Trial run
I conducted a trial run to see if the method I will use has any problems that will prevent me from answering my aim. For this trial run, I increased the surface area in increments of 5cm², and I recorded the volume of oxygen produced every 10 seconds, for a total of 90 seconds (one and a half minutes). For the trial run I used 40cm³ of H2O2; this amount is constant throughout the experiment. I will use this volume of H2O2 in the final experiment.
The results were not too bad, but the line for each surface area is clearly still increasing. This means that I did not conduct the experiment for long enough. I will make some changes for the final experiment. These are:
- Increase the range of surface areas that I will use, from 6cm² to 100cm²
- Increase the time in between results, from 10 to 30 seconds. This will enable me to take more accurate results as 10 seconds was too hurried, and with the smallest surface areas, often there was no change, as the oxygen production was quite slow – due to the small surface area, and because of the inaccurate method, a smaller volume than the larger surface areas.
- I will take results every 30 seconds for a total of 3 minutes. This is a total of 6 results
- I will conduct the experiment at each surface area twice, to eliminate anomalies. This will also enable me to average the results. This will mean that the line on the graphs I draw will represent both results, using a mean average.
- By taking results for longer, I will be able to see if the graph begins to plateau; by this I mean, I will be able to see if the line on the graph representing oxygen produced over time begins to flatten out. If this does happen, I expect all the lines to flatten out eventually at the same volume of oxygen – explained in my prediction. If this does not happen, it means that I did not take results for long enough. If the graph represents that produced by the results from the trial run, it will show that oxygen is still being produced after 3 minutes. Due to limitations I will not be able to conduct the experiment for any longer, as I have to take 18 sets of results, two for each surface area. If I had more time, I would be able to take readings until the oxygen production stopped. By doing this I could prove my prediction – that at each surface area, the same amount of oxygen will be produced. In my evaluation I will provide some suggestions that would enable me to overcome this.
- I could take results using a syringe; this would dramatically improve the accuracy of the results, as the measuring cylinders that I will use, have a volume of 100ml, and increase in 1ml increments. Because of this, when conducting the experiment for smaller surface areas, I can only measure, accurately the volume of oxygen in 0.5cm³ increments. A syringe would increase this accuracy, but due to limitations this will not be possible as I do not have the resources necessary.
Variables
In this investigation, the variables that affect the activity of the enzyme, Catalase, were considered and controlled so that they would not disrupt the success of the experiment.
- Temperature – As temperature increases, molecules move faster (kinetic theory). In an enzyme catalysed reaction, such as the decomposition of hydrogen peroxide, the temperature increases the rate at which the enzyme and substrate molecules meet and therefore the rate at which the products are formed. As the temperature continues to rise, however, the hydrogen and ionic bonds, which hold the enzyme molecules in shape, are broken. If the molecular structure is disrupted, the enzyme ceases to function as the active site no longer accommodates the substrate. The enzyme is denatured.
- To control this variable, the temperature was maintained at a fairly constant level that allowed the enzyme to work effectively (room temperature, approximately 21°C). This is slightly lower than average, as I conducted the experiments in October, where the temperature outside was around 18°C. One problem I encountered was that the heat from my hands, holding the test tube would have affected the results. I could have overcome this by using a test tube rack and tongs to handle the apparatus so that the heat from my hands did not affect the Catalase.
- pH – Any change in pH affects the ionic and hydrogen bonding in an enzyme and so alters its shape. Each enzyme has an optimum pH at which its active site best fits the substrate. Variation either side of pH results in a denaturising of the enzyme and a slower rate of reaction.
- In this experiment, I could have used a pH buffer to keep the pH constant; by using a buffer of pH 7, I would be able to maintain a pH level suited to the enzyme by being equal to the natural environment of the enzyme (potato tissue).
- Substrate Concentration – When there is an excess of enzyme molecules, an increase in the substrate concentration, produces a corresponding increase in the rate of reaction. If there are sufficient substrate molecules to occupy all of the enzymes’ active sites, the rate of reaction is unaffected by further increases in substrate concentration as the enzymes are unable to break down the greater quantity of substrate.
- To control the substrate concentration, I will use 40cm³ of substrate in each experiment. To ensure that this was measured precisely, I used a measuring cylinder to gauge precisely 40cm³ of substrate.
- Inhibition – Inhibitors compete with the substrate for the active sites of the enzyme (competitive inhibitors) or attach themselves to the enzyme, altering the shape of the active site so that the substrate is unable to occupy it and the enzyme cannot function (non-competitive inhibitors). Inhibitors therefore slow the rate of reaction. They should not have affected this investigation, however, as none were added.
