V23. The graph below is a graph of Vmax(a) Label the graph clearly with both the Vmax and the Km. Estimate the Km from thisgraph giving the correct units.v/Vmax1.00.50.00.0vs [S] for an enzyme.Enzyme Activity vs [Substrate]:0.20.4Substrate (HM):0.6(b) If the Vmax = 25 mmoles per minute per µmole of enzyme calculate the Keat andthe specificity constant."

As given in the question, V/Vmax was plotted with substrate concentration. V= rate of a reaction…

Answered: 23. The graph below is a graph of Vmax…

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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Similar questions

At what substrate concentration would an enzyme with a Km of 0.005M operate at one quarter of its Vmax? What would be the Km for an enzyme if it is operating at 90% of its Vmax when the substrate concentration is 0.01M
5.50 1/V, min/umol 5.00 4.50 4.00 y = 0.9474x + 2.6649 y = 0.9997x + 2.032 0.00 1.00 2.00 2.50 3.00 1/[S], uM -1 Looking at the double reciprocal plot for an enzyme in the absence of inhibitor and in the presence of two concentrations of inhibitor, what would be the Vmax for the uninhibited enzyme? (bottom graph) Equation is given. Choose the one best answer. 3.50 3.00 2.50 2.00
The typical Michaelis-Menten equation mathematically describes the overall rate of the reaction as V (this is because biologists don't like math). What does V actually mean? (write the definition of V in differential equation form). V= d( )/dt Reaction rate Substrate concentration V max ·½V TURK
D 1. 0.25 Short answer questions, write answers on blank paper. A plot of enzyme activity with and without an inhibitor present gave the following plot. 0.2 0.15 1/V v in mM/s 0.1 0.05 0 0 0.05 0.1 y=0.67x+0.02 0.15 y=0.2x+0.02 0.2 0.25 0.3 1/[S]. [S] in mM Without the inhibitor, what are the values of Vmax and Km? In the presence of the inhibitor, which of these items has changed and what is (are) the new value(s)? What type of inhibitor is present (explain why you have arrived at this conclusion)? Be sure to include units as necessary--answers that do not contain required units will be marked incorrect
a. What is the Vmax of this enzyme WITHOUT inhibitor? Please show your work. b. What is the Km of this enzyme WITHOUT inhibitor? Please show your work. c. The specificity constant of enzyme X is 8 x 10^7 /(M * seconds) What is the kcat of enzyme X WITHOUT inhibitor? Please show your work d. What was the concentration of enzyme used for measuring the kinetics of enzyme X WITHOUT inhibitor? Please show your work
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (µM). 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 Vmax of the inhibited enzyme reaction?
You are evaluating the kinetics of an enzyme catalyzed reaction containing 5.5 μM total enzyme and 11.2 μM substrate. At this substrate concentration, you determine that the Vo = 88.6 μmol mL-¹. s-¹. If the Vmax 833.3 mM s the KM is: . == " 30.9 μΜ 10.4 μΜ Ο 124.6 μΜ 234 μΜ 94.5 μM
Answer #2
The following data have been obtained for two different initial enzyme concentrations for an enzyme-catalyzed reaction. [E]=0.015 g/l) (g/l-min) [S] (g/l) v([E]=0.00875 g/l) (g/l-min) 1.14 20.0 0.67 0.87 10.0 0.51 0.70 6.7 0.41 0.59 5.0 0.34 0.50 4.0 0.29 0.44 3.3 0.39 2.9 0.35 2.5 a. Find K b. Find V for [E]=0.015 g/l. c. Find V for [E]=0.00875 g/l. d. Find k2. Use Eddie-Hostee plot to find your answer. Compare your result using Hanes- Wolf plot.

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