CHI Energy Functions | Force Field

The Carbohydrate Intrinsic (CHI) Energy Functions were developed to score oligosaccharide structures based on glycosidic linkage conformations. Quantum mechanical calculations were used to obtain the torsional energy profiles of glycosidic torsion angles using several tetrahydropyran-based disaccharide models. Similar energy profiles were grouped, and an average energy curve for each group was calculated, resulting in final four CHI energy equations described below. Currently, CHI energy functions can be applied to all standard two-bond glycosidic linkages between pyranoses in the chair conformation.

Equations

Summary tables of the equations are given below this section.

CHI energy equation for the $\phi$ torsion angle of disaccharides in which the C₁O₁ bond of the non-reducing saccharide is in the axial position:
$E(\phi|\alpha) = 2.977 * e^{-\frac{(\phi+1.9949*10^2)^2}{6.7781*10^2}} + 1.0225 * 10^2 * e^{-\frac{(\phi-1.706*10^2)^2}{1.6968*10^3}} + 1.0745 * 10 * e^{-\frac{(\phi+1.0531* 10^2)^2}{4.7246*10^3}} + 3.6735 * e^{-\frac{(\phi-6.2012)^2}{1.3477*10^3}} + 2.061 * e^{-\frac{(\phi-9.1655*10)^2}{1.5*10^3}} + 6.1939 * e^{-\frac{(\phi+2.2979*10)^2}{2.1223*10^3}} - 2.1115 * e^{-\frac{(\phi-8.3602*10)^2}{1.2541*10^3}} - 9.8001 * 10 * e^{-\frac{(\phi-1.7001*10^2)^2}{1.5987*10^3}}$
where, -180 < $\phi$ < 180

CHI energy equation for the $\phi$ torsion angle of disaccharides in which the C₁O₁ bond of the non-reducing saccharide is in the equatorial position:

$E(\phi|\beta) = 4.5054 * 10^2 * e^{-\frac{(\phi+3.3077*10^2)^2}{4.4498*10^3}} + 2.3712 * 10 * e^{-\frac{(\phi-3.0463*10^2)^2}{8.3752*10^3}} + 5.9353 * e^{-\frac{(\phi+1.5208*10^2)^2}{6.0498*10^3}} + 2.2467 * 10 * e^{-\frac{(\phi+2.3516*10)^2}{6.0690*10^2}} + 1.0036 * 10 * e^{-\frac{(\phi-1.2096*10^2)^2}{4.038*10^3}} - 1.8141 * 10 * e^{-\frac{(\phi+2.4268*10)^2}{5.4305*10^2}} + 5.8823 * e^{-\frac{(\phi-1.9632*10)^2}{8.9793*10^2}} - 2.1283$
where, -180 < $\phi$ < 180

CHI energy equation for the $\psi$ torsion angle of disaccharides in which the C_2^‘O_2^‘/C_4^‘O_4^‘ bond of the reducing saccharide is in the axial position or the C₃O₃ bond of the same saccharide is in the equatorial position: $E(\psi|2a3e) = 4.6237 * e^{-\frac{(\psi-5.0456)^2}{5.0058*10^3}}+4.6139 * e^{-\frac{(\psi-3.6249 * 10^2)^2}{2.0906*10^3}} + 4.9419 * e^{-\frac{(\psi-1.212 * 10^2)^2}{2.0938*10^3}}+ 4.029 * 10^{-1} * e^{-\frac{(\psi-2.4143 * 10^2)^2}{4.5683*10^2}} + 7.9888 * 10^{-1} * e^{-\frac{(\psi-6.8425 * 10)^2}{6.7881*10^2}} + 2.2299 * 10^{-1} * e^{-\frac{(\psi-1.9293 * 10^2)^2}{3.4725*10^2}} - 1.2565 * 10^{-1}$
where, 0 < $\psi$ < 360

CHI energy equation for the $\psi$ torsion angle for disaccharides in which the C_2^‘O_2^‘/C_4^‘O_4^‘ bond of the reducing saccharide is in the equatorial position or the C₃O₃ bond of the same saccharide is in the axial position: $E(\psi|2e3a) = 4.4681 * e^{-\frac{(\psi-10^{-30})^2}{1.2796*10^3}} + 4.382 * e^{-\frac{(\psi-3.5777 * 10^2)^2}{6.0501*10^3}} + 2.8495 * 10^2 * e^{-\frac{(\psi-1.4664 * 10^2)^2}{1.5518*10^3}} + 4.7613 * e^{-\frac{(\psi-2.2068 * 10^2)^2}{5.8929*10^3}} - 1.692 * 10^2 * e^{-\frac{(\psi-1.4737 * 10^2)^2}{1.7425*10^3}} - 1.1844 * 10^2 * e^{-\frac{(\psi-1.4606 * 10^2)^2}{1.3598*10^3}} + 1.0220$
where, 0 < $\psi$ < 360

Summary of Equations

The CHI energy equations are of the generalized form:

$E(\theta) = \sum_i a_i e^{-\frac{(\theta-b_i)^2}{2c_i^2}} + d$

The parameters for each of the four equations are given in the following tables:

