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Does the EC50 equals Kd for a partial agonist?

SiMan

Does EC50 = Kd for a partial agonist? And if so why?

Reply 1

SiMan

Is someone able to help?

Reply 2

thegodofgod

Original post by SiMan

Does EC50 = Kd for a partial agonist? And if so why?

EC₅₀ and K_D are related.

If you plot Michaelis-Menten (non-logarithmic) curves of [Drug] vs % Maximal Effect, you will see that:

Effect = (Effect_max [Drug]) / (EC₅₀ + [Drug])

where Effect_max is the maximal response of a system to a drug, and EC₅₀ is the [Drug], at which 50% of the maximal response is produced.

This is important to understand, because drug-receptor binding is characterised by saturation kinetics, where the following equilibrium exists:

k_on
R + L <=======> R.L
k_off

where L is the ligand (agonist), R is the receptor that the ligand binds to, and R.L is the ligand-receptor complex.

The equilibrium constant (K_eq) for forming R.L is described as:

[R.L] / ([R] + [L]) = k_on / k_off = K_eq.

The proportion (fraction) of the receptor that is bound (R.L) is:

f_R.L = [R.L] / ([R] + [L]) = [L] / ((k_off / k_on + [L]).

When you plot f_R.L vs [L] (non-logarithmically), you will see that f_R.L = 1 represents [R_total], and the [L] corresponding to 0.5 [R_total] is the [L] concentration at which k_off / k_on, also known as K_D.

Thus, 50% of available receptors are occupied by the ligand when [Drug] = k_off / k_on.

This shows that EC₅₀ and K_D are related.