Chemistry Investigation
An investigation to find the optimum temperature that will produce the
maximum amount of oxygen from hydrogen peroxide using the enzyme
catalase
Introduction
Hydrogen peroxide is poisonous to all living things if it builds up in
the cells but when it is introduced to the enzyme catalase the react
between the two makes water and oxygen as byproducts. There are many
factors witch affect the production of oxygen these are listed below.
For this experiment the enzyme will be held in potato.
Prediction
I predict that the enzyme catalase will have a temperature at which it
will work best this is called its optimum but this could be anything,
I have chosen temperatures at a ten degrees difference starting at ten
going up to eighty degrees. If the temperature gets too low the
reaction will slow down until the reaction stops this is because of
collision theory this tells us that the more energy a partial has the
fast it goes and there for there is more chance of a successful
collision. At a higher temperature the reaction will speed up but the
enzymes active site will change shape so the enzyme and the partial do
not fit in just right like this
[IMAGE][IMAGE][IMAGE]The enzyme at optimum: just above:
far above
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and eventually the enzymes will denature. To find out the best
conditions I need to consider more then just temperature. These are
possible factors that determine how much oxygen is given off.
Independent Variables
1) The amount of enzyme used (fixed)
2) The concentration of enzyme in the potato (fixed)
3) The surface area of the potato (fixed)
4) Temperature (experimental)
5) Amount of hydrogen peroxide (fixed)
6) The concentration of the hydrogen peroxide (fixed)
7) Freshness of the vegetables (fixed)
8) The pH (fixed)
The amount of enzyme used
I must use the same amount of enzyme each time I do an experiment so I
3.) Divide your 30g of white substance into the 4 test tubes evenly. You should put 7.5g into each test tube along with the water.
Test-tube C had the best concentration according to the results. Three test-tubes were labelled A-C. Test-tube A had 1ml enzyme solution which was added to test-tube B which had 4ml buffer (pH 5 was used). 1ml of the solution from test-tube B was then added to the test-tube C which also had 4ml buffer (pH 5). Test-tube C was used as the enzyme in all the reactions. Nine test-tubes were taken out of them one was used as the the blank, labelled as test-tube 9. The blank had 5ml buffer (pH 5), 2ml hydrogen peroxide, 1ml guaiacol and no enzyme. Then, 3ml of buffer (pH 3) and 2ml of enzyme were added to test-tube 1. Test-tube 2 had 2ml hydrogen peroxide and 1ml guaiacol. Test-tube 1 and 2 were mixed. The same procedure was used for test-tube 3 as test-tube 1, but this time the buffer was pH 5. Test-tube 4 was prepared the same way as test-tube 2. Then, Test-tube 3 and 4 were mixed. Test-tube 5 was prepared as test-tube 1 but with buffer of pH 7 and test-tube 6 was prepared as test-tube 2. Next, test-tube 5 and 6 were mixed. Last but not the least, test-tube 7 was prepared as test-tube 1 but with buffer of pH 9 and test-tube 8 was prepared as test-tube 2. Then, test-tube 7 and 8 were mixed. The spectrophotometer was set to 470nm and using the blank it was set to zero. The four test-tubes with different pH’s (pH 3, pH 5, pH 7, pH 9) were read
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second test tube also add 6 mL of 0.1M HCl. Make a solution of 0.165
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