What is the Vmax of an enzyme at 1 µM with an activation free energy (AG#) of 55000 J/mol at 298K?-AG #RTkcat= 6.2× 10¹²(Ideal gas constant R = 8.314 J K¯1 mol-¹1)0.14 μM s-1-1S-11.4 μM s14 μM s-10.14 mM s-11.4 mM s-1

Catalytic efficiency of enzymatic reaction (turnover)K values tell the speed or rate with which a…

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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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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?
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What is the Vmax of an enzyme at 1 µM with an activation free energy (AG*) of 55000 J/mol at 298K? -4G* RT k=6.2x 10¹² cat (Ideal gas constant R = 8.314 J K¹ mol-¹) 0.14 μM S-1 1.4 μM s-1 14 μM s-1 0.14 1.4 mM S-1 0000 mM s-1

What is the Vmax of an enzyme at 1 µM with an activation free energy (AG) of 55000 J/mol at 298K? -4G RT k=6.2x 10¹² cat (Ideal gas constant R = 8.314 J K¹ mol-¹) 0.14 μM S-1 1.4 μM s-1 14 μM s-1 0.14 1.4 mM S-1 0000 mM s-1