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.

See more: At The Mouth Of A Cave - What Is The Entrance To A Cave Called

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.