Investigating the Effect of Substrate Concentration on Catalase Reaction

Investigating the Effect of Substrate Concentration on Catalase Reaction
Planning -Aim: The aim of the experiment is to examine how the concentration of the substrate (Hydrogen Peroxide, H2O2) affects the rate of reaction of the enzyme (Catalase). -Background information: Enzyme Enzymes are protein molecules that act as the biological catalysts. A catalyst is a molecule which can speed up chemical reaction but remains unchanged at the end of the reaction. Enzymes catalyze most of the metabolic reactions which take place within a living organism. They speed up the metabolic reactions by lowering the amount of energy needed to activate the reacting molecules. They are specific that usually act on only one type of substrate, so each of them just perform one particular reaction. Furthermore, only small amount of enzyme is needed every time to speed up a reaction. Enzymes are globular proteins that have a precise three-dimensional shape. Their hydrophilic side-chains on the outside of the molecule make them soluble in water. Enzymes can catalyze both anabolic and catabolic reactions within an organism.
That means by the interaction between the side-chains of the enzyme and the atoms of the substrate, the enzyme can encourage the formation or breaking of bonds in a substrate molecule. Each enzyme possesses an active site. The active site is a region of enzyme which allows a substrate to bind with it. The configuration of the active site gives the specificity of enzyme. That means the active site and the substrate should be exactly complementary so that the substrate can fit in perfectly. Once they collide, the substrate and some of the side-chains of the enzyme?s amino acids form a temporary bond so that the substrate can be held in the active site. They combine to from an enzyme-substrate complex and the enzyme can start its work. It is called the ?lock and key? hypothesis. (Lock: enzyme, key: substrate) There is another hypothesis called ?induced fit?. That is when the substrate molecule combines with the enzyme it may induce a small change of shape in the enzyme molecule, forcing the substrate molecules to combine. The resulting end product is then released by the enzyme which returns to its normal shape, ready to receive more. This means enzyme is a wobbly and flexible molecule. However, enzymes are still extremely specific as most of them are specific to one particular substrate. (From Biology?A Functional Approach, Nelson) (From New Understanding Biology by Glenn Toole and Susan Toole, Stanley Thornes) (From Advanced Biology 1 by Richard Allan and Tracey Greenwood, Biozone) Hydrogen Peroxide Hydrogen peroxide is formed continually as a by-product of various metabolic reactions in living cells. Oxidation of lactic acid in liver cells and photorespiration in mesophyll cells are the examples. It is toxic and it is a strong oxidizing agent. If it was not immediately broken down, it will accumulate and kills cells. Catalase Catalase is an enzyme found in both plant and animal cells. It is present in potato, apple and liver. It catalyzes the decomposition of hydrogen peroxide into water and oxygen: 2H2O2 Catalase 2H2O + O2 [image]Catalase behaves like other enzymes. It speeds up the catabolic reaction by lowering the amount of energy needed to activate the molecules of hydrogen peroxide. Catalase is extremely fast acting because it has 4 active sites per molecule which is quite unusual. It can break down hydrogen peroxide at a rate of 107 molecules per second. The fast action of Catalase ensures that the hydrogen peroxide would not accumulate in the body cells. -Predictions: I believe the catalase will break the hydrogen peroxide down very quickly at the beginning of the experiment as catalase is one of the fastest enzymes known. Oxygen gas will be collected from the reaction. I predict that as the substrate concentration increases, the initial rate of reaction will increase. At low concentration, the rate of reaction increases direct proportionally with the substrate concentration. It will come to a point that the rate of reaction becomes constant when the substrate concentration is getting higher and keeping the enzyme concentration constant. It is because every enzyme active site is working continuously. They cannot work faster if more substrates are added so the rate levels