- Enzyme Concentration – Provided there is an excess substrate, an increase in enzyme concentration will lead to a corresponding increase in rate of reaction. Where the substrate is in short supply (i.e. it is limiting) an increase in enzyme concentration has no effect. This last point will not be a problem, as 40cm³ of substrate is sufficient to occupy all the catalase molecules from the potato. This factor – enzyme concentration – will help me to answer my aim, as I am trying to find how the surface area and therefore the concentration of the catalase in the substrate affect the rate of reaction.
I varied the enzyme concentration by altering the surface area of the potato cuboids that contain the Catalase, in the reaction, the greater the surface area of the potato, the greater the enzyme concentration. The concentration of the catalase molecules in the hydrogen peroxide solution is determined by the amount of molecules per a certain measure. So as the hydrogen peroxide remains constant at a volume of 40cm³, the concentration of the catalase increases as more catalase is present in the H2O2. This would not be the case if the amount of substrate was altered, as concentration is dependent upon the amount of molecules in a certain volume.
Method
I set up the equipment as shown in the diagram below:
Procedure
- I cut the potato into strips of increasing surface area. This method is problematic and leads me to record inaccurate results. As I increased the surface area of the potato, I also increased the volume (the problem with the method), increasing the amount of catalase available, and making the results unfair. This problem prevents me achieving my aim completely, as my aim is to see how the action of catalase is affected by the surface area. In my evaluation, I will suggest how to overcome this problem.
- I measured the surface area of each strip using a ruler, and then calculated the surface area using the formula.
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I put 40cm³ of H2O2 into a conical flask. In the evaluation I have suggested how to improve the method, in such a way that might have enabled me to achieve my aim more effectively and in a more conclusively.
- I then added the potato strips to the hydrogen peroxide solution and placed a bung onto the conical flask, with a delivery tube that directed the gas into a water bath, which I collected using the method of displacement – as the gas was produced, it filled up the measuring cylinder, displacing the water which was in the measuring cylinder. In my evaluation, I have suggested a problem with this method of gas collection.
- I recorded the volume of oxygen produced every 30 seconds for 180 seconds. This is a total of 6 readings. I conducted each experiment twice at each surface area, and then averaged these results. Finally I graphed the results.
Safety
To prevent accidents, I tucked in any loose clothing, and wore goggles, to prevent me getting any peroxide in my eyes, as it is a corrosive substance. Care was taken when handling the peroxide.
Results
Gradient – Rate of reaction
To calculate the rate of reaction, I need to calculate the gradient of the point on the graph of 180 seconds. This will give me the overall oxygen production over the 180 seconds. To do this, I will take the amount of oxygen produced at 180 seconds, and divide it by 180. This will give me the rate of reaction in the form of cm³ O2/sec.
This table of results shows the rate of reaction at 180 seconds for each surface area. The graph on the previous page shows these results in the form of a histogram. There is one anomaly. At a surface area of 40cm² both the results 1 and 2 therefore the mean average produced less oxygen than at 36cm², and therefore had a lower rate of reaction. This could have been due to a problem with the method. Or in this instance, an inaccurate amount of substrate could have been used, the temperature could have been too low, or the bung may not have been totally tight, and therefore some of the oxygen could have escaped.
Graphs
I have drawn three graphs to show the overall results. Graph 1 shows the average volume of oxygen produced at each time reading for each surface area. Graph 2 shows the average volume of oxygen produced for each surface area and at each time reading. Graphs 3-11 show the individual results and the average volume of oxygen produced at each time reading and for each surface area. Graph 12 shows the average volume of oxygen produced at each time reading for each surface area. The results are plotted on a shorter x axis, and a longer y axis; this helps to show the fact that the lines are still rising. In my evaluation, I will look at this and sketch a graph to show the results I would have expected if I had taken results for a much longer time period and explain why I expected this. Also on the sketch I will indicate the section of the line that represents that of the results that I collected.
Analysis
The graphs show the results that I expected. The only major anomaly, in respect to all the other results that I collected (due to my method, all my results were anomalous to a certain extent) was at 40cm² the overall amount of oxygen produced was less than that at 36cm². This goes against my prediction, but can be reasoned due to a problem in my method that only affected the results recorded at 40cm². It is obvious that at 100cm² the amount of oxygen produced was a lot more than at 64cm², in relation to the gap of 36cm², the increase was not proportional to the increase in surface area. This is explained below.
