Transcript Calculation of Reduction potential of FAD in MCAD

Calculation of Reduction Potential
of FAD in MCAD using Combined
DFTB/MM Simulations
Sudeep Bhattacharyay, Marian Stankovich,
and Jiali Gao
Overview

Objective

Methods of calculation and strategies

Results

Future Directions
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
Acyl-CoA Dehydrogenases (CAD) in
Electron Transfer
FAD is reduced
FAD is reoxidized
Medium chain acyl-CoA
Dehydrogenase (MCAD)
Ghisla, S. et al. Eur J Biochem, 2004. 271, 494-508.
MCAD: Structural Information
N
All -domain
Extends into
the other dimer
Two orthogonal
-sheets
All -domain
C
 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
A Tale of Two Quasi-independent
Processes
Scheme 1
R256Q
R
H
CH3
N
O
N
transient intermediate
T168A
wt-
N
N
N
CH3
H
H
O
H
H
O
C
C
C5H11
C
CoA
S
H
 Both proton and hydride transfer CONTRIBUTE TO
THE OVERALL CATALYTIC RATE in wild-type
H
(-)
O
 Potentials of mean force computation of MCAD
demonstrate a STEPWISE mechanism
O
Glu376
-proton abstraction
by the catalytic base
Glu376
-hydride transfer on
to the flavin nitrogen
 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
Reduction of Flavin
blue (neutral)
semiquinone
red (anionic)
semiquinone
FAD + 2e + 2H+  FADH2
Experimental Mid-point Potentials Values
MCAD-bound FAD
yellow
-98.6 kcal/mol
-197.6 kcal/mol
blue (neutral)
semiquinone
red (anionic)
semiquinone
-99.3 kcal/mol
Gustafson et al.
(1986) J. Biol. Chem. 261, 7733-7741
Mancini-Samuelson et al.
(1998) Biochemistry, 37, 14605-14612
Methods and Strategies
 Hybrid QM/MM methods; calculation of
electron and proton affinities
 Thermodynamic integration through FEP
 Dual topology single coordinate method
 Boundary condition
O
H
Electron and proton affinities
Model Reactionsa,b
O
N
N
N
CH3
N
CH3
CH3
B3LYP/
6-31+G(d,p)
H (kcal/mol)
SCC-DFTB
AM1(Gaussian)
H (kcal/mol)
H (kcal/mol)
FAD + e¯  FAD¯
-44.3
-37.3
-63.5
FAD¯ + e¯  FAD2-
+53.4
+58.2
48.3
FAD + 2e¯  FAD2-
+9.1
+20.9
-15.1
FAD2- + 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
-333.7
-314.3
FADH¯+ H+  FADH2
-331.7
-331.4
-320.2
FAD + 2e¯ + 2H+  FADH2
-759.1
-752.8
-755.3
Free Energy Perturbation
State A
Potential energy of a hybrid system: Uhybrid = (1-λ)UA + λUB
λ is a coupling parameter varied from 0-1 (0.1, 0.2, ….)
Thermodynamic integration method
∫
1
Free energy change ΔG = (∂G(λ)/ ∂λ) dλ =
0
1
∂U(λ)/ ∂λ)dλ
∫
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 B
QM/MM Interactions
With same number of atoms in the two chemical states
Utot (R) = │ĤQM+ĤelQM/MM │ + UvanQM/MM (R) + UbondedQM/MM (R) + UMM (R)
