Diagram of the lock and key theory.
4.Describe in detail, how the experiment will be carried. Someone else following your plans should be able to carry it out exactly as you would do. Include an apparatus list.
The experiment will be carried out by one or two person (yet unknown).
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We will be adding 15 cm3 of Hydrogen Peroxide into the flat bottomed tube (specimen tube)
- Next we will be cutting out small circles of filter paper, the filter paper will be all the same size because we will be using the same whole puncher to make these holes.
- One circle of the filter paper is dipped into the catalase solution, which is done using forceps.
- The filter paper is then held over the catalase solution for a short while, so that any extra catalase solution is dropped off.
- The Filter paper that was soaked into the catalase solution is then put into the Hydrogen peroxide solution at a angle that is best suitable e.g. Flat or side on. But I will use the flat side to drop the filter paper into the hydrogen peroxide solution.
- The filter paper will sink to the bottom of the hydrogen peroxide solution.
- As it sinks the catalase comes in contact with the hydrogen peroxide breaking the substrate down into water and oxygen.
- The oxygen that is produced will be around the filter paper, this will make the filter paper float back to the top.
- Now if the filter paper always has the same area, then it will take the same amount of oxygen gas to make it rise back up to the top.
- During the experiment we measure the amount of time it takes the filter paper to sink down to the bottom and float back up to the top.
Apparatus List:
Solution of Hydrogen Peroxide
Solution of the Catalase
Specimen Tube
Forceps
Stopwatch
Filter paper
Whole Puncher
5.Which factors are you keeping constant so as to make the test fair?
Below are the factors, which are kept constant.
- Concentration of Hydrogen peroxide - use 1% concentration
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Concentration of Catalase - use 50eu/cm3 of Catalase
- Size of the paper (surface area)
- Volume of Catalase – Use the same size and the same thickness of the filter paper
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Volume of Hydrogen peroxide – Measure 10cm3 of H2O2
- Put the paper in at the same angle – Using Forceps
- Accurate timing – Use Stopwatch
- pH - Use buffers which keep the pH constant
6. If there are any factors that you would like to keep constant but cannot then explain.
There are no particular factors that I would like to keep constant but can’t because the factor that I am investigating in this experiment is only the temperature. I am keeping everything constant except for the temperature.
7. How many results are you planning to take and over what range?
I am planning to take between 5-10 readings because this will make results more reliable. I will be having a range of temperature from 10°c to 60°c each increasing by 10°c each time. These are the ranges of temperature that I will be investigating: 10°c
20°c
30°c
40°c
50°c
60°c
I will be using these ranges of temperatures to plan and gain the results.
If the enzyme doesn’t denature at these temperatures then I am intending to use higher temperatures to find out at which temperature the enzyme will denature. For example I could use temperatures such as 70°c, 80°c or maybe 90°c to find out at which temperatures the enzyme denatures.
8. How do you think you should check the accuracy and reliability of your results?
To get reliable results it’s best to repeat the experiment several times for example I will be repeating the experiment 6 times; this will give you reliable results because you have done the same test on the same temperature more than once, which allows you to gain some accurate results of the experiment. Another way you can gain accurate results is by comparing your results to another set of results and check if they come out the same. If these results do not come out the same then there are some anomalous results, which were gained during the experiment.
9. What safety Precautions should you take?
These are the safety precautions that you should take when doing a Practical experiment.
- Hydrogen peroxide is a corrosive solution, use gloves and handle it with care.
- If spilled onto your skin then rinse it off with loads of water
- When doing practical experiments always wear Lab Coat and Goggles
- Hydrogen peroxide can also bleach your clothes and skin
- Catalase is a protein and it can produce some allergic reactions, if spilled rinse it off with loads of water.
- When filling the specimen tube with the hydrogen peroxide I will stand it in a beaker for safety so that it does not spill very easily.
Preliminary Work
Before hand doing the actual experiment, I have done some preliminary work to show how the enzyme catalase reacts with the hydrogen peroxide. The first preliminary work that I done was to show if catalase was present in living tissues such as liver, and kidney. Below are the results that I have collected from the preliminary experiment.
The amount of catalase varies in different tissues. The liver, kidney, carrot and the yeast all contained catalase, but some tissues contained more catalase than others e.g. the liver. The potato tissue did not contain much of the catalase because it did not light up the splint. During the reaction some froths was produced. The whole point I lighted up the splint was to check if the oxygen was present and can it be detected by relighting up the splint.
