Problem solving steps

Transcription

1 Problem solving steps Determine the reaction Write the (balanced) equation ΔG K v Write the equilibrium constant v Find the equilibrium constant using v If necessary, solve for components K K = [ p ] ν i i i products ν [ r ] j j reactants ΔG = exp RT j 0 K ΔG v Find the equilibrium constant using v Find ΔG using 0 ΔG = RT ln K K = [ p ] ν i i i products ν [ r ] j j reactants j [r] [p] v Use equilibrium constant definition

2 Reactions: phase partition, LogP Two immiscible phases, e.g.: v cellular membrane (lipid) and extracellular fluid (aqueous) v cellular membrane (lipid) and cytoplasm (aqueous) v octanol (non-polar) and water (polar) experimental model system Reaction of spontaneous transfer of the compound from non-polar to polar phase: v cpd dissolved in non-polar phase same cpd dissolved in polar phase K = Coct/CH2O (ratio of two concentrations) Popular compound characteristic: v logp = log10 (Coct/CH2O) K = 10logP May need to use molar amt = conc x volume

3 Phase Partition Equilibrium I Problem: 4.02 mmol of a compound with logp = 2.3 is added to a container with 0.5L octanol and 0.5L water. Find the concentrations of the compound in water and octanol after the system equilibrates. Solution: v Reaction: Compound in H 2 O Compound in Octanol v logp = 2.3 K = C oct / C H2O = , so C oct 200 C H2O Molar amount Concentration Volume Water x [mol] x / 0.5 L = 2x [M] 0.5 L Octanol 200x [mol] 200 C H2O = 400x [M] 0.5 L v 200x + x = 4.02 mmol 201x = mol x = 20 µmol v 20 µmol in water, 4 mmol in octanol v Concentrations: 20 µmol / 0.5 L = 40 µm in water, 4 mmol / 0.5 L = 8 mm in octanol (sanity check, please?) Answer: 40 µm and 8 mm

4 Phase Partition Equilibrium II Problem: 12.3 µmol of a compound with logp = 2.3 is added to a container with 0.5L octanol and 2.5L water. Find the concentrations of the compound in water and octanol after the system equilibrates. Solution: v Reaction: Compound in H 2 O Compound in Octanol v logp = 2.3 K = C oct / C H2O = and C oct 200 C H2O Molar amount Concentration Volume Water x mol x / 2.5 L = 0.4x M 2.5 L Octanol 200*(0.5/2.5)= 40x mol 200 C H2O = 80x M 0.5 L v 40x + x = 12.3 µmol 41x = mol x = 0.3 µmol v 0.3 µmol in water, 12 µmol in octanol v Concentrations: 0.3 µmol / 2.5 L = 0.12 µm in water, 12 µmol / 0.5 L = 24 µm in octanol (sanity check, please?) Answer: 0.12 µm and 24 µm

5 Phase Partition, K to ΔG Problem: 4.6 mmol of a compound was added to a container with 1L octanol and 3L water. After the system equilibrated, the aqueous concentration of the compound was found to be 0.2 mm. Find the compound logp and molar ΔG of its transition from octanol to water. Solution: Molar amount Concentration Volume Water 0.2 mm 3L = 0.6 mmol 0.2 mm 3L Octanol = 4 mmol 4 mmol / 1L = 4mM 1L v logp = log C oct / C H2O = log (4mM / 0.2 mm) = log 20 ~ 1.3 v ΔG = RT ln C H20 /C oct v ΔG= 0.6 ln 1/20 = 0.6 ln 20 = = 1.8 kcal/mol (sanity check?) Answer: logp = 1.3, ΔG (octanol water) = 1.8 kcal/mol Useful hint: ln 20 ~ 3.

6 Favorite reactions: conformational transition Compound has two distinct stereo-isomers, e.g.: v (if there is a chiral center) (S) and (R) v (if there is a double bond) cis- and trans- v state A and state B Reaction of spontaneous transition: v A B K = C B /C A (ratio of two concentrations) Enantiomers: Cis- and Trans- isomers: trans- cis- (S)-thalidomide Teratogenic (R)-thalidomide Safe, effective against morning sickness

7 Problem: Isomerisation, ΔG to K The Gibbs free energies of formation of two stereoisomers, X and Y, of a compound at T = 300K are G f (X)=93.2 kcal/mol and G f (Y) = 96.2 kcal/mol. Estimate the molar fraction of [Y] in the (equilibrium) racemic mixture at this temperature. Solution: v Reaction Y X v ΔG = G f (X) G f (Y) = 3 kcal/mol v Find the equilibrium constant, K = [X]/[Y] v K = exp ( ΔG / RT) = exp (3 / 0.6) = exp 5 ~ 149 v [X]/[Y] ~ 149/1 (sanity check, please?) v Molar fraction of Y is [Y]/([Y]+[X]) ~ 1 / 150 Answer: ~ 1/150

