Chapter 35, Problem 6P

Interpretation Introduction

(a)

Interpretation:

The standard free energy of binding is to be determined.

Concept introduction:

The thermodynamic quantity that measures the free energy available in the system to do the work at a constant temperature is called Gibbs free energy. It is represented by G. It determines the feasibility of the chemical reaction.

The formula for $\Delta G$ is,

$\Delta G=RT\ln K_{\text{d}}$

Where,

• $\Delta G$ is the change in the Gibbs free energy.
• $R$ is the universal gas constant.
• $T$ is the absolute temperature.
• $K_{\text{d}}$ is the dissociation constant.

The conversion factor to convert $°\text{C}$ to Kelvin is:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273$

Answer to Problem 6P

The standard free energy of binding is $-37\text{kJ}{\text{mol}}^{-1}$.

Explanation of Solution

The formula for $\Delta G$ is,

$\Delta G=RT\ln K_{\text{d}}$   ....... (1)

Where,

• $\Delta G$ is the change in the Gibbs free energy.
• $R$ is the universal gas constant.
• $T$ is the absolute temperature.
• $K_{\text{d}}$ is the dissociation constant.

Convert the temperature from $°\text{C}$ to Kelvin as follows:

$\begin{array}{c}T\left(\text{K}\right)=25°\text{C}+273\\ =298\mathrm{K}\end{array}$

The value of $R$ is $8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$.

The value of $T$ is $298\mathrm{K}$.

The value of $K_{\text{d}}$ is $3\times {10}^{-7}\text{M}$.

Substitute the values in equation (1).

$\begin{array}{c}\Delta G=\left(8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(\frac{1\text{kJ}}{1000\text{J}}\right)\left(298\mathrm{K}\right)\ln \left(3\times {10}^{-7}\text{M}\right)\\ =-37.211\text{kJ}{\text{mol}}^{-1}\\ \approx -37\text{kJ}{\text{amol}}^{-1}\end{array}$

Conclusion

The standard free energy of binding is $-37\text{kJ}{\text{mol}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The affinity of ${\text{F}}_{\text{ab}}$ is to be determined.

Concept introduction:

The affinity of an antibody is the strength of the bond formed between an antibody and an antigen. High affinity means the bond is formed rapidly between the two and vice-versa.

The formula for the affinity of the antibody is:

$K_{\text{a}}=\frac{1}{K_{\text{d}}}$

Where,

• $K_{\text{a}}$ is the affinity of the antibody.
• $K_{\text{d}}$ is the dissociation constant.

Answer to Problem 6P

The affinity of ${\text{F}}_{\text{ab}}$ is $3.3\times {10}^{-8}{\text{M}}^{-1}$.

Explanation of Solution

The formula for the affinity of the antibody is,

$K_{\text{a}}=\frac{1}{K_{\text{d}}}$   ....... (2)

Where,

• $K_{\text{a}}$ is the affinity of the antibody.
• $K_{\text{d}}$ is the dissociation constant.

The value of $K_{\text{d}}$ is $3\times {10}^{-7}\text{M}$.

Substitute the value in equation (2).

$\begin{array}{c}{K}_{\text{ab}}=\frac{1}{3\times {10}^{-7}\text{M}}\\ =3.333\times {10}^{-8}{\text{M}}^{-1}\\ \approx 3.3\times {10}^{-8}{\text{M}}^{-1}\end{array}$

Conclusion

The affinity of ${\text{F}}_{\text{ab}}$ is $3.3\times {10}^{-8}{\text{M}}^{-1}$.

Interpretation Introduction

(c)

Interpretation:

The rate constant for the association is to be determined. Also, the implication of its magnitude about the extent of structural changes in an antibody on a hapten should be determined.

Concept introduction:

The formula for the dissociation constant of the reaction is,

$K_{\text{d}}=\frac{K_{\text{off}}}{K_{\text{on}}}$   ....... (3)

Where,

• $K_{\text{d}}$ is the dissociation constant.
• $K_{\text{off}}$ is the rate for the dissociation reaction.
• $K_{\text{on}}$ is the rate for the association reaction.

Answer to Problem 6P

The rate constant for the association is $4\times {10}^{-8}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$. There is a very less structural change in the antibody on the binding hapten.

Explanation of Solution

Rearrange equation (3) for $K_{\text{on}}$ is,

$K_{\text{on}}=\frac{K_{\text{off}}}{K_{\text{d}}}$   ....... (4)

Where,

• $K_{\text{d}}$ is the dissociation constant.
• $K_{\text{off}}$ is the rate for the dissociation reaction.
• $K_{\text{on}}$ is the rate for the association reaction.

The value of $K_{\text{d}}$ is $3\times {10}^{-7}\text{M}$.

The value of $K_{\text{off}}$ is $120\text{}{\text{s}}^{-1}$.

Substitute the values in equation (4).

$\begin{array}{c}{K}_{\text{on}}=\frac{120{\text{s}}^{-1}}{3\times {10}^{-7}\text{M}}\\ =4\times {10}^{-8}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}\end{array}$

The value of the association rate constant is $4\times {10}^{-8}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$ that is very close to the diffusion-controlled limit for the combination of a small molecule with the protein molecule. The extensive conformational transitions take time and therefore the extent of the structural change is very small.

Conclusion

The rate constant for the association is $4\times {10}^{-8}\text{mol}{\text{L}}^{-1}{\text{s}}^{-1}$. There is a very less structural change in the antibody on the binding hapten.