I will;
Keep the volume and concentration of catalase (yeast) constant
Keep the temperature constant at 40 oC by using a water bath.
Keep the volume and concentration of Hydrogen peroxide constant and the time
Result:
Experiment 5
Aim: is to find out the effect of inhibitors on the rate of reaction.
Plan: copper sulphate used ranged from 0.0m -0.2m in increments of 0.02
Variables:
I will;
Keep the volume of catalase (yeast) constant
Keep the temperature constant at 40 oC by using a water bath.
Keep the volume of Hydrogen peroxide constant
Result:
Conclusion
Catalase is made of a central heme and four polypeptide chains. Its active site binds to hydrogen peroxide and decomposes it to oxygen and water. My data suggests that enzyme activity depends on several modifiable variables. Factors that increase binding of H2O2 to catalase, such as increasing substrate and enzyme amounts, and increasing temperature, will increase the rate of reaction. After all active sites are occupied, adding more hydrogen peroxide will not affect the rate of reaction. Factors that cause catalase denaturation, such as high temperatures, extreme pH, and non-competitive inhibitors, can change its 3-dimensional structure, rendering it less active.
Fair tasting
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.
Controlled variable
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pH :In this experiment, the pH will be kept constant using pH 7 buffer solutions, this is because change in pH affects the ionic and hydrogen bonding in an enzyme and so alters it shape. Each enzyme has an optimum pH at which its active site best fits the substrate. Variation either side of pH results in denaturation of the enzyme and a slower rate of reaction.
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Temperature: In this experiment, the temperature will be kept constant at 40 ◦c (optimum temperature) using a water bath. This is because as temperature increases up to the optimum, the rate of enzyme activity also increases. In other word the substrate molecules gain more kinetic energy and the collisions between active sites of enzymes and substrate molecule become more frequent. The rate of enzymes substrate complex formation increases so enzymes activity speeds up. Beyond the optimum temperature, the rate of enzyme activity decreases because the high temperature causes the enzyme molecule to lose its specific 3D shape (due to breaking of Hydrogen bonds and ionic bonds). The active sites cannot bind with substrate, so enzyme substrate complexes cannot form.
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Volume of yeast solution (catalase): The volume of yeast solution (catalase) will also be kept constant at 10ml using a 10 cm3measuring cylinder to ensure the accuracy.
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Concentration of yeast solution (catalase) enzyme :In this experiment, the Concentration of yeast solution (catalase) will be kept constant at 5% this is because as the enzyme concentration increases (at constant substrate concentration) the rate of reaction increases until it reaches a maximum rate (V max). This is because there will be more number of free active sites, at any given time. So the rate of enzyme substrate complex formation increases. Thus rate of reaction increases. The rate doesn’t increase beyond the V max because the substrate concentration becomes a limiting factor. Even though there will be many free active sites, there will not be enough substrate molecules to bind with them. So, rate of enzyme substrate complex formation remains constant at V max. But, increasing substrate concentration would further increase the rate of reaction.
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Volume of hydrogen peroxide (substrate): The volume of hydrogen peroxide (substrate) will also be kept constant at 10ml using a 10 cm3syringe to ensure the accuracy.
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Concentration of hydrogen peroxide (substrate):In this experiment, the Concentration of hydrogen peroxide (substrate) will be kept constant at 20Vol this is because as the substrate concentration increases (at constant enzyme concentration) the rate of reaction increases until it reaches a maximum rate (V max). This is because there will be less number of free active sites and more number of substrate molecules to bind with the enzyme active site , at any given time. So the rate of enzyme substrate complex formation increases. Thus rate of reaction increases. The rate doesn’t increase beyond the V max because the enzyme concentration becomes a limiting factor. Even though there will be many substrate molecules, there will not be enough enzyme molecules to bind with them. So, rate of enzyme substrate complex formation remains constant at V max. But, increasing enzyme concentration would further increase the rate of reaction.
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Time: The time will be recorded every 10 seconds interval for 60 seconds using digital stop watch.
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Volume of copper sulphate (inhibitor): The volume of copper sulphate (inhibitor) will also be kept constant at 10ml using a 10 cm3 syringe to ensure the accuracy.
Independent variables
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Concentration of copper sulphate (inhibitor): This is the variable that will change and affects the volume of oxygen that is produced. I will add different volumes of distilled water to the copper sulphate solution to make different concentrations copper sulphate solutions. The concentrations to be used are as followed; 0.00 moldm-3, 0.02 moldm-3, 0.04 moldm-3, 0.06 moldm-3, 0.08 moldm-3, 0.1 moldm-3, 0.2 moldm-3.
Dependent variable
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Volume of oxygen: In this investigation, the dependent variable will be the volume of oxygen produced. This volume oxygen produced will depend on the Concentration and volume of copper sulphate (inhibitor) and the rate of activity of the catalase’s ability, to covert the hydrogen peroxide to water and oxygen. The volume of oxygen produced will be measured using a gas syringe every 10 seconds for 60 seconds. Since I was using 100cm3 of gas syringe, to ensure accurate results the volume of the gas will be measured to 1d.p.
