Biochemistry Online: An Approach Based on Chemical Logic

Biochemistry Online




Last Update:  3/25/16

  • Learning Goals/Objectives for Chapter 5A:  After class and this reading, students will be able to

    • write equations for the dissociation constant (KD), mass balance of total macromolecule (M0), total ligand (L0), and [ML]  as a function of L or Lo ([ML] = [M0][L]/(KD+ [L])  (when Lo >> Mo or when free [L] is known) and Y = fractional saturation =  Y = ([ML]/[M0] = [L]/(KD+ [L])
    • decide which of two given equations for [ML] should be used under conditions when the above conditions for L0 and L are given
    • based on the equation ([ML] = [M0][L]/(KD+ [L]) draw qualitative graphs for different given L0, L, and Kd values
    • determine fraction saturation given relatives values of Kd and L, assuming L0 >> M0
    • compare relative % bound for covalent binding of protons to an acid and noncovalent binding of a ligand to a macromolecule given pka/pH and Kd/L values
    • describe differences in binding curves for binding of a ligand to a macromolecule and the dimerization of a macromolecule
    • derive an equation which shows the relationships between the rate constant for binding (kon), dissociation (koff) and the thermodynamic dissociation (Kd) or equilibrium constant (Keq).
    • describe the structural and mathematic differences between specific and nonspecific binding
    • given a Kd, estimate t1/2 values for the lifetime of the ML complex.
    • describe techniques used to determine ML for given L or L0 values, including those that do and do not require separation of ML from M , so that Kd values for a M and L interaction can be determined
    • List advantages of isothermal titration calorimetry and surface plasmon resonance in determination of binding interaction parameters

    A4.  The Binding Continuum

    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 Kd reflects binding strength. Remember, Kd = 1/Keq. Also, we know that Keq is related to ΔGo, by the equations: ΔG o = - R T ln Keq = RT ln Kd. Given these simple equations, you should be able to interconvert between Keq, Kd, and ΔG o. (Keep your units straight.).

    NO INTERACTION: One end of the binding continuum represents no interaction. Let's assume that Keq is tiny (Kd large), for example Keq ~  2.4 10-72. Plugging this into the equation ΔG o = - R T ln Keq, 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 ΔGoform. Using General Chem. thermodynamics, we can calculate ΔGo for H-H formation. (ΔGo = ΣGoform prod. - ΣGoform 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.

    M + L <-===> 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
    vf = kf [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 vr = kr [ML], where kr is the rate constant for the reverse reaction. Notice that the units of kf are M-1s-1, while units of kr are s-1. At equilibrium, vf = vr, or kf [M][L] = kr [ML]. Rearranging the equation gives [ML]/[M][L]= kf/ kr = Keq. Hence Keq 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 108-9 ) and kr must be very small (10-2 - 10 -4 s-1).

    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 M-1s-1. Knowing this, the Kd and the fact that kon/ koff = Keq = 1/Kd, 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  Lo << Kd. 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 koff.

    Now you will remember from General Chemistry that from a first order rate constant, the half-life of the reaction can be calculated by the expression: k = 0.693/t1/2. 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 ΔGo or Kd, and assuming a kon (108 M-1s-1 ), 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.

    Calculated koff and t1/2 for binary complexes assuming diffusion-controlled kon


    KD (M)

    koff (s-1)

    t ½


    1 x 10-71

    1 x 10-63

    2 x 1055 yr

    RtV3 : Rt'L3(a)


    1 x 10-9

    2 yr



    1 x 10-7

    80 days


    5 x10-14

    5 x 10-6

    2 days


    1 x 10-13

    1 x 10-5

    0.8 days



    1 x 10-3

    700 s




    7 s


    2 x 10-9

    2 x 10-1

    3 s


    4 x 10-9

    4 x 10-1

    2 s

    LDH (pig): NADH(g)

    7.1 x 10-7(j)

    7.1 x 101

    10 ms

    profilin: CaATP-G-actin

    1.2 x 10-6

    1.2 x 102

    6 ms

    TBP: DNAnonspec(h)

    5 x 10-6

    5 x 102

    1 ms

    TCR(i): cyto C peptide



    100 us


    1 x 10-4

    1 X104

    70 us

    uridine-3P: RNase

    1.4 x 10-4 (j)


    50 us

    Creatine Kinase: ADP

    8.2 x 10-4 (j)


    10 us


    1.2 x 10-3

    1.2 x 105

    6 us

    no interaction

    4 x 1073

    4 x 1081


    1. Trivalent Vancomycin derivative RtV3 + Trivalent D-Ala-D-Ala deriv, Rt'L3'

    2. Hirudin is a potent thrombin inhibitor from leach saliva

    3. lac rep is the E. Coli lac operon repressor protein, and DNAoper is the specific DNA binding region in the E. Coli genome that binds to the repressor

    4. Zif268 is a mouse zinc-finger binding protein

    5. GroEL is a chaperone protein; r-lactalbumin is the reduced form of lactalbumin

    6. TBP is the TATA Binding Protein which binds to the TATA box consensus sequence

    7. LDH is lactate dehydrogenase

    8. DNAnonspec is DNA which does not contain the specific DNA sequence region involved in specific
      binding to a DNA binding protein

    9. TCR is the T-cell receptor

    10. calculated from equation:  KD = koff/kon.

    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 co-localized in the same environment and for which a biological function is elaborated upon binding.



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