Biological Function
Equilibrium Binding
Many processes in biochemistry and pharmacology involve the
reversible binding of one molecule to another and the aim of all
these studies is to characterize the strength of the interaction and
the number of sites that the target molecule has for the other
molecule. The target molecule is generally a biological
macromolecule, which we well designate as A, and the molecule
that binds to it is called the ligand and we shall call it B.
These binding experiments are done in essentially the same
way no matter the nature of A or B.
The concentration of the target molecule is kept a constant in a
series of test tubes, and the amount of B is varied in each tube.
The solutions are allowed to react until equilibrium is
established and then by chemical or spectral characterization,
the amount of B bound to A is measured in each tube.
Binding Experiment
[A] known
Buffer
[B] known
All have exactly the same volume. [A]o is same but [B]o is different.
1. Single Binding site.
Let us examine the simplest case where A and B combine to form AB. At equilibrium we can
for this system write the following:
A + B  AB
Since this is an association reaction, we know from the law of mass action that the concentrations
of A, B and AB at equilibrium are related through the equilibrium constant – in this case an
association equilibrium constant.
Kassoc =
[AB]__
[A][B]
(1.0)
The Kassoc can be looked as an index of attraction between A and B. The larger the Kassoc,
the higher the concentration of AB relative to free A and free B at equilibrium. In other words
the bigger Kassoc the more AB is present at equilibrium.
We could have just as easily written the chemical equation in the reverse way. That is, in the form
of a dissociation. In this case,
AB
 A + B
Again from the law of mass action we would be able to show that the equilibrium constant for this
dissociation, Kd is given by:
Kd = [A][B]
[AB]
(1.1)
This is just the reciprocal of the expression we had previously. Thus, Kd is equal to 1/Kassoc.
Kd can be looked upon as a repulsion index. The larger is Kd, the less tendency for A to combine
with B and the lower the amount of AB at equilibrium.
As a rule in biological studies, the binding equilibrium is represented as a dissociation and
the binding given by the “repulsion” index, Kd.
Now the concentration terms for the equilibrium expression refer to the concentrations
present at equilibrium and NOT TO THE INITIAL CONCENTRATIONS.
The equilibrium concentrations can be related to the initial concentrations by conservation of mass
equations since it is obvious that:
and
[A]o = [A] + [AB]
(1.2)
[B]o = [B] + [AB]
(1.3) .
Now recall that in the experimental procedure used, we made sure that [A]o was the same in every
tube but the total B concentration, that is [B]o, was varied. Thus, a knowledge of the concentration
of any one of the three species existing at equilibrium will allow for the determination of Kd since
the concentrations of the other two can be determined by substitution into equations (1.2) and (1.3).
Since it is [A]o which is maintained constant, it is easily shown (and you must be able to derive it)
that the amount of AB present at equilibrium is related to the concentration of free B present, [B]
and NOT [B]o by the function:
[AB]
=
[A]o [B]
Kd + [B]
(1.4)
or in its alternative form,
[AB] =
[A]o x ____1_______
1 + (Kd/[B])
(1.5).
Let us pause here and see what these two equations exactly represent. These mathematical equations
represent the chemical equilibrium process for an A plus B forming an AB complex. They predict
that at equilibrium, the amount of bound B (given by [AB]) is related to the free B present at
equilibrium (given by [B]) by these functions where [A]o is a constant and Kd is obviously a
constant also.
Thus, if we carry out the generic binding experiment, we can determine the amount of AB and free B
in each tube allowing us to write a Table for these paired values. If we now plot B bound (ie., [AB])
versus free B (ie [B]), these functions give us some idea of how these paired data points would look
like on a direct plot.
Q1. What would such a plot look like?
If you have difficulty with this expression rewrite the two equations as:
y = ax / (kd + x)
or as,
y = a * _____1______
1 + (Kd/x)
and indicate what the plot of y versus x would look like. This is known as a direct plot. In our
binding case y is [AB] and x is free [B].
The equation (1.3 or 1.4) is known as the direct binding function.
