Download (ΔG) Hydrogen Bonds

Amino Acids – High Chemical Diversity, Low Structural Diversity
There are 20 common naturally occurring amino acids termed -amino acids because both the amino- and
carboxylic acid groups are connected to the same (α) carbon atom. Of the 20 common residues 19 have
the general structure shown below:
R
H3N+
C
CO2-
R=H (Gly), CH3 (Ala), etc.
H
The exception is the amino acid proline, whose side chain is bonded to the nitrogen atom to give a
cyclic imino acid called proline:
CH2
H2C
H2N+
CH2
C
CO2-
H
Because each side chain group attached to Cα is different (except for glycine, in which R=H), Cα is
asymmetric and, in nature, is always the L-enantiomer.
On the basis of the gross physical properties of the R-groups it is possible to divide the amino acids into
classes, namely, hydrophobic, charged, and polar. Further divisions may be made on the basis of the
chemical natures of the R-groups.
Proteins: One Linkage Position and Configuration
The Peptide Bond
H
C
N
C
O
x
C
N
trans-Peptide
C-N-C-C torsion angle = 180
C
N
O
H
C
O
cis-Peptide
C-N-C-C torsion angle = 0
C
N
O
With few exceptions, peptide groups assume a trans-conformation, in which successive Cα atoms are on
opposite sides of the peptide bond joining them.
The observation that the peptide bond prefers to be planar is best explained with reference to the
molecular orbitals. Maximum orbital overlap (and so maximum stabilization) is achieved when the p-type
orbital of the N-atom is parallel to the p-orbitals of the π-bond. During rotation around the C—N bond the
overlap decreases and becomes nearly zero when the orbitals are perpendicular to each other. This is a
higher energy conformation, because it lacks the stabilizing orbital interactions. Thus, rotation is
energetically disfavoured, and the bond remains close to planarity.
These characteristics are important because they indicate that the backbone of a protein is a linked
sequence of rigid planar peptide groups. The protein backbone conformation can therefore be
specified by the torsion (or dihedral) angles about the Cα—N bond (φ-torsion angle) and Cα—C
bond (ψ-torsion angle) of each of its amino acid residues.
6
4
OH
5
OH
6
O
5
4
1

2
3
 -D-pyranose
3
-D-pyranose
1
OH
OH
OH
O
O
OH
HO
OH
-D-glucopyranose, -D-Glc
3
OH
HO
HO
O
1-3 linkage ()
OH
O
O
O
1

2
OH
HO
O
OH
OH
OH
OH
1
NHAc
2-N -acetyl--D -glucopyranose,  -D-GlcNAc
1-4 linkage ()
4
O
O
OH
OH
OH
O
OH
OH
OH
HO
HO
OH
9
 -D-galactopyranose, -D-Gal
HO
HO
HO
OH
O
OH
-D-mannopyranose, -D-Man
8
CO2
6
O
5
HO
-
O
1
2
7
AcHN
OH
1
4
OH
3
N -acetyl- -neuraminic acid, -Neu5Ac
1-6 linkage ()
O
6
O
OH
O
OH
O
O
O
HO
Glc-α-(1-4)-Glc (Starch)
Glc-β-(1-4)-Glc (Cellulose)

