Biochemistry II: Binding of ligands to a macromolecule(or the secret of life itself...)

Karsten RippeKirchoff-Institut für PhysikMolekulare BiophysikIm Neuenheimer Feld 227Tel: 54-9270e-mail: Karsten.Rippe@kip.uni-heidelberg.de

• http://www.dkfz.de/kompl_genome/rippe/pdf-files/Rippe_Futura_97.pdf

• Principles of physical biochemistry, van Holde, Johnson & Ho, 1998.

• Slides available at http://www.dkfz.de/Macromol/lecture/index.html

The secret of life, part I

“The secret of life is molecular recognition; the

ability of one molecule to "recognize" another

through weak bonding interactions.”

Linus Pauling at the 25th anniversary of the Institute of

Molecular Biology at the University of Oregon

The secret of life, part II

A new breakthrough in biochemical research... and paradise is to be mine! It was strange the way it happened... suddenly you get a break... and whole pieces seem to fit into place ... and that's how I discovered the secret to life... itself!

Frank’n Further in the Rocky Horror Picture Show

The secret of life, part III

In the end it is nothing but equilibrium binding and kinetics

Karsten Rippe, Biochemistry II Lecture

Binding of dioxygen to hemoglobin(we need air ...)

Binding of Glycerol-Phosphate to triose phosphate isomerase(we need energy...)

Complex of the HIV protease the inhibitor SD146(we need drugs ...)

Antibody HyHEL-10 in complex with Hen Egg White Lyzoyme

(we need to protect ourselves ...)

X-ray crystal structure of the TBP-promoter DNA complex(transcription starts here...)

Nikolov, D. B. & Burley, S. K. (1997). RNA polymerase II transcription initiation: a structural view. PNAS 94, 15-22.

Structure of the tryptophan repressor with DNA(what is this good for?)

Biochemistry II: Binding ofligands to a macromolecule

• General description of ligand binding– the esssentials

– thermodynamics

– Adair equation

• Simple equilibrium binding– stoichiometric titration– equilibrium binding/dissociation constant

• Complex equilibrium binding– cooperativity

– Scatchard plot and Hill Plot

– MWC and KNF model for cooperative binding

The mass equation law for binding of a protein P to its DNA D

Dfree+Pfree

→← DP K1=Dfree⋅Pfree

DP

binding of the first proteins with the dissociation constant K1

Dfree, concentration free DNA; Pfree, concentration free protein

binding constantKB = 1dissociation constantKD

What is the meaning of the dissociation constant forbinding of a single ligand to its site?

2. KD gives the concentration of ligand that saturates 50% of the

sites (when the total sit concentration ismuch lower than KD)

3. Almost all binding sites are saturated if the ligand

concentration is 10 x KD

1. KD is a concentration and has units of mol per liter

4. The dissociatin constant KD is related to Gibbs free energy

∆G by the relation ∆G = - R T KD

KD values in biological systems

Allosteric activators of enzymes e. g. NAD have KD 0.1 µM to 0.1 mM

Site specific binding to DNA KD 1 nM to 1 pM

Movovalent ions binding to proteins or DNA have KD 0.1 mM to 10 mM

Trypsin inhibitor to pancreatic trypsin protease KD 0.01 pM

Antibody-antigen interaction have KD 0.1 mM to 0.0001 pM

What is ∆G? The thermodynamics of a system

• Biological systems can be usually described as having constant pressure P and constant temperature T– the system is free to exchange heat with the surrounding to remain

at a constant temperature– it can expand or contract in volume to remain at atmospheric

pressure

Some fundamentals of solution thermodynamics

• H is the enthalpy or heat content of the system, S is the entropy of the system

• a reaction occurs spontaneously only if ∆G < 0• at equilibrium ∆G = 0• for ∆G > 0 the input of energy is required to drive the reaction

ΔG=ΔH −TΔSG≡H −TS

• At constant pressure P and constant temperature T the system is described by the Gibbs free energy:

the problem

In general we can not assume that the total free energy G of a

solution consisting of N different components is simply the sum of the

free energys of the single components.

Why is it necessary to use partial specific orpartial molar quantities?

the problem: In general we can not assume that the total volume V of a

solution consisting of N different components is simply the sum of the

volume of the single components. For example this is the case for mixing

the same volumes of ethanol and water.

