Chapter 6, Problem 60RE

Interpretation Introduction

Interpretation:

From the given experimental data both in absence and presence of its inhibitor a Lineweaver–Burk plot is to be drawn and values of ${\text{K}}_{\text{m}}$ and ${\text{V}}_{\text{max}}$ for both inhibited and non-inhibited reaction and type of the inhibition either it is competitive or noncompetitive is to be determined.

Concept introduction:

An inhibitor may be defined as a molecule that binds with the enzymes and decreases the activity of enzyme toward the substrate and this decrease in enzyme activity is called enzymatic inhibition.

Enzymatic inhibition can be of many types. When substrate and inhibitor of an enzyme compete with each other to bind with the enzyme and both of them are structurally same it is known as competitive inhibition. On the other hand, when inhibitors don’t compete with the substrate and both of them can bind to the enzyme at their specific site present on the substrate it is known as noncompetitive inhibition.

To determine the rate of the reaction, experimental data must be put into Lineweaver Burk plot where the x-axis represents the values of $\frac{1}{\left[s\right]}$ ([s]= substrate concentration) and the y-axis represents the values of $\frac{1}{v}$ (v=velocity).

Answer to Problem 60RE

Plot the reciprocal of substrate concentration along the x-axis and the reciprocal of velocities (both in the presence and absence of inhibitors) along the y-axis.

Before drawing the plot, the data needs to be modified so that it can be put into Lineweaver–Burk plot. The modification of the data is

$\begin{array}{rrr}\hfill 1/\text{[S]mM}& \hfill 1/{\text{V, No inhibitor (mmol min}}^{-1})& \hfill 1/{\text{V, inhibitor present(mmol min}}^{-1})\\ \hfill .333333& \hfill 0.218341& \hfill 0.273224\\ \hfill .2& \hfill .15625& \hfill .195313\\ \hfill .142857& \hfill .129534& \hfill .161812\\ \hfill .111111& \hfill .114679& \hfill .143266\\ \hfill .090909& \hfill .105263& \hfill .131579\end{array}$

The Lineweaver–Burk plot from above data is as follows:

From this plot, ${\text{V}}_{\text{max}}$ of the non-inhibited reaction can be determined from y-axis intercept. It is equal to $\frac{1}{{\text{V}}_{\text{max}}}$. The y-intercept is 0.0629

${\text{V}}_{\text{max}}=\frac{1}{0.0629}$

Or, ${\text{V}}_{\text{max}}=15.9\text{mmol/min}$

The value ${\text{K}}_{\text{m}}$ is determined from x-intercept. In the case of non-inhibited reaction x-intercept is 0.14. So, ${\text{K}}_{\text{m}}$ is $\frac{1}{0.14}$

So, the value of ${\text{K}}_{\text{m}}$ for the reaction is 7.14mM.

The ${\text{V}}_{\text{max}}$ for the inhibited reaction will be

${\text{V}}_{\text{max}}=\frac{1}{0.0784}$

${\text{V}}_{\text{max}}=12.75\text{mmol/min}$

The value of ${\text{K}}_{\text{m}}$ is the same as the value of non-inhibited reaction so ${\text{K}}_{\text{m}}$ is 7.14mM.

By comparing these two plots it can be said that it is a type of noncompetitive inhibition.

Explanation of Solution

In the case of competitive inhibition, the substrate and inhibitor compete with each other to bind with the active site of the enzyme. By increasing substrate concentration, the effect of competitive inhibition can be discarded. In high substrate concentration ${\text{V}}_{\text{max}}$ (maximal rate of the enzymatic reaction) of the competitive inhibition never changes. Although inhibitors decrease the ${\text{K}}_{\text{m}}$ (substrate concentration in which reaction rate is half of ${\text{V}}_{\text{max}}$). ${\text{K}}_{\text{m}}$ value also represents the substrate affinity of the enzyme. A high ${\text{K}}_{\text{m}}$ value means an enzyme has low substrate affinity and vice-versa. In case Lineweaver Burk for noncompetitive inhibition ${\text{V}}_{\text{max}}$ value remains the same but ${\text{K}}_{\text{m}}$(x-axis) decreases or shifts to the right.

In the case of non-competitive inhibition, inhibitors bind with the allosteric site of the enzyme and enzyme itself can bind with both substrate and inhibitors. It is possible because it has two different sites for both inhibitor and substrate. When the inhibitor binds with an enzyme it blocks its activity even if the substrate is also bound to the enzyme. These inhibitors do not change the substrate affinity of the enzyme so the ${\text{K}}_{\text{m}}$ will be the same in non-competitive inhibition, only the value of ${\text{V}}_{\text{max}}$ changes. It actually decreases and in Lineweaver Burk plot velocity of reaction shifts to the top.

In this Lineweaver Burk plot, ${\text{K}}_{\text{m}}$ of both reactions remains the same but ${\text{V}}_{\text{max}}$ changes 15.9mm/min to 12.75 mm/min.

Conclusion

The ${\text{K}}_{\text{m}}$ and ${\text{V}}_{\text{max}}$ value for the non-inhibited reaction are 7.14 mM and 15.9 mm/min. The ${\text{K}}_{\text{m}}$ and ${\text{V}}_{\text{max}}$ value for the inhibited reaction is 7.14 mm and 12.75mm/min. As ${\text{K}}_{\text{m}}$ of both reactions remains the same and only ${\text{V}}_{\text{max}}$ changes, it can be said this is the example of noncompetitive inhibition.