Human immunodeficiency virus (HIV) is a retrovirus that causes acquired immunodeficiency syndrome (A — Physical Chemistry — Thermodynamics Chemistry Question
HIV protease
Human immunodeficiency virus (HIV) is a retrovirus that causes acquired immunodeficiency syndrome (AIDS). AIDS is a condition in which the afflicted patient’s immune system fails progressively, allowing otherwise benign infections to be life-threatening. The life cycle of HIV relies on the enzyme HIV-1 protease. As this enzyme plays a crucial role in the replication of the virus, HIV-1 protease has been a prominent target for therapy, with drugs being designed to inhibit the action of the enzyme. An HIV-1 protease inhibitor binds to the active site of the enzyme more strongly than the substrate that it mimics, thus disabling the enzyme. Consequently, without active HIV-1 protease present, viral particles do not mature into infectious virions.
Several inhibitors of HIV-1 protease have been licensed as drugs for HIV therapy. A detailed thermodynamic and kinetic study of seven then-available inhibitors of HIV-1 protease was performed in 2003 in Uppsala (interested readers are referred to the original publication in J. Mol. Recognit. DOI: 10.1002/jmr.655). The molecular structures of six of them are shown below.
The affinity of the selected compounds to the HIV-1 protease was measured in terms of equilibrium constants for the dissociation of the protease–inhibitor complex, in a range of temperatures from 5 °C to 35 °C under otherwise identical conditions including pH. The obtained data are presented below; dissociation constants KD are in units of nM, i.e. 10−9 mol dm−3.
Temperature °C | Amprenavir | Indinavir | Lopinavir | Nelfinavir | Ritonavir | Saquinavir
5 | 1.39 | 3.99 | 0.145 | 6.83 | 2.57 | 0.391
15 | 1.18 | 2.28 | 0.113 | 5.99 | 1.24 | 0.320
25 | 0.725 | 1.68 | 0.101 | 3.67 | 0.831 | 0.297
35 | 0.759 | 1.60 | 0.0842 | 2.83 | 0.720 | 0.245
The dissociation rate constants kD (in the units of 10−3 s−1) of the protease–inhibitor complexes for each inhibitor at two temperatures are presented below:
Temperature °C | Amprenavir | Indinavir | Lopinavir | Nelfinavir | Ritonavir | Saquinavir
5 | 1.85 | 1.88 | 0.506 | 0.912 | 1.93 | 0.146
25 | 4.76 | 3.44 | 0.654 | 2.17 | 2.59 | 0.425
Which of the compounds binds most strongly to the protein at 35 °C?
Model Answer
Lopinavir binds most strongly, as illustrated by its smallest dissociation constant KD.
Calculate the standard Gibbs energy of binding (i.e. association) for each compound at each temperature. It may be of advantage to use a spreadsheet application.
Model Answer
Apply ΔG° = −RT lnKD, and consider that the dissociation and the binding are opposite reactions. Thus, ΔG°(bind.) = −ΔG°(dissoc.) = RT lnKD, or in a slightly different way, ΔG°(bind.) = −RT lnKA = −RT ln[1 / KD] = RT lnKD. See below for the numerical results.
Use the temperature-dependent data to calculate the standard enthalpy and entropy of binding of each of the compounds. Consider the enthalpies and entropies to be independent of temperature in the interval 5–35 °C.
Model Answer
Consider ΔG° = ΔH° − TΔS°. Thus, perform a linear regression of the temperature dependence of ΔG°. This can be done in at least two simplified ways: (i) Plot the dependence and draw a straight line connecting the four data points in the best way visually. Then, read off the slope and intercept of the straight line, which correspond to −ΔS° and ΔH°, respectively. (ii) Alternatively, choose two data points and set up and solve a set of two equations for two unknowns, which are ΔS° and ΔH°. The most accurate result should be obtained if the points for the lowest and highest temperatures are used.
Note 1: ΔS° and ΔH° may also be obtained from a fit of KD or KA, without considering ΔG°. Here, a straight line would be fitted to the dependence: ln KA = −lnKD = ΔS° / R − ΔH° / R × 1/T.
Note 2: It is evident that the binding is entropy-driven for all the inhibitors. The entropic gain stems from the changes in the flexibility of both the protease and the inhibitors, and also involves solvent effects. However, a molecular picture of those changes is rather complex.
Identify the inhibitor with the slowest dissociation from the protease at 25 °C.
Model Answer
The slowest dissociation is observed for the compound with the smallest dissociation rate constant, i.e. Saquinavir.
Calculate the rate constants of association (i.e. binding) of the protein–inhibitor complexes, kA, at 25 °C for all inhibitors. Which of the inhibitors exhibits the fastest association with the protease?
Model Answer
Using the relation for the dissociation constant KD = kD / kA and the data at 25 °C, we obtain for Amprenavir: kA = kD / KD = 4.76 × 10−3 s−1 / (0.725 × 10−9 mol L−1) = 6.57 × 106 L mol−1 s−1. Analogous calculations performed for the other inhibitors yield the following numerical results. The fastest association is exhibited by the compound with the largest association rate constant, i.e. Amprenavir.
Amprenavir: 6.57 × 10^6 dm3 mol−1 s−1
Indinavir: 2.05 × 10^6 dm3 mol−1 s−1
Lopinavir: 6.48 × 10^6 dm3 mol−1 s−1
Nelfinavir: 0.59 × 10^6 dm3 mol−1 s−1
Ritonavir: 3.12 × 10^6 dm3 mol−1 s−1
Saquinavir: 1.43 × 10^6 dm3 mol−1 s−1
Using the Arrhenius equation, calculate the activation free energy of dissociation ΔG‡ (or Ea) of Lopinavir as well as that of the slowest dissociating inhibitor from question 3.4, and the compound with the largest association rate constant (which were obtained in question 3.5). Assume that the activation free energy is constant in the respective temperature range.
Model Answer
The Arrhenius equation for the rate constant reads k = A × exp[−ΔG‡ / RT]. For two known rate constants of dissociation k1 and k2 determined at temperatures T1 and T2, respectively, we obtain a system of two equations,
k1 = A × exp[−ΔG‡ / RT1]
k2 = A × exp[−ΔG‡ / RT2],
from which the activation energy of dissociation results as ΔG‡ = (ln k1 / k2) / (1 / RT2 − 1 / RT1). Numerically, the activation energy is 8.9 kJ mol−1 for Lopinavir, 32.6 kJ mol−1 for Amprenavir (which has the fastest association rate constant) and 36.8 kJ mol−1 for Saquinavir (which has the lowest dissociation rate constant).
Is the inhibitor with the largest activation energy for dissociation the same compound as the strongest binder which was identified in question 3.1? We may extrapolate this finding: What relationship is there between the strength of binding expressed by the dissociation constant and the rate of dissociation expressed by the activation energy of dissociation?
Model Answer
No, these are two different compounds. The strongest protease binder is not the same inhibitor as the one with the slowest dissociation. This observation may seem counter-intuitive if the distinction between thermodynamics (here, the strength of binding expressed by the equilibrium constant) and kinetics (the rate of binding represented by the rate constant or activation energy for dissociation) is not understood properly. While the equilibrium constant of dissociation captures the thermodynamic stability of the respective protein–inhibitor complex, the rate constant describes the kinetics of the process. These are two different sets of properties and they only become related if the rates of both dissociation and association are considered, KD = kD / kA.