Reconciling Experimental and Theoretical Energies

Theory Experiment Energy Minimization (Internal Thermodynamic Data (ΔG, ΔH, Energy) ΔS) MM/GBSA (average internal Kinetic Data energy with solvation (KA, KD, kon, koff) included)

FEP/TI (relative free energies, Inhibition Data (Ki, IC50) ΔG)

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): P + L PL

P L PL the dissociation constant KD for the complex is: [P][L] K  where [P], [L] and [PL] represent molar concentrations of D [PL] 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

The smaller the KD , the more tightly bound the ligand is, or the P + L PL higher the affinity between ligand and protein. Or conversely, the larger the KA, the more tightly the ligand binds

For example: a ligand with a nanomolar (nM) ΔGBinding = -RT lnKA = RT lnKD dissociation constant binds more tightly to a particular protein than a ligand with a micromolar (μM) dissociation constant.

-3 1) Convert the KD units to moles: 1mM = 1x10 M, 1nM = What is the difference in 1x10-9M binding free energy for 2) Plug the K values into the formula two drugs, one that D binds with mM affinity Drug 1 Drug 2 and one that binds with K 1x10-3 M 1x10-9M nM affinity? Assume D room temperature (RT = ΔG -4.1 kcal/mol -12.4 kcal/mol 0.6). 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.

KD ΔG Direct Binding Surface Plasmon Resonance Isothermal Titration Method (KD, kon, koff) Calorimetry (ΔH, and ΔS) NMR Spectroscopy Titration- Spectrophotometry Based Potentiometry Methods 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.

kon P + L C koff

koff KD  kon

 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” kon to the SPR sensorgram koff

d[PL]  [P][L]k [PL]k dt on off

 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 IC Cheng Y, Prusoff WH. Relationship between the K  50 inhibition constant (Ki) and the concentration of P + L + I i inhibitor which causes 50 per cent inhibition [L] (IC50) of an enzymatic reaction. Biochem 1+ Pharmacol 1973; 22:3099–108. KD KI PI

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 K D IC K  50 i [L] How can we relate IC50 values 1+ to free energies of binding? KD 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 

-9 -9 ΔΔGBCV-RTV = -RT ln 0.4x10 /1641x10 = -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 K D IC K  50 i [L] How can we relate IC50 values 1+ to free energies of binding? KD

2) Alternatively, it is sometimes assumed that Ki ≈ IC50 Therefore ΔG ≈ RT ln IC50  -9 ΔGBCV = RT ln(0.4x10 ) = -13.0 kcal/mol

-9 ΔGRTV = RT ln(1641x10 ) = -7.99 kcal/mol

ΔΔGBCV-RTV = 5.01 kcal/mol Example: Design of HIV Transcriptasae Inhibitors Example: Theoretical Background

-6 ΔGN01= RT ln(0.125x10 ) = -9.5 kcal/mol

-6 ΔGN14= RT ln(15.3x10 ) = -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