When the catalase enzyme loosens slightly the substrate molecule (the hydrogen peroxide molecule) is pulled apart producing two new substances. The substances formed from the decomposition of the hydrogen peroxide are oxygen and water. The two new substances are separated and released. This process may be repeated many times until the hydrogen peroxide solution cannot be broken down any further. The harmful hydrogen peroxide is decomposed to form harmless product such as water and oxygen. The diagram below shows the process of the decomposition of the substrate.
The word equations and the symbol equations for this reaction are listed below.
Hydrogen peroxide oxygen + water
Breakdown
2H2O2 (aq) O2 (g) + 2 H2O (l)
Catalase
This reaction can also be shown as:
Reaction
Substrate (A) Products (B + C)
Catalase
2H2O2 O2 (g) + 2 H2O (l)
Hydrogen peroxide oxygen + water
The catalyst in this experiment, which is catalase, remains unchanged after this reaction. Catalysts also work by providing an easier route for the reaction to follow such as having lower activation energy. If the activation energy is lower, many more of the collisions will have the necessary minimum energy and so produce a reaction.
The catalyst can change a solution due to the enzymes a catalyst contains. Enzymes work best at temperatures higher than body heat (which is 370C). Enzymes are an important group of chemicals that speed up chemical reactions. They act as catalysts, which break things down. An enzyme is special, because it is unchanged at the end of a reaction that it has speeded up. Enzymes also have five main properties.
- They are made out of protein
- They are specific, speeding up one reaction only
- They work best within a small pH range which is different for each enzyme
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They work fastest when warm, usually at body temperature (370C)
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Above 500C, their shape is changed and is denatured causing them to stop working.
If I was to use the enzyme catalase in this experiment, I would make sure that the pH and temperature was controlled. This is because as explained above, the enzyme can only work properly if the pH has a suitable range and also if the temperature is just right so that the active site of the enzyme is not damaged.
Potatoes belong to the nitrate group. The biological enzyme catalase is found in potatoes, which can be used for the decomposition of hydrogen peroxide. So therefore, the pH of the enzyme catalase has to be appropriate so that it can decompose the substrate quickly and also does not damage the active site of the enzyme either. The pH of a potato is pH 7 as you can see from the graph below.
As you can see from the graph above, if the pH of the experiment is between pH 1 and 6 (acidic), then the catalase enzyme can not work as fast compared to when the pH is 7, causing less products to be made. The acidic solution will damage the active site not allowing the substrate to be decomposed as fast. Also, when the pH of the solution is between pH 8 and 14, the decomposition becomes slower compared to the solution when the pH is 7. Therefore, if the decomposition of the substrate hydrogen peroxide should be fast, the pH of the enzyme should be 7, neutral; therefore water should be used because it has a pH of 7.
The temperature of the enzyme should also be controlled because enzymes work best at 370C. However, if the temperature is above 500C, this causes the enzyme to be damaged preventing the enzyme from working. Below is a graph showing the temperature of the enzyme against the amount of product produced.
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As you can see if the temperature is 370C more product is produced. However, if the temperature is 500C, the amount of product produced decreases and nothing is produced. To lead a good experiment where the amount of product produced is a lot, the suitable temperature should be 370C.
If the amount of product produced (the rate of the reaction) were drawn against the time, I would expect my results to e in a similar shape as the graph drawn below.
This graph shows that as the enzyme and the substrate are left together, generally the amount of product produced increases to a point. The steeper the line is the faster the reaction is. Where the line is horizontal, the reaction has stopped. At that horizontal point, the enzyme cannot decompose the substrate anymore. I would expect this type of graph for my results because at fist the rate at which the oxygen is produced will be high then at a certain point, the line will be horizontal showing hat no more hydrogen peroxide can de decomposed.
Another thing I may discover if I plot the rate of reaction against the surface area of the potato (also the amount of catalase), I am likely to find a graph similar to the one below.
As you can see from this graph as the amount of catalase enzyme (the surface area of the potato) increases, so does the amount of oxygen produced. The straight line shows that there is a direct link so therefore; we can say that the surface area (or the amount of catalase) is directly proportional to the rate of reaction (amount of oxygen produced).
