Catalase Chemistry Experiment

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Discussion The efficiency of enzymes depends on a number of factors, such as temperature, pH and the concentration of the enzyme just to name a few. This experiment was conducted to show the effects of different pH levels on the activity of the catalase enzyme, the data obtained in this experiment supports the initial hypothesis of the experiment which states that the catalase enzyme will function optimally at a pH of 7 and efficiency will decrease as the pH of solutions moves further away from 7. This is true because in the experiment the test tube containing the distilled water which has a pH of 7, produced the highest amount of foam out of the four test tubes. The efficiency of the reaction can be determined by the amount of foam (foam = oxygen, which is the result of the breaking down of the hydrogen peroxide molecule into water and oxygen) produced, or the level of the fizzing that takes place when hydrogen peroxide is added. The pH of a solution that an enzyme has to function in is very important because if the pH turns out to be too low or too high the enzyme will denature making it unable to function, however, the pH of a solution may be far from the neutral level of 7 but the enzyme might not fully denature which means that the enzyme catalase would not lose its catalytic ability completely whereas a fully denatured enzyme’s shame will have been altered completely causing the active site and substrate to not fit together the way they’re supposed to. In certain reactions, the active site and enzyme substrate are still able to fit together which means that the reaction will continue to take place but at a very low rate, this seemed to be the case in one of the test tubes during the experiment. The foam produced in the react... ... middle of paper ... ...f denaturing the enzyme, thus making it unable to function because its active site had been reshaped into a shape that did not correspond with the substrate, to allow for them to interlock. The denaturing of the enzyme took place in Test tube B at 7 ½ minutes. The rest of the reactions in the other test tubes were continuing to react. All reactions maintained a very opaque colour and only when the hydrogen peroxide was added to the solution did the contents of the test tube clear up slightly and become more transparent in colour. The transparency of the solutions within the test tubes at the end of the experiment was because of the water that had been produced from the breaking down of the hydrogen peroxide which forms water (H2O) and oxygen (O2). The water diluted the solutions making them more transparent. These results are evidence in supporting my hypothesis.

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