Chapter 12, Problem 12.4IA

Interpretation Introduction

Interpretation:

The plot of ${\left[\text{I}\right]}_{\text{0}}$ against ${\left(\text{δ}\nu \right)}^{-1}$ is a straight line with the slope ${\left[\text{E}\right]}_{\text{0}}\Delta \nu$ and $y$-intercept $-K_1$.

Concept introduction:

Many nuclei and electrons have spin, due to this spin magnetic moment arises.  The energy of this magnetic moment depends on the orientation of the applied magnetic field.  In NMR spectroscopy, every nucleus has a spin.  There is an angular momentum related to the spin.  The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus.  Tetramethylsilane (TMS) is taken as reference.

Answer to Problem 12.4IA

The plot of ${\left[\text{I}\right]}_{\text{0}}$ against ${\left(\text{δ}\nu \right)}^{-1}$ is a straight line with the slope ${\left[\text{E}\right]}_{\text{0}}\Delta \nu$ and $y$-intercept $-K_1$ has been rightfully shown.

Explanation of Solution

If the plot of ${\left[\text{I}\right]}_{\text{0}}$ against ${\left(\text{δ}\nu \right)}^{-1}$ is a straight line with the slope ${\left[\text{E}\right]}_{\text{0}}\Delta \nu$ and $y$-intercept $-K_1$, then the relation given below will hold true.

    ${\left[\text{I}\right]}_{\text{0}}=\frac{{\left[\text{E}\right]}_{\text{0}}\Delta \nu }{\text{δ}\nu }-K_1$

The dissociation reaction of the enzyme-inhibitor complex is shown below.

    $\text{EI}\rightleftharpoons \text{E}+\text{I}$

The expression for the resonance frequency ($\nu$) of the single peak in the NMR spectrum is shown below.

    $\nu =f_{\text{I}}{\nu }_{\text{I}}+f_{\text{EI}}{\nu }_{\text{EI}}$        (1)

Where,

• ${\nu }_{\text{I}}$ is the frequency of free inhibitor, I.
• ${\nu }_{\text{EI}}$ is the frequency of bound inhibitor, EI.
• $f_{\text{I}}$ is the frequency of free inhibitor, I.
• $f_{\text{EI}}$ is the frequency of bound inhibitor, I.

The expression for $f_{\text{I}}$ is given below.

    $f_{\text{I}}=\frac{\left[\text{I}\right]}{\left[\text{I}\right]\text{+}\left[\text{EI}\right]}$

The expression for $f_{\text{EI}}$ is given below.

    $f_{\text{I}}=\frac{\left[\text{EI}\right]}{\left[\text{I}\right]\text{+}\left[\text{EI}\right]}$

Where,

• $\left[\text{I}\right]$ is the concentration of free inhibitor, I.
• $\left[\text{EI}\right]$ is the concentration of bound inhibitor, EI.

Substitute the value of $f_{\text{I}}$ and $f_{\text{EI}}$ in equation (1).

    $\nu =\frac{\left[\text{I}\right]}{\left[\text{I}\right]\text{+}\left[\text{EI}\right]}{\nu }_{\text{I}}+\frac{\left[\text{EI}\right]}{\left[\text{I}\right]\text{+}\left[\text{EI}\right]}{\nu }_{\text{EI}}$        (2)

The expression for $\left[\text{I}\right]$ is given below.

    $\begin{array}{c}{\left[\text{I}\right]}_0=\left[\text{I}\right]+\left[\text{EI}\right]\\ \left[\text{I}\right]={\left[\text{I}\right]}_0-\left[\text{EI}\right]\end{array}$

Where,

• ${\left[\text{I}\right]}_0$ is the initial concentration of the inhibitor, I.

Substitute the value of $\left[\text{I}\right]+\left[\text{EI}\right]$ and $\left[\text{I}\right]$ in equation (2).

    $\nu =\frac{{\left[\text{I}\right]}_0-\left[\text{EI}\right]}{{\left[\text{I}\right]}_0}{\nu }_{\text{I}}+\frac{\left[\text{EI}\right]}{{\left[\text{I}\right]}_0}{\nu }_{\text{EI}}$        (3)

Equation (3) was solved further as shown below.

