The following data were obtained in a study of an enzyme that is known to follow the Michaelis-Menten kinetics. Approximately, what is the value of Km tor this enzyme?Substrate added,umol/LVo uM/sec48971751871921972161001000198OA 1 uMO8.2 pMOC 4 uM.OD. 8 uMOE 16 IMOF 1000 uMOG 97 uM/secOH 175 UM/sec

Michaelis menten constant, Km is the substrate concentration required to produce half maximum…

Answered: The following data were obtained in a…

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

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The following were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: Reaction Velocity (mmol/min) Substrate added (mmol/L) 217 0.8 325 2 433 4 488 6 647 20 652 1000 The Km for this enzyme is approximately _____________. (Round to the nearest integer)
1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062 1.2) If the Km of an enzyme for it's substrate remains constant as the concentration of the inhibitor icreaces, what can be said about the mode of inhibition and why? 1.3) calculate the turnover number for an enzyme, assuming Vmax is 0.5M.sec-1 and the concentration of the enzyme used is 0.002M . why is it usefull to know this? 1.4) discuss the mechanism of the bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues. make use of relavant graphs and diagrams to explain your answer.
For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.
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? Vmax = 5 umol min^-1, Km = 2.5 mM? Vmax = 5 mmol min^-1, Km = 25 M? Vmax = 5 umol min^-1, Km = 25 mM? Vmax = 5 mol min^-1, Km = 2.5 mM? Vmax = 5 mol min^-1, Km = 25 mM?
You obtain a calculated Vmax of 4.26uM/s and a Km of 122.5uM from a kinetics experiment performed using 0.5uM enzyme. What is the catalytic efficiency of this enzyme?
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.
For an enzymatic reaction, the following data were obtained for two different initial enzyme concentrations: [S](g/L) v ([E0] = 0,015 g/L)(g/L.min) v ([E0] = 0,00875 g/L) (g/L.min) 40 2,28 1,34 20 1,74 1,02 13,4 1,4 0,82 10 1,18 0,68 8 1 0,58 (a)Calculate Vmax (in g/(L.min)) for the initial enzyme concentration equal to 0.015 g/L. (b)Calculate Vmax (in g/(L.min)) for the initial enzyme concentration equal to 0.00875 g/L. (c)Calculate Km (in g/L) for the initial enzyme concentration equal to 0.015 g/L. (d)Calculate Km (in g/L) for the initial enzyme concentration equal to 0.00875 g/L.
The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.
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?
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 following kinetic data were obtained for an enzyme in the absence of an inhibitor (1) and in the presence of two different inhibitors (2) and (3) at 5mM concentration. [S], mM (mmol/mL/sec) (mmol/mL/sec) (mmol/mL/sec) (1) (2) (3) 1 12 4.3 5.5 2 20 8 9 3 29 14 13 8 35 21 16 12 40 26 18 Determine the Vmax and KM of the enzyme Determine the type of inhibition.
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?

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The following data were obtained in a study of an enzyme that is known to follow the Michaelis-Menten kinetics. Approximately, what is the value of K for this enzyme?