Enzyme Catalysis 牋牋牋牋Introduction and Overview- In this lab, we will observe the conversion of hydrogen peroxide (H2O2) into water and oxygen gas. An enzyme known as catalase facilitates this conversion reaction. The catalase enzyme acts as catalysis, helping lower the energy needed to activate the reaction while the enzyme itself is not affected. The reaction could take place without the help of catalase, but it would happen a lot more slowly because more energy is needed for the reaction.牋牋牋牋The substance being acted on by the enzyme is known as the substrate, and in this case is hydrogen peroxide. The enzyme changes these substrates into products, in this case water and O2 gas. The purpose of this lab is to observe and measure the rates of conversion of hydrogen peroxide to water and oxygen by the enzyme catalase over different periods of time.牋牋牋牋To measure the rate of the reaction, we measure the rate of disappearance of the substrate of over time. Potassium permanganate (KMn04) is used to indicate the presence of any remaining H2O2 in the solution after the reaction. The amount of remaining substrate is an indication of how often the conversion reaction took place.牋牋牋牋After allowing an enzyme to act on a substrate solution for different periods of time and measuring the amount of remaining substrate, a rate of reaction can be calculated. And that is what the experiment boils down to, figuring out the rate of an enzyme catalyzed reaction. I predict that the rate for reaction will remain fairly constant over time.牋牋牋牋Materials- 路 Burette 牋牋牋牋路 2% KmnO4 牋牋牋牋 牋牋牋牋路 1M H2SO4 (Sulfuric acid) 牋牋牋牋路 H2O (Water) 牋牋牋牋路 Potato 牋牋牋牋路 1.5% H2O2 (Hydrogen Peroxide) 牋牋牋牋路 Catalase 牋牋牋牋路 1 ml syringe 牋牋牋牋 牋牋牋牋路 10 ml syringe 牋牋牋牋路 Stopwatch 牋牋牋牋路 Six 50 ml beakers 牋牋牋牋路 2 more beakers (any fairly large size) to store acid and hydrogen peroxide 牋牋牋牋Procedure- Part A: Demonstration of Presence of Catalase in Living Tissue.Note: Demonstration done by teacher. No procedure to be recorded, observations in Data section.牋牋牋牋Part B: Determination of Rate of an Enzyme Catalyzed Reaction.1. 牋牋牋牋Determine baseline- In this experiment, was figured by someone else. The baseline is the measure of a substance in a solution. In this experiment, the amount of H2O2 in a 5 ml solution was figured to be 3.2 ml. This baseline will be used for the duration of the experiment.2. 牋牋牋牋Prepare beakers- Extract 10 ml of H2O2 in syringe and dispense in beaker.3. 牋牋牋牋Start reaction- Extract 1 ml of catalase solution with syringe and dispense into beaker w/ H2O2. Swirl solution. Start stopwatch.4. 牋牋牋牋Stop reaction- After 10 seconds has elapsed, stop reaction by adding 10 ml of H2SO4, also known as sulfuric acid. This denatures the enzyme and reactions stop. The amount of remaining substrate can now be measured.5. 牋牋牋牋Measure amount of remaining H2O2- First, 5 ml of the solution must be extracted from the beaker w/ a syringe and placed into a clean beaker. The beaker is position underneath the burette and KMnO4 is slowly added to the solution until the solution no longer remains clear. The amount of KmnO4 used is calculated by finding the difference between the initial and final readings of the burette.6. 牋牋牋牋Repeat the experiment- Repeat steps 2 through 5 for 30, 60, 120, 180, and 360 seconds.Data Collected- Part A- After the H2O2 was added to a piece of potato, the potato appeared to release small bubbles.牋牋牋牋Part B- Below is a table of the data collected in this portion of the experiment: Table 1: Recorded Data of Part B Recorded and Calculated Information in Part B 牋牋牋牋 KmnO4 (ml) 牋牋牋牋 TIME (Seconds) 牋牋牋牋 牋牋牋牋10 牋牋牋牋30 牋牋牋牋60 牋牋牋牋120 牋牋牋牋180 牋牋牋牋360 牋牋牋牋 A. Baseline 牋牋牋牋3.2 牋牋牋牋3.2 牋牋牋牋3.2 牋牋牋牋3.2 牋牋牋牋3.2 牋牋牋牋3.2 牋牋牋牋 B. Final Reading 牋牋牋牋15.2 牋牋牋牋12.4 牋牋牋牋18.4 牋牋牋牋21.7 牋牋牋牋24.1 牋牋牋牋25 牋牋牋牋 C. Initial Reading 牋牋牋牋12.4 牋牋牋牋8.7 牋牋牋牋15.2 牋牋牋牋18.4 牋牋牋牋21.7 牋牋牋牋24.1 牋牋牋牋 D. Amount of KmnO4 Consumed (B minus C) 牋牋牋牋2.8 牋牋牋牋3.7 牋牋牋牋3.2 牋牋牋牋3.3 牋牋牋牋2.4 牋牋牋牋.9 牋牋牋牋 E. Amount of H202 used (A minus D) 牋牋牋牋.4 牋牋牋牋-0.5 牋牋牋牋0 牋牋牋牋-0.1 牋牋牋牋.8 牋牋牋牋2.3 牋牋牋牋 Important Note: There was some confusion in the beginning of the experiment, and we eventually got the recordings for the 30 second beaker before the recordings of the 10 second beaker (don't ask how). So that is why some of the numbers (i.e. B. Final Reading) are not in a logical order. This switch only occurred between 10 and 30 seconds. That explains why the initial and final readings of the burette are lower for the 30-second beaker than for the 10-second beaker.牋牋牋牋Summary/Conclusion- Part A: The bubbles being released from the potato were oxygen bubbles. The point of the demonstration was to show that all living organisms have the enzyme catalase to breakdown hydrogen peroxide, a waste product produced during metabolic functions. When the H2O2 was poured onto the potato, it reacted with enzymes in the potato which broke the H2O2 down into water and O2 gas, forming fizzing and bubbles.牋牋牋牋Part B: Some considerations before beginning with the actual conclusion: The baseline is measured to show the actual amount of H2O2 in a solution. Because not all 5 ml of a 1.5% H2O2 solution is going to be hydrogen peroxide, the amount of "榯rue' H2O2 in the solution must be figured to base other calculations on. This is the part of the solution actually reacting, and its initial amount must be figured so that calculations of change can be made.牋牋牋牋Potassium permanganate, or KMnO4, was used to figure the amount of H2O2 remaining in the solution after the reaction had taken place. The KMnO4 reacts w/ H2O2, so after all after enough KMnO4 is added and all of the H2O2 is consumed, the solution will turn pink or purple if any more KMnO4 is added. The amount of H2O2 consumed during a reaction can be calculated by doing the following: 1. 牋牋牋牋Finding baseline 2. 牋牋牋牋Starting reaction and stopping after a period of time.3. 牋牋牋牋Adding KMnO4 to solution until it turns color and figuring amount of KMnO4 used by finding the difference between initial and final readings of the burette.4. 牋牋牋牋Subtracting the amount of KMnO4 used from the baseline.Adding H2SO4, or sulfuric acid stops the reaction. This stops the reaction because it lowers the pH of the solution and denatures the enzyme. When an enzyme is denatured, it looses its three-dimensional shape, meaning that it can not perform its task and react w/ substrates.牋牋牋牋Rate of reaction: 牋牋牋牋