This information was collected from a book called ‘Biology’ by Jones and Jones published by Cambridge.
PREDICTION: I predict that as the surface area of the potato increases, more oxygen will be produced as shown in my background theory and variables. The (the amount of oxygen produced will be directly proportional to the surface area (the amount of catalase). This is because since there are more cubes in the solution, this provides a larger surface area of the potatoes.
As more cubes are added, this increases the amount of catalase in the solution, which decomposes the hydrogen peroxide quicker since there are more enzymes available in the solution. Therefore, as I will increase the surface area of the potato, which also increases the catalase enzyme in the solution, more oxygen will be decomposed.
I also predict that the graph that I will plot with my results will show that the surface area of the potatoes is directly proportional to the amount of oxygen produced. This is because as the surface area increases there is more catalase in the experiment therefore the substrate will be decomposed faster producing more oxygen in less time.
Also the rate of reaction will increase the amount of cubes are increased, which also increase the surface area. This is because as there are more potatoes in the experiment, there is more surface area, which also increases the amount of catalase in the experiment. As there is more catalase, the hydrogen peroxide will decompose faster as the surface area of the cubes increase so therefore, the rate of reaction increases with the surface area. Therefore, less time will be needed to produce the same amount of oxygen as before when there were fewer cubes, meaning that there was a smaller surface area of the potato.
I will repeat my experiment three times and gain an average to plot an overall graph. I will only be changing the surface area of the potatoes and keeping the time constant in every experiment because I am investigating the surface against the time taken for the product to be made. Repeating the experiment three times will allow me to gain accurate and reliable results to base a conclusion on.
EQIUPMENT: In this experiment, I will be using, the following equipment:
- A conical flask
- Bung (stopper on top of a conical flask)
- A plastic tub
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Potatoes (cut into cubes measuring 1cmx1cmx6cm = 6cm3 each)
- Measuring cylinder
- Water
- Knife
- Stop watch
- Hydrogen Proxide (concentration of 6%)
- Ruler
Here is a diagram of the apparatus:
BASIC METHOD: Firstly, before I carried out the actual experiment, I carried out a preliminary test to find a certain time to carry out the experiment up to. This is because for my actual experiment I would keep the time of each experiment the same and only change the surface area of the potato. My preliminary test for this experiment started off by cutting small cubes (of which each cube had a surface area of 6cm3, 1cmx1cmx6cm) and putting them into a conical flask.
I then filled up a tub of water. After, I attached one side of a tube onto the conical flask and the other side I left into the tub of water. I then filled up a measuring cylinder and put my thumb on the top and placed it upside down in to the tub of water making sure the water in the measuring cylinder was full.
I then put the other side of the tube, which was in the tub into the measuring cylinder like the one below.
This method is known as ‘collecting over water’. I then poured 75ml of hydrogen peroxide into the conical flask, whilst anther member of the class held the measuring cylinder, upside down in the tub making sure that they did not press onto the tube stopping the oxygen from being collected. When I poured the hydrogen peroxide solution into the conical flask, I started to time the reaction and noted the results. I then repeated these results up to fourteen minutes making sure that there was an interval of two minutes each time. Then I increased the surface area of the potato by 6 cm3 (by adding another cube each time) and repeated the same method as above until I had results for eight cubes of potatoes giving a total surface area of 48 cm3. Below is a table of results, which were collected during my preliminary test.
In this reaction, the oxygen was produced as a product when the catalase enzyme had decomposed the hydrogen peroxide. My preliminary work allowed me to make a few changes to my final method. Firstly, I will only use 50 ml of hydrogen peroxide instead of 75 ml. This is because, the experiment is so big that there is not enough substrate for everyone to test with so therefore, I will have to use 50ml of it. I also noticed that after ten minutes of collecting the oxygen, the oxygen would stop being collected. This was because the hydrogen peroxide could not be decomposed anymore.