V≠ ni⋅Vi0

i=1

N

∑ ni: moles of component i

Vi0: molar volume of pure component i

v i =∂v∂mi

Vi =∂V∂ni

partial specific volume= increase of volume upon addition of mass

partial molar volume= increase of volume upon addition of mol particles

∂m

∂n

the solution: use of partial specific or partial molar quantities

Increase of the solution volume upon adding solute

Solute added (moles)

Vol

ume

of s

olut

ion

V i =∂V∂ni

real solutions

volumes additive

slope =

n1⋅V 10

V 2

V 20

volume ofpure solvent

slope at indicated concentration =

V= ni⋅V ii=1

N

∑The total volume of the solution is thesum of the partial molar volumes:

The chemical potential µ of a substance is the partial molar Gibbs free energy

G i =∂G∂ni

=μi G= ni⋅μi

i=1

n

∑

μi =μi0+RTlnCifor an ideal solution it is:

μi =μi0 forCi =1 mol/ l

Ci is the concentration in mol per liter

is the chemical potential of a substance at 1 mol/lμi0

Changes of the Gibbs free energy ∆G of an reaction

ΔG=G(final state)−G(initial state)

from μi =μi0+RTlnCi it follows:

aA+bB+... →

← gG+hH...

ΔG=gμG +hμH +...−aμA −bμB −...

ΔG=gμG0 +hμH

0 +...−aμA0 −bμB

0−...+RTln[G]g[H]h...[A]a[B]b...

ΔG=ΔG0+RT ln[G]g[H]h...[A]a[B]b...

∆G of an reaction in equilibrium

0=ΔG0+RT ln[G]g[H]h...[A]a[B]b...

⎛

⎝ ⎜

⎞

⎠ ⎟

Eq

ΔG0=−RTln[G]g[H]h...[A]a[B]b...

⎛

⎝ ⎜

⎞

⎠ ⎟

Eq

=−RT lnK

K =[G]g[H]h...[A]a[B]b...

⎛

⎝ ⎜

⎞

⎠ ⎟

Eq

=exp−ΔG0

RT⎛

⎝ ⎜

⎞

⎠ ⎟

aA+bB+... →

← gG+hH...

Titration of a macromolecule D with n binding sitesfor the ligand P which is added to the solution

ν = [boundligand P][macromolecule D]

ΔXΔXmax

=νn

=θ (fraction saturation)

free ligand Pfree (M)

degr

ee o

f bi

ndin

g

n n binding sitesoccupied

∆Xmax

∆X

0

Analysis of binding of RNAP·54 to a promoter DNA sequenceby measurements of fluorescence anisotropy

Rho

Rho

+

RNAP·54

promoter DNA

RNAP·54-DNA-Komplex

Kd

free DNA with a fluorophorewith high rotational diffusion

-> low fluorescence anisotropy rmin

RNAP-DNA complexwith low rotational diffusion

-> high fluorescence anisotropy rmax

θ =Ptot[ ]

Ptot[ ]+KD

=rmeasured−rmin

rmax−rmin

How to measure binding of a protein to DNA?One possibility is to use fluorescence anisotropy

z

y

x sample

I

verticalexcitation

filter/mono-chromator

polarisatorIII

filter/monochromator

polarisator

measured fluorescence

emission intensity

r =III −I⊥

III +2I⊥

Definition of fluorescenceanisotropy r

The anisotropy r reflects the rotational diffusion of a fluorescent species

0

0.2

0.4

0.6

0.8

1

0.01 0.1 11 0 100 1000

nifH nifLglnAp2

RNAP 54 (nM)

200 mM KAc

Measurements of fluorescence anisotropy tomonitor binding of RNAP·54 to different promoters

0

0.2

0.4

0.6

0.8

1

0.01 0.1 11 0 100 1000

50 mM 150 250 mM

RNAP 54 (nM)

Vogel, S., Schulz A. & Rippe, K.

