For an enzyme obeying the Michaelis-Menten equation with Km = 5 µM, kcat = 10 s-¹and a total enzyme concentration of 1 nM,Calculate Vmax.Calculate the substrate concentration at which v = 0.1 VmaxCalculate the substrate concentration at which v = 0.9 VmaxWhat 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 axeson your plot.Label on the graph:i) the point on the graph at which S=Kmii) VmaxAt low substrate concentration, v = Keat [Etot] [S]KmCircle 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 thefastest enzymes?

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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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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.
Make a v vs. [S] plot for an enzyme that obeys Michaelis-Menten kinetics with a Vmax of 150 mM/min and Km of 1.5 mM.
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?
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
An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-s
At what substrate concentration would an enzyme with a kcat of 33.0 s-1 and a Km of 0.0046 M operate at one-tenth of its maximum rate? Assume the enzyme obeys Michaeli s-Menten kinetics.
An enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :
An enzyme catalyzes a reaction at a velocity of 20 micromole/min when the concentration of substrate (S) is 0.01 M. The Km for this 1 X 10^-5 M. Assuming that Michaelis-Menten Kinetics are followed, what will the reaction velocity be when the concentration of S is 1.0 X 10^-6 M?
An enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is 1 ×10-6 M?
An enzyme that follows Michaelis-Menten kinetics has a KM value of 3.00 µM and a keat value of 181 s1. At an initial enzyme concentration of 0.0100 µM, the initial reaction velocity was found to be 1.07 x 10-0 µM/s. What was the initial concentration of the substrate, S, used in the reaction ? Express your answer in micromolar to three significant figures. > View Available Hint(s) ? [S] !! µM Submit
Vmax and Km were calculated for enzyme substrate data. Results Y=0,0063x+0,3929 Km =62,4 mikromol/L Vm =159 mikromol/L.min But two additional data pairs were obtained at low concentrations and new graph equation Y=0,0063x+0,391 Recalculate Vmax and Km and compare and comment with previous results.
An enzyme that follows Michaelis-Menten kinetics has a KM value of 10.0 μM and a kcat value of 206 s−1 . At an initial enzyme concentration of 0.0100 μM , the initial reaction velocity was found to be 1.07×10−6 μM/s . What was the initial concentration of the substrate, [S] , used in the reaction ? Express your answer in micromolar to three significant figures..

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