# Solvation energy of a protein with a continuum electrostatic model¶

## Background¶

The energy required to dissolve a molecule (that is, moving it from vacuum into a solvent) is known as the solvation energy. This quantity is of interest in several biomolecular applications. In a continuum electrostatic description, the dissolved biomolecule can be modeled with two dielectric regions interfaced by a molecular surface: an outer solvent region, usually comprised of water and salt ions, and an inner protein region, where the solvent cannot access, which contains point charges at the locations of the atoms.

<img src="sketch_protein.png", style="width:300px">

The protein region ($\Omega^-$) is governed by a Poisson equation with a source term due to the partial charges of the atoms, whereas the solvent region ($\Omega^\text{+}$) uses a modified Helmholtz equation to account for the presence of salt. The electrostatic potentials in the two regions, $\phi^-$ and $\phi^+$, are given by $$\Delta \phi^- (\mathbf{x}) = \sum_k q_k \delta (\mathbf{x}_k) \quad\text{in }\Omega^-,$$ $$\Delta \phi^+ (\mathbf{x}) - \kappa \phi^+ (\mathbf{x}) = 0 \quad\text{in }\Omega^+,$$ $$\phi^- = \phi^+\quad\text{on }\Gamma$$ $$\epsilon^- \frac{\partial \phi^-}{\partial \nu} = \epsilon^+ \frac{\partial \phi^+}{\partial \nu}\quad\text{on }\Gamma$$ where $q_k$ and $\mathbf{x}_k$ are the charge and position of each molecule atom, $\kappa$ is the inverse of the Debye length.

Following the work by Yoon and Lenhoff, the system of PDEs with interface conditions can be formulated in terms of boundary operators, as follows:

$$\begin{bmatrix} \tfrac12\mathsf{Id} + \mathsf{K}_L & -\mathsf{V}_L \\ \tfrac12\mathsf{Id} - \mathsf{K}_H & \frac{\epsilon_1}{\epsilon_2}\mathsf{V}_H \end{bmatrix} \left\{ \begin{matrix} \phi^- \\ \frac{\partial \phi^-}{\partial \mathbf{\nu}} \end{matrix} \right\} = \left\{ \begin{matrix} \sum_k q_k \\ 0 \end{matrix} \right\}$$

where $\mathsf{V}_L$ and $\mathsf{K}_L$ are the single and double layer boundary operators for the Laplace kernel (Poisson equation) and $\mathsf{V}_H$ and $\mathsf{K}_H$ are the corresponding operators for the modified Helmholtz kernel (Poisson-Boltzmann or Yukawa equation).

From there, we can compute the reaction potential ($\phi_\text{reac} = \phi^--\phi_\text{Coulomb}$) at the locations of the atoms using

$$\phi_\text{reac} = \mathcal{K}_L\left[ \frac{\partial \phi}{\partial n} \right] - \mathcal{V}_L \left[ \phi \right],$$ where $\mathcal{V}_L$ and $\mathcal{K}_L$ are the single and double layer potential operators for the Laplace kernel. We then obtain the solvation energy using

$$\Delta G_\text{solv} = \tfrac12 \sum_{k=1}^{N_q} q_k \phi_\text{reac}(\mathbf{x}_k)$$

## Example¶

### Molecular structure and mesh¶

As an example, we will calculate the solvation energy of the bovine pancreatic trypsin inhibitor. The molecular crystal structure is available the protein data bank (PDB code 5PTI), and we obtained the atomic radii and charges with pdb2pqr, resulting in a .pqr file.

There are several molecular surface definitions which can be used. In this case, we use the solvent-excluded surface (SES), which is the result of rolling a spherical probe of the size of a water molecule around the protein, and tracking the contact points. We mesh the SES with msms, and reformat the result to a .msh file.

First, let's import the required libraries and mesh and read the .pqr file for the charges.

In [1]:
import bempp.api
bempp.api.set_ipython_notebook_viewer()

grid = bempp.api.import_grid('5pti_d1.msh')
grid.plot()

import numpy as np
q, x_q = np.array([]), np.empty((0,3))

ep_in = 4.
ep_ex = 80.
k = 0.125

# Read charges and coordinates from the .pqr file
for line in molecule_file:
line = line.split()
if len(line)==0 or line[0]!='ATOM': continue
q = np.append( q, float(line[8]))
x_q = np.vstack(( x_q, np.array(line[5:8]).astype(float) ))


Next, we compute the potential due to the charges on the boundary, required for the right-hand side

In [2]:
def charges_fun(x, n, domain_index, result):
global q, x_q, ep_in
result[:] = np.sum(q/np.linalg.norm( x - x_q, axis=1 ))/(4*np.pi*ep_in)

def zero(x, n, domain_index, result):
result[:] = 0

dirichl_space = bempp.api.function_space(grid, "DP", 0)
neumann_space = bempp.api.function_space(grid, "DP", 0)

charged_grid_fun = bempp.api.GridFunction(dirichl_space, fun=charges_fun)
zero_grid_fun = bempp.api.GridFunction(neumann_space, fun=zero)


Next, we generate the $2\times 2$ block matrix with the single and double layer operators of the Laplace and modified Helmholtz kernels.

In [3]:
# Define Operators
from bempp.api.operators.boundary import sparse, laplace, modified_helmholtz
identity = sparse.identity(dirichl_space, dirichl_space, dirichl_space)
slp_in   = laplace.single_layer(neumann_space, dirichl_space, dirichl_space)
dlp_in   = laplace.double_layer(dirichl_space, dirichl_space, dirichl_space)
slp_out  = modified_helmholtz.single_layer(neumann_space, dirichl_space, dirichl_space, k)
dlp_out  = modified_helmholtz.double_layer(dirichl_space, dirichl_space, dirichl_space, k)

# Matrix Assembly
blocked = bempp.api.BlockedOperator(2, 2)
blocked[0, 0] = 0.5*identity + dlp_in
blocked[0, 1] = -slp_in
blocked[1, 0] = 0.5*identity - dlp_out
blocked[1, 1] = (ep_in/ep_ex)*slp_out


We now use gmres to solve the system.

In [4]:
sol, info, it_count = bempp.api.linalg.gmres(blocked, [charged_grid_fun, zero_grid_fun],
use_strong_form=True, return_iteration_count=True, tol=1e-3)

print("The linear system was solved in {0} iterations".format(it_count))

The linear system was solved in 273 iterations


The two following GridFunction calls store the calculated boundary potential data (separated by $\phi$ and $\frac{\partial \phi}{\partial n}$) for visualization purposes.

In [5]:
solution_dirichl, solution_neumann = sol
solution_dirichl.plot()

/usr/lib/python3/dist-packages/numpy/core/numeric.py:747: ComplexWarning: Casting complex values to real discards the imaginary part
arr = array(a, dtype=dtype, order=order, copy=False, subok=subok)


To compute $\phi_\text{reac}$ at the atoms' locations, we use operators.potential to then multiply by the charge and add to compute $\Delta G_\text{solv}$. The 332.064 term is just a unit conversion constant.

In [6]:
slp_q = bempp.api.operators.potential.laplace.single_layer(neumann_space, x_q.transpose())
dlp_q = bempp.api.operators.potential.laplace.double_layer(dirichl_space, x_q.transpose())
phi_q = slp_q * solution_neumann - dlp_q * solution_dirichl

# total dissolution energy applying constant to get units [kcal/mol]
total_energy = 2 * np.pi * 332.064 * np.sum(q * phi_q).real
print("Total solvation energy: {:7.2f} [kcal/Mol]".format(total_energy))

Total solvation energy: -352.82 [kcal/Mol]