free energy calculations and the potential of mean force · free energy calculations and the...
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Free energy calculations and the potential of mean force
Mark TuckermanDept. of Chemistry
and Courant Institute of Mathematical Science100 Washington Square East
New York University, New York, NY 10003
IMA Workshop on Classical and Quantum Approaches in Molecular Modeling
Free Energy
( , ) ( , , )3 ( )
1( , , ) = !
N N H A N V TN D V
Q N V T d d e eN h
β β− −= ∫ ∫ p rp r
Canonical Ensemble (Helmholtz free energy):
( ( , ) ) ( , , )3 0 ( )
0
1( , , ) = !
N N H PV G N P TN D V
N P T dV d d e eV N h
β β∞ − + −∆ = ∫ ∫ ∫ p rp r
Isothermal-Isobaric Ensemble (Gibbs free energy):
( , , ) ln ( , , )A N V T kT Q N V T= −
( , , ) ln ( , , )G N P T kT N P T= − ∆
Free Energy (cont’d)
P
VT
1
2
State function:
21 2 1A A A∆ = −
Free energy and work
• If an amount of work W is required to change the thermodynamic state of the system from 1 to 2, then
• Equality holds when the work is performed infinitely slowly or reversibly.
• Jarzynski’s equality [PRL, 78 2690, (1997)] shows how to relate irreversible work to the free energy difference. Let W21(x) be a microscopic function whose ensemble average is the thermodynamic work W21.
21 21W A≥ ∆
21 21
1
Ae eβ β− − ∆=W
1 2q d d= −
d1 d2
Free energy profiles
Ak e βκ −= ‡
A‡
Protein Folding EnergeticsFrom G. Bussi, et al. JACS 128, 13435 (2006)
1
2
( , ) ( ) ( )( , ) ( ) ( ) ( , )
E I E E I I
E I E E I I EI E I
U U UU U U U
= += + +
r r r rr r r r r r
1 2( , , ) ( ) ( , ) ( ) ( , )E I E I E IU f U g Uλ λ λ= +r r r r r r(0) 1 (1) 0(0) 0 (1) 1
f fg g
= == =
[ ][ ][ ]
bindGi
E IK e
EIβ− ∆= =
1
bind 0
UG dλ
λλ
∂∆ =
∂∫
Binding Free Energies
Inhibition constant:
Thermodynamic state potentials:
Meta-potential:
Thermodynamic integration (Kirkwood, 1935)
Binding free energies: Thermodynamic perturbation
( , )3 3( )
( ) 2
( )
1 ( , , )( , , ) ! !
( , , ) / 2
N N HN ND V
N U
D V
Z N V TQ N V T d d eN h N
Z N V T d e h m
β
β
λ
λ β π
−
−
= =
= =
∫ ∫
∫
p r
r
p r
r
2 221
1 1
ln lnQ ZA kT kTQ Z
∆ = − = −
2 1 2 1
2 1
( ) ( ) ( ( ) ( ))2
1 1 1
( )
1
1 1
U U U U
U U
Z d e d e eZ Z Z
e
β β β
β
− − − −
− −
= =
=
∫ ∫r r r rr r
Free energy difference related to partition function ratio:
Perturbation formula:
Need sufficient overlap between two ensembles
λ dynamics methods
Use molecular dynamics to sample λ via a Hamiltonian:2 2
1
2 2
1 1 2 1
( ,..., , )2 2
( ) ( ,..., ) ( ) ( ,..., )2 2
iN
i i
iN N
i i
pH Um m
p f U g Um m
λλ
λ
λ
λ
λ
λ λ
= + +
= + + +
∑
∑
p r r
p r r r r
Free energy from probability distribution of λ:
( , )( ) UP d e β λλ −= ∫ rr
21
( ) ln ( )(1) (0)
A kT PA A Aλ λ= −
∆ = −
Need to have best sampling at the endpoints of the λ-path, which arenormally the most difficult to sample.
λ dynamics methods
( )A λ
0λ = 1λ =
Aim for a profile with a barrier:
In order to generate such a profile, we need:
1. A high temperature Tλ >> T to ensure barrier crossing2. An adiabatic decoupling between λ and other degrees of freedom3. Choose mλ >> mi.
