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- 7. You obtain the following set of data for two enzymes that follow Michaelis-Menten kinetics: Enzyme A Initial velocity (µM/min) 38 Enzyme B Initial velocity (пМ/min) 140 Substrate concentration (µM) 233 350 2 73 4 133 8 229 467 20 400 583 150 706 682 300 750 691 786 798 697 699 1000 3000 6000 800 700 (a) What are the approximate Km values for each enzyme based on these data? Briefly justify your answer. (b) In the above experiment, 0.1 µM of enzyme A or enzyme B were used. Calculate kçat and specificity constant for each enzyme. Which enzyme is more efficient? (c) The kinetics of Enzyme B were re-measured in the presence of a small molecule inhibitor. The Km increased while the Vmax remained unchanged. What type of inhibition is this and why does the Km increase?3.An enzyme catalyzed reaction is studied and the following kinetic analysis is obtained: |[S), mM 0.050 0.075 v, (µM min") 0.93 1.264 1.77 2.14 3.7 0.125 0.175 0.935 a. Using Excel, make a fully labeled Lineweaver-Burk plot and determine the Km and Vmax for the enzyme b. The reactions were set up dissolving 1 mg of the enzyme (MW= 100000 Da) in 100 ml of final reaction buffer. Determine the turnover number for the enzyme assuming 1 active site exists per enzyme molecule? c. Determine the catalytic efficiency for the enzyme d. The same reactions are performed in presence of an inhibitor A and the resulting velocities determined: v plus inhibitor, (µM min) 0.272 0.37 0.518 0.626 |1.08 |[S), mM 0.050 0.075 0.125 0.175 0.935 Plot these data on the same graph as above and determine the new Km and Vmax and the type of inhibitor (competitive, non-competitive). e. Can the effects of the inhibitor be over-ridden by adding more substrate? Why?1. a. Calculate the physiological DG of the reaction shown below at 37°C, as it occurs in the cytosol ofneurons, with phosphocreatine at 4.7 mM, creatine at 1.0 mM, ADP at 0.73 mM, and ATP at 2.6mM. The standard free energy change for the overall reaction is –12.5 kJ/mol. Phosphocreatine + ADP ® creatine + ATP b. The enzyme phosphoglucomutase catalyzes the conversion of glucose 1-phosphate to glucose6-phosphate. Calculate the standard free energy change of this reaction if incubation of 20 mMglucose 1-phosphate (no glucose-6 phosphate initially present) yields a final equilibrium mixtureof 1.0 mM glucose 1-phosphate and 19 mM glucose 6-phosphate at 25°C and pH 7.0. c. If the rate of a nonenzymatic reaction is 1.2 x 10–2 μM s–1, what is the rate of the reaction at 37℃ inthe presence of an enzyme that reduces the activation energy by 30.5 kJ/mol?
- 2. Initial rate data for an enzyme following Michaelis-Menten kinetics are shown below when the enzyme concentration is 3 nmol/mL. A Lineweaver-Burk plot of this data set gives a line with a y-intercept of 0.00426 (ml-sec/umol). Substrate Concentration, [S] (uM) Ve (µmol/ml-sec) 320 169.0 132.0 160 80 92.0 40 57.2 20 32.6 10 17.5 Calculate kcat for the reaction and KM for the enzyme When the reactions using the enzyme are repeated in the presence of 12 uM of an uncompetitve inhibitor, the y-intercept of the resulting Lineweaver-Burk plot is 0.352 (ml-sec/umol). Calculate the K", for this inhibitor. a. b.The enzyme triosephosphate isomerase (TIM) catalyzes the following reaction in glycolysis, where it converts dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (GAP). CH₂OH C=O CH₂OPO²- DHAP triose phosphate isomerase [DHAP] (MM 1.00 2.00 3.00 6.00 O V. (mM/s) 3.700 6.727 9.250 14.800 = H HCOH CH₂OPO²- Kinetics experiments were performed on TIM. Enzyme activity (initial velocity, V.) was measured at varying concentrations of DHAP. The enzyme kinetics were also measured in the presence of two inhibitors, A and B. The enzyme concentration used in all experiments was 25.00 μM. The data are shown below. GAP V. (mM/s) + A 1.452 2.794 4.038 7.281 V₁ (mM/s) + B 0.755 1.379 1.905 3.077An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-s
- 4. Using a fluorescent model substrate, you study the kinetics of an enzyme-catalyzed reaction. You observe the below data. Write an equation describing reaction velocity (v) versus [S]. Define and provide a numerical value for any constants that you include in your equation. 16 14 12 4 2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 [S] / mM `N' at 50 mM and observe the following In pilot experiments, you add the putative inhibitor imidazole Doesn't affect anything 1 reaction velocities: [S] = 1 mM5 µM min-1 [S] = 2 mM9 9 µM min-1 [S] = 4 mM 911 µM min-1 Is imidazole a competitive or non-competitive inhibitor? Provide a brief explanation. > 9 00 v/ µM min-Enzyme A catalyzes the reaction S → P and has a KM of 50 μM and a Vmax of 100 nM s–1. EnzymeB catalyzes the reaction S → Q and has a KM of 5 mM and a Vmax of 120 nM s–1. When 100 μM ofS is added to a mixture containing equal amounts of enzymes A and B, which reaction product (Por Q) will be more abundant after 1 minute of reaction?1. An enzyme catalyzes a reaction at a velocity of 20 µmol/min when the concentration of substrate (S) is 0.01 M. The Km for this substrate is 1 x 10-5 M. Assuming that Michaelis-Menten kinetics are followed, what will the reaction velocity be when the concentration of S is (a) 1 x 10-5 M and (b) 1 x 10-6 M?
- 10) The graph shown below is the pre-steady state data for the serine protease (chymotrypsin) enzyme. Explain how this result suggests the formation of a covalent enzyme intermediate in the active site during the first step in the pathway. 3.0 2.0 1.0 1 3 Time (min) p-Nitrophenol (mol/mol of enzyme) 2.6. Examine the plot of the enzyme-catalyzed reaction below: Wheat-germ acid phosphatase shows Michaelis-Menten kinetics when acting on para-nitrophenol phosphate 16 14 12 10 8 4 2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 (S] / mM a) Is this reaction better characterized by Michaelis-Menten kinetics or simple mass action? b) What is the approximate value of Vmax? Km? c) Assuming that this plot was made at 100 nM enzyme, draw the curve, including axis labels, for 300 nM enzyme. d) Imagine that this enzyme is inhibited by a competitive inhibitor with Ki (for inhibitor) = Km (for substrate). Note that v = (kcat [E] tot [S]/Km) / (1 + [S]/Km + [I]/Ki). Draw the curve of reaction velocity (v) versus [S] for [I] = 1 mM. e) Now consider an experiment where you simultaneously introduce substrate and competitive inhibitor at equal concentrations. Make a plot of v versus concentration of [S] = [I], i.e. the X-axis should be the concentration of both I and S, which are equal. v/ uM min-An enzyme catalyzes a reaction with a Km of 7.50 mM and a Vmax of 2.90 mM - s-1. Calculate the reaction velocity, un, for each substrate concentration. [S] = 2.75 mM mM · s- * TOOLS x10 [S] = 7.50 mM mM · s-! [S] = 11.0 mM mM - s-