme and substrate. Therefore beyond a certain temperature, the rate of reaction will decrease. To control the temperature, each system must be heated to the same temperature. For this experiment, I have chosen to keep each system at 30oC. This is because room temperature can vary in a lab, as some days can be warmer than others, and the room’s ventilation can also affect the temperature of the system. This is also to allow the reactions to happen at an efficient enough rates to collect results. pH: Any change in pH affects the ionic and hydrogen bonding in an enzyme and so alters it shape. Each enzyme has an optimum pH at which its active site best fits the substrate. Variation either side of this pH results in denaturation of the enzyme and a slower rate of reaction. The optimum pH for catalase in potatoes is generally 7, but can differ depending on the acidity of the soil it was grown in. In this experiment a pH 7 buffer was used. This is because the optimum pH of most types of catalase is 7, and so that the systems can all be kept constant. Concentration of solutions: the concentration both the enzyme and substrate have to be taken into consideration. The concentration of substrate for example will affect the rate of reaction in a positive way until all the active sites are occupied. The same applies to