Chapter 32, Problem 9P

Interpretation Introduction

Interpretation:

The fraction of the receptor bonded with the hormone needs to be calculated.

Concept Introduction :

While considering the simple equilibrium binding of receptor to hormone the below equilibrium could be used:

  $\left[\text{HR}\right]\rightleftarrows \left[\text{H}\right]\text{+}\left[\text{R}\right]$

Within this equilibrium, [HR] is representative of the hormone’s concentration receptor complex. [H] signifies the free hormone’s concentration while [R] signifies the noncomplexed receptor’s concentration. The equilibrium dissociation constant for this specific reaction could be found to be as:

  ${\text{K}}_{\text{D}}\text{=}\frac{\left[\text{H}\right]\left[\text{R}\right]}{\left[\text{HR}\right]}$ ......(1)

Answer to Problem 9P

  ${\text{f}}_{\text{bound}}=\text{0}\text{.496}$

Explanation of Solution

The equilibrium equation can also be written as follows:

  $\begin{array}{l}{\text{K}}_{\text{D}}\text{=}\frac{\left[\text{H}\right]\left[\text{R}\right]}{\left[\text{HR}\right]}\\ {\text{K}}_{\text{D}}\left[\text{HR}\right]=\left[\text{H}\right]\left[\text{R}\right]\\ \left[\text{H}\right]=\left[{\text{H}}_{\text{Total}}\right]-\left[\text{HR}\right]\text{ and }\left[\text{R}\right]=\left[{\text{R}}_{\text{Total}}\right]-\left[\text{HR}\right]\end{array}$

After substituting,

The quadratic equation to solve for the variables:

  $\left[\text{HR}\right]=\frac{\left[{\text{H}}_{\text{Total}}\right]+\left[{\text{R}}_{\text{Total}}\right]+{\text{K}}_{\text{D}}\pm \sqrt{{\left(\left[{\text{H}}_{\text{Total}}\right]+\left[{\text{R}}_{\text{Total}}\right]+{\text{K}}_{\text{D}}\right)}^2-4\left[{\text{H}}_{\text{Total}}\right]\left[{\text{R}}_{\text{Total}}\right]}}{2}$

Substituting in the initial hormone and receptor concentration gives:

  $\begin{array}{l}\left[\text{HR}\right]=\frac{\left[{\text{H}}_{\text{Total}}\right]+\left[{\text{R}}_{\text{Total}}\right]+{\text{K}}_{\text{D}}\pm \sqrt{{\left(\left[{\text{H}}_{\text{Total}}\right]+\left[{\text{R}}_{\text{Total}}\right]+{\text{K}}_{\text{D}}\right)}^2-4\left[{\text{H}}_{\text{Total}}\right]\left[{\text{R}}_{\text{Total}}\right]}}{2}\\ \text{ = }\frac{\text{1nM+0}\text{.1nM+1nM}\pm \sqrt{{\left(\text{1nM}+\text{0}\text{.1nM}+\text{1nM}\right)}^2-\text{4}\left(\text{1nM}\right)\left(\text{0}\text{.1nM}\right)}}{2}\\ \text{ = }\frac{2.1\text{nM}\pm \sqrt{{\left(\text{2}\text{.1nM}\right)}^2-\text{0}{\text{.4nM}}^2}}{2}\\ \text{ = }\frac{2.1\text{nM}\pm \sqrt{4.1{\text{nM}}^2-0.4n{M}^2}}{2}\end{array}$

Since [HR] must be 0 when $\left[H_{\text{Total}}\right]$ and $\left[H_{\text{Total}}\right]$

  $\begin{array}{l}\left[\text{HR}\right]=\frac{2.1\text{nM}\pm \sqrt{4.01{\text{nM}}^2}}{2}\\ \text{ = }\frac{2.1\text{nM}\pm 2.0025\text{nM}}{2}\\ \text{ = }\frac{0.0975}{2}\\ \text{ = }0.048\text{8 nM}\end{array}$

To determine the fraction receptor bound divide [HR] by $\left[R_{\text{Total}}\right]$

  $\begin{array}{l}{\text{f}}_{\text{bound}}=\frac{[\text{HR}]}{\left[{\text{R}}_{\text{Total}}\right]}\\ \text{ = }\frac{0.0488\text{nM}}{0.1\text{nM}}\\ \text{ = 0}\text{.488}\end{array}$

These calculations can be repeated for when the concentration of receptor is decreased to 0.33 nM.

  $\begin{array}{l}\left[\text{HR}\right]=\frac{\left[{\text{H}}_{\text{Total}}\right]+\left[{\text{R}}_{\text{Total}}\right]+{\text{K}}_{\text{D}}\pm \sqrt{{\left(\left[{\text{H}}_{\text{Total}}\right]+\left[{\text{R}}_{\text{Total}}\right]+{\text{K}}_{\text{D}}\right)}^2-4\left[{\text{H}}_{\text{Total}}\right]\left[{\text{R}}_{\text{Total}}\right]}}{2}\\ \text{ = }\frac{\text{1nM+0}\text{.033nM+1nM}\pm \sqrt{{\left(\text{1nM}+\text{0}\text{.033nM}+\text{1nM}\right)}^2-\text{4}\left(\text{1nM}\right)\left(\text{0}\text{.033nM}\right)}}{2}\\ \text{ = }\frac{2.033\text{nM}\pm \sqrt{{\left(\text{2}\text{.033nM}\right)}^2-\text{0}{\text{.132nM}}^2}}{2}\\ \text{ = }\frac{2.1\text{nM}\pm \sqrt{4.133089{\text{nM}}^2-0.132{\text{nM}}^{\text{2}}}}{2}\end{array}$

Taking the negative root again:

  $\begin{array}{l}\left[\text{HR}\right]=\frac{2.033\text{nM}\pm \sqrt{4.001089{\text{nM}}^2}}{2}\\ \text{ = }\frac{2.033\text{nM}\pm 2.003\text{nM}}{2}\\ \text{ = }\frac{0.0327}{2}\\ \text{ = }0.0164\text{ nM}\end{array}$

Finding the fraction receptor bound gives:

  $\begin{array}{l}{\text{f}}_{\text{bound}}=\frac{[\text{HR}]}{\left[{\text{R}}_{\text{Total}}\right]}\\ \text{ = }\frac{0.0164\text{ nM}}{0.033\text{ nM}}\\ \text{ = 0}\text{.496}\end{array}$

Conclusion

The fraction of the receptor bonded with the hormone is 0.496.