However, there will be a delay in heating up the hydrogen peroxide solution (in the boiling tube) because the heat transfer takes a while. The boiling tube must stay in the water bath for while before the experiment can begin.
A number of other factors (besides temperature) can have an effect enzyme activity. In order to make the investigation a fair test all these factors have to be constant.
Firstly, the cores should all be from the same potato, because in theory the amount of catalase should be evenly distributed throughout the potato (i.e. the concentration of catalase is the same throughout the potato). The potato cores should all be the same shape and size so that the same amount of catalase is used in each of the experiment. Having the same shape and size means that the cores will all have the same surface area. The surface area of the potato cores affects the rate of oxygen production as the larger the surface area, the more catalase molecules make contact with the hydrogen peroxide molecules and thus the more oxygen produced. Keeping the same surface area constant eliminates this effect and makes the experiment a fairer test. All of the above is to keep the enzyme concentration and amount the same.
Another factor that affects the rate of oxygen production is the amount and the concentration of the substrate, hydrogen peroxide. In the experiments, both the concentration and amount of hydrogen peroxide used should be kept constant. This again makes the experiment more of a fair test as it ensures that only one variable, the temperature, is changed while the others, one being the concentration and volume of hydrogen peroxide, are kept constant.
NOTE: The hydrogen peroxide used in the experiments will be diluted since pure hydrogen peroxide is very corrosive. The concentration of hydrogen peroxide will be 20 volume for all of the experiments. (Whatever that means!)
All enzymes, catalase included, are affected by the pH in which the enzyme works in (i.e. the pH of the enzymes environment). As this could consequently affect the rate of oxygen production, it must also be kept constant. In an earlier investigation (earlier in the year), it was shown that potato catalase operates best at pH 7. Hydrogen peroxide is neutral (pH 7) therefore the pH is not a major concern in this experiment, however, it is still advisable to check that the pH of the hydrogen peroxide just in case.
Inhibitors can also affect the rate of enzyme activity, but I do not know of any specific potato catalase inhibitors. Hydrogen peroxide cannot be an inhibitor for the obvious reason that catalase is designed to deal with hydrogen peroxide. Water cannot be an inhibitor (of potato catalase) because water accounts for a substantial amount of the potato’s mass. Oxygen cannot be an inhibitor either because cells are constantly handling it since it plays a vital role in aerobic respiration.
The Method
- Gather the equipment shown in Fig. 1 and the following additional equipment:
●Stands and Clamps ●Measuring Cylinder/Syringe
●Bunsen Burner ●Heat-proof Mat
●Ice ●Thermometer(s)
●Core Borer ●Scalpel
●Cutting Tile ●Ruler
●Stopwatch(es)
- Set up the equipment so that it resembles the apparatus diagram on Fig 1. Use the stands and clamps to hold the equipment in place. Make sure that the beaker is placed high enough for a Bunsen burner to fit easily under it. (NOTE: When using a Bunsen burner, always place a heat-proof mat under)
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Pour 15cm3 of the hydrogen peroxide solution into the boiling tube (This can done by measuring out 15cm3 of H2O2 with a measuring cylinder or using a syringe).
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Place the boiling tube into the water bath. Adjust the temperature of the water bath by either adding ice or by heating it up with a Bunsen burner (or neither, depending on the room temperature). Using a thermometer, check that the temperature of the water bath and the temperature of the hydrogen peroxide is the desired temperature (by simply putting one thermometer in the water bath and another is the boiling tube). The experiments should begin with 10 ºC.
- After the desired temperature is obtained, and only after, bore out a 6 cm long core, 1 cm in diameter (the overall shape being a cylinder). Further cut the core into 12 smaller pieces (creating small discs), each with a length of 0.5 cm. Such discs fit easily into the boiling tube and provide an ample surface area.
The core should only be cut after the desired temperature is reached because if the potato cores are cut beforehand, and are therefore left in the open air, they dry up and this could damage the enzyme or restrict its activity. Therefore it is advisable to only cut the cores out when the hydrogen peroxide is ready.
- Place the 12 discs into the boiling tube and immediately place the bung over the boiling tube. This prevents any of the oxygen gas escaping. Once the bung is placed on top of the boiling tube, begin timing the experiment (i.e. start the stopwatch).
Mark where the dyed water, in the U-shaped manometer, began. The apparatus is designed so that as oxygen is produced from the breakdown of H2O2, it remains in an enclosed system. As more oxygen gas is made, the pressure in the system increases. This increase in pressure will push the dyed water out of the U-shaped manometer, so the water line will go down on the left hand side of the manometer and rise in the right hand side.
Measuring the distance travelled by the dyed water will give a comparative measure of how much oxygen is made. Therefore the further distance travelled by the dyed water in the manometer, the more oxygen is made.
The water in the U-shaped manometer is dyed (or should be) to make it easier to measure how far the water line moves during the course of the experiment.
- After 15 minutes, record the length/distance travelled by the water line for the 10 ºC on the Results Table.
- Empty the contents of the boiling tube.
- Repeat Steps 3-8, except in Step 4, instead of adjusting the temperature to 10 ºC, it should be adjusted to 20 ºC, then 30 ºC, 40 ºC. 50 ºC and finally 60 ºC.
