Chapter 5, Problem 5.6IA

(a)

Interpretation Introduction

Interpretation:

The relation, $K=\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}$ has to be proved.

Concept introduction:

The equilibrium constant of a reaction represents the rate of the forward and backward reactions.  If the equilibrium constant is small, the rate of the backward reaction will be more than the rate of the forward reaction.  If the equilibrium constant is large, the rate of the forward reaction is more than the rate of backward reaction.

Answer to Problem 5.6IA

The relation, $K=\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}$ is proved

Explanation of Solution

The average number ($\nu$) of small molecule A bound to macromolecule M is given by the expression below.

    $\nu =\frac{{\left[\text{A}\right]}_{\text{bound}}}{\left[\text{M}\right]}=\frac{{\left[\text{A}\right]}_{\text{in}}-{\left[\text{A}\right]}_{\text{out}}}{\left[\text{M}\right]}$        (1)

Where,

• ${\left[\text{A}\right]}_{\text{bound}}$ is the concentration of the small molecule A bound to macromolecule.
• ${\left[\text{A}\right]}_{\text{in}}$ is the concentration of the small molecule A inside the dialysis bag.
• ${\left[\text{A}\right]}_{\text{out}}$ is the concentration of the small molecule A outside the dialysis bag.
• $\left[\text{M}\right]$ is the concentration of the macromolecule M.

The equilibrium reaction is given below.

    $\text{M+A}\rightleftharpoons \text{MA}$

The equilibrium constant ($K$) for binding is written as shown below.

    $K=\frac{\left[\text{MA}\right]c^{\Theta }}{{\left[\text{M}\right]}_{\text{free}}{\left[\text{A}\right]}_{\text{free}}}$        (2)

Where,

• ${\left[\text{A}\right]}_{\text{free}}$ is the unbound small molecule A.
• ${\left[\text{M}\right]}_{\text{free}}$ is the unbound macromolecule M.

The equation to be proved is shown below.

  $K=\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}$        (3)

Substitute the equation (1) in the left hand side of equation (3).

  $\begin{array}{c}\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}=\frac{\left(\frac{{\left[\text{A}\right]}_{\text{in}}-{\left[\text{A}\right]}_{\text{out}}}{\left[\text{M}\right]}\right)c^{\Theta }}{\left(1-\frac{{\left[\text{A}\right]}_{\text{in}}-{\left[\text{A}\right]}_{\text{out}}}{\left[\text{M}\right]}\right){\left[\text{A}\right]}_{\text{out}}}\\ =\frac{\left(\frac{{\left[\text{A}\right]}_{\text{bound}}}{\left[\text{M}\right]}\right)c^{\Theta }}{\left(1-\frac{{\left[\text{A}\right]}_{\text{bound}}}{\left[\text{M}\right]}\right){\left[\text{A}\right]}_{\text{out}}}\end{array}$        (4)

The value of ${\left[\text{A}\right]}_{\text{out}}={\left[\text{A}\right]}_{\text{free}}$.

Substitute the above in equation (4).

    $\begin{array}{c}\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}=\frac{\left(\frac{{\left[\text{A}\right]}_{\text{bound}}}{\left[\text{M}\right]}\right)c^{\Theta }}{\left(1-\frac{{\left[\text{A}\right]}_{\text{bound}}}{\left[\text{M}\right]}\right){\left[\text{A}\right]}_{\text{free}}}\\ =\frac{\left(\frac{{\left[\text{A}\right]}_{\text{bound}}}{\left[\text{M}\right]}\right)c^{\Theta }}{\left(\frac{\left[\text{M}\right]-{\left[\text{A}\right]}_{\text{bound}}}{\left[\text{M}\right]}\right){\left[\text{A}\right]}_{\text{free}}}\\ =\frac{\left({\left[\text{A}\right]}_{\text{bound}}\right)c^{\Theta }}{\left(\left[\text{M}\right]-{\left[\text{A}\right]}_{\text{bound}}\right){\left[\text{A}\right]}_{\text{free}}}\end{array}$        (5)

As the small molecule A is bound to the macromolecule M. the value of ${\left[\text{A}\right]}_{\text{bound}}=\left[\text{MA}\right]$.

The value of $\left[\text{M}\right]-{\left[\text{A}\right]}_{\text{bound}}={\left[\text{M}\right]}_{\text{free}}$

Therefore, equation (5) can be written as shown below.

  $\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}=\frac{\left[\text{MA}\right]c^{\Theta }}{{\left[\text{M}\right]}_{\text{free}}{\left[\text{A}\right]}_{\text{free}}}$        (6)

Substitute equation (2) in (6).

    $\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}=K$

Therefore, left hand side is equal to right hand side in equation (3).

Therefore, the relation, $K=\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}$ is proved.

(b)

Interpretation Introduction

Interpretation:

In case each macromolecule behaves as $N$ separate molecules, then the Scatchard equation has to be obtained.

Concept introduction:

The equilibrium constant of a reaction represents the rate of the forward and backward reactions.  If the equilibrium constant is small, the rate of the backward reaction will be more than the rate of the forward reaction.  If the equilibrium constant is large, the rate of the forward reaction is more than the rate of backward reaction.

