Biochemistry I
Lecture 14
February 15, 2013
Lecture 14: Analysis of Cooperative Binding
Key Terms:





Hill coefficient, nh
1. Positive cooperativity: nh >1
2. Negative cooperativity: nh < 1
3. Non-cooperative: nh = 1
A homotropic positively cooperative system will show a _____________ in affinity as ligand binds, due
to the stabilization of the _______ state.
A homotropic negatively cooperative system will show a ____________ in affinity as ligand binds, due
to the stabilization of the _______ state.
Hill Plot: log(Y/(1 - Y)) versus log[L]
x-intercept = logKD
Review of Types of Binding:
1
protein with a single site must show noncooperative binding.
Homotropic positive cooperativity: Multiple
interacting ligand binding sites required, binding
at one increases affinity at another by increasing
R state.
Fractional Saturation (Y)
Non-cooperative: No interaction between sites. A
0.8
0.6
0.4
Homotropic negative cooperativity: Multiple
0
interacting ligand binding sites required, binding
0
 Heterotropic activator increases R-state.
Binding affinity of ligand for one or more noninteracting sites increases.
 Heterotropic inhibitor increases T-state.
Binding affinity of ligand for one or more noninteracting sites decreases.
Fractional Saturation (Y)
at one decreases affinity at another by
increasing T state.
Allosteric control - non-cooperative binding:
5
10
15
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
20
[L] (free)
No Act/Inh
Activator
Inhibitor
0
5
10
15
20
[L] free
Fractional Saturation (Y)
Allosteric control with cooperative binding:
Heterotropic activator increases R-state,
increasing average affinity. Heterotropic
inhibitor increases T-state, reducing average
affinity. Ligand binds to multiple interacting
states (homotropic) with some form of
cooperative binding, neg or positive (positive
cooperativity for the ligand is shown here).
Non-coop
Pos-Coop
Neg-Coop
0.2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
It is possible to quantify the degree of cooperativity by
0
analysis of the binding data using a Hill
Hill Coefficient
plot. The outcome of this analysis is
<1
the Hill coefficient, which has the
=1
characteristics summarized to the right:
>1
Behavior of Cooperative Systems:
Single
Site
Binding
Non- = n, number of binding sites.
cooperative:
1
1
[ M ][ L]
[ L]
KD
KD
[ ML]
[ L]
Y



1
1
[ M ]  [ ML]
K D  [ L]
[M ] 
[ M ][ L] 1 
[ L]
KD
KD
No Act/Inh
Activator
Inhibitor
5
10
15
20
[L] free
Interpretation
Negative cooperativity.
Non-cooperative
Positive cooperativity.
Infinitely strong positive cooperativity.
When Y = 0.5 what is [L]?
1
Biochemistry I
Lecture 14
February 15, 2013
Consider a two step binding:
A1
[ M ]  [ L] K
[ ML ]
K D1 
i) Non-cooperative. The binding
constant remains the same for
both binding events.
A2
[ ML ]  [ L] K

[ ML2 ]
[ M ][ L]
[ ML][ L]
K D2 
[ ML]
[ ML2 ]
ii) Positive cooperativity if KD2 <
iii) Negative cooperativity if KD2 >
KD1 (or KA2 > KA1) i.e. the
KD1 (or KA2 < KA1) i.e. the second
second binding is higher in
binding is lower in affinity.
affinity.
Two Binding Sites - Infinitely Positive Cooperativity:
Consider an infinitely positive cooperative system such that the second
dissociation constant (KD2) is much lower than the dissociation constant for
binding the first ligand (KD1). Then the only species present in solution are
[M] and [ML2]. The fractional saturation under these conditions is:
1
1
[ M ][ L][ L]
[ ML2 ]
K D2
K D1 K D 2
[ L]2
Y



2
1
1
[ M ]  [ ML2 ]
[M] 
[ ML][ L] [ M ] 
[ M ][ L][ L] K D1 K D 2  [ L]
K D2
K D1 K D 2
[ ML][ L]
When Y = 0.5 what is [L]?
Generalize to N-Ligands - Infinitely Positive Cooperativity:
Y
[ L]n
K D1K D1   K Dn  [ L]n
When Y = 0.5 what is [L]?
N-Ligands – General Cooperative Binding:
For less cooperative systems, the fractional
saturation can be approximated by:
Y



