Free energy calculations
Molecular simulations can be carried out to gain either dynamical or
energetical information on the simulated system. Today we'll focus on
different ways to extract energies and free energies from simulations.
Solvation free energy of a sodium ion in water
One way to gain access to free energies by simulation is to perform
so-called free-energy perturbation (FEP) simulations. In such
simulations, the system is gradually "morphed" from one state to
another (let's say from state "A" to state "B"), during which the
effect of this perturbation onto the free energy is monitored. In
practice, this is achieved by defining a morph parameter lambda, which
is defined to be zero in the initial state (A), and one in the final
state (B). By collecting the derivative of the potential energy with
respect to lambda during the simulation, and by integrating over
lambda afterwards, we gain access to the free energy. We are now going
to use this procedure to calculate the free energy of solvation for a
sodium ion in water. For this, we'll slowly create a sodium ion in a
water box. In order to start the simulation, download the
topology, and the MD parameter file first.
First have a look at the starting structure:
In rasmol, type:
Note that the sodium ion is already "present" in the starting
structure. Now have a look at the topology "na.top", with 'xedit',
'nedit', 'more' or 'less', and look for the line containing "Na" under
"[atoms]". You'll notice that two states (A and B) are defined, with
charge 0 and 1, respectively. This means that during the simulation,
the charge of the sodium ion will be slowly switched on.
Now start the simulation:
grompp -f fep -c start.pdb -p na
mdrun -v -c charged.gro
Analyse the free energy change by:
In order to compute the free energy, we must integrate dG/dlambda over
dlambda. Instead, we have dG/dlambda as a function of time. This is
not a problem, as we know that lambda changed from0 to 1 during this
time (10 ps). So instead, we can integrate the curve over time, and
divide the obtained value by ten to derive the desired free energy.
To integrate, under "Data", select "Transformations", followed by
"Integration", and press "Accept".
The experimentally observed solvation free energies for sodium range
from -365 to -372 kJ/mol. How does the obtained value correspond to
Instead of integrating dG/dlambda, we can also compare the initial and
final values of the potential energy of the system:
Select "Potential", followed by "0" (and, if necessary, an extra "enter").
Subtract the initial value from the final value.
What value do you get? Why is this value so much larger (in absolute
value) than the free energy difference obtained by integrating
One important check in simulations in general and in free energy
simulations in particular is to make sure that the answer has
converged. In order to do so, we can either perform a longer (or
shorter!) simulation, and compare the result to the original, or we
can perform the backward transition to see if the free energy
difference is the opposite of the forward transition. In this
practical, we'll do both. First, we'll carry out the backward
transition. Copy the MD parameter file to incorporate the changes:
cp fep.mdp back.mdp
and open "back.mdp" in your favorite editor (xedit, nedit, emacs,
to make the following changes:
First search for "Free energy", then change "init-lambda" to 1 and
change delta-lambda to "-0.0002". Now we can start the backward
simulation. For this, we'll use the final structure of the previous
simulation (charged.gro) as initial structure:
grompp -f back -c charged.gro -p na
And analyse the free energy change using xmgrace. Remember that we now
changed lambda from one to zero, meaning that if we integrate over
time, we should divide the obtained value by -10 instead of by 10.
What value do you get? How does this value compare to the free energy
change for the forward simulation? Would you consider the obtained
value for the solvation free energy sufficiently converged?
If we take the forward and backward
simulation together, is the total free energy change positive or
negative? Which of the two should it have been?
As stated above, another way to check convergence is to perform a
longer simulation and see if the free energy change remains the same
as compared to a shorter simulation. For this, open "fep.mdp" in your
favorite editor. Change "nsteps" from 5000 to 50000, and
"delta-lambda" from 0.0002 to 0.00002, and repeat the forward
grompp -f fep -c start.pdb -p na
Is the obtained free energy very different as compared to the shorter
simulations? What would you consider an appropriate length for this
type of simulation?
