Investigate how changing the concentration of hydrogen peroxide (substrate) affects the rate of reaction of the enzyme Catalase.

Biology Coursework

Introduction:

Catalase, like all enzymes, is made up of protein molecules. It can be found in the cytoplasm of living tissue. It speeds up the decomposition of Hydrogen Peroxide, a metabolic waste product, into water and oxygen that can safely be removed from the cell.

The type of reaction involved is known as a catabolic reaction (i.e. substrate broken down.)This is simply because the substrate enters the active site and is broken down, and leaves as 2 separate products, in this case water and oxygen:

2H2O2 > 2H20 + O2

Like all enzymes, the rate at which the enzyme works is affected by many variables. These are:

Temperature: As increase in temperature and therefore heat energy reaching the enzymes and substrate molecules causes them to increase random movement. The more heat energy the more the molecules move and so collide more often. The more collisions between molecules the greater chance there is that the substrates will fit into the active site. Therefore there is an increase in rate of reaction. However if the temperature continues to increase above the optimum temperature of the enzyme (what it works best at) the polypeptide bonds that are responsible for the specific shape of the active site, start to break. This means that the specific shape of the active site starts to distort. The active sight can become permanently damaged and at this point the enzyme is said to be denatured. The substrates will no loner fit into the active sights and so a reaction does not take place.
PH Value: Enzymes also have an optimum PH value at which they work best. If the PH value of the medium is changed it can have the same affect as a temperature change. The polypeptide bonds start to break and the shape of the active sight begins to distort meaning the substrate will no longer fit. Therefore the reaction slows down and eventually stops.
Enzyme Concentration: The optimum rate of reaction is reached when all the active sites are in use. Therefore an increase in enzyme concentration when it is the limiting factor will lead to an increase in reaction rate. An increase in enzyme concentration when the amount of substrate is the limiting factor will have no affect.
Substrate Concentration: As with the enzyme concentration when all active sights of enzymes are being used, an increase in substrate concentration will have no affect on the rate of reaction, as the enzyme is unable to break down the excess of substrate. When there is an excess of enzyme molecules, an increase of substrate concentration produces a corresponding increase in the rate of reaction.
Inhibitors:

Competitive: These are molecules of similar shape to the substrate and are able to bind to the active site. They do not react with the site and leave after a time without a product forming. The rate of reaction is increased when these are present because when the inhibitor is in the active sight no substrate can enter the sight. These substrate molecules compete for the active sight and therefore the higher proportion of competitive inhibitor the slower the rate of reaction: (see diagrams below):

Non-Competitive: These are molecules, which bind to a different part of the enzyme, not the active site. They are known as allosteric molecules. They have a different shape to the normal substrate, and when they enter the alternative sight they change the shape of the active sight. This can have 2 effects:

It can stimulate the reaction if the active site becomes a better shape for the substrate to bind with. (allosteric activation).
It can also inhibit the reaction if the active site becomes an inappropriate shape that the substrate will not bond with. (allosteric inhibition).

In allosteric inhibition the substrate molecules may reach the active site before the non-competitive inhibitor does. The rate of reaction is much reduced when non-competitive inhibitors of this nature are present.

Aim:

To investigate how changing the concentration of hydrogen peroxide (substrate) affects the rate of reaction of the enzyme Catalase.
To investigate whether the chosen nitrate solution is an inhibitor and if so whether it is competitive or non-competitive.

Preliminary Work:

After doing a lesson of preliminary work I have to the following conclusions about how I will conduct my experiment:

I will use 10ml of solution each time in my experiment. As this is enough for the celery to be fully covered, it produces good results, and also does not waste the hydrogen peroxide. When I change the concentrations of the hydrogen peroxide I will still use 10ml of solution, but change the proportions of the solution. So if I wanted a concentration of 90% the original concentration of hydrogen peroxide, I would simply use 9ml of this, 1 ml of water.
I will use 10ml syringes to measure out my quantities, and 1ml syringes when I am changing the concentrations of the hydrogen peroxide, as this is far more accurate.
I will measure how much oxygen is given off in a 30 second time period using a digital stopwatch. This gives plenty of time for the oxygen to be given off and accurate results taken, and simply means I will be able to carry out more experiments. In the preliminary work I tried doing it for 1 minute but found that too much oxygen was produced, and the measuring cylinder was full.
I have decided to use 3g of celery in each experiment. I tried using 1.5g in my preliminary work, but found that little oxygen was given off, and few results could be obtained. On the other scale I tried to use 5g of celery. Not only would this just about fit in the text tube, but also so much oxygen was produced the measuring cylinder became full. I will weigh my celery using a digital weighing machine to the nearest 0.01g where possible. I will then carefully cut my celery into thin wafer –like slices. I will do the same for each piece of celery as it gives a larger surface area for the enzyme to work off, therefore giving better results. I felt that this was more accurate than using an electric mincer, as the celery came out different each time, i.e. sometimes it was finely shredded and other times it came out as a large lump of mush.
Results from preliminary work as follows:

