Question

Just wondering what equation to use to work about a and c? 3. A competitive antagonist of the CB1 receptor, SR141716A, has an equilibrium binding affinity constant (KN) for the CB1 receptor of approximately 5.6 x 109 M-1. (This is the information from the question prior: THC binds to the CB1 cannabinoid receptor and causes receptor activation, leading to a range of intracellular signals through an array of G proteins.   THC binds to CB1 with a dissociation constant (Kd) of 0.0353 mM)

In the presence of 50 nM SR141716A, what is the fraction of the target sites that are bound by THC when the concentration of THC is 85 nM?

What would be the required concentration of SR141716A to reduce the THC binding to CB1 to 15% of the level of binding expected from 120 nM THC in the absence of SR141716A?

Answer 1

Answer:

The answer could be attempted by using the dissociation constants of both substrate and inhibitor. The dissociation constant (Kd) quantifies the equilibrium between a ligand (S) being free in solution and bound to a site in a protein (ES).

           

                                                            [ES] --> [E] [S]         Here, dissociation constant Kd = [E] [S]/ [ES]

In the question, dissociation constant of substrate (THC) is given, Kd = 0.0353 X 10^-3 M

The dissociation constant of the inhibitor (Ki) would be the inverse of binding affinity constant (KN).

Ki = 1/KN = 1/5.6 X 10⁹ M, answer is => Ki = 1.8 X 10^-10 M.

Now that we have the values of dissociation constants for both substrate and inhibitor, we can get an estimate of their relative binding in the following manner:

               Kd = [E] [S]/ [ES];    Ki = [E] [I]/ [EI]

hence, Kd/Ki = [E] [S] [EI]/ [E] [I] [ES]

i.e.,        Kd/Ki = [EI] [S]/ [ES] [I]

hence, [EI]/ [ES] = Kd [I]/Ki [S]

Now, let’s solve for a.) [EI]/ [ES] = Kd [I]/Ki [S]

                                        So, [EI]/ [ES] = (0.0353 X 10^-3)(5 X 10^-8) / (1.8 X 10^-10)(85 X 10^-9)

                                       i.e., [EI]/ [ES] = 1.15 X 10⁵

It is important to consider that the total amount of enzyme present [Et], if we assume saturation, would be [Et] = [EI] + [ES]. Also, from the result just obtained, [EI] = 1.15 X 10⁵ [ES].

So, the fraction of enzyme occupied by substrate would be [ES] / [Et] = [ES] / [EI] + [ES]

  Hence,   [ES] / [Et] = [ES] / (1.15 x 10⁵ + 1) [ES]

  So, [ES] / [Et] = 1 / 1.15 X 10⁵    (as 1.15 X 10⁵ >>> 1, we can disregard 1)

                                                                                 Hence, [ES] / [Et] = 8.7 X 10^-6  [percentage of [ES] / [Et] = 0.00087%]

I am unable to suggest a method to solve part c. This is because to estimate how much inhibitor would be required, we would have to find out the level of saturation attained at the given substrate concentration. It can not be done without knowing certain other parameters such as Km/Kcat, which are not provided here.