What this analysis tells us is whether the data conform to the expected behavior based on the
chemical mechanism. If it does then the data points should lie on a line whose shape conforms to
that predicted by the direct binding function. In the event that it does not, then another chemical
mechanism must be invoked to explain the deviant data.
If we plot the data with an appropriate scale we could use the plot to determine the Kd.
To see how we could do this we could go to the direct binding function and see what the free [B]
at equilibrium would be when the Bbound is equal to 1/2 of Bboundmax.
So do this and tell me what you get and then how you would use this information and get the Kd
from your direct plot.
Let us pause and see what this all means. If we were to do the binding experiment and analyze the
contents of each tube at equilibrium, we could set up a table of the B bound versus free [B] each row
being the data from one test-tube. If we plot these data as a direct plot and use an appropriate scale
then we should be able to get some estimate for the Kd of the binding. It would require estimating
what the Bboundmax would be from the asymptote, and then finding the value of the free [B] when B bound
is 1/2 of the Bboundmax.
The main problem here is the accuracy in estimating theBboundmax.
There are other ways of treating the data. These treatments involve rearrangement of the direct function.
For example, starting with the direct function you should be able to show that:
Bbound
=
Bboundmax / (1 + 10(pB-pKd))
This is known as the semi-logarithmic plot. pB is equal to -log[B] and pKd = -logKd.
So what would the plot of Bbound versus pB look like? See if you can figure that out.
Thus, if we take the raw data of Bbound versus free[B] and convert the free [B] to pB, we can then plot
Bbound versus pB. These data should when plotted in this way have the same general shape as predicted
by this function and can be used to get Kd also.
Again the requirement here is that we are able to estimate Bboundmax with reasonable accuracy.
Since the Kd can be determined by getting the pB where Bbound is equal to 1/2 of Bboundmax.
You can very easily determine what pB is for this situation.
The limit on the accuracy is how well one can estimate the Bboundmax .
So let us recap for a minute. We have seen now that the mechanism for reversible binding yields the
direct function, enabling the Kd to be estimated graphically from the direct plot.
We have also see that the direct binding function can be rearranged in the form of a semi-logarithmic
function and that when the data are plotted in this way the Kd can also be determined from the pKd.
The main factor limiting the accuracy of the Kd is how well we can estimate Bboundmax .
There are two other transformations of the direct function which can be used to overcome this
difficulty.
The first of these is known as a double reciprocal function. To get it you have to reciprocate the
LHS and the RHS of the direct binding function. So see if you can do this.
This then gives the following equation:
1/ Bbound = (1/ Bboundmax) + (Kd/ Bboundmax)(1/( [B]))
So if we plot 1/ Bbound versus 1/[B] what will the graph look like?
How could you determine Bboundmax and Kd from such a plot?
Thus, we see that using the double reciprocal form we can, in general, get a better idea of what is the
Bboundmax value, and also what is the Kd.
Again let us recap and see what this means. The double reciprocal function enables us to take the
raw data and plot their reciprocals to evaluate Kd and Bboundmax .
So now we have three different ways of graphing the experimental data.
We now have one other way to treat the experimental data. This is known as a Scatchard plot.
In this case we plot Bbound /[B] versus Bbound. So starting with the direct function determine the
relationship between these parameters and determine what the plot of Bbound /[B] would look like.
You should also be able to determine how you would evaluate Kd and Bboundmax from this Scatchard
plot.
Thus, if we have Bbound versus free [B] data, we can rearrange it in the form required for the Scatchard
function and from the graph evaluate Kd and Bboundmax .
Moreover, if we know the initial concentration of A ( [A]o) then dividing B boundmax by [A]o yields
the number of sites present on A for the binding of B.
So let us recap. We have seen that the mechanism for reversible binding equilibrium for the dissociation
of a binary complex AB can lead to an expression (the direct function)relating B bound versus free [B]
in terms of Kd and themaximum amount of B which can be bound (Bboundmax = [A]o). The latter two
parameters define the binding process.
Furthermore, we saw that this direct function could be rearranged three other ways offering alternate
methods of analyzing the data. Since the double reciprocal and the Scatchard functions yield linear
plots of the rearranged data, these are known as linear transforms.