Thermodynamics of reversible peptide folding
For the case of a sodium ion in water we could compute the free energy
difference between two states because we defined a path connecting the
two states, and we had access to the free energy gradient along the
path. In the case of the ion in water, the path was an artificial path
(hence such free energy perturbation studies are also called
"computational alchemy"). Likewise, a free energy difference can also
be computed along a real path, e.g. by forcing an ion across an
electrostatic gradient and integrating the force required to displace
the ion along the path.
There are also cases, however, where such a path, artificial or real,
can not be easily defined. Peptide or protein folding is a good
example. The problem of defining a path is that there are multiple
folding (and unfolding) pathways that connect the (well-defined)
folded state and the degenerate unfolded state. If we would like to
compute the folding free energy
by integrating the force along a pathway connecting the two, we'd
therefore have to follow multiple pathways, and take a weighted
average of the results. In the case, however, where we have a
simulation in which folding and unfolding takes place spontaneously
(i.e. without forcing the system to fold or unfold) and reversibly
we can also access the free energy difference between the folded and
unfolded state in a different way, namely simply by monitoring the
fraction of time the system resides in each state. In the equilibrium
case (i.e. if the simulation was long enough to visit all relevant
configurations multiple times) the probability of finding the system
is exactly proportional to its Boltzmann factor. Therefore, for any
two states, the following equation holds:
with DeltaG = G1 - G2
Therefore, if we know the respective probability to find the system in
a particular pair of states, we can calculate the free energy
difference between these two states. Note that for this type of free
energy calculation we solely need the conformational information
(coordinates, trajectory) from the simulation, no energies!
To illustrate the above, we are going to calculate the free energy of
PDB file of the peptide and the MD trajectories at four different
First view part of one trajectory with pymol:
trjconv -s pep.pdb -f 340.xtc -o movie.pdb -e 10000 -fit rot+trans
and select "4" followed by "1" when asked to select a group.
in pymol, type:
and press the "play" button on the bottom right of the main pymol
window. You see the first folding transition during the simulation at
What kind of structure is the folded structure?
For a more quantitative analysis of the folding/unfolding
transitions, we'll analyse the root mean square deviation of each
structure in each trajectory with respect to the folded
structure. Here we will call the structure "folded" if it has a root
mean square deviation (RMSD) of less than 2 Angstroms (0.2 nm) of the
backbone atoms with respect to the reference structure in pep.pdb.
g_rms -s pep.pdb -f 340.xtc -o rmsd_340.xvg
when asked for a selection of groups, press "4" and "4" again.
View the result with xmgrace:
How many folding/unfolding transitions do you observe?
Repeat the same procedure for the other 3 temperatures to generate
rmsd_298.xvg, rmsd_350.xvg and rmsd_360.xvg. With a command like the
following, we can count how many configuration from each trajectory
are "folded" or "unfolded":
tail +11 rmsd_340.xvg | awk '$2
if the counting didn't work, then try:
tail +11 rmsd_340.xvg | sed 's/0\./0,/g' | awk '$2
(never mind if you didn't understand that command completely. The
"tail" part throws away the header of the rmsd_340.xvg file, the
output is then "piped" to the "awk" command that filters the output and
only prints the lines of which the second column (the RMSD) is smaller
than 0.2; the "wc -l" simply counts the number of lines). The sed command,
finally, replaces the dots in the file with comma's, for the case your
computer has expects german input, which uses comma's for floating point numbers.
The obtained number is the number of "folded" configurations in the 340K
simulation. Repeat the procedure, now to count the number of
"unfolded" conformations. You'll see, at this temperature the system
is folded about one third of the time.
What is the difference in free energy between the folded and unfolded
state at this temperature? (Hint: kT = 2.5 kJ/mol at 300K)
What about the other three temperatures? What would you estimate to be
the folding temperature?
Assuming that the folding enthalpy is the same for the different
temperatures, how can the entropy difference between the folded and
unfolded state be estimated from the set of free energies? How large
do you estimate the entropy of folding in this case? Is it positive or