Apparatus
Results
Analysing
Graph 1
Notice on this graph the production oxygen increased by 43.8cm3 during the first 20 seconds, after 40 seconds the amount of oxygen produced was 61.3 cm3 and after 60 seconds the amount of oxygen produced was 67.3 cm3.
The production of oxygen increased proportionally the first 10 seconds, slight increase between 10-40 seconds and after 40 seconds the amount of oxygen produced starts to levels off. Overall the production oxygen increased through out 60 seconds
Graph 2
As you can see on this graph the production oxygen increased by 41.5 cm3 during the first 20 seconds, after 40 seconds the amount of oxygen produced was 57.5 cm3 and after 60 seconds the amount of oxygen produced was 63.5 cm3.
The amount of oxygen produced increased proportionally the first 10 seconds, slight increase between 10-30 seconds and after 30 seconds the amount of oxygen produced starts to levels off. Overall the production oxygen increased through out 60 seconds
Graph 3
On this graph the production oxygen increased by 37.5 cm3 during the first 20 seconds, after 40 seconds the amount of oxygen produced was 52.8 cm3 and after 60 seconds the amount of oxygen produced was 58.8 cm3.
Again the amount of oxygen produced increased proportionally the first 10 seconds, slight increase between 10-30 seconds and after 30 seconds the amount of oxygen produced starts to levels off. Overall the production oxygen increased through out 60 seconds
Graph 4
As you can see on this graph the production oxygen increased by 22.8 cm3 during the first 20 seconds, after 40 seconds the amount of oxygen produced was 43.3 cm3 and after 60 seconds the amount of oxygen produced was 51.3 cm3.
The amount of oxygen produced increased proportionally the first 30 seconds this time, slight increase between 30-40 seconds and after 40 seconds the amount of oxygen produced starts to levels off. Overall the production oxygen increased through out 60 seconds
Graph 5
On this graph the production oxygen increased by 23.3 cm3 during the first 20 seconds, after 40 seconds the amount of oxygen produced was 38.8 cm3 and after 60 seconds the amount of oxygen produced was 43.5 cm3.
Again the amount of oxygen produced increased proportionally the first 30 seconds, slight increase between 30-50 seconds and after 50 seconds the amount of oxygen produced starts to levels off. Overall the production oxygen increased through out 60 seconds
Graph 6
You can see on this graph the production oxygen increased by 20 cm3 during the first 20 seconds, after 40 seconds the amount of oxygen produced was 37.5 cm3 and after 60 seconds the amount of oxygen produced was 38.3 cm3.
Again the amount of oxygen produced increased proportionally the first 30 seconds, slight increase between 30-40 seconds and after 40 seconds the amount of oxygen produced starts to levels off. Overall the production oxygen increased through out 60 seconds
Graph 7
As you can see on this graph the production oxygen increased by 12.3 cm3 during the first 20 seconds, after 40 seconds the amount of oxygen produced was 21.3 cm3 and after 60 seconds the amount of oxygen produced was 23.5 cm3.
Again the amount of oxygen produced increased proportionally the first 40 seconds, slight increase between 40-50 seconds and after 50 seconds the amount of oxygen produced starts to levels off. Overall the production oxygen increased through out 60 seconds
Explanation
On graph 1 the production of oxygen increased proportionally the first 10 seconds sand slight increase between 10-40 seconds and this is because there was equal amount enzyme and substrate number of free active sites at any given time. So the rate of enzyme substrate complex formation increases until it reaches the maximum rate (Vmax). After 40 seconds the amount of oxygen produced starts to levels off this is because both enzyme and substrate concentration becomes a limiting factor so the rate of reaction remains constant at (Vmax).
On graph 2 and 3 the production of oxygen increased proportionally the first 10 seconds sand slight increase between 10-30 seconds this is because there was equal amount enzyme and substrate molecules sites during the first 30 seconds and less time for the inhibitor (0.02 moldm-3 of copper sulphate was used) to work because it was less concentrated than the substrate molecules. So the rate of enzyme substrate complex formation increases until it reaches the maximum rate (Vmax). After 30 seconds the amount of oxygen produced starts to levels off this is because enzyme concentration becomes a limiting factor as most of enzyme molecules are permanently denatured by copper ions. Even though there are plenty of substrate molecules, there are no enough free active sites to bind with the substrate molecules, at any given time. So enzyme substrate complex formations remain constant at (Vmax).
On graph 7 the production of oxygen increased proportionally the first 20 seconds sand slight increase between 20-25 seconds this is because there was equal amount enzyme and substrate molecules sites during the first 25 seconds and less time for the inhibitor (0.2 moldm-3 of copper sulphate was used) to work. So the rate of enzyme substrate complex formation increases until it reaches the maximum rate (Vmax). After 25 seconds the amount of oxygen produced starts to levels off this is because enzyme concentration becomes a limiting factor as most of enzyme molecules are permanently denatured by copper ions. Even though there are plenty of substrate molecules, there are no enough free active sites to bind with the substrate molecules, at any given time. So enzyme substrate complex formations remain constant at (Vmax).
