Question

Propranolol binds to $\mathit{\beta }$-adrenergic receptors with a ${\mathbf{\text{K}}}_{\mathbf{I}}$ of $\mathbf{\text{8.9}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{- 9}}}\mathbf{\text{M}}$. What concentration of propranolol would be required to achieve a 50% reduction in the binding of the receptor agonist isoproterenol if the agonist concentration is 10 nM and its dissociation constant for the receptor is $\mathbf{\text{4.8}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{- 8}}}\mathbf{\text{M}}$?

Short answer

Concentration needed to achieve 50% reduction in binding is $\text{1.1}\times {\text{10}}^{\text{- 7}}\text{M.}$

Step-by-step solution

Description

Beta-blockers are a class of medications that includes propranolol. It helps with anxiety, treats cardiac issues, and shields against migraines.Propranolol is a medication that can be used to treat high blood pressure if you have a heart condition.

Calculation of the concentration

The expression that connects the binding affinity of the competing ligand and the binding affinity of the bonded ligand is:

${\text{K}}_{\text{I}}\text{=}\frac{{\text{[I]}}_{\text{50}}}{\text{1 +}\frac{\text{[L]}}{{\text{K}}_{\text{L}}}}$

${\text{K}}_I$ of propranolol is$\text{8.9}\times {\text{10}}^{\text{- 9}}\text{M}$ , the concentration of isoproterenol is${\text{[A] = 10 nM = 10}}^{\text{- 8}}\text{M}$,and${\text{K}}_L$of isoproterenol is $\text{4.8}\times {\text{10}}^{\text{- 8}}\text{M}$.

Final result

So, the result is

$\begin{array}{rcl}& & {\text{I}}_{\text{50}}{\text{=K}}_{\text{I}}\left(\text{1 +}\frac{\text{[A]}}{{\text{K}}_{\text{L}}}\right)\\ & & \text{= 8.9}\times {\text{10}}^{\text{- 9}}\left(\text{1 +}\frac{{\text{[10}}^{\text{- 8}}\text{]}}{\text{4.8}\times {\text{10}}^{\text{- 8}}}\right)\\ & & \text{= 1.1}\times {\text{10}}^{\text{- 7}}\text{M}\end{array}$

Therefore, the concentration needed to achieve 50% a reduction in binding is$\text{1.1}\times {\text{10}}^{\text{- 7}}\text{M.}$