In this experiment I am going to use the enzyme catalase. Catalase is the fastest known enzyme and has a turnover number of 6 million. Catalase is specific to reacting with Hydrogen Peroxide. It decomposes the poisonous hydrogen peroxide into water and oxygen, which are harmless.
catalase + H₂O₂ → catalase H₂O + O₂
The catalase is the same at the end of the reaction as it was at the beginning.
Plan
Using this background knowledge, I am going to use 5 different concentrations of hydrogen peroxide (substrate) for the catalase to decompose. I will then measure the amount of oxygen given off by each reaction for a certain amount of time. I am measuring the oxygen because it is a product of this reaction and it is easier to see how much oxygen has been produced than how much water has been produced. The catalase will come from a potato. Having done this I will work out the rate of reaction for each reaction by: dividing the amount of oxygen collected by the time it took to collect it
Factors I will keep the same to make it a fair test
- The concentration of the enzyme
- The temperature
- Same cork borer so the potato is the same size for each experiment
- Same apparatus
I am keeping the above factors the same and the only factor I’m changing is the concentration of the substrate.
I will repeat all my results once to ensure reliability. I will be doing the experiment on one day because I am dealing with living cells and if I left the potato to do repeats on another day, they would be inaccurate because the potato would be rotten. The results would also be inaccurate because I would be using different apparatus and the temperature would be different.
I am going to use the following apparatus and set it up as shown:
- 5 boiling tubes
- Test Tube rack
- Potato
- Cork Borer
- Scalpel
- Ruler
- White tile
- Bung and tube
- Delivery tube
- Tub
- 2cm syringe
- 10cm³ measuring cylinder
- 50cm³ hydrogen peroxide
- Stopclock
- Marker pen
The 5 boiling tubes are for each experiment with the 5 different concentrations and are where the reactions will take place. The test tube rack is to hold these tubes. The potato is where I will get the catalase from and the cork borer is to cut the potato into tubers, which I will then make the same length using the scalpel and ruler. I am using the white tile to cut the potatoes on so as not to make marks on the table and so it’s easy to get rid of the potato peelings.
The bung and is to seal each boiling tube so oxygen doesn’t escape and the tube is then connected to the delivery tube which the oxygen will travel along during the reaction. The tub will hold water and will be where the measuring cylinder is placed (full of water) to collect the oxygen given off from the delivery tube, which will also be in the tub. I am using a 10cm³ measuring cylinder because it is a sufficient size to collect the oxygen. I am using a syringe to measure and transfer the hydrogen peroxide from its container because it’s more accurate and also prevents spillage, which would be dangerous because H₂O₂ is an irritant. I am using 50.0cm³ hydrogen peroxide because it’s specific to catalase and is what the catalase will decompose. I am using this volume because it’s what we are given but I will dilute it to get the desired concentrations. The stop clock is to time for a certain time while the oxygen is being produced. The marker pen is to label the boiling tubes.
To make my method safe:
- Hydrogen peroxide is an irritant so wear goggles while using and wash hands after use
- Be careful with cork borer and scalpel while cutting potato
- Wipe any spills to keep surface clean and so potato doesn’t react before experiment has started
- Wear hair tied back
- Keep boiling tubes in test tube rack so they don’t smash- preventing injury to others and myself.
I am going to measure the volume of water and hydrogen peroxide to 1 decimal place and the concentration of the hydrogen peroxide to the nearest 5 vol. I am going to measure the volume of oxygen produced and the rate of reaction to 2 decimal places for maximum accuracy. I will measure the length of the potato to the nearest 0.1cm and I will measure the time at exactly 3 minutes to keep the time constant and so the test is fair.
Prediction
Using my background knowledge, I predict that as the concentration of the hydrogen peroxide increases the rate of reaction will also increase. I think this will happen because if you increase the concentration of the substrate there are more molecules per cm³ and they are more densely packed. Therefore there is a much higher chance of the molecules colliding and reacting. I predict that the rate of reaction will reach a plateau because the higher the concentration the more saturated the active sites of the enzymes are and so the enzymes are working as hard and as fast as they can to break the substrate molecules into products. Therefore the enzymes can’t work any harder and so the rate of reaction won’t increase any more and will reach a plateau. I will display my results on a graph:
Rate of reaction cm³/ min
Concentration of hydrogen peroxide/ vol
I expect the graph to look like this because it shows the rate of reaction increasing as the concentration of hydrogen peroxide increases, as predicted.
The graph levels off because the enzymes are overworked in higher concentrations, which is what I’m predicting. I also expect the line of best fit on the graph to go through the origin because when there’s no hydrogen peroxide; there will be no rate of reaction.
To help plan my method, I carried out some preliminary experiments to see if the experiment worked best with a whole tuber or one, which was cut up into pieces. To do this I:
- Cut 2 pieces of potato using the cork borer.
- Used the scalpel to make them 3cm and cut one tuber into 3 1cm pieces.
- Used the syringe to put 20cm³ of hydrogen peroxide in a boiling tube and placed tube in rack.
- Attached bung and tube to delivery tube and filled the tub with cold water. Put delivery tube in water
- Then filled the measuring cylinder with water and turned it upside-down, trying not to get air bubbles inside it.
- Put potato inside tube with hydrogen peroxide and placed bung in boiling tube
- Quickly put delivery tube inside measuring cylinder and started stop clock
- Stopped the stop clock after 3 minutes and recorded the amount of oxygen produced, by looking at the reading where water level had gone down in the cylinder.
