4. Enzymes with a kcat / Km ratio of about 10³ M-¹s-¹ are considered to show optimal catalytic efficiency.Fumarase, which catalyzes the reversible-dehydration reactionfumarate + H₂0malatehas a ratio of turnover number to the Michaelis-Menten constant, (kcat / Km) of 1.6 × 108 for the substratefumarate and 3.6 x 107 for the substrate malate. Because the turnover number (Kcat) for both substrates isnearly identical, what factors might be involved that explain the different ratio for the two substrates?

The Kcat/Km ratio is a measure of the catalytic efficiency of an enzyme. Kcat (also known as…

Answered: 4. Enzymes with a kcat / Km ratio of…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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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-
7. Glucose-6-phosphate dehydrogenase catalyzes the reaction shown below. Kinetic data for the enzyme isolated from the thermophilic bacterium T. maritima are presented in the following table. CH₂OPO H но CH₂OPO OH H H OH Н OH Glucose-6-phosphate Substrate NADP+ NAD+ NADP+ Glucose-6-phosphate NADPH + H+ glucose-6- phosphate dehydrogenase H 12.0 НО H OH H Η OH 6-Phosphoglucono-8-lactone KM (mM) Vmax (Umg-¹) 0.15 20 0.03 20 6 a. NADPH inhibits glucose-6-phosphate dehydrogenase. What kind of inhibitor is NADPH? b. How are the values of KM and Vmax likely to differ from those listed in the table if NADPH is present? c. Does glucose-6-phosphate dehydrogenase prefer NAD+ or NADP+ as a cofactor?
6. Sweetzyme® IT Extra is a trademarked immobilized enzyme formulation of glucose isomerase sold by the Danish company Novozymes. The following reaction is carried out in a 4.0 L reactor in the presence of an non-competitive inhibitor, I: E + G EG = 2 => E + F EG+1 EIG G=glucose, F = fructose The Sweetzyme beads added to the reactor have a total surface area, a = 342 cm². Kinetic Parameters k₁ = 0.00820 M₁¹ S¹ k.₁ = 0.00190 s¹ k₂ = 0.570 s¹ Initial/Total Enzyme concentration: Initial Bulk Substrate Concentration: Mass transfer coefficient, Eo 2.30 x 105 M Sb = 0.152 M k₁ = 9.15 x 10 cm/s A. What is the maximum reaction velocity with no inhibitor (in mol/L-s)? (recall that Rapid Equilibrium and Quasi steady state approaches predict the same value for Vmax for this mechanism) B. What is the maximum reaction velocity (in mol/L-s) with the non-competitive inhibitor, I, at a concentration of [1] = 0.0029 M, with K₁ = 1.8 x 10³ M. C. Calculate the Dahmköhler number to determine if the…
Many metabolites are maintained at steady-state concentrations that are far from equilibrium. A comparison of Kéq and Q, the mass-action ratio, can determine whether a metabolic reaction is far from equilibrium. The equation for this equilibrium is, fructose 6-phosphate + ATP = fructose 1,6-bisphosphate + ADP Calculate Kg for this reaction at T = 25.0 °C. Ka 1.8 x10-6 AG' = -14.2 kJ/mol Incorrect Calculate the mass-action ratio, Q, from the approximate physiological concentrations for rat heart tissue shown in the table. Metabolite Concentration (µM) Q = 3 fructose 6-phosphate 81.0 fructose 1,6-bisphosphate 15.0 Incorrect ATP 12,500 ADP 1,340
11. Calculate the equilibrium constant for the hydrolysis of glucose-1-phosphate at 25°C AG ° for the reaction, Glucose -1-phosphate+ water Glucose + Pi is - 20.9 KJ.mol-¹1 12. Calculate the Agº for the reaction Phosohoenolpuruvate ADP + Pi ===== Phosphoenolpyruvate + H₂O + ADP ==== Puruvate + ATP ATP + H₂0 AG° = 30.5 KJ.mol-¹ 13. Calculate the AG for the reaction 0 A + B ==== Pyruvate + Pi AG° = -61.9 KJ.mol-¹ C + D at 25°C When equilibrium concentrations are (A) = 10µM (B) = 15μM (C) = 3џM (D) = 5µM. Is the reaction exergonic and endergonic under standard conditions? 14. For the reaction A---B at 298K, the change in enthalpy is -7KJ.mol-¹ and the change in entropy is -25 J.K-¹.mol-¹¹. Is the reaction spontaneous? If not, the temperature be increased or decreased to make the reaction spontaneous? 15. A spherical bacterium with a diameter of 1μm contains 2 molecules of a particular protein. What is the molar concentration of the particular protein?
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.
3. The succinate dehydrogenase reaction in the direction it proceeds within the CAC is shown below (left to right): FAD FADH, H-C-H H-C-H coo- succinate dehydrogenase Succinate Fumarate AG" - O kl/mol a) Calculate the standard reduction potential for the overall reaction above. b) Based on the reduction potential, calculate and compare your standard transformed free energy change to the textbook value (above), and venture to explain any discrepancies. c) Assuming the temperature of the cell is 37 °C, calculate the equilibrium ratio of oxidized to reduced cofactor when the ratio of oxidized product to reduced substrate is 1.2. .
when saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.
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 3.077
13. Calculate the equilibrium constant K'eg, for each of the following reactions at pH 7.0 and 25°C. glucose + Pi a. Glucose 6-phosphate + H20 enz. Glucose 6-phosphatase; AG'O=-13.8kJ/mol
Enzyme X exhibits maximum activity at pH = 6.3. X shows a fairly sharp decrease in its activity when the pH goes much lower than 5.8. One likely interpretation of this pH activity is that: a Glu residue on the enzyme is involved in the reaction. a Tyr residue on the enzyme is involved in the reaction. a His residue on the enzyme is involved in the reaction the enzyme uses NADH has a cofactor. the enzyme uses coenzyme A has a cofactor.
1. Hydrolysis of 1 M glucose 6-phosphate catalyzed by glucose 6-phosphatase is 99% complete at equilibrium (i.e., 1% of the substrate remains). Which of the following statements is most nearly correct? (R= 8.315 J/mol-K; T = 298 K) A AG" is-11 kl/mol. B) AG" is +5 kJ/mol. C) AG" is 0 kJ/mol. DAG" is +11 kJ/mol. E) AG" cannot be determined from the information given. Explain your choice in one line: -2.16x298 (2) - 11386.15 -> 11.4 КТ 5 The high state conculto alle My negative and st" positive (either for D) then onditions, one

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