Ptot = KD

= 0.5 = 0.5

Ptot = KD

Example: binding of a protein P to a DNA-fragment D with one or two binding sites

Dfree+Pfree

→← DP K1=Dfree⋅Pfree

DPbinding of the first proteins with

the dissociation constant K1

Dfree, concentration free DNA; Pfree, concentration free protein;

DP, complex with one protein; DP2, complex with two proteins;

binding constantKB = 1dissociation constantKD

DP+Pfree

→← DP2 K2=DP⋅Pfree

DP2

binding of the second proteins withthe dissociation constant K2

D+2Pfree

→← DP2 K 2

* =Dfree⋅Pfree2

DP2

K 2* =K1⋅K2 alternative expression

Definition of the degree of binding

for n binding sites (Adair equation)

ν =

i⋅1Ki

⋅Dfrei⋅Pfreii

i=1

n

∑1K i

⋅Dfrei⋅Pfreii

i=0

n

∑=

i⋅1K i

⋅Pfreii

i=1

n

∑1Ki

⋅Pfreii

i=0

n

∑mitK0=1

ν1= DPDfree+DP

ν2= DP+2⋅DP2Dfree+DP+DP2

ν = [boundligand P][macromolecule D]

degree of binding for one binding site for two binding sites

Binding to a single binding site: Deriving an expressionfor the degree of binding or the fraction saturation

KD =Dfree⋅Pfree

DPDfree+Pfree

→← DP

ν1=θ =

1KD

⋅Pfree

1+1

KD

⋅Pfree

⇔ ν1=θ= Pfree

KD+Pfree

from the Adair equation we obtain:

ν1=Dtot+Ptot+KD− Dtot+Ptot+KD( )

2−4⋅Dtot⋅Ptot

2⋅Dtot

Pfree=Ptot−ν1⋅Dtot

Often the concentration Pfree can not be determined but the total concentration of added protein Ptot is known.

Stoichiometric titration to determinethe number of binding sites

To a solution of DNA strands with a single binding site small amounts of protein P are added. Since the binding affinity of the protein is high (low KD value as compared to the total

DNA concentration) practically every protein binds as long as there are free binding sites on the DNA. This is termed “stoichiometric binding” or a “stoichiometric titration”.

or

0

0.2

0.4

0.6

0.8

1

0 1·10 -10

Ptot (M)

equivalence point1 protein per DNA

2·10 -10

Dtot = 10-10 (M)

KD = 10-14 (M)

KD = 10-13 (M)

KD = 10-12 (M)

νn

=θ

for n=1

ν =θ

Binding to a single binding site. Titration of DNA with aprotein for the determination of the dissociation constant KD

ν1= Pfree

Pfree+KD

≈ Ptot

Ptot+KD

if Pfree≈Ptot d. h. 10×Dtot≤KD

Ptot (M)

or

0

0.2

0.4

0.6

0.8

1

0 2 10-9 4 10-9 6 10-9 8 10-9 1 10-8

Ptot = KD

or = 0.5

Dtot = 10-10 (M)

KD = 10-9 (M)

KD = 10-9 (M)

ν1= Pfree

Pfree+KD

ν1= Ptot

Ptot+KD

Increasing complexity of binding

all binding sites areequivalent and independent

cooperativity heterogeneity

all binding sites areequivalent and not independent

cooperativityheterogeneity

all binding sites arenot equivalent and not independent

all binding sites areindependent but not equivalent

simple

difficult

verydifficult

Binding to n identical binding sites

ν1= Pfree

Pfree+KDbinding to a single binding site

νn = n⋅Pfree

kD+Pfree

binding to n independent andidentical binding sites

D+n⋅Pfree

→← DPn Kn =Dfree⋅Pfree

n

DPn

νn = n⋅Pfreen

Kn+Pfreen

strong cooperative binding to n identical binding sites

νn = n⋅PfreeαH

KαH +PfreeαH

approximation for cooperative binding ton identical binding sites, H HiIl coefficient

Difference between microscopic andmacroscopic dissociation constant

1 2

1 2 1 2

1 2

kD kD

kD kD

Dfree

DP

DP2

K1

K2

=kD 22⋅kD

=14

K1=kD

2

K2 =2⋅kD

microscopic binding macroscopic binding

2 possibilities for the formation of DP

2 possibilities for the dissociation of DP2

Cooperativity: the binding of multiple ligandsto a macromolecule is not independent

2

P free (M)