- Worked out rate of reaction
- Repeated steps 3-9 with the tuber which had been cut into 3 1cm pieces
Results from prelims
The results in black are for the whole tuber and the results in red are for the tuber that was cut into pieces. The tuber that was cut into pieces produced more oxygen and so I have decided to cut the tuber into 3 1cm pieces for the experiment.
Method
I am going to do 5 experiments and repeat them once for reliability. I am going to use the following concentrations of hydrogen peroxide: 20, 15, 10, 5 and 0 and the unit used is vol. I will dilute the hydrogen peroxide with water to get these concentrations:
- Label 5 boiling tubes with their concentration, using marker pen
- Cut 5 potato tubers using cork borer. Cut each tuber into 3 1cm pieces using ruler and scalpel.
- Dilute hydrogen peroxide into desired concentration and put solution into designated boiling tube.
- Do first experiment: repeat steps 4-9 from prelims.
- Repeat steps 4-9 from prelims for remaining 4 experiments and do repeats.
Results
Table to show the amount of oxygen produced and the rates of reaction when varying the concentration of hydrogen peroxide
The result in black is the original reading and the result in red is the repeat reading.
I worked out the rate of reaction by: Amount of product (cm³)
Time taken (minute)
For example the rate of reaction for 20vol would be: 7.35 = 2.45cm³/min
3
Conclusion ( see graph on p )
From my graph, I can conclude that as the concentration of hydrogen peroxide increases the rate of reaction increases, which is what I predicted. This is because as the concentration of the hydrogen peroxide increased, the more densely packed the molecules were per cm³ and so there was more chance of the molecules colliding and reacting. Therefore the rate of reaction also increased. The line of best fit goes through the origin, which is also what I predicted. However, in my prediction I said that the line of best fit would reach a plateau because the active site of the enzyme becomes saturated in higher concentrations so as soon as one substrate molecule is turned into products, the enzyme is immediately taken up by another substrate molecule. Therefore the enzyme is working as hard as it can so the rate of reaction can’t increase anymore. Instead my graph shows a straight line, and a straight line directs direct proportion. By direct proportion, I mean that if the concentration of the hydrogen peroxide doubles, the rate of reaction doubles. If this is correct then when there is 10vol of hydrogen peroxide on my graph, the rate of reaction should be 1.27 (I got this by seeing what the rate of reaction was at 20vol, using the line of best fit and halving that number). The rate of reaction on my graph at this point is 1.25 which is very accurate and so proves my graph is directly proportional. The shape of my graph is a straight line because the highest concentration of hydrogen peroxide I used was 20vol so the hydrogen peroxide wasn’t in excess. Therefore the active sites of the enzymes had not yet become saturated, the enzymes weren’t working as hard as they could and so the rate of reaction continued to increase. Because the hydrogen peroxide was not yet in excess my results do not fully match my prediction and so I don’t have enough or sufficient evidence to support my conclusion. (See evaluation)
Evaluation (see graph on p )
Even though my results and graph don’t fully match my prediction, my experiment still went well because there were no anomalous results. By anomalous, I mean results that don’t fit in with the rest of the readings.
The only result that was less reliable than the others was 10vol of hydrogen peroxide and this is because the original and repeat results are further apart from each other on the graph than the others are. This unreliable reading could be due to my efficiency at taking the delivery tube out of the measuring cylinder at exactly 3 minutes and so the test wouldn’t have been fair because some of the experiments could have had slightly more time to collect gas than others, and so the rate of reaction would be affected. The original reading’s potato might have had a slightly smaller surface area so less of it would have been available to decompose the hydrogen peroxide and so it had a slower of reaction than the repeat result.
My results were mostly accurate due to using the same potato and using the same equipment for original readings and repeat results so it was a fair test. However because I am dealing with live cells the catalase won’t always be the same for each experiment and so this could also contribute to unreliable results. The graph shows the accuracy and preciseness of my results because most of my results are close together and close to the line of best fit.
I think the procedure I used was suitable because I did all the original readings first and then the repeat ones as opposed to doing one original reading and then a repeat and so on. By doing this it made the experiment much more efficient and I achieved good results. The equipment I used was suitable and I would not change it. If I did the experiment again I would cut all my potatoes before starting the experiment - for the original readings and the repeats-so that the surface area of the potatoes were the same. I would try and stop collecting gas at the exact time for each experiment to prevent inaccuracies. I would also do more repeats if I had more time to make my results really reliable.
Because my results didn’t fully match my prediction I would have to get additional and sufficient evidence for the conclusion. To do this I could use higher hydrogen peroxide concentrations to see if the active sites of the enzymes became saturated and so see if the line of best fit reached a plateau on my graph and therefore try and prove my prediction. Because higher concentrations are not available to me, I could instead try the same concentrations as I have done for this experiment but use smaller bits of potato so there would be fewer enzymes to decompose the hydrogen peroxide. The line of best fit would then level off because there will be a point when the substrates are queuing up to get into the active sites, and so the reaction will be taking place as quickly as possible, and so the rate of reaction will remain constant
I could also try using concentrations of hydrogen peroxide with smaller intervals (10, 10.2vol, 10.4vol) so I had more points to plot on the graph and so the line of best fit would be more definite. I could use a larger potato, like 6cm, to see if more enzymes mean a quicker rate of reaction. The rate of reaction would be quicker because there is more enzyme and so more molecules per cm³, which means there’s a higher chance of he molecules colliding with the substrate molecules.