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# Biol 304 Week 3 Equilibrium Binding Multiple - PowerPoint PPT Presentation

Biol 304 Week 3 Equilibrium Binding Multiple Multiple Binding Sites Last week we discussed the mechanism for the reversible binding to form a binary complex AB from its constituents A and B. The mechanism was treated as a dissociation of the complex into its constituents.

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### Biol 304 Week 3

Equilibrium Binding Multiple

Multiple Binding Sites

Last week we discussed the mechanism for the reversible binding to form a binary complex AB from

its constituents A and B. The mechanism was treated as a dissociation of the complex into its constituents.

We noted that the basic experiment was done so that the total concentration of A in the mixture

was kept constant and that the total amount of B in the mixture was varied from tube to tube.

Given the equilibrium expression:

AB  A + B

Then at equilibrium, the concentration of the species present in solution are related through the

Law of Mass Action by the equation:

Kd = [A][B] / [AB]

From the mass conservation equations and the above equation, we were able to derive the direct

function which relates the concentration of AB in terms of the free B concentration, Kd and the

total amount of A present:

[AB] = [A]o x [B] = [A]o x 1

Kd + [B] { 1 + (Kd/[B]) }

Since the maximum binding that can occur is when [AB] = [A]o and since [AB] is the concentration

of B bound, we can rewrite[AB] as Bb and the equation as:

Bb = Bbmax x [B] = Bbmax x 1

Kd + [B] { 1 + (Kd/[B]) }

Thus a plot of Bb versus [B], wil have the shape of a right hyperbola with the asymptote

being Bbmax. The direct function also shows that when Bb is equal to 1/2 Bbmax then

[B] at that point is equal to Kd.

We also saw that we could rearrange this function in three alternative ways to yield the semilog,

the double reciprocal and the Scatchard functions. The last two being particularly useful since they

yielded linear plots and allowed for the determination of Kd and Bbmax by simple extrapolations.

Thus, all four different functions can be used to determine the Kd and Bbmax using experimental

data of the form [AB] versus [B].

We now have to consider what happens if A has multiple binding sites for B.

We shall start with the simplest case where A has only two sites for binding B.

There are a few different possibilities. For example, the two sites may be independent and have

different affinities for binding B. Alternatively, the two sites could have the same affinity (identical)

and be independent, or that they are identical but dependent.

So how do we proceed? and since [AB] is the concentration

The first thing is that we have to do is find a chemical mechanism that represents two site binding.

Obviously treating the binding as an association, we can represent the binding by a two step mechanism:

A + B  AB

and

AB + B  AB2

and for each we can write an equilibrium association constant.

However, we shall deal with these as if they were dissociation constants and so we represent them

in the reverse direction:

AB  A + B

and

AB2  AB + B

and we can now write the equilibrium dissociation constants Kd1 and Kd2 in terms of the equilibrium

concentrations of the participants.

K1d = [A][B] and K2d = [AB][B]

[AB] [AB2]

The experiment is done as before. The total A concentration is kept constant in every tube and the

total B concentration is varied but is known. The system is allowed to come to equilibrium and then

by some chemical or spectral analysis the concentrations of B bound is obtained and from the

conservation equations the free concentration of B can be evaluated.

The thing to note here is how do we measure B bound? We have to be careful since we want a measure

of how many moles of B are bound. To do this we look at the two equilibria which must be occurring

simultaneously.

By doing this you should be able to determine which species are the ones which have B bound to them

and from this determine an expression which yields the molar concentration of B bound in the system.

Clearly, the two species which have B bound are AB and AB2 so that the molar amount of B bound

is given by:

Bb = [AB] + [AB2]

or is it

Bb = [AB] + 2[AB2]

Since we are counting the amount of B bound it is the latter expression, since each mole of [AB2] has two

moles of B bound to it.

We can get the ratio of Bb / Bbmax and relate it to [B] and K1d and K2d. Recall that Bbmax is just

equal to 2 [A]o since each A has two sites for binding B. Try and derive this expression.

If you have done this correctly, you should have obtained the following expression:

Bb / 2[A]o = Bb / Bbmax = {( [B]/K1d ) + (2 [B]2/K1dK2d) }

2 { 1 + ( [B]/K1d) + ( [B]2 / K1dK2d) }

Now let us assume, for the time being, that the 2 sites on A to bind B are identical and independent.

This means that the affinity for B at site 1 is the same as that for site 2 and, moreover, the binding

at one or the other site has no effect on the binding of the other.

Q. If this is the case then are K1d and K2d equal? Think about it and make sure you can defend your

Although the quick response is that K1d and K2d are equal, since we are assuming that the 2 sites

are identical and independent. The correct answer is that they are unequal. To see this we have to

understand exactly what K1d and K2d are measuring and remember that the 2 sites are identical

and independent. So think about this for a minute or so and see if you can see what K1d and K2d

are actually measuring.

K1d is not measuring dissociation from site 1 and K2d is not measuing the dissociation from site 2.

Q. What are they measuring?

K1d is measuring the dissociation of B from an AB molecule whether the B resides on site 1 or site 2.

K2d is measuring the dissociation of B from an AB2 molecule and this could be from either site 1 or 2.

In fact, because K1d and K2d are not measuring dissociation from individual sites they are actually

average probability factors and are called MACROSCOPIC equilibrium constants.

Since we are interested in determining the equilibrium at each site, we need to find the INTRINSIC

dissociation constant (sometimes called the MICROSCOPIC dissociation constant) at each site.

