Sweetzyme IT Extra is a trademarked immobilized enzyme formulation of glucose isomerase sold by the Danish company Novozymes. The following reaction is carried out in a 4.0 L reactor in the presence of an non-competitive inhibitor, I: E + G <=1=> EG =2=> E + F EG+1 <=3 => EIG G = glucose, F = fructose The Sweetzyme beads added to the reactor have a total surface area, a = 342 cm². Kinetic Parameters k₁= 0.00820 M-¹ s¹ k.₁ 0.00190 s¹¹ k₂ = 0.570 s¹ Initial/Total Enzyme concentration: Initial Bulk Substrate Concentration: Mass transfer coefficient, Eo 2.30 x 105 M Sb = 0.152 M k₁= 9.15 x 10 cm/s A. What is the maximum reaction velocity with no inhibitor (in mol/L-s)? (recall that Rapid Equilibrium and Quasi steady state approaches predict the same value for Vmax for this mechanism) B. What is the maximum reaction velocity (in mol/L-s) with the non-competitive inhibitor, I, at a concentration of [1] = 0.0029 M, with K₁= 1.8 x 10³ M.

Sweetzyme IT Extra is a trademarked immobilized enzyme formulation of glucose isomerase sold by the Danish company Novozymes. The following reaction is carried out in a 4.0 L reactor in the presence of an non-competitive inhibitor, I: E + G <=1=> EG =2=> E + F EG+1 <=3 => EIG G = glucose, F = fructose The Sweetzyme beads added to the reactor have a total surface area, a = 342 cm². Kinetic Parameters k₁= 0.00820 M-¹ s¹ k.₁ 0.00190 s¹¹ k₂ = 0.570 s¹ Initial/Total Enzyme concentration: Initial Bulk Substrate Concentration: Mass transfer coefficient, Eo 2.30 x 105 M Sb = 0.152 M k₁= 9.15 x 10 cm/s A. What is the maximum reaction velocity with no inhibitor (in mol/L-s)? (recall that Rapid Equilibrium and Quasi steady state approaches predict the same value for Vmax for this mechanism) B. What is the maximum reaction velocity (in mol/L-s) with the non-competitive inhibitor, I, at a concentration of [1] = 0.0029 M, with K₁= 1.8 x 10³ M.

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

See similar textbooks

Similar questions

What type of inhibition is being shown in the following graphs?
Just 15-3
You are working on an enzyme that obeys standard Michaelis-Menten kinetics. What variable is the V, dependant on if the concentration of the substrate is substantially higher than the concentration of the enzyme? [S] [E] [ES] O [P] O not enough information provided
Please provide Typed solution in details, you can use only handwritten diagram
Many enzyme -catalyzed reactions are consistent with a modified version of the Michaelis -Menten mechanism in which the second step is also reversible. For this mechanism obtain an expression for the rate of formation of product and find its limiting behavior for large and small concentrations of substrate using steady state approximation
. The optimal conditions for salivary lysozyme (hydrolyzing glycoproteins of bacterial wall) are 37 C - temperature and pH is 5.2. Explain the decrease in this enzyme activity if the temperature will rise up to 60 °C and pH will be changed to 8.0. To answer the question: a) draw the graph of the velocity dependency on temperature and pH; b) calculate the relative enzyme activity if 10 mg of lysozyme catalyzes the formation of 5 uM of the product per 2 minutes. Concidor NH3: 5.
Which strategies listed does this enzyme use to facilitate cleavage of α-galactoside? Choice 1 of 6:Dehydration synthesis Choice 2 of 6:acid base catalysis Choice 3 of 6:covalent catalysis Choice 4 of 6:metal ion catalysis Choice 5 of 6:stabilizing the transition state Choice 6 of 6:binding with high affinity to the product
(b) You are investigating the effects of several agents on the activity of alcohol dehydrogenase. The enzyme activity data are shown in the table below. Construct a [substrate] vs. activity plot and a double-reciprocal plot for this enzyme. Be sure to label all axes. Determine the Vmax and KM for AD from the graphs in each type of plot. AD activity (nM/min) AD activity + agent A (nM/min) AD activity + agent B (nM/min) [Alcohol] (nM) 0.1 14 2 0.5 50 7 8. 1.0 65 10 30 2.0 72 12 45 4.0 80 14 62 8.0 85 15 75 32.0 90 16 90
enzyme-inhibitor complex requires 450 kJ.mol-1 to dissociate and that it displays kinetics somehow similar to non-competitive inhibition, does this is inhibitor good to use to inhibit toxanthine oxidase in the case of hyperuricemia and gout?
Calculcate Kcat for PNP substrate for both enzyme concentrations. enzyme volume: 20 ul Bovine Intensince Alkaline phosphatase molecular weight: 140,000 Bovine intenstine Alkaline phosphatase activity: 300 units/ml and 14 units/mg extinction coefficient PNP: 18.5 abs (mM-1 cm-1) Vmax: 0.332 moles/sec a) enzyme 1 concentration: undiluted b) enzyme 2 concentration: 1:1 dilution
A purified protease enzyme from the fungus Aspergillus sp. Is tested in a laboratory. This enzyme was lyophilized as a white powder. When reconstituting with phosphate buffer pH 7.2 the active enzyme is obtained. To check its purity, an electrophoresis is performed where a single band of approximately 70,000 molecular weight is observed. The solution with enzymatic activity was stored at 4ºC for later use. A few days later it was found that the enzymatic activity had been lost and in the electrophoretic analysis, instead of a single band there were three bands of weights 40,000, 20,000 and 10,000. Come up with a reasoned explanation of what might have happened to the enzyme
KINETIC CONSTANT No Na2HPO4 25mM Na2HPO4 50mM Na2HPO4 Vmax nmol p-NP. Min- 20.3252 14.30615 17.30104 Km mM -0.819106 -0.46495 -0.352941 1. What does this suggest about the structure of the active side of the enzyme?