Method:

Apparatus needed:

10ml & 1ml syringe (x2)
Boiling tubes
Plastic ice-cream tub
Delivery tube
10 cubic centimetres glass measuring cylinder
Stick of celery
Scalpel
Electric scales
Stopwatch
Test-tube rack

Using my previous knowledge and guidance from preliminary work I am going to use the following method to carry out my experiment:

Using a 10ml syringe measure out 10ml of the hydrogen peroxide supplied (making sure there are no air bubbles in the syringe) and carefully squirt into boiling tubes. Then place in a test tube rack.

This is a preview of the whole essay

Method:

Apparatus needed:

10ml & 1ml syringe (x2)
Boiling tubes
Plastic ice-cream tub
Delivery tube
10 cubic centimetres glass measuring cylinder
Stick of celery
Scalpel
Electric scales
Stopwatch
Test-tube rack

Using my previous knowledge and guidance from preliminary work I am going to use the following method to carry out my experiment:

Using a 10ml syringe measure out 10ml of the hydrogen peroxide supplied (making sure there are no air bubbles in the syringe) and carefully squirt into boiling tubes. Then place in a test tube rack.
Fill a plastic ice-cream tub with water from tap, so the water level rests about 1.5cm from the lip.
Collect an accurate 10cc glass measuring cylinder and a delivery tube, making sure that the bung at the end of the delivery tube fits over the end of the text tube.
Weigh out 3g of celery using electric scales to an accuracy of the nearest 0.01g where possible.
Cut the celery up into small wafer-like slices using a scalpel, making sure that when you repeat the experiment you cut all of the celery pieces in the same way.
Submerge the glass-measuring cylinder into the water-filled ice cream tub; making sure that it is full of water with no air bubbles. Stand in up in the corner of the ice cream tub, making sure that it does not fall over, and that it is still full of water.
Next take the delivery tube and place the glass-tube end in the ice cream tub so that the end is inside the measuring cylinder and that any air will be caught in the cylinder.
Carefully drop the celery slices into the boiling tube containing the hydrogen peroxide and quickly place the bung of the delivery tube tightly over the top of the test tube so that any air given off will travel up the delivery tube.
Using a digital clock/timer, leave the oxygen collecting in the measuring cylinder for 30seconds. When this time is up quickly remove the glass end of the delivery tube out of the water, being careful not to knock over the measuring cylinder, and consequently losing what oxygen has been collecting.
Measure how far down the tube the water has gone, and record in a suitable table.

Use the same procedure when repeating the experiment. When changing the concentrations of hydrogen peroxide repeat the procedure but do the following instead of step 1:

E.g. 5parts water: 5 parts hydrogen peroxide to get a 50% concentration of the original hydrogen peroxide=Using a 10ml syringe measure out 5ml of hydrogen peroxide and squirt into boiling tube. Using a clean syringe measure out 5ml of water and add to boiling tube. Repeat changing the amounts of water and hydrogen peroxide to get desired concentrations. When measuring small amounts such as 1ml-3ml use a 1ml syringe to get more accurate measurements.

Follow the same procedure as above when testing inhibitor using nitrate solution supplied instead of water. E.g. to get a concentration of 50% Hydrogen peroxide and 50% inhibitor = Use a 10ml syringe to measure out 5ml of hydrogen peroxide and squirt into boiling tube. Using a clean 10 ml syringe measure out 5ml of the nitrate solution and squirt into boiling tube with hydrogen peroxide.