8 Favorite reactions: 1:1 protein-ligand binding Compound (ligand) binds to its target in 1:1 stoichiometry Reaction: P + L PL K a = [PL] / [P][L] (association constant, binding constant, affinity constant, binding affinity ) K d = [P][L] / [PL] (dissociation constant) = 1/K a ΔG bind = RT ln K a AND ΔG bind = RT ln K d

9 Protein-ligand binding, K to ΔG In the solution at equilibrium, the concentrations of unbound drug and protein are 13.5 nm and 0.5 nm, respectively, while the concentration of protein/drug complex is 4.5 nm. Find ΔG binding. Solution: v Reaction: P + L PL v Kd = [P][L] / [PL] = / ( ) = 1.5 nm v Ka = 1 / ( ) ~ M 1 v ΔG = RT ln Ka = RT ln Kd v ΔG = 0.6 ( 9 ln 10 + ln 1.5) kcal/mol Answer: ΔG kcal/mol

10 K 2 / K 1 = 10 Shortcut: K to ΔG ΔG 2 ΔG 1 = RT ln K 2 + RT ln K 1 = RT ln (K 2 / K 1 ) ΔG 2 ΔG 1 = 0.6 ln kcal/mol K increases by a factor of 10 when ΔG decreases by 1.4 kcal/mol For example: correspondence between ΔG and Kd for protein/ligand binding: ΔG bind Kd mm µm nm pm fm

11 Problem: Protein/drug binding, using K to ΔG shortcut A drug candidate was chemically optimized to reduce the therapeutic concentration 1000 times. Estimate the binding energy improvement required to reach that goal. Solution: v 10-fold Kd improvement 1.4 kcal/mol decrease in ΔG v 100-fold Kd improvement 2.8 kcal/mol decrease in ΔG v 1000-fold Kd improvement 4.2 kcal/mol decrease in ΔG Answer: The binding energy needs to be decreased by 4.2 kcal/mol.

12 Protein-ligand binding equilibrium Problem: In the solution at equilibrium, the concentrations of unbound drug and protein are 13.5 nm and 0.5 nm, respectively. Given the Kd of 1.5 nm, estimate the fraction of total protein which is bound (the binding reaction has 1:1 stoichiometry). Solution: v Reaction: P + L PL, v Kd = [P][L] / [PL] = M v [PL] = [P][L] / Kd = / ( ) = M v Unbound protein: 0.5 nm, bound protein 4.5 nm, total 5 nm v Fraction bound = 4.5 / 5 = 90% Answer: 90% of the protein is bound.

13 Protein-Ligand Binding Equilibration Simplest case, 1:1 binding stoichiometry P + L PL Eq. constant of the forward reaction: Ka = [PL]/[P][L] Eq. constant of the reverse reaction: Kd = [P][L]/[PL] dissociation constant, widely used to describe proteinligand binding v At equilibrium, Kd = [P][L] / [PL] x Kd = (P 0 x)(l 0 x) v x Kd = x 2 (P 0 +L 0 )x + P 0 L 0 v x 2 (P 0 +L 0 +Kd) x + P 0 L 0 = 0 - quadratic v a = 1; b = (P 0 +L 0 +Kd); c = P 0 L 0 v Solve ax 2 + bx + c = 0 Protein Ligand Complex Start (no equilibrium) [P] = P 0 [L] = L 0 0 Equilibration [P] = P 0 x [L] = L 0 x [PL] = x

14 Example 0.30 µm of protein is mixed with 0.36 µm of drug. The dissociation constant is K d = 0.01 µm. Evaluate the bound protein concentration after the system equilibrates. Solution: Protein Ligand Complex Start (no equilibrium) [P] = P 0 [L] = L 0 0 Equilibration [P] = P 0 x [L] = L 0 x [PL] = x v x 2 (P 0 +L 0 +K d ) x + P 0 L 0 = 0 quadratic v Assuming that x, P 0, L 0, and K d are all measured in the same units (e.g. µm), we can cancel out the prefix factor (e.g )

15 Example cntd. v a = 1 v b = ( ) = 0.67 v c = = v Solve ax 2 + bx + c = 0 v x = ( b ± Sqrt(b 2 4ac)) / 2a = 0.27 µm or 0.40 µm v x cannot exceed P 0 or L 0, so x = 0.27 µm (use the solution with ) Protein Ligand Complex Start (no equilibrium) [P] = 0.3 µm [L] = 0.36 µm 0 Equilibration [P] x = 0.03µM [L] x = 0.09µM x = 0.27µM v And, BTW, (0.03µM 0.09µM) / 0.27µM = 0.01 µm = Kd Answer: 0.27µM

16 Equilibrium [PL] as a function of total ligand Given a test tube with the initial protein concentration P 0, how much complex is formed upon addition of L 0 (concentration) of ligand with a given Kd? x = L bound = P bound = P 0 + L 0 + K d P 0 + L 0 + K d ( ) 2 4P 0 L 0 2

