In: Biology
As discussed during the lecture, 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 [D2]tot that is needed to bind 99% of the HIV-RT at the given concentration [E]tot. Provide your answer as a numerical expression with 3 significant figures in the unit mol/l.
You do NOT have to consider competition betwwen the drugs D1 and D2! They are administered separately.
We have the drug D2 binding HIV-RT (enzyme E) to form the complex ED2:
The dissociation constant is given by:
where [E]u is the concentration of unbound enzyme and [D2]u is the concentration of unbound D2 at equilibrium.
The fraction of [E] that is bound by [D2] =
where [E]tot is the total concentration of E.
Now,
Similarly,
where [D2]tot is the total concentration of D2.
So,
i.e.
Given, fraction of bound [E] = 99% = 0.99
So,
Given Kd = 1 x 10-7mol/l for E and D2
So,
Hence, for a total concentration of D2 of 1.18 x 10-5mol/l, 99% of the given concentration of HIV-RT will be bound to D2.