To make my results as accurate as possible and to eliminate errors I will repeat my tests 3 times for each different test, including the control test where I will simply be using 10ml Hydrogen Peroxide. When testing the different concentrations of hydrogen peroxide and inhibitor I will try to get 3 readings for 20,30,40,50,60,70,80,90% of the original solutions. But this is very dependant on whether time allows me. If I am short for time I will have to miss out some of the concentrations. I will, where possible take readings to an accuracy of 0.1 cubic centimetres.

To make my test as reliable as possible I must use all the equipment as accurately as possible and I will do the following:

When measuring out the solutions I will always use syringes and where possible 1ml syringes instead of 10ml.
Use new syringes for measuring out each substance so that they are not contaminated.
Use the same size, but clean boiling tubes for each test.
Use the same celery where possible.
Use the same measuring cylinder to measure the oxygen given off each time. I will try to find the most accurate measuring cylinder with measurements to every 0.1 cubic centimetres.
Use the same electronic scales each time, as electronic scales are more accurate, and different scales vary in accuracy.
Use the roughly the same size delivery tube. As they are of different lengths and it takes longer for the oxygen to travel up one than it does to travel up another.

Safety: I will wear safety goggles throughout my experiment as the hydrogen peroxide we are using, although not of a strong concentration, could cause serious harm if it gets into my eyes. I will also be careful not to spill any as it could bleach skin and clothes.

Variables:

As discussed in the introduction of my essay there are many variables, which could affect the rate at which the enzyme works. I will be keeping these constant apart from the ones which I am going to investigate which are: Substrate concentration & Inhibition. To do this I will make sure of the following:

To keep handling of the text tube to a minimum as the heat from my hand could warm up the solutions by using a test-tube rack. Also to keep handling of the celery to a minimum where possible. All of the reactions will take place in the laboratory at room temperature somewhere between 230C & 250C.
The PH value will remain unchanged, and therefore has little affect in the experiment. Distilled water will be used when changing the concentrations of hydrogen peroxide where possible.
Enzyme concentration will also remain unchanged. I will be using the same amount of celery for each experiment, and will be slicing it up in the same way also.
Adding water to the hydrogen peroxide will change substrate concentration. I.e. 9mls hydrogen peroxide, 1 ml water to give a concentration of 90% the original hydrogen peroxide.
Inhibition: I will be adding nitrate to the hydrogen peroxide in the same way as with the water. This nitrate solution may be either no inhibitor, competitive or non-competitive inhibitor. Throughout the rest of the experiment this variable will remain unchanged.

Prediction:

In my experiment I will be studying how the varying concentrations of Hydrogen Peroxide affects the rate of reaction/amount of oxygen given off. I will also study how the varying concentrations of nitrate solution affect the rate of reaction/amount of oxygen given off. I can then compare the 2 sets of results. From my background knowledge gained from my A-S Course can make some predictions on what I think will happen. I think that when I add water to the hydrogen peroxide, therefore decreases the concentration the rate of reaction will fall quite a bit in comparison to when there is 100% concentration. As the concentration of hydrogen peroxide increases, the rate of reaction will also increase at a corresponding rate. This is because there are less substrate molecules to react with the enzyme and therefore less oxygen will be given off. I also think that as the concentration increases the rate at which the rate of reaction increases will decrease. This is because all of the enzyme molecules are in use and it is the enzyme concentration that is the limiting factor, not the substrate concentration.

When I use nitrate solution to vary the hydrogen concentration instead of water, one of two things could happen. If the nitrate is an inhibitor, the rate of reaction/amount of oxygen given off will decrease more dramatically as more nitrate is added compared to when water is added to the hydrogen to dilute the solution. This is because not only is the hydrogen peroxide becoming more dilute, but also an inhibitor is replacing it. Therefore the inhibitor will spend time in the active site instead of the substrate and slow down the reaction (see above). If the inhibitor is a non-competitive it will have a more dramatic affect than if it was a competitive. This is because substrate molecules compete for the active site with the competitive inhibitor. Therefore it is a simple case of who gets there first, and waiting for the other to leave The reaction can reach its maximum velocity, known as Vmax, but it takes longer than if there were no inhibitor present. In non-competitive inhibition the inhibitor attaches itself to the enzyme and changes the shape so no substrate can attach itself to the active site. Finally they leave the allosteric site, but substrate molecules do not compete for these, so they have a greater inhibitory effect. In this case the reaction will never be able to reach its maximum rate. Using my knowledge from the course I predict that I should get a graph that looks like the following:

Results:

Amount of oxygen given off at varying hydrogen peroxide concentrations using water to dilute the solution.