Anomalous result
On graph 4, 5 and 6 there are some anomalous results. (Explained under evaluation section)
Initial Rate
In this investigation an initial rate graph has to be plotted to show the difference in initial rate as concentration increases. It is used to work out the initial rate at the beginning of the reaction so I can compare the initial rates of all the concentrations to see if they change.
Initial rate is worked out by calculating the slope closes to the tangent to the curve, as close to 0 as possible. This means I will measure from just before the graph starts to curve.
Sample calculation
e.g. On graph 1 starts to curve on about 20 second. This means I will take 20seconds. I then look on the graph to see the amount of gas produced in this time. It says 56 cm3 of oxygen. This mean 56 cm3 are produced per 20sec.
I then do a calculation using the following formula:-
Initial rate = Volume 56 cm3
Time 20 = 2.8 cm3/sec
A table to show the average rate of the volume of oxygen given off as the copper sulphate concentration changes
Graph 9
Notice on this graph, the average rate of reaction was 2.8cm3/s during the controlled experiment (0.0 mol/dm3of copper sulphate), the average rate of reaction reduced to 2.63 cm3/s when 0.02 mol/dm3 of copper sulphate was used, again the average rate of reaction remains 2.63cm3/s when 0.04 mol/dm3 of copper sulphate was used.
When 0.06 mol/dm3 of copper sulphate was used the average rate of reaction was 1.2cm3/s. Again the average rate of reaction remains 1.2cm3/s when 0.08 mol/dm3 of copper sulphate was used. The average rate of reaction reduced to 1.18cm3/s when 0.1 mol/dm3 of copper sulphate was used. When 0.2 mol/dm3 of copper sulphate was used the average rate of reaction was 0.63cm3/s. The average rate of reaction halved compared to 1.18cm3/s (when 0.1 mol/dm3 of copper sulphate was used). Overall as the concentration of copper sulphate increase the average rate of reaction decreases and the plotted results on the graph produce a strong negative correlation.
Trends
As you noticed from all the graphs there are several patterns and trends which can be seen clearly.
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The initial rate of reaction (production of oxygen) decreased as the concentration of copper sulphate (see graph 9).
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The initial rate reaction was 0.63cm3/s when 0.2 mol/dm3 of copper sulphate is used but the initial rate of reaction doubled (it was 1.18cm3/s) when 0.1 mol/dm3 of copper sulphate is used. As you can see from the result, when the concentration of copper sulphate decreased by 0.1mol/dm3 the initial rate of reaction doubles.
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Graph 1&6 has a similar trend, when the concentration of copper sulphate decreased by 0.1mol/dm3 the initial rate of reaction doubles.
- All, the three graphs (Graph 1, 2& 3) shows, the production of oxygen increased proportionally during the 10 seconds of each experiment. This is because both the 2 concentration of copper sulphate were week and their effect on the activity of catalase was negligible during the first 10 seconds of each experiment.
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On both (graph 2 and 3) the graph starts to curve at 30 seconds. This is because the rate of enzyme substrate complex formation can only increase up until it reaches the maximum rate (Vmax). After 30 seconds the amount of oxygen produced starts to levels off this is because enzyme concentration becomes a limiting factor as most of enzyme molecules are permanently denatured by copper ions.
Conclusion
My prediction was,” the greater the concentration of copper sulphate the lower the production of oxygen and less, the activity of catalase”. The overall result supports my prediction .
Evaluation
Sources of error:
There are several sources of error in this experiment. Since the investigation takes a very long time to complete, the experiment had to be continued over several days. As a result, different H2O2 and catalase (yeast) solutions were used on each day. Some of the solutions may have been older than the original solutions used which may have had an effect on their reactivity. Also, different concentrations of solutions may have been used. For example, when the original yeast solution ran out, a new solution had to be made and probably resulted in a slightly different concentration.
Limitation of experimental techniques and uncertainties associated with the equipment:
First I will calculate the uncertainties associated with the measurements that I have taken. I will choose the lowest value where I have a choice to illustrate the ‘worst case scenario’
Measuring cylinder:
The uncertainty associated with the measuring cylinder reading is 0.5cm3
The % uncertainty in a measuring cylinder reading of 50cm3 =
0.5 × 100 =1.0%
50
Percentage error = 1.0%
Thermometer:
The thermometer I used in this experiment is accurate to 0.5 oC
The uncertainty associated with the thermometer reading is 0.5oC
The % uncertainty in a thermometer reading 40oC
= 0.5 × 100 = 1.25%
40
Percentage error = 1.25%
Gas syringe:
The gas syringe I used in this experiment is accurate to 0.5 cm3
The uncertainty associated with the gas syringe reading is 0.5cm3
The % uncertainty in a gas syringe reading of 100cm3
0.5 × 100 = 0.5%
100
Percentage error = 1.25%
I have listed some of the limitations of the experimental techniques in the following table
Improvement:
If I was given a chance to conduct another similar experiment, I would change some of the apparatus, improve the procedures and techniques that were during the experiment.
My experiment could be improved in numerous ways.
Further investigation
This experiment can be extended to investigate the type of inhibitor that was used through out the experiment i.e. (Competitive or Non- competitive inhibitor)
