For an enzyme catalyzedreaction of the form: S +E → P + E, the rate ofproduct formation, [P], isgiven by:d[P]/dt = k2[E}total [S] /(Km+ [S]) =For the enzymaticallycatalyzed hydrolysis ofATP at 25 °C and pH 7.0,the Michaelis Mentenconstant, Km was found tobe equal to 16.8 μmol L1 and the value ofk2[E]total was found to be0.220 μmol L-¹s¹.Find the initial rate at aninitial ATP concentration of30.0 μmol L-1

Km is Michael's constant which is equal to substrate concentration at half of maximum velocity of…

Answered: For an enzyme catalyzed reaction of the…

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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The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: Vo (micromol/min) Substrate added (mM) 200 1.0 400 2.0 800 4.0 1600 8.0 3200 16.0 The enzyme concentration used in all reactions is 106 M. Which of the following is true: O A) The Km for the reaction must be less than 1 mM ○ B) It is possible to determine Km for the reaction, but not Vmax or kcat OC) It is possible to determine only kcat, not Vmax or Km D) It is not possible to determine Vmax for the reaction OE) The ratio Vmax/Km for the reaction cannot be determined
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.
The accompanying figure shows three Lineweaver–Burk plots for enzymereactions that have been carried out in the presence, or absence, of aninhibitor. Indicate what type of inhibition is predicted based on eachLineweaver–Burk plot. For each plot indicate which line corresponds to thereaction without inhibitor and which line corresponds to the reaction withinhibitor present.
The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. Vmax [S] Km + [S] where v is the velocity, or rate, Vmax is the maximum velocity. K is the Michaelis-Menten constant, and [S] is the substrate concentration. A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (ro) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. Next, move the line labeled 1/2 Vmax to its correct position. Then, move the line labeled Km to its correct position. Estimate the values for Vmax and Km- Vmax= µM/min v (µM/min) 300 275 250 225 200 175 150 125 100 75 50 Km = 25 K 0 10 20 30 V max 40 50 [S] (M) 1/2 V max Michaelis-Menten curve 60 70 80 90 100 HM
Consider an enzyme that follows standard Michaelis-Menten kinetics and has the following kinetic constants: %3| k2 = 1.5 x 10? s1 Еo 3D 1 х 104 М = 1 x 104 M a. What is the value of the maximum rate VM? b. Prepare a hand-drawn quantitative plot on graph paper (not a simple sketch nor an EXCEL-generated graph) of the enzymatic reaction rate versus substrate concentration (like Fig 3-3) using the kinetic parameters given above. Be sure to label the axes and include numeric values on the axes. C. Based upon your hand drawn saturation plot, at what substrate concentration is the enzymatic reaction rate 75% of Vm?
The turnover number for an enzyme is known to be 5000min-1. Given the following set of data, Substrate concentration(mM) 1, 2, 4, 6, 100, 1,000 Initial Rate(micromol/min) 167, 250, 334, 376, 498, 499 a) What is the Km of the enzyme for the substrate? (do this without using a calculator) b) What is the total amount of enzyme present in the assay?
The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)
From a kinetics experiment, kcat was determined to be 295sec-1. For the kinetic assay, 0.3mL of a 0.25mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 3mL. Calculate Vmax (µM∙min-1) for the enzyme and catalytic efficiency ( in M-1sec-1) for the enzyme. The Km for the enzyme was determined to be 2.55 x 10-2M.
Consider a general reaction enzyme A(aq) B(aq) The AGo of the reaction is -9.150 kJ mol-. Calculate the equilibrium constant for the reaction at 25 °C. Keg = What is AG for the reaction at body temperature (37.0 °C) if the concentration of A is 1.9 M and the concentration of B is 0.50 M? 9:26 PM 35% 77°F 6/14/2021
Calculate the overall ΔG° (report up to two decimal places) for the net reaction (see attached image). Answer: _____ kcal/mol Note: R = 1.98 x 10 -3 kcal/mol-K
ATP can be hydrolyzed into ADP + inorganic phosphate (P.) with a AG of -30500 J/mol. If, galactose-1-phosphate can be hydrolyzed into galactose + P, with a AG of -15.3 kJ/mol what is the Gibbs free energy change for the reaction of galactose + ATP = galactose-1-phospate + ADP? O a. -45.8 kJ/mol O b. 15.2 kJ/mol Oc. -30484.7 J/mol O d. -15.2 kJ/mol Oe. -1520 J/mol

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