7. You obtain the following set of data for two enzymes that follow Michaelis-Menten kinetics:Enzyme AInitial velocity(µM/min)38Enzyme BInitial velocity(пМ/min)140Substrateconcentration (µM)2333502734133822946720400583150706682300750691786798697699100030006000800700(a) What are the approximate Km values for each enzyme based on these data? Brieflyjustify your answer.(b) In the above experiment, 0.1 µM of enzyme A or enzyme B were used. Calculate kçatand specificity constant for each enzyme. Which enzyme is more efficient?(c) The kinetics of Enzyme B were re-measured in the presence of a small moleculeinhibitor. The Km increased while the Vmax remained unchanged. What type of inhibitionis this and why does the Km increase?

Michaelis Menton kinetics describes the relation between rate of reaction and substrate…

Answered: 7. You obtain the following set of data…

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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Vo 60 50 40 30 20 mm/sec 0100 0 O 0 0 50 100 150 200 250 300 Substrate (mM) Use the Michaelis-Menten plot to answer this question. What is the estimated value of Vmax of the enzyme catalyzed reaction (square data points)? (choose the one best answer) 10
The enzyme triosephosphate isomerase (TIM) catalyzes the following reaction in glycolysis, where it converts dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (GAP). CH₂OH C=O CH₂OPO²- DHAP triose phosphate isomerase [DHAP] (MM 1.00 2.00 3.00 6.00 O V. (mM/s) 3.700 6.727 9.250 14.800 = H HCOH CH₂OPO²- Kinetics experiments were performed on TIM. Enzyme activity (initial velocity, V.) was measured at varying concentrations of DHAP. The enzyme kinetics were also measured in the presence of two inhibitors, A and B. The enzyme concentration used in all experiments was 25.00 μM. The data are shown below. GAP V. (mM/s) + A 1.452 2.794 4.038 7.281 V₁ (mM/s) + B 0.755 1.379 1.905 3.077
An enzyme catalyzes a reaction with a Km of 7.50 mM and a Vmax of 2.90 mM - s-1. Calculate the reaction velocity, un, for each substrate concentration. [S] = 2.75 mM mM · s- * TOOLS x10 [S] = 7.50 mM mM · s-! [S] = 11.0 mM mM - s-
8 7- 6 2- 180 225 270 315 [Substrate] (nM) 45 90 135 360 405 450 495 540 58! What is the KM of the enzyme data shown above? rate (nM/sec) 3.
From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.
For an enzyme-catalyzed reaction, the velocity was determined at two different concentrations of the substrate. Estimate the value of Vmax. [S] (MM) 10 20 88 nmol/min 79 nmol/min 67 nmol/min V(nmol/min) 51 nmol/min 27 Not enough information is given to form a reasonable estimation. 48
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (uM) 1.3 - Inhibitor + Inhibitor 2.50 0.62 2.6 4.00 1.42 6.5 6.30 2.65 13.0 7.60 3.12 26.0 9.00 3.58 What is the Km of the inhibited enzyme reaction?
Consider the Michaelis-Menten equation, below: Vmaz (S V. k + [S] %3D What is the relationship between changes in substrate concentration and velocity when the concentration of substrate, [S), is well below k7 The S terms cancel out in this equation, so there is no effect of changing substrate concentration. The S] term in the numerator is negligible, so there is no impact of changing substrate concentration. Because the enzyme has reached Vr there is no effect of changing substrate concentrations on enzyme velocity. OThe [S term in the denominator is negligible compared to k There is a linear relationship between substrate concentration and velocity.
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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.
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 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

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