Peter Fajer
Biophysical Methods in Biology
Lecture 10
Ligand Binding
Consider ligand A binding to macromolecule P:
A  P  AP
[1]
Equilibrium constant is defined as:
Kd 
 P A
 PA
[2]
(note that Kd is often referred wrongly as binding constant: Kb = 1/Kd)
define fractional occupancy (moles of ligand/mole of macromolecule) r:
r
 A bound
 PA

 P total  P   PA
[3]
substitute [P][A]/Kd for [PA]:
r
 A
Kd   A
[4]
this hyperbolic dependence is called binding isotherm or (Langmuir isotherm).
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Peter Fajer
Biophysical Methods in Biology
Lecture 10
binding isotherm
1.2
1
0.8
r
Kd=1
Kd=5
0.6
Kd=5
Kd=10
0.4
0.2
0
0
5
10
15
20
25
[A]
Note that although the isotherm has the concentration for free ligand [A] as well as bound ligand ([A]bound
in r ) you have to measure only one (bound or free) because conservation of mass the total ligand
[A]total = [A]+[PA] (and total macromolecule [P]total = [P]+[PA]).
Measure the concentrations of free or bound ligand by variety of methods: chromatography, equilibrium
dialysis, ultrafiltration, spectroscopy.
Multiple binding sites
Macromolecule can have more than one site for binding the ligand. These sites can be independent:
binding of one ligand does not influence binding of the next, or cooperative (binding of one affects
binding of another).
Independent
A  P  A1 P  A2 P .......

identical sites: A1P=A2P or

different sites: A1PA2P
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[5]
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Peter Fajer
Biophysical Methods in Biology
Lecture 10
Identical sites
site 1
site 2
macromolecule
all sites are independent and identical so that total number of sites is n[P]total rather than [P]total :
rall _ sites  nrsin gle _ site  n
A
K d  A
[6]
Scatchard plot
plot the ratio of r/[A] versus r:
r
n
r


 A K d K d
[8]
For identical binding this results in a linear plot. Deviation from linearity implies non-identical sites.
r/[A]
r
Different sites
to be very general: each class of sites with different Kdi can have ni identical sites
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Peter Fajer
Biophysical Methods in Biology
Lecture 10
site class 1
site class 2
macromolecule
The binding isotherm is a sum of the single binding isotherms, each corresponding to different class of
sites:
r  r1  r2 ..... 
n1  A
n2  A

........
Kd ,1   A Kd ,2   A
[7]
Cooperativity
Binding to one site on a molecule modulates binding of another.
For strong cooperativity, binding of one ligand triggers binding to n sites:
nA  P  An P
Kd
[9]
n

P  A

An P 
Abound
r
Ptotal
[10]
n

PAn 
nA
n

P  PAn  K d  An
[11]
or measure the ratio of filled sites Y (Y=r/n) to the unfilled sites (1-Y):
A
Y
r


1Y n  r
Kd
n
[12]
(Hill’s equation)
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Peter Fajer
Biophysical Methods in Biology
Lecture 10
Hill’s Plot
Plot log(Y/(1-Y) v. log[A], the slope is n; an intercept is 1/Kd.
positive cooperativity n=2
log[r/(n-r)]
no cooperativity n=1
negative cooperativity n=0.5
log [A]
Kinetics
Time course of reaction, kinetics give information about the rates rather than equilibria. Of course rates
and equilibria are related Keg = k+1/k-1, however the ratio of rates tells us nothing about the value of each
rate i.e. is it fast or is it slow.
Two sorts of kinetics are usually considered:

steady-state: in which the forward flux = backward flux resulting in no change of concentration;

transient kinetics: non-equilibrium kinetics when concentrations are changing.
steady-state (Michaelis-Menten)
k1
k2
k 1
k2
E S X E  P
[13]
for a reaction with an intermediate X the rate of intermediate’s production is equal to that of its
disappearance:

d X 
  k 1  k 2  X   k1  E  S   k 2  E  P  0
dt
[14]
The rates of substrate disappearance and the appearance of the product are identical:
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Peter Fajer

Biophysical Methods in Biology
d  S  d  P

 k1  E  S   k 1  X 
dt
dt
Lecture 10
[15]
From mass conservation we have:
E    E   X 
 S    S    P
o
[16]
o
At the beginning of the reaction:
 P  0
[17]
thus the initial velocity v can be solved since we have five equations with 4 unknowns (concentrations of
E, S, X, P)
v
dS 
Vmax  S 

dt
S   K Michaelis
[18]
(dirty shortcut: realize that initial velocity is a maximum velocity reduced by the partial enzyme
occupancy, i.e.:
v  rVmax
[18]
then substitute v for r in the equation for Langmuir isotherm.)
turnover
k cat 
Vmax
Eo
[19]
non-steady state
Steady-state approximation is not always applicable, for general reaction:
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Peter Fajer
Biophysical Methods in Biology
Lecture 10
k1
E  S  X1
[20]
k 1
the rate equations are:
 
d E 
d S  d X 1


 k1  E  S   k 1 X 1
dt
dt
dt

 
[21]
with two mass conversation relationships:
E    E   X 
S    S    X 
o
1
o
[22]
1
The differential (rate) equation can be solved but it is a mess since the equation is non-linear (the rate
depends on product of [E] and [S]), better approximate that So >> Eo so that concentration of substrate
never changes (pseudo-first order approximation):
 k
d X1
dt
1
 E    X S   k  X 
o
1
1
o
[23]
1
This equation is a linear in X1 and it can be integrated to give the concentration of X1 as function of time:
ln
E   E
o
  k S   k t

equilibrium
 E    E equilibrium
1
o
[24]
1
Presence of two intermediates complicates the affairs considerably:
k1
k2
k 1
k 2
E  S  X1  X 2
[25]
conservation of mass gives:
E    E   X    X 
S    S    X    X 
o
o
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[26]
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Peter Fajer
Biophysical Methods in Biology
Lecture 10
the rate of enzyme disappearance is:

d E 
 k1  E  S   k 1 X 1
dt
 
[27]
the rate of product appearance is:

 k
d X2
dt
2
X  k X 
2
2
[28]
1
At this point you are advised to give up. What you have is the set of simultaneous, non-linear differential
equations which even your grandma can’t solve.
Bill Gates to the rescue; integrate the set using a PC and any mathematical package with numerical
integration routines e.g. Mathcad, Mathematica, Matlab.
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