Enzyme Catalysis Lab
.docx
School
Barrington High School**We aren't endorsed by this school
Course
BIOLOGY 101
Subject
Biology
Date
Dec 17, 2024
Pages
14
Uploaded by AgentElectron11085
Enzyme Catalysis LabGroup Members: Colin McDermott, Nathan Gravelle, Indie Lamb, Keith Hartman, Cole Ornstedt,Lillyana FlorianiPre-lab Questions1.Explain what it means to be a catalyst.a substance that speeds up a reaction and decreases the activation energy of the reaction2.Explain what this equation means? E + S ES E + penzyme + substrate forms a temporary union (ES), then creates enzyme and products 3.Identify the Enzyme, substrate and product in our experimentThe enzyme is a catalyst, the substrate is hydrogen peroxide, and the products are water (H2O) and oxygen (O2)4.Explain how salt, pH and temperature might affect an enzyme?These are all factors that could lead to the denaturing of an enzyme, for example, if an environment is too hot, it will denature and break apart and will not catalyze anymore. They can all effect the structure of the enzyme, thus hurting its function.5.Explain how activators and inhibitors affect an enzyme?Activators bind to the enzymes and enhance their ability to bind substances, increasing the rate of the reaction. Inhibitors do the opposite, decreasing the enzyme activity. Competitive inhibitors compete with the substrate for the active site, whole non-competitive inhibitors bind to an allosteric site, changing the shape of the enzyme and therefore reducing its function. Uncompetitive inhibitors bind to the enzyme-substrate complex, preventing the reaction from completing.6.Describe Catalase.It is an enzyme found in many living organisms exposed to oxygen which helps catalyze hydrogen peroxide to water and oxygen. It is a clear liquid that is not very dense and behaves similarly to water.7.What does the curve line from the graph on pg. 8 indicate about the reaction rate?1
The curve on this graph represents the slowing down of reactions after 3 minutes. 8.Why is the initial reaction rate usually faster than the rate as the reaction proceeds?It is faster because at the beginning of the reaction, the concentrations of the reactants are at their highest. According to the rate law, the rate of a reaction is directly proportional to the concentration of reactants. As the reaction progresses and reactants are consumed so the rateis slower.9.What is the formula for determining initial rate of a reaction? Explain in words what thismeans.Formula: Change in reaction concentration/change in time, this means that the concentration of the measured solution in the reaction is divided by the change in the time it took to reach that state.10.In our experiment what will we measure to determine the rate of the reaction?The disappearance of hydrogen peroxide is measured by measuring the amount of potassium permanganate that is added to the solution until it changes colors (pink/brown) permanently (which is proportional to the amount of hydrogen peroxide that is remaining). 11.What happens when sulfuric acid is added to a solution of catalase and hydrogenperoxide? Why does this happen?It will create an acidic environment that can affect the activity of the catalase enzyme. This occurs because the enzyme structure is sensitive to pH12.Potassium permanganate is a bright pink color. What happens to the substance when it isadded to hydrogen peroxide and sulfuric acid?It would go from bright pink and then come back to a colorless solution13.If additional potassium permanganate is added and it does not break down and turncolorless, what does that tell us about the amount of hydrogen peroxide still present in thesolution.It tells us that the hydrogen peroxide is fully saturated with potassium permanganate
14.If I need to add 50 drops of potassium permanganate to solution A and 10 drops ofpotassium permanganate to solution B before both turn pink, which solutions mustcontain the most hydrogen peroxide? Solution A would have the most hydrogen peroxide because the hydrogen peroxide gets oxidized by the potassium permanganate. So the more drops you need means the more hydrogen peroxide in a solutionoriginal PDF3
5
7
Procedure for Establishing a Base Line1.Put 10 mL of 1.5% H202 into a clean glass beaker.2.Add 1 mL of H20 (instead of enzyme solution).3.Add 10 mL ofH2S04 (1.0110. Use extreme care in handling reagents. Your teacherwill instruct you about the proper safety procedures for handling hazardousmaterials.4.Mix well.5.Remove a 5-mL sample. Place this 5-mL sample into another beaker and assay forthe amount of H202 as follows. Place the beaker containing the sample over a pieceof white paper. Use a burette, a syringe, or a 5-mL pipette to add KMn04, a drop ata time, to the solution until a persistent pink or brown color is obtained. Rememberto gently swirl the solution after adding each drop. Check to be sure that youunderstand the calibrations on the burette or syringe (see Figure 2.4). Record yourreading in the box below.