electronic energy
of the QM system
+ the electrostatic
interaction energy
van der Waals
bonded
energy of the
MM atoms
Li, G. et al. J. Phys. Chem. B (2003) 107, 8643
G
(R;λ) =
λ
UA/MM(RQM ,RMM ) + UB/MM(RQM ,RMM )
Only the electrostatic term contributes to the free energy
derivative as the two states have same cartesian coordinate
Thermodynamic Schemes
FEP for reduction potential calculation
ΔGRd/Ox
E-FAD (ox)
E-FAD(red)
ΔGRd/Ox is obtained from a single FEP calculation
FEP for pKa calculation
AH.E(aq)
G(1)
[A¯
G
E
AH/A¯
GH+solv
-D].E(aq)
G E = G(1) + G(2) + Gsolv
+
G(2)
Aˉ .E(aq) + D(g)
Aˉ .E(aq) + H+ (aq)
AH/A¯
G = 0.0
H
Aˉ.E(aq) + H+ (g)
Li, G. et al. J. Phys. Chem. B (2003) 107, 14521
Representations of Atoms
O4
H
4
H
4a
N3
N5 5a
2
CH3
6
7
9 8
CH3
N1 10a N10 9a
O2
H
HO
2'
 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
B 1'
3'
OH
4'
HO
5'
O5'
(-)
O
P1
O
O
O
P2
H
HO
HO
3*
2*
(-)
O
O5*
4*
5*
H
O
8
H
H
1*
N9
N7
4
5
6
N3
2
N1
NH2
Stochastic Boundary Conditions
Reaction zone
upto 24 Å
Buffer zone
24 - 30 Å
Reservoir zone
beyond 30 Å
30 Å
Reaction center
average of the
coordinates of
atoms treated
by QM
45 Å
•
Reaction zone : Newtonian Mechanics
• Buffer zone: Langevin’s equation of
• 30 Å water sphere added
•
around the active site center
Deleting all atoms beyond 45 Å
motion
Friction coefficient and a harmonic
restoring force with a gradiant: Scaled
to 0 at reaction zone boundary
• Reservoir zone provides a static
forcefield
Which Route ?
FAD + 2e +
2H+
Model Reactionsa,b
FADH2 FAD
e
B3LYP/
6-31+G(d,p)
H
(kcal/mol)
SCC-DFTB
AM1(Gaussian)
H
(kcal/mol)
H (kcal/mol)
FAD + e¯  FAD¯
-44.3
-37.3
-63.5
FAD¯ + e¯ 
FAD2-
+53.4
+58.2
48.3
FAD + 2e¯  FAD2-
+9.1
+20.9
-15.1
FAD2- + H+  FADH¯
-436.5
-442.3
-420.0
FAD¯ + H+  FADH
-333.7
-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+  FADH2
-331.7
-331.4
-320.2
FAD + 2e¯ + 2H+ 
FADH2
-759.1
-752.8
-755.3
FADˉ
•
e
FAD2ˉ
H+
H+
FADH•
e
FADHˉ
H+
FADH2
Calculations using Stochastic Boundary
Condition
FAD + e  FADˉ•
ΔG1Rd/Ox (FEP) = -79.64 kcal/mol
ΔG (Born Correction) =-5.46 kcal/mol
ΔG1Rd/Ox = -85.1 kcal/mol
FADˉ• + e  FAD2-
ΔG1Rd/Ox (FEP) = -69.7 kcal/mol
ΔG (Born Correction) =-16.4 kcal/mol
ΔG2Rd/Ox = -86.1 kcal/mol
Calculated Reduction Potential
Enzyme mid-point reduction potential for MCAD-bound FAD
FAD + 2e + 2H+
FADH2
Em = -145 mV G = -197.9 kcal/mol
ΔGtotal = ΔG1
-85
Rd/Ox
+ ΔG2
-86
Rd/Ox
(E-FAD2- /E-FADH-)
+ G
-56
+
G
(E-FADH- /E-FADH2)
-30
Estimated ΔGtotal = -256 kcal/mol; Overestimated energy ~ 60 kcal/mol
Possible reason for this overestimation

