Amino Acids – High Chemical Diversity , Low Structural Diversity. - PowerPoint PPT Presentation
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 () carbon atom. Of the 20 common residues 19 have the general structure shown below:The exception is the amino acid proline, whose side chain is bonded to the nitrogen atom to give a cyclic imino acid called proline:
Because each side chain group attached to C is different (except for glycine, in which R=H), C is asymmetric and, in nature, is always the L-enantiomer.R=H (Gly), CH3 (Ala), etc.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 ConfigurationThe Peptide Bond
With few exceptions, peptide groups assume a trans-conformation, in which successive C 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 bond. During rotation around the CN 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 CN bond (-torsion angle) and CC bond (-torsion angle) of each of its amino acid residues.xCarbohydrates Huge Structural Diversity, Low Chemical Diversity
Oligosaccharides: Multiple Linkage Positions and Configurations
The same two amino acids 1 possible peptideThe same two monosaccharides 20 possible disaccharides
Avian Flu ReceptorHuman Flu Receptor-(2-3)-Galversus-(2-6)-GalGlc--(1-4)-Glc (Starch)Glc--(1-4)-Glc (Cellulose)BacteriaToxinsVirusesOther CellsHormonesCellCarbohydrate-Protein Interactions in Cell Life and Death
Electrostatic Interactions (Hydrogen-bonds, charge-charge, charge-dipole, dipole-dipole)Dispersive Interactions (Van der Waals attractions and repulsions)
H < 0 reaction is exothermic, tells us nothing about the spontaneity of the reaction H > 0 reaction is endothermic, tells us nothing about the spontaneity of the reaction Examples:
Oxidation of glucose: C6H12O6 + 6O2 6CO2 + 6H2O H = -2803 kJmol-1Just because a reaction is exothermic (that is because H < 0) does not mean that it is spontaneous.
And what about
Dissolving salt: NaCl(s) + H2O(l) Na+(aq) + Cl-(aq) H = 4 kJmol-1Just because this reaction is endothermic (H > 0) does not mean that it doesnt happen.
Enthalpy alone is not sufficient to decide whether a reaction will occur. The missing factor is called Entropy or S.
Solute Related (conformational entropy)Solvent Related (ligand and receptor desolvation)
S < 0 reaction leads to order, tells us nothing about the spontaneity of the reaction S > 0 reaction leads to disorder, tells us nothing about the spontaneity of the
Enthalpy (H) and Entropy (S)Remember:Greaction = Gproducts GreactantsG = H - T S, The reaction is favourable only when G < 0Ligand Binding Energy is also computed as if it were a reaction:
Ligand + Receptor Complex
GBinding = GComplex GLigand GReceptor
= (HComplex T SComplex) (HLigand T SLigand) (HReceptor T SReceptor)
There is a temptation to draw conclusions only from the structure of the complex, but:
GBinding 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 cant be trusted!
Thermodynamics of Ligand-Protein Interactions (G)
A molecule which has a weakly acidic proton (OH, NH) 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).
A typical hydrogen bond between polar uncharged groups has its maximum stability at an interatomic (A D) separation of 2.7-3.1 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 DH A angle to be linear. X-ray crystallographic studies of sugar-protein complexes can provide detailed structural information pertaining to hydrogen bonding in the binding site.DH AEnergetic Contributions to Ligand Binding
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. Typical hydrogen bonds between a sugar and a protein. B. Between a sugar and water.Hydrogen Bonds, ContinuedThe 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) = 0.00198398ln(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 isnt counting H-bonds a good measure of affinity?Effect of Loss of a Hydrogen Bond on Binding EnergiesSince 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.
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 M1.The Role of Water
Antigenic oligosaccharides from S. flexneri Y with antibody SYAJ6
Vyas, N.K., et al., Biochemistry, 2002. 41: p. 13575-13586.
Example: Effect of loss of OH on affinityVan 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 ). 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.Other Enthalpic Contributions to BindingOHOMeOEtOPrHG>0-7.6-6.2>0>0Ka----3.9 x 1053.6 x 104--------Antigenic oligosaccharides from V. cholerae with antibodyS20-4
OgawaInabaWang, J., et al., J. Biol. Chem., 1998. 273(5): p. 2777-2783.Af