Therefore, after ten minutes no more oxygen would be collected. From this finding, I will time the experiment up to ten minutes instead of fourteen and take no results in between because the I am only testing how the surface area affects the rate of reaction, not the time against the rate of reaction. I will time my experiments up to ten minutes in my final method because in my preliminary test I used 75 ml of hydrogen peroxide, and after ten minutes less oxygen was being collected. For my actual method, I will be using 50ml of hydrogen peroxide so therefore; less oxygen will be collected after ten minutes because there is less substrate compared to my preliminary test.
In the actual test I am using 50 ml of hydrogen peroxide so therefore, the decomposition of the substrate will finish quicker because there is fewer hydrogen peroxide molecules in the solution so therefore, the decomposition of the substrate will also finish quickly. The quicker the experiment ends the easier it is for me to repeat. Consequently, I will time the experiment up to ten minutes with no intervals and each time; increase the surface area by 6cm3, until I have results at ten minutes for 48cm3 (from 1 to 8 potato cubes).
If I time the whole experiment less than two minutes, then there will not be enough time which will allow the oxygen to build up. Therefore, if I time the experiment for ten minutes, this gives enough time for the substrate to be decomposed and the oxygen to be collected which will give me a better result. This will allow me to get enough results to plot a graph showing the surface area against the rate of reaction.
I would repeat the whole experiment three times to ensure that my results are accurate and reliable to come to a conclusion. I will gain an average of these results as well to plot them on a graph. These changes will help me lead a good experiment, allowing me to get reliable results to base a conclusion on. I will also round off the oxygen collected to the nearest millilitre because that way the results are more accurate and reliable. During the experiment, the catalyst, the potato will remain unchanged.
FAIR TESTING: To ensure that the investigation that I carry out is a fair one many factors the same for the whole investigation.
- I will always use the same volume and the same concentration of the hydrogen peroxide. However, I will change the surface area of the potato used every time.
- I will always take the measurements when collecting the oxygen to the nearest full millilitre, as they are more accurate results.
- I will always time each experiment for 10 minutes and then note down the amount of oxygen collected.
- I will also make sure that each result is repeated three times to allow me to gain an average at the end of the investigation to make my results more accurate and reliable.
- I will also keep the following key factors constant. I will always make sure that the measuring cylinder is clear so that the amount of oxygen produced at the end can be seen properly.
- The amount of oxygen produced also needs to be measured at eye level to gain a reliable result.
- The shape of all the flasks needs to be the same so that all the reactions have a fair advantage.
- The thickness of the tube should be the same for each experiment so that the speed of which the amount of oxygen produced at the end is fair.
- The same person should watch the reaction so that they can standardise the judgement to suggest how much oxygen is produced at the end.
- The size and the shape of the potato cubes should be the same.
- I will use water, so that the pH of the experiment is 7, neutral so that the catalase enzyme is not affected by the pH of the substances the experiment uses.
- The experiment will be conducted at room temperature so that the temperature does not affect the amount of oxygen made in each experiment because certain temperatures either increase the time taken to produce the oxygen or decrease it. Each set will have a fair advantage if the experiment is conducted at room temperature.
- The measuring cylinder when collecting over water should not press the tube that is connected from the flask to the plastic tub as this can lead to the amount of oxygen being produced to decrease and make the results unreliable.
- The measuring cylinder used to collect the oxygen released needs to be clear so that I can view the experiment properly and gain an accurate result to base a conclusion on.
If all these factors are controlled it will make my test a fair one and also lead me to gain reliable results.
SAFETY: To ensure that a safe experiment is carried out, I will:
- Be careful when using the hydrogen peroxide because it tends to bleach clothes and damage the skin
- I will also be careful when using the hydrogen peroxide and always close the lid because the hydrogen peroxide can decompose if it is not bottled in dark bottles or closed.
- I will wear protective footwear to protect my feet from spills
- I will wear goggles so that my eyes are protected from the substrate, which is the hydrogen peroxide.
RESULTS:
SET 1
SET 2
SET 3
AVERAGE SET OF ALL 3 RESULTS
AVERAGE RESULTS = SET 1 + SET 2 + SET 3
3
OBSERVATIONS AND OBTAINING EVIDENCE: The main purpose of this investigation was to find out whether increasing the surface area of the potato (increasing the amount of catalase) affected the rate of the reaction (the rate at which the oxygen is formed. When carrying out this experiment I saw that the hydrogen peroxide attacked the surface area of the potato and formed oxygen. Also, I saw bubbles being produced in the conical flask meaning that water was also being produced in this experiment.
