Lecture 9: Theory of Non-Covalent Binding Equilibria

Dr. Ronald M. Levyronlevy@temple.edu

Statistical Thermodynamics

General chemical reaction: ν A A+νB B νC C+ν D D

Very important type of “reaction”: bimolecular non-covalent binding

R(sol) + L(sol) RL(sol)

Small molecule dimerization/association Supramolecular complexes Protein-ligand binding Protein-protein binding/dimerization Protein-nucleic acids interactions ...

*Note*: We are implicitly assuming above that we can describe the system as being composed of 3 distinct chemical “species”, R, L, and RL (quasi-chemical description).

If interactions between R and L are weak/non-specific then it would be more appropriate to treat the system as a non-ideal solution of R and L.

Based on the earlier results for general reactions:

K b (T )=Bindingconstant

=e−Δ Ab∘ (T )/ kT=

(C RL /C∘ )eq.

(C R /C∘ )eq. (C L /C

∘ )eq.

ϕ i (T )=1

∏j

nu

Λuj3nu

∫ d xu3nu−3 e−βΨ i( xu)

Ψ i ( x u) Effective potential energy of solute i in solution.

ν A=ν B=1 ; νC=1 ; νD=0

K b(T )=(V∘)νC+νB−νA−νB

ϕ C (T )νCϕ D(T )

νD

ϕ A (T )νA ϕ B(T )

νB=C ∘ ϕRL (T )

ϕ R (T )ϕ L(T )

If the solution is isotropic ( invariant upon rotation of solute),

ϕ i (T )=8π2ϕ θ i=8π2

∏j

nu

Λuj3nu

∫ d xu3nu−6 e−βΨ i(x u)=

8π2 Z ' i

∏j

nu

Λuj3nu

Ψ i ( x u)integrate analytically over rotational degrees of freedom (ignoring roto-vibarational couplings, OK at physiological temperatures.

Internal coordinates

When inserting into the expression for Kb(T), Λ's cancel because nRL=nR+nL

We get:

K b (T )=C ∘

8 π2

Z ' RL

Z ' R Z ' L

Z ' R=∫ d x R3n R−6 e−βΨ R (x R) Z ' L=∫ d x L

3n L−6 e−β Ψ L(x L)

Z ' RL=???

[Gilson et al. Biophysical Journal 72, 1047-1069 (1997)]

In the complex, RL, the “external” coordinates (translations, rotations) of the ligand become internal coordinates of the complex.

Z ' RL=∫"BOUND"d xR

3n R−6 d x L3n L−6 d ζ L

6 e−βΨ RL(xR , x L , ζ L)

R

L

ζ L6= {(r ,θ ,ϕ ) ,(ω1,ω2,ω3)}

Position of the ligandrelative to receptor frame

Orientation of the ligandrelative to receptor frame

External coordinates of the ligandrelative to receptor

It is up to us to come up with a reasonable definition of “BOUND”. That is we need to define the RL species before we can compute its partition function. The binding constant will necessarily depend on this definition. Must match experimental reporting.If the binding is strong and specific the exact definition of the complexed state is often not significant.

It is convenient to introduce an “indicator” function for the complex:

I (ζL)={1 ligand is in binding site0 ligand is outside binding site

Z ' RL=∫"BOUND"…=∫ d x R

3n R−6 d xL3n L−6 d ζ L

6 I (ζ L)e−β ΨRL (xR , x L , ζ L)

then:

Next, define “binding energy” of a conformation of the complex:

u=u (xR , x L , ζL )=Ψ RL(x R , x L , ζ L)−Ψ R ( xR )−Ψ R (x L)

basically, change in effective energy for bringing ligand and receptor together at fixed internal conformation:

+

In terms of binding energy:

then:

Ψ RL (xR , x L , ζ L)=Ψ R (x R)+Ψ R ( xL)+u (xR , xL , ζ L)

K b(T )=C∘

8π2

Z 'RL

Z 'R Z 'L

=

C∘

8π2

∫ d xR3nR−6 d xL

3nL−6 d ζL6 I (ζL)e

−βΨR (xR)e−βΨ L(xL)e−βu (xR , xL , ζL)