off. The expected graph is shown below: [image] My predictions are based on the following science knowledge: The more substrate molecules there are available, the more often one will bump into the active site of an enzyme, so the more rapidly they will be converted to product. Therefore, increasing substrate concentration increases the rate of reaction. However, if you only have a limited amount of enzyme available, this pattern will not continue for ever. At high substrate concentrations, each enzyme will be working as fast as it can, with substrate molecules virtually ?queuing up? for empty active sites. The curve of the graph of the reaction therefore flattens out at high substrate concentrations. This is when the rate of reaction does not increase obviously and almost as a straight line. The point that the reaction reaches is called Vmax (its maximum rate). At that time, all active sites are being engaged, leaving the substrate molecules to ?queue up? for an active site to become vacant. (From Advanced Biology by Mary and Geoff Jones, Cambridge University Press) -Variable: I?ve chosen the concentration of substrate as the independent variable. I varied the concentration of substrate by diluting the hydrogen peroxide solution into 5 different concentrations (20%, 40%, 60%, 80% and 100%). I simply diluted 10cm3 to vary the concentration and it?s shown in the following table. Concentration of H2O2 / % Volume of H2O2 / cm3 Volume of distilled water / cm3 100 10 0 80 8 2 60 6 4 40 4 6 20 2 8 The concentration of substrate (variable) should cause changes in the dependent variable. The dependent variable is the volume of oxygen gas released over a same period of time. The rate of reaction can be calculated by this. I have also set up a control experiment to confirm that there is no unknown variable that is responsible for any observed changes in the responding variable. So it helps to make experiment fair. The experiment is basically the same except the concentration of H2O2 is zero and only distilled water is added into the enzyme solution. -Factors that must be keep constant to make the experiment fair: Temperature I didn?t choose temperature as my independent variable because it is difficult to control or vary. It is also very hard to maintain different range of temperature. When temperature increases, the molecules gain more kinetic energy and they move faster. It is the same in a enzyme-catalyzed reaction. The enzyme and substrates move faster as the temperature increases and they have a higher chance to collide and combine thus the rate of reaction will be higher. In order to control this variable, I have to keep the temperature at a constant level which allows the catalase work effectively. I kept the experiment at room temperature, approximately 20oC. This could be achieved by using a conical flask instead of a boiling tube. Then the heat of my hands will not affect the temperature because the flask can stand on its own. pH I didn?t choose this as my variable because buffers are imperfect. They will affect the results and make the experiment unfair. Changing pH away from the optimum pH of enzyme affects the ionic charge of the amino and carboxyl groups and hydrogen bonds in enzyme which make up the active sites. This alters the shape of the active site. To keep the pH constant, all apparatus are washed with tap water then distilled water. This can keep the experiment under a neutral condition. Enzyme concentration The enzyme concentration has to be kept constant because if more enzymes are present, more active sites will be available for the substrate to combine with it. In this experiment, the catalase concentration was kept constant by homogenising the liver used. Same source of enzyme Liver or potato must be cut out from the same one because there are different amount of catalase present in different sources. Range of time between each reading The aim of the experiment is to compare the volume of oxygen within a same time, so it was kept constant. The same apparatus used It was done in order to allow for consistency. -Apparatus l 250 cm3 beakers x 6 l 100 cm3 burette l 10 cm3 syringe l 1 cm3 syringe l Conical flask and bung l Delivery tube l Basin l Liquidizer l Knife l Digital stop watch l Electrical balance l Forceps l Clamp + stand l waterproof pen l White tile l Safety goggles l Laboratory coat -Chemicals l Hydrogen Peroxide l Liver