To investigate how temperature affects the rate of reaction of the enzyme catalase on its substrate hydrogen peroxide Essay Example

To investigate how temperature affects the rate of reaction of the enzyme catalase on its substrate hydrogen peroxide Essay Enzymes are biological catalysts which increase the rate of reactions by lowering the activation energy needed for the reaction to tale place. The activation energy is the amount of energy needed for molecules to react when they collide. Molecules need to collide in order to react, this is known as the collision theory. When they collide they may not react as a certain amount of energy is required to break bonds, this energy is the activation energy.Enzymes are made of a long amino acid chain, within this some molecules are attracted to each other, so the chain folds in on itself to form a 3D shape.How enzymes are shaped.An area on the surface of the enzyme is known as the active site. This is where reactions take place to form or break down substances. Enzymes are specific which means a particular enzyme only works on one substance known as its substrate. For example, the substrate of amylase is starch and the substrate of lipase is fats. They only have one substrate because the act ive site is formed in a different shape for each enzyme, where only one substance can fit. The lock and key hypothesis states that the enzyme is like a lock which will only have one key.Lock and Key hypothesisThe substrate shown is the only substance that fits the enzyme. An enzyme substrate complex is the compound formed when the substrate is attached to the active site, it is only in this form for a short time while the substrate is being broken down.Enzymes can break own substances, known as catabolism, or can join substances together, known as anabolism. Together they form metabolism which is every chemical reaction in the body.catabolism and anabolism.Enzymes are affected by four factors which are1. Temperature2. pH3. Enzyme concentration4. Substrate concentrationA temperature increase gives ore energy to gives more energy to the substrate and the enzyme so they are more likely to collide and react. The frequency of the collisions with the right activation energy will increase so the rate of reaction will increase. The rate of increase is shown by a mathematical coefficient known as Q10, which states that a ten degree rise in temperture will cause the rate of reaction to approximately double. However at high temperatures enzymes will begin to denature. This means the attractions holding together the shape of the enzyme will begin to break so the active site loses its unique shape and is unable to react with its substrate. The optimum temperature for most enzymes is 37?C, after this they begin to denature. The enzymes in the body have this optimum temperature and the body has adapted to control its temperature so the enzymes are working at there best.Enzymes also have an optimum pH level, where they work best, any changes to this level will cause the enzymes to begin to denature.Pepsin works best in acidic conditions because it is used in the stomach along with stomach acid. Lipase works best in alkali conditions because it works with bile in the intestine s.Increasing the concentration of either the enzyme or the substrate will increase the number in the solution meaning there is more chance of collisions and reactions. There is a limit to the rate of reaction. For example if the enzyme concentration is increased from the same concentration of enzyme and substrate then the rate of reaction will not increase as there are not enough subsrate molecules to react with.HypothesisI predict that a rise in temperature will cause a rise in the rate of reaction until 40?C, after which enzymes will denature so the rate will fall. This will happen because a rise in temperature will mean the are moving faster and are more likely to collide with the catalase on the potato resulting in a greater frequency of collisions. A higher temperature will also mean more hydrogen peroxide molecules will have an energy above the activation energy, so there will be more collisions with the right activation energy. This will result in the rate of reaction increas ing.After 40?C the rate will fall because catalase will denature. His means the attractions between amino acid molecules in the enzyme will break and the enzyme will lose its shape. The active site of the enzyme changes so it can not break down hydrogen peroxide. As the temperature rises further the catalase will denature more quickly and the rate of reaction will fall further.A graph of rate of reaction against temperature may look like this.The rise of rate of reaction is governed by the Q10 coefficient, which states that a 10?C rise will result in an approximate double of the rate of reaction.MethodPreliminary work was undertaken to determine the amount of hydrogen peroxide and potato to use, and what temperature differences to use. The results are as follows;Volume of Hydrogen peroxide (ml)Length of Potato (cm)Temperature (?C)Volume of gas produced in 5 min (ml)112022202140322032401cm potato was too small to handle, and 2cm potato was not fully covered by 2cm3 hydrogen peroxide. So 2cm of potato was used with of hydrogen peroxide. There will not be enough time for all the experiments to be left for five minutes. So they will be left for four minutes, leaving enough time to complete all experiments.Pour 3cm3 of hydrogen peroxide