A competitive inhibitor changes both the K_M and vmax of a reaction.
A.True | |
B.False |
A competitive inhibitor changes both the K_M and vmax of a reaction.
False
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In competitive inhibition, the maximum velocity (V_\max) of the reaction is unchanged, while the apparent affinity of the substrate to the binding site is decreased (the K_d dissociation constant is apparently increased). The change in K_m (Michaelis-Menten constant) is parallel to the alteration in K_d. Any given competitive inhibitor concentration can be overcome by increasing the substrate concentration in which case the substrate will outcompete the inhibitor in binding to the enzyme
hi
it is FALSE
Vmax is the maxim initial velocity (Vo) that an enzyme can
achieve. Initial velocity is defined as the catalytic rate when
substrate concentration is high, enough to saturate the enzyme, and
the product concentration is low enough to neglect the rate of the
reverse reaction. Therefore, the Vmax is the maximum catalytic rate
that can be achieved by a particular enzyme. Km is determined as
the substrate concentration at which 1/2 Vmax is achieved. This
kinetic parameter therefore importantly defines the affinity of the
substrate for the enzyme.
These two parameters for a specific enzyme defines:
Vmax - the rate at which a substrate will be converted to product
once bound to the enzyme.
Km - how effectively the enzyme would bind he substrate, hence
affinity.
A competitive inhibitor changes both the K_M and vmax of a reaction. A.True B.False
The inhibitor Draw a sketch on where a(n). inhibitor binds to the enzyme: The reaction in the presence of a(n)inhibitor can be written In the Michaelis-Menten graph, the inhibited reaćtion would look this as compared tot he uninhibited reaction (label both axes and draw both curves). In the Lineweaver-Burk graph, the inhibited reaction would look like this as compared tot he uninhibited reaction (label both axes and draw both curves). inhibitor, the apparent changes of Vmax and Km are as...
b. Look at the graph below of how a competitive inhibitor affects the kinetics of an enzyme C. Rate of reaction is the Vmax of the enzyme affected? Why or why not: explain in terms of substrate concentration and enzyme active site saturation) Without inhibitor With competitive inhibitor d. is Vmax/2 affected? Why or why not: explain in terms of Vmax. Substrate concentration e. Is Km affected? Explain in terms of the active site. Hint a competitive inhibitor is competing...
Which is the most inappropriate description of enzyme inhibition? Select one: a. A competitive inhibitor increases Km only. b. A noncompetitive inhibitor decreases Vmax only. c. An uncompetitive inhibitor decreases both Km and Vmax. d. All of the above e. None of the above
In the presence of 1.4 uM of a competitive inhibitor, how will the kinetics of an enzyme change? KM = 10 UM, Vmax = 2 nM/min, K; = 1 UM Km will be: In the presence of 1.4 uM of a competitive inhibitor, how will the kinetics of an enzyme change? KM = 10 UM, Vmax = 2 nM/min, K; = 1 UM Vmax will be:
In a competitive inhibition, Km is increased while Vmax is unchanged. An enzyme is being assayed in the presence of a fixed amount of a competitive inhibitor. How could the rate be increased in this reaction?
How do competitive inhibitors affect the KM and Vmax of an enzyme? Draw a plot of velocity as a function of substrate concentration, both with and without inhibitor added.
An enzyme-catalyzed reaction to the presence of 5 nM of reversible inhibitor yields a Vmax value that is 80% of the value in absence of the inhibitor. The KMvalue is unchanged. a) what type of inhibition is occurring? b) what proportion of the enzyme molecule will have bound inhibitor? c) Draw the Lineweaver-Burk (known as double-reciprocal plot) for uninhibited and inhibited reaction. SHOW ALL YOUR WORK PLEASE
c. Describe the properties of i, competitive inhibitor and ii, noncompetitive inhibitor for this enzyme. Draw Lineweaver Burk plots for each and indicate where you can obtain Km and Vmax values for each plot and how they change with the addition of each type of inhibitor 3
An enzyme catalyzes a reaction with a Km of 9.50 and a Vmax of
1.75
An enzyme catalyzes a reaction with a K_m of 9.50 mM and a of 1.75 mM- s^-1. Calculate the reaction velocity, V_0, for the following substrate concentrations. 2.50 mM 9.50 mM 12.0 mM
What is the Km of this graph if 1/2 the Vmax with inhibitor-
yellow dot- is 1.96. And Kmax 1/2 with out inhibitor(grey dot) is
2.34.
V/Inhibitor V/Reaction Rate (mol/min) V/No Inhinitor km(no inhibitor) Km/w/inhinitor) 0.00033 0.000033 0.000133 0.000133 0.00133 0.002 0.000667 [S] (mol/liters)