87-6-2-1-4590135180225 27031536040545049554058![Substrate] (nM)What is the Vmax of the enzyme data shown above?rate (nM/sec)3.

Michaelis Menton kinetics: This is well known model for Enzyme kinetics. This model relates rate of…

Answered: 8- 2- 315 405 [Substrate] (nM) 45 90…

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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Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km? Vmax = 5 umol min^-1, Km = 2.5 mM? Vmax = 5 mmol min^-1, Km = 25 M? Vmax = 5 umol min^-1, Km = 25 mM? Vmax = 5 mol min^-1, Km = 2.5 mM? Vmax = 5 mol min^-1, Km = 25 mM?
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?
If you want to determine the KM for lactate, what protocol do you set up? Discuss the significance of the following kinetic parameters that are used to characterize enzyme activity: KM, Vmax, kcat, and kcat / KM.
How to identify the Vmax and the Michaelis constant (Km) in an enzyme kinetics graph?
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
The following were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: Reaction Velocity (mmol/min) Substrate added (mmol/L) 217 0.8 325 2 433 4 488 6 647 20 652 1000 The Km for this enzyme is approximately _____________. (Round to the nearest integer)
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"
From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.
The turnover number for an enzyme that approximates Michaelis-Menten kinetics is known to be 500 min^-1. From the results shown in the table, enumerate Km and total amount of enzyme present. What is the Km for this enzyme? What is the Vmax for this enzyme? And what is the [E]T for this enzyme?
The following kinetic data were obtained for an enzyme in the absence of an inhibitor (1) and in the presence of two different inhibitors (2) and (3) at 5mM concentration. [S], mM (mmol/mL/sec) (mmol/mL/sec) (mmol/mL/sec) (1) (2) (3) 1 12 4.3 5.5 2 20 8 9 3 29 14 13 8 35 21 16 12 40 26 18 Determine the Vmax and KM of the enzyme Determine the type of inhibition.
NOTE: the enzyme-inhibitor complex requires 450 joule/mol to dissociateJOULE/MOL
An enzyme catalysed reaction has a Km of 8 mM and a Vmax of 13 nM.s-1. Use the Michaelis-Menten equation to calculate the reaction velocity when the substrate concentration is 18 mM.

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8- 2- 315 405 [Substrate] (nM) 45 90 135 180 225 270 360 450 495 540 58! What is the Vmax of the enzyme data shown above? 7. 6. 4. rate (nM/sec)