Chapter 8, Problem 30P

Interpretation Introduction

(a)

Interpretation:

The values of $V_{\max }$ and $K_{\text{M}}$ in the presence of inhibitorneeds to be determined.

Concept introduction:

The Michaelis-Menten equation relates the concentration of substrate with the velocity of the reaction. A constant which expresses the substrate’s concentration if the velocity of reaction equals to the half of the maximum velocity of the reaction is known as Michaelis-Menten constant. It is denoted by $K_{\text{M}}$.

Answer to Problem 30P

The values of $V_{\max }$ and $K_{\text{M}}$ in the presence of inhibitor are $9.5\text{μmol/min}$ and $10.2\text{μmol}$ respectively.

Explanation of Solution

The given concentration of enzyme is $100\text{μM}$.

The given data is showing the different values of velocity at different concentrations of enzymes in the presence and absence of inhibitor.

The value of $V_{\max }$ without inhibitor is $47.6\text{μmol/min}$.

The reciprocal values of substrate concentration $\left[\text{S}\right]$ that is $1/\left[\text{S}\right]$, velocity with no inhibitor, $1/\left[V\right]$ and velocity with inhibitor, $1/\left[V\right]\text{'}$ (velocity with inhibitor)are calculated in the table given below.

$\left[\text{S}\right]$	$1/\left[\text{S}\right]$	Velocity with no inhibitor	$1/\left[V\right]$	Velocity with inhibitor	$1/\left[V\right]\text{'}$
3	$1/3=0.33$	$10.4$	$0.097$	$2.1$	$0.48$
5	$1/5=0.2$	$14.5$	$0.069$	$2.9$	$0.34$
10	$1/10=0.1$	$22.5$	$0.044$	$4.5$	$0.22$
30	$1/30=0.033$	$33.8$	$0.029$	$6.8$	$0.15$
90	$1/90=0.011$	$40.5$	$0.024$	$8.1$	$0.12$

The graph between $1/\left[\text{S}\right]$ and $1/\left[V\right]\text{'}$ is plotted as follows.

Figure 2

The slope of the above graph is shown by $K_{\text{M}}/V_{\max }$ which is calculated by using the expression as follows.

$\text{slope}=\frac{y_2-y_1}{x_2-x_1}$

(1)

The values of $y_1,y_2,x_1$ and $x_2$ obtained from the graph are $0.48,0.34,0.33$ and $0.2$ respectively.

Substitute the respective values in equation (1).

$\begin{array}{c}\text{slope}=\frac{0.34-0.48}{0.2-0.33}\\ =1.07\end{array}$

Thus, the value of slope is $1.07$. Thus, the expression, of $K_{\text{M}}/V_{\max }$ equals to $1.07$.

$K_{\text{M}}/V_{\max }=1.07$

(2)

It is shown in the graph that $y-\text{intercept}$ is equal to $-1/V_{\max }=0.105$. Thus, the value of $V_{\max }$ is calculated as follows.

$\begin{array}{c}{V}_{\max }=\frac{1}{0.105}\\ =9.5\text{μmol/min}\end{array}$

Substitute the value of $V_{\max }$ in equation (2).

$\begin{array}{c}{K}_{\text{M}}/9.5\text{μmol/min}=1.07\\ =1.07\times 9.5\text{μmol/min}\\ =10.2\text{μmol}\end{array}$

Thus, the values of $V_{\max }$ and $K_{\text{M}}$ in the presence of inhibitor are $9.5\text{μmol/min}$ and $10.2\text{μmol}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The type of inhibition for the given enzymeneeds to be determined.

Concept introduction:

A molecule that disturbs or inhibits the activity of an enzyme is known as enzyme inhibitor. The enzyme inhibition stops the generation of enzyme-substrate complex which, in turn, inhibits the formation of the product. The three main types of inhibition are competitive inhibition, uncompetitive inhibition and noncompetitive inhibition.

Answer to Problem 30P

The type of inhibition for the given enzyme is noncompetitive inhibition.

Explanation of Solution

In case of noncompetitive inhibition, there is no competition between inhibitor or substrate. Both substrate and inhibitor are able to bind to the active of enzyme simultaneously. In case of noncompetitive inhibition, the addition of inhibitor only decreases the value of $V_{\max }$ without changing the value of $K_{\text{M}}$.

According to the calculation for the predicting the values of $V_{\max }$ and $K_{\text{M}}$ in the presence of inhibitor, the value of $K_{\text{M}}$ remains constant but there is a decrease in the value of $V_{\max }$. Thus, the type of inhibition for the given enzyme is noncompetitive inhibition.

