Assume that an enzyme-catalyzed reaction follows the scheme shown:E + SESE + Pk₁ = 1 x 10%/M-s k-1 = 2.5 x 10%/s k2= 3.4 x 107sWhat is the dissociation constant for the enzyme-substrate, Ks?What is the Michaelis constant, Km, for this enzyme?What is the turnover number, Kcat, for this enzyme?What is the catalytic efficiency for the enzyme?If the initial Et concentration is 0.25mM, what is Vmax?

Michaelis Menten postulated that free enzyme reacts with the substrate reversibly to form an…

Answered: Assume that an enzyme-catalyzed…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter23: Fatty Acid Catabolism

Section: Chapter Questions

Problem 21P: Using the ActiveModel for enoyl-CoA dehydratase, give an example of a case in which conserved...

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Using the ActiveModel for enoyl-CoA dehydratase, give an example of a case in which conserved residues in slightly different positions can change the catalytic rate of reaction.
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"
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?
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
In enzyme kinetics, for the reversible with two central complexes mechanism, please provide complete proof that the rate equation is the equation below. The variables denoted with f indicate forward direction while the variables denoted with b indicate backward direction.
For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.
when saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.
When enzyme solutions are heated, there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at 45 °C lost 50% of its activity in 12 min, but when incubated at 45 °C in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 min. Suggest why thermal denaturation of hexokinase was retarded in the presence of one of its substrates.
If a 0.1 M solution of glucose 1- phosphate at 25 °C is incubated with a catalytic amount of phosphoglucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are Calculate Keq and ΔG′° for this reaction.
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.
When enzyme solutions are heated, there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at 450C lost 50% of its activity in 12 minutes, but when incubated at 450C in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 minutes. Suggest why thermal denaturation of hexokinase was retarded in the presence of one substrates
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?

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