Results
Any points below 1ml could not be measured on the measuring cylinder therefore I have included them as a /. The average value has been calculated using the equation for the mean.
Mean = Sum of all the values added together
Total number of items
(E.g. 1.0 + 1.1 / 2 = 1.05)
All / values should be taken into account as 0ml. For the plotting of my graphs I will use the average values only. These averages are more reliable as they are taken from two experiments rather than just one.
All my controls released 0ml of oxygen. The control was boiled potato when the enzyme is boiled it becomes denatured, the active site is lost and the enzyme can no longer work. This proves that it was the enzyme catalase, which was having an effect on the experiment.
Water temperature =10°C
All the average results are highlighted. These are the ones I will use for my graphs.
A Table to show the Effect different Amounts of Hydrogen Peroxide can have on its breakdown by the enzyme Catalase.
Analysis
Catalase speeds up the decomposition of hydrogen peroxide because the shape of its active site is specified to the shape of the hydrogen peroxide substrate. This type of reaction where a molecule is broken down into smaller components is called an anabolic reaction.
On the first graph the reaction steadily increases until it reaches 6ml concentration and 6.1 of oxygen gas has been given off. Therefore as the concentration of hydrogen peroxide increases so does the rate of reaction. As more enzymes active sites are being occupied more oxygen gas is being produced. Until another factor such as enzyme concentration restricts the rate of reaction. This is because once all the enzymes active sites are occupied (said to be saturated) the enzymes cannot work any faster. The maximum numbers of active sites are being used. Therefore the time to enter and then leave the active site becomes a limiting factor. Any extra substrate molecules have to wait for an active site to be empty.
Thus at the beginning of the graph we can see that the enzyme is in excess and therefore as more substrate is added the faster the reaction can go. When the graph levels off the substrate is in excess and therefore the enzyme is limiting. Therefore this experiment reflects the concentration theory. According to my results the fastest reaction was at 6 ml of hydrogen peroxide however the result might have been different had I used more concentrations for example every 0.5 ml intervals instead of 2ml intervals. The graph would have been more accurate had I done this.
Conclusion
The oxygen production increases as the concentration does until another factor limits its production.
Evaluation
The concentration at 8ml is an anomaly therefore I have ignored it in my curve of best fit. And have highlighted this on my graph. This anomaly could be due to several factors (see sources of error). 10ml also has a slight anomaly, as the graph should have levelled off straight when it rose by 0.1 ml this could have perhaps be due to temperature fluctuations. However as this anomaly is only small I have included it in the curve of best fit anyway. Graph two demonstrates (for 2, 4 & 6) that the higher the concentration of hydrogen peroxide the faster the reaction will work and the more gas which is given off. After 6 ml however the concentration does not seem to make much significance to the rate of reaction.
The limitations of this experiment are all the possible variables. For example:
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Temperature, ideally the temperature should be at its optimum. This can be achieved using a water bath. However even then temperature can fluctuate causing problems and inaccurate results.
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pH the pH should also be at its optimum for the enzyme to work at it’s best. As with all variables (except the one being used) this needs to be kept constant.
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Enzyme concentration how much enzyme (potato) is used should be the same for each experiment. The ruler used to cut the potato into 5mm slices is only accurate to the nearest mm. The experiment would continue to rise if more potato was added. Therefore it is important that the amount of potato remains the same.
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Potato Tissue I should use the same potato for every experiment. However for biological reasons this is not always possible. E.g. there might not be enough tissue on one potato for all the experiments.
To improve this experiment I could add 0.5ml intervals of hydrogen peroxide concentrations instead of 2ml intervals. I would also try to use the optimum pH and temperature and keep these constant.
Sources of Error