I suggest that I could do further investigation on the enzyme catalase to find out the optimum temperature in different species. For example I could investigate the different tissue e.g. liver and try to find out the optimum temperature of the catalase enzyme within a mammalian body. However there are many different enzymes, which are always present in a mammals and in our body, all these enzymes have a different chemical properties. However this investigation will very much link to the investigation that I am doing, because in my investigation I am trying to find out the optimum temperature of the enzyme catalase, when it reacts with hydrogen peroxide forming the products water and oxygen. I am using different range of temperatures from 10°c-60°c to find out the optimum temperature of the enzyme catalase. Since all these tissues contain the same type of enzyme catalase I will investigate to find out the optimum temperature in these species.
Another preliminary experiment that I did was to check how the strength of the catalase solution used, affects the rate at which hydrogen peroxide breaks down into water and oxygen. I did this experiment to find out the best strength of the catalase solution so that I could use this strength in my temperature investigation. What I expect from the experiment was the higher the concentration of the catalase solution used, the faster the hydrogen peroxide will be broken down as long as there is enough hydrogen peroxide present.
Reactions occur when reacting molecules collide with enough energy. The more molecules there are of the enzyme catalase, the higher the chances of collisions with hydrogen peroxide molecules and so there is an increase chance of the oxygen being produced. By doing the preliminary test this gave me a brief idea of how I can investigate the temperature when using these two same substances. The preliminary test helped me work out the method of my own investigation with the temperature. For this experiment this is the method that I carried.
- Filled the specimen tube with hydrogen peroxide solution, stand it in a beaker for safety reasons.
- I cut out identical filter paper discs using a hole punch
- I obtain different strength of the catalase solution in a beaker
- I used forceps to dip the filter paper in the catalase solution.
- I then lowered the filter paper flat onto the top of the hydrogen peroxide solution
- I measured the time it takes for the filter paper to sink to the bottom and rise back up to the top
- Repeated this six times
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Repeated the investigation using different strength of the catalase solution e.g. 20, 40, 60, 80, 100eu/cm3
- I worked out the average and the rate of reaction for each of the different strength of catalase solution.
Below are the results I obtained from the experiment.
From the results that I have collected, it shows that the catalase enzyme with the 100eu/cm3 had the fastest reaction with the hydrogen peroxide. The rate at which the catalase enzyme broke down the hydrogen peroxide into oxygen and water was very quick. This shows that using the catalase enzyme with the strength of 100eu/cm3 it will break down the hydrogen peroxide molecules very quickly forming the two products water and oxygen.
I think that if I was to do the 100 eu/cm3 I would gain inaccurate results because if I look at the average for each strength it shows that the 100 eu/cm3 is the fastest, but when I do my experiment it will be hard to record the results down because if you look at the range of temperature that I am using between 40°c-60°c the reaction will be very fast and it will be very difficult to record the results. If I increase the temperature each time by 10°c it will half the time it takes for the reaction to finish. For example at 20°c if the temperature is 13s then it will half the time it takes when it’s at 30°c 6.5s. So therefore if I was to record the temperature at 60°c it would be approximately 0.82s. If the enzyme catalase does not denature at these temperatures then this suggests that I will have to investigate further on by using higher temperatures to find out at which temperature the enzyme will denature. So therefore this suggests that I would have to use temperatures such as 70°c or 80°c to investigate this problem. So it would be very difficult to record these results. If I used the catalase enzyme with a strength of 60 eu/cm3 it will break down the hydrogen peroxide quiet quickly forming the products water and oxygen. I could record the results very easily and this will give me some accurate results.
RESULTS TABLE
Below is the result table that I have produced to show all my readings from the experiment.
ANALYSIS
As from the results table above it shows that as the average time for each of the temperature decreases, the rate at which these reaction occurs increases. So therefore a simple pattern in the results table could be explained as being, the average time decreases while the rate of reaction increases. The reason why the average time are decreasing is because the temperatures are getting higher so therefore the time it takes for the filter paper to sink to the bottom and rise back up to the top happens very fast due to the change in temperature, and the reason why the rate of reaction increases is because, the temperatures are getting higher each time, and the rate at which these reactions occur are getting faster, so therefore if the reactions are occurring faster this suggests that there must be a lot of collisions between the catalase and the hydrogen peroxide particles.
