Below are shown three Lineweaver-Burk plots for enzyme reactions that have been brought out in the presence, or absence, of one inhibitor.
You are watching: Indicate what type of inhibition is predicted based on lineweaver-burk plot.
Part A
Indicate what form of inhibition is predicted based on Lineweaver-Burk plot.

mixed inhibition | |
competitive inhibition | |
uncompetitive inhibition |
Part B
Indicate which line synchronizes to the reaction there is no inhibitor and which line coincides to the reaction with inhibitor present.

red - there is no inhibitor, blue - v inhibitor | |
blue - there is no inhibitor, red - v inhibitor |
Part C
Indicate what kind of inhibition is predicted based on Lineweaver-Burk plot.

mixed inhibition | |
competitive inhibition | |
uncompetitive inhibition |
Part D
Indicate i beg your pardon line synchronizes to the reaction there is no inhibitor and also which line synchronizes to the reaction through inhibitor present.

Indicate i beg your pardon line corresponds to the reaction without inhibitor and also which line synchronizes to the reaction
blue - without inhibitor, red - v inhibitor | |
red - without inhibitor, blue - v inhibitor |
Part E
Indicate what kind of inhibition is predicted based on Lineweaver-Burk plot.

mixed inhibition | |
competitive inhibition | |
uncompetitive inhibition |
Part F
Indicate i beg your pardon line synchronizes to the reaction without inhibitor and which line coincides to the reaction with inhibitor present.

blue - without inhibitor, red - with inhibitor | |
red - there is no inhibitor, blue - through inhibitor |
Solutions
Expert Solution
Part-A
The lineweaver-Burk equation because that the mixed inhibition is, \(\frac1v_0=\left(\frac\alpha K_mV_\max \right) \frac1+\frac\alpha^\primeV_\max \)
Here, \(v_0\) is the early stage velocity, \(\) is the substrate concentration,
$$ \alpha^\prime=\left(1+\frac<\mathrmI>K_I^\prime\right), \alpha=\left(1+\frac<\mathrmI>K_I\right) $$
\(<\mathrmI>\) is the inhibitor concnetration, \(V_\max \) is the preferably velocity \(K_I^\prime\) and also \(K_I^\prime\) space the dissociation constants, and \(K_m\) is the Michael"s constant.
The corresponding lineweaver-Burk plot because that the mixed inhibition is shown below:

Plot 1: Lineweaver-Burk plot in the existence of mixed inhibitor
The curled arrow which is in anti-clock way direction indicates boosting inhibitor concentration.
Compare the specified plot with the plot-1. Thus, based on the plot the enzyme reaction is a mixed inhibition type.
Part-B
Compare the specified plot with plot-1, Thus, the blue line indicates the line equation there is no inhibitor and, the red line shows the line equation v inhibitor.
Part-C
The lineweaver-Burk equation for the vain inhibition is, \(\frac1v_0=\left(\frac\alpha K_mV_\max \right) \frac1+\frac1V_\max \)
The matching lineweaver-Burk plot for the compete inhibition is shown below:

Plot 2: Lineweaver-Burk plot in the existence of a competitive inhibitor
Compare the specified plot with the plot-2. Thus, based on the plot the enzyme reaction is compete inhibition type.
Part-D
Compare the stated plot with plot-2, Thus, the blue line shows the line equation without inhibitor and, the red line suggests the heat equation with inhibitor.
Part-E
The lineweaver-Burk equation for the uncompetitive inhibition is, \(\frac1v_0=\left(\fracK_mV_\max \right) \frac1+\frac\alpha^\primeV_\max \)
The equivalent lineweaver-Burk plot because that the uncompetitive inhibition is presented below:

Plot 3: Lineweaver-Burk plot in the presence of the uncompetitive inhibitor
Compare the specified plot with the plot-3. Thus, based upon the plot the enzyme reaction is an uncompetitive inhibition type.
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Part-F
Compare the mentioned plot v plot-3, Thus, the blue line indicates the line equation without inhibitor and, the red line indicates the line equation v inhibitor.