Long-range electrostatic interactions
Test Calculations
 Stochastic boundary set up with net charge of the complex set
to zero
 General solvent boundary condition
 Matching with experimental results
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
Setup of Zones in GSBP
Inner zone upto 16 Å
Buffer zone upto 18 Å
Secondary buffer (18-20) Å
Reaction center
Protein atoms which have 1-3 connections
with reservoir zone are kept fixed
Reservoir zone > 20 Å
 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
Compared Values of Free Energy Changes
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-FAD2-
-53.07
-46.4
E-FAD2-  E-FADHˉ
-47.9
-53
Reaction
 dU/dl (kcal/mol)
FEP for Electron Additions
 dU/dl (kcal/mol)
0
-20
-40
-60
-80
-100
-120
0
0.2
0.4
0.6
0.8
1
l
E-FAD  E-FADˉ·
ΔG1Rd/Ox (FEP) = -51.0 kcal/mol
ΔG (Born Correction) =-5.46 kcal/mol
ΔG1Rd/Ox = -57.46 kcal/mol
20
0
-20
-40
-60
-80
-100
-120
0
0.2
0.4
0.6
0.8
1
l
E-FADˉ·  E-FAD2ΔG2Rd/Ox (FEP) = -36.67 kcal/mol
ΔG (Born Correction) =-16.4 kcal/mol
ΔG2Rd/Ox = -53.07 kcal/mol
 dU/dl (kcal/mol)
du/dl (kcal/mol)
Calculations for Proton Additions
260
220
180
140
100
0
0.2
0.4
0.6
0.8
1
260
220
180
140
100
0
0.2
l
l
0.6
0.8
1
E-FADH- + H+  E-FADH2
E-FAD2- + H+  E-FADHˉ
(E-FAD2- /E-FADH-)
0.4
(E-FADH-- /E-FADH2)
ΔG1
(FEP) = -184.8 kcal/mol
E(H+) a, self interaction energy of H-atom
= -141.9 kcal/mol
b
+
ΔG ( H solvation) = 262.4 kcal/mol
ΔG (Born Correction) =16.4 kcal/mol
ΔG1 (total) = -47.9 kcal/mol
ΔG2
(FEP) = -164.1kcal/mol
E(H+) a, self interaction energy of H-atom
= -141.9 kcal/mol
b
+
ΔG ( H solvation) = 262.4 kcal/mol
ΔG (Born Correction) =5.46 kcal/mol
ΔG2 (total) = -38.15 kcal/mol
a
b
Zhou, H. et al. Chemical Physics (2002) 277, 91
Zhan, C. et al. P. Phys. Chem. A (2001) 105, 11534
e
FAD
√
FADˉ •
e
√
H+
FAD2ˉ
H+
√
 dU/dl (kcal/mol)
FEP Electron Addition
0
-20
-40
-60
-80
-100
-120
FADH•
e
H+
FADH2
0.2
0.4
0.6
0.8
1
l
FADHˉ
?
√
0
√
E-FADH·  E-FADHˉ
ΔG1Rd/Ox (FEP) = -55.47 kcal/mol
ΔG (Born Correction) =-5.46 kcal/mol
ΔG1Rd/Ox = -60.93 kcal/mol
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-FAD2-
-53.07
E-FAD2- + H+  E-FADH¯
-47.9
E-FADH¯ + H+  E-FADH2
-38.15
Enz-FADH·  E-FADH2
-60.93
Gcomput
= -196.6 kcal/mol
-6.3 kcal/mol
Summary
Gexpt
-57.46 kcal/mol
Enz-FAD
Gexpt
e
-98.6 kcal/mol
Gcomput
= -97.4 kcal/mol
-7.0 kcal/mol
Overestimation
~ 6 kcal/mol
Overestimation
~ 6 kcal/mol
-197.9 kcal/mol
-53.07 kcal/mol
Enz-FAD¯•
e
Enz-FAD2-
H+ - 39.97 kcal/mol
Enz-FADH•
e
H+
-47.9 kcal/mol
Enz-FADH¯
-60.93 kcal/mol
H+
Enz-FADH2
-38.15 kcal/mol
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 FADH2 through a two electrons/two protons reduction
pKa calculation of E-FADH¯ and E-FADH2 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
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
Converting FAD Reduction Potential
FAD + 2e + 2H+
FADH2
(FAD2- /FADH2)
Total = S1 + S2 + GS
Em = -(Total)/nF + NHE ; n = number of electrons = 2
F= Faraday constant = 23.06 kcal/mol
Total - nFNHE = -nFEm , F = 23.06 kcal/(mol.V)
For FAD bound MCAD:
nFNHE = -n4.43 F V = -2x 4.43 x 23.06 kcal/mol = -204.31 kcal/mol
Mid point
potential, E0 = -0.145V
For FAD-MCAD:
E0 = -0.145V
Lenn, N. D. Stankovich, M. T. Liu, H. Biochemistry, 1990, 29, 3709
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
Calculating Absolute Reduction Potential
Standard Hydrogen Electrode (Normal Hydrogen Electrode) =
Free energy change, EH0 in the reaction
H+ (s) + e¯  ½H2 (g)
EH0
For a general reduction process:
M (s) + e¯  M¯(s)
EMM¯
Then combining the above 2 equations
½H2 (g) + M(s)  M¯(s) + H+ (s)
=> EMM¯ = (E0 + EH0)
E0
E0H = -4.43 eV
Irregular Solvent-Protein Interfaces
Rinner
Rexact
ΔRdiel
= fixed atoms
Im, W. et al. J. Chem. Phys. 2001, 114, 2924