As referring to the graph I can say that as the temperature increases from 10°c to 60°c, the rate at which the reaction the reaction occurs also increases. Above 60°c the rate of reaction gradually decreases. The reason why the rate of reaction increases is due to the amount of collisions that is taking place. A collision occurs when two or more particles hit or collide against each other. The rate of reaction between any substances depends upon two factors:
- The rate at which the particles frequently collide – the faster the particles move about the faster they will collide with other molecules and this will make the reaction occur faster.
- How much energy the particles have when they collide – energy is needed to break existing bonds in the reacting particles and so if they do not have enough energy then they will not react when they collide.
The harder and the more often the particles collide the faster the rate of reaction. For example, in this experiment the catalase reacted with the hydrogen peroxide, the reaction takes place only when the two particles collide against each other. At 10°c the collisions between the catalase particles and the hydrogen peroxide particles occurred at a very slow rate. The two different particles collided against each other transferring a small amount of energy between them. As the particles collide more often and faster they will gradually increase the rate of reaction.
Above is the diagram for the high temperature, as we can see that at a high temperature the particles collide more often at a faster rate, this is because of kinetic energy, the heat supplies kinetic energy to the reacting molecules giving them energy to collide faster. So the reason why the rate of reaction increased is because of the number of particles that were taking place. So therefore at a high temperature I will expect the rate of reaction to increase because there will be a high number of particles present and there will be a lot of collisions between each particles speeding up the rate of reaction.
As referring back to my planning I have stated that as I change or increase the temperature, it would speed up the rate of the chemical reaction. When I heated up the hydrogen peroxide solution, the heat supplies kinetic energy to the reacting molecules that are present in the hydrogen peroxide solution. This kinetic energy allows the molecules to move rapidly and collide at a faster rate. If there is more kinetic energy inside the molecules then there is an increased chance of the molecules colliding into each other. As the particles collide they transfer energy between the molecules. As I increase the temperature the kinetic energy was gradually increasing. As the kinetic energy increases, the particles collide at a faster rate with a stronger force. This increases the rate of reaction because the particles have energy to react. However heat energy also increases the vibration of the atoms, which make up the enzyme molecule. If the vibration becomes too violent then the chemical bonds in the enzyme break. As the vibration becomes violent and the chemical bonds break up this means that the enzyme has denatured and will change shape at a high temperature. As the chemical bonds in the enzyme break, then this suggests that the enzyme has changed shape so therefore the key will not fit into the lock, in other words the substrate hydrogen peroxide will not fit into the enzyme catalase. For example from 60°c onwards the rate of reaction gradually decreased, because the enzyme are slowly changing shape because it is coming into a higher temperature.
This diagram above shows that as the enzyme catalase becomes into the higher temperature, it will change shape and become denatured, and the substrate hydrogen peroxide will not fit into the enzyme catalase due to the enzyme changing shape. Below is the list of how the enzyme and the substrate worked in this experiment.
- The enzyme catalase does only one job and this is to react with a specific substrate hydrogen peroxide. The shape of the enzyme catalase is specific and only the hydrogen peroxide substrate can fit into this enzyme and not any other.
- Once the catalase enzyme reacted with the hydrogen peroxide substrate, it releases the product oxygen and water. This process is repeated over again with the same enzyme molecule, which is the catalase.
- The enzyme catalase does one of the two things, it either splits a complex molecule into two simple ones or it either joins the two complex substrate into a one large molecule. In this experiment I found out that the catalase enzyme breaks the substrate hydrogen peroxide down into two simple ones, which form water and oxygen.
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The part of the enzyme catalase that reacts with the substrate hydrogen peroxide is known as the active centre. It was very important that this part of the enzyme does not change shape. As I know from the experiment that the active centre of the enzyme catalase did change shape, the reason why this part of the enzyme changed shape is because it was at a high temperature, and this is why above 60°c onwards the rate of reaction gradually decreased is because the enzyme molecule has changed shape and denatured at 80°c, which shows that the lock will not fit into the key in other words the substrate hydrogen peroxide will not fit into the enzyme catalase.
Due to all this, the reaction was also slowed down because of the amount of particles that were present in the reaction, the reason why the reaction started to decrease is because the enzyme changed shape and denatured at a high temperature.
