Chapter 8, Problem 28P

Interpretation Introduction

(a)

Interpretation:

The plot of ${\text{V}}_0$ versus $\left[\text{S}\right]$ and a plot $\frac{1}{{\text{V}}_0}$ versus $\frac{1}{\left[\text{S}\right]}$ is to be drawn. Whether penicillinase appears to follow the Michaelis-Menten equation or not is to be stated. The value of $K_{\text{M}}$ is to be calculated.

Concept introduction:

Kinetics is the study of the rate of chemical reactions. Enzyme kinetics is the more specific study of the rate of the enzyme-catalyzed reaction. In the Michaelis-Menten equation, the rate of the reaction is plotted as a function of the concentration of the substrate.

Answer to Problem 28P

The plot of ${\text{V}}_0$ versus $\left[\text{S}\right]$ is shown below.

The plot of $\frac{1}{{\text{V}}_0}$ versus $\frac{1}{\left[\text{S}\right]}$ is shown below.

Yes, Penicillinase follows the Michaelis-Menten equation. The value of $K_{\text{M}}$ is $5.21\times {10}^{-6}.$.

Explanation of Solution

The given data is shown below.

[Penicillinase]   $\text{μM}$	Amount hydrolyzed ( $\text{nmol}$)
1	0.11
3	0.25
5	0.34
10	0.45
30	0.58
50	0.61

The concentration of the penicillinase is taken as the substrate concentration, $\left[\text{S}\right]$. The amount hydrolyzed is taken as the rate of reaction, ${\text{V}}_0$.

The plot of ${\text{V}}_0$ versus $\left[\text{S}\right]$ is shown below.

Figure 1

The plot of $\frac{1}{{\text{V}}_0}$ versus $\frac{1}{\left[\text{S}\right]}$ is shown below.

Figure 2

The expression for the Lineweaver Burk plot from the Michaelis Menten equation is shown below:

$\frac{1}{{\text{V}}_0}=\frac{K_{\text{M}}}{{\text{V}}_{\max }}\times \frac{1}{\left[\text{S}\right]}+\frac{1}{{\text{V}}_{\max }}$  (1)

Where,

• ${\text{V}}_{\max }$ is the maximum rate.
• $K_{\text{M}}$ is the Michaelis Menten constant.

The plots obtained correspond to the equation of the line from the Lineweaver Burk plot. Therefore, penicillinase appears to follow the Michaelis-Menten equation.

The equation of the line is shown below.

$y=7.62\times {10}^{3}x+1.46\times {10}^9$

The intercept of the Lineweaver Burk plot on the $x$ axis is equal to $-\frac{1}{K_{\text{M}}}$.

The $x$ intercept can be determined by setting $y$ equal to $0$ as shown below:

$\begin{array}{c}0=7.62\times {10}^{3}x+1.46\times {10}^9\\ 7.62\times {10}^{3}x=-1.46\times {10}^9\\ x=\frac{-1.46\times {10}^9}{7.62\times {10}^3}\\ =-1.92\times {10}^5\end{array}$

The value of $K_{\text{M}}$ can be calculated as shown below:

$\begin{array}{c}-1.92\times {10}^5=-\frac{1}{K_{\text{M}}}\\ K_{\text{M}}=\frac{1}{1.92\times {10}^5}\\ =5.21\times {10}^{-6}\end{array}$

Therefore, the value of $K_{\text{M}}$ is $5.21\times {10}^{-6}$.

Interpretation Introduction

(b)

Interpretation:

The value of ${\text{V}}_{\max }$ is to be calculated.

Concept introduction:

Kinetics is the study of the rate of chemical reactions. Enzyme kinetics is the more specific study of the rate of the enzyme-catalyzed reaction. In the Michaelis-Menten equation, the rate of the reaction is plotted as a function of the concentration of the substrate.

Answer to Problem 28P

The value of ${\text{V}}_{\max }$ is $0.685\times {10}^{-9}\text{mol}/\min$.

Explanation of Solution

The given data is shown below:

[Penicillinase]   $\text{μM}$	Amount hydrolyzed ( $\text{nmol}$)
1	0.11
3	0.25
5	0.34
10	0.45
30	0.58
50	0.61

The concentration of the penicillinase is taken as the substrate concentration, $\left[\text{S}\right]$. The amount hydrolyzed is taken as the rate of reaction, ${\text{V}}_0$.

The plot of $\frac{1}{{\text{V}}_0}$ versus $\frac{1}{\left[\text{S}\right]}$ is shown below.

Figure 2

The expression for the Lineweaver Burk plot from the Michaelis Menten equation is shown below.

$\frac{1}{{\text{V}}_0}=\frac{K_{\text{M}}}{{\text{V}}_{\max }}\times \frac{1}{\left[\text{S}\right]}+\frac{1}{{\text{V}}_{\max }}$  (1)

Where,

• ${\text{V}}_{\max }$ is the maximum rate.
• $K_{\text{M}}$ is the Michaelis Menten constant.

The equation of the line is shown below.

$y=7.62\times {10}^{3}x+1.46\times {10}^9$

The intercept of the Lineweaver Burk plot on the $y$ axis is equal to $\frac{1}{{\text{V}}_{\max }}$.

The $y$ intercept can be determined by setting $x$ equal to $0$ as shown below.

$\begin{array}{c}y=7.62\times {10}^3\left(0\right)+1.46\times {10}^9\\ y=1.46\times {10}^9\end{array}$

The value of ${\text{V}}_{\max }$ can be calculated as shown below.

$\begin{array}{c}1.46\times {10}^9=\frac{1}{{\text{V}}_{\max }}\\ {\text{V}}_{\max }=\frac{1}{1.46\times {10}^9}\\ =0.685\times {10}^{-9}\text{mol}/\min \end{array}$

Therefore, the value of ${\text{V}}_{\max }$ is $0.685\times {10}^{-9}\text{mol}/\min$.

Interpretation Introduction

(c)

Interpretation:

The turnover number of penicillinase is to be calculated.

Concept introduction:

Enzyme kinetics is the more specific study of the rate of the enzyme-catalyzed reaction. In the Michaelis-Menten equation, the rate of the reaction is plotted as a function of the concentration of the substrate. Turnover number is defined as the number of substrate molecules that can be converted into product per second.

Answer to Problem 28P

The value of turnover number of penicillinase is $203{\text{s}}^{-1}$.

Explanation of Solution

The turnover number, $k_2$, is calculated by the formula shown below.

$k_2=\frac{V_{\max }}{{\left[\text{E}\right]}_{\text{T}}}$  (2)

Where,

• $V_{\max }$ is the maximum rate.
• ${\left[\text{E}\right]}_{\text{T}}$ is the total enzyme concentration.

The mass of one enzyme is $29.6\text{kDa}$.

The conversion of $1\text{kDa}$ to $\text{Da}$ is shown below.

$1\text{kDa}={10}^3\text{Da}$

Therefore, the conversion of $29.6\text{kDa}$ to $\text{Da}$ is shown below.

$29.6\text{kDa}=29.6\times {10}^3\text{Da}$

The formula to calculate the total number of enzymes is shown below.

$\text{Total}\text{number}\text{of}\text{enzymes}=\frac{\text{Total}\text{mass}\text{of}\text{enzyme}}{\text{Mass}\text{of}\text{one}\text{enzyme}}$  (3)

The total mass of the enzyme is ${10}^{-9}\text{g}$.

Substitute the mass of one enzyme and the total mass of the enzyme in equation (3).

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{enzymes}=\frac{\text{Total}\text{mass}\text{of}\text{enzyme}}{\text{Mass}\text{of}\text{one}\text{enzyme}}\\ =\frac{{10}^{-9}\text{g}}{29.6\times {10}^3\text{Da}}\times \frac{1\text{Da}}{1.66\times {10}^{-24}\text{g}}\\ =2.04\times {10}^{10}\end{array}$

The number of moles is calculated by the formula shown below.

$\text{Number}\text{of}\text{moles}=\frac{\text{Number}\text{of}\text{enzymes}}{N_{\text{A}}}$  (4)

Where,

• $N_{\text{A}}$ is Avagadro’s number. ( $6.022\times {10}^{23}$)

Substitute the value of the number of enzymes and $N_{\text{A}}$ in equation (4).

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{2.04\times {10}^{10}}{6.022\times {10}^{23}}\\ =3.38\times {10}^{-14}\text{moles}\end{array}$

The volume of the solution is $10\text{mL}$.

The conversion of $1\text{mL}$ to $\text{L}$ is shown below.

$1\text{mL}={10}^{-3}\text{L}$

Therefore, the conversion of $10\text{mL}$ to $\text{L}$ is shown below.

$10\text{mL}={10}^{-2}\text{L}$

The expression to determine the concentration of the enzyme is given below.

${\left[\text{E}\right]}_{\text{T}}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}}$  (5)

Substitute the number of moles and the volume of solution in equation (5).

$\begin{array}{c}{\left[\text{E}\right]}_{\text{T}}=\frac{3.38\times {10}^{-14}\text{moles}}{{10}^{-2}\text{L}}\\ =3.38\times {10}^{-12}\text{mol}/\text{L}\\ =3.38\times {10}^{-12}\text{mol}/\text{L}\times \frac{\text{M}}{\text{mol}/\text{L}}\\ =3.38\times {10}^{-12}\text{M}\end{array}$

The value of $V_{\max }$ is $0.685\times {10}^{-9}\text{mol}/\min$.

Substitute the value of $V_{\max }$ and ${\left[\text{E}\right]}_{\text{T}}$ in equation (2).

$\begin{array}{l}{k}_2=\frac{0.685\times {10}^{-9}}{3.38\times {10}^{-12}}\\ =203\end{array}$

Therefore, the value of turnover number is $203{\text{s}}^{-1}$.