1. Calculation of Reduction Potential of FAD in MCAD using Combined DFTB/MM Simulations Sudeep Bhattacharyay, Marian Stankovich, and Jiali Gao

2. Overview Objective Methods of calculation and strategies Results Future Directions

3. Flavoenzymes  Mediates electron transfer  Flavin ring shuttles between reduced and oxidized states  Protein environment controls the reduction potential of FAD  Coupled electron-proton transfer  pKa from experiment often misleading if that is observed through a observable signature belonging to a particular redox state  Need to predict correctly through simulation  Requires accuracy in a) reduction potential calculation and b) pKa calculation

4. Acyl-CoA Dehydrogenases (CAD) in Electron Transfer Ghisla, S. et al. Eur J Biochem, 2004. 271, 494-508. Medium chain acyl-CoA Dehydrogenase (MCAD) FAD is reoxidized FAD is reduced

5. MCAD: Structural Information All  -domain Extends into the other dimer N C All  -domain Two orthogonal  -sheets  Forms homotetrameric structure  Active site is formed at protein-protein interface  One FAD (cofactor) and one acyl-CoA (substrate) bind to the active site  Each active site work independent of the other  Passes electron to electron transfering protein (ETF) when it binds to MCAD ETF

6.  -hydride transfer on to the flavin nitrogenN  -proton abstraction by the catalytic base Glu376O  Potentials of mean force computation of MCAD demonstrate a STEPWISE mechanism  Both proton and hydride transfer CONTRIBUTE TO THE OVERALL CATALYTIC RATE in wild-type  Effect of PROTEIN ENVIRONMENT on the two steps can be investigated independently  A very attractive enzyme system to work with i.e. to TUNE the two reaction barriers  Need to know the effect of protein environment on the two processes Bhattacharyya S. et al. (2005) Biochemistry,44,16549 A Tale of Two Quasi-independent Processes R256Q T168A wt- transient intermediate

7. Reduction of Flavin blue (neutral) semiquinone red (anionic) semiquinone FAD + 2e + 2H +  FADH 2

8. Experimental Mid-point Potentials Values MCAD-bound FAD blue (neutral) semiquinone red (anionic) semiquinone -98.6 kcal/mol -99.3 kcal/mol Gustafson et al. (1986) J. Biol. Chem. 261, 7733-7741 -197.6 kcal/mol Mancini-Samuelson et al. (1998) Biochemistry, 37, 14605-14612 yellow

9. Methods and Strategies  Hybrid QM/MM methods; calculation of electron and proton affinities  Thermodynamic integration through FEP  Dual topology single coordinate method  Boundary condition

10. Electron and proton affinities Model Reactions a,b B3LYP/ 6-31+G(d,p)  H (kcal/mol) SCC-DFTB  H (kcal/mol) AM1(Gaussian)  H (kcal/mol) FAD + e¯  FAD¯-44.3-37.3-63.5 FAD¯ + e¯  FAD 2- +53.4+58.248.3 FAD + 2e¯  FAD 2- +9.1+20.9-15.1 FAD 2- + H +  FADH¯-436.5-442.3-420.0 FAD + 2e¯ + H +  FADH¯-427.4-421.4-435.1 FADH + e¯  FADH¯-49.4-50.3-57.2 FAD¯ + H +  FADH-333.7 -314.3 FADH¯+ H +  FADH 2 -331.7-331.4-320.2 FAD + 2e¯ + 2H +  FADH 2 -759.1-752.8-755.3

11. Free Energy Perturbation Potential energy of a hybrid system: U hybrid = (1-λ)U A + λU B λ is a coupling parameter varied from 0-1 (0.1, 0.2, ….) Free energy change ΔG = ∫ (∂G(λ)/ ∂λ) dλ = ∫  ∂U(λ)/ ∂λ)  dλ 0 1 1 0 Cartesian coordinates of the QM system kept invariant in the two states Change of chemical state of the system without any major change of the cartesian coordinate State AState B Thermodynamic integration method

