Question

K, = dissociation constant kon R-receptor L ligand off RIIL K, Total Rec:R,-[R]+[RL] = - Kd ILI QUESTIONS You've identified a drug that has the potential to bind and inhibit 5 different histamine receptors found on human mast cells. The drug if ingested orally typically reaches a concentration of 1 x 108 Molar in the blood stream. Given the K of the drug for each of the 5 receptors listed below, calculate the percentage of receptors occupied with a drug concentration of 1 X 10-8 Molar Percentage of Receptors Occupied (with 108 Molar drug concentration) RECEPTOR Receptor A K, 1 micromolar Receptor B K 10 micromolar Receptor C K, 100 micromolar Receptor D K1 nanomolar Receptor E K, 10 nanomolar

Answer 1

In the question, it is mentioned that ([RL]/R_T)= [ 1 / {1+(K_d / [L])} ]

([RL]/R_T) represents the fraction of receptors occupied with a drug.

So,

The fraction of receptors occupied with a drug= [ 1 / {1+(K_d / [L])} ]= [1 / {([L] + K_d) / [L]}] = {[L] / ([L]+K_d)} Where, [L]= Drug concentration, K_d= Dissociation constant.

To solve these problems, we can use the above-mentioned formula.

Calculation for receptor A:

[L]= 1 $\times$ 10^-8 Molar= 1 $\times$ 10^-8 $\times$ 10⁶ micromolar [As 1 Molar= 10⁶ micromolar] = 10^-2 micromolar.

K_d= 1 micromolar

So, the percentage of receptor A occupied with the drug= {[L] / ([L]+K_d)}= {10^-2/(10^-2+1)}

= 9.9 $\times$ 10^-3= (9.9 $\times$ 10^-3 $\times$ 100) %= 0.99 % [Answer]

Therefore, 0.99% of receptor A is occupied with the drug.

Calculation for receptor B:

[L]= 1 $\times$ 10^-8 Molar= 1 $\times$ 10^-8 $\times$ 10⁶ micromolar [As 1 Molar= 10⁶ micromolar] = 10^-2 micromolar.

K_d= 10 micromolar

So, the percentage of receptor B occupied with the drug= {[L] / ([L]+K_d)}= {10^-2/(10^-2+10)}

=9.9 $\times$ 10^-4= (9.9 $\times$ 10^-4 $\times$ 100) %= 0.099 % [Answer]

Therefore, 0.099% of receptor B is occupied with the drug.

Calculation for receptor C:

[L]= 1 $\times$ 10^-8 Molar= 1 $\times$ 10^-8 $\times$ 10⁶ micromolar [As 1 Molar= 10⁶ micromolar] = 10^-2 micromolar.

K_d= 100 micromolar

So, the percentage of receptor C occupied with the drug= {[L] / ([L]+K_d)}= {10^-2/(10^-2+100)}

= 9.9 $\times$ 10^-5= (9.9 $\times$ 10^-5 $\times$ 100) %= 0.0099 % [Answer]

Therefore, 0.0099% of receptor C is occupied with the drug.

Calculation for receptor D:

[L]= 1 $\times$ 10^-8 Molar= 1 $\times$ 10^-8 $\times$ 10⁹ nanomolar [As 1 Molar= 10⁹ nanomolar] = 10 nanomolar.

K_d= 1 nanomolar

So, the percentage of receptor D occupied with the drug= {[L] / ([L]+K_d)}= {10/(10+1)}= 0.91 =(0.91 $\times$ 100) %= 91% [Answer]

Therefore, 91% of receptor D is occupied with the drug.

Calculation for receptor E:

[L]= 1 $\times$ 10^-8 Molar= 1 $\times$ 10^-8 $\times$ 10⁹ nanomolar [As 1 Molar= 10⁹ nanomolar] = 10 nanomolar.

K_d= 10 nanomolar

So, the percentage of receptor E occupied with the drug= {[L] / ([L]+K_d)}= {10/(10+10)}= 0.5 =(0.5 $\times$ 100) %= 50% [Answer]

Therefore, 50% of receptor E is occupied with the drug.