Find kcat for a reaction in which Vmax is 4 × 10-4 mol-min-1 and the reaction mixturecontains one nanogram of enzyme. The molecular weight of the enzyme is 200,000Da (1 Da = 1 g/mol).2 x 10-13 min-12 x 10-11 min-¹80 min-¹8 x 107 min-¹8 x 10¹0 min-¹

Kcat or the turnover number is the the number of substrate molecules converted into product per…

Answered: Find kcat for a reaction in which Vmax…

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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An enzyme that catalyzes the reaction X Y is isolated from two bacterial species. The enzymes have the same Vmax but different Km values for the substrate X. Enzyme A has a Km of 2.0 µM, and enzyme B has a Km of 0.5 µM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 µM. Which curve corresponds to which enzyme? Explain. | Time
Given the line-weaver Burke plot below for Enzyme Y, identify the Vmax for Enzyme Y. -10 30 1/v 1/[S] 5 10 Enter the numerical value for Vmax only, no letters or symbols. Round answer to 2 significant figures.
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If a data from an enzyme experiment is plotted as a Lineweaver - Burk plot, and the Vmax is 0.02 sec/mol, and x-intercept is -2.5 mM-1, then what is the KM value?
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Find kcat for a reaction in which Vmax is 4 x 10-4 mol-min-1 and the reaction mixture contains one nanogram of enzyme. The molecular weight of the enzyme is 200,000 Da (1 Da 1 g/mol). = O2 x 10-13 min-1 2 × 10-11 min-1 80 min-1 8 × 107 min-1 8 x 1010 min-1