Bacteria  

Toxins  

Viruses  

Other  Cells  

Hormones  

Cell  

Carbohydrate-­‐Protein  interac;ons  are  Cri;cal  in  Life  and  Death  

How  to  Model  Protein-­‐ligand  interac;ons?  

Protein  –  Protein  Protein  –  DNA/RNA  

Protein  –  Carbohydrate  Protein  –  Drug  

An;body  –  An;gen  Enzyme  –  Substrate  

 

What  do  you  want  to  know?  

•  The  3D  structure  of  the  complex!    •  But  why  do  you  need  this?  •  To  iden;fy  cri;cal  interac;ons!  •  But  why?  

–  To  understand  the  mechanism!  •  How  will  you  prove  it?  •  Compare  to  experimental  data  (such  as  from  muta;ons)!  

–  To  design  an  inhibitor!  •  Must  be  able  to  compute  interac;on  energies  •  What  level  of  accuracy  is  required?    

–  To  guide  protein  engineering!  •  Do  you  need  to  know  interac;on  energies?  •  What  level  of  accuracy  is  required?  

How  to  generate  the  ini;al  model?  

Co-­‐Complex  Source   Suitable  for    

X-­‐ray  structure  of  the  protein  -­‐  ligand  complex  (so  called  “co-­‐complex”)    

Guiding  ligand  design  Insight  into  binding  mechanism  Guiding  protein  engineering  

X-­‐ray  structure  of  the  protein  with  docking  of  the  ligand    

Insight  into  binding  mechanism  Guiding  protein  engineering  experiments    

Homology  model  of  the  protein  with  docking  of  the  ligand  

Guiding  protein  engineering  experiments  

Lower  Accuracy  

Higher  Accuracy  

Amino  Acids:  High  Chemical  Diversity,  Low  Structural  Diversity  

CαH3N+

H

CO2-

R

H2CCH2

CH2

CO2-

H

H2N+ Cα

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:

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 Cα is different (except for glycine, in which R=H), Cα 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.

O

OH

O

OH

O

OH

O

OH

O

OH

HO

HOOH

OH

HOOH

HO

HO

OH

OH OH

OH

OH

OH

O

O

OH

O

OH

OH

O

O

OH

O

OH

OH

O

O

OH

O

OH

αβ1

23

4 5

6

123

4 56

β-D-glucopyranose, β-D-Glc

β-D-pyranose α-D-pyranose

β-D-galactopyranose, β-D-Gal

β-D-mannopyranose, β-D-Man

1-4 linkage (α)

1-3 linkage (β)

1-6 linkage (β)

1

1

1

4

3

6

O

OHHO

HONHAc

OH

2-N -acetyl-β-D-glucopyranose, β-D-GlcNAc

O

CO2-

OH

HOHO

AcHN

OHHO

5−N -acetyl-α-neuraminic acid, α-Neu5Ac

5

1

2

34

67

89

Carbohydrates:  High  Structural  Diversity,  Low  Chemical  Diversity  

O

O

HO

HOOH

OHO

O

O

OO

O

α/β

α/β

The same two amino acids à 1 possible peptide The same two monosaccharides à 20 possible disaccharides

O

OH

AcHN

OH

HO CO2-

HO

OHO

OH

O

OH

O-GlcNAc2

3

O

OH

AcHN

OH

HO CO2-

HOO

OH

HO

OH

O-GlcNAc

O2 6

Avian Flu Receptor Human Flu Receptor

α-(2-3)-Gal versus α-(2-6)-Gal

Glc-α-(1-4)-Glc (Starch) Glc-β-(1-4)-Glc (Cellulose)

Oligosaccharides:  Mul;ple  Linkage  Posi;ons  and  Configura;ons  

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) 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

Enthalpy  (ΔH)  and  Entropy  (ΔS)  

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!

Thermodynamics  of  Ligand-­‐Protein  Interac;ons  (ΔG)  

1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75O...H Separation (Angstroms)

-6

-4

-2

0

2

4

6

8

10

Tota

l Ene

rgy

(kca

l/mol

)

Total EnergyDipole/Dipolevan der Waals

Hydrogen Bond Energy

roh

R"

HO

H

R'

O

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.

X-ray crystallographic studies of sugar-protein complexes can provide detailed structural information pertaining to hydrogen bonding in the binding site.