l Distilled water l Tap water -Reasons: After my preliminary study and experiment, I chose the following apparatus: Apparatus Reasons for choice Liver - As a source of catalase. - From my preliminary experiment, I found out that the rate of reaction of potato is too slow. - It can produce the best observable changes within a small range of time. Burette - To measure the volume of oxygen gas produced from the reaction - It allows me to collect gas by downward displacement. It is important because oxygen gas has a lower density than water. - I have chosen the burette because it is more accurate than the measuring cylinder and the syringe as we can take the reading up to 0.05 cm3. - Also, It can collect 100 cm3 of gas which is large enough that the displacement of water would not exceed its volume. Conical flask - I have chosen the conical flask instead of the boiling tube because it can stand on its own and the heat of my hands will not affect the temperature of the experiment thus will not affect the rate of reaction. Liquidizer - I have chosen this, instead of using a knife to cut to liver into pieces, to homogenise the liver. - That will make the experiment more likely to be a fair test because the amount of liver extract added is more accurate by homogenizing them than cutting them into piece. - If we cut the liver into pieces, some enzymes located in the innermost part of the liver may not involved in the experiment. Electrical balance - To weigh out the amount of liver Stop watch - To measure the time interval 10 cm3 syringe - To measure the amount of hydrogen peroxide(substrate) which are added and transfer it to the conical flask 1 cm3 syringe - To measure the amount of liver extract(enzyme) which are added and transfer it to the conical flask White tile - The liver can be cut on it Several assumptions must be made before the experiment: 1. I assume all results are reliable. 2. I assume all pieces of apparatus are accurate. -Preliminary experiments: I carried out three experiments for the preliminary work. I have chosen two different concentrations (1:4 and 1:2) and volumes (0.5 cm3 and 10 cm3 ) for liver and potato extract. It is because there is liver has a higher concentration of catalase, only small amount of liver is needed. If I add too many liver extract into the flask the reaction will be very fast and the oxygen will be released too fast for taking the reading. Also, in the preliminary tests, we calculate the volume of gas released from the burette reading instead of counting the bubbles released because the bubbles are not the same and they have different volumes. This was achieved by recording the initial and final reading from the burette in every 20 seconds. The readings were taken at eye level and were observed from the bottom of the meniscus to make sure the results are taken accurately. Firstly, I used liver as the source of enzyme Liver : water = 1:4 I added 0.5cm3 of liver extract into 10cm3 of different concentrations of hydrogen peroxide solution. The following table shows the results: Concentration of H2O2 / % Time/sec Initial reading/cm3 Final reading/cm3 Volume of O2 /cm3 100 20 48.5 31.5 17.0 80 20 46.0 32.4 13.6 60 20 38.1 26.3 11.8 40 20 49.3 39.2 10.1 20 20 49.5 41.6 7.9 0 20 47.2 47.2 0.0 Secondly, I used potato the source of enzyme Potato : water = 1:2 I added 10cm3 of potato extract into 10cm3 of different concentrations of hydrogen peroxide solution. The following table shows the results: Concentration of H2O2 / % Time/sec Initial reading/ cm3 Final reading/cm3 Volume of O2 /cm3 100 20 38.2 34.6 3.6 80 20 49.2 46.7 2.5 60 20 44.1 42.1 2.0 40 20 37.7 35.6 2.2 20 20 32.1 31.3 0.8 0 20 42.3 42.3 0.0 Thirdly, I the liver as the source of enzyme This time I added 0.1 cm3 of 1:4 liver extract into 100% concentration of hydrogen peroxide. I had chosen a 10-second time interval to record the results and I recorded the results within 360 seconds. Time /sec Volume of O2 produced/ cm3 Time /sec Volume of O2 produced/ cm3 0 0 190 30.2 10 4.8 200 31.5 20 7.6 210 32.7 30 9.7 220 33.7 40 10.7 230 34.8 50 12 240 35.9 60 13.2 250 36.95 70 14.65 260 38.05 80 16.1 270 39.1 90 17.6 280 40.2 100 19.12 290 41.3 110 20.6 300 42.35 120 21.9 310 43.4 130 23.4 320 44.5 140 24.5 330 45.4 150 25.7 340 46.45 160 26.9 350 47.5 170 28 360 48.2 180 29.2 -Changes I have made after the preliminary experiments: From my first and second