An investigation into the effect of Hydrogen Peroxide concentration on Yeast Catalyse activity

Variables

As a measure of initial reaction rate I will collect the oxygen (in ml) given off over a period of 20 seconds - this is the dependant variable. I determined 20 seconds after several preliminary tests as being the most suitable for the equipment I will use. My independent or manipulated variable is HP concentration, measured in moldm-3. Variables which will remain fixed are :

: Incubation time. 20 seconds.

2: The temperature of the substrate and product. Catalysts reduce activation energy, therefore the energy that substrate and product have in the first place will affect how readily they react together. I will perform all my experiments at the same ambient temperature, which I will measure initially and keep an eye on throughout.

3: pH of the mixture. Enzymes have an optimal pH at which they work fastest. For Catalase this is around neutral, i.e. pH 7. Therefore I will add 5cm3 of pH 7 buffer to 200cm3 of yeast solution, (a proportion suggested as suitable by a scientist), and stir the mixture to ensure an even pH throughout.

4: Quantity of Hydrogen Peroxide and Yeast solution. Obviously the amount of HP solution will effect the number of H2O2 molecules, which will effect the amount of oxygen produced. I will therefore use 5ml of HP solution in all the experiments. The amount of Catalase also determines the amount of O2 given off, so I will use 5ml of Yeast solution throughout the investigation. These figures were arrived at after preliminary experiments. They were chosen because they tended to produce amounts of oxygen in the range of the equipment I am to use, improving the accuracy of my experiment.

Method

Equipment Required:

Large test tube & rubber bung with tube through centre.

Rubber hose

Large water container

Honeycomb ceramic 'spacer'

00ml measuring cylinder

Beakers for yeast and H2O2

Measuring pipettes for yeast and H2O2

Clamps

Test tube rack

Thermometer

Stopwatch

. I will set up the equipment as above, then take 200cm3 of yeast solution and add 5cm3 of pH7 buffer and stir. I will then take a reading of the ambient temperature.

2. I will fill the measuring cylinder with water and invert it. I will take a measure of the amount of gas remaining in the cylinder.

3. I will stir the yeast solution and put 5ml of it into the test tube. Ready with the bung, I will start the stopwatch. After 5s have elapsed I will add 5ml of Hydrogen Peroxide to the test tube using a pipette, and 2s later seal the test tube with the bung.

4. After 20s, I will remove the bung from the test tube, and record the amount of gas in the cylinder. My measurement of the amount of oxygen released over the incubation time will come from the subtraction of the initial amount of gas from the final amount.

5. I will repeat steps 2 through 4, three times for each of the concentrations of H2O2.

6. To achieve concentrations of 16, 8, 4, and 2moldm-3 is simply a matter of adding water to the previous higher concentration in a proportion of 1:1 therefore halving it's concentration. To get 28moldm-3 I will add 14ml of water to 100ml of 32moldm-3, because...

H2O2 is toxic and caustic, and therefore must be handled with care - it is also colourless so I will label the beaker to ensure it's not confused with water.

Sources of error

I decided to use the inverted cylinder method for measuring the oxygen produced because I think it would provide more reliable and accurate results than the gas syringe method. This is because the gas syringes require a certain amount of pressure to build up before the plunger will move which may effect the accuracy of readings at lower concentrations which I expect will produce little oxygen.

Oxygen produced by the enzyme/Hydrogen Peroxide mixture will likely escape before I have been able to fix the bung in the test tube. In step 3 I have tried to eliminate errors arising from sometimes being quicker or slower in affixing the bung by ensuring the delay is always 2s.

The concentrations I will use (in moldm-3) are: 32, 28, 16, 8, 4 and 2. These I believe cover a decent range, yet it is practical to perform three repetitions of each experiment in the time available.

Results

Ambient temperature: 24C

Concentration (mol)

Repetition Number

Initial Cylinder Reading (ml)

Final Cylinder Reading (ml)

Difference

(ml)

Average

(ml)

Initial rate of reaction

(ml of O2 min-1)

2

9.0

21.0

2.0

2.5

7.5

2

2

21.0

24.0

3.0

2

3

24.0

26.0

2.0

4

0.0

7.0

7.0

6.5

9.5

4

2

7.0

3.0

6.0

4

3

3.0

9.0

6.0

8

0.0

3.5

3.5

3.5

40.5

8

2

3.5

27.5

4.0

8

3

27.5

31.0

3.5

6

0.0

22.0

22.0

23.5

70.5

6

2

22.0

48.0

26.0

6

3

0.0

22.5

22.5

28

0.0

26.0

26.0

24.5

73.5

28

2

0.0

24.5

24.5

28

3

24.5

47.5

23.0

32

0.0

45.0

45.0

41

23

32

2

0.0

39.0

39.0

32

3

0.0

39.0

39.0

All measurements of amount are to the nearest half ml.