∫ d xR3nR−6 d xL

3nL−6 e−βΨR(xR)e−βΨ L(xL)

Now: we are not very good at computing partition functions. We are much better at computing ensemble averages:

⟨O ⟩=∫ dx O ( x )e−βU (x )

∫ dx e−βU (x )=∫ dx O ( x )ρ( x )

To transform the expression for Kb so that it looks like an average:

multiply and divide by: ∫ d ζ L6 I ( ζ L)=V siteΩsite

(Ω site=8 π 2 if any ligand orientation is considered "bound" )

K b(T )=C ∘

8π2

Z 'RL

Z ' R Z ' L

=

C ∘

8π2

∫ d x R3n R−6 d x L

3n L−6 d ζL6 I (ζL)e

−βΨ R(x R)e−βΨ L( xL)e−βu( x R , x L , ζL)

∫ d x R3nR−6 d x L

3nL−6 e−βΨ R(x R) e−βΨ L(x L)=

V site

V ∘Ωsite

8π2

∫ d x R3n R−6 d x L

3n L−6 d ζL6 I (ζL)e

−βΨ R(x R) e−βΨ L(x L)e−βu( xR , x L , ζL)

∫ d x R3nR−6 d x L

3nL−6 d ζL6 I (ζL)e

−βΨR( xR)e−βΨ L (x L)

then:

or:

K b(T )=V site

V ∘Ωsite

8 π2⟨e−βu(xR , xL ,ζ L) ⟩R+ L+ I ()

We can see that binding constant can be expressed in terms of an average of the exponential of the binding energy over the ensemble of conformations of the complex in which the ligand and the receptor are not interacting while the ligand is placed in the binding site.

K b(T )=V site

V ∘Ωsite

8 π2⟨e−βu(xR , xL ,ζ L) ⟩R+ L+ I ()

Standard free energy of binding: Δ Ab∘ (T )=−kT ln K b (T )

Δ Ab∘ (T )=−k T ln

V site

V ∘ −k T lnΩsite

8 π2−k T ln ⟨e−βu (x R , xL , ζ L) ⟩R+ L+ I ( )

Δ Ab∘ ( translational) Δ A b ( rotational) Δ Ab (excess)

(analytic formulas) (numerical computation)

Interpretation in terms of binding thermodynamic cycle:

RLR + L

R(L)“Virtual” state in which ligand is in binding site without interacting with receptor

Ligand in solutionat concentration Cº

Ligand bound to receptor

Loss of translational, rotational freedom(to fit binding site definition)

Binding while in receptor site(independent of concentration)

Δ Ab∘ ( t+ r )

Δ Ab (exc.)

Δ Ab∘

Δ A b∘=Δ A b

∘ ( t+ r )+Δ A b (exc.)

A future lecture (BEDAM) and computer exercise will focus on the computation of by computer simulations. Δ Ab (exc.)

Entropy/Enthapy of binding:

Δ S b∘=−

∂Δ Ab∘ (T )

∂TStandard entropy of binding

Δ Ab∘ (T )=−k T ln

V site

V ∘ −k T lnΩsite

8 π2+Δ Ab (exc.)

From the definition of entropy:

with

get

Δ S b∘ (T )=k ln

V site

V ∘Ωsite

8 π2−∂Δ Ab(exc.)

∂T

the first term is the entropy loss due to to the loss of translational and rotational freedom of the ligand:

Δ S b , t+ r∘ =k ln

V site

V ∘Ωsite

8 π2

Note that it's temperature-independent. It depends only only on the standard concentration and the definition of the binding site.

Δ S b(exc.)=−∂Δ Ab(exc.)

