A. What are the Vmax and Km Values for the enzyme-catalyzed reaction without inhibitor? Please give units and explain your reasoning. (See note.) Vmons
A. What are the Vmax and Km Values for the enzyme-catalyzed reaction without inhibitor? Please give units and explain your reasoning. (See note.) Vmons
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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Please work this problem out thoroughly with detailed step by step explanation. Thank you!
Transcribed Image Text:ENZYME KINETICS; PATHWAYS & FEEDBACK INHIBITION
3-1. The kinetics of an enzyme are measured as a function of substrate concentration in the presence or absence of
an inhibitor. The following results are obtained:
UnitsiKm> (mm)
units: Vmon > (nm/mn)
Velocity (nanomoles/min)
No inhibitor
55.8 x 103
54.9 x 10
51 x 10
47 x 10
40 x 103
28 x 10
18.7 x 10
11.2 x 103
2.7 x 10°
Substrate Concentration
in millimoles (mM)
With inhibitor
55.7 x 10
53.3 x 10
45 x 10
37.3 x 10
28 x 103
16 x 10
9.3 x 10
100
10
1
0.5
0.2
0.1
0.05
Not tested
0.01
Not tested
0.2 0°5
2.
A. What are the Vmax and Km Values for the enzyme-catalyzed reaction without inhibitor? Please give
units and explain your reasoning. (See note.)
Vmons >
Km →
B. (i). What are the apparent Vmax and Km values plus inhibitor?
(ii). Is this a case of competitive or noncompetitive inhibition? How did you know?
Note: You do not need a calculator or graph paper to figure any of this out. However you may want to plot the results on
graph paper to get a better feel for what is going on. If you do it that way, try to do it again without the graph, using the
numbers shown.
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