This means we have to take a microscopic view of A and designate it as -A- where the LH - represents

site 1 and the RH - represents site 2. Using this notation we see that [AB] is actually comprised of two

microscopic species B-A- and -A-B so that:

[AB] = [B-A-] + [-A-B]

However, for there is only one microscopic species and that is B-A-B.

To proceed we will now define the microscopic dissociation constants at each of the two sites as

K’1 and K’2 for sites 1 and 2 respectively. So see if you can write these two dissociation constants

in terms of the microscopic species present.

You should have something like this: measuing the dissociation from site 2.

B-A-  -A- + B and B-A-B  -A-B + B

so that K’1 = [-A-] [B] = [-A-B] [B]

[B-A-] [B-A-B]

and -A-B  -A- + B and B-A-B  B-A- + B

so that K’2 = [-A-] [B] = [B-A-] [B]

[-A-B] [B-A-B]

We now have to try and relate the microscopic K’1 and K’2 to the macroscopic dissociation constants

K1d and K2d. To do this we take each of the macroscopic equilibria and convert the macroscopic species

in terms of the microscopic species.

So let us do the first of the two equilibria:

K1d = [A] [B] = [A] [B]

[AB] {[B-A-] + [-A-B] }

Now divide top and bottom of the RHS of this equation by the numerator which is [A] [B] and this yields:

K1d = measuing the dissociation from site 2.[A] [B] = [A] [B] = _______1_______

[AB] {[B-A-] + [-A-B] } { [B-A-] + [-A-B] }

[A] [B] [A] [B]

and substituting the microscopic equilibrium constants for the two expressions in the numerator

yields:

The solution is:

K1d = _____________1___________

( 1/ K’1) + (1/K’2)

and this can be simplified to yield:

K1d = _____K’1 K’2______

K’1 + K’2

In a similar fashion you should be able to show that:

K2d = K’1 + K’2

These equations enable us to relate the microscopic K’s to the macroscopic Kd’s. We have so far

kept these relationships general and have designated the two microscopic constants differently.

Now in the case where we assume that the 2 sites are identical and independent, we then can write:

K’1 = K’2 = K’ where K’ is the microscopic dissociation constant which is the same for

both sites.

For this situation of identical and independent binding sites, we then have the following relationships

between the macroscopic and microscopic binding constants:

K1d = K’ / 2 and K2d = 2 K’

So we see that the macroscopic binding constants are not equal for 2 identical and independent sites.

We can now substitute the microscopic binding constants for the macroscopic binding constants in

the direct binding function, and you should be able to show that:

Bb / Bbmax = ______[B]_______

( K’ + [B] )

and since Bbmax = 2 [A]o we could multiply both sides of the above equation and obtain,

Bb = 2 [A]o x ______[B]______ = Bbmax x _______1_______

( K’ + [B] ) { 1 + (K’ / [B] ) }

Thus, if we carry out the binding experiment in the normal way where we keep total A constant

and equal to [A]o in every tube but vary the [B]o, then at equilibrium analyses of the concentrations

of the constituents would again yield a Table of Bb versus free [B].

The function derived from the mechanism for two identical and independent binding sites indicates

that it would be in the form of a right hyperbola just like we saw for the simplest case for AB.

Since the mathematical function is essentially identical, except for the factor 2 in the numerator of the

direct binding function, a plot of Bb versus [B] would asymptotically approach Bbmax (2 [A]o)

and the K’ (microscopic binding constant) would be equal to the value of Bb where the plot is equal

to 1/2 Bbmax.

Because of the identity of the direct function for two (or n identical and independent sites) to that of

a single site, the direct function in this case can also be rearranged to give essentially the same

semilog, double reciprocal and Scatchard rearrangements and used in the same manner to get K’

and Bbmax. A knowledge of Bbmax gives the number of binding sites since Bbmax = n [A]o

where n is the number of binding sites per mole of A.

Thus, the situation where we have n identical and independent binding sites leads to a direct

function which is essentially the same as that for the simple single binding case. The determination

of the intrinsic (microscopic) dissociation constant follows the same graphical procedures.

The number of binding sites is obtained by dividing Bbmax by [A]o.

It is informative to look at the situation where the two sites are independent but NOT identical.

In this situation the two microscopic constants are not equal so that K’1 is not equal to K’2.

So in this case we have to substitute these for the macroscopic dissociation constants in the

direct binding equation.

You should be able to show that in this case the equation becomes a function of both

microscopic dissociation constants.

Recall that

K1d = _____K’1 K’2______

K’1 + K’2

and that

K2d = K’1 + K’2

The answer is either of the following expressions:

Bb/ Bbmax = [B] { (K’1 + K’2) + 2[B] }

2 { K’1K’2 + [B] {(K’1 + K’2) + [B] }}

Since Bbmax = 2 [A]o we can multiply both sides of the previous equation to get:

Bb = [A]o {[B] (K’1 + K’2 + 2 [B] )}

{K’1K’2 + [B] {K’1 + K’2 + [B] }

Now recall that the independent variable is [B] and the dependent variable is Bb,

[A]o, K’1 and K’2 are constants.

We could write this equation as:

Bb = [A]o x {[B] (K’1 + K’2 + 2 [B] )}

{K’1K’2 + [B] {K’1 + K’2 + [B] }

We can now see what happens if we change the ratio of K’1/K’2. See if you can plot Bb versus [B]

when you change the ratio from 1, to 0.1 and 0.01. That is the same as making K’2 > or < K’1?