Looking at the graph, I have found out the optimum temperature of the enzyme catalase. The optimum temperature where the enzyme works suitably is at 60°c. The reason why this is the optimum temperature is because the rate of reaction occurs at a faster rate at this temperature. It occurs at a faster rate because there are more particles of the hydrogen peroxide, which are heated at a very high temperature, which suggests that these particles have a high kinetic energy. As these two different particles collide the catalase particles and the hydrogen peroxide particles they transfer a high energy between them, which also suggests that they collide with a strong force. As this is the optimum temperature there were more collisions between the particles, which speeded up the rate of reaction.
From this experiment I have also found out the temperature where the enzyme catalase becomes completely denatured. The temperature where the enzyme becomes denatured is at 80°c. The reason why this is at 80°c is because, when I done this temperature I found out that the filter paper took very, very long to rise back up to the top, and the rate of reaction for this temperature is approximately zero. From the graph that I have produced I can see that this is where the enzyme becomes denatured because above 60°c onwards the rate of reaction was gradually decreasing, this occurred because the shape of the enzyme has changed, the reason why the enzyme changed shape is because the active site of the enzyme has got damaged at a high temperature.
From the experiment I can see that the product oxygen and water was formed. How I know that the product oxygen was formed is because when I dipped the filter paper into the catalase solution first, the catalase particles will be present in the filter paper, these particles inside the filter paper move about randomly colliding against each other transferring energy. Once the filter is dipped into the solution it was then taken out and left over the top of the catalase solution for a while so that the extra catalase is dropped off. The filter paper is then put into the hydrogen peroxide solution, as the catalase solution comes in contact with the hydrogen peroxide both of the particles from the different solution combine together colliding against each other transferring heat energy. Once the catalase solution is in contact with the hydrogen peroxide the gas oxygen is formed. I can tell that the oxygen was formed because bubbles were given off when the catalase reacted with the hydrogen peroxide. As I drop the filter paper into the hydrogen peroxide solution, the filter paper sinks to the bottom of the hydrogen peroxide forming oxygen. The oxygen that is produced will be round the filter, it’s the gas oxygen that makes the filter paper rise back up to the top. Now if the filter paper always has the same area, then it will take the same amount of oxygen gas to make it rise back up to the top.
Looking back at my planning I predicted that increasing the temperature would speed up the rate of chemical reaction of the enzyme catalase. The higher the temperature the faster the reaction would occur. This supports the prediction that I made here, because from the results that I have gathered I have found out that if I increase the temperature then there will be faster rate of reaction. So therefore the prediction that I have made in the planning links to my conclusion because, from doing the experiment I have found out that as I gradually increase the temperature the rate at which the reaction occurs between the hydrogen peroxide solution and the catalase enzyme is very fast. The reason why the reaction is very fast is due to the amount of particles that are taking place when they react. The particles that are present in a high temperature are highly supplied with kinetic energy when they are heated up. The particles will move about more quickly and they will collide more frequently and more often with other molecules. The particles in the reaction will collide against each other at a strong force with a high amount of kinetic energy. If the particles collide more often and faster they will gradually increase the rate of reaction. Increasing the temperature of the enzyme catalase gave me the optimum temperature where the enzyme works suitably. This is the temperature where the rate of reaction was at its greatest, at this temperature the reaction occurred very fast between the catalase enzyme and the hydrogen peroxide; the particles that were present were colliding at its greatest force with enough kinetic energy. Also by having an increased temperature I have found out the temperature where the enzyme catalase denatures. The temperature where the enzyme denatured is at 80°c so therefore this shows that I had to do further investigation on the enzyme catalase by using higher temperatures to find out at which temperature the enzyme denatures. This occurred because; above 60°c onwards the rate of reaction was gradually decreasing due to the enzyme changing shape at a high temperature. The prediction that I made about increasing or having a high temperature in the planning does support my conclusions that I have made here.
EVALUATION
The practical experiment that I did worked successfully, because what I was expecting from the experiment was that as I increases the temperature of the enzyme catalase, the rate at which the reaction occurs should also increase. Due to this practical experiment, my aim was also to find out the optimum temperature of the enzyme catalase and also to find out the temperature where the enzyme becomes denatured. From the practical experiment that I have done I have gained all the results for all the different temperature, I have also found out the optimum temperature where the enzyme catalase works suitably and also found out the temperature where the enzyme catalase denatures.