Because I did not cut up strips of potato of equal volume, my results were not accurate. To overcome this I should have cut up strips of equal volume into different surface areas. Because I did not, as the surface area of the potato increased so did the amount of catalase available, as the volume was greater. This provided inaccurate results.
This table and graph show the relationship between the surface area of a cuboid of potato, like I used in the experiment, and its volume. To calculate the surface area of a cuboid, given three lengths, A, B and C I used a formula, which calculates the surface area of the two sides, top and bottom, and the two ends, shown in the diagram below.
The formula is: ((AxB) x2) + ((BxC) x2) + ((AxC) x2)
Conclusion
The reaction was fastest when the surface area was 100cm². This would have meant the highest enzyme concentration and therefore the fastest rate of reaction. At this enzyme concentration there was the greatest number of free active sites available to the substrate molecules so that they could be broken down. The rate increased steadily from 6cm² to 64cm², and then a larger jump in rate from 64cm² to 100cm² due to the increase in volume as well as surface area, and one anomaly at 40cm².
Evaluation
Because enzymes are not used up during a reaction, only a certain concentration is required to decompose a substrate. At the point where there is enough catalase, and therefore active sites to accommodate all the substrate or hydrogen peroxide, then any more catalase or an increase in the concentration, or surface area will be superfluous and will have no affect upon the rate of reaction. At a certain surface area, with 40cm³ of substrate, any increase in surface area will react at the same rate.
The shape of all the graphs, as I expected show that as the surface area increases so does the oxygen production. This occurs at an exponential rate. Because my results were so limited, they only show the very first part of the results that I would have received if I had carried out the experiment over a much larger time scale. It is not possible to take accurate readings from the graph in between the readings that I recorded, so the line is only a very rough ‘line of best fit’. To increase its accuracy, I needed to increase the number of surface areas that I took readings from. Also, by keeping the volume constant would have reduced the number of anomalies I recorded, and given more accurate results, resulting in a more accurate, and ‘curved’ graph. This sketch below shows the results that I would expect to record if I had taken results for a longer time period.
The blue line represents a high concentration of catalase, the red line represents a lower concentration, and the black line represents a low concentration of catalase. The blue line shows a very fast initial rate of reaction, which slowly reaches a plateau. All the substrate is reacted quite quickly, so the line increases exponentially, the red and black lines show slower reactions, the black slower than the red, due to the gentler gradient. The small box represents the results that I collected. The black line, can be clearly compared to the results I collected where the surface area was lower, as, like on this representation, the line is relatively straight, showing a slower rate of reaction. This was observed in the results that I collected. The blue line, representing the result where the surface area was 100cm², shows a steep initial incline; this is very similar to all the graphs showing the result for a surface area of 100cm². If I had taken readings for long enough my graph would have represented this graph very accurately, as it is very clear the graphs I have drawn represent the first section of this graph. This helps me to prove my prediction, as this graph shows that if the surface area of the potato strip is larger, the catalase concentration is greater therefore the greater the initial reaction rate, but overall the graphs will all reach the same point, where the reaction has ended and oxygen production has ceased. This small sketch helps me to explain why on the hand drawn graph (12) and graph (1) the gap between the line for 64 and 100cm² is disproportionately large in comparison to the difference, as the increase in reaction rate in relation to surface area is one of exponential increase. This also shows my prediction is correct.
A problem I encountered was that I used too much substrate, as the reaction did not finish within an acceptable time limit, or in the time that I had to carry out all the experiments. As there was so much substrate (hydrogen peroxide) it did not all get reacted with the catalase; by using a lesser amount of substrate the catalase would have catalysed the substrate faster. The reaction rate would have remained the same, but oxygen production would finish a lot sooner, and would not reach the same level as when I used 40cm³. I think an acceptable volume of substrate to use would be 5cm³.
My results would look like this (5cm³ H2O2): Not like this (using 40cm³ H2O2):
Another factor that may have affected the results was where the potato came from, because if I used a strip of potato from the centre, rather than the edge, there may have been more catalase present. This could account for the anomaly in results at 40cm². For all the other results, I may have used potato from the edge, then for this reading I may have used potato from the centre. I do no know if this is true, but it is a possibility.
One problem with the method of gas collection was that I had to fill the cylinder with water, and then put it into the trough. Sometimes a small amount of air was left in the tube; this would have created a bias result. To over come this I could have used a syringe to collect the gas, but I did not have the resources available.
Overall I feel that I answered my aims conclusively and I feel that my prediction was correct to a large extent, although due to inaccuracies in my results I cannot draw totally complete conclusions.
