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

Chapter25: Nitrogen Acquisition And Amino Acid Metabolism

Section: Chapter Questions

Problem 22P

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The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1
The rate constants of an enzyme-catalyzed reaction, obeying the Michaelis-Menten kinetics, have been determined : E + S K₁ = 2 x 108 M-¹ S-¹ -1 -1 -1 K-₁= 1 x 10³ S K₂ = 5 x 10³ S-1 K₁ 1 K-1 ES K₂ E +P 1- Determine the Michaelis constant Km of the enzyme. 2- Determine the catalytic constant (kcat) of the enzyme. 3- Determine the catalytic efficiency of the 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
For an enzyme obeying the Michaelis-Menten equation with Km = 5 µM, kcat = 10 s-¹ and a total enzyme concentration of 1 nM, Calculate Vmax. Calculate the substrate concentration at which v = 0.1 Vmax Calculate the substrate concentration at which v = 0.9 Vmax What fraction of the enzyme is bound to substrate when v = 0.9 Vmax? Sketch a graph showing the Michaelis-Menten plot. Make sure you label the axes on your plot. Label on the graph: i) the point on the graph at which S=Km ii) Vmax At low substrate concentration, v = Keat [Etot] [S] Km Circle on your graph where this equation applies. What name is used to refer to keat/Km? : Calculate keat/Km for this enzyme. How does this value compare with the fastest enzymes?
The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V for an enzyme-catalyzed, single-substrate reaction E + SES →E + P. The model can be more readily understood when comparing three conditions: [S] > Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity V, where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] > Km Reaction rate is independent of [S]. Not true for any of these conditions The rate is half of the maximum rate.
Given the following reaction and equation for the initial velocity of the reaction: k₁ k3 E+S ES E + P V=Keat [ES] = k3 [ES] k₂ where keat is the rate constant for the reaction which forms the product from the ES complex. Explain in words why the velocity is directly proportional to theamount of enzyme added in the presence of saturating substrate levels.
The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate Vo for an enzyme-catalyzed, single-substrate reaction E + S ES → E + P. The model can be more readily understood when comparing three conditions: [S] > Km- Match each statement with the condition that it describes. Note that "rate" refers to initial velocity Vo where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] > Km Not true for any of these conditions Almost all active sites will [ES] is much lower than [Efree]. be filled. The rate is directly proportional to Increasing [Etotal] will increase [S]. Km: Adding more S will not increase [Efree] is equal to [ES]. the rate.
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.
An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min-1 when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme in µmol⋅min−1?
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?
For an enzyme catalyzed reaction of the form: S + E → P + E, the rate of product formation, [P], is given by: d[P]/dt = k2[E}total [S] /(Km + [S]) = For the enzymatically catalyzed hydrolysis of ATP at 25 °C and pH 7.0, the Michaelis Menten constant, Km was found to be equal to 16.8 μmol L 1 and the value of k2[E]total was found to be 0.220 μmol L-¹s¹. Find the initial rate at an initial ATP concentration of 30.0 μmol L-1
k_1 E + SES E + P +0 k₂ S ↓↑ KIS ESS Based on this model, please answer the following questions: i) This is the model for what? Uncompet tive ii) Show the rate equation for an enzyme reaction of this type. iii) Show the Lineweaver-Burk plot for this case. A

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