Amount of oxygen given off at varying hydrogen peroxide

Concentrations using an unknown nitrate to dilute the solution

In my results all of the concentrations are based on the fact that the original Hydrogen Peroxide Concentration was 3%. So 90% of that is actually 2.7% conc., etc…

Conclusion:

Based on the results I obtained and the graphs I have produced I deduct some clear conclusions about how the enzyme Catalase works.

In my first experiment I investigated how changing substrate concentration would have an affect on the rate of reaction. From my graph it is clear that as substrate concentration increases, the rate of reaction increases at a corresponding rate. As substrate concentrations are low the rate of reaction is low simply because there are few substrate molecules bonding to the active sites of the enzymes. The substrate concentration is the limiting factor. As more substrate concentration continues to increase so does the rate of reaction, but at a slower rate compared to the beginning of the graph. This is because nearly all of the active sites of the enzymes are in use; therefore adding more substrate will only make a marginal difference. It is at its maximum rate and the enzyme concentration is now the limiting factor. E.g. If I added the control results (100% concentration) onto my graph, they were nearly identical to my 90% concentration results. 90% concentration average: 3.6, 100% concentration average: 3.7. Referring back to my prediction this is what I thought would happen. The only thing that did not work as I predicted was the 100% concentration results. There was very little difference between these results and the 90%, 80% results (see above). This is because the enzyme concentration becomes the limiting factor, not the substrate concentration. If there were an excess of Catalase, there would have been a larger drop from the control results.

In my second experiment I investigated what affect adding an unknown nitrate to the hydrogen peroxide to dilute the solution instead of water, would have on the rate of reaction. As anticipated the nitrate was an inhibitor. The rate of reaction, slowed down more rapidly as more nitrate was added than it did in comparison to when more water was added. Looking at the graph comparing the 2 experiments it is clear to see that far less oxygen is produced when the nitrate is present. This is because the solution was not only getting more dilute, but also more inhibitor molecules were being added. The inhibitor molecules were binding with the active site of the enzymes so that the substrate molecules could not get in and react and therefore very little oxygen was produced. The graph of nitrate solution concentration shows typical features of a competitive enzyme other than it never reaches maximum velocity. Using results from my first experiment this is roughly around 4cc of oxygen produced (in the time limit, if I would have left it for 10mins, it would have been much higher). The maximum produced in my second experiment was 2.6. But this can be explained by how long the enzyme had to react. It takes much longer to reach maximum velocity with a competitive inhibitor present and therefore I believe that if I had left the experiment going the same levels of oxygen would have been produced as in the first experiment. These findings follow my prediction, other than the 90% readings. According to my prediction, even with an inhibitor present the 90% concentration readings should have been higher than the 80% concentration readings. As stated above this could be because that the enzyme concentration is the limiting factor. But the figures actually drop from 80% concentration to 90%. This could be explained because there are so many substrate molecules competing for the active sites of the enzyme molecules that they actually start to block the site. Therefore there is a slight decrease in oxygen produced.

I believe that the inhibitor in the nitrate solution was a competitive one. This because if it was a non-competitive the reaction rate would have risen as concentration rises only for a short while, and then it would have remained constant at the same level, reaching its maximum velocity which is usually about half that of 100% concentration would achieve. It is difficult to tell which inhibitor it is simply because of the few results available. If the reaction was left for a longer time, I believe that it would have reached the same maximum velocity. Then I could be positive that the inhibitor was competitive and not non-competitive. (see evaluation)

Evaluation:

My experiment went very well and I got a good set of results, which I could base my conclusion on.

The accuracy and observations of my experiment, I felt was consistently good. All of my readings were to a degree of accuracy that I felt was adequate. All weight measurements of the celery were 3g, +/- 0.15g. I felt this was accurate enough for my purposes; a size difference of 0.3g at the most would not have made considerable difference. My measurements of oxygen given off were all to 0.1 cc and throughout my experiment I used the most accurate glass measuring cylinder available.

