I have a choice of using either a concentration of 10% hydrogen peroxide or 20% hydrogen peroxide. I tested both of these substrate concentrations and found that visibly the 20% concentration of hydrogen peroxide produced a higher rate of reaction with in a time period that the 10%. This was due to the 10% containing less substrate molecules with the solution which lowers the chance that they will bind with the catalase active site and react and produce oxygen. I have chosen the 20% because in my experiment I think that this will be more accurate to measure if there is a more noticeable production of oxygen.
Another independent variable which I have to control is the pH, is a measure of the acidity or hydrogen ion concentration of a solution. It is measured on a scale of 0-14 with pH values below 7 being acidic, values above 7 being basic and a value around 7 is neutral. Enzymes have a specific range of pH at which they are most active and therefore productive; above this specific range (generally acidic) the enzyme tends to gain hydrogen ions from the solution. Below this range (normally alkaline) the enzyme tends to lose hydrogen ions to the solution. In both cases the changes produced in the chemical bonds of the enzyme molecule result in a change in conformation that decreases enzyme activity.
Therefore it is vital that I keep this variable constant, to control the pH concentration I have chosen to use a biological buffer. Buffers are important because most Biochemical reactions are especially sensitive to pH. Most biological molecules contain groups of atoms that may be charged or neutral depending on pH, and whether these groups are charged or neutral has a significant effect on the biological activity of the molecule. The use of a buffer enables the pH condition of the reaction in my catalase experiment to be controlled. I am going to add 10ml of buffer at a pH of 7 (this pH known as neutral) to each experiment to ensure the pH does not affect the shape of the active site of the catalase enzyme. To test the pH of the solution I will use a universal indicator paper before and after the experiment.
The next decision I have to make is the amount of time in which the experiment will take place. Time is an important variable because it can have a major effect on the amount of water and oxygen measured during the experiment. Clearly the more time the enzyme (catalase) has to meet other substrates the more likely the will react and produce oxygen to be measured. Therefore this must be kept constant to correctly measure the amount of oxygen. I’m going to test the oxygen produced during 1 minute and 2 minutes at the apparent most active temperature of catalase. I found that very little oxygen was produced during one minute this shows the time is not long enough to produce measurable accurate results. So in my experiment I am going to time the reaction for two minutes.
Another variable which must be kept under control to make my experiment a fair test is enzyme concentration. The effect this has on a reaction is the more enzymes the solution contains there is an increased chance the active site will bind with a substrate thus increasing the product and the rate of reaction.
Therefore the concentration of catalase in my experiment must be kept constant. I’m going to obtain the enzyme catalase by using potato cubes. It is essential I use exactly 1cm cubes of potato to ensure the surface area in which the catalase enzyme diffuses through is kept constant. I’m going to do various preliminary experiments to discover at what enzyme concentration it is the most accurate to measure the affect temperature has on the rate of reaction. The preliminary results proved that 5 1cm cubes of potato did not produce enough oxygen to differentiate between the changing rates of reactions at different temperatures.
In theory Hydrogen peroxide doesn’t decompose within the time I have allocated for my experiment on its own. However to ensure the results are accurate I am going to perform the experiment without any catalase at room temperature and 60 degrees centigrade, only a small amount of hydrogen peroxide in a beaker attached to a gas syringe and measure the air which is produced. I performed the experiment at room temperature and at 60 degrees centigrade and found that no gas was produced from the hydrogen peroxide ‘decomposing’. From these results I’m now sure that whilst I undertake the actual experiment no gas is produced and the hydrogen peroxide doesn’t decompose on its own at any of the temperatures I am testing, the maximum or the minimum. Therefore my results will be accurate, the only gas that I will account for is the oxygen produced, during the reaction between hydrogen peroxide and catalase.
List of equipment:
- 3 Conical flasks
- 3 tubes
- 1 gas syringe
- 20ml measuring cylinder
- 1 scalpel
- 1 Ruler
- 1 thermometer
- 1 water bath
- 1 clamp and stand
- 1 stop watch
- Hydrogen peroxide (20% concentration)
- PH7 buffer
- Potato
- Potato chip cutter
Method
- Set up water bath at 5 degrees centigrade, and leave for a few minutes so the device can reach the correct temperature. Then test the water in the bath with a thermometer, ensure it is at the correct temperature.
- Set up the rest of the equipment; attach clamp to stand and secure the syringe to the clamp with the tubing attached. Above the water bath in preparation.
- Measure 20ml of hydrogen peroxide (20% concentration) in a 20ml measuring cylinder. Then pour this substance into a clean conical flask ensuring not to spill any or leave any in the cylinder.
- Measure 10ml of PH7 buffer in a 20ml measuring cylinder, add this substance to the hydrogen peroxide in the conical flask. (use a universal indicator to check the PH is neutral)
- Dip the end of the universal indicator paper into the solution. And compare the colour against a colour pH chart, to ensure the solution is at the correct pH.
- Leave the substance for a minute, so that it can acclimatise to the temperature and ensure there is no air being produced from the hydrogen peroxide decomposing on its own.
- Place the conical flask in the water bath
- Place a potato in the chip cutter, to get long 1cm cubed width chips. Then ensure you use the pieces which do not have skin on them. Then line the potato chip up to the ruler’s edge, and then cut directly downwards (so that it is not at an angle) at the 1 cm mark, repeat this ten times.
- Place the conical flask in the water bath so that the solution becomes the right temperature before the experiment.
- Add the ten 1cm cubes of potato to the conical flash and attach the bung securely with tube and syringe on the end.
