Other Cells. Hormones. Viruses. Toxins. Cell. Bacteria
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1 Other Cells Hormones Viruses Toxins Cell Bacteria
2 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: And what about Oxidation of glucose: C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O ΔH = kjmol -1 Just because a reaction is exothermic (that is because ΔH < 0) does not mean that it is spontaneous. Dissolving salt: NaCl(s) + H 2 O(l) Na + (aq) + Cl - (aq) ΔH = 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 reaction
3 Remember: ΔG reaction = ΔG products ΔG reactants ΔG = ΔH - T ΔS, The reaction is favourable (spontaneous) only when ΔG < 0 Ligand Binding Energy is also computed as if it were a reaction: Ligand + Receptor Complex ΔG Binding = ΔG Complex ΔG Ligand ΔG Receptor = (ΔH Complex T Δ S Complex ) (ΔH Ligand T ΔS Ligand ) (ΔH Receptor T ΔS Receptor ) There is a temptation to draw conclusions only from the structure of the complex, but: ΔG Binding MM Energy Complex MM Energy is often just the potential energy from a force field calculation. MM Energy is not a free energy, often ignores entropy and desolvation, and is often NOT computed as a difference between reactants and products! Bad modeling can t be trusted!
4 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 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. D H A X-ray crystallographic studies of sugar-protein complexes can provide detailed structural information pertaining to hydrogen bonding in the binding site, but not about energies
5 If each hydrogen bond could stabilize a ligand protein interaction by as much as 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 ΔG Binding (L1) -20 kcal/mol L2 + Receptor Complex2 ΔG Binding (L2) -15 kcal/mol ΔΔG = ΔG Binding (L1) ΔG Binding (L2) = -5 kcal/mol (favoring the binding of L1) How much would the loss of a single hydrogen bond change the binding affinities? Recall: ΔG = RT ln(k) ΔΔG = RTln(K 1 ) RTln(K 2 ) = RT(ln(K 1 ) ln(k 2 )) = RTln(K 1 /K 2 ) So for the two ligands, the ratio of their binding affinities (at 293 K) is: -5 = RT ln(k 1 /K 2 ) = ln(K 1 /K 2 ) = 0.59 ln(k 1 /K 2 ) Therefore K 1 /K 2 = e 8.47 = 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?
6 Consider the case of glycans (sugars) binding to proteins 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: K a = M 1.
7 Vyas, N.K., et al., Biochemistry, : p
8 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. 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.
9 Antigenic oligosaccharides from V. cholerae with antibodys20-4 Too few favorable interac:ons or too many unfavorable ones will hurt binding Ogawa Inaba Affinities of Vibrio cholerae O:1 Serotypes Ogawa and Inaba for mab S20-4 OH OMe OEt OPr H ΔG > >0 >0 K a x x Wang, J., et al., J. Biol. Chem., (5): p
10 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. This release of water results in an increase in the entropy of the system, i.e. ΔS Binding > 0 and so -TΔS Binding < 0. But the desolvation free energy may still be unfavorable (>0) depending on ΔH ΔG Desolvation = ΔH Desolvation - TΔS Desolvation
11 In the crystal structures of bound drug-protein complexes it has frequently been observed that aromatic residues, such as Tyr, Trp and Phe, are present in the binding site. Similarly, for bound sugars, these residues appear to stack against the back face of the sugar: How does the hydrophobic effect differ from van der Waals contacts? How does it differ from orbital overlap? 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 (ΔS > 0). Moreover, it exposes its hydrophilic face to the solvent and so helps promote good solvent-ligand hydrogen bonding.
12 Carbohydr. Res., (2005) 340, 1007 PNAS, (2006) 103, 8149
13 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
14 Component Energy Theoretical Contribution Electrostatic Interactions Van der Waals Interactions Total Molecular Mechanical Energy Desolvation Energy Entropy 77.6 Total Binding Energy -4.9 Kadirvelraj, R., et al., PNAS, (21): p
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