I did not have many anomalous results. But both experiment numbers 2, in the two variable tests (shown in red in the results table) seem to be of a generally lower value than results 1&3. Both these experiments were carried out on the same day. There could have been a number of reasons why these anomalous results occurred:

My teacher supplied the celery I used for my experiments. It had been purchased from various supermarkets in the area. It is possible that on the day I did the experiments with the anomalous results the celery had been purchased from a different supermarket, or simply that it had less enzyme in. If the celery was not fresh and had been opened the day before this could have also affected the concentration of enzyme in the celery.
It could have also been that I used different parts of the celery than in experiments 1&2. There could have been a higher concentration of enzyme in the top, or bottom part of the celery and vice versa. Unfortunately due to the shortage of celery I was unable to pick and choose what I had.
I could have used a different size delivery tube, as again it was difficult to try and use the same equipment and it was based on first come, first serve. If the delivery tube was longer than the one I used in experiments 1&3, it would have taken longer for the oxygen to travel up the deliver tube, pushing out the water as it goes, and therefore less would have been collected in the measuring cylinder.
It could also have the Hydrogen peroxide supplied on that day for use. It could have been of a different concentration, or more likely it could have been contaminated from previous experiments.

After carrying out the experiment it became obvious that there were limitations in some of the experimental procedures:

The use of celery as a source of enzyme was not very accurate, and the amount of enzyme present in the celery could not be measured.
The fact that each time I weighed the celery it was cut by hand into what I thought was wafer thin pieces was also a source of inaccuracy. As each time they would have been different sizes and therefore would have had bigger/smaller surface area for the enzyme to work off.
The experiment was carried out over a series of lessons and the celery used was different each time. Sometimes the celery may have more or less Catalase in it. Also as mentioned above the shortage of celery meant that you could not pick and choose what parts you wanted to use. So sometimes it would be the end of the celery, and sometimes the middle and there could have been more or less Catalase in the different parts.
The time delay between trying to place the celery in the Catalase, making sure that it was all covered in hydrogen peroxide and reacting, and then placing the bung of the delivery tube tightly over the top to collect the oxygen was also a source of inaccuracy. Sometimes it would take longer to do this, and therefore more oxygen was lost as a result.
The water that collects in the delivery tube when it is submerged under the water has to be pushed out first by the oxygen before it can start collecting in the measuring cylinder.
The time delay between setting up the apparatus and starting the timer for 30seconds meant that sometimes it would have more time to react if it took longer to attach the bung to the test tube.
The fact that I only had time to take readings for 30, 40, 50, 70, 80, 90, &100 % of the original concentration of hydrogen peroxide left a gap in my results.
Also because I only took readings for 30seconds this meant that I could never be too sure what the reactions maximum velocity was, and whether this would be reached with the nitrate present, as it would have taken longer.

To improve my results and procedure I could have done the following:

Taken readings from 5% up to 100% increasing the concentration each time by 5 only, to give a larger more accurate range of results with which to plot my graph and make any conclusions.
I could have crushed and then drained the celery, so that I could use just the juices of the celery as a source of enzyme. This would have been more accurate than simply using blocks of celery because the solution could have been mixed up, and therefore the concentration of enzyme used it each experiment would have been the same.

It would have been even more accurate to use a known molar solution as a source of enzyme.

To minimise the errors of the timing between placing the bung over the end of the boiling tube to start collecting the oxygen and starting the timer could have been minimised by 2 people doing the experiment. That way one person could have placed the celery in the text tube, the other could have put the bung on the end of the test tube, and the first person starts the timer.
I could have also, if more time was available taking readings using a larger measuring cylinder, for a longer time, to see what the maximum velocity was, and whether the experiment using the nitrate solution would have reached it.

I believe the fact that I did not get results for all the concentrations going up in 10%’s, has not affected my overall conclusion. But I do believe that because I did not know what the maximum velocity was-this has affected my overall conclusion. If I knew what the maximum velocity was I could have made more accurate decisions on what the nitrate solution was. If I had tested using the hydrogen peroxide, and added 3g of celery; using a larger measuring cylinder to find the Vmax, I could have done the same with the experiments using the nitrate solution. If they had reached Vmax, but taken longer to do so, I could then be definite that the nitrate solution was a competitive one. But if they did not reach Vmax, then I would have known that it was a non-competitive one. This can be shown in the earlier drawn graph of predicted results.

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