From the practical experiment that I have done, I think that I have got fairly reliable results. I think that my results are reliable because I have used several filter paper discs to repeat the same temperature over and over again, in other words I did six runs on the same temperature to check if I get the same timings for that temperature. The reason why I think that my results are reliable is because I have repeated the same temperature by doing six different runs by using a different and a fresh filter paper for each of the runs. By doing these repeated runs for each of the temperatures it shows that the timings that I have gained are fairly close to each other which suggests that the results that I have gained could also be accurate. But then again I can say that the results that I have gained may not be accurate, the reason why I think this is because if I look back at the results table I can see that most of the timings are overlapping each other. For example at 10°c I have 3 number 8s which are overlapping each other in the first temperature, this also occurs at the second temperature. If I look at the results table, for every temperature that I have done, I have gained several overlaps, so there is not a lot of difference between the timings. So the overlaps that have occurred in the temperatures may have an affect on the overall practical experiment, which may lead to inaccurate and non-reliable results. So therefore the evidence that I have obtained from the practical experiment allowed me to come to a conclusion about whether or not my results are reliable or accurate. The conclusion that I have come to is that I am still unsure whether the results that I have gained are accurate or reliable. The thing that I should do to enable me to come to a more sure conclusion is to repeat the whole experiment again, because this may improve the results that I have gained from the previous experiment and give me some accurate results.
As I plotted the graph with the rate of reaction against the temperature, I have found out that there was one anomalous result that occurred in the graph. The reason why this was the anomalous result is because it did not fit into the graph when I did the curve of best fit. When I did the curve of best fit I ignored this point and then continued to join up the other points. This anomalous result occurred at 20°c. So therefore the point was actually quite of from the actual curve of best fit. The reason why this anomalous result could have occurred may be due to the errors if there were any in the practical experiment. The possible reasons could be that this error occurred due to my physical reaction on the timing.
Looking back at the results, the times that I have gained for the slow temperatures e.g. from 10°c to 40°c seem to be looking quiet fast. The timings are very fast because there might have been some catalase present in the hydrogen peroxide solution. When I did the first run the solution that I used was all fine without any trace of catalase, but when I did the second run the time that I gained was very fast this is probably due to some of the traces of catalase left in the hydrogen peroxide solution, if there was some traces of catalase left in the hydrogen peroxide solution then this suggests that as the filter paper was dropped inside the hydrogen peroxide solution the products oxygen was formed very quickly due to the traces of catalase inside the hydrogen peroxide. The reason why there maybe some traces of catalase inside the hydrogen peroxide solution is because I did not use a fresh hydrogen peroxide every time for the different temperatures that I did, so therefore this may have effected the overall experiment.
Referring back at my planning I have predicted that if I increase the temperature each time by 10°c it will half the time for the reaction to finish in the next temperature for example at 10°c the time was 8.2seconds so therefore at 20°c the timing should have been 4.1seconds a half of the timing that occurred at 10°c, but this did not happen. The times that I have gained from the practical experiment did not fit into the actual pattern that I have predicted in the planning, so therefore I think that the times that I have gained from the practical experiment maybe wrong, to improve these results I should do the whole experiment again.
The likely sources of errors in the experiment could have been using the equipments. We might have gained errors in the practical experiment due to the equipments that we used, in other words the equipments that we used were they accurate enough to measure the solutions of the catalase and the hydrogen peroxide. The errors can also occur due to our ability of using these sorts of equipments, because if we don’t know how to use some equipment then we may gain errors in the experiments because of our ability of using them. We must of the ability of using these sorts of equipments in an appropriate experiment. The likely errors that could have occurred in the experiment is that did we measure out the solution of hydrogen peroxide nearest to the top of the beaker, in other words was it almost full? To reduce these sorts of errors we could use more accurate equipments to measure out the solution of hydrogen peroxide for example we could use a burette or a graduated pipette to measure out the hydrogen peroxide or any other solutions.
If I was to repeat the whole experiment again, this might enable me to gain better results from the different temperature. As I have stated above that I used the same hydrogen peroxide solution for every temperature that I did, so therefore there may have been some traces of catalase inside the hydrogen peroxide solution when I did different runs on the different temperatures, which may have effected the results so therefore to improve this I will repeat the whole experiment by using a fresh solution of hydrogen peroxide for every different temperatures that I do. By doing this I will hopefully gain some better results. If I repeated the experiment I could use more accurate equipments such as thermostatic water bath to keep the temperature constant, so therefore I will not have to wait for the temperature to come down to the right degrees each time when I do a different temperature, and also use a burette or a graduated pipette to measure out the solutions of hydrogen peroxide.