ES → ES'v = k,[ES]ES' → E+ Pv = k,[ES']Show that the rate of formation of product has the same form as that shown inegn 17F.16, but with vmar and K, given bykk[E]oV max =k.(K+k,)KM =K,(k, +k. )P17F.6 For many enzymes, the mechanism of action involves the formation oftwo intermediates:E+S→ ESv = k,[E][S]ES → E +Sv = k, [ES]

Answered: Image /qna-images/answer/62ddb7fb-b57a-4d1b-8f9e-b1bb69e0b873.jpg

Answered: ES → ES' v = k,[ES] ES' → E+ P v =…

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter11: Chemical Kinetics: Rates Of Reactions

Section: Chapter Questions

Problem 83QRT

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An enzyme-catalyzed reaction has a Km of 1 mM and a Vmax of 4 nmole/L sec-1. The reaction velocity (nmole/L sec-1) when the substrate concentration is 0.25 mM is: A. 1.25 B. 10.0 C. 5.0 D. 0.50 E. 1.0 F. 100 G. 150 H. 55 I. 75
The enzyme-catalysed conversion of a substrate at 25 oC has a Michaelis constant of 0.015 mol dm-3 . and a max-imum velocity of 4.25 x 10-4 mol dm-3 s-1 when the enzymeconcentration is 3.60 x 10-9 mol dm-3. Calculate koat and the catalytic efficiency η. Is the enzyme 'catalytica lly perfect'?
An enzyme kinetics experiment is carried out by adding 1.00mg of a 50.0kDa enzyme into a total volume of 50uL of buffer, and measuring the initial rates while adding increasing amounts of substrate (1 Da = 1 g/mol). The calculated Vmax from this series of experiments was 30.5mM/s. Calculate the turnover number for the enzyme.
The enzyme-catalysed conversion of a substrate at 298 K has KM = 0.032 mol dm−3 and vmax = 4.25 × 10−4 mol dm−3 s−1 when the enzyme concentration is 3.60 × 10−9 mol dm−3. Calculate the catalytic efficiency of the enzyme, as defined in Exercise E17F.5(a).
Determine the value of the turnover number of the enzyme catalase, given that RmaxRmax (Vmax)(Vmax) for catalase is 41 mmol⋅L−1⋅s−141 mmol⋅L−1⋅s−1 and [E]t[E]t equals 4.8 nmol⋅L−14.8 nmol⋅L−1 . Catalase has a single active site.
An enzyme catalyzes a reaction with a KmKm of 7.507.50 mM and a VmaxVmax of 3.00 mM⋅s−1.3.00 mM⋅s−1. Calculate the reaction velocity, v0,v0, for each substrate concentration. a. [S]=4.00mM*s^-1 b. [S]=7.5mM*s^-1
An enzyme with a single active site has a Michaelis-Mentin constant (KM ) of 25 mM and aturnover number of 4.0 x 107 s –1 .(a) Calculate the initial rate of the reaction in μM/s if the total enzyme concentration is0.016 μM and the initial substrate concentration is 4.32 μM.(b) Calculate the value of Rmax
For an enzyme that follows Michaelis –Menten kinetics, k1=1X106 M-1 sec-1, k-1= 2x103 sec-1 and k2=2X 102 sec-1. a. What is the Km for the enzyme ? If the enzyme concentration is 10 nanomoles in 1 ml, what is Vmax for the enzyme ? What is the catalytic efficiency for this enzyme?
A An enzyme that follows Michaelis-Menten kinetics has a KM value of 16.0 uM and a kcat value of 181 s-1. At an initial enzyme concentration of 0.0100 uM, the initial reaction velocity was found to be 1.07 x 10- uM/s. What was the initial concentration of the substrate, [S], used in the reaction ? Express your answer in micromolar to three significant figures
The first-order rate constant, k1, for the decomposition of ampicillin at pH 5.8 and 35?C is k1 = 2 × 10-7 sec-1. The solubility of ampicillin is 1.1 g/100 mL. If it is desired to prepare a suspension of the drug containing 2.5 g/100 mL, calculate the zero-order rate constant, k0, and the shelf-life, that is, the time in days required for the drug to decompose to 90% of its original concentration (at 35?C) in solution. Note: 100 mL = 1 deciliter = 1 dL.
The equilibrium constant for the attachment of a substrate to the active site of an enzyme was measured as 200. In a separate experiment, the rate constant for the secondorder attachment was found to be 1.5 x 108 dm3 mol-1 s- 1. What is the rate constant for the loss of the unreacted substratefrom the active site?
For a simple reaction, S 1 P, k1 is the rate of the forward reaction and k-1 is the rate of the backwards reaction. The k1 in the absence of any enzyme was 0.01/s. The k1 in the presence of enzyme was 5,000,000/s. Calculate the rate enhancement (k1 catalyzed divided by k1 uncatalyzed) provided by the enzyme. If the equilibrium constant in the absence of enzyme is 1 x 104, what is the equilibrium constant (Keq) in the presence of enzyme?

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