Biology: Asample 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?

The answer is Ahope this helpshave a good day :)Explanation:...

Asample of cells with a total receptor concentration

josmanu235

08.01.2022

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,

$K_f=\frac{1}{K_d}$

The expression of K_f is:

$K_f=\frac{[RL]}{[R][L]}$

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:

$K_f=\frac{[RL]}{[R][L]}$

$K_f=\frac{(35)}{(15)(65)}$

$K_f=0.0358\mu M^{-1}$

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

$K_f=\frac{1}{K_d}$

$0.0358\mu M^{-1}=\frac{1}{K_d}$

$K_d=27.9\mu M$

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

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