∂T

Now let's turn to the “interesting” piece:

recall that:

Δ Ab(exc.)=−k T ln ⟨e−βu ⟩R+L+ I ( )=−k T lnZ ' RL

Z ' R+L+ I ( )where:

is the binding energy (think of it as the ligand-receptor interaction energy) of a given conformation of the complex (U is the potential energy and W is the solvent potential of mean force ~solvation free energy).

and: ⟨…⟩R+ L+ I ()

denotes an average over the ensemble of conformations of the complex in which the receptor and ligand are not interacting, but the ligand is restrained to within the binding site.

u=u ( x R , x L , ζ L ;T )=Ψ RL ( x R , x L , ζ L ;T )−Ψ R ( x R ; T )−Ψ R ( x L ; T )

Ψ(x ;T )=U (x )+W (x ;T ):with:

−∂Δ Ab(exc.)

∂T=k ln

Z 'RL

Z 'R+L+ I ( )+kT ∂

∂Tln

Z 'RL

Z 'R+L+ I ( )=

−Δ Ab(exc.)

T− 1

T [ 1Z 'RL

∂Z 'RL

∂β − 1Z 'R+L+ I ( )

∂ Z 'R+L+ I ()

∂β ]

so:

Δ Ab (exc.)=−k T ln ⟨e−βu ⟩R+L+I ()=−k T lnZ 'RL

Z 'R+L+ I ( )

(k T ∂∂T

=− k TK T 2

∂∂β=−

1T∂∂β )generally:

1Z∂Z∂β =

1Z∂∂β∫ dx e−βU (x )=−

∫ dx U (x )e−βU (x )

Z=−⟨U ⟩

so:

−∂Δ A b (exc.)

∂ T=−

Δ A b(exc.)T

+ 1T [ ⟨Ψ ⟩

RL− ⟨Ψ ⟩R+L+ I ( ) ]=−

Δ A b(exc.)T

+ Δ ⟨Ψ ⟩T

Nice, but … WRONG!

… because Ψ is temperature-dependent!

1Z∂Z∂β =

1Z∂∂β∫dx e−βΨ(x ;T )=−

∫ dx [Ψ+β ∂Ψβ ]e−βΨ (x ; T )

Z

Only the solvent potential of mean force part, W, of Ψ is temperature-dependent:

The derivative has an additional piece:

=− ⟨Ψ ⟩−β ⟨ ∂Ψ∂β ⟩=− ⟨Ψ ⟩+T ⟨ ∂Ψ∂T ⟩

1Z∂Z∂β=−⟨U+W ⟩+T ⟨ ∂W

∂T ⟩So we get:

Δ Sb (exc.)=−∂Δ A b(exc.)

∂T=−

Δ A b(exc.)T

+ Δ ⟨U+W ⟩T

−Δ ⟨ ∂W∂T ⟩

Δ ⟨ ∂W∂T ⟩=⟨ ∂W

∂T ⟩RL

−⟨ ∂W∂T ⟩

R+L+ I ()

where:

we finally obtain the following formula for the standard entropy of binding:

Including the ideal term:

Δ S b∘=−

Δ Ab∘

T+Δ ⟨U +W ⟩

T⏟Δ S configurational

∘

−Δ ⟨ ∂W∂T ⟩⏟

Δ S solvation

Δ S b , t+ r∘ =k ln

V site

V ∘Ωsite

8 π2

The entropy of binding measures the reduction of conformational freedom of receptor/ligand upon binding + the change of entropy of the solvent (hydrophobicity, etc.).

Δ Sb∘=Δ S conf.

∘ +Δ S solv.

The total binding entropy can be measured from, for example, the temperature variation of the binding free energy. However, the conformational and solvation contributions can not be measured individually (although people have designed NMR experiments to try to probe ) Δ S conf.

∘

Energy of binding (=enthalpy of binding apart from pΔV, which is often negligible):

Change in internal energy due to receptor-ligand interactions.

Δ E b=Δ Ab∘ +T Δ S b

∘

From the definition of free energy ...

substituting ...

Δ S b∘=−

Δ Ab∘

T+Δ ⟨Ψ⟩

T−Δ ⟨ ∂W

∂T ⟩we get:

Δ Eb=Δ⟨Ψ⟩−T Δ ⟨ ∂W∂T ⟩

Change in internal energy of the solvent.

