Becker's World of the Cell (9th Edition)
9th Edition
ISBN: 9780321934925
Author: Jeff Hardin, Gregory Paul Bertoni
Publisher: PEARSON
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Chapter 6, Problem 6.12PS
Summary Introduction
To derive: The Michaelis-Menten equation for the enzyme-catalyzed reaction in which substrate is converted into product.
Introduction: For any emzyme-catalysed reaction, the reaction proceeds by the conversion of substrate into product with an enzyme-substrate intermediate. Most of these reactions follow Michaelis-Menten equation for enzyme kinetics.
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molecule
A Plot of velocity versus substrate
B Lineweaver-Burk plot
1/v
Km 1
Vmax (S) Vmx
1
V max
1/2Vmax
1/Vmax
-1/Km
Km
[S]
1/[S]
fppt.com
molecule
Exercise
The following data describe an enzyme-catalyzed reaction. Plot these results
using the Lineweaver-Burk method, and determine values for KM and Vinax-
The symbol mM represents millimoles per liter; 1 mM = 1 × 10 3 mol L.
(The concentration of the enzyme is the same in all experiments.)
Velocity
(mM sec-)
Substrate Concentration
(тм)
2.5
0.024
5.0
0.036
10.0
0.053
15.0
0.060
20.0
0.061
fppt.com
not true about the Michaelis-Menten equation?
The equation that gives the rate, v, of an
the substrate concentration [S] is the Michaelis-Menten equation
= Vmax[S]/(Km + [S]), where V,
enzyme-catalyzed reaction for all values of
max and Km are constants. Which of the following is
a)
for [S] << Km, V = Vmax
applies to most enzymes, but allosteric enzymes have different kinetics
when [S] = Km, then v =
Vmax/2
gives the rate when the enzyme concentration, temperature, pH, and ionic
strength are constant
for very high values of [S], v approaches Vmax
e)
Which is correct about the constant Km in the Michaelis-Menten equation?
also called the catalytic constant or turnover number
equal to the number of product molecules produced per unit time when the
enzyme is saturated with substrate
it is the constant in the first order rate equation v = k[A]
it is the constant in the second order rate equation v =
equal to the substrate concentration at which the velocity or rate of a reaction is
½ the…
6-25
substrate-band enzyme concentrations. The the
turnover number is equal to umax-
b)
V=Umax •57(Km+S)
anstont
For an enzyme that displays Michaelis-Menten kinetics, what is
the reaction velocity, V (as a percentage of Vmax), observed at
the following values?
a)
[S] = KM
C)
d)
e)
[S] = 0.5KM
[S] = = 0.1KM
[S] = 2KM
[S] = 10KM
w
reactores
-maximumrate of reaction
boteles conc.
Would you expect the structure of a competitive inhibitor of a
given enzyme to be similar to that of its substrate?
Chapter 6 Solutions
Becker's World of the Cell (9th Edition)
Ch. 6 - Gasoline is highly combustible yet doesnt burst...Ch. 6 - How can an enzyme recognize and bind one specific...Ch. 6 - Prob. 1QCh. 6 - You work at a biotechnology company and are...Ch. 6 - Why do enzymes need to be regulated? By what...Ch. 6 - The Need for Enzymes. You should now be in a...Ch. 6 - Prob. 6.2PSCh. 6 - Prob. 6.3PSCh. 6 - Temperature and pH Effects. Figure 6-4 illustrates...Ch. 6 - MichaelisMenten Kinetics. Figure 6-16 represents a...
Ch. 6 - Prob. 6.6PSCh. 6 - QUANTITATIVE More Enzyme Kinetics. The galactose...Ch. 6 - QUANTITATIVE Turnover Number. Carbonic anhydrase...Ch. 6 - Inhibitors: Wrong Again. For each of the following...Ch. 6 - What Type of Inhibition? A new mucinase enzyme was...Ch. 6 - Prob. 6.11PSCh. 6 - Prob. 6.12PSCh. 6 - Prob. 6.13PS
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