preliminary work, I found out that the rate of reaction of potato is too slow and the results obtained from the second experiment were not conspicuous to show how the concentration of hydrogen peroxide affects the volume of gas released(i.e. rate of reaction). So I decided to use liver instead of potato in order to get a better result. Furthermore, I found out that using 1:4 liver extract gave out oxygen at a rate which was so high. So I decided to dilute the liver extract to 1:20(liver : water) and also I have decided to use only 0.1 cm3 of liver extract instead of of liver extract. After the preliminary work, I decided to use 10s as the time interval. From my third preliminary work, the results showed that around 120s to 150s, the curve started to level off slightly. Therefore, I have chosen 120 second as the time limit. -Procedures: 1. To start off with the experiment, I set up the apparatus as follows: 2. 2 g of liver was cut out by a knife and weighed by a electrical balance. 3. I added 40 cm3 of water to make a 1:40 liver extract. 4. I used the liquidizer to homogenize the liver. Then poured out the liver extract into a 250 cm3 beaker. 5. By using a 10 cm3 syringe and a 1 cm3 syringe, correct concentrations of hydrogen peroxide were mixed up and were placed into 5 beakers separately. I mixed up the solution according to the following table: Concentration of H2O2 / % Volume of H2O2 / cm3 Volume of distilled water / cm3 100 10 0 80 8 2 60 6 4 40 4 6 20 2 8 6. The solutions were swirled up gently to ensure the H2O2 and distilled water mixed thoroughly. 7. The beakers were marked by a waterproof pen in order to identify the beakers containing different concentration of hydrogen peroxide. 8. Record the initial reading of the burette. 9. 10 cm3 of 100% hydrogen peroxide solution was added into the conical flask by using a 10 cm3 syringe. 10. 0.1 cm3 of liver extract was added into the conical flask by using the 1 cm3 syringe. At the same time, the bung was used to close the mouth of the conical flask. 11. Start timing as soon as the first bubble appeared in the burette. 12. Volume of oxygen was recorded in a 10s interval until 120s. 13. Wash the conical and the 10 cm3 syringe with water. 14. Repeat with concentration of hydrogen peroxide 80%, 60%, 40%, 20% and 0% (as a control). 15. In order to make my result reliable, I repeated the experiment until I got two concordant results. 16. All apparatus were cleared and washed up after the experiment. 17. Plot graphs from the results obtained. - Range of readings I decided to use 6 concentrations of Hydrogen Peroxide in the experiment. They are 100%, 80%, 60%, 40%, 20% and 0% which acts as a control. It is a suitable range as it covers both high and low concentrations. Also, the differences between them are the same. There are two ways to measure the results. One is to measure the time needed to evolved a certain amount of gas. Another one is to measure the amount of oxygen gas produced within a time which is the one I have chosen. I measured the volume of oxygen released per 10 seconds, up to 120 seconds. It is the initial rate of reaction that is taken. This rate is the highest rate that the enzymes work under the condition of the experiment. -Risk assessment: - Wear safety goggles because the hydrogen peroxide used in the experiment is toxic and it is a strong oxidizing agent. - Wear laboratory coat because the hydrogen peroxide can bleach our cloths. - We should not pour out the hydrogen peroxide solution at a great height - We should not keep the hydrogen peroxide under light for a long time. We should store the hydrogen peroxide in a dark bottle because it will slowly decompose under light. - Follow all school laboratory rules. - Wash hands before and after the experiment. - Use all chemicals in the laboratory carefully. - Take care when using the liquidizer. - All benches and floor surfaces should be kept clear and clean throughout the experiment - Be careful of using all glassware. -Bibliography l Books Author Title Publisher Notes Mary and Geoff Jones Advanced Biology Cambridge University Press Background of enzymes Effects of substrate concentration mbv Roberts Biology?A functional Approach Nelson Background of catalase and Hydrogen Peroxide Richard Allan and Tracey Geenwood Advanced Biology 1 Biozone International Mechanism of enzyme action Glenn and Susan Toole New Understanding Biology Stanley