Equilbrium constants

Biochemistry/MCB 568 -- Fall 2007
John W. Little--University of Arizona

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A useful concept in biochemistry is that of the equilibrium constant. You should have a good working knowledge and intuition about this. Consider the following binding reaction:

 In this reaction, A and B could be any molecules that interact. Examples are two proteins; a specific DNA-binding protein and its binding site;
an RNA molecule binding to another RNA; or a small molecule (like the "inducer" in the models) binding to a protein.

One can define an equilbrium constant K_d:

K_d = [A] [B] / [A:B]

where [A], [B] and [A:B] represent the concentrations of A, B, and A:B respectively. This is a "dissociation constant". An "association constant" is simply the inverse of this. I like to use dissociation constants because the units are Molar as opposed to M^-1. This makes it easier (at least for me) to think about affinities in terms of molar concentrations. Importantly, the equilibrium constant has a constant value (under a defined set of conditions), so that if one knows the concentrations of two components the concentration of the third component is fixed. If one knows the concentration of one component, the ratio of the other two is determined.

How is this concept useful? Let's take an example, discussing specific protein-DNA interactions. Suppose that the value of K_d is 10 nM. Call the protein A and the free binding site B. Hence, if the free protein concentration is 10 nM, half the operator sites will be bound; [B] = [A:B]. The "fractional occupancy" is [A:B] / ([B] + [A:B])--that is, the fraction of the total operators that are in a complex. What fraction do you think will be bound if the free protein concentration is 1 nM, 100 nM or 1000 nM? There's no need to be exact about this; a rough number is good enough. Answer is here. In each case, you can obtain it by a) setting [A] in the equation defining K_d equal to the free protein concentration; b) determining the ratio of [B]/[A:B]; c) expressing [B] in terms of [A:B]; and d) substituting into the equation for fractional occupancy. For instance, at 100 nM free A,

10 = 100 [B] / [A:B]

[B] = 0.1 [A:B]

fractional occupancy is [A:B] / ([A:B] + 0.1 [A:B]) or (cancelling out [A:B]) 1/1.1 = 0.91 .

Biochemists often refer to this effect of raising the reactant concentration as "driving the equilibrium towards binding".

It is important to realize that, at equilibrium, it is not the case that all the molecules are just sitting there. In most cases, except very tight binding, the reactions continue to go in both directions. As I said about DNA binding proteins, they are always going on and coming off. It helps to understand this by realizing that, at equilibrium, the rates of the forward and back reactions are the same. They aren't zero. Since these rates are the same, the net change over time in the concentrations of A, B and A:B is zero, but the rates are not zero.

Now let's imagine that you have a mutation in the binding site that weakens binding--as discussed in class, this would be equivalent to "lower affinity". Let's say the dissociation constant is now 100 nM. What fraction of the operators will be bound at free protein concentrations of 1, 10, 100 and 1000 nM? Answer is here.

 By this example, you can see that even with a weak binding site you can obtain high fractional occupancy simply by raising the concentration of protein high enough. Again, you're driving the equilibrium towards binding. Looking at both answers, you can see that at low concentrations the fractional occupancy is 10 times higher for the stronger binding site; but at 1000 nM A both sites are almost fully occupied.

Understanding equilibria is helpful in thinking about the method of co-immunoprecipitation (or "co-IP"). This technique is designed to ask whether two (or more) molecules are present in a complex. Consider the simple complex discussed above. Antibody directed against A is added to a mixture of A and B (this could also be done in a cell-free crude extract). Assume that antibody binds tightly to A, so that the antibody-antigen complex does not dissociate. After time to allow antibody binding, antibody-antigen complexes are precipitated, washed a few times, and analyzed, typically using a Western. If the interaction between A and B is relatively weak, the A:B complex may dissociate during the washing steps--that is, since free A and B have been removed, the back reaction A:B --> A + B can occur, and some portion of B will be lost. If this rate is relatively rapid, the complex will dissociate before you can complete the washes.

It's worth being aware that one can have the same value for the equilibrium constant if both forward and back reactions are rapid, or if they are slow. As stated above, at equilibrium these two rates are equal. But knowing the value of the equilibrium constant does not tell you the rates of the forward and back reactions. Even for relatively tight binding, if the back reaction is rapid the complex may dissociate (as in the co-IP example).

Cooperativity

A more complicated situation is observed in cases in which a large complex is held together by a number of weak interactions. "Weak interactions" simply means that the individual interactions are not strong--i.e. the dissociation constant is high (let's say 1 micromolar or more). Individually they are too weak to make a stable complex, but adding several together creates a strong complex:

Picture from Frankel and Kim, Cell 65, 717-719 (1991).

In this case, the goal is to assemble the three-part complex on the right. The initial interactions are very weak, as symbolized by the heavy left-ward arrows (back reaction), but the second reaction is very strong. As noted in the legend, cooperativity of the second interaction makes it stronger by a factor m, the cooperativity parameter. m could have a large value, e.g. 1000, representing 1000-fold tighter binding. For example in the bottom pathway, the second step is adding the "ball" to the complex; the ball interacts weakly with either component alone (top two pathways) but much more strongly to the binary complex. This is because it can make many more interactions, and because the other two components are already fixed in space relative to each other, so entropy has already been lost.

You can think in a similar way about binding of a dimer of a DNA-binding protein to DNA, it's far more favored than binding of two monomers separately. Cooperative interactions among proteins already bound to DNA are another example.

You can get more information about equilibria, and how to relate it to free energy, in any biochemistry textbook.

 

Answer for 10 nM: Return

Concentration

Concentration	Fractional occupancy
1 nM	~0.1 (actual value 0.091)
100 nM	~0.9 (actual value 0.91)
1000 nM	~0.99 (actual value 0.9901)

Rough rule of thumb: If you're 10 times below the dissociation constant, you have 10% occupancy; 10 times above it, you have 90% occupancy.

Answer for 100 nM: Return

Concentration

Concentration	Fractional occupancy
1 nM	~0.01 (actual value 0.0099)
10 nM	~0.1 (actual value 0.091)
100 nM	0.5
1000 nM	~0.9 (actual value 0.91)