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1 Lecture 8 Binding and kinetics Antoine van Oijen BCMP201 Spring 2008 Donald T. Hayne Biological Thermodynamics James Goodrich, Jennifer Kugel Binding and Kinetics for Molecular Biologists
21

Binding and Kinetics

Oct 24, 2014

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Page 1: Binding and Kinetics

1

Lecture 8

Binding and kinetics

Antoine van Oijen

BCMP201 Spring 2008

Donald T. HayneBiological Thermodynamics

James Goodrich, Jennifer KugelBinding and Kinetics for Molecular Biologists

Page 2: Binding and Kinetics

2

Goals

- Quantitative measurements of biological binding reactions

- Affinities- Cooperativity in binding- Kinetics

Practical use!!!

Assays: how much is bound?

• Assays that separate complexes from a solution- Filter-binding (or cell-binding)- Gel-filtration chromatography- Electrophoretic mobility shift assays (EMSAs/ gel-shift)

• Assays that detect complexes in solution- Fluorescence (quenching, anisotropy, FRET)- Protection assays (Rnase, Dnase footprinting)

• Assays in which a biomolecule is bound- Affinity resins- Surface plasmon resonance

(More details later in the semester)

Protein-protein, protein-DNA, protein-ligand, …

Page 3: Binding and Kinetics

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Bimolecular interactions

A + B ABkon

koff

Binding is not all-or-nothing:

Portion of A and B will be bound, portion will be free

X Ykon

koff

!

d[Y]

dt= [X ] " k

on# [Y] " k

off= 0

Equilibrium

Reaction is in equilibrium when concentrations do not change:

(unimolecular reaction)

(mass action law)

Page 4: Binding and Kinetics

4

Equilibrium

Reaction is in equilibrium when concentrations do not change:

!

d[AB]

dt= [A] " [B] " k

on# [AB] " k

off= 0 (mass action law)

A + B ABkon

koff

Equilibrium is reached when:

!

[A] " [B] " kon

= [AB] " koff

Equilibrium is still dynamic!!!

Binding (bimolecular reaction):

Equilibrium dissociation constant KD

Rearrange to define equilibrium dissociation constant KD:

!

KD

=k

off

kon

=[A] " [B]

[AB]

When [A]=Keq, 50% of B is bound to A

Equilibrium is reached when:

!

[A] " [B] " kon

= [AB] " koff

Page 5: Binding and Kinetics

5

Units

Units:

!

KD

=[A] " [B]

[AB]

!

{M} ={M} " {M}

{M}

(Conversely, equilibrium binding constant, KB, is defined as:

!

KB

=[AB]

[A] " [B]

!

{M"1} ={M}

{M} # {M})

!

KD

=k

off

kon

=[A] " [B]

[AB]

koff: {s-1}

kon: {M-1·s-1 }

Rate constants:

Where does this KD come from?

From Lecture 5:

Page 6: Binding and Kinetics

6

How to measure KD ?

!

KD

=[A] " [B]

[AB]Measure [A], [B], and [AB]?

Introducing [A]Total=[A]+[AB]:

!

[AB]

[A]Total

=[B]

KD

+ [B]

D

Figure from: G

oodrich, Kugel

Experimental considerations

• [A] constant; titrate B• Measure fraction bound

!

[AB]

[A]Total

=[B]free

Keq + [B]free

If [A]Total << KD, then [B]≈[B]+[AB]

No need to measure [B],Just take [B]Total!

Figure from: G

oodrich, Kugel

Page 7: Binding and Kinetics

7

Logarithmic versus linear display

As a corollary: Choose your titrations logarithmically!

1, 3, 10, 30, 100, 300 nM, or2, 4, 8, 16, 30, 60, 180, 360 nM, instead of50, 100, 150, 200, 250, 300 nM

Figure from: G

oodrich, Kugel

Example: Repressor binding to DNA

DNA + R DNA-Rkon

koff

In E. coli, how much repressor is bound non-specifically to DNA and how much is free?

[non-operator DNA] ≈ 106 / 1 µm3 ≈ 10 mM (107 bp/genome; 10 bp/site; volume. E.coli 1 µm3)

!

F =[R]

[R] + [R "DNA]=

[R][DNA

non]

[R "DNA]

[R][DNA

non]

[R "DNA]+ [R "DNA]

[DNAnon

]

[R "DNA]

=K

D

KD

+ [DNAnon

]=

10#4M

10#4M +10

-2M

= 0.01

Hardly any free repressor; almost all bound to nonspecific DNA!

