BIOCHEMISTRY  DR. JAKUBOWSKI
Last Update: 3/25/16
Learning Goals/Objectives for Chapter 5A: After class and this reading, students will be able to

We have studied macromolecule structure. Now it is time to impart function to these molecules. It is simple to imagine that before these molecules can perform a function, they must interact with specific molecule(s) or ligand(s) in their environment. In fact, binding and subsequent release of a ligand might be the sole function of the macromolecule (example myoglobin binding oxygen). Binding is the first step necessary for a biological response (with the exception of visual transduction in which photoinduced isomerization of rhodopsin initiates the response). To understand binding, we must consider the equilbria involved, how binding is affected by ligand and macromolecule concentrations, and how to experimentally analyze and produce binding curves.
These derivations can be made and interpreted
using simple principles from General Chemistry, which you reviewed and
strengthened in Analytical Chemistry, with some slight differences.
Biochemists rarely talk about equilibrium or association constants, but
rather their reciprocals  the dissociation constants, Kd. For
the reactions M + L ↔ ML, where M is free macromolecule, L is free
ligand, and ML is macromoleculeligand complex (which is held together by
intermolecular forces, not covalent forces), the K_{d} is given by
[M]eq[L]eq/[ML]eq.
Figure: M is free macromolecule, L is free ligand, and ML is macromoleculeligand complex
Notice the unit of Kd is molarity, M. The lower the Kd (i.e. the higher the [ML] at any given M and L), the tighter the binding. The higher the Kd, the looser the binding. Kd's for biological molecules are finely tuned to their environments. They vary from about 1 mM (weak interactions) for some enzymesubstrate complex, to pM  fM levels. Examples of very tight, noncovalent interactions include the avidin (an egg protein)biotin (a vitamin) and thrombin (enzyme initiating clotting)hirudin (a leech salivary protein) complexes.
M = macromolecule; L = ligand
For a simple equilibrium M + L <> ML
where M = free macromolecule, L = free ligand, and ML = bound M and L (a complex)
3 equations can be written:
Equation 1  Dissociation constant: K_{d} = ([M]eq[L]eq)/[ML]eq = ([M][L])/[ML] (units of molarity)
Equation 2  Mass Balance of M: Mo = M + ML
Equation 3  Mass Balance of L: Lo = L + ML
We would like to derive equations which give ML as a function of known or measurable values. The Kd equations shows that ML depends on free M and free L. From Equations 13, two different and equally valid equations can be derived for two different cases.
EXPERIMENTAL CASE 1: USE THIS FORM OF THE EQUATION WHEN L IS MEASURABLE OR WHEN Lo >> Mo (i.e. L= Lo)
Equation 4  Substitute 2 into 1: K_{d} = ([M][L])/[ML] = [MoML][L])/[ML]
(ML)Kd = (Mo)L  (ML)L
(ML)Kd + (ML)L = (Mo)L
(ML)(Kd+L) = (Mo)L
Equation 5: ML = MoL/(Kd + L)
This equation is ALWAYS TRUE for the chemical equation written above. L is the free ligand concentration at equilibrium.
Wolfram Mathematica CDF Player  Interactive Graph of ML vs L at different Mo and Kd values (free plugin required)
Interactive SageMath Graph: ML vs L at different Mo and Kd values
If Lo >> Mo, then the equations simplifies to:
Equation 6: ML = MoLo/(Kd + L)..
Dividing Equation 5 by Mo gives the fractional saturation of the macromolecule M, where
Equation 7: Y = θ= [ML]/Mo = L/(Kd + L)
where Y can vary from 0 (when L = 0) to 1 (when L >> Kd)
Wolfram Mathematica CDF Player  Interactive Graph of Y vs L at different Kd values (free plugin required)
Graphs of ML vs L (equation 5) and ML vs Lo
(equation 6), when Lo >> Mo, and Y vs L (equation 7) are all HYPERBOLAs
Equations 5. ML = MoL/(Kd + L) (and by analogy 6 and 7) can be understood best by examining three cases:
Case 1: L = 0, ML = 0
Case 2: L = Kd, ML = MoL/(L + L)= MoL/2L = Mo/2
which indicates that M is half saturated. In fact the operational definition of Kd is the ligand concentration at which the M is half saturated.
Case 3: L >> Kd, ML = Mo
EXPERIMENTAL CASE 2 (more general): USE THIS FORM OF THE EQUATION WHEN FREE L IS NOT KNOWN (such as when Lo is not >> Mo) OR YOU WISH TO CALCULATE ML FROM JUST Lo, Mo AND KD
Equation 8  Substitute 2 AND 3 into 1: K d = ([M][L])/[ML] = [MoML][LoML]/[ML]
(ML)Kd = (Mo  ML)(Lo  ML)
(ML)Kd = (Mo)(Lo)  (ML)(Lo)  (ML)(Mo) + (ML)2 or
Equation 9: (ML)2  (Lo + Mo +Kd)(ML) + (Mo)(Lo) = 0, which is of the form
ax^{2} + bx + c = 0, where
a = 1
b =  (Lo + Mo +Kd)
c = (Mo)(Lo)
which are all constants, and
x = {b +/ (b2  4ac)^{1/2}}/2a or
Equation 10: ML = {(Lo+Mo+Kd)  ((Lo+Mo+Kd)2  4MoLo)^{1/2}}/2
Wolfram Mathematica CDF Player  Interactive Graph of ML at various Lo, Mo, and Kd values (free plugin required)
A graph of ML calculated from this formula vs free L (or Lo if Lo >> Mo) give a A HYPERBOLA.
Play around with the sliders. If you set Kd to a very low number and vary Mo, you will see a curve very much like a titration curve with a sharp rise and abrupt plateau that occurs when Mo is approximately equal to Lo.
Interactive SageMath Graph: ML at various Lo, Mo and Kd values
In the derivations, we came up with two equations for ML:
Both equations are valid. In the first you must known free L which is often Lo if Mo << Lo. In the second, you don't need to know free M or L at all. At a given Lo, Mo, and Kd, you can calculate ML, which should be the same ML you get from the first equation if you know free L.
Equations 5 and 10 are useful in several circumstances. They can be used to
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