Variety of potato used
The variety of the potato used can affect the rate of activity because different varieties of potatoes may have different concentrations of catalase. I also need to use the same potato for all my samples because there maybe variations in the amount of catalase present in different potatoes of the same variety.
Mass of potato cylinder
The mass of the potato cylinder can affect the rate of activity because the larger the mass the more catalase there may be present which will mean the rate of activity will increase.
How the potato is handled
The way the potato is handled could affect the rate of activity because if the potato cylinder is touched a lot it could remove catalase from the exposed broken cells at the surface, which could affect the investigation.
Amount of time potato cylinders left in solution
The amount of time the potato cylinders are left the hydrogen peroxide solution can affect the rate of activity, because more oxygen could be released.
For this experiment to be a fair I need to keep all but one of the variables the same, I am going to vary the concentration of hydrogen peroxide, because this is what my investigation is about, ‘to see the effect of changing the concentration of the substrate hydrogen peroxide on the rate of activity of the enzyme Catalase’. I am going to use six different concentrations, from 0% of original solution of hydrogen peroxide to 100% of original solution of hydrogen peroxide, going up in 20% each time. On the following page is a table of the dilution series I am going to use.
To keep all the other variables the same I am going to:
Size, shape and surface area of the potato cylinder: use the same cork borer and the same ruler to measure the length.
Volume of hydrogen peroxide and distilled water: use 10cm3 of solution each time.
Temperature of hydrogen peroxide: do the experiment at room temperature on the same day, so the temperature doesn’t change.
pH of hydrogen peroxide: the pH should not need to be controlled in this experiment.
Variety of potato used: use one potato so the concentration of catalase is the same.
Mass of potato cylinder: keep all the masses as similar as possible.
How the potato is handled: use metal forceps, which have been washed in distilled water.
Amount of time potato cylinders left in solution: leave all cylinders in solution for five minutes and record the amount of oxygen produced.
Also to make my investigation more reliable I am going to repeat the experiments twice, I am also going to compare the results and to take the average from both of them.
For the experiment the equipment I will need is:
- Ten test tubes - experiment will be carried out in these.
- Distilled water – to be used to create a dilution series with hydrogen peroxide.
- Tile - to cut the potato on.
- Scalpel - to cut the ends of the potato cylinders.
- Knife - to cut one end of the potato off.
- Balance – to weigh the mass of the potato cylinders.
- Potato – source of catalase.
- Cork borer – to cut the potato cylinders.
- Ruler – to measure the length of potato cylinders.
- 20 volume (1.7M)-hydrogen peroxide – to react with catalase.
- Gas syringe – to measure the amount of oxygen produced.
- Graduated pipette – to measure out the volume of solutions.
- Rubber bung with delivery tube – to deliver the oxygen to the gas syringe.
- Clamp and stand – to hold the gas syringe.
- Test tube holder – to hold the test tubes.
- Stop watch – to time the experiment.
- Forceps – to handle the potato cylinders.
- Vaseline – to help reduce the loss of oxygen.
Before I carry out my investigation I am going to do some preliminary work to find out the best range of concentrations to use. In this research I am going to experiment with the 20% of original solution of hydrogen peroxide and the 100% of original solution of hydrogen peroxide. But before I carry out this preliminary work I am going to predict that in the experiment, I would expect that the higher the percentage of original solution of hydrogen peroxide, the more oxygen produced in a certain time. This is because as you increase the concentration of the substrate hydrogen peroxide the rate of activity should increase too because there will be more substrate molecules which can react with the enzyme catalase. This is shown in the two diagrams below, the one to the left shows a solution with a low concentration of substrate molecules, and the one to the right shows a solution with a higher concentration of substrate molecules. You can see that in the diagram to the left there are plenty of enzyme molecules that could react with a substrate molecule, but because the amount of substrate is low, there are few enzyme-substrate complexes, which means little product will be produced, and the rate of reaction is low. You can see that in the diagram to the right there are more substrate molecules, and more enzyme-substrate complexes, which means more product will be produced, and the rate of reaction increases because there are more chances of a successful collision, increasing the chances of successful collisions increases the chances of an enzyme-substrate complex being formed. I also predict that if you continue to increase the concentration of the substrate the rate of reaction will not continue to increase because all the active sites on the enzyme will be used up.