into a test tube, place this in a water bath of 20?C to warm up. Use a cork borer to retrieve a strip of potato from a potato, cut this to 2cm using a cutting board and a knife. Pour water into a beaker and place and measuring cylinder, full of water, into it, as shown below. Take care not to allow air into the cylinder. Place the end of a delivery tube into the measuring cylinder as shown below. Place the potato on the side of the test tube and close it with the bung of the delivery tube. Allow the potato to drop into the hydrogen peroxide, which should be the temperature of the water bath, and start a timer.Set up of equipmentAs oxygen is produced in the reaction it will displace the air trapped in the test tube, this will be forc ed through the delivery tube into the measuring cylinder. The air will rise to the top as the measuring cylinder is full of water and its volume can be measured. Take readings of gas produced every 30 seconds for four minutes.Repeat the experiment twice for reliability and verification of results. Conduct similar experiments with water bath temperatures of 10?C, 30?C, 40?C, 50?C, 60?C and 70?C. In each case make sure the temperature of the substrate has reached the temperature of the water bath before adding the potato.Other methods which can be used are counting the number of bubbles produced, this would not be accurate as the bubble sizes are not the same and the volume is not measured. Measuring the mass of the gas lost is a better method as readings on the scale will be accurate, however this would require equipment which is not available.To make the test fair all other factors affecting the rate of reaction must be kept constant. This includes surface area of the potato and the concentration of the hydrogen peroxide. The experiment will be kept a fair test by:? Using the same length of potato, to keep the surface area constant,? Using the same volume of hydrogen peroxide,? Washing the test tube out with water and drying it, this will prevent concentration changes in hydrogen peroxide,? Using the same potato, as different potatoes will have different levels of catalase present,? Using the same concentration of hydrogen peroxide,? Using the same cork borer to cut the potato, to keep the surface area constant.The experiment will be safe by:? Always wearing safety goggles, as hydrogen peroxide an damage your eyes,? Not spilling the hydrogen peroxide as it is an irritant, and bleaches,? Taking care when cutting the potato,? Taking care when handling hot water.ResultsThe results obtained are as follows:There was not enough time to conduct two repetitions, however, one repetition was conducted. The rest of the method was followed as planned.1st Set of ResultsGas collected (ml) with the following temperatures (?C)Time (s)10203040506070300.050.100.200.400.100.100.20600.100.100.400.800.200.200.30900.100.200.601.300.300.400.301200.200.500.901.800.400.400.301500.200.501.102.200.400.400.301800.250.601.402.700.400.400.302100.300.701.603.000.600.400.302400.300.801.803.400.600.400.302nd Set of ResultsGas collected with (ml) the following temperatures (?C)Time (s)10203040506070300.050.050.100.200.300.300.25600.050.050.200.600.350.400.25900.200.100.300.900.400.400.251200.100.200.601.300.450.400.251500.150.400.801.800.450.400.251800.200.451.002.200.450.400.252100.200.601.302.700.450.400.252400.300.701.603.000.450.400.25AveragesGas collected (ml) with the following temperatures (?C)Time (s)10203040506070300.050.080.150.300.200.200.23600.080.080.300.700.280.300.28900.100.150.451.100.350.400.281200.150.350.751.550.430.400.281500.180.450.952.000.430.400.281800.230.531.202.450.430.400.282100.250.651.452.850.530.400.282400.300.751.703.200.530.400.28Rate of reactionTemperature (?C)Total gas collected (ml)Rate of Reaction (ml/s)/103100.301.25200.753.13301.707.08403.2013.33500.532.19600.401.67700.281.15AnalysisThe amount of gas produced is proportional to the rate of reaction because if the rate of reaction doubles then twice as many reactions are occurring per second so the amount a gas produced is doubled. The rate of reaction can be found by dividing the gas produced by the time. This has been done in the results.From the graph of temperature against rate of reaction we can see the highest rate of reaction, of 0.013 /s, occurs at 40?C. The rate is slow, 0.00125 /s, at 10?C it then rises with temperature until it reaches its maximum at 40?C. This happens because at low temperatures the hydrogen peroxide has less energy a moves more slowly. It will collide with the catalase less often, meaning the frequency of collisions is low. They are less likely to have the right activation energy so there are less collisions resulting in reactions. This will mean the rate of reaction will be low. At higher temperatures the hydrogen peroxide has enough energy to reach its activation energy and it is also colliding more often, so the rate of reaction will be higher.After 40?C the rate of reaction falls , this is because the catalase begins to denature. When enzymes denature the attractions between amino acids in the enzyme break and the enzyme begins to return to its original shape. The shape of the active site also changes so it cannot break down the hydrogen peroxide.The rise in rate of reaction between 10?C and 40?C complies with the Q10 coefficient, as a 10?C rise causes the rate of reaction to approximately double. This can be shown by dividing the higher rate of reaction by the lower one. For