12. QM/MM Interactions With same number of atoms in the two chemical states U tot (R) =  │ Ĥ QM +Ĥ el QM/MM │  + U van QM/MM (R) + U bonded QM/MM (R) + U MM (R) electronic energy van der Waals bonded energy of the of the QM system MM atoms + the electrostatic interaction energy Only the electrostatic term contributes to the free energy derivative as the two states have same cartesian coordinate GλGλ (R;λ) =  U A/MM (R QM,R MM ) + U B/MM (R QM,R MM )  Li, G. et al. J. Phys. Chem. B (2003) 107, 8643

13. Thermodynamic Schemes FEP for reduction potential calculation E-FAD (ox) E-FAD(red) FEP for pK a calculation ΔG Rd/Ox ΔG Rd/Ox is obtained from a single FEP calculation AH.E(aq) [A ¯ -D].E(aq) Aˉ.E(aq) + D(g) Aˉ.E(aq) + H + (aq) Aˉ.E(aq) + H + (g)  G solv H+H+  G E AH/A¯  G(1)  G(2)  G = 0.0  G E =  G(1) +  G(2) +  G solv AH/A¯ H + Li, G. et al. J. Phys. Chem. B (2003) 107, 14521

14. Representations of Atoms  QM atoms are treated by SCC-DFTB  MM atoms by CHARMM forcefield  QM/MM boundary treated with generalized hybrid orbital (GHO) method or link atom method  Stochastic boundary or general solvent boundary

15. Stochastic Boundary Conditions Reaction center average of the coordinates of atoms treated by QM Reaction zone upto 24 Å Buffer zone 24 - 30 Å Reservoir zone beyond 30 Å 30 Å water sphere added around the active site center Deleting all atoms beyond 45 Å Reaction zone : Newtonian Mechanics Buffer zone: Langevin’s equation of motion Friction coefficient and a harmonic restoring force with a gradiant: Scaled to 0 at reaction zone boundary Reservoir zone provides a static forcefield 45 Å 30 Å

16. Which Route ? FAD + 2e + 2H + FADH 2 FAD FADˉ FAD 2 ˉ FADH FADHˉ ee H+H+ H+H+ e FADH 2 H+H+ Model Reactions a,b B3LYP/ 6-31+G(d,p)  H (kcal/mol) SCC-DFTB  H (kcal/mol) AM1(Gaussian)  H (kcal/mol) FAD + e¯  FAD¯-44.3-37.3-63.5 FAD¯ + e¯  FAD 2- +53.4+58.248.3 FAD + 2e¯  FAD 2- +9.1+20.9-15.1 FAD 2- + H +  FADH¯-436.5-442.3-420.0 FAD¯ + H +  FADH-333.7 -314.3 FADH + e¯  FADH¯-49.4-50.4-57.2 FAD¯ + e¯ + H +  FADH -383.1-384.1-371.7 FADH¯+ H +  FADH 2 -331.7-331.4-320.2 FAD + 2e¯ + 2H +  FADH 2 -759.1-752.8-755.3

17. Calculations using Stochastic Boundary Condition FADˉ + e  FAD 2- FAD + e  FADˉ ΔG 1 Rd/Ox (FEP) = -79.64 kcal/mol ΔG (Born Correction) =-5.46 kcal/mol ΔG 1 Rd/Ox = -85.1 kcal/mol ΔG 1 Rd/Ox (FEP) = -69.7 kcal/mol ΔG (Born Correction) =-16.4 kcal/mol ΔG 2 Rd/Ox = -86.1 kcal/mol