λ dynamics methodsUnder adiabatic conditions, we generate a free energy profile at Tλ
( ; ) ( ; ) ( , ) A A Ue e d eλλ
λ
βββ λ β β λ β β λ ββ− − − = = ∫ rr
Free energy profile at temperature Tfrom probability distribution generated under adiabatic conditions:
adb( ; ) ln ( ; , )A kT Pλ λλ β λ β β= −
Chemical Potential of Lennard-Jones Argon
( ) 24 ]1[ −= λλf 24 ]1)1[()( −−= λλg
2000 200m m T Tλ λ= =
TI
[ ]bins
exact1bins
1( ) ( ; ) ( )N
i ii
t P x t P xN
ς=
= −∑
HO
CH
H
0.145
0.06
0.06-0.683
0.418
Backbone
HO
C
H
HH
0.085
0.06
0.06-0.683
0.418
(Serine) (Methanol)
0.06
H
CH
H
-0.27
0.09
0.090.09
Backbone
H
C
H
HH
-0.36
0.09
0.090.09
(Alanine) (Methane)
0.09
Solvation free energies of amino acid side-chain analogs
1 Solute (CHARMm22 Parameters)
• 256 TIP3P Water molecules• Cubic Simulation Box (L = 19.066 A)• Periodic Boundary Conditions• Ewald Summation Technique for charges• System Temperature: 298 K• NVT via GGMT Thermostats (Liu,MET 2000)
λ-AFED Parameters:• kTλ = 12,000 K = 50 kT• mλ = 16,000 g.mol-1
a. Wolfenden, et al. Biochem. 20, 849 (1981)b. Shirts, et al. JCP 119, 5740 (2003); JCP 122, 134508 (2005)c. Yin and Mackerell J. Comp. Chem. 19, 334 (1998)
The Free Energy Profile
( , )1
1( ) ( ( ,..., ) ) N N HNP q d d e q q
Qβ δ−= −∫ p rp r r r
Probability distribution function:
Free energy profile in q:
( ) ln ( )F q kT P q= −
Bluemoon Ensemble Approach
1( ,..., )Nq q=r r
1 1( ,..., , ,..., ) 0iN N i
i ii i i
q qqm
∂ ∂= = =
∂ ∂∑ ∑ pr r p p rr ri i
Impose a constraint of the form:
However, constraints also require:
But in constrained MD, what we are actually computing is:
( , )1 1 1
1( ) ( ( ,..., ) ) ( ( ,..., , ,..., ))N N HN N NP q d d e q q q
Qβ δ δ−= −∫ p rp r r r r r p p
E. A. Carter, et al.Chem. Phys. Lett. 156. 472 (1989); M. Sprik and G. Ciccotti, J. Chem. Phys. 109, 7737 (1998).
Bluemoon Ensemble
[ ]1/ 211
1/ 211
Z kTdFdq Z
λ−
−
− + ℑ=
**When using SHAKE/RATTLE, λ must be the SHAKE multiplier!! M. Sprik and G. Ciccotti, J. Chem. Phys. 109, 7737 (1998).
Using the Lagrange multiplier to compute free energy:
2
2,
1 1 1 i i ji i i i j i i j j
q q q q qzm z m m
∂ ∂ ∂ ∂ ∂= ℑ =
∂ ∂ ∂ ∂ ∂ ∂∑ ∑r r r r r ri i i
[ ]1/ 2
constr1/ 2
constr
z kTdFdq z
λ−
−
− + ℑ=
1 2d dδ = −
d1 d2
Free energy profiles
‡
A‡
From D. Marx and MET, PRL 86, 4946 (2001).
