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 \( v_i \) and the degree of inhibition caused by a competitive inhibitor under the following conditions:
(a) \([S] = 2 \times 10^{-3} \, \text{M and} \, [I] = 2 \times 10^{-3} \, \text{M}\)
(b) \([S] = 4 \times 10^{-4} \, \text{M and} \, [I] = 2 \times 10^{-3} \, \text{M}\)
(c) \([S] = 7.5 \times 10^{-3} \, \text{M and} \, [I] = 10^{-5} \, \text{M}\)
Assume that \( K_m = 2 \times 10^{-3} \, \text{M}, \, K_i = 1.5 \times 10^{-4} \, \text{M} \) and \( V_{max} = 270 \, \text{nmoles} \times \text{liter}^{-1} \times \text{min}^{-1} \).
The degree of inhibition is the percent of the uninhibited velocity reached in the presence of the inhibitor.
15. (a) What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration of \( 1.5 \times 10^{-3} \, \text{M} \) if \( K_m = 2.9 \times 10^{-4} \, \text{M} \) and \( K_i = 2 \times 10^{-5} \, \text{M} \)?
(b) To what concentration must the substrate be increased to reestablish the velocity at the original uninhibited value?
**Note:** The questions are designed to test understanding of enzyme kinetics, focusing on competitive inhibition, the Michaelis-Menten equation, and the concept of \( K_i \), the inhibition constant.](/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 \( v_i \) and the degree of inhibition caused by a competitive inhibitor under the following conditions:
(a) \([S] = 2 \times 10^{-3} \, \text{M and} \, [I] = 2 \times 10^{-3} \, \text{M}\)
(b) \([S] = 4 \times 10^{-4} \, \text{M and} \, [I] = 2 \times 10^{-3} \, \text{M}\)
(c) \([S] = 7.5 \times 10^{-3} \, \text{M and} \, [I] = 10^{-5} \, \text{M}\)
Assume that \( K_m = 2 \times 10^{-3} \, \text{M}, \, K_i = 1.5 \times 10^{-4} \, \text{M} \) and \( V_{max} = 270 \, \text{nmoles} \times \text{liter}^{-1} \times \text{min}^{-1} \).
The degree of inhibition is the percent of the uninhibited velocity reached in the presence of the inhibitor.
15. (a) What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration of \( 1.5 \times 10^{-3} \, \text{M} \) if \( K_m = 2.9 \times 10^{-4} \, \text{M} \) and \( K_i = 2 \times 10^{-5} \, \text{M} \)?
(b) To what concentration must the substrate be increased to reestablish the velocity at the original uninhibited value?
**Note:** The questions are designed to test understanding of enzyme kinetics, focusing on competitive inhibition, the Michaelis-Menten equation, and the concept of \( K_i \), the inhibition constant.
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