KD≈10-10 M for operator DNA (specific binding)KD≈10-4 M for non-operator DNA (non-specific binding)

Page 8: Binding and Kinetics

8

Protein

B BB

000 + BK

00B

Not Cooperative

00B + BK

0BB

Protein

B BB Cooperative

000 + BK

00B

00B + BτK

0BB

Non-cooperative versus cooperative

τ can be positive or negative (positive or negative cooperativity)

Cooperative binding

A + nB ABn

kon

koff

(perfect cooperativity)

Simplification:

!

KD

=[A] " [B]n

[ABn]

Rearrange (next Problem Set???):

!

logY

1"Y

#

$ %

&

' ( = n

H) log[B] " logK

D ,

where Y=[ABn]/[A]total

Page 9: Binding and Kinetics

9

Cooperative binding

!

logY

1"Y

#

$ %

&

' ( = n

H) log[B] " logK

D ,

where Y=[ABn]/[A]total

Figure from: G

oodrich, Kugel

Hemoglobin

Page 10: Binding and Kinetics

10

Reaction kinetics

Equilibrium thermodynamics does not provide anyinformation on rates of chemical changes!

Energy profile for ageneric chemical reaction:

Gibbs free energy (ΔG0) determines ratio of reactants/products(thermodynamic properties), activation energy (ΔG++) determinesrates (kinetics)

(dynamite versus nitroglycerin)

Figure from: H

aynie, Biological Therm

odynamics

++

Rate of reaction

Reaction rate = a measure of how fast the concentration of reactants /products changes with time

Example: hydrolysis of ATP into ADP

ATP ADP + Pi

Reaction rate:

!

J = "d[ATP]

dt= +

d[ADP]

dt= +

d[Pi]

dt

Figure from: H

aynie, Biological Therm

odynamics

Page 11: Binding and Kinetics

11

Rate constant and order of reaction

!

J = k[A]n

Reaction rate/velocity is related to concentration of reactant:

n is order of reaction (often identical to stoichiometry)k is rate constant (don’t confuse with binding constant)

We saw that , so k will have:

Per second (s-1) as unit for 1st order reaction,Per molar per second (M-1s-1) as units for 2nd order reaction

!

J = "d[A]

dt

1st order reaction

1st order reaction A P

!

J = k[A]

!

J = "d[A]

dtCombining with gives:

!

"d[A]

dt= k[A]

!

1

[A]d[A] = "kdt

Integrate ( ):

!

ln[A] = ln[A]0 " kt

!

1

x" dx = ln x + C

!

[A]

[A]0

= e("kt)

Figure from: H

aynie, Biological Therm

odynamics

Page 12: Binding and Kinetics

12

2nd order reaction

2nd order reaction 2A P

!

J = k[A]2

!

J = "d[A]

dtCombining with gives:

!

"d[A]

dt= k[A]2

!

1

[A]2d[A] = "kdt

Integrate ( ):

!

1

[A]=

1

[A]0

+ kt

!

1

x2

" dx = #1

x+ C

!

[A]

[A]0

=1

1+ kt

Figure from: H

aynie, Biological Therm

odynamics

1st and 2nd order reactions

1st order:

!

[A]

[A]0

=1

1+ kt

!

[A]

[A]0

= e("kt)

2nd order:

Figure from: H

aynie, Biological Therm

odynamics

Page 13: Binding and Kinetics

13

Half-times and rate constants

Half time t1/2 is not the same as k-1 :

!

[A]

[A]0

= 0.50 = e("kt1/ 2 )

# " ln2 = "kt1/ 2 # t1/ 2 =ln2

k$

0.693

k

Temperature effects

Rates depend on temperature

Arrhenius:

!

k = Ae("#G

++ / RT)

A+B ABkon

koff

!

lnk = ln A "#G++

/ RT

Page 14: Binding and Kinetics

14

Reversible reaction

A + B ABkon

koff

!

d

dt[A "B] = k

on[A][B] # k

off[A "B]

Formation(2nd order)

Dissociation(1st order)

Under equilibrium, equals zero:

!

d

dt[A "B]

!

[A][B]

[A "B]=

koff

kon

= KD

In terms of free energies:

!

