Example: Relative Binding Affinity

Free Energy Fundamentals
Theory Definitions Exponential Averaging / Zwanzig Thermodynamic Integration Bennett Acceptance Ratio WHAM MBAR
Methods of Free Energy Simulations Thermodynamic Cycle Intermediate States (Constructing) Soft Core Potentials Running Simulations Extracting Simulation Information Analysis / Bootstrap Sampling
Improving Sampling and Accelerating Simulations Umbrella Sampling Hamiltonian Replica Exchange [math]\displaystyle{ \lambda }[/math]-dynamics / Expanded Ensemble

Free Energy How-to's
Example Calculations Absolute Solvation Free Energy Relative Binding Affinity Absolute Binding Affinity
Tutorials Ethanol Solvation - GROMACS Expanded Ensemble - GROMACS Absolute Binding Free Energy - GROMACS

This example will be a bit more complex than the solvation example since we are now transforming one molecule to another within an environment. Again, the purpose of this page is not to take you through step-by-step the process of simulation, but instead guide you through the logical decisions that can be applied to any simulation package of your choosing. We will work through the same questions that were asked on the previous example as these should be asked for any free energy simulation.

For this particular example, we will look at the difference in the relative binding free energies of toluene and phenol in the apolar cavity of T4 Lysozyme.

Toluene in the pocket...

... becomes phenol in the pocket.

What is the Thermodynamic Cycle?

Example of using a thermodynamic path to find which molecule has a stronger binding affinity. We could either remove both molecules from the pocket which is time prohibitive, or we could simply transform one molecule to another both in and out of pocket to find the difference that way.

If you have already read through the thermodynamic cycle page, then you will have already seen this picture on the left. For our simulation, we can define the thermodynamic cycle either by directly converting toluene to phenol (which would be very costly due to the large number of changing interactions and box size), or we can recognize that free energy is a state function and we can use this relation to solve for it as well:

[math]\displaystyle{ \Delta \Delta A_{\mathrm{bind}} = \Delta A_{\mathrm{bind}}^B - \Delta A_{\mathrm{bind}}^A = \Delta A_{A\rightarrow B}^{\mathrm{bound}} - \Delta A_{A\rightarrow B}^{\mathrm{unbound}} }[/math]

which does not require an excessively large box and can be done relatively simply by choosing the correct intermediate states to transform the ligands.

To provide the visuals for this particular system, please see the rendered images below; note that the solvent can be either implicit or explicit and the cycle will still be valid.

	A [math]\displaystyle{ \Delta A_{\mathrm{bind}} }[/math]
Bound [math]\displaystyle{ \displaystyle \Delta A_{A\rightarrow B} }[/math]		Unound [math]\displaystyle{ \displaystyle \Delta A_{A\rightarrow B} }[/math]
	B [math]\displaystyle{ \Delta A_{\mathrm{bind}} }[/math]

What are the End States?

There are two real end states we need here. Since the T4 lysozyme and solvent are not changing in either case, they will be present in both end states.

• State A) T4 lysozyme in solvent with an intermediate molecule that looks like toluene in the pocket.
• State B) T4 lysozyme in solvent with an intermediate molecule that looks like phenol in the pocket.

[[Image:Toluene_Phenol_Topology.png.png|right|thumb|300px|Different topology choices. Even though the intermolecular forces will be for the real atoms, there are still dummy intramolecular forces. Please see [[the full page on topologies for a more in depth description.]] What does it mean to have an "intermediate state look like" a molecule as an end state? This intermediate state could well be just toluene or phenol, however, there will still be the dummy atoms on it based on the topology choice. Recall from the figure on the right that, even though the interacting molecule will be a physical molecule, there are still the dummy interactions that must be built in and considered.

For this example, a dual state topology is shown on the right, but it is by no means the only set of end states one could chose; an ortho- or meta- arrangement could just as easily be selected with the dual topology scheme. There is also not a limitation on just dual topologies as single topologies will also work in any of the arrangements.

Although beyond the scope of this example, it is also worth mentioning there is also research ongoing in predicting how a real molecule will behave and the end point will not have the full interactions of a real molecule, but some unphysical intermediate. The free energies of the real molecule are extrapolated from these results. This is particularly helpful for doing more than one transformation as you would not need a unique set of intermediate states for each pair of real molecule end points.

What are the Intermediate States?

For this example, we will assume that the phenol is the initial state, although we could have just as easily assumed it was the toluene. Observe that the OH moiety must disappear and the methyl must appear to make the transformation complete. A good first approach sample set of intermediates can be constructed that follow this process:

1. Turn the charge on the OH and the para- hydrogen to zero. Recall it is important to make sure the overall system is at the same total charge at each intermediate.
2. Next, you will want to turn the Lennard-Jones [math]\displaystyle{ \epsilon }[/math] on the hydroxyl and para- hydrogen turned to zero.
   • You can simultaneously turn off angle, torsional, and bonded terms linearly.
3. Now turn on the methyl's Lennard-Jones [math]\displaystyle{ \epsilon }[/math] and its corresponding para- hydrogen.
   • The corresponding angle, torsional, and bonded terms will also need turned on.
4. Lastly, turn on the charges of the methyl and corresponding para- hydrogen.

Independent of which pathway you select, you will still need a sufficient number of intermediates. By choosing BAR or MBAR as our method, we may need 2-4 intermediates for the charge and 4-6 for the Lennard-Jones changes. This is less than the ethanol solvation because there are fewer atoms changing, corresponding to better phase space overlap. TI would take roughly 2-3 times the number of intermediates. You could run both Lennard-Jones transformations at once, but you would have to first disable improper torsions first otherwise the atoms will collide with each other.