A medicinal chemist is doing some computer modeling to predict the best inhibito

Question

A medicinal chemist is doing some computer modeling to predict the best inhibitor for the HIV protease enzyme. He/She models the transition state of a drug undergoing peptide-bond cleavage in the active site of HIV protease. He/She obtains the following computed absolute energies (even though absolute energies don't exist, they can be calculated from theory). All energies are in kJ/mol and entropies in J/mol.K. Note that the reaction is Peptide (in enzyme active site) rightarrow Peptide Transition State (TS, in enzyme) rightarrow Cleaved peptide (in active site) The chemist studied two molecules, first a normal peptide (PEP), and second a drug candidate (DRG). Calculate delta H, delta S, and delta G from this data. One set of data from PEP and one set for the drug molecule DRG. Myoglobin unfolding was studied in urea solution.

Explanation / Answer

Solution:

(a) For HIV protease enzyme, PEP:

The enthalpy/entropy of activation, H‡/S‡ are the change in entropy when the reactants change from their initial state to the activated complex or transition state

Here, PEP having energy (Enthalpy, H1)= -5775000 kJ/mol

            PEP-TS(‡), Enthalpy, H2 = -5774953 kJ/mol

H‡ = H2 – H1 = -5774953 + 5775000 = 47 kJ/mol = 47000 J/mol

S‡ = S2 – S1 = 400 – 475 = -75 J/mol.K

We know, G‡ = H‡ - TS‡ = 47000 – 298(-75) = 69350 J/mol

= 69.35 kJ/mol

(Assuming room temperature reaction, i.e. T = 298 K)

(b) For the drug, DRG:

DRG having energy (Enthalpy, H1)= -3855671 kJ/mol

DRG-TS(‡), Enthalpy, H2 = -3855612 kJ/mol

H‡ = H2 – H1 = -3855612 + 3855671 = 59 kJ/mol = 59000 J/mol

S‡ = S2 – S1 = 240 – 310 = -70 J/mol.K

We know, G‡ = H‡ - TS‡ = 59000 – 298(-70) = 79860 J/mol

= 79.86 kJ/mol

(Assuming room temperature reaction, i.e. T = 298 K)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.