Peter FajerBiophysical Methods in BiologyLecture 10

Ligand Binding

Consider ligand A binding to macromolecule P:

[1]

Equilibrium constant is defined as:

[2]

(note that Kd is often referred wrongly as binding constant: Kb = 1/Kd)

define fractional occupancy (moles of ligand/mole of macromolecule) r:

[3]

substitute [P][A]/Kd for [PA]:

[4]

this hyperbolic dependence is called binding isotherm or (Langmuir isotherm).

Note that although the isotherm has the concentration for free ligand [A] as well as bound ligand ([A]bound in r ) you have to measure only one (bound or free) because conservation of mass the total ligand [A]total = [A]+[PA] (and total macromolecule [P]total = [P]+[PA]).

Measure the concentrations of free or bound ligand by variety of methods: chromatography, equilibrium dialysis, ultrafiltration, spectroscopy.

Multiple binding sites

Macromolecule can have more than one site for binding the ligand. These sites can be independent: binding of one ligand does not influence binding of the next, or cooperative(binding of one affects binding of another).

Independent

[5]

  • identical sites: A1P=A2P or
  • different sites: A1PA2P
Identical sites

all sites are independent and identical so that total number of sites is n[P]total rather than [P]total :

[6]

Scatchard plot

plot the ratio of r/[A] versus r:

[8]

For identical binding this results in a linear plot. Deviation from linearity implies non-identical sites.

Different sites

to be very general: each class of sites with different Kdi can have ni identical sites

The binding isotherm is a sum of the single binding isotherms, each corresponding to different class of sites:

[7]

Cooperativity

Binding to one site on a molecule modulates binding of another.

For strong cooperativity, binding of one ligand triggers binding to n sites:

[9]

[10]

[11]

or measure the ratio of filled sites Y (Y=r/n) to the unfilled sites (1-Y):

[12]

(Hill’s equation)

Hill’s Plot

Plot log(Y/(1-Y) v. log[A], the slope is n; an intercept is 1/Kd.

Kinetics

Time course of reaction, kinetics give information about the rates rather than equilibria. Of course rates and equilibria are related Keg = k+1/k-1, however the ratio of rates tells us nothing about the value of each rate i.e. is it fast or is it slow.

Two sorts of kinetics are usually considered:

  • steady-state: in which the forward flux = backward flux resulting in no change of concentration;
  • transient kinetics: non-equilibrium kinetics when concentrations are changing.

steady-state (Michaelis-Menten)

[13]

for a reaction with an intermediate X the rate of intermediate’s production is equal to that of its disappearance:

The rates of substrate disappearance and the appearance of the product are identical:

[15]

From mass conservation we have:

[16]

At the beginning of the reaction:

[17]

thus the initial velocity v can be solved since we have five equations with 4 unknowns (concentrations of E, S, X, P)

[18]

(dirty shortcut: realize that initial velocity is a maximum velocity reduced by the partial enzyme occupancy, i.e.:

[18]

then substitute v for r in the equation for Langmuir isotherm.)

turnover

[19]

non-steady state

Steady-state approximation is not always applicable, for general reaction:

[20]

the rate equations are:

[21]

with two mass conversation relationships:

[22]

The differential (rate) equation can be solved but it is a mess since the equation is non-linear (the rate depends on product of [E] and [S]), better approximate that So > Eo so that concentration of substrate never changes (pseudo-first order approximation):

[23]

This equation is a linear in X1 and it can be integrated to give the concentration of X1 as function of time:

[24]

Presence of two intermediates complicates the affairs considerably:

[25]

conservation of mass gives:

[26]

the rate of enzyme disappearance is:

[27]

the rate of product appearance is:

[28]

At this point you are advised to give up. What you have is the set of simultaneous, non-linear differential equations which even your grandma can’t solve.

Bill Gates to the rescue; integrate the set using a PC and any mathematical package with numerical integration routines e.g. Mathcad, Mathematica, Matlab.

Binding_Kinetics.doc3/4/04 11:08 AM1