    $\begin{array}{c}\nu =\frac{{\left[\text{I}\right]}_0{\nu }_{\text{I}}-\left[\text{EI}\right]{\nu }_{\text{I}}+\left[\text{EI}\right]{\nu }_{\text{EI}}}{{\left[\text{I}\right]}_0}\\ \nu {\left[\text{I}\right]}_0={\left[\text{I}\right]}_0{\nu }_{\text{I}}+\left[\text{EI}\right]\left({\nu }_{\text{EI}}-{\nu }_{\text{I}}\right)\\ \frac{\nu {\left[\text{I}\right]}_0-{\left[\text{I}\right]}_0{\nu }_{\text{I}}}{{\nu }_{\text{EI}}-{\nu }_{\text{I}}}=\left[\text{EI}\right]\\ \left[\text{EI}\right]=\frac{{\left[\text{I}\right]}_0\left(\nu -{\nu }_{\text{I}}\right)}{{\nu }_{\text{EI}}-{\nu }_{\text{I}}}\end{array}$        (4)

The dissociation constant ($K_1$) is given by the expression shown below.

    $K_1=\frac{\left[\text{E}\right]\left[\text{I}\right]}{\left[\text{EI}\right]}$        (5)

Where,

• $\left[\text{E}\right]$ is the concentration of the enzyme.

The expression for $\left[\text{E}\right]$ is given below.

    $\left[\text{E}\right]={\left[\text{E}\right]}_0-\left[\text{EI}\right]$

Substitute the value of $\left[\text{E}\right]$ and $\left[\text{I}\right]$ in equation (5).

    $K_1=\frac{\left({\left[\text{E}\right]}_0-\left[\text{EI}\right]\right)\left({\left[\text{I}\right]}_0-\left[\text{EI}\right]\right)}{\left[\text{EI}\right]}$        (6)

As the initial concentration of the inhibitor is greater than the concentration of the bound inhibitor, the equation (6) can be written as shown below.

    $\begin{array}{c}{K}_1=\frac{\left({\left[\text{E}\right]}_0-\left[\text{EI}\right]\right){\left[\text{I}\right]}_0}{\left[\text{EI}\right]}\\ K_1\left[\text{EI}\right]={\left[\text{E}\right]}_0{\left[\text{I}\right]}_0-\left[\text{EI}\right]{\left[\text{I}\right]}_0\\ {\left[\text{E}\right]}_0{\left[\text{I}\right]}_0=K_1\left[\text{EI}\right]+\left[\text{EI}\right]{\left[\text{I}\right]}_0\\ {\left[\text{E}\right]}_0=\frac{\left[\text{EI}\right]\left(K_1+{\left[\text{I}\right]}_0\right)}{{\left[\text{I}\right]}_0}\end{array}$        (7)

Substitute the value of $\left[\text{EI}\right]$ from equation (4) in equation (7).

    $\begin{array}{c}{\left[\text{E}\right]}_0=\frac{{\left[\text{I}\right]}_0\left(\nu -{\nu }_{\text{I}}\right)}{{\nu }_{\text{EI}}-{\nu }_{\text{I}}}\times \frac{\left(K_1+{\left[\text{I}\right]}_0\right)}{{\left[\text{I}\right]}_0}\\ =\left(K_1+{\left[\text{I}\right]}_0\right)\frac{\nu -{\nu }_{\text{I}}}{{\nu }_{\text{EI}}-{\nu }_{\text{I}}}\end{array}$        (8)

The expression for $\nu -{\nu }_{\text{I}}$ is given below.

    $\nu -{\nu }_{\text{I}}=\text{δ}\nu$

The expression for ${\nu }_{\text{EI}}-{\nu }_{\text{I}}$ is given below.

    ${\nu }_{\text{EI}}-{\nu }_{\text{I}}=\Delta \nu$

Substitute the value of $\nu -{\nu }_{\text{I}}$ and ${\nu }_{\text{EI}}-{\nu }_{\text{I}}$ in equation (8).

    $\begin{array}{c}{\left[\text{E}\right]}_0=\left(K_1+{\left[\text{I}\right]}_0\right)\frac{\text{δ}\nu }{\Delta \nu }\\ {\left[\text{E}\right]}_0\Delta \nu =K_1\text{δ}\nu +{\left[\text{I}\right]}_0\text{δ}\nu \\ {\left[\text{I}\right]}_0\text{δ}\nu ={\left[\text{E}\right]}_0\Delta \nu -K_1\text{δ}\nu \\ {\left[\text{I}\right]}_0=\frac{{\left[\text{E}\right]}_0\Delta \nu }{\text{δ}\nu }-K_1\end{array}$        (9)

The equation (9) is in the form of an equation for straight line.  Therefore, the plot of ${\left[\text{I}\right]}_{\text{0}}$ against ${\left(\text{δ}\nu \right)}^{-1}$ is a straight line with the slope ${\left[\text{E}\right]}_{\text{0}}\Delta \nu$ and $y$-intercept of $-K_1$.