Notes 9
Derivation of the Scatchard Eqn.
Relates to the Binding of Ligands and Metal Ions with proteins and specific
membrane receptor sites and equilibrium dialysis
Extensively covered biochemical phenomenon
Apply Equilibrium Ideas to binding of ligands and metal ions to macromolecules
(proteins, etc.) in solution
Sometimes proteins may have only one binding site, but may have more.
Consider two cases:
a. One binding site
b. n equivalent and independent binding sites
A. One Binding Site:
P+ L
->
<-
PL
The equilibrium constant for this is:
Ka =
Instead of equil for association, may often work with equil for dissociation
Kd =
where Ka * Kd =
What does a small value of Kd indicate
So we would like to determine the value of Kd
Helps to define a quantity, Y
fractional saturation of sites = Y = (conc. of L bound to P / conc. of all forms of P)
Y = ([PL]) / ([P] + [PL])
What is the range on Y?
To determine Kd:
Rearrange the Kd expression
Plug in the expression for [PL]
get
Y =
or
Y =
Remember this is the equilibrium expression so these are the equilibrium concentrations!
Rearranging again by taking the reciprocal one gets
(1)
OR rearranging this by multiplying by Y/Kd
(2)
Get two equations that are linearized and can be used to find Kd and Ka
Based on (1), a plot of 1/Y versus
Based on (2), a plot of Y/[L] versus
Now suppose there is more than one binding site, say n equivalent - independent binding
sites. Lets consider the case initially where n = 2
Get binding equilibria like:
Association P + L - > <- PL
Kd1 = K1 +
Association PL + L - > <- PL2
Kd2 = K2
Now define Y as conc. of L bound to P / total concentration of all forms of P
So
Y = ([PL] + 2[PL2]) /
(why 2 [PL]?)
Substituting [PL] = [P] [L] / K1
Can write:
Y=
and [PL2] = ?
Are K1 and K2 equal to each other?
Define an Intrinsic Dissociation Constant K
The relationship between the it h dissociation constant Ki and K is
Ki = ( i / (n-i-1)) K
so if i= 1, 2 and n = 2 then:
and K = (K1K2)1 / 2
Now get an expression for Y in terms of K for two sites:
Y=
And for n sites generalize to
Can rearrange this into several forms that will allow a graphical solution for K etc.
1. Direct Plot
2. Double Reciprocal Plot
3. Scatchard Plot
Starting with eqn above and multiplying both sides by [L] + K one gets:
Y[L] + KY = n[L]
and dividing by K:
now rearranging
and again
Plot of
and y intercept of
will give a straight line with slope of
and x intercept of
For Example
The red blood cell membrane can act as an osmotic membrane and it also acts as a
binding center for ligands. Small molecules, including ions may pass, but large
macromolecules cannot.
How?
Ligands in contact with the membrane will diffuse across the dialysis membrane, and
move across until equilibrium is reached. Stirring speeds the equilibration.
The binding site inside of the membrane or bag, maybe a macromolecule like
Hemoglobin (Hb) or myoglobin (Mb) serves to concentrate the ligand by binding it ( ie
O2).
Alternatively if a protein has salted-out (for instance due to the presence of ammonium
sulfate), the protein can be freed of the salt by:
a. dissolving the precipitate in water or a buffer solution
b. place the protein solution in a cellophane bag,
c. Put the cellophane bag in a beaker containing the same buffer
Because both the NH4+ and SO42 - ions are both small enough to diffuse through the
membrane, but the protein molecules are not, the ions in the bag will begin to move
through the membrane into the outside solutons, where the chemical potentials of those
species are lower.
Flow will continue until equil is reached which means that the chemical potentials of the
sulfate and ammonium ions are ??
If desired, all of the ammonium sulfate could be removed by continually changing the
buffer solution in the beaker.
This procedure may be reversed to study the binding of ligands to proteins such as O2 to
Hb or Mb by placing the protein without ligands into a buffer solution in the cellophane
bag and then immersing the bag into a similar buffer solution that contains the ligand (say
O2) of known concentration.