Thornes Mechanism of enzyme action Effects of temperature and pH on enzyme-catalysed reaction Michael Kenf Advanced Biology Oxford University Press Mechanism of enzyme action CJ Clegg and DG Mackean Advanced Biology Principles and Applications John Murray Mechanism of enzyme action Effects of substrate concentration Endorsed by ocr Advanced Science Biology 1 Cambridge University Press Effects of substrate concentration l Websites - www.bbc.co.uk - www.h2o2.com - http://web.indstate.edu/thcme/mwking/enzymekinetics.html#intro - http://www.worthington-biochem.com/introBiochem/intoEnzymes.html - http://www.clunet.edu/BioDev/omm/catalase/cat1.htm - http://user .rnc.com/jkimball.ma.ultranet/BiologyPages/E/Enzymes.html Implementing -Results: The following are the results I obtained by carrying out the above experiment: First reading Total volume of O2 produced / cm3 [image] Concentration of H2O2 / % Time / sec 100 80 60 40 20 0 0 0.0 0.0 0.0 0.0 0.0 0.0 10 5.2 4.6 3.1 2.8 2.5 0.0 20 8.0 7.5 5.3 3.7 3.4 0.0 30 10.5 9.2 7.3 4.6 4.7 0.0 40 14.3 10.7 9.1 5.7 5.4 0.0 50 18.5 12.4 12 6.6 6.3 0.0 60 22.2 15.0 13.2 7.8 6.9 0.0 70 24.8 17.3 14.7 9.2 7.7 0.0 80 28.2 19.5 16.1 10.5 8.1 0.0 90 30.9 22.1 17.6 12.0 8.6 0.0 100 32.4 23.9 19.1 13.6 9.3 0.0 110 33.4 25.0 20.6 15.2 9.8 0.0 120 34.1 25.7 21.9 16.8 10.3 0.0 Second Reading Total volume of O2 produced / cm3 [image] Concentration of H2O2 / % Time / sec 100 80 60 40 20 0 0 0.0 0.0 0.0 0.0 0.0 0.0 10 4.4 3.6 3.3 3.0 2.4 0.0 20 5.5 5.1 4.7 4.5 4.1 0.0 30 7.0 6.8 6.6 6.0 5.3 0.0 40 8.7 8.2 8.4 7.6 6.3 0.0 50 11.1 9.9 10.3 9.3 7.3 0.0 60 13.1 11.4 12.1 11.2 8.1 0.0 70 15.2 13.1 14.0 13.2 8.9 0.0 80 16.8 15.1 15.7 15.1 9.5 0.0 90 18.3 17.1 17.4 17.2 10.4 0.0 100 19.6 19.5 18.8 19.1 11.0 0.0 110 21.8 21.1 20.3 20.9 11.5 0.0 120 24.4 23.1 21.6 22.0 12.1 0.0 Third Reading Total volume of O2 produced / cm3 [image] Concentration of H2O2 / % Time / sec 100 80 60 40 20 0 0 0.0 0.0 0.0 0.0 0.0 0.0 10 5.2 3.8 3.2 3.1 2.7 0.0 20 7.3 5.3 4.8 4.5 3.9 0.0 30 9.4 6.7 6.2 5.6 4.9 0.0 40 11.8 8.0 7.6 6.8 5.6 0.0 50 14.3 9.6 8.8 7.9 6.2 0.0 60 16.8 11.1 10.4 9.2 6.8 0.0 70 19.6 13.0 11.9 10.3 7.3 0.0 80 22.1 14.7 13.7 11.5 7.8 0.0 90 24.8 16.3 15.5 12.5 8.0 0.0 100 27.4 17.8 17.3 13.4 8.6 0.0 110 30.1 19.2 19.3 14.4 8.9 0.0 120 32.3 21.4 20.7 15.4 9.3 0.0 Please refer to graph 1-3 Analysing Average reading I calculate the average of the concordant of volume of oxygen evolved at 120 seconds. Here is the table: Concentration/% First reading/cm³ Second reading/cm³ Third reading/cm³ Average/cm³ 100 34.1 32.2 24.4 33.2 80 25.7 21.4 23.0 24.4 60 21.9 20.7 21.6 21.4 40 16.8 15.4 27.0 16.1 20 10.3 9.3 12.1 9.8 0 0.0 0.0 0.0 0.0 Please refer to graph 4 - Patterns and trends observed from the graph 1-3: The results in the first and third reading follow the trend in my prediction. That is at the beginning of the reaction, the higher concentration of hydrogen peroxide, the more oxygen it evolves. The highest concentration of hydrogen peroxide (100%) collected the largest volume of oxygen gas, and then 80% and so on. As seen in the tables and the graphs, there are fluctuations between the readings, especially those in the second reading. So I took an average between concordant results. I expected this will give me a more reliable result. From the graphs, the amount of oxygen gas released rapidly at the beginning of the experiment. Although there were some points shown in a graph do not lie exactly on the curve because of the experimental error, there were enough points for me to draw a best fit line between them. The best fit line was a straight line through the origin. This showed that without the present of catalase, no reaction took place. Also, the volume of oxygen released was directly proportional to time as shown. However, as the time goes on, the volume of oxygen gas released decreases. It eventually starts to level off when the time reaches about 120 seconds. This is mainly because when more hydrogen peroxide molecules are being concerted by catalase, there are fewer hydrogen peroxide molecules to combine with the active sites in catalase. The amount of oxygen gas evolved decreases. Thus, the rate of reaction decreases. I had decided to use the average readings at 120 seconds to calculate the rate of reaction graph as I found out some readings before 120 seconds were not following the trend that I had predicted. Also, I found out