17 Two types of titration curves Plot [PL] vs total ligand Straight line, fast saturation strong binding Differentiate Slope 0 strong binding [PL] (µm) Δ[PL] Slope << 0 weak binding Curved line, slow saturation weak binding P 0 = 10 µm, K d = 10 µm P 0 = 10 µm, K d = 1 µm P 0 = 10 µm, K d = 0.1 µm Total Ligand (µm) Total Ligand (µm)

18 If a drug could be profiled against all proteins in the body Primary target # of proteins in the body Secondary target(s), close homologs of primary target Plasma proteins (e.g. albumin) ΔG bind (kcal/mol) (simulated histogram)

19 Total protein vs Kd: important cases P tot << K d [P] << K d [P] / K d << 1 [PL] << [L] Ligand is not depleted by binding to the protein True for most biological targets in vivo P tot > K d [P] ~ K d [P] / K d ~ 1 [PL] ~ [L] Ligand is depleted by binding to the protein True for albumin, antitrypsin and other abundant plasma proteins) Only unbound fraction exhibits pharmacological action.

20 If ligand is not depleted P tot << K d and [PL] << [L] [L tot ] [L] Target bound/unbound ratio (from definition of K d ): [PL] / [P] = [L] / K d ~ [L tot ] / K d When [L tot ] K d, [PL] = [P], i.e. K d is the ligand concentration at which 50% target is bound. Similarly, [L tot ] Kd [PL]/[P] for any bound/unbound ratio.

21 Kd ligand concentration at which 50% target is bound? P 0 = 1 nm Rule applicable P 0 = 10 µm Rule not applicable Kd = 10 µm Kd = 1 µm Kd = 10 µm Kd = 1 µm [PL] (µm) 50% protein bound [PL] (µm) 50% protein bound Works only when P 0 << Kd!!! 1 µm 10 µm Total Ligand (µm) 1 µm 10 µm Total Ligand (µm)

22 Example: using total Lig Kd [PL]/[P] The concentration of the target protein in the patient s body is 5 pm. Given a drug with Kd of 10 nm, what concentration of the drug is needed for 80% of the protein to be bound? Solution: v [P] << Kd v Desired [PL]/[P] ratio is 80/20 = 4/1 v Total ligand = [PL]/[P] Kd 40nM Answer: 40 nm Note: we need 90 nm drug for 90% bound protein, and 190 nm for 95% bound protein.

23 If ligand is depleted P tot > K d ; [PL] ~ [L] If protein is in excess, i.e. [PL] << [P] and [P tot ] [P] Ligand bound / unbound ratio (from definition of K d ): [PL] / [L] = [P] / K d ~ [P tot ] / K d [P tot ] / K d defines bound/unbound ratio for the ligand Fraction bound ligand = [P tot ] / ([P tot ] + K d )

24 High affinity drug-albumin binding Problem: Kd (Albumin, warfarin) is ~5 µm, calculate albumin binding in %, assuming albumin in physiological range. Solution: v [P tot ] / ([P tot ] + K d ) v with albumin concentration at 450 µm: 450 / 455 ~ 98.9% v with albumin concentration at 750 µm: 750 / 755 ~ 99.3% Note: unbound warfarin varies between 0.7% and 1.1%, i.e. 57% increase for a skinny fasting person For drugs with high plasma protein binding, small changes in plasma protein can dramatically affect free drug concentration.

25 Low affinity drug-albumin binding Example: Kd (Albumin, drug B) is ~5 mm, calculate albumin binding in %, assuming albumin in physiological range. Solution: v [P tot ] / ([P tot ] + K d ) v with albumin concentration at 450 µm: 450 / 5450 ~ 8.3% v with albumin concentration at 750 µm: 750 / 5750 ~ 13% Unbound drug B varies between 87% and 91.7%, i.e. only 5.5% increase for a skinny fasting person Variations of plasma concentrations of unbound drug B are not so dramatic.

26 Direct experimental determination of Kd Simultaneous monitoring of [P], [L], and [PL] is unfeasible Use several test tubes with the same P 0, add different total ligand concentrations OR: use a single test tube, add ligand in small injections Monitor one of the following: v [PL] v [P] (unbound protein concentration) v [L] (unbound ligand concentration) Plot the monitored value against total ligand Find Kd by fitting the curve into the equation (software available) ( ) P0 + L0 + Kd P0 + L0 + Kd 4P0 L0 x = 2 A precise method: Isothermal Titration Calorimetry, ITC 2

27 Isothermal Titration Calorimetry Drug is injected to the sample cell in equal doses The amount of heat on each injection is proportional to x = Δ[PL] The shape of the curve defines Kd ΔG The values on the curve define ΔH ΔS can be calculated from ΔG = ΔH TΔS ΔG ΔH TΔS