Note that does not depend on standard state concentration.Δ E b

It follows that the effective energy and configurational entropy pair is an equally valid decomposition of the free energy as the energy/entropy pair.

Δ A b∘=Δ E b−T Δ S b

∘=Δ ⟨Ψ ⟩−T Δ S conf.∘

Combining the formulas for and we see that the solvation term cancels in the expression for the free energy of binding:

Δ E b Δ S b∘

One important distinction is that are directly measurable whereas are not.

(ΔEb ,Δ S b∘ )

(Δ ⟨Ψ⟩ ,Δ S conf.∘ )

Given that they are easier to compute are encountered more often in theoretical models than

(Δ ⟨Ψ⟩ ,Δ S conf.∘ )

(ΔE b ,Δ S b∘ )

Historical note: the fact that the solvation term cancels in the expression of the binding free energy has caused endless confusion in the literature. Some have even claimed that the solvent does not affect the enthalpy/entropy of binding, which is of course not true. The effect of the solvent is included implicitly in the values of (Δ ⟨Ψ ⟩ ,Δ S conf.

∘ )

A side note: why is

called the “configurational” entropy change?

Δ S conf∘ =−

Δ Ab∘

T+ Δ ⟨Ψ ⟩

T

− AT+ ⟨Ψ ⟩

T=k ln Z + 1

T ∫ dxΨ( x ) exp−βΨ(x )

Z=k ln Z+ 1

T ∫ dxΨ(x )ρ(x )

Consider the configurational entropy (of the solute) :

We can write:

Ψ( x)=−kT ln e−βΨ(x )=−kT lne−βΨ(x )Z

Z=−kT ln ρ( x )−kT ln Z

S conf=k ln Z−k ( ln Z )∫ dx ρ( x)−k∫ dx ρ(x ) lnρ( x )

So:

S conf

1

S conf=−k∫ dx ρ(x ) lnρ( x )Shannon's information theory definition of the entropy of a probability distribution.

S=−k∫ dxρ( x ) ln ρ( x )

Notice that to compute Sconf all we need is the distribution of conformations of the solute in solution (no energies, etc.). In that sense it is a “configurational” entropy.

The entropy expression:

Is one of the most fundamental relations of statistical physics.

Reorganization free energy of binding

Δ S b∘=−

∂Δ Ab∘ (T )

∂TStandard entropy of binding

Δ Ab∘ (T )=−k T ln

V site

V ∘ −k T lnΩsite

8 π2+Δ Ab (exc.)

Consider the following thermodynamic cycle:

with

get

Δ S b∘ (T )=k ln

V site

V ∘Ωsite

8 π2−∂Δ Ab(exc.)

∂T

the first term is the entropy loss due to to the loss of translational and rotational freedom of the ligand:

Δ S b , t+ r∘ =k ln

V site

V ∘Ωsite

8 π2

Note that it's temperature-independent. It depends only only on the standard concentration and the definition of the binding site.

loss of conformational freedom, energetic strain

translational/rotationalentropy loss

G reorg°

Δ E b

ΔG b∘

Interatomic interactions

Δ S b , t+ r∘ =k ln

V site

V ∘Ωsite

8π2

Reorganization free energy of binding

Consider the following thermodynamic cycle:

ΔG conf∘

Binding = reorganization + interaction ΔG b∘=ΔG reorg

∘ +Δ E b

Δ E b≡⟨u ⟩RL=⟨Ψ RL ( x R , x L , ζ L ; T )−Ψ R ( x R ; T )−Ψ R ( x L ; T )⟩RL

This is usually straightforward to compute from, say, a MD trajectory of the complex

If we know ΔG b∘

then ΔG reorg∘

can be obtained by difference:

ΔG reorg∘ =ΔG b

∘−Δ E b

Key facts about reorganization:

Reorganization always opposes binding (because it involves shifting the unbound conformational ensembles of binding partners away from equilibrium).

The only favorable thermodynamic driving force towards binding is the formation of receptor-ligand interactions.

The relative binding affinity of two complexes can be understood

ΔG reorg∘ >0