O

O
HO
O
O
OH
The same two amino acids  1 possible peptide
The same two monosaccharides  20 possible disaccharides
OH
OH
CO 2-
HO
OH
O
HO
O
AcHN
OH HO
CO 2-
HO
2
AcHN
O
O-GlcNAc
O
OH
2
O
6
O
OH HO
O-GlcNAc
HO
3
OH
OH
Avian Flu Receptor
α-(2-3)-Gal
Human Flu Receptor
versus
α-(2-6)-Gal
Other Cells
Hormones
Viruses
Toxins
Cell
Bacteria
Enthalpy (ΔH) and Entropy (ΔS)
Electrostatic Interactions (Hydrogen-bonds, charge-charge, charge-dipole, dipole-dipole)
Dispersive Interactions (Van der Waals attractions and repulsions)
ΔH < 0 reaction is exothermic, tells us nothing about the spontaneity of the reaction
Δ H > 0 reaction is endothermic, tells us nothing about the spontaneity of the reaction
Examples:
Oxidation of glucose: C6H12O6 + 6O2  6CO2 + 6H2O ΔH = -2803 kJmol-1
Just because a reaction is exothermic (that is because ΔH < 0) does not mean that it is
spontaneous.
And what about
Dissolving salt: NaCl(s) + H2O(l)  Na+(aq) + Cl-(aq) H = 4 kJmol-1
Just because this reaction is endothermic (ΔH > 0) does not mean that it doesn’t
happen.
Enthalpy alone is not sufficient to decide whether a reaction will occur. The missing factor is
called Entropy or ΔS.
Entropic Contributions:
Solute Related (conformational entropy)
Solvent Related (ligand and receptor desolvation)
ΔS < 0 reaction leads to order, tells us nothing about the spontaneity of the reaction
ΔS > 0 reaction leads to disorder, tells us nothing about the spontaneity of the
Thermodynamics of Ligand-Protein Interactions (ΔG)
Remember:
ΔGreaction = ΔGproducts – ΔGreactants
ΔG = ΔH - T ΔS,
The reaction is favourable only when ΔG < 0
Ligand Binding Energy is also computed as if it were a reaction:
Ligand + Receptor → Complex
ΔGBinding = ΔGComplex – ΔGLigand – ΔGReceptor
= (ΔHComplex – T Δ SComplex) – (ΔHLigand – T ΔSLigand) – (ΔHReceptor – T ΔSReceptor)
There is a temptation to draw conclusions only from the structure of the complex, but:
ΔGBinding ≠ MM EnergyComplex
MM “Energy” is often just the potential energy from a force
field calculation.
MM Energy often ignores entropy and desolvation!
and is often NOT computed as a difference between reactants and products!
Bad modeling can’t be trusted!
Energetic Contributions to Ligand Binding
Hydrogen Bonds
A molecule which has a weakly acidic proton (O—H, N—H) may function as a proton donor (DH)
in a hydrogen bond with another molecule in which an electronegative atom (O, N) is present to
act as an acceptor (A).
D—H ···A—
A typical hydrogen bond between polar uncharged groups has its maximum stability at an
interatomic (A ···D) separation of 2.7-3.1 Å and may contribute up to approximately 5 kcalmol-1 in
the gas phase. Hydrogen bonds show a high dependence on the orientation of the donor and
acceptor groups, with a tendency for the D—H ··· A angle to be linear.
Hydrogen Bond Energy
Total Energy (kcal/mol)
10
8
R"
Total Energy
Dipole/Dipole
van der Waals
6
4
D—H ··· ··· ··· ··· A
O
2
H
R'
O
0
roh
-2
-4
-6
1.25
1.75
2.25
2.75
3.25
3.75
4.25
H
4.75
O...H Separation (Angstroms)
X-ray crystallographic studies of sugar-protein complexes can provide detailed structural
information pertaining to hydrogen bonding in the binding site.
Hydrogen Bonds, Continued
The hydrophilic nature of sugars arises from the presence of hydroxyl groups attached typically to
5 out of 6 of the carbon atoms of the sugar:
The polyhydroxylated structure of a sugar has often been cited in support of the importance of
hydrogen bonding in the interaction between the sugar and either a receptor or with solvent.
For example, in the case of arabinose binding protein, the arabinopyranose is involved in
approximately 54 hydrogen bonds either with the protein, or with coordinated water molecules.
A
B
H
H2N
+
H
O
NH2
O
H
H
OH
OH
HO
H
NH2
O
O
HO
HO
O
OH
H
N
H
N
+
HO
H
OH
NH3+
O
HO
O
O
OH
OH
H
O
H
O
H
H
H
O
H
A. Typical hydrogen bonds between a sugar and a protein. B. Between a sugar and water.
Effect of Loss of a Hydrogen Bond on Binding Energies
The presence of hydrogen bonding is essential to the binding of a sugar to a protein. If there are
not at least as many hydrogen bonds in the complex as there are between the sugar and the
solvent, the binding will not be ENTHALPICALLY favored.
If each hydrogen bond stabilizes the interaction by 5 kcal/mol, the loss of a single hydrogen bond
would severely diminish the binding affinity.
Consider two ligands: one makes 4 hydrogen bonds to the receptor, the other makes 3 hydrogen
bonds.
L1 + Receptor → Complex1
GBinding (L1)  -20 kcal/mol
L2 + Receptor → Complex2
GBinding (L2)  -15 kcal/mol
G = GBinding (L1) – GBinding (L2) = -5 kcal/mol
(favoring the binding of L1)
How much would the loss of a single hydrogen bond change the binding affinities?
Recall: G = –RT ln(K)
G = –RTln(K1) – RTln(K2) = –RT(ln(K1) – ln(K2)) = –RTln(K1/K2)
So for the two ligands, the ratio of their binding affinities (at 293 K) is:
-5 = –RT ln(K1/K2) = –0.00198∙398ln(K1/K2) = –0.59 ln(K1/K2)
Therefore K1/K2 = e8.47 = 4788, so the net loss of a hydrogen bond decreases affinity ~ 5000 fold.
But why isn’t counting H-bonds a good measure of affinity?
Since many sugars all have the same number of hydroxyl groups and differ only in the
configuration of the hydroxyl groups, they all can exhibit very similar hydrogen bonding patterns if
they can physically fit into the receptor site.
O
OH
H
H
H
O
HO
H
HO
H
H
The Role of Water
O
OH OH
H O
H
O
O

O
O
H
O
H
H
H
H
O
H
H
+
S > 0
+
H
H
H
H
H
H
O
O
OH
OH OH
O
O
H
H
OH
OH
H
HO
OH
HO
OH
Each hydroxyl group in a sugar may act as both a proton acceptor and a proton donor in
hydrogen bonds.
In solution it is possible for two water molecules to orient themselves along each sugar hydroxyl
group lone-pair axis and so an optimum hydrogen bonding network is present. However in the
complex it may not be possible to orient the protein side chains as optimally.
Since for every hydrogen bond the sugar forms with the protein, it must break at least one
with the water, thus the net ENTHALPIC gain from hydrogen bonding may be relatively
small.
Consequently, while hydrogen bonding is essential to the binding of the sugar, it is not sufficient to
generate very tight binding, or to discriminate between different sugars. This may explain why
monosaccharide-protein interactions are often very weak: Ka=102-3 M-1.
Example: Effect of loss of OH on affinity
Antigenic oligosaccharides from
S. flexneri Y with antibody SYAJ6
Vyas, N.K., et al., Biochemistry,
2002. 41: p. 13575-13586.
Other Enthalpic Contributions to Binding
Van der Waals Interactions (instantaneous dipole - induced dipole)
As any pair of atoms approach each other a weak attraction develops that is called a van der
Waals interaction.
In order to provide a noticeable ENTHALPIC benefit the atoms must be no further apart than the
sum of their van der Waals radii (typically less than ~ 4 Å). In a carbohydrate protein complex
there may be many such interactions, and although each one provides very little energy, their
sum may be significant.
R'
R"
O
H
O
H
roh
The maximum energetic contribution from vdw interaction is small (only about 0.2 kcal/mol) per
interacting atom and decreases with distance as a function of 1/r6.
The combination of vdw and hydrophobic effects can be significant. For example: in the antigenic
oligosaccharide from the bacterial polysaccharide of V. cholerae types Ogawa (OMe) and Inaba
(OH). Antibodies against Ogawa do not bind to Inaba!
Because of the extreme sensitivity of the energies of van der Waals contacts to interatomic
distance, a slight change in ligand shape or binding orientation can greatly alter the number of
van der Waals contacts. Thus, carbohydrate ligand specificity depends very highly on van der
Waals contacts.
Example: Effect of loss of van der Waals and hydrophobic
contacts on affinity
Antigenic oligosaccharides from V. cholerae with antibodyS20-4
Too few favorable
interactions or too
many unfavorable
ones will hurt binding
Ogawa
Inaba
Affinities of Vibrio cholerae O:1 Serotypes Ogawa and Inaba for mAb S20-4
ΔG
Ka
OH
>0
----
OMe
-7.6
3.9 x 105
OEt
-6.2
3.6 x 104
OPr
>0
----
Wang, J., et al., J. Biol. Chem., 1998. 273(5): p. 2777-2783.
H
>0
----
Effect of Entropy on Binding Energies
A large contributor to the ENTROPY of binding is from the release of water molecules. This
arises from two contributions, desolvation entropy and the hydrophobic effect.
Desolvation Entropy
As already seen, when the sugar binds to the protein, it displaces water molecules that were
previously present in the binding site. It also must release water molecules that were directly
coordinated to the sugar itself.
O
OH
O
OH OH
H
H
H
O
HO
H
HO
H
H
H O
H
O
O

O
O
H
O
H
H
H
H
O
H
H
+
S > 0
+
H
H
H
H
H
H
O
O
OH
OH OH
O
O
H
H
OH
OH
H
HO
OH
HO
OH
This release of water results in an increase in the entropy of the system, i.e.
SBinding > 0 and so -TSBinding < 0.
But the desolvation free energy may still be unfavorable (>0) depending on ΔH
GDesolvation = HDesolvation - TSDesolvation
The Hydrophobic “Effect”
While it is obvious that a sugar is highly hydrophilic (typically being soluble only in water), sugars
are also capable of hydrophobic interactions.
In the crystal structures of bound sugar-protein complexes it has frequently been observed that
aromatic residues, such as Tyr, Trp and Phe, are present in the binding site. Moreover, these
residues appear to stack against the “back” face of the sugar:
OH
Hydrophilic Top Face
O
O
HO
HO
H
H
OH
OH
H
+
H
H
O
H
HO
O
H
H
O
H
O
H
OH
HO
H
O
OH
OH
Hydrophobic
Back Face
How does the hydrophobic
effect differ from van der
Waals contacts?
How does it differ from
orbital overlap?
O
H
Aromatic residues on the surface of the protein are not able to hydrogen bond effectively with the
solvent and so they force the nearby water into non-ideal orientations.
When the sugar places its hydrophobic face against the aromatic residues, it releases the waters
from their non-ideal orientation.
This results in a gain in ENTROPY (S > 0). Moreover, it exposes its hydrophilic face to the
solvent and so helps promote good solvent-ligand hydrogen bonding.
The Origin of Hydrophobicity – “why water and oil don’t mix”
The solubility of a molecule in water depends on a balance between the energy needed to create
a cavity in the water and the energy gained by the resulting interactions.
Thermodynamic data indicate that it is not enthalpy, but rather entropy that drives the non-polar
molecules to avoid water.
if ΔH > 0 then this implies that energy must be added to get the reaction to occur
if ΔS > 0 (- TΔS < 0) then this implies that the reaction favours disorder.
In all cases ΔG is negative indicating that hydrocarbons will spontaneously separate from water.
BUT it is enthalpically more favorable for small hydrocarbons to dissolve in water than in large
non-polar solvents!
Process
CH4 in H2O versus CH4 in C6H6
C2H6 in H2O versus C2H6 in C6H6
C2H4 in H2O versus C2H4 in C6H6
ΔH
kJ/mol
11.7
9.2
6.7
- TΔS
kJ/mol
-22.6
-25.1
-18.8
ΔG
kJ/mol
-10.9
-15.9
-12.1
Entropy is a measure of disorder in a system. It decreases with increasing order. If
-T ΔS is negative as in the above table then ΔS must be positive.
Rationalization: Because the non-polar group can not hydrogen bond to water, the water
molecules at the surface of the non-polar molecule have fewer ways in which to hydrogen bond to
each other. That means they have less freedom, or that they must reorient themselves into a
more ordered structure at the surface of the cavity. This causes the entropy to decrease.
So the preference for a hydrophobic group to avoid water is because otherwise it would
force the water into entropically unfavorable orientations.
Carbohydr . Res., (2005) 340, 1007
PNAS , (2006) 103, 8149
Ligand Flexibility and Conformational Entropy Change on Binding
Entropy may also change as a function of the properties of the ligand. In flexible sugars,
particularly oligosaccharides and polysaccharides, binding reduces the flexibility (entropy) of the
ligand and so disfavors binding.
Thus, certain regions of the ligand may introduce beneficial entropies upon binding while other
regions may not.
Free Ligand (in solvent)
Bound Ligand
OH
OH
O
HO
HO
O
-
S < 0
OH
O
O
HO
HO
Rigid linkage
OH
O
O
O
HO
HO
H2N
O
OH
Flexible linkage
O
HO
HO
OH
OH
NH3
+
OH
How it all fits together: Polysaccharide – antibody binding analysis
Component Energy
Theoretical Contribution
Electrostatic Interactions
-167.5
Van der Waals Interactions
-126.9
Total Molecular Mechanical Energy
-294.4
Desolvation Energy
211.9
Entropy
77.6
Total Binding Energy
-4.9
Kadirvelraj, R., et al., PNAS, 2006. 103(21): p. 8149-8154.