Answer to Problem 5.6IA

When each macromolecule behaves as $N$ separate molecules with the same value of $K$, the Scatchard equation, $\frac{\nu c^{\Theta }}{{\left[\text{A}\right]}_{\text{out}}}=KN-K\nu$ is obtained.

Explanation of Solution

The equilibrium constant ($K$) can be written as shown below.

    $K=\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}$        (3)

Where,

• ${\left[\text{A}\right]}_{\text{out}}$ is the concentration of the small molecule A outside the dialysis bag.
• $\nu$ is the average number of small molecule A bound to macromolecule M.

The equation to be proved is shown below.

    $\frac{\nu c^{\Theta }}{{\left[\text{A}\right]}_{\text{out}}}=KN-K\nu$        (7)

Substitute the value of $K$ from equation (3) in the left hand side of equation (7).

    $\begin{array}{c}KN-K\nu =\left(\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}\right)N-\left(\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}\right)\nu \\ =\frac{\nu c^{\Theta }}{\left(1-\nu \right){\left[\text{A}\right]}_{\text{out}}}\left(N-\nu \right)\end{array}$        (8)

Each macromolecule behaves as $N$ separate molecules with the same value of $K$ .

Therefore, equation (8) can be written as shown below.

    $KN-K\nu =\frac{\nu c^{\Theta }}{{\left[\text{A}\right]}_{\text{out}}}$

Therefore, the left hand side and the right hand side of equation (7) are equal.

Therefore, when each macromolecule behaves as $N$ separate molecules with the same value of $K$, the Scatchard equation, $\frac{\nu c^{\Theta }}{{\left[\text{A}\right]}_{\text{out}}}=KN-K\nu$ is obtained.

(c)

Interpretation Introduction

Interpretation:

A Scatchard plot has to be obtained from the given data.  The intrinsic equilibrium constant, $K$ and the total number of sites per DNA molecule have to be determined.  Whether the identical and independent site for binding model is applicable has to be stated.

Concept introduction:

The equilibrium constant of a reaction represents the rate of the forward and backward reactions.  If the equilibrium constant is small, the rate of the backward reaction will be more than the rate of the forward reaction.  If the equilibrium constant is large, the rate of the forward reaction is more than the rate of backward reaction.

Answer to Problem 5.6IA

The Scatchard plot is shown below.

The intrinsic equilibrium constant, $K$ is $\underset{\_}{1.167d{m}^3\mu mo{l}^{-1}}$ and the total number of sites per DNA molecule is $\underset{\_}{5.24}$.  The identical and independent site for binding model is applicable.

Explanation of Solution

The data for the total concentration of EB is given below.

Side without DNA	Side with DNA
0.042	0.292
0.092	0.590
0.204	1.204
0.526	2.531
1.150	4.150

The Scatchard equation for ethidium bromide is given below.

    $\frac{\nu c^{\Theta }}{{\left[\text{EB}\right]}_{\text{out}}}=KN-K\nu$        (9)

The formula to calculate average number ($\nu$) is shown below.

    $\nu =\frac{{\left[\text{EB}\right]}_{\text{bound}}}{\left[\text{M}\right]}$        (10)

The value of ${\left[\text{EB}\right]}_{\text{bound}}$ is calculated by the expression given below.

    ${\left[\text{EB}\right]}_{\text{bound}}={\left[\text{EB}\right]}_{\text{in}}-{\left[\text{EB}\right]}_{\text{out}}$        (11)

Where,

• The side with DNA is ${\left[\text{EB}\right]}_{\text{in}}$.
• The side without DNA is ${\left[\text{EB}\right]}_{\text{out}}$.
• $\left[\text{M}\right]$ is the concentration of the macromolecule M.
• ${\left[\text{EB}\right]}_{\text{bound}}$ is the concentration of ethidium bromide bound to the DNA.
• $K$ is the intrinsic equilibrium constant
• $N$ is the total number of sites per DNA molecule.

Therefore, the value of $\frac{\nu c^{\Theta }}{{\left[\text{EB}\right]}_{\text{out}}}$ is calculated in the table shown below.

The concentration of $\left[\text{M}\right]$ is $1.00\text{μmol}{\text{dm}}^{-3}$

${\left[\text{EB}\right]}_{\text{out}}/\text{μmol}{\text{dm}}^{-3}$	${\left[\text{EB}\right]}_{\text{in}}/\text{μmol}{\text{dm}}^{-3}$	${\left[\text{EB}\right]}_{\text{bound}}$	$\nu$	$\frac{\nu c^{\Theta }}{{\left[\text{EB}\right]}_{\text{out}}}$
0.042	0.292	0.250	0.250	5.95
0.092	0.590	0.498	0.498	5.41
0.204	1.204	1.000	1.000	4.90
0.526	2.531	2.005	2.005	3.81
1.150	4.150	3.000	3.000	2.61

The graph between $\frac{\nu c^{\Theta }}{{\left[\text{EB}\right]}_{\text{out}}}$ and $\nu$ is plotted below.

Figure 1

On comparing the graph with the equation (9), the slope of the graph is equal to negative of the intrinsic equilibrium constant.

Therefore,

$K=\underset{\_}{1.167d{m}^3\mu mo{l}^{-1}}$

At intercept, $v=0$, the value of total number of sites per DNA molecule is given below.

$N=\underset{\_}{5.24}$

The identical and independent site for binding model is applicable as the graph is a close fit.