I also made sure that tested fairly whilst obtaining these results. I gained all the results that I had intended to collect and I timed each experiment for ten minutes, taking the volume of oxygen collected after the ten minutes ensuring that enough oxygen was collected to note down just like I had planned.
I also repeated the experiment three times to get an average set of results so that they are reliable and accurate. I did this because in case one set of the results were odd the other two could be accurate results, which could still lead me to gain n average to get reliable results. I also rounded up the amount of oxygen collected to the nearest whole number so that the results we reliable and accurate. Therefore, there are enough results in my results section, for me to comment on.
Whilst conducting this investigation, I made sure that I always controlled the main variables. I always viewed the amount of oxygen being collected. I also made sure that I standardised how much oxygen had been collected fairly every time. I also made sure that I stood the same distance away from the experiment to make it fair on all the experiments when I had to standardise the amount of oxygen collected. I always repeated the experiment in the same conical flask so therefore; every experiment I lead had the conical flask of the same width making it fair on every experiment.
I also made sure that the measuring cylinder was clear every time so that I could read clearly the amount of oxygen that had been released. I also, used the same tube so that the thickness of the tube was always the same because if a tube is thicker, it can lead to more oxygen being collected because the thicker tube allows more oxygen to pass through.
Therefore, to make it fair I always made sure that I kept the same tube. I also made sure that the beaker that I measured the hydrogen peroxide in was clear so that I could see that 50 ml of hydrogen peroxide was always being used. I also used water when I was collecting the evidence so that the pH of the substances used in the experiment did not affect the catalase enzyme, which could change the amount of oxygen collected.
Also, I always made sure that each cube was cut into a shape of a proper cube and was 6cm3. Each cube was cut into 1cmx1cmx6cm, the surface area equalling 6cm3 so that it was fair on every experiment. I made sure that I noted the amount of oxygen collected at the end of the ten minutes so that each time the results were reliable. Controlling these variables allowed me to get reliable results.
ANALYSIS: In this investigation, generally I found out that different surface areas (which is also the different amounts of catalase in the experiment) decomposed the hydrogen peroxide at different speeds producing different amounts of oxygen at ten minutes. Specifically, I realised that as I increased the surface area of the potato (which also increased the amount catalase in the experiment), more oxygen was produced at ten minutes.
Therefore, the speed at which the hydrogen peroxide is decomposed increases as the surface area increases, thus more oxygen is produced in the same time as before when there was a smaller surface area. As you can from my results and graphs in the results sections, the rate of reaction, the amount of oxygen produced was the highest when 8 cubes were added which was the most amount of surface area available in the experiment.
I have experimented three times and therefore, have drawn three graphs for each separate experiment and I have also worked out an average. Below, I have also plotted a graph comparing the three sets of results. As you can see from this graph, generally all sets of results fit into the best fit line and prove my prediction, which was as the surface area increases, which also causes the amount of catalase in the experiment to increase, the hydrogen peroxide is decomposed faster causing more oxygen to be produced after ten minutes.
As you can see from these results, sometimes the results did not prove my prediction. When I plotted the results for set 1, I realised that there were four odd results, which are circled on the graph and highlighted on the table below. These results do not fit in with the best-fit line and therefore do not back up my prediction. The graph should show that the surface area is directly proportional to the amount of oxygen collected. When 1 cube was added, less oxygen was collected than expected; therefore this result fell below my line of best fit.
This also happened when 2 cubes of potato were added. However, when I added 3 cubes, the amount of oxygen produced increased dramatically and fitted with my line of best fit. However, when I added 7 cubes I expected the amount of oxygen produced to increase but it did not. The same thing happened when I added 8 cubes. Therefore, these results did not fit in with my line of best fit. This was because previously the amount of oxygen produced was proportional to the surface area. Therefore the results for 7 and 8 cubes were odd because the amount of oxygen produced was not as proportional to the surface area as it was when 6 cubes were added.
