Given that a receptor responds to epinephrine in a classical

Given that a receptor responds to epinephrine in a classical manner (no homotropic allostery and % occupancy = % efficacy), the cells show 50% activity when the hormone level is 10 nM. What is the % activity at 10 nM epinephrine of cells with a mutated form of the receptor where the Kd is shifted to 150nM?

Solution

From the given data, % occupancy = % efficacy. So,

10 nm = 50%

The [E] free = 10 nm

[E]bound = 10 nm

Kd = 150

[10]/[150]*100

=6.6%