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.