Analysis and Interpretation

The graph of my results does not look how I expected it to. While the first four points follow quite a smooth upward gradient, as one would expect with a steadily increasing concentration of HP, the last two points throw off this smooth curve completely. I am inclined to suspect that one of these points is therefore anomalous.

Disregarding the result for the Hydrogen Peroxide concentration at 28moldm-3 (shown by the red line on the graph), would give a pretty straight line, showing that the initial rate of reaction increases constantly with H2O2 concentration up to 32moldm-3. This would indicate that concentrations of substrate didn't get high enough for the enzyme molecules to reach their maximum rate of working, as there is no flattening of the graph at the higher concentrations; a proportionally higher amount of oxygen is produced by the enzyme at 32moldm-3 compared to 16moldm-3. Vmax isn't reached because up to 32moldm-3 the enzyme can work faster and faster (and perhaps beyond 32moldm-3).

Disregarding the result at 32moldm-3, would give a smooth curve which flattens out at around 16moldm-3 (shown in blue on the graph). This flattening would suggest that the Catalase has reached Vmax, it's maximum rate of decomposition of Hydrogen Peroxide. Increasing the concentration any further shouldn't increase the initial rate of reaction noticeably because all the active sites of the Catalase available in the 5ml of Yeast solution, are all occupied pretty much constantly, hence the flattening of the curve.

Evaluation

The experimental procedure that was used was generally suitable for the investigation, because the range of measurements I took suited the range of the equipment, that is the whole range of the 50ml measuring cylinder was used. This increased the reliability and accuracy of the data because using a larger cylinder (or less yeast and Catalase) may have meant that smaller values (e.g. a change of 2ml) wouldn't have registered.

Sources of error could've arisen from the equipment in that some of the oxygen produced remained in the rubber hose after each repetition, the amount being dependant on the position of the hose which it was necessary to alter when refilling the cylinder with water.

The fact that the 3 repetitions for all the experiments at each concentration were within a maximum of 6ml (the difference between repetition 1 and 2 at 32moldm-3) of each other and usually less, suggests that the procedure I used was fairly good. It also suggests that the anomalous result I got wasn't due to random errors. To increase the reliability of my experiment I would perform 5 trials at each concentration, reducing the risk of human error significantly distorting the results.

I believe that perhaps the order in which I performed the actual experiments may have produced the anomalous result at 28moldm-3. Due to the practicalities of diluting the H2O2 solution, I performed the 32moldm-3 experiments first, then the 16, 8, 4, and 2, then the 28moldm-3. Over the period of time that it took to do 3 repetitions of each, the yeast solution could have weakened in some way- perhaps through cells dying and the Catalase denaturing, or a Catalase inhibitor being produced in the cells themselves. This would mean more oxygen would've been produced for the first experiments (i.e. at 32 and 16 moldm-3) than the last (i.e. 28moldm-3). Lowering the oxygen measurement for the 32 and 16 moldm-3 experiments and raising the measurement for the 28moldm-3 would flatten the large dip in the graph at 28moldm-3. To account for this possible weakening of the yeast solution, one would have to repeat the whole investigation, but instead of using one batch of premixed yeast solution for all the experiments, an amount of yeast solution would have to be mixed (from inert dry yeast) just before each trial. This would mean each batch of yeast used would have been in solution for the same amount of time, reducing the suspected error arising from the effective yeast Catalase concentration diminishing over time.

Another possible reason for the error could be a mistake in mixing the H2O2 solution. The difference in average measurements between those for 16moldm-3 and 28moldm-3 is only 1 ml, with the measurement of 26 ml of O2 being seen at both concentrations. To determine if this is in fact the reason I would merely have to repeat the experiment for the 28moldm-3 concentration.

The anomalous result is very significant in terms of the conclusions that can be drawn. Without further experimentation it is impossible to say for definite whether the result for 32moldm-3 is wrong and conclude Vmax was in fact reached at lower concentrations, (around 28moldm-3), or whether the result at 28moldm-3 is wrong and concentrations up to 32moldm-3 of H2O2 can be decomposed per second by Catalase.

Bibliography:

Catalase Kinetics: http://www.science-projects.com/catalasekinetics.htm

Yeast Treatise: http://home.earthlink.net/~ggda/biology_of_yeast_cells_simplified.htm

Enzymes: http://www.ultranet.com/~jkimball/BiologyPages/E/Enzymes.html

Cambridge Advanced Sciences Biology 1

Cambridge Advanced Sciences Chemistry 1