Interpretation Introduction

(c)

Interpretation:

The dissociation constant for noncompetitive inhibitorneeds to be determined.

Concept introduction:

A molecule that disturbs or inhibits the activity of an enzyme is known as enzyme inhibitor. The enzyme inhibition stops the generation of enzyme-substrate complex which, in turn, inhibits the formation of the product. The three main types of inhibition are competitive inhibition, uncompetitive inhibition and noncompetitive inhibition.

Answer to Problem 30P

The dissociation constant for noncompetitive inhibitor is $2.5\times {10}^{-5}\text{M}$.

Explanation of Solution

The given concentration of enzyme is $100\text{μM}$.

The value of $V_{\max }$ without inhibitor is $47.6\text{μmol/min}$.

The calculated value of $V_{{}_{\max }}^{\text{app}}$ in the presence of inhibitor is $9.5\text{μmol/min}$.

The value of dissociation constant for the noncompetitive inhibitor is calculated by the formula given below.

$V_{{}_{\max }}^{\text{app}}=V_{\max }\left(1+\frac{\left[\text{I}\right]}{{\text{K}}_{\text{i}}}\right)$

(3)

Here,

• $V_{{}_{\max }}^{\text{app}}$ is the maximum velocity in the presence of inhibitor.
• $V_{\max }$is the maximum velocity in the absence of inhibitor.
• ${\text{K}}_{\text{i}}$ is the dissociation constant.
• $\left[\text{I}\right]$ is the concentration of enzyme.

Substitute the values of velocity in the presence and absence of inhibitor, value of kinetics in the above expression.

$\begin{array}{c}9.5\text{μmol/min}=47.6\text{μmol/min}\left(\frac{{\text{K}}_{\text{i}}+100\text{μM}}{{\text{K}}_{\text{i}}}\right)\\ 9.5\text{μmol/min}=\left(\frac{47.6{\text{μmol/minK}}_{\text{i}}+\left(47.6\text{μmol/min}\right)100\text{μM}}{{\text{K}}_{\text{i}}}\right)\\ {\text{K}}_{\text{i}}=2.5\times {10}^{-5}\text{M}\end{array}$

Thus, the dissociation constant value is $2.5\times {10}^{-5}\text{M}$.

Interpretation Introduction

(d)

Interpretation:

The fraction of the molecules of enzyme that contains a bound substrate in the absence and in the presence of $100\text{μM}$ inhibitorneeds to be determined.

Concept introduction:

A molecule that disturbs or inhibits the activity of an enzyme is known as enzyme inhibitor. The enzyme inhibition stops the generation of enzyme-substrate complex which, in turn, inhibits the formation of the product. The three main types of inhibition are competitive inhibition, uncompetitive inhibition and noncompetitive inhibition.

Answer to Problem 30P

The fraction of the molecules of enzyme that contains a bound substrate in the absence and in the presence of $100\text{μM}$ inhibitor is $0.71$.

Explanation of Solution

Given Information:

The given concentration of enzyme is $100\text{μM}$.

The given concentration of substrate is $30\text{μM}$.

The value of velocity, in the absence of inhibitor at $\left[\text{S}\right]=30\text{μM}$ is $33.8$$\text{μmol/min}$.

The value of velocity, in the presence of inhibitor at $\left[\text{S}\right]=30\text{μM}$ is $6.8$$\text{μmol/min}$.

The value of $V_{\max }$ without inhibitor is $47.6\text{μmol/min}$.

The calculated value of $V_{{}_{\max }}^{\text{app}}$ in the presence of inhibitor is $9.5\text{μmol/min}$.

Calculation:

The value of fraction of the molecules of enzyme that contains a bound substrate in the absence is calculated by the formula given below.

${\text{f}}_{\text{ES}}=\frac{V}{V_{\max }}$

(4)

Here,

• ${\text{f}}_{\text{ES}}$ is the fraction of the filled active site.
• $V$ is the velocity in the absence of inhibitor.
• $V_{\max }$is the maximum velocity in the absence of inhibitor.

Substitute the values of velocity and maximum velocity in the absence of inhibitor in the equation (4).

$\begin{array}{c}{\text{f}}_{\text{ES}}=\frac{33.8\text{μmol/min}}{47.6\text{μmol/min}}\\ =0.71\end{array}$

Substitute the values of velocity and maximum velocity in the presence of inhibitor in the equation (4).

$\begin{array}{c}{\text{f}}_{\text{ES}}=\frac{6.8\text{μmol/min}}{9.52\text{μmol/min}}\\ =0.71\end{array}$

Thus, the fraction of the molecules of enzyme that contains a bound substrate in the absence and in the presence of $100\text{μM}$ inhibitor is $0.71$.