0

0.5

1

1.5

2

0 2 10-9 4 10-9 6 10-9 8 10-9 1 10-8

Adair equation: ν2= K2⋅Pfree+2⋅Pfree2

K1⋅K2+K2⋅Pfree+Pfree2

independent bindingmicroscopic binding constantkD = 10-9 (M)

macroscopic binding constantsK1 = 5·10-10 (M); K2 = 2·10-9 (M)

cooperative bindingmicroscopic binding constantkD = 10-9 (M)

macroscopic binding constantsK1 = 5·10-10 (M); K2 = 2·10-10 (M)

Logarithmic representation of a binding curve

P free (M)

0

0.5

1

1.5

2

10-11 10-10 10-9 10-8 10-7

2

• Determine dissociation constants over a ligand concentration of at least three orders of magnitudes• Logarithmic representation since the chemical potential µ is proportional to the logarithm of the concentration.

independent bindingmicroscopic binding constantkD = 10-9 (M)

macroscopic binding constantsK1 = 5·10-10 (M); K2 = 2·10-9 (M)

cooperative bindingmicroscopic binding constantkD = 10-9 (M)

macroscopic binding constantsK1 = 5·10-10 (M); K2 = 2·10-10 (M)

Visualisation of binding data - Scatchard plot P

fre

i

0

1 109

2 109

3 109

0 0.5 1 1.5 2

intercept = n/kD

slope = - 1/kD

intercept = n

νn = n⋅Pfree

kD+Pfree

⇔ νn

Pfree

= nkD

−νn

kD

independent bindingmicroscopic binding constantkD = 10-9 (M)

macroscopic binding constantsK1 = 5·10-10 (M); K2 = 2·10-9 (M)

cooperative bindingmicroscopic binding constantkD = 10-9 (M)

macroscopic binding constantsK1 = 5·10-10 (M); K2 = 2·10-10 (M)

Visualisation of binding data - Hill plot

ν2= K2⋅Pfree+2⋅Pfree2

K1⋅K2+K2⋅Pfree+Pfree2

⇔ ν2

2−ν2

= θ1−θ

= K2⋅Pfree+2⋅Pfree2

2⋅K1⋅K2+K2⋅Pfree

Pfree→ 0 ⇒ logν2

2−ν2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =logPfree( )−log2K1( )

Pfree→ ∞ ⇒ logν2

2−ν2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =logPfree( )−log

K2

2⎛ ⎝ ⎜ ⎞

⎠ ⎟

Pfree≈K ⇒ logν2

2−ν2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =αH ⋅logPfree( )−αH ⋅logK( )

νn = n⋅PfreeαH

KαH +PfreeαH

log (

2–)

]= lo

g (

1–)

]

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

10-11 10-10 10-9 10 -8

strongbinding

limitslope H =1.5

log (P free)

weakbinding

limit 2·K1 K2/2

Binding of dioxygen to hemoglobin

The Monod-Wyman-Changeau (MWC)model for cooperative binding

+P

T0 T1 T2

R0 R1 R2

L0 L1 L2

kT

+P

kR

+P

kR

+P

kT

T conformation (all binding sites

are weak)

R conformation (all binding sites

are strong)

• in the absence of ligand P the the T conformation is favored

• the ligand affinity to the R form is higher, i. e. the dissociation constant kR< kT.

• all subunits are present in the same confomation

• binding of each ligand changes the T<->R equilibrium towards the R-Form

P P P P P

P

P P

P P

+P

kT

+P

kT

T3 T4

P P P P P

P P

P P

P

L3 L4

+P

kR

+P

kR

R3 R4

The Koshland-Nemethy-Filmer (KNF)model for cooperative binding

+P

k1

+P

k2

• Binding of ligand P induces a conformation change in the subunit to which it binds

from the into the -conformation (“induced fit”).

• The bound ligand P facilitates the binding of P to a nearby subunit

  in the -conformation (red), i. e. the dissociation constant k2 < k’2.

• subunits can adopt a mixture of confomations.

-conformation-conformation(induced by ligand binding)

+P

P

P

k’2

P PP

-conformation(facilitated binding)

Summary

• Thermodynamic relation between ∆G und KD

• Stoichiometry of binding

• Determination of the dissociation constant for simple systems

• Adair equation for a general description of binding

• Binding to n binding sites

• Visualisation of binding curves by Scatchard and Hill plots

• Cooperativity of binding (MWC and KNF model)