14. Calculate v, and the degree of inhibition caused by a competitive inhibitor under the following conditions: (a) [S] = 2 x 10-3 M and [I] = 2 x 103 M (b) [S] = 4 x 104 M and [I] = 2 x 10-3 M (c) [S] = 7.5 x 10-3 M and [I]= 10-$ M Assume that Km = 2 x 10-3 M, K; = 1.5 x 104 M and Vmax = 270 nmoles x liter x min-. The degree of inhibition is the percent of the uninhibited velocity reached in the presence of the inhibitor. %3D
14. Calculate v, and the degree of inhibition caused by a competitive inhibitor under the following conditions: (a) [S] = 2 x 10-3 M and [I] = 2 x 103 M (b) [S] = 4 x 104 M and [I] = 2 x 10-3 M (c) [S] = 7.5 x 10-3 M and [I]= 10-$ M Assume that Km = 2 x 10-3 M, K; = 1.5 x 104 M and Vmax = 270 nmoles x liter x min-. The degree of inhibition is the percent of the uninhibited velocity reached in the presence of the inhibitor. %3D
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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![Problems 14 and 15: some of the exponents are unclear. Here they are:
14. Calculate V; and the degree of inhibition caused by a competitive inhibitor
under the following conditions:
(a) [S]=2 x 103 M and [I] = 2 x 10-3 M
(b) [S] = 4 x 104 M and [I] = 2 x 10-3 M
(c) [S]= 7.5 x 10-3 M and [I] = 10-$ M
Assume that Km= 2 x 10-3 M, K; = 1.5 x 104 M
and Vmax = 270 nmoles x liter x min-.
The degree of inhibition is the percent of the uninhibited velocity
reached in the presence of the inhibitor.
%3D
15. (a) What concentration of competitive inhibitor is required to
yield 75% inhibition at a substrate concentration of 1.5 x 10-3 M if
Km = 2.9 x 104 M and K; = 2 x 10-5 M? (b) To what concentration
must the substrate be increased to reestablish the velocity at the
original uninhibited value?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe77e44c7-8f97-4332-8d23-eb24d773472e%2F8b817005-3f1e-458e-b06f-ec6e3f986fb7%2F5q0yny7b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problems 14 and 15: some of the exponents are unclear. Here they are:
14. Calculate V; and the degree of inhibition caused by a competitive inhibitor
under the following conditions:
(a) [S]=2 x 103 M and [I] = 2 x 10-3 M
(b) [S] = 4 x 104 M and [I] = 2 x 10-3 M
(c) [S]= 7.5 x 10-3 M and [I] = 10-$ M
Assume that Km= 2 x 10-3 M, K; = 1.5 x 104 M
and Vmax = 270 nmoles x liter x min-.
The degree of inhibition is the percent of the uninhibited velocity
reached in the presence of the inhibitor.
%3D
15. (a) What concentration of competitive inhibitor is required to
yield 75% inhibition at a substrate concentration of 1.5 x 10-3 M if
Km = 2.9 x 104 M and K; = 2 x 10-5 M? (b) To what concentration
must the substrate be increased to reestablish the velocity at the
original uninhibited value?
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