As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT)
ID: 187093 • Letter: A
Question
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.
Explanation / Answer
Answer.
The binding and dissociation of D2 from HIV-RT can be represented as;
Binding of enzyme and drug, D2
HIV-RT + D2 ----> HIV-RT.D2 .It has a binding constant, K1 , which is not given.
Dissociation of enzyme drug complex
HIV-RT.D2 ----> HIV-RT + D2 it has a dissociation constant, K2 which is given as 100 nM-1. (It is wrongly given as 100nM. Since it is a dissociation, K2 will have units of reverse nM).
Lets try to understand what is happening. We had a 2uM concentration of HIV-RT and we added D2 so that 99% of D2 has bound to the enzyme. In otherwords, there is only 0.0198% of free enzyme left [99% of 2uM = 0.0198uM].
Applying law of conservation of mass,
[HIV-RTTot] = [HIV-RTfree] + [HIV-RT.D2]
[HIV-RTfree] = [HIV-RTTot] - [HIV-RT.D2]
0.0198 uM = 2 uM - [HIV-RT.D2]
[HIV-RT.D2] = 2 uM - 0.0198 uM
[HIV-RT.D2] = 1.9802 uM
We are also given, K2 ,
K2 = [HIV-RT.D2] / [HIV-RTTot] x [D2]
0.1 uM-1 = 1.9802 uM / 2 uM x [D2]
[D2] = 1.9802 uM / 2 uM x 0.1uM-1
[D2] = 0.396 uM
or in mole/l
[D2] = 0.396 x 10-6 mol/l
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