D—H ··· ··· ··· ··· A  

Energe;c  Contribu;ons  to  Ligand  Binding  

NH2H2N

N

N

H

NH2 OOHO

HO OHOH

OH HO

NH3+

H

+

O

O

-+

H

O

H

H

O

H

H

O

H

OHO

HO OHOH

OH

H

O

H

H

O

H

HO

H

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

A. H-bonds between a sugar and a protein. B. H-bonds between a sugar and water.

Hydrogen  Bonds,  Con;nued  

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 ΔGBinding (L1) ≅ -20 kcal/mol

L2 + Receptor → Complex2 ΔGBinding (L2) ≅ -15 kcal/mol

ΔΔG = ΔGBinding (L1) – ΔGBinding (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(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?

Effect  of  Loss  of  a  Hydrogen  Bond  on  Binding  Energies  

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.

H O

HH

OH

O

OHOH

HOHO

OH

HOH

HO

H

H

OH

OH

H

OH

OH

++

H

OH

H

OH

H

OH

H

OH

H

OH

H

OH

ΔΗ = ?

O

OHOH

HOHO

OH

OHOH

ΔS > 0

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~103 M-1.

The  Role  of  Water  in  Ligand  Binding  Thermodynamics  

Bacterial surface 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  Affinity  

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). VdW energies decrease with distance as a function of 1/r6. In a ligand protein complex there may be many such interactions, and although each one provides very little energy, their sum may be significant.

R"

HO

H

R'

O

roh

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, ligand specificity depends very highly on van der Waals contacts.

Other  Enthalpic  Contribu;ons  to  Binding  

The maximum energetic contribution from vdw interaction is small (only about 0.2 kcal/mol) per interacting atom, but can add up to a significant contribution

OH OMe OEt OPr H

ΔG >0 -7.6 -6.2 >0 >0

KA ---- 3.9 x 105 3.6 x 104 ---- ----

Bacterial surface oligosaccharides from V. cholerae with antibodyS20-4

Wang, J., et al., J. Biol. Chem., 1998. 273(5): p. 2777-2783.

Affinities of Vibrio cholerae binding to mAb S20-4

Too  few  favorable  interac;ons,  or  too  many  unfavorable  ones,  will  hurt  binding  

Example:  Effect  of  loss  of  van  der  Waals  and  hydrophobic  contacts  on  affinity  

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.

H O

HH

OH

O

OHOH

HOHO

OH

HOH

HO

H

H

OH

OH

H

OH

OH

++

H

OH

H

OH

H

OH

H

OH

H

OH

H

OH

ΔΗ = ?

O

OHOH

HOHO

OH

OHOH

ΔS > 0

This release of water results in an increase in the entropy of the system, i.e. ΔSBinding > 0 and so -TΔSBinding < 0.

But the desolvation free energy may still be unfavorable (>0) depending on ΔH

ΔGDesolvation = ΔHDesolvation - TΔSDesolvation

Effect  of  Entropy  on  Binding  Energies  

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 “lower” face of the sugar:

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.

How does the hydrophobic effect differ from van der Waals contacts? How does it differ from orbital overlap?

The  Hydrophobic  “Effect”  

Process ΔH

kJ/mol - TΔS

kJ/mol ΔG

kJ/mol CH4 in H2O versus CH4 in C6H6 11.7 -22.6 -10.9

C2H6 in H2O versus C2H6 in C6H6 9.2 -25.1 -15.9

C2H4 in H2O versus C2H4 in C6H6 6.7 -18.8 -12.1

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. It is enthalpically more favorable for small hydrocarbons to dissolve in water than in large non-polar solvents!

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.

The  Origin  of  Hydrophobicity  is  Entropic!  

Entropy may also change as a function of the properties of the ligand or the protein. In flexible ligands, particularly oligosaccharides and polysaccharides, binding reduces the flexibility (entropy) of the ligand, which disfavors binding.

Thus, certain regions of the ligand (or protein) may introduce unfavorable entropies upon binding.

Conforma;onal  Entropy  

Carbohydr  .  Res.,  (2005)  340,  1007  PNAS  ,  (2006)  103,  8149  

Changes  in  Flexibility  Affect  Binding  Energy  

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.

Simula;ons  Can  Quan;ty  Each  Energy  Contribu;on