Physical chemistry Question: From the Lecture: HIV-RT will exist in two forms: -

Question

Physical chemistry Question:

From the Lecture:

HIV-RT will exist in two forms:

- free enzyme [E]

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).

Part A - Difference in binding free eenergies

Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol.

The difference in binding free energy between both complexes is ______ kJ/mol

Part B - Compare difference in free energy to the thermal energy

Divide the difference between the binding free energies of the ED1 and ED2 complexes by the thermal energy (at the physiological temperature). Provide a numerical expression with 3 significant figures.

Part C - Required drug concentrations 1

Compute the total concentration of [D1]tot that is needed to bind 90% 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.

______mol/l

Explanation / Answer

Part A) acc to given eq binding energy delta G = RTlndeltaKdiss

we have dissociation constabt for both the drugs so putting them in seperate eq we get the value of delta G.

for D1, deltaG = 8.314×310× ln 10-9

   = 0.732

similarly for D2, deltaG =5.411

difference is 5.411-0.732= 4.679 kj/mol

part B) dividing the answer in A by RT

4.679/8.314×310= 1.81×10-3

Part C) [D]tot= Kdiss + 1uM

= 1×10-9 + 1×10-6

   = 10-6mol/l

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects