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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Although graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction, sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained:
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.
Q is an analog of substrate A that binds to enzyme X and produces the following kinetics:[A] V0 (µmole/ml/min) [Q] = 0 [Q] = 0.5 µM [Q] = 2 µM1 µM 10 7 43 µM 20 16 1010 µM 35 31 2330 µM 43 41 3680 µM 47 46 43a) Plot the data in Lineweaver Burk plot form (Hint: there should be three plots of dataon your graph) and determine the following: the KM and Vmax of the enzyme X in theabsence of Q, and the KMapp the Vmaxapp at each concentration of Q. b) What type of inhibition is this?c) Calculate the inhibition constant (Ki) for Compound Q
Suppose that the data shown in the margin are obtained for an enzyme -catalyzed reaction. [S] (mM) V (mmol/ml min) 0.1 3.33 0.2 5.0 0.5 7.14 0.8 8.00 1.0 8.33 2.0 9.09 a.) Determine the Km and Vmaxb.) Assuming htat the enzyme present in the system had a concentration of 10-6M, calculate the turn -over number
1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062 1.2) If the Km of an enzyme for it's substrate remains constant as the concentration of the inhibitor icreaces, what can be said about the mode of inhibition and why? 1.3) calculate the turnover number for an enzyme, assuming Vmax is 0.5M.sec-1 and the concentration of the enzyme used is 0.002M . why is it usefull to know this? 1.4) discuss the mechanism of the bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues. make use of relavant graphs and diagrams to explain your answer.
Given the following data in enzyme-catalyzed reaction, what is the Vm, Km and type of inhibition of Experiment A?
You obtain a calculated Vmax of 4.26uM/s and a Km of 122.5uM from a kinetics experiment performed using 0.5uM enzyme. What is the catalytic efficiency of this enzyme?
Given the following information, calculate the catalytic efficiency of the enzyme. Step by step please [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμM
At what substrate concentration would an enzyme with a kcat of 25.0 s-1 and a KM of 3.5 mM operate at 25% of its maximal rate? How many reactions would the enzyme catalyze in 45 seconds when it is fully saturated with substate, assuming the enzyme has one active site?
(a) If the total enzyme concentration 1 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute?(b) Calculate kcat/KM for the enzyme reaction. Is this a fairly efficient enzyme? (as shown.)
The Km value of an enzyme-catalyzed reaction and its Vmax is 70 mmol/min. What is the rate of the reaction of the reaction when the substrate concentration is 7 × 10−2 mmol/min? a. 35 mmol/min b. 50 mmol/min c. 60 mmol/min d. 70 mmol/min
Based on the kinetic constants below, which enzyme will most efficiently catalyze conversion of the substrate into product? A) Vmax = 10 uM s-1, KM = 10 µM B) Vmax = 10 uM s-1, KM = 0.01 µM C) Vmax = 1000 uM s-1, KM = 500 µM D) Vmax = 1 uM s-1, KM = 1 µM E) Vmax = 200 uM s-1, KM = 10 µM

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