that the first reading (first 10 seconds) was always slightly above the best fit line. This might because pressure was produced inside the conical flask when I put the bung into the mouth of the conical flask. - Patterns and trends observed from graph 4 (the average graph): The average graph showed the hypothesis that when the concentration increases, the rate of reaction will increases. At lower substrate concentration the rate of reaction increases in direct proportion to the substrate concentration, but at a higher substrate concentration the rate of reaction becomes constant, and shows no increase. This is what I have predicted. The reason for that is when I increased the hydrogen peroxide (substrate) concentration, the more hydrogen peroxide molecules present, the higher chance for them to collide with the active sites in enzymes. So the reaction rate is fast. However, when the curve reached 80% concentration, the rate of reaction seems to level off. The rate of reaction did not increase obviously and further increase in the substrate concentration had little effect on the rate of reaction. This is the point that the reaction reaches Vmax (its maximum rate). All active sites were engaged and the substrate molecules had to ?queue up? for an active site. Evaluating I have proved above that the scientific knowledge and the hypothesis can back up the results and the graph obtained in the analysis. The results and graphs obtained follow the hypothesis so I think my method was suitable and effective. Although the experiment didn?t show levelling off, I still think it was right to choose liver because I expected it would be more difficult to use potato to show the hypothesis. Maybe I should make the time limit longer in order to show the level off of the volume of oxygen gas prodused. Nevertheless, there were still some errors that would have affected the experimental results. -Anomalous results From the line of best fit on graph 4, it is clear that points of 80% and 100% do not fit exactly. These are anomalous results. But comparing to the raw data, the repeating readings are similar. From graph 1 to 4, the first readings (first 10 seconds) from every substrate concentration are always above the line of best fit. This might because of some pressure was produced inside the conical flask when I inserted the bung into the mouth of the conical flask. From the graphs, there are some odd data. This indicates the errors were made in the experiment. The following table shows the sources of error and improvements: Source of error Effects on results Improvements Starting timing This is the biggest source of error. I might have different reaction time when I saw the first bubble in the burette. But as the same method was used every time, so this can minimize the error. Data log to record pressure within the reaction vessel. Measuring of the volume of oxygen released in the burette It is a procedural error and it is subjective. Water level is different each time so the readings are depended on the person who read the volume from the burette. The scale of the burette can show the measurement up to 0.05cm3 .All measurements that I used should be corrected to the nearest 0.1cm3. It is impossible to measure precisely the amount of oxygen gas evolved each time. The reading should be taken from the bottom of the meniscus each time. Also, by repeating the experiment until I get 2 concordant results and averaging the concordant results can increase accuracy. Time of putting the bung to cover the conical flask. After adding the enzyme extract into the conical flask, there may be some oxygen gas leaked out before putting the bung in the right place. This may slightly decrease the initial rate for each concentration. However, as I carried out the method in the same way every time, it could make less difference to the overall result. Put the bung closer to the conical flask so as to minimize the time needed to fit in the conical flask and the volume of oxygen leaks out. Diluting hydrogen peroxide solution When I was making the dilutions, some solution might leave inside the syringe. This would lower the expected substrate concentration. But as the same method is used each time, this can minimise the error. Also, the hydrogen peroxide solution and water might leave inside the syringe when I started the next dilution. This would slightly increase the concentration of