18. Calculated Reduction Potential Enzyme mid-point reduction potential for MCAD-bound FAD Em Em = -145 mV  G = -197.9 kcal/mol -85 -86 -56 -30 Estimated ΔG total = -256 kcal/mol; Overestimated energy ~ 60 kcal/mol Possible reason for this overestimation  Long-range electrostatic interactions FAD + 2e + 2H + FADH 2 (E-FAD 2- /E-FADH - )(E-FADH - /E-FADH 2 ) ΔG total = ΔG 1 Rd/Ox + ΔG 2 Rd/Ox +  G +  G

19. Test Calculations  Stochastic boundary set up with net charge of the complex set to zero  General solvent boundary condition  Matching with experimental results

20. Treating Solvation with General Solvent Boundary Potential (GSBP)  Inner region atoms (ligand + part of enzyme + solvent) treated explicitly  Outer region: Atoms of enzyme are treated explicitly but the solvent is represented as a continuous dielectric medium Im, W. et al. J. Chem. Phys. 2001, 114, 2924

21. Setup of Zones in GSBP Reaction center  Inner region atoms (ligand + part of enzyme + solvent) treated explicitly (< 16 Å)  Water atoms deleted beyond 18 Å  Charges of residues beyond 20 Å set to 0  Protein atom fixed beyond 20 Å  Outer region: Atoms of enzyme are treated explicitly but the solvent is represented as a continuous dielectric medium Inner zone upto 16 Å Buffer zone upto 18 Å Secondary buffer (18-20) Å Reservoir zone > 20 Å Protein atoms which have 1-3 connections with reservoir zone are kept fixed

22. Compared Values of Free Energy Changes Reaction Stochastic boundary with neutralized charge  G (kcal/mol) General solvent boundary potential  G (kcal/mol) E-FAD  E-FADˉ · -57.46-64.1 E-FADˉ ·  E-FAD 2- -53.07-46.4 E-FAD 2-  E-FADHˉ-47.9-53

23. FEP for Electron Additions E-FAD  E-FADˉ· E-FADˉ·  E-FAD 2- ΔG 1 Rd/Ox (FEP) = -51.0 kcal/mol ΔG (Born Correction) =-5.46 kcal/mol ΔG 1 Rd/Ox = -57.46 kcal/mol ΔG 2 Rd/Ox (FEP) = -36.67 kcal/mol ΔG (Born Correction) =-16.4 kcal/mol ΔG 2 Rd/Ox = -53.07 kcal/mol

24. Calculations for Proton Additions E-FAD 2- + H +  E-FADHˉ E-FADH - + H +  E-FADH 2 ΔG 2 (FEP) = -164.1kcal/mol E(H + ) a, self interaction energy of H-atom = -141.9 kcal/mol ΔG b ( H + solvation) = 262.4 kcal/mol ΔG (Born Correction) =5.46 kcal/mol ΔG 2 (total) = -38.15 kcal/mol (E-FADH -- /E-FADH 2 ) ΔG 1 (FEP) = -184.8 kcal/mol E(H + ) a, self interaction energy of H-atom = -141.9 kcal/mol ΔG b ( H + solvation) = 262.4 kcal/mol ΔG (Born Correction) =16.4 kcal/mol ΔG 1 (total) = -47.9 kcal/mol (E-FAD 2- /E-FADH - ) a Zhou, H. et al. Chemical Physics (2002) 277, 91 b Zhan, C. et al. P. Phys. Chem. A (2001) 105, 11534

25. FEP Electron Addition FAD FADˉ FAD 2 ˉ FADH FADHˉ ee H+H+ H+H+ e FADH 2 H+H+ √√ √ √ ? E-FADH·  E-FADHˉ ΔG 1 Rd/Ox (FEP) = -55.47 kcal/mol ΔG (Born Correction) =-5.46 kcal/mol ΔG 1 Rd/Ox = -60.93 kcal/mol √

26. Computed Free Energy Changes Reaction Stochastic Boundary with 0 charge  G (kcal/mol) E-FAD + e  E-FAD¯·-57.46 E-FAD¯· + H +  E-FADH · -39.97 E-FAD¯· + e  E-FAD 2- -53.07 E-FAD 2- + H +  E-FADH¯-47.9 E-FADH¯ + H +  E-FADH 2 -38.15 Enz-FADH·  E-FADH 2 -60.93

