Population distribution with multiple...

Population distribution with multiple sites

The probability of finding a macromolecule with j ligands bound is:

p( j )=[PL j ]

[P ]T

For a given [L], a fraction θ of all sites will be bound and a fraction 1-θ will be free.

=[L ]

K d[L]

f =1−=K d

K d[L ]

The probability of finding a particular combination of j sites bound and n-j sites free is (from the binomial distribution):

p j sites filled ,n− j sites free=j f n− j


But we need to multiply that by the total number of combinations for filling j sites to get the probability of picking a macromolecule with j sites filled:

We can instead, express this in terms of Kd and [L]:

Which simplifies to:

p j =n !

j ! n− j !

j f n− j

p( j )=n !

j ! (n− j )! ([L]

K d+ [L ])j

( K d

K d+ [L ])n− j

p( j )=n !

j ! (n− j )!

[L] j K dn− j

(K d+ [L ])n


Or to:

p( j )=n !

j ! (n− j )!

([L]/K d)j

(1+ [L ]/K d )n

If we define α = [L]/Kd and recall that p(j) = [PL

j]/[P]

T:

[PL j ]

[P ]T=

n!j ! n− j !

j

1n

This provides the relative distribution of [PLj] as a function of j, n,

[L] and Kd. In other words it tells the fraction of macromolecules

with j ligands bound and therefore the population distribution of species for all j.


So now we can illustrate this distribution as a function of [L] for a macromolecule with a K

d of 10nM for binding L.

[L](nM)

α [P]/[P]T

[PL1]/[P]

T[PL

2]/[P]

T[PL

3]/[P]

T[PL

4]/[P]

Tθ

0 0 1.0 0 0 0 0 0

2 0.2 0.482 0.386 0.116 0.015 0.001 0.667

10 1 0.063 0.250 0.375 0.250 0.063 2.0

30 3 0.004 0.047 0.211 0.422 0.316 3.0

100 10 0.000 0.003 0.041 0.273 0.683 3.636

The value of θ can be calculated also by adding up the relative fractional contributions of each state:

=[PL1]

[PL ]T2

[PL2]

[PL ]T3

[PL3]

[PL ]T4

[PL4]

[PL ]T


How does this look if we graph [Pj]/[P]

T versus [L]?

Significance of the population distribution

Suppose you are working with an enzyme that depends on binding 4 Mg2+ ions (with a K

d=10nM) for its function. The 4

metal binding sites are independent, but the enzyme is only functional when all 4 sites are filled.

What sort of curve would you see for activity as a function of [Mg2+]?

Measurement of Binding

There are a variety of methods by which binding can be measured. We'll discuss just a few:

● Equilibrium dialysis

● Surface plasmon resonance

● Measurement of “signals” such as fluorescence

Equilibrium Dialysis

Semi-permeable dialysis membrane

Chamber 1 Chamber 2

L

P+L

PL

● Ligand is placed in chamber 1 ● Macromolecule is placed in chamber 2. ● The ligand equilibrates across the membrane. ● At equilibrium, the free ligand concentration is the same in

both chambers, but chamber 2 also contains the bound ligand.● The difference is the amount of ligand bound.

Equilibrium Dialysis

Hence we can calculate θ via:

=[LT ]2−[LT ]1

[P ]T

and [L] is simply equal to [LT]1.

Thus one simply measures the total ligand concentrations in both chambers at several [L

T].

Surface plasmon resonance

Buffer

Gold chip

Incident light generates an evanescent wave at the gold surface. The evanescent wave interacts with the electron density of the gold atoms and induces plasmons (electron charge density waves), reducing the intensity of the reflected light. The angle at which this occurs is dependent on the refractive index of the solution near the gold layer.

Reflected light (reduced intensity)

Evanescent wave

Incident polarized light

BiacoreTM


Buffer flow

Gold chip

● The refractive index of the solution near the gold layer is affected by the amount of material bound to the surface of the gold layer.

● The resonance signal changes in proportion to the amount of material bound.

● Flowing a ligand that can bind to the immobilized molecule produces a change in the resonance signal.

Evanescent wave

Immobilized molecule (e.g. protein)


● With a continuous flow of ligand, the signal eventually reaches a steady state.

● The buffer solution is then changed to one containing no ligand.

● The resonance signal now falls in proportion to the dissociation of the ligand.

● Hence both kon

and koff

are directly measured.

● One can then calculate Kd

Buffer

Sig

na l

Ligand

Time

Bufferonly


The dissociation phase allows the direct calculation of the koff

rate constant from the change in signal (ΔR) starting from the point where buffer alone is added (ΔR

0).

R=R0 exp−k off t

Calculation of kon

is somewhat more complicated since ligand can bind and dissociate during the time that ligand is flowing over the chip. In other words (where s represents sites):

d [sL ]dt

=kon [s ][L ]−k off [sL]

Ultimately from this equation and knowing that [sL] = αΔR, where α is a proportionality constant, one can derive the equation:

R=k on[s ]T [L ]

kon[L]k off 1−exp −kon [L ]k off t

Knowing koff

from the dissociation phase, allows calculation of kon

by a non-linear least-squares fit to this equation.


Alternatively, one can calculate Kd from the plateau value by

noting that the plateau is reached where t →∞, thus making the exponential term approach zero. Resulting in:

RP=[