Again at equilibrium what will be true?
In terms of the activities this means
and if the solutions are dilute use concentrations instead
And now the total concentration of the ligands inside the bag is
And the conc. of ligand bound to protein molecule is
This is how both the intrinsic dissociation constant K and the number of binding sites n
can be obtained using the equil. dialysis technique
Suppose start with protein soln. of known concentration in phase 1 and a ligand solution
of known concentration in phase 2. At equil. the quantity Y is given by where [P]total is
the concentration of the original protein solution.
The experiment can be repeated by using different concentrations for the protein and
ligand solutions and the values of K and n can be determined from the Scatchard plot.
Remember that the quantity [L] refers to the conc. of unbound ligands at equilibrium,
which is [L]unbound here
Keep in mind that the ligand has been assumed to be an non-electrolyte,
If it is an electrolyte, need Donnan Effect for successful treatment of this problem.
Doonan Effect and Donnan Potential
Equilibrium dialysis, was an easy way to measure binding by a macromolecule. But it
did not completely work if the ligand and/or macromolecule are charged.
Proteins can be positive, negative, or neutral, depending on the pH of the solution.
Complicated by the fact that there can be an increase in binding of a ligand with opposite
charge and decrease if same charge. In fact is the binding effect caused completely by
charge differences or are there effects just from binding. The effect of the net charge on
the macromolecule can be minimized by using high concentration of a salt not involved
in binding, so that the total ions in solution is governed by the salt.
When equilibrium is established for a charged species, a voltage is developed across the
membrane, leaving an assymetric distibution. DONNAN EFFECT,
The transmembrane potential is called the DONNAN Potential.
Take the case where the negative protein Pz - is placed in a dialysis bag with Na+
ions and then the dialysis bag is surrounded by a solution of NaCl
Fig.
To attain equilibrium, some of the NaCl will move. It will move from the outer
solution through the membrane into the bag with the protein in it. In effect, the added
salt reduces and at high enough salt concentration, eliminates the Donnan effect.
There is a change in free energy when a molecule is transferred from a solution at one
concentration to one at a different concentration in the same solvent.
 = A(phase 2) - A(phase 1)
These apply to the case of uncharged molecules, but in this case we are talking about
charged molecules and ions. If there is a potential difference between the two phases
V (volts) then an additional term is necessary here.
 = A(phase 2) - A(phase 1) = Z F V
where Z is the charge in the ion
F is Faraday’s constant
V is potential difference between the two phases in volts
So in the case where the macromolecule is charged then this will affect the binding of the
ligand.
There is a requirement for electrical neutrality.
So for a positively charged macromolecule the concentration of positive ions inside will
be ______________ than that outside.
And the concentration of negative ions inside will be _____________ than outside.
For a negatively charged macromolecule the concentration of positive ions
is greater inside than outside.
The assymmetry in Na+ or Cl- concentration on the two sides of the membrane is due to
the Donnan potential across the membrane and with the requirement G = 0 when the
inside and the outside are in equilibrium with each other we obtain
RT ln aNa+(inside)/aNa+(outside) + ZFV
V = -RT/ZF ln (aNa+(inside)/aNa+(outside))
Equilibrium Dialysis is the equilibrium process of exchanging small ions and other solute
molecules from a protein solution through a semipermeable membrane. Movement
through the membrane involves Diffusion largely and the Viscosity may also play a role.
First Describe Membranes
Biological Membranes
Fig. shows widely accepted model of cell membrane structure called the fluid mosaic
model.
Phases?
Contain lipids, proteins, glycolipids, and other amphiphilic molecules. Constitute a
___________ separate from the rest of the cell.
Proteins lie at or near the inner and/or outer membrane surface, or they may partially or
totally penetrate the membrane surface. The extent of interaction between the proteins
and the lips depends on the ypes of intermolecular forces and thermodynamic
considerations.
Generally membranes are not rigid, but have high electrical insulating properties and
great physical strength.
