Question

The enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.

The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9 mol/l). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7 mol/l).

Compute the total concentration of [D1]tot that is needed to bind 90% of the HIV-RT at the given concentration [E]tot.You do NOT have to consider competition betwwen the drugs D1 and D2! They are administered separately.

Compute the total concentration of [D2]tot that is needed to bind 90% of the HIV-RT at the given concentration [E]tot. You do NOT have to consider competition between the drugs D1 and D2! They are administered separately.

Answer 1

[ED] = [E] [D] / (K_D + [D])

D1:

For D1, [ED1] = [E] [D1] / (K_D1 + [D1]) where K_D1 = 1.00 x 10^-9 mol/l; [E] = 2 x 10^-6 mol/l

If 90% of D1 is bound to ED1, then [ED1] = 0.9 [D1]

=> 0.9 [D1] = 2 x 10^-6 x [D1] / (1 x 10^-9 + [D1])

=>10^-9 + [D1] = 2 x 10^-6 x [D1] / 0.9 [D1]

=> 10^-9 + [D1] = 2 x 10^-6 / 0.9

=> [D1] = (2 x 10^-6 / 0.9) - 10^-9

  => [D1] = 2.22122 x 10^-6

Total concentration of D1 that is needed to bind 90% of the HIV-RT at the given concentration = 2.22122 x 10^-6 mol/L

D2:

For D2, [ED2] = [E] [D2] / (K_D2 + [D2]) where K_D2 = 1.00 x 10^-7 mol/l; [E] = 2 x 10^-6 mol/l

If 90% of D2 is bound to ED2, then [ED2] = 0.9 [D2]   

=> 0.9 [D2] = 2 x 10^-6 x [D2] / (1 x 10^-7 + [D2])

=>10^-7 + [D2] = 2 x 10^-6 x [D2] / 0.9 [D2]

=> 10^-7 + [D2] = 2 x 10^-6 / 0.9

=> [D2] = (2 x 10^-6 / 0.9) - 10^-7

  => [D2] = 2.1222 x 10^-6

Total concentration of D2 that is needed to bind 90% of the HIV-RT at the given concentration = 2.1222 x 10^-6 mol/L