example dividing the rate of reaction at 20?C by the rate of reaction at 10?C should give a figure close to 2.These figures are all close to 2, they will not be exact because Q10 is only an approximate and the results are not perfect .From the table of results showing the average volume of gas produced every 30 seconds we can see that at 50?C the enzymes denature within 2 minutes, as gas is no longer produced. This happens because the enzyme takes time to heat up, while it is still reacting with the substrate. Once it is heated to the temperature of the hydrogen peroxide not all the enzymes are denatured. At 60?C the potato is heated faster and it takes 90 seconds for the enzymes to denature, at 70?C the potato is heated even faster and it takes 30 seconds to denature.The results agree with my hypothesis because I have predicted that the rate of reaction will rise between 10?C and 40?C, and the rate will fall after 40?C. The graph obtained for the results is also similar to the one predicted, and the results seem to follow as predicted.From conducting the experiment and gathering data I can conclude that the rate of reaction between catalase and hydrogen peroxide rises as the temperature of the mixture rises. Th is happens until 40?C, after which the rate of reaction falls because the catalase begins to denature. When enzyme denature attractions between the amino acids break so the enzymes loses its shape. The active site will no longer have its unique shape and the enzyme will be unable to react with its substrate.This is because only the right active site shape can break down hydrogen peroxide, according to the lock and key hypothesis, which suggests the substrate, like a key, will only have one lock, enzyme, it fits into. The reaction follows Q10 until 40?C, because a 10?C rise will give enough energy to the substrate to increase the number of collisions and give more molecules the right activation energy to react when the collide. This will double the rate of reaction.Two anomalous results occurred during the experiment.1. During the repeat reading of the experiment at 10?C the reading at 90 seconds is higher than that of 120 seconds. This does not affect the analysis as the reading was ignored when taking averages.2. During the 50?C experiment the first time the volume of gas produced stopped increasing between 120 and 180 seconds. Gas was then released, it may have been trapped in the delivery tube. When drawing the best fit line this was taken into account, so it shouldnt affect the analysis.EvaluationThe experiment was conducted successfully, the results obtained indicates a clear pattern which can be used to draw and support a valid conclusion. The experiment could not be conducted as planned because there was not enough time to repeat each experiment twice. However, one repetition was conducted which did make the results more reliable.The results are reliable because the experiment was a fair test. This was done by keeping all variables constant. The concentration of hydrogen peroxide was not changed, however, it did vary as it naturally decomposed into water and oxygen. It also decomposed more during the higher temperature experiments because the hydrogen p eroxide had more energy. This was a slight change and could not vary the volume of the gas produced significantly. The same volume oh hydrogen peroxide was used. The surface area of the potato was kept constant by using the same size cork borer and cutting it to the same size.The results are accurate because a narrow measuring cylinder was used, so the volume measured is more accurate. Hot and cold water were mixed to achieve accurate temperatures.An anomalous results occurred during the repeat reading of the 10?C experiment, the reading at 90 seconds is higher than that at 120 seconds. This reading was ignored when taking the average so It does not affect the analysis. It occurred because the measurement was misread, it may have been 0.1 cm3, instead of 0.2 cm3. Another anomalous result occurred during the first taking of the 50?C experiment, gas was released at 210 seconds when the experiment seemed to have stopped. The gas may have been trapped in the delivery tube and should hav e been released earlier in the experiment. This was taken into account when drawing the best fit line on the graph, so does not affect the analysis.The method used was good enough to achieve reliable readings , but it can be improved by measuring the mass of the gas lost, this would be more accurate as digital readings would be taken. Using a smaller frequency and a larger range of temperatures would give more evidence for the conclusion. However would require more time and equipment.Other improvements are, using thinner measuring cylinders, to measure out the hydrogen peroxide and the gas produced. Using a thermocouple thermometer to accurately measure the temperature. The hydrogen peroxide took time to heat up as the test tube is glass and is insulated. Using a better material would save time and would have allowed the plan to be completed. Using electronic equipment to take readings a exactly 30 seconds, would eliminate human error.Further work that would extend the investigation and give more evidence to the conclusion would be, to use different concentrations of hydrogen peroxide and lengths of potato, to see how these affect the rate of reaction. Using other substances with catalase, like liver, to see how enzyme concentration affects the rate.