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The Effect of Enzyme and Substrate Concentrations On the Rate of a Reaction

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The effect of enzyme and substrate concentrations on the rate of a reaction Requirements: CARE: Eye protection must be worn. * burette (50cm ) * trough or bowl * boiling tube with bung and delivery tube * graduated pipette (5cm ) and safety filler * measuring cylinder (10cm ) * hydrogen peroxide solution, 5 vol. (25cm ) * yeast suspension (20cm ), made from 2g dried yeast in 160cm water aerated for several hours * stopwatch Introduction: In this experiment I will follow the progress of a catalysed reaction by measuring the volume of gas produced as the reaction proceeds. I will use the initial rates of a series of experiments to try and find the orders of reaction with respect to enzyme and substrate. The enzyme in this reaction will be catalase, and the substrate will be hydrogen peroxide (H O ). Catalase is an enzyme and so catalyses the decomposition of hydrogen peroxide, thus producing water and oxygen. This reaction is written as a balanced equation as follows; 2H O (aq) - 2H O(l) ...read more.


After organising ourselves into taking turns in recording readings of volume at 10 second intervals for 4 minutes. 4: Then I added 5cm of the hydrogen peroxide to the yeast suspension and by quickly replacing the bung to the boiling tube, began the stopwatch as the first bubble reached the top of the burette. Recording the volume of gas in the burette every 10 seconds for 4 minutes was relatively straightforward. 5: After washing out the boiling tube, and refilling the burette with 50cm of water, I repeated steps 1-4 four more times using the following; 4cm H O + 1cm distilled H O 3cm H O + 2cm distilled H O 2cm H O + 3cm distilled H O 1cm H O + 4cm distilled H O in the measuring cylinder. Everything else was kept the same in each of the experiments in order to make sure that the test was a fair one. 6: Finally I plotted a graph from my results, with the volume of O given off on the vertical y axis, against time on the horizontal x axis for each experiment. ...read more.


There only explanation that I can account for is the probability that there was some extra H O present or maybe a slight difference in the volume of enzyme. Another common problem is that the person whom is recording the data may have made a slight misjudgement. From part two in the experiment, where the volume of yeast was altered in ratio to the distilled water, the results were much more straightforward, and fitted the expected outcome. Thus being that the less yeast added along with the more distilled water, the volume of O given off was decreasingly less. The differences amongst the separate experiments was quite large, as the amount of yeast obviously had a big impact upon the oxygen. I have learnt from doing this experiment that you should always make sure you have the absolute correct measurements, and that the conditions are all kept at a constant to ensure a fair test, and hopefully disable any strange results. I will now construct two graphs, one for each part of the experiment. This should give a better indication of the visual effect that the variables whence changed, have an impact upon the outcome of the volume of oxygen given out. ...read more.

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