* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Lecture6_RW
Discovery and development of non-nucleoside reverse-transcriptase inhibitors wikipedia , lookup
Cannabinoid receptor antagonist wikipedia , lookup
Discovery and development of ACE inhibitors wikipedia , lookup
Pharmacokinetics wikipedia , lookup
DNA-encoded chemical library wikipedia , lookup
Discovery and development of neuraminidase inhibitors wikipedia , lookup
Discovery and development of angiotensin receptor blockers wikipedia , lookup
Discovery and development of tubulin inhibitors wikipedia , lookup
5-HT3 antagonist wikipedia , lookup
Plateau principle wikipedia , lookup
Discovery and development of direct Xa inhibitors wikipedia , lookup
Nicotinic agonist wikipedia , lookup
Discovery and development of integrase inhibitors wikipedia , lookup
Neuropharmacology wikipedia , lookup
NK1 receptor antagonist wikipedia , lookup
Discovery and development of antiandrogens wikipedia , lookup
Reconciling Experimental and Theoretical Energies Theory Experiment Energy Minimization (Internal Energy) Thermodynamic Data (ΔG, ΔH, ΔS) MM/GBSA (average internal energy with solvation included) Kinetic Data (KA, KD, kon, koff) FEP/TI (relative free energies, ΔG) Inhibition Data (Ki, IC50) Activity Data (EC50) How do you compare theoretical ligand (drug) binding energies with experimental data? What is Affinity? Affinity: The strength of noncovalent chemical binding between two substances as measured by the dissociation constant (KD) of the complex. For a protein (P), binding to a ligand (L) to for a complex (PL): the dissociation constant KD for the complex is: KD [P][L] [PL] P P+L L PL PL where [P], [L] and [PL] represent molar concentrations of the protein, ligand and PL-complex, at equilibrium, respectively. The units of KD are moles (M) The dissociation constant is the concentration of ligand [L] at which the binding site on a particular protein is half occupied, i.e. KD is the concentration of ligand, at which the concentration of protein with ligand bound [PL], equals the concentration of protein with no ligand bound [P]. KD and ΔG are Easily Related P+L PL The smaller the KD , the more tightly bound the ligand is, or the higher the affinity between ligand and protein. Or conversely, the larger the KA, the more tightly the ligand binds ΔGBinding = -RT lnKA = RT lnKD What is the difference in binding free energy for two drugs, one that binds with mM affinity and one that binds with nM affinity? Assume room temperature (RT = 0.6). For example: a ligand with a nanomolar (nM) dissociation constant binds more tightly to a particular protein than a ligand with a micromolar (μM) dissociation constant. 1) Convert the KD units to moles: 1mM = 1x10-3 M, 1nM = 1x10-9M 2) Plug the KD values into the formula Drug 1 Drug 2 KD 1x10-3 M 1x10-9M ΔG -4.1 kcal/mol -12.4 kcal/mol ΔΔG -8.3 kcal/mol Why Do Experiments Measure KD and not ΔG? Dissociation constants can be determined from many different types of experiment. ΔG is extremely difficult to measure experimentally. Direct Binding Method KD ΔG Surface Plasmon Resonance (KD, kon, koff) Isothermal Titration Calorimetry (ΔH, and ΔS) NMR Spectroscopy TitrationBased Methods Spectrophotometry Potentiometry enzyme-linked immunosorbent assays (ELISA) ΔGBinding = RT lnKD What about kon and koff Some experimental methods, notably surface plasmon resonance (SPR) can measure the kinetic on and off rates for a binding reaction. P+L kon koff koff KD k on C What about kon and koff Some experimental methods, notably surface plasmon resonance (SPR) can measure the kinetic on and off rates for a binding reaction. Depending on the quality of the data and the rate of the kinetics, kon and koff can be determined by fitting the “mass action equation” to the SPR sensorgram kon koff d[PL] [P][L]kon [PL]koff dt What about IC50 values and Ki The IC50 is the concentration of an inhibitor that reduces the binding of the native ligand by half (50%). The IC50 is a measure of how well a drug (or inhibitor) competes with the natural or native ligand for binding to a receptor. KD PL P+L+I KI PI IC 50 Ki [L] 1+ KD Cheng Y, Prusoff WH. Relationship between the inhibition constant (Ki) and the concentration of inhibitor which causes 50 per cent inhibition (IC50) of an enzymatic reaction. Biochem Pharmacol 1973; 22:3099–108. IC50 does not necessarily say anything about the thermodynamics of binding. IC50 tells you how concentrated your solution with respect to ligand has to be in order for 50% of the receptors to be bound. If you have poor binding, you will need a large concentration to start filling up receptor binding sites. If you have very good binding, you won't need very much of your inhibitor to start filling up binding sites. What about IC50 values and ΔG It is often easier to measure IC50 than Ki, since to measure Ki requires IC50 as well as [L] and KD IC 50 Ki [L] 1+ KD How can we relate IC50 values to free energies of binding? 1) If two inhibitors are assayed against the same protein with the ligand at the same concentration then ΔΔG = -RT ln Ki,1/Ki,2 and therefore ΔΔG = -RT ln IC50,1/IC50,2 ΔΔGBCV-RTV = -RT ln 0.4x10-9/1641x10-9 = -RT ln 0.0002437 = -0.6 (-8.32) ΔΔGBCV-RTV = 4.99 kcal/mol XV Intl HIV Drug Resistance Workshop June 13-17, 2006, Sitges, Spain "Resistance-Associated Amino Acid Substitutions and Drug Susceptibility Analysis of Virus from Subjects Entering the Phase II Dose-Ranging Study of a New Protease Inhibitor (PI), Brecanavir, HPR20001 (STRIVE)" What about IC50 values and ΔG It is often easier to measure IC50 than Ki, since to measure Ki requires IC50 as well as [L] and KD IC 50 Ki [L] 1+ KD How can we relate IC50 values to free energies of binding? 2) Alternatively, it is sometimes assumed that Ki ≈ IC50 Therefore ΔG ≈ RT ln IC50 ΔGBCV = RT ln(0.4x10-9) = -13.0 kcal/mol ΔGRTV = RT ln(1641x10-9) = -7.99 kcal/mol ΔΔGBCV-RTV = 5.01 kcal/mol Example: Design of HIV Transcriptasae Inhibitors Example: Theoretical Background ΔGN01= RT ln(0.125x10-6) = -9.5 kcal/mol ΔGN14= RT ln(15.3x10-6) = -6.6 kcal/mol ΔΔGN01,N14 = 2.9 kcal/mol ΔΔGN01,N14 = -RT ln IC50,N01/IC50,N14 = 0.6 ln (0.008) = 2.9 kcal/mol Example: Computational Setup Example: Initial Model Example: Checking Data Convergence Example: Comparison with Experimental Energies