Variable transformations and statistical mechanics
Adiabatic dynamics and free energy profiles
Hamiltonian from transformation:
2 23
1 31 1
( ,..., )2 2
n Ni i
Ni i ni i
p pH U q qm m= = +
= + +∑ ∑
Adiabatic conditions:
k k
q
m mT T
L. Rosso, P. Minary, Z. Zhu and MET, J. Chem. Phys. 116, 4389 (2000); Maragliano, Vanden Eijnden CPL 426, 168 (2006)
Free energy surface:
1 1( ,..., ) ln ( ,..., )n q nA q q kT P q q= −
Conformational sampling of the solvated alanine dipeptide
ψφ
AFED Tφ,ψ = 5T, Mφ,ψ = 50MC 4.7 nsUmbrella Sampling 50 ns
CHARMm22αR
β
[L Rosso, J. B. Abrams and MET J. Phys. Chem. B 109, 2099 (2005)]
C7ax αL
7.64.50.00.2
4.43.91.410.0
7.44.80.01.0Meta1
US2
AFED
Time FF β αR C7ax αL
5ns
5ns
400ns
CHm27
CHm22
CHm22
1. Ensing, et al. ACR 39, 73 (2005)2. Smith, JCP 111, 5568 (1999)
NH
NH O
CH3
N
O
H CH3
φ
ψ
Why NATMA?• Small compared to actual proteins and can be easily studied
• Tryptophan side-chain gives rise to a free energy landscape endemic of actual proteins
• Experimental1 and DFT1 Minimization data available
1. Dian, B.C., et al. Science, 296, 2369 (2002); J. Chem. Phys. 117, 10688. (2002)
NATMA (gas phase)(N-acetyl-Tryptophan-methyl-amide)
C7eq
C5(AP)
C5(AΦ)
kT(φ,ψ)=20kT, m(φ,ψ)=600mC t=2.5 ns
AFED Minima Predictions:
Conf. Energy Location (φ,ψ)C5(AP) 0.0 kcal/mol (-160, 160)C5(AΦ) +1.03 kcal/mol (-140, 140)C7eq +2.18 kcal/mol (-100, 100)
Ab initio energies1
Conf. Rel. EnergyC5 (AP) 0.0 kcal/molC5(AΦ) +0.65 kcal/molC7eq +2.28 kcal/mol
φ ψ
F(φ,ψ)
Metadynamics
Add bias potential to Hamiltonian:
A. Laio and M. Parrinello, PNAS 99, 12562 (2002); A. Laio, et al. JPCB 109, 6714 (2005)
2
1 1( ,..., ) ( ,..., , )2
iN G N
i i
H U U tm
= + +∑ p r r r r
( )2
1 2,2 ,... 1
( ) ( ( )( ,..., , ) exp
2G G
nk k G
G Nt k
q q tU t W
τ τ= =
−= −
∆ ∑ ∑
r rr r
Free energy is negative of bias potential:
( )2
1 2,2 ,... 1
( ( )( ,..., ) exp
2G G
nk k G
nt k
q q tA q q W
τ τ= =
−= − −
∆ ∑ ∑
r
REPSWA (Reference Potential Spatial Warping Algorithm)
No Transformation Transformation
5kT
10kT
10kT
V‡
How it works
V‡=kT
V‡=5kT
V‡=10kT
Forces:
( )
ref
ref
( )( )
( ) ( )
V V xF ux u
xF x F xu
∂ − ∂= −
∂ ∂∂
= −∂
/x u∂ ∂ becomes large in the barrierregion!
Barrier Crossing Transformations (cont’d)
‘
‘
‘
‘
ref ( )V φ
ref tors 1 inter 4 5 2 4 5( ,{ }) ( ) ( ) (| ( ,{ }) |) (| ( ,{ }) |)V V S V Sφ φ φ α φ φ= + − −r r r r r r r
P. Minary, G. J. Martyna and MET SIAM J. Sci. Comp. (accepted)
Comparison for 50-mer using TraPPE with all interactionsPT replicas = 10; PT exchange prob. = 5%, REPSWA α = 0.8; Every 10th dihedral not transformed
Comparison to parallel tempering and CBMCSiepmann and Frenkel, Mol. Phys. 75, 59 (1992)
End-to-end distance fluctuations
Comparison for 50-mer using CHARMM22 all interactions
Comparison of 50-mer using CHARMM22 all interactions
Honeycutt and Thirumalai
Model sheet protein βNo TransformationParallel TemperingSDC-REPSWA
PT replicas = 16; PT exchange prob. = 5%