A4. The Binding Continuum
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 132764
Binding affinities give us a way to measure the relative strength of binding between two substances. But how "tight" is tight binding? Weak binding? Let us exam that issue by considering a binding continuum. Consider two substances, \(A\) and \(B\) that might interact. Over what range of strengths can they actually bind to each other? It would helpful to set up the extremes of the binding continuum. At one end is no binding at all. At the other end, consider two things that bind covalently. We have discussed how \(K_d\) reflects binding strength. Remember,
\[K_d = \dfrac{1}{K_{eq}}.\]
Also, we know that \(K_{eq}\) is related to \(ΔG^o\), by the equations:
\[\Delta G^o =  RT \ln K_{eq} = RT \ln K_d\]
Given these simple equations, you should be able to interconvert between \(K_{eq}\), \(K_d\), and \(ΔG^o\). (Keep your units straight.).
NO INTERACTION
One end of the binding continuum represents no interaction. Let's assume that \(K_{eq}\) is tiny (Kd large), for example \(K_{eq} ~ 2.4 1072\0. Plugging this into the equation
\[\Delta\,G^o = RT\, \ln K_{eq}\]
where R = 2.00 cal/mol.K, and T is about 300K, the ΔGo ~ +100 kcal/mol. That is, if we add A + B, there is no drive to form AB. If AB did form, then it would immediately fall apart.
COVALENT INTERACTION
At the other end of the continuum consider the interaction of 1H atom with another to form H2. From a general chemistry book we can get \(ΔG^o_{form}\). Using General Chem. thermodynamics, we can calculate \(ΔG^o\) for HH formation.
\[ΔG^o = \sum G^o_{form} prod.  \sum G^o_{form} react.\]
Doing this gives a value of 97 kcal/mol.
SPECIFIC AND NONSPECIFIC BINDING
Consider the interaction of a protein, the lambda repressor (R), with a small oligonucleotide to which it binds tightly (called the operator DNA, O). This is an example of a biologically tight, but reversible interaction. R can bind to many short oligonucleotides due to electrostatic interactions and H bonds from the positively charged protein to the negatively charge nucleic acid backbone. The tight binding interaction, however, involves oligonucleotides of specific base sequence. Hence we can distinguish between tight binding, which usually involves specific DNA sequence and weak binding which involves nonspecific sequences. Likewise, we will speak of specific and nonspecific binding. R and O, which bind with a Kd of 1 pM, is an example of specific binding, while R and nonspecific DNA (D), which bind mostly through electrostatic interactions with a Kd of 1 mM, is an example of nonspecific binding. You might expect any positively charged protein, like mitochondrial cytochrome C, would bind negatively charged DNA. This nonspecific interaction would have no biological significance since the two are localized in different compartments of the cell. In contrast, the interaction between positively charged histone proteins, bound to DNA in the nucleus, would be specific.
RATE CONSTANTS FOR ASSOCIATION AND DISSOCIATION
When the reaction \(M + L \rightleftharpoons ML\) is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. From General Chemistry, the forward reaction is biomolecular and second order. Hence the vf, the rate in the forward direction is proportional to [M][L], or \(v_f = k_f [M][L]\), where kf is the rate constant in the forward direction. The rate of the reverse reaction, vr is first order, proportional to [ML], and is given by \(v_r = k_r [ML]\), where kr is the rate constant for the reverse reaction. Notice that the units of kf are M1s1, while units of kr are s1. At equilibrium, \(v_f = v_r\), or \(k_f [M][L] = k_r [ML]\). Rearranging the equation gives
\[ \dfrac{[ML]}{[M][L]} = \dfrac{k_f}{k_r} = K_{eq}.\]
Hence \(K_{eq}\) is given by the ratio of rate constants. For tight binding interactions, Keq >> 1, Kd << 1, and kf is very large (in the order of 1089 ) and kr must be very small (102  10 4 s1).
