Question

A sample of cells with a total receptor concentration of 50 M was incubated with a ligand concentration of 100 M. Following the incubation period, the concentration of unbound receptors was determined to be 15 M. What is the K d for the receptor-ligand interaction?

Answer 1

Answer : The value of
`K_d` for the receptor-ligand interaction is
`27.9\mu M`

Explanation :

The equilibrium reaction between receptor and ligand will be:

`R+L\overset{K_f}\rightarrow [RL]` (forward reaction)

`[RL]\overset{K_d}\rightarrow R+L` (backward reaction)

where,

R = receptor

L = ligand

[RL] = receptor-ligand complex

As we know that,

`K_f=(1)/(K_d)`

The expression of
`K_f` is:

`K_f=([RL])/([R][L])`

As we are given that:

Total [R] =
`50\mu M`

Total [L] =
`100\mu M`

Free [R] =
`15\mu M`

Bound [R] =
`50-15=35\mu M`

Bound [L] = Bound [R] =
`35\mu M`

Free [L] =
`100-35=65\mu M`

Now put all the given values in above expression, we get:

`K_f=([RL])/([R][L])`

`K_f=((35))/((15)(65))`

`K_f=0.0358\mu M^(-1)`

Now we have to calculate the value of
`K_d` for the receptor-ligand interaction.

`K_f=(1)/(K_d)`

`0.0358\mu M^(-1)=(1)/(K_d)`

`K_d=27.9\mu M`

Therefore, the value of
`K_d` for the receptor-ligand interaction is
`27.9\mu M`