KD

=k

off

kon

=Ae

("#Goff

++ / RT)

Ae("#Gon

++ / RT)= e

("(#Goff

++"#Gon

++ ) / RT)= e

("#G0 / RT)

ΔG0= ΔGoff*-ΔGon

*

Figure from: H

aynie, Biological Therm

odynamics

!

[A][B]

[A "B]=

koff

kon

= KD

Relation between KD, kon/off, and ΔG

A+B

AB

++++ ++

Page 15: Binding and Kinetics

15

Rates of binding and dissociation

A+B ABkon

koff

Association rate for two objects with diffusion coefficients D1 and D2and diameter r1 and r2:

kdiff=4πNA(D1+D2)(r1+r2) (units: {mol-1}{cm2s-1}{cm} = {M-1s-1} )

For a small ligand and protein: kdiff ≈ 109 M-1s-1,for two proteins: kdiff ≈ 106 - 107 M-1s-1

This rate can be further slowed down if a conformational changeneeds to take place before binding

Example: Repressor binding to DNA

DNA + R DNA-Rkon

koff

!

d

dt[R "DNA] = k

on[R][DNA] # k

off[R "DNA]

Formation(2nd order)

Dissociation(1st order)

It takes 0.1 seconds to switch off gene expression in E.coli afterlactose depletion. What is kon?

!

d

dt[R "DNA] # k

on[R][DNA]

With ~10 repressors per E.coli and [DNA]≈10-9 M (1 operator sequence in 1 µm3 cell),kon needs to be at least 109 M-1s-1 (is actually measured to be 1010 M-1s-1)

How come this is much faster than diffusion limit???

Page 16: Binding and Kinetics

16

1D sliding along DNA to speed up kon

BWH (Berg, Winter, von Hippel) model:Combine 3D diffusion (‘hopping’) with1D diffusion (‘sliding’). Scan short stretch of DNA by 1Dsearch, then jump to different area.

!

"L(# ) = D1D#

Length explored by one 1D sliding event:

(1D random walk)

Typical duration will be τ=1/knonsp.off:

!

"L = D1D / knonsp.off

Remember, repressors spend 99% of time on nonspecific DNA:

!

L(t) = tknonsp.off D1D / knonsp.off Total length explored L(t) is linear with time!

1D sliding: the numbers

D1D ≈ 10-9 cm2/s (limited by rotational drag)knonsp.off ≈ 10 s-1

L(τ) ≈ 100 nm (300 bp)100 kb of DNA is searched by singlerepressor in half a minute

Searching 100 kb with only 1D sliding would takeTtotal = L2

total/D1D ≈ 3 hours!

Now we understand why 99% of repressor is bound to nonspecific DNA:They’re actively involved in the search process.

Page 17: Binding and Kinetics

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Folding revisited: a riboswitch

• Folded RNA that binds small molecule (aptamer)• Plays role in regulation of gene expression

How does it fold?

?

Single-molecule probing of RNA folding

Liphardt et al., Science (2001)

Page 18: Binding and Kinetics

18

Pulling at an RNA hairpin

Force-extension curve ofsingle RNA unfolding/folding

Liphardt et al., Science (2001)

Along the reaction coordinate, anamount of energy equal to forcetimes displacement is added

Force tilts free-energy diagrams

ΔG = -F·Δx

ΔG = ΔG0 - F·Δx

Page 19: Binding and Kinetics

19

!

P(unfolded)

P(folded)= exp "

#G

kT

$

% &

'

( ) = exp "

#G0 "F#x

kT

$

% &

'

( )

Liphardt et al., Science (2001)

Pulling at an RNA hairpin

Liphardt et al., Science (2001)

Single-molecule kinetics:Direct observation of kopen and kclose

Pulling at an RNA hairpin: kinetics

Page 20: Binding and Kinetics

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Unfolding a riboswitch

Unfolding a riboswitch

Page 21: Binding and Kinetics

21

Take-home message

Equilibrium constant K is related to free energy difference ΔG0

between initial and final state, rates k are related to free energydifferences ΔG‡ between initial/final state and transition state

!

[A][B]

[A "B]=

koff

kon

= KD

!

KD

=k

off

kon

=Ae

("#Goff

++ / RT)

Ae("#Gon

++ / RT)= e

("(#Goff

++"#Gon

++ ) / RT)= e

("#G0 / RT)

A+B

AB

++++ ++ΔG0= ΔGoff

*-ΔGon*