To get a more intuitive understanding of Kd's, it is often easier to think about the rate constants which contribute to binding and dissociation. Let us assume that kr is the rate constant which describes the dissociation reaction. It is often times called koff. It can be shown mathematically that the rate at which two simple object associate depends on their radius and effective molecular weight. The maximal rate at which they will associate is the maximal rate at which diffusion will lead them together. Let us assume that the rate at which M and L associate is diffusion limited. The theoretical kon is about 108 M1s1. Knowing this, the Kd and the fact that \(\dfrac{k_{on}}{k_{off}} = K_{eq} = \dfrac{1}{K_d}\), we can calculate koff, which remember is a first order rate constant..
We can also determine k off experimentally. Imagine the following example. Adjust the concentrations of M and L such that Mo << Lo and Lo>> Kd. Under these conditions of ligand excess, M is entirely in the bound from, ML. Now at t = 0, dilute the solution so that \(L_o \ll K_d\). The only process that will occur here is dissociation, since negligible association can occur given the new condition. If you can measure the biological activity of ML, then you could measure the rate of disappearance of ML with time, and get koff. Alternatively, if you could measure the biological activity of M, the rate at which activity returns will give you \(k_{off}\).
Now you will remember from General Chemistry that from a first order rate constant, the halflife of the reaction can be calculated by the expression: \(k = \dfrac{0.693}{t_{1/2}}\). Hence given koff, you can determine the t1/2 for the associated species existence. That is, how long will a complex of ML last before it dissociates? Given Δ< face="Arial">Go< face="Arial"> or Kd, and assuming a kon (108 M1s1 ), you should be able to calculate koff and t1/2. Or, you could be able to determine koff experimentally, and then calculate t1/2. Applying these principles, you can calculate the parameters below.
Complex  KD (M)  k_{off} (s^{1})  t ½ 

H2  1 x 10^{71}  1 x 10^{63}  2 x 10^{55} yr 
RtV3 : Rt'L3(a)  10^{17}  1 x 10^{9}  2 yr 
Avidin:biotin  10^{15}  1 x 10^{7}  80 days 
thrombin:hirudin(b)  5 x10^{14}  5 x 10^{6}  2 days 
lacrep:DNAoper(c)  1 x 10^{13}  1 x 10^{5}  0.8 days 
Zif268:DNA(d)  10^{11}  1 x 10^{3}  700 s 
GroEL:rlactalbumin(e)  10^{9}  0.1  7 s 
TBP:TATA(f)  2 x 10^{9}  2 x 10^{1}  3 s 
TBP:TBP  4 x 10^{9}  4 x 10^{1}  2 s 
LDH (pig): NADH(g)  7.1 x 10^{7}(j)  7.1 x 10^{1}  10 ms 
profilin: CaATPGactin  1.2 x 10^{6}  1.2 x 10^{2}  6 ms 
TBP: DNAnonspec(h)  5 x 10^{6}  5 x 10^{2}  1 ms 
TCR(i): cyto C peptide  7X10^{5}  7X10^{3}  100 us 
lacrep:DNAnonspec(h)  1 x 10^{4}  1 X10^{4}  70 us 
uridine3P: RNase  1.4 x 10^{4} (j)  1.4X10^{4}  50 us 
Creatine Kinase: ADP  8.2 x 10^{4} (j)  8.2X10^{4}  10 us 
Acetylcholine:Esterase  1.2 x 10^{3}  1.2 x 10^{5}  6 us 
no interaction  4 x 10^{73}  4 x 10^{81}   

What is usually measured is Kd and/or koff (if the koff is reasonable). This analysis is very simplified. Electrostatic forces and other orientation factors may significantly change kon, while conformational changes in the complex may prevent ready unbinding of the bound ligand, dramatically altering koff.
The structure of one of the tighest binding complexes, avidin and biotin, is shown below.
Jmol: Updated Avidin:Biotin Complex (1AVD) Jmol14 (Java)  JSMol (HTML5)
It is important to note that even reactions characterized by high Kd can be specific. Specificity is ultimately defined as a binding interaction between a macromolecule and ligand that can be colocalized in the same environment and for which a biological function is elaborated upon binding.