data from group experiment - faultyBaseline calculationFinal reading of burrette_______2.8 (26 drops)__________ mLInitial reading of burrette________5_________ mLBase line (Final - Initial)_______2.2__________ mL KMn04data from class - used for data table and discussion of results:Baseline calculationFinal reading of burrette______________5.0___ mLInitial reading of burrette__________1.6_______ mLBase line (Final - Initial)_________3.4________ mL KMn04Figure 2.4: Proper Reading of a BuretteInitialReadingFinalReading9
The base line assay value should be nearly the same for all groups. Compare your results to another team's before proceeding.Remember, the amount of KMn04 used is proportional to the amount of H202 that was in the solution.Note: Handle KMn04 with care. Avoid contact with skin and eyes.Part B: An Enzyme Catalyzed Rate of Hydrogen Peroxide DecompositionIn this experiment you will detennine the rate at which a 1.5% H202 solutiondecomposes when catalyzed by the purified catalase To do this, you should determinehow much H20 has been consumed after 10, 30, 60, 90, 120, 180, and 360 seconds.Procedure for a Time-Course DeterminationTo determine the course of an enzymatic reaction, you will need to measure howmuch substrate is disappearing over time. You will measure the amount of substratedecomposed after 10, 30, 60, 90, 120, 180, and 360 seconds. To use lab time moreefficiently, set up all of these at the same time and do them together. Stop eachreaction at the proper time.1. 10 secondsa.Put 10 mL of 1.5% H202 in a clean 50-mL glass beaker.b.Add I mL of catalase extract.c.Swirl gently for 10 seconds.d.At 10 seconds, add 10 ofH2S04 (1.0 M).2. 30, 60, 90, 120, 180, and 360 secondsEach time, repeat Steps 1—4 as described above, except allow the reactions to proceedfor 30, 60, 90, 120, 180, and 360 seconds, respectively, while swirling gently.Note: Each time, remove a 5-mL sample and assay for the amount of H202 in the sample.Use a burette to add KMnOa, a drop at a time, to the solution until a persistent pink orbrown color is obtained. Should the end point be overshot, remove another 5-mL sampleand repeat the titration. Do not discard any solutions until the entire lab is completed.Record your results in Table 2.1 and Graph 2.1.
10: 4mL30: 3.7mL-3.8ml60: 7mL3.8ml90: 120:180: 360:Table 2.1 (data from group experiment - faulty)Time (seconds)(time that catalyse had to react with Hydrogen Peroxide before Sulfuric Acid was added)KMn04 (ml)10306090120180360a) Base Line*2.22.22.22.22.22.22.2b) Final Reading33.43.43.63.844.2c) initial Reading7777777d) Amount of KMn04 Consumed (B minus C)43.63.63.43.232.8e) Amount of 8202 Used (A minus D)1.81.41.41.21.00.80.6Table 2.1 (data from class - used for graph)TimeKMn04 (ml)10306090120180300a) Base Line*3.43.43.43.43.43.43.4b) Final Reading8.610.211.211.511.611.6c) initial Reading6.18.610.211.211.511.511.6d) Amount of KMn04 Consumed (B minus C)2.31.61.0.3.1.11 drop11
e) Amount of H2O2 Used(A minus D).91.82.43.13.33.33.43. Record the base line value, obtained in Exercise 2D, in all of the boxes on line A in Table 2.1.* Remember that the base line tells how much 1-1202 is in the initial 5-mL sample. 'Ihe difference between the initial and final readings tells how much H002 is left after the enzyme-catalyzed reaction. The shorter the time, the more H202 remains and therefore the more KMn04 is necessary to titrate to the endpoint. If syringes are used. KMn04 consumed maybe calculated as c—b.4. Graph the data for enzyme-catalyzed H202 decomposition. For this graph you will need to determine the following:a.The independent variable: TimeUse this to label the horizontal (x) axis. b.The dependent variable: Amount of H2O2 usedUse this to label the vertical (y) axis.WITH CLASS DATA:
Analysis of Results1. From the formula described earlier recall that rate = Δy / ΔxDetermine the initial rate of the reaction and the rates between each of the time points.Record the rates in the table below.Tlme Intervals (seconds)initial 0-1010-30 30-60 60-9090-120 120 -180 180-360Rates*0.09.045.02.0233.00670.0000832. When is the rate the highest? Explain why.The rate from 0-10 seconds is the highest because it is the initial rate of the reaction when the concentration of substrate is at its highest and all the enzyme active sites are readily available.3.When is the rate the lowest? For what reasons is the rate low?The rate from 120-180 is the lowest, however this is due to a discrepancy in the date. The rate from 180-300 should be the lowest, as the concentration of H2O2 (the substrate) decreases over time as it is converted into its products (H2O and O2), thus leading to fewer opportunities for the catalase to tobind and react with the H2O2.
4.Explain the inhibiting effect of sulfuric acid on the function of catalase. Relate this toenzyme structure and chemistry.The sulfuric acid lowers the pH, effectively denaturing the catalase and inhibiting its function (it modifies the chemical structure of the amino acids forming the protein, which causes the chemical bonds that contribute to the tertiary structure to break which in turn causes the protein to unfold) 5.Predict the effect that lowering the temperature would have on the rate of enzyme activity.Explain your prediction.Lowering the temperature would decrease the rate of enzyme activity, as the catalase and H2O2 would have less kinetic energy, thus leading to fewer collisions between the molecules and reducing the rate of the chemical reactions. 6.Design a controlled experiment to test the effect of varying pH, temperature, or enzymeconcentration.To test the effect of varying enzyme concentration, a student could run multiple trials of an experiment in which they would add catalase (the enzyme) to H2O2 (the substrate) for a certain period of time, then stop the reaction using sulfuric acid and measure the amount of H2O2 remainingthrough the addition of KMnO4 until the solution changes colors (this will help determine the rate of reaction, as the amount of KMnO4 added before the solution changes color is proportional to the amount of H2O2 remaining). This experiment would use various amounts of catalase (trials of 1 mL, 2mL, 3 mL, 4 mL and 5 mL) while holding the interval of time before the addition of the sulfuric acid to stop the reaction (60 seconds) constant for each trial as well as holding the amount of H2O2 used constant as well (20 mL). The temperature would be held constant as well.