[ L]nh
n
K Dh ave  [ L]nh
Where nh is the Hill coefficient. This is a
measure of the degree of cooperativity.
KD-AVE is the "average" KD; when [L]=KDAVE, Y=1/2.
The KD-AVE is the root of a polynomial of the individual dissociation constants (Note: for two binding
sites KD-AVE= 2√𝐾𝐷1 𝐾𝐷2 , regardless of the degree of cooperativity.)
2
Biochemistry I
Lecture 14
Single Site - Noncooperative
Y
Two sites - infinitely
positive cooperative
Y
N-sites - infinitely
positive
cooperative:
N-sites - just plain
pos. cooperative:
Y
Y
Y=0.5 when
[ L]  K D
[ L]1
K D  [ L]1
[ L]2
K D1 K D 2  [ L]2
[ L] N
K D1 K D 2 K Dn[ L] N
[ L]nh
n
K Dh ave  [ L]nh
Y=0.5 when
[ L]  2 K D1 K D2
[ML] = 0
[ML2] > 0
Y=0.5 when
[ L]  N K D1 K D2 ...K DN
[ML]…[MLN-1] = 0
[MLN] > 0
Y=0.5 when [ L]  K D ave
Hill Equation and Plot: The Hill coefficient, and the "average" KD
can be obtained from a Hill Plot. The Hill plot is based on the
following transformation of the above binding equation:
 Plot of log [Y/(1 - Y)] versus log[L]
 The Hill coefficient, nh, is the slope as the line crosses the x-axis.
 The logKD-ave is the intersection of the Hill curve with the x-axis.
Non-Cooperative Systems (n =1):
Y 

[ L]
K D  [ L]
log
Y
 log K D  log[ L]
(1  Y )
This is a straight line with a unit slope.
Intersection with x-axis (Y = 0.5) gives the true KD.
0  log
February 15, 2013
1
 log[L]
KD
Y
[ML]…[MLN-1] > 0
[MLN] > 0
[ L ]n h
n
K Dh ave  [ L]nh
Y
[ L]nh

1  Y K nh
D  Ave
 Y 
   nh log K D  ave  nh log[ L]
log 
 (1  Y ) 
Log
Y
1-Y
0
1
 log[L]
KD
log K D  log[L]
K D  [ L]
 log
Log[L]
Cooperative Systems.
Low ligand: At very low ligand concentration,
the binding appears non-cooperative because
most of the macromolecule is in the [M]
form. Therefore the Hill plot is initially
linear, with a slope =1, intersecting x-axis at
logKD1.
High ligand: At very high ligand concentration,
the binding also appears non-cooperative
because most of the macromolecule is in the
[MLn] form. Therefore the Hill plot is again
linear, with a slope = 1, intersecting the xaxis at log KDn.
Intermediate Ligand Concentration (@Y=0.5):
Slope: Hill coefficient (0 ↔ 1 ↔ n)
Intercept: Ligand concentration to give Y=0.5 (True KD for non-cooperative binding.)
3
Biochemistry I
Lecture 14
February 15, 2013
Microscopic and Macroscopic Binding constants:
Cooperativity requires a change in the affinity of the binding site.
 In the case of positive cooperativity the affinity increases (KA increases, KD decreases).
 In the case of negative cooperativity the affinity decreases (KA decreases, KD increases).

For proteins that bind multiple ligands to identical sites the experimentally determined affinity constants
will change even if there is no cooperativity. For a two binding sites the affinity for the second site will
be lower by a factor of two. This difference is due entirely to statistical factors – there are no
differences between the molecular interactions in the two sites.
Microscopic KA: This is the association constant for a single site,
and is just the ratio of the on- and off-rates: KA=kON/kOFF. It
reflects the intrinsic affinity between the protein and the ligand:
Go = -RT ln KA
kon
koff
KA
Macroscopic KA: This is the observed KA based on the
concentrations of the various species, i.e. KA1=[ML]/[M][L], i.e.
what you would typically consider the equilibrium constant.
One-binding site: The macroscopic and microscopic are the same.
M
Microscopic
kon
kon
Two binding sites: For the first binding event this is 2KA since
there are two ways to form the [ML] species, i.e. KA1=2kON/kOff.
For the second binding event there are two ways for the ligand to
leave, so KA2=kON/2kOff.
Three binding sites
ML
koff
kon
koff
kon
koff
koff
Macroscopic
2kon
kon
koff
2koff
KA1
M
ML
KA1
M
KA2
ML
KA2
ML2
ML2
KA3
ML3
4