hydrogen peroxide solution (substrate concentration). The syringe should be washed by distilled water before each dilution. -Other possible sources of error: - 0.1cm3 of liver extract taken out each time may have different enzyme concentration. So I could not ensure the catalase concentration in each experiment was the same. - Pressure might be produced inside the conical flask when I put the bung into the mouth of the conical flask. This might slightly increase the reading of the first 10 seconds. - Although minimum contact to the flask was made, the change in room temperature may affect the rate of reaction. This can increase rate of reaction and the gas inside the burette to expand and displace water out of the burette. - Some hydrogen peroxide solution might leave on the sides of the conical flask due to splashing of solution when it was transferred into it. This might decrease the number of substrate molecule in the solution and affect the oxygen gas produced. - The bung might not fit in the conical flask perfectly; some oxygen gas might leak out. -Other improvements that can be made: -Wash all apparatus with tap water and then distilled water before every attempt to ensure there is no contamination to reduce the accuracy of the results. Dry the apparatus with tissue paper thoroughly before carrying out the experiment every time as the water inside the flask may affect the substrate concentration of the next dilution. -Assessment Some points on the graph do not fit in the best fit line. This shows the sources of error mentioned above did have influences on the result. Nevertheless, the trend of the graphs followed my predictions and the science knowledge. So the overall sources of error did not affect the conclusion of the investigation. So the conclusion is the same as I have predicted. -Conclusion The volume of oxygen gas produced increases with the hydrogen peroxide concentration. That means the rate of reaction increases with the substrate concentration. The reaction was fastest when the substrate concentration is 80%. The rate increased steadily form 0% up to 100% and slowed beyond this point to give a maximum level. The rate of reaction started to level off after reaching the maximum level and the rate became constant. A table showing the rate of enzyme reaction on different concentrations of hydrogen peroxide: Total volume of O2 produced / cm3 [image] Concentration of H2O2 /% Time / sec 100 80 60 40 20 0 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 10 5.2 4.4 5.2 4.6 3.6 3.8 3.1 4.4 3.2 2.8 3.0 3.1 2.5 2.4 2.7 0.0 0.0 0.0 20 8.0 5.5 7.3 7.5 5.1 5.3 5.3 5.5 4.8 3.7 4.5 4.5 3.4 4.1 3.9 0.0 0.0 0.0 30 10.5 7.0 9.4 9.2 6.8 6.7 7.3 7.0 6.2 4.6 6.0 5.6 4.7 5.3 4.9 0.0 0.0 0.0 40 14.3 8.7 11.8 10.7 8.2 8.0 9.1 8.7 7.6 5.7 7.6 6.8 5.4 6.3 5.6 0.0 0.0 0.0 50 18.5 11.1 14.3 12.4 9.9 9.6 12 11.1 8.8 6.6 9.3 7.9 6.3 7.3 6.2 0.0 0.0 0.0 60 22.2 13.1 16.8 15.0 11.4 11.1 13.2 13.1 10.4 7.8 11.2 9.2 6.9 8.1 6.8 0.0 0.0 0.0 70 24.8 15.2 19.6 17.3 13.1 13.0 14.7 15.2 11.9 9.2 13.2 10.3 7.7 8.9 7.3 0.0 0.0 0.0 80 28.2 16.8 22.1 19.5 15.1 14.7 16.1 16.8 13.7 10.5 15.1 11.5 8.1 9.5 7.8 0.0 0.0 0.0 90 30.9 18.3 24.8 22.1 17.1 16.3 17.6 18.3 15.5 12.0 17.2 12.5 8.6 10.4 8.0 0.0 0.0 0.0 100 32.4 19.6 27.4 23.9 19.5 17.8 19.1 19.6 17.3 13.6 19.1 13.4 9.3 11.0 8.6 0.0 0.0 0.0 110 33.4 21.8 30.1 25.0 21.1 19.2 20.6 20.3 19.3 15.2 20.9 14.4 9.8 11.5 8.9 0.0 0.0 0.0 120 34.1 24.4 32.3 25.7 23.1 21.4 21.9 21.6 20.7 16.8 22.0 15.4 10.3 12.1 9.3 0.0 0.0 0.0 A table showing the average volume of oxygen produced within 120 seconds(rate of reaction) when different concentration of hydrogen peroxide is used: Total volume of O2 produced / cm3 [image] Concentration of H2O2 / % Time / sec 100 80 60 40 20 0 0 0.0 0.0 0.0 0.0 0.0 0.0 10 5.0 3.7 3.2 3.0 2.6 0.0 20 7.7 5.2 5.2 4.2 3.8 0.0 30 10.0 6.8 7.2 5.8 5.0 0.0 40 13 8.1 8.5 7.7 5.8 0.0 50 12.7 9.8 11.6 7.3 6.6 0.0 60 15.0 11.3 13.2 8.5 6.9 0.0 70 17.4 13.1 15.0 9.8 7.5 0.0 80 25.2 14.9 16.5 11 8.5 0.0 90 27.9 16.7 18.0 12.3 8.3 0.0 100 29.9 18.7 19.4 13.5 9.0 0.0 110 31.8 20.2 20.0 14.8 9.4 0.0 120 33.2 24.4 21.4 16.1 9.8 0.0 A table of rate of reaction at the 120th second: Concentration of H2O2 / % Time/ s Volume of oxygen produced/ cm3 Rate of reaction/ cm3s-1 0 120 0.0 0.0 20 120 9.8 0.08 40 120 16.1 0.13 60 120 21.4 0.18 80 120 24.2 0.20 100 120 33.2 0.28

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