- And finally, repeat each of the experiments (i.e. each of the different temperatures) twice, preferably three times, in order to get averages and to eliminate any anomalous results.
Safety
As this investigation involves a corrosive substance, hydrogen peroxide, it must be handled with care (See Step 8 of Method). As a mandatory rule, one must wear eye protection when handling hydrogen peroxide. If the hydrogen peroxide makes contact with the skin, it should be rinsed off immediately.
The investigation also uses flames, from the Bunsen burner, when heating up the water bath. Again, eye protection must be worn when dealing with flames. Shirt sleeves should be rolled back, to prevent the shirt from catching fire.
The Results
The Conclusion
The (average) results show that as the temperature increases, the amount of oxygen gas produced also increases i.e. the rate of enzyme activity increases. However, this does not continue indefinitely. At around 30 ºC, the amount of oxygen produced gradually decreases.
This pattern is supposedly present in all enzymes. To explain this pattern, we must refer to an enzyme’s structure, how it related to the enzyme’s function and how it is affected by temperature.
As with all chemical reactions, when the temperature increases the rate of reaction increases due the fact that collision between the reactants are more frequent and occur with more force, increasing the probability that the two reactants react.
In an enzyme-controlled reaction, the enzyme itself does not react with the substrate; instead it forms an enzyme-substrate complex, in which the enzyme aids the reaction. As the temperature increases, the probability (and number) of collisions between enzyme and substrate (in this practical, catalase and hydrogen peroxide) will increase. This results in more enzyme-substrate complexes occurring. This means more of the substrate molecules being acted upon by an enzyme. This will, in turn, result in more substrate being converted into product (water and oxygen) and thus the rate of activity increases. This is evident by the fact the amount of oxygen gas produced increases with temperature, up to a certain limit.
To explain the gradual decline in enzyme activity we need to know what enzymes are. All enzymes are globular proteins. They are large polymers of amino acids (monomers) or commonly referred to as polypeptides. The amino acids in the polypeptide chain are organised in a certain order (the primary structure). The order in which the amino acids occur give the enzyme a particular shape, with a special area on it called the active site. The active site is the area involved in speeding up (catalysing) a reaction.
Enzymes maintain their shape by having certain areas with cross linkages or bridges between different amino acids (the most common of which is the disulphide bridge).
When exposed to high temperatures, enzymes can deform in the following ways:
- The active site itself could be directly damaged. If the shape of the active site changes, the substrate will no longer fit into it. If the substrate can’t get into the active site, the enzyme cannot act upon it and thus the enzyme is now useless. The amount of product made will therefore decrease.
- The cross linkages in the enzyme could be broken and the shape of the enzyme could change significantly. This will, in turn, change the shape of the active site and the enzyme will not be able to function properly.
- Other bonds in the enzyme could be broken, affecting the whole of the enzyme (e.g. some of the amino acids could be separated from the enzyme). If these parts (amino acids) happen to be vital in the enzyme (especially if they are present in the active site), then again, the enzyme will be unable to catalyse the reaction.
Any or all of the above scenarios could occur when enzymes are put in high temperatures. This would mean that the enzyme can no longer function properly and it will not catalyse the reaction. As a result, less oxygen gas will be made and this is why the graph shows a decline in the amount of oxygen gas produced. When any of the above deformations occur in an enzyme, it has denatured.
The temperature at which the rate of enzyme activity is maximum (i.e. where the graph peaks) is known as the optimum temperature. From the average results, it seems that the optimum temperature of potato catalase is around 30 ºC. (Note: The intervals of 10 ºC do not allow this experiment to precisely pinpoint the optimum temperature).
The Evaluation
Firstly, the practical I planned and the one I carried out were totally different! Due to lack of equipment, all of the students had to use one standardised method.
Instead of using a manometer, we had to collect the gas in a measuring cylinder and record the actual volume of gas produced as the cylinder is wider that the manometer, it would have lead to slight inaccuracies.
Another problem with the practical is that the range was smaller that that I proposed. In plan, I had a temperature range between 10 ºC and 60 ºC with 10 ºC intervals. In the actual practical the range was reduced (from 20 ºC to 60 ºC, the intervals were kept the same). This gives fewer results to investigate and so there is less evidence to build a firm conclusion on.
From the results, the optimum temperature is around 30 ºC. However the 10 ºC intervals make it difficult to find the exact optimum temperature. Having intervals of 5 ºC would
There was another problem with the timing of the experiment. Originally, I planned to keep the potato cores in the hydrogen peroxide solution for 15 minutes and in the procedure we had to keep the experiment running for only 5 minutes. In the higher temperatures, not much oxygen was made. In fact, so little oxygen was made that it was difficult to measure the volume produced as the scale on measuring cylinder began at 1 cm3. If the experiments were to last 15 minutes, more oxygen could have been made and it would have been easier to measure the volume of oxygen produced.
Although there is a general pattern in the results, they do vary. This does not mean that they are anomalous. These variations could be caused by many factors. For example, not all of the students used the same potato. This would mean that there are two possible differences, being the type of catalase and its concentration per disc.
To eliminate these variations, it would have been better for the group of students to use the same potato, as it is more likely that the concentration and type of catalase will be similar if not the same. This would make the procedure a fairer test.