Membrane proteins have multiple functions:
a. act as recetors
b. act as enzymes
c. act as carriers for ions or other molecules across the membrane.
Lipid Molecules
Amphiphilic molecules ?
Surfactants-
Sodium Dodecylsulfate
Zwitterionic?
Lipid Bilayers
Phase Transitions in Lipids, Bilayers, and Membranes
DSC shows peaks before the melting point for lipid crystals like
phosphatidylethanolamine or phosphatidylcholine
The lowest temperature hydrocarbon tails
Next highest
Factors influence the transition temperatures are:
1) the length of the hydrocarbon tail,
2) nature of the headgroup
3) presence of unsaturation (double bonds) i the hydrocarbons,
4) additional molecules like cholesterol that are incorporated into the bilayer
Active and Passive Membrane Transport
Passive Membrane Transport
Simple Diffusion
Facilitated Diffusion
Example:
___________________ Transport
The presence of a protein inside seaweed that strongly binds iodide ion ensures that the
iodide concentration in the seaweed is always higher than in the seawater.
Active Transport
In active transport transfer occurs from a lower to a higher chemical, against a
concentration gradient.
potential, so it must occur in concert with another process that has a large negative G.
Example:
Sodium-Potassium ion pump
Divide into three parts:
1) Transport of Na+ ions out
2) Transport of the potassium ions in
3) Hydrolysis of ATP
Transport
Three types of transport properties to discuss.
Diffusion
Mass transfer by movement of molecules and mass due to a concentration gradient
Viscosity
Momentum transfer restricted by a velocity gradient
Heat
Energy transfer as heat through conductive, convective, or radiative processes.
Kinetic Theory
Look at Molecular Motion and Transport Properties
Maxwell Boltzmann Distribution of Molecular Speeds (Derive from Ideal Gases)
Diffusion and Effusion (Concentration Gradient)
Sedimentation
Viscosity
Electrophoresis
.....................................................................................Kinetic Theory of Gases
Model for Perfect Gas – Molecules in continuous chaotic motion; kinetic energy of the
molecules is the only energy
Based on several assumptions:
1. Gas consists of
2. Size of molecules is
3. The molecules have no attractive or
Actually a distribution of speeds rather that a single speed. Maxwell-Boltzmann
distribution of speeds
Diffusion
Random Walk - 2 d - Drunken Sailor
Randomly walks either to the left or the right
Brownian Motion
Mean displacement
Root mean square displacement (measures the average displacement per unit time.
Fick’s First Law
The diffusion occurs whenever there is a ________________ concentration gradient in
the container.
Rate of diffusion will depend on the size and shape of the molecule and on the properties
like the viscosity of the solvent.
Measure diffusion by looking at the molecules that cross a certain cross sectional area in
some plane.
Define the flux Jx as the net amount of solute that diffuses through this unit area, per unit
time, in a direction perpendicular to the plane
Think about what happens if there are solute molecules on one side of the plane than on
the other. (A Concentration Gradient Exists)
The steeper the concentration gradient, the larger is the net flux. This leads to the
equation for Fick’s First Law:
Jx = -D (dc/dx)
at some instant in time
D is just a constant called _______ _____________________ ________________.
Negative sign means that the transport by diffusion is in a direction _____________ to
the ____________ ________________.
As diffusion continues expect:
So need to describe how the concentration gradient changes with time.
Ficks Second Law
(c/t)x = D(2c/x2) t
How would you determine the Diffusion Coefficient from Ficks First Law or Fick’s
Second Law?
The diffusion coefficient is related tot he Mean Square Displacement:
Can solve Fick’s Second Law to find how the concentration changes as the diffusion
progresses.
c = w/[(4Dt)1/2] e-x2/4Dt
In general to find <x>, you need can integrate the product of
x times the distribution function integrated over all space.
or <x2> is the integral of the product of x2 and the distribution over all space.
Applying this here for <x2> to get in terms of D:
gives:
D = <x2>/(2t)
A diffusion coefficient can be measured fro the disappearance of any concentration
gradient caused by diffusion.
