It takes 9.4 msec for Enzyme X to catalyse a single reaction when it is fully saturated with substrate. Calculate the Vmax ofthe reaction if the concentration of Enzyme X is 2.4 mM.Select one:a. 26 mM/sec Xb. 3.9/secc. 0.26 M/secd. 0.039 M/sece. 106.4 M/sec

A chemical reaction is a process in which the reactant molecules convert to product molecules. These…

Answered: It takes 9.4 msec for Enzyme X to…

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

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Problem 1P

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An enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :
Suppose that the data shown in the margin are obtained for an enzyme -catalyzed reaction. [S] (mM) V (mmol/ml min) 0.1 3.33 0.2 5.0 0.5 7.14 0.8 8.00 1.0 8.33 2.0 9.09 a.) Determine the Km and Vmaxb.) Assuming htat the enzyme present in the system had a concentration of 10-6M, calculate the turn -over number
During a test of kinetics of an enzyme-catalyzed reaction, the following data were recorded: E, () T(C) I (mmol/ml) S (mmol/ml) V (mmol/ml-min) 1.6 1.6 30 0.1 2.63 1.92 147 30 0.033 1.6 30 0.02 0.01 0.005 16 30 0.96 0.56 5.13 1.6 30 1.6 49.6 0.1 1.6 49.6 0.033 3.70 1.89 1.6 1.6 1,6 49.6 0.01 0.0067 49.6 1.43 49.6 0.005 LII 0.92 30 0.1 1.64 0.02 0.01 0.1 0.90 0.92 0.92 30 30 0. 0.58 0.92 30 0.6 0.92 0.92 30 30 0.6 0.6 0.033 0.02 1.33 0.80 0.57 a. Determine the Michaelis-Mentcn constant for the reaction with no inhibitor present at 30°C and at 49.6°C. b. Determine the maximum velocity of the uninhibited reaction at 30°C and an enzyme concentration of 1.6 g/l. c. Determine the KI for the inhibitor at 30*C and decide what type of inhibitor is being used.
You are evaluating the kinetics of an enzyme catalyzed reaction containing 5.5 μM total enzyme and 11.2 μM substrate. At this substrate concentration, you determine that the Vo = 88.6 μmol mL-¹. s-¹. If the Vmax 833.3 mM s the KM is: . == " 30.9 μΜ 10.4 μΜ Ο 124.6 μΜ 234 μΜ 94.5 μM
Given the following enzyme kinetics data, determine the maximal velocity and the Michaelis constant of the enzyme. Show your solutions
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The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)
An enzyme is present at 100 nM (nanomolar) and has a Vmax value of 25 uM/s (micromolar/second). The Km for the substrate is 5.2 uM. What is the initial velocity (V0) at a substrate concentration of 15.2 uM? Report your answer to three significant figures in units of uM/s.
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The following results were obtained for an enzyme-catalysed reaction Substrate concentration (mmol r'): 5.0 Initial velocity (umol l' min'): 6.67 10.0 20.0 40.0 147 182 233 323 400 Calculate Km and Vmax
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It takes 9.4 msec for Enzyme X to catalyse a single reaction when it is fully saturated with substrate. Calculate the Vmax of the reaction if the concentration of Enzyme X is 2.4 mM. Select one: a. 26 mM/sec x b. 3.9/sec c. 0.26 M/sec d. 0.039 M/sec e. 106.4 M/sec