The initial rate for an enzyme-catalyzed reaction has been determined at several substrate concentrations.Data are as follows:[S] µmol/LV[(umol/L) min1]5221039206550102100120200135If the total enzyme concentration was 1nmol/L, how many molecules of substrate can a molecule ofenzyme process in each minute?A. 175000B. 325000C. 900000D. 10000E. 375000F. 166000G. 3750H. 1000000I. 12500

Enzyme-substrate reactions that follow Michaelis-Menten Kinetics show a characteristic relation…

Answered: The initial rate for an…

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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The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.
Below is kinetic data obtained for an enzyme-catalyzed reaction. The enzyme concentration is fixed at 100 nM. Using a Lineweaver-Burke plot, calculate the kcat value for this reaction. Report your answer to three significant figures in units of 1/sec.
From the chart below, comment on the temperature dependence of this enzyme catalyzed reaction.
At what substrate concentration would an enzyme with a Km of 0.005M operate at one quarter of its Vmax? What would be the Km for an enzyme if it is operating at 90% of its Vmax when the substrate concentration is 0.01M
You run a series of assays at 25°C on enzyme A. You measure the velocity for a range of S concentrations. What is the Km (in mM) for Enzyme A?
Given the following equations of the line, a) determine the KM of the substrate that binds the strongest with the enzyme. b) determine which of the following substrates binds the least with the enzyme
The enzyme, fumarate, has the following kinetic constants: k 1 k 2 k -1 where k 1 = 10 9 M -1 s -1 k -1 =4.4 x 10 4 s -1 k 2 = 10 3 s -1 a. What is the value of the Michaelis constant for this enzyme? b. At an enzyme concentration of 10 -6 M, what will be the initial rate of product"
In Lineweaver Burk plot: O a. The slope of the plot is the Vmax/Km O b. The intercept on X-axis is 1/Km. O c. The intercept on y-axis is 1/Vmax O d. double-reciprocal plot of enzyme kinetics is generated, where 1/VO is plotted as a function [S]. O e. All answers are correct
The KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 104 M. At a substrate concentration of 0.20 M, v = 43 μmol/min. Calculate the rate of reaction when the substrate concentration is tenfold lower, 0.020M. 43 μmol min rate of reaction → Vmax Км substrate concentration →
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (µM). 1.3 - Inhibitor + Inhibitor 2.50 0.62 2.6 4.00 1.42 6.5 6.30 2.65 13.0 7.60 3.12 26.0 9.00 3.58 What is the Vmax of the inhibited enzyme reaction?
Compound A is the substrate for two enzymes, El and E2, their reaction rates, r1 and r2,Determine the Km and rmax for both enzymes, with respect to the concentration of A. Which set of data is more likely to be for El and which for E2, and why? Concentration of 0.2 0.6 1.2 3 4 5 6 8 12 15 A (mM) Reaction rate (r.) (mmol/L'min) 3.33 4.29 4.62 4.76 4.84 4.88 4.9 4.92 4.94 4.95 4.96 4.97 Reaction rate (r,) (mmol/L*min) 0.09 0.23 0.38 0.5 0.6 0.67 0.71 0.75 0.8 0.82 0.86 0.80
From a series of flasks with a constant concentration of enzyme the following initial velocities weretaken, they were obtained as a function of the concentration of the substrate.a) Calculate the KM and Vmax kinetic parameters of the three forms (Lineweaver-Burk, Eadie-Hofstee, Dixon).b) Analyze which are the atypical data that cause a low correlation, which can be eliminated and explain youranswer.

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