Also I predict that a lot of oxygen will be produced in the first minute, because when the substrate is first added none of the enzymes will be in any enzyme-substrate complexes, so therefore all of them can start to break down the substrate. In addition I predict that if the solution has 0% of the original solution of hydrogen peroxide I predict that no oxygen will be produced because there is no hydrogen peroxide to be broken down.
I first tried the 100% of original solution of hydrogen peroxide. The potato cylinder I used weighed 1.07g and measured 2.0cm in length. I checked the amount of oxygen produced after every minute for ten minutes. Below is a table with my results.
I then tried the 20% of original solution of hydrogen peroxide. The potato cylinder I used weighed 1.06g and measured 2.0cm in length. I checked the amount of oxygen produced after every minute for eight minutes. Below is a table with my results.
Below is a diagram of how I am going to set up my equipment:
I have set out the equipment like this so there are no loops in the delivery tube and also I can see both the gas syringe and the test tube. On the diagram above I have not drawn several parts, I have not drawn the clamp that will attach the syringe to the stand, nor I have I drawn the test tube holder that will hold the test tube in place. I did not draw these because in this drawing they will obstruct the view of the syringe and test tube.
To carry out the investigation I will first wash my equipment with distilled water, to make sure they are clean. Then I will cut the bottom of the potato off, with a knife so I have a flat surface to work with, this way the potato can’t slip and I can’t hurt anyone. I will then push the cork borer down and get my potato cylinders, I will also stand upright to do this so it is easier to cut and I am less likely to hurt anyone else because I will have more control over the equipment. I will then use a scalpel to cut the potato cylinders to 2cm in length with the aid of a ruler, I will use forceps to handle the potato cylinders at all times. Next I will make up a dilution of hydrogen peroxide and distilled water in a test tube. I will carry out this by using two graduate pipettes, one for the hydrogen peroxide and one for the distilled water, this way I will reduce the risk of contaminating any equipment. I am going to use graduated pipettes instead of measuring cylinders because they are more accurate. After I have made up the solution in the test tube I will place it in a test tube holder. I will then set up the gas syringe and delivery tube with bung, with the help of a clamp and stand. After that is complete I will use the forceps to pick up the potato cylinder, and I drop it into the test tube and replacing the bung with delivery tube attached very quickly. When I drop the potato cylinder in, I will ask someone else to start the stopwatch, as I don’t have enough hands. I will time the experiment for five minutes, and then I will record the reading on the gas syringe in a table. I will then repeat the experiment.
To make sure I do not harm anyone or myself else I must be very careful when using the knife and scalpel and I must use them away from everyone else so accidents do not happen. I will wear goggles to prevent getting hydrogen peroxide in my eyes. Also I should be careful that the solutions are not consumed as hydrogen peroxide is poisonous, but apart from these there are not any bigger matters to be cautious of.
There were no problems when carrying out the experiment and I tried to make sure that all my tests were as fair as possible by keeping all the variables except the concentration of the hydrogen peroxide the same, below is a table of my results.
Now that I have got this table showing all of my results I am going to work out how much oxygen is produced per gram for each of the tests, to work this out I am going to use the equation below.
Amount of oxygen produced per gram = Amount of oxygen produced (cm3)
(cm3/g) Mass (g)
Test 1 Results
0%
Amount of 02 per gram = 0 = 0 cm3/g
1.60
20%
Amount of 02 per gram = 0.7 = 0.432 cm3/g
1.62
40%
Amount of 02 per gram = 1.4 = 0.875 cm3/g
1.60
60%
Amount of 02 per gram = 2.6 = 1.615 cm3/g
1.61
80%
Amount of 02 per gram = 3.2 = 1.975 cm3/g
1.62
100%
Amount of 02 per gram = 4.1 = 2.563 cm3/g
1.60
Test 2 Results
0%
Amount of 02 per gram = 0 = 0 cm3/g
1.64
20%
Amount of 02 per gram = 0.6 = 0.370 cm3/g
1.62
40%
Amount of 02 per gram = 1.3 = 0.802 cm3/g
1.62
60%
Amount of 02 per gram = 2.4 = 1.463 cm3/g
1.64
80%
Amount of 02 per gram = 3.3 = 2.025 cm3/g
1.63
100%
Amount of 02 per gram = 3.9 = 2.438 cm3/g
1.60
Average Results
0%
Amount of 02 per gram = 0 = 0 cm3/g
1.62
20%
Amount of 02 per gram = 0.7 = 0.432 cm3/g
1.62
40%
Amount of 02 per gram = 1.4 = 0.870 cm3/g
1.61
60%
Amount of 02 per gram = 2.5 = 1.534 cm3/g
1.63
80%
Amount of 02 per gram = 3.3 = 2.025 cm3/g
1.63
100%
Amount of 02 per gram = 4.0 = 2.500 cm3/g
1.60
After working out how much oxygen is produced per gram for each of the tests I am going to put all of this information into a table so it is easier to read and understand.