These results from set 1 show that sometimes the amount of oxygen produced as the surface area increased was less than what I should have gained allowing the results not to fit in with the line of best fit. These results do not seem as reliable as the other two sets.
However, even in set 2, there are two results, which do not fit with the best-fit line but do support my prediction. These results are highlighted on the table below and on the graph.
These results in general do prove that when the surface area of the potato increases so does the amount of oxygen produced. However, the amount of oxygen produced should not be as much as the results indicate. Sometimes there was more oxygen produced than what was expected meaning that the results did not fit the line of best fit but generally proving my prediction. The highlighted results above and the graph show that as the surface are of the potato increased, sometimes the amount of oxygen produced increased too much which then did not fit onto the line of best fit. This also happened when 2 cubes were added to the solution. Therefore, these results did not justify my prediction.
Even in my third set of results, there were also four results, which did not fit in my line of best fit, but did prove my prediction. These results are circled on my graph and also are highlighted on the table below.
These results, which are highlighted, in general do prove my prediction, which was as the surface area increases so, does the amount of oxygen produced. However, some results like the one for 8 cubes fall below the best-fit line and does not prove my prediction. Also, the result for 6 cubes fell above the best-fit line. However, this generally proves my prediction but does not fit in with my best-fit line.
The same thing happens when 5 cubes (30cm3 of surface area) are added in set 3. Also I did not except so much oxygen to be produced however, previously less oxygen was produced so therefore now that there has been an increase in the surface area increasing the rate at which the catalase decomposes the hydrogen peroxide to produce more oxygen. Sometimes less oxygen was produced leading to unreliable results.
Out of all the results the set 2 results seemed more reliable than the rest. This is because they fitted in with the line of best fit and also proved that as the surface area of the potatoes is increased the amount of oxygen produced as increases. Also, as the surface area (the amount of catalase in the experiment) increase the rate of the reaction also increases therefore, less time is needed to get the same amount of oxygen as before.
In conclusion, as I had predicted, the amount of oxygen collected increased as the surface area of the potatoes increased. This proves my background theory and my prediction, because as the surface area of the potatoes increased, more catalase was added into the solution (because the potato contained the catalase).
Therefore, as the hydrogen peroxide was added, it attacked the surface area causing the catalase enzyme to change shape and make the substrate (the hydrogen peroxide) fit into it’s active site. Then the catalase locked the hydrogen peroxide and decomposed the hydrogen peroxide into oxygen and water, which were released. As I added more cubes of potato, which increased the surface area of the potato, which also increased the catalase in the experiment, more hydrogen peroxide was decomposed. Therefore, as the surface area increased so did the amount of oxygen made. We can say that the surface area is directly proportional to the amount of oxygen produced.
This is because from the results I plotted four graphs. Three graphs showed the three different experiments and the fourth graph shows the average results from the three experiments. As you can see, the results generally form a straight line. My best-fit line fits in many of the results proving them to be reliable. As you can see from the average results graph below, the best-fit line is generally very steep throughout the graph, proving that as you increase the surface area of the potato, the amount of oxygen collected will also increase. There were only a few odd results that did not fit my line of best fit but the majority of the results supported my prediction.
As you can see, since the best fit line gets steeper as the surface area of the potato increases, if I were to draw a gradient, no matter where I draw the gradient, the answer be the same. This means that there is no connection between the surface area and the amount of oxygen produced except that as the surface area increases so does the amount of oxygen produced. Therefore the amount of oxygen produced is directly proportional to the surface area, proving my prediction and that the more catalase there is in the solution, the faster the substrate, the hydrogen peroxide will be decomposed forming the oxygen. Therefore, as the surface area of the potatoes increase, more catalase is added to the experiment which increases the rate at which the two products are formed therefore, it takes less time to produce the same amount of oxygen when there were fewer potato cubes and when there was less catalase in the experiment. It takes a certain amount of catalase to decompose the substrate because of the active site in the catalase enzyme.