27. Summary  G expt H+H+ Enz-FAD Enz-FADH Enz-FAD ¯ Enz-FAD 2- Enz-FADH ¯ H+H+ ee -57.46 kcal/mol -53.07 kcal/mol -47.9 kcal/mol e -60.93 kcal/mol -98.6 kcal/mol - 39.97 kcal/mol H+H+ Enz-FADH 2 -197.9 kcal/mol -38.15 kcal/mol  G expt  G comput = -97.4 kcal/mol -7.0 kcal/mol Overestimation ~ 6 kcal/mol  G comput = -196.6 kcal/mol -6.3 kcal/mol Overestimation ~ 6 kcal/mol

28. Conclusions  The two-electron two-proton reduction potential of FAD in MCAD was calculated  Both reduction potential calculations yield results that are consistent to the experimental values  Computed 2e/2H + reduction potential for FAD bound MCAD is -180 mV, which is 35 mV more negative than the experimental value  The first reduction potential of MCAD-bound FAD was calculated to be ~ -103 kcal/mol compared to the experimentally observed value of -98.6 kcal/mol  Computed second reduction potential for MCAD-bound FAD was ~ -105 kcal/mol, about 6 kcal/mol more negative than the experimentally observed value of -99.3 kcal/mol  This calculation shows that at neutral pH MCAD-bound FAD will be converted to the hydroquinone form FADH 2 through a two electrons/two protons reduction  pKa calculation of E-FADH ¯ and E-FADH 2 show that both values are quite high: 35 and 27, respectively. The pKa of E-FADH was calculated to be ~30 Future Directions  Using the method to compute the reduction potential and pKa of FAD and FMN in aqueous solution  pka calculation of acyl-CoA substrate bound to MCAD

29. Acknowledgements $ NIH Professor Jiali Gao Professor Don G. Truhlar Dr. Kowangho Nam Dr. Alessandro Cembran Dr. Marian Stankovich Dr. Qiang Cui Dr. Haibo Yu Minnesota Supercomputing Institute

30. FAD + 2e + 2H + FADH 2  Total =  S1 +  S2 +  G S (FAD 2- /FADH 2 ) E m = -(  Total )/nF +  NHE ; n = number of electrons = 2 F= Faraday constant = 23.06 kcal/mol  Total - nF  NHE = -nFE m, F = 23.06 kcal/(mol.V) nF  NHE = -n4.43 F V = -2x 4.43 x 23.06 kcal/mol = -204.31 kcal/mol For FAD-MCAD: E 0 = -0.145V Thus  Total - nF  NHE = -nF x (-0.145) = -2 x 23.06 x (-0.145) kcal/mol = 6.687 kcal/mol Thus  Total = (6.687 + nF  NHE ) kcal/mol = (6.687 -204.31) kcal/mol = -197.6 kcal/mol Converting FAD Reduction Potential For FAD bound MCAD: Mid point potential, E 0 = -0.145V Lenn, N. D. Stankovich, M. T. Liu, H. Biochemistry, 1990, 29, 3709

31. Calculating Absolute Reduction Potential Standard Hydrogen Electrode (Normal Hydrogen Electrode) = Free energy change, E H 0 in the reaction H + (s) + e ¯  ½H 2 (g) E H 0 For a general reduction process: M (s) + e ¯  M¯(s) E M  M¯ Then combining the above 2 equations ½H 2 (g) + M(s)  M¯(s) + H + (s) E 0 => E M  M¯ = (E 0 + E H 0 ) E 0 H = -4.43 eV

32. Irregular Solvent-Protein Interfaces R inner R exact ΔR diel = fixed atoms Im, W. et al. J. Chem. Phys. 2001, 114, 2924