Now using both sets of results in the tables I am going to draw six graphs to represent this information (the graphs are at the end). The first three graphs show the percentage of original solution of hydrogen peroxide against amount of oxygen produced. All three graphs show the same shaped line, and there are no anomalous results. The first graph shows the results of test one, the second graph shows the results of test two and the third graph shows the average results.
The second set of three graphs shows the amount of oxygen produced per gram. These results are more accurate. The first graph shows the results from test one of oxygen produced per gram, the second graph shows the results of test two of oxygen produced per gram and the third graph shows the average results of oxygen produced per gram. All three graphs show the same shaped line.
In conclusion as you increase the concentration of hydrogen peroxide, the rate of reaction increases, because as you increase the concentration of the substrate hydrogen peroxide there will be more substrate molecules in a given volume, which can react with the enzyme catalase to produce oxygen. The rate of reaction increases because there are more chances of a successful collision, increasing the chances of successful collisions increases the chances of an enzyme-substrate complex being formed. This only happens up to a certain point, when all the active sites on an enzyme are being used, after this the rate of reaction cannot increase anymore because the substrate has nothing to react with, and therefore stays at a constant rate. On the next page are two diagrams, the one to the left shows a solution with a low concentration of substrate molecules, and the one to the right shows a solution with a higher concentration of substrate molecules. You can see that in the diagram to the left there are plenty of enzyme molecules that could react with a substrate molecule, but because the amount of substrate is low, there are few enzyme-substrate complexes, which means little product will be produced, and the rate of reaction is low. You can see that in the diagram to the right there are more substrate molecules, and more enzyme-substrate complexes, which means more product will be produced, and the rate of reaction increases because there are more chances of a successful collision, increasing the chances of successful collisions increases the chances of an enzyme-substrate complex being formed. Also if you continue to increase the concentration of the substrate the rate of reaction will not continue to increase because all the active sites on the enzyme will be used up. This diagram is just showing my conclusion.
Below is a graph that summarises this, my graphs do not look exactly like this but they look like the first part of the graph (the slope).
My results do not show that all the active sites are occupied, my results may not have shown this because the hydrogen peroxide was not concentrated enough. My graphs do begin to start to show the curve, where the rate of reaction starts to slow down, which indicates that the active sites are beginning to get used up.
Although the results have validated satisfactory graphs of amount of oxygen produced against concentration of hydrogen peroxide, and established a conclusion that is acceptable, the results were not perfect. One error occurred when the potato was put into the hydrogen peroxide because the bung needed to be replaced at the same time to prevent the loss of oxygen, and the stopwatch to be started. This transaction was done as concurrently as rationally possible, but there was certainly some delay, so a little gas would have been lost, also gas could have been lost at other connections. This would have slightly affected the results, but as the procedure was repeated on each occasion, the error will be the same throughout the experiment. Another error occurred when readings were taken, these were taken by eye, and although they were taken by looking at the scale as near to right angles as possible, there may well have been possible errors. Errors could also have occurred when trying to get all the tissue the same size and the same mass, this would have come through problems with cutting tissues to the same length and also cutting the tissue at right angles each time. Some cylinders may have been cut at a slope and therefore they have increased in surface area. I did not monitor any changes in the temperature of the hydrogen peroxide and this may have altered my results slightly, this was very had to control because we had a rubber bung in the end of the test tube, so there was no where for a thermometer to go. I could have controlled the temperature of the hydrogen peroxide better by putting it in a water bath, to try and keep the temperature more constant.
The procedure of this experiment could be improved, although the results do give adequate conclusions. Improvements would simply remove certain errors, and improve the accuracy of results. Using slightly different procedures could reduce the delay between putting in the hydrogen peroxide and replacing the bung and starting the stopwatch. The best solution would be to find a method that doesn’t require the bung to be removed and replaced each time.
The study of the enzyme catalase could be further investigated by analysing the effects of temperature upon the rate of the reaction between the enzyme and hydrogen peroxide. Although catalase can withstand reasonably high temperatures, it would probably denature at extreme temperatures. It would be interesting to investigate at what temperature catalase stops working.