EVALUATION: In this investigation, I should have timed the experiment less than ten minutes because it took too long for me to observe and conduct each set of results for ten minutes before taking a result, wasting a lot of time. I should have timed up to six minutes each time I increased the surface area in the experiment and then taken the amount of oxygen collected so that less time was wasted in this experiment. The results would have ended plotting a similar graphs because as the surface increases so does the amount of oxygen produced. Six minutes would also have allowed the oxygen to build up giving reliable results.
There were a few odd results in this investigation because some results did not fit along my line of best fit; therefore they were not accurate or reliable. Firstly, when I took three sets of results, all three sets of results showed an odd result as shown in the results and analysis section. Also, the average results graph also showed two odd results as shown on the graph in my analysis section. The two-circled results show that the points do not fit with my line of best fit and therefore are incorrect. I controlled all the key variables such as using the same conical flask, the same tube, making sure that the measuring cylinder and beaker are clear, observing the experiment from the same distance each time and using the same amount and concentration of the hydrogen peroxide.
I also made sure that I judged how much oxygen was collected fairly for every experiment and I always timed each experiment up to 10 minutes. I also made sure that each cube was correctly cut and measured by using a ruler and I also made sure that I water, which had a pH of 7 which would not damage the enzyme and stop it from working. I kept the temperature of the experiments at room temperature so it was fair and also because the experiment was fast enough at room temperature.
However, I still gained two odd results in my average result’s graph and few other odd results in my original three sets of results. A reason for this could be that different potatoes have different enzymes. If an old potato was used, the catalase enzyme may have old tissue, which is not active as the new tissue therefore causing less hydrogen peroxide to be decomposed causing an odd result. Therefore, it does not decompose the substrate as fast as the other cube of potato, which is from a younger potato containing a younger tissue, which can react faster.
Also, another reason why I may have gained odd results is because the tube in which the oxygen was transported to the measuring cylinder was very thin. Therefore, it may have been clogged at some points with a rush of oxygen at one point therefore, some times there was not a lot of oxygen collected because the tube may have been blocked with oxygen, whereas other times there was a lot of oxygen collected because the tube was not blocked. The results below show this.
Also, it was quite difficult to set up and keep the equipment consistent whilst measuring the oxygen. It was not always easy to hold the measuring cylinder making sure that all of the key factors such as the temperature; concentration and volume of the substrate and the clarity of the measuring cylinder were controlled. However, the results were repeated three times so therefore, provided with the conditions, the results were as reliable as can be.
If I were to do this experiment again I would make sure that I do not time the experiment for so long. Instead of timing each experiment for 10 minutes I would time it for around 6 minutes because that way the experiment can be completed more quickly and enough oxygen can be produced to comment on. I would also use a thicker tube so that the oxygen released can pass through the tube faster giving accurate results because previously I found out that a thin tube could delay the oxygen being collected.
This is because the tube got clogged with the oxygen at some points enabling the oxygen to be collected fairly. Also, I would use a gas syringe rather than try collecting over water because I found the experiment harder to conduct when I had to collect the oxygen over water because there were so may things to control A gas syringe would have been easier to observe and set up.
I would also check the temperature of the experiment because at different places in the lab, the temperature varied and sometimes that allowed the catalase enzyme to work faster than other times. This is because enzyme work fastest at 370C and therefore, if the room temperature was higher at one point or one day, that may have allowed the catalase enzyme to work faster allowing more oxygen to be made.
Also, not all the faces of the potato cube were attacked by the substrate easily. This is because, one of the faces of the cube, was always facing the bottom of the conical flask, not allowing the substrate to attack it. Therefore, this was not fair and not all of the six faces were attacked by the substrate. Therefore, if I was to do this experiment again, I would stick some pins or cocktail sticks into the potato cubes so that all the faces of the cubes had a fair chance for the substrate to attack and none were facing the bottom of the conical flask. A diagram below shows this.
In future if I were to do this experiment again, I would repeat more results to get an even better average result. I would also try using other catalyst such as magnesium oxide and compare the results when the potato was the catalyst. I would also try using more hydrogen peroxide to see how increasing the substrate affects the same amount of catalyst. Finally, I would try testing with different pH’s, to see the effect it has on decomposing the substrate.