Diffusion Coefficient is related to Molecular Parameters
Frictional Force
What should D be related to?
Size and Shape of material diffusing?
frictional force that opposes movement
Ffriction = -f u
u = velocity
dynamic force that increases with velocity
eventually driving force and frictional force balance one another and reach terminal
velocity
f = the frictional coefficient characterizes this. In diffusion the driving force is due sot
the chemical potential associate with the concentration difference and the diffusion and
the diffusion coefficient are related to it theoretically by
D = kT/f
(Einstein's Relationship)
f = 6r
(Stokes Eqn.
where k is
 is
r is
calculated f is usually smaller than that measured by experiment. Why?
Solvation
Shape Factors
Polymer Coils and Diffusion
for large polymer coils of N monomeric units
the root mean square radius of the polymer is proportional to (N)1 / 2 so also Mm1 / 2
Sedimentation
Used for purification separation and analysis of cellular species like proteins and nucleic
acids to viruses, mitochondria etc.
Sedimentation or sinking is caused by difference between gravitational force and buoyant
force.
driving force = mg - mv2  g
From F = ma = m du/dt
Viscosity
When a force F is applied to a solid particle to make it move through a liquid with a
velocity ut, the molecules of the liquid at the interface with the particle move a the same
velocity ut, due to the intermolecular forces between them. However at the wall the
molecules must ______________ _____________ and in the center of the stream they
are likely moving ________________ ___________________.
A velocity gradient is set up in the stream, with molecules from below having a lower
velocity and lower momentum so there is a net transfer of momentum from above to
below. The steeper the velocity gradient, the larger the net momentum transfer.
Jmu ≈ -constant dux/dy
Jmu = -  dux/dy
or
Jmu is the rate of momentum transfer per unit time per unit cross sectional area.
 is the viscosity coeficient of viscosity SI kg/(m s) also poise (g/(cm s)
Newtonian Fluid  is a constant independent of dux/dy. If  depends on dux/dy then the
fluid is non-Newtonian
Measuring Viscosity
Free Fall of a Particle Through a Viscous Medium
Volume Rate of Flow Through a Capillar;y
For solutions define the specific viscosity
sp = (’
 is the viscosity of the solvent and ’ is the viscosity of the solution.
The intrinsic viscosity is related to the molecular properties.
ELECTROPHORESIS
Molecules that are biologically important are often charged. This means that they will
move in the presence of an electric field:
For a particle in a nonconducting solvent, the velocity of migration u in an electric field is
u = (Z e E) / f
Z is # of charges
e = charge on electron
E is the electric field in volt/m
f is the frictional coefficient in kg/s
The velocity of migration per electrinc field is called the electrophoretic mobility, .
Applications of Electrophoresis to the Characterization of Macromolecules
Gel Electrophoresis
DNA Sequencing
Double-Stranded DNA
DNA Fingerprinting
Nucleic Acid Conformations
Protein Charge
Diffusion
Fick's Laws describe diffusion
Fick's 1st Law
J = - D (dc/dx)t
Fick's 2nd law
(dc/dt)x = D (d2 c/dx2 )
time dependence
Flux at some instant in time
Mean square displacement
1 dimensional <x2 > = 2Dt
D is diffusion constant
3 dimensional <d2 > = 6Dt
( <d2 >)1 / 2 related to Cav/z1 / 2
Viscosity - momentum transfer
When a force F is applied to a fluid to make it move through a pipe, the molecules move
at a velocity ut, but near the wall, the velocity is different - molecules mus be slower or
completely stopped at the wall. In the center the stream there much the a higher velocity.
Jmomentum = -  dux/dy
here the viscosity coefficient, , is independent of dux/dy
If  depends on dux/dy then the fluid is non-Newtonian
How is viscosity measure?
1) Free Fall of Particle through a medium -
2) Volume Rate of Flow through a Capillary.
Can also define a specific viscosity
and an intrinsic viscosity