$E(\phi\|\alpha)$	$E(\phi\|\beta)$
$a_1=2.977,b_1=1.994910^2,c_1=6.778110^2$	$a_1=4.505410^2,b_1=-3.307710^2,c_1=4.4498*10^3$
$a_2=1.022510^2,b_2=1.70610^2,c_2=1.6968*10^3$	$a_2=2.371210^1,b_2=3.046310^2,c_2=8.3752*10^3$
$a_3=1.074510^1,b_3=-1.053110^2,c_3=4.7246*10^3$	$a_3=5.9353,b_3=-1.520810^2,c_3=6.049810^3$
$a_4=3.6735,b_4=6.2012,c_4=1.3477*10^3$	$a_4=2.246710^1,b_4=-2.351610^1,c_4=6.0690*10^2$
$a_5=2.061,b_5=9.165510^1,c_5=1.510^3$	$a_5=1.003610^1,b_5=1.209610^2,c_5=4.038*10^3$
$a_6=6.1939,b_6=-2.297910^1,c_6=2.122310^3$	$a_6=-1.814110^1,b_6=-2.426810^1,c_6=5.4305*10^2$
$a_7=-2.1115,b_7=8.360210^1,c_7=1.254110^3$	$a_6=5.8823,b_6=1.963210^1,c_6=8.979310^2$
$a_8=-9.800110^1,b_8=1.700110^2,c_8=1.5987*10^3$	$d = -2.1283$
$d = 0.0$

$E(\psi\|2a3e)$	$E(\psi\|2e3a)$
$a_1=4.6237,b_1=5.0456,c_1=5.0058*10^3$	$a_1=4.4681,b_1=10^{-30},c_1=1.2796*10^3$
$a_2=4.6139,b_2=3.624910^2,c_2=2.090610^3$	$a_2=4.382,b_2=3.577710^2,c_2=6.050110^3$
$a_3=4.9419,b_3=1.21210^2,c_3=2.093810^3$	$a_3=2.849510^2,b_3=1.466410^2,c_3=1.5518*10^3$
$a_4=4.02910^{-1},b_4=2.414310^2,c_4=4.5638*10^2$	$a_3=4.7613,b_3=2.206810^2,c_3=5.892910^3$
$a_5=7.988810^{-1},b_5=6.842510^1,c_5=6.7881*10^2$	$a_4=-1.69210^2,b_4=1.473710^2,c_4=1.7425*10^3$
$a_6=2.229910^{-1},b_6=1.929310^2,c_6=3.4725*10^2$	$a_5=-1.184410^2,b_5=1.460610^2,c_5=1.3598*10^3$
$d = -1.2565 * 10^{-1}$	$d = 1.0220$

Usage
In order to obtain the total CHI energy for a given oligosaccharide conformation, the CHI energies of the individual glycosidic torsion angles comprising the molecule are first calculated by substituting the corresponding torsion angle values in the appropriate CHI energy equations. The summation of the individual CHI energies gives the total CHI energy of the molecule.

Publication
A. K. Nivedha, S. Makeneni, B. L. Foley, M. B. Tessier , R. J. Woods, J. Comput. Chem. 2014, 35, 526–539. DOI: 10.1002/jcc.23517

$E(\phi\|\alpha)$	$E(\phi\|\beta)$
$a_1=2.977,b_1=1.994910^2,c_1=6.778110^2$	$a_1=4.505410^2,b_1=-3.307710^2,c_1=4.4498*10^3$
$a_2=1.022510^2,b_2=1.70610^2,c_2=1.6968*10^3$	$a_2=2.371210^1,b_2=3.046310^2,c_2=8.3752*10^3$
$a_3=1.074510^1,b_3=-1.053110^2,c_3=4.7246*10^3$	$a_3=5.9353,b_3=-1.520810^2,c_3=6.049810^3$
$a_4=3.6735,b_4=6.2012,c_4=1.3477*10^3$	$a_4=2.246710^1,b_4=-2.351610^1,c_4=6.0690*10^2$
$a_5=2.061,b_5=9.165510^1,c_5=1.510^3$	$a_5=1.003610^1,b_5=1.209610^2,c_5=4.038*10^3$
$a_6=6.1939,b_6=-2.297910^1,c_6=2.122310^3$	$a_6=-1.814110^1,b_6=-2.426810^1,c_6=5.4305*10^2$
$a_7=-2.1115,b_7=8.360210^1,c_7=1.254110^3$	$a_6=5.8823,b_6=1.963210^1,c_6=8.979310^2$
$a_8=-9.800110^1,b_8=1.700110^2,c_8=1.5987*10^3$	$d = -2.1283$
$d = 0.0$

$E(\psi\|2a3e)$	$E(\psi\|2e3a)$
$a_1=4.6237,b_1=5.0456,c_1=5.0058*10^3$	$a_1=4.4681,b_1=10^{-30},c_1=1.2796*10^3$
$a_2=4.6139,b_2=3.624910^2,c_2=2.090610^3$	$a_2=4.382,b_2=3.577710^2,c_2=6.050110^3$
$a_3=4.9419,b_3=1.21210^2,c_3=2.093810^3$	$a_3=2.849510^2,b_3=1.466410^2,c_3=1.5518*10^3$
$a_4=4.02910^{-1},b_4=2.414310^2,c_4=4.5638*10^2$	$a_3=4.7613,b_3=2.206810^2,c_3=5.892910^3$
$a_5=7.988810^{-1},b_5=6.842510^1,c_5=6.7881*10^2$	$a_4=-1.69210^2,b_4=1.473710^2,c_4=1.7425*10^3$
$a_6=2.229910^{-1},b_6=1.929310^2,c_6=3.4725*10^2$	$a_5=-1.184410^2,b_5=1.460610^2,c_5=1.3598*10^3$
$d = -1.2565 * 10^{-1}$	$d = 1.0220$