- Set the stop watch.
- After two minutes stop the experiment measure the amount of gas in the syringe and record this.
- Dip a thermometer into the conical flask at the end of the experiment to check the temperature is still the same as at the beginning of the experiment.
- Repeat this experiment three times at the same temperature.
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Repeat at 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 degrees centigrade by going back to the first point and setting the water bath at these specific temperatures instead.
Results
Analysis
The graph shows a steady curve, as the temperature increases so does the rate of reaction this is shown by the amount of oxygen produced in two minutes. Then at the optimum temperature which is thirty degrees centigrade the amount of oxygen produced begins to decrease. Then past the point of the optimum temperature the graph shows the amount of oxygen produced proceeds to decease steadily.
At the lowest temperature which is five degrees centigrade the graph shows there is very little oxygen produced this suggests the rate of reaction is also low. This can be explained using enzyme kinetics as there is very little heat energy available to the enzyme and the substrate to convert into kinetic energy there is a very small rate of reaction. As the substrate tends to stay in the same place they rarely come into contact with substrates in turn causing very few enzyme substrate complexes, these occur when an enzyme and a substrate join together at the active site. There are many theories in how these join together the most accepted is the lock and key model.
The oxygen produced between 5 and 10 degrees centigrade was only 2 centimetres cubed. This shows the enzymes rate of reaction is very low at this temperature. Then as the temperature so does this amount of oxygen suggesting the rate of reaction is increasing. This is due to the extra kinetic energy the higher temperature increasing the enzymes and substrates movement making it more likely they will create enzyme substrate complexes as shown above. This increases the rate of reaction.
At 30 degrees centigrade the graph shows the enzymes optimum temperature the optimum temperature is when the temperature at which the enzyme can use the kinetic energy and move around fast but not be affected by the heat. Past the optimum temperature the enzyme begins to denature this occurs when, the rate of reaction decreases, as the increasing kinetic energy causes the hydrogen bonds, holding the tertiary structure of the enzyme together, to break. This causes the active site to change shape. The substrate can therefore no longer fit the active site and the rate of reaction decreases until all the enzyme molecules are denatured. This is shown on the graph because after the point of thirty degrees centigrade the amount of oxygen produced begins so fall indicting the rate of reaction is also decreasing.
Evaluation
My results as shown on the graph appear accurate and consistent; they show a structured correlation. I can make this assumption due to the smooth curve which the graph shows and the repeats of the experiments being within normal range of each other also the absence of anomalies proves my method was fairly accurate.
However after completely the experiment in retrospect I can see that my method was not full proof and there were possible sources of error which might lead to inconsistent results. For example the way in which I measured and cut the cubes of potato was left to human error; this is often very inaccurate. This could have lead to a larger amount of enzymes within the cube due to a larger mass, which would cause a higher rate of reaction at a temperature which does not correspond with the other results. In addiction the some of the cubes may have had a larger surface area, making diffusion faster and more effective, this would speed up the rate of reaction, making the results inaccurate. To improve this part of my method I could get another person to check the measurements of the cubes of potato or I could use some sort of machine, to limit errors that could occur.
Also another issue with the enzyme concentration was the time period in which I conducted the experiment. I did the experiment in over a few days and it was unfeasible to use the exact same potato for all of cubes which I used, therefore the enzyme concentration may have varied from each temperature experiment, this could lead to uneven results. To amend this mistake in my method I could simply conduct the experiment using the same potato on the same day.
There were also concerns with the timing of the experiment, visible in the method. There was a slight delay in between putting the potato in the solution and placing the bung on top of the conical flask, this may have cause gas (oxygen) to escape and not entering the gas syringe. Decreasing the amount of gas syringe lowering the rate of reaction. I can improve my method in this aspect by placing the potato into the buffer then adding the hydrogen peroxide through a tube in the bung of the conical flask would reduce the amount of gas lost, making the method more accurate.
Also my method lacks could have been improved by checking the controls more often. For example I could have checked the pH on various occasions throughout the experiment rather than on just once, to keep this variable constant to ensure the affect of the pH was not confused with the effect of temperature on the rate of reaction.
The temperatures which I chose were ideal for this broad experiment, to gage the general reaction of the enzyme catalase to see the effect temperature had on its rate of reaction. However it would be very useful to decipher the specific rates of reaction around the optimum temperature. The optimum temperature is a vital stage in the enzymes, therefore in the next experiment I should concentrate around the area of 25 to 30 degrees centigrade. This would show the rate at which the hydrogen bonds in the protein of the enzyme were broken and the active site was distorted. Also the interval between the temperatures that I tested at was fairly large, so the rate of reaction could have varied between these temperatures. This means that the shape of my graphs could be altered by an anomalous result at the temperature at which I tested. If I had tested at intervals of 1oC, an incorrect result would not affect the shape of my graph a great deal. However, it was impossible for me to conduct the experiment at temperatures that increase by only 1oC because I would not have had time and because I would not have been able to accurately maintain the temperature.
It is also possible that the catalase did not have time to acclimatise to the change in temperature. This would have resulted in the reaction occurring more slowly at the beginning of the experiment than it would have if the catalase enzyme started at the correct temperature. This error could have been avoided by allowing the solution and the enzyme time to acclimatise to the temperature.
Finally the last misjudgement I have made in my experiment was the subjective process of determining the temperature at which to end the experiment. I stopped at 60 degrees centigrade, although from my results you can see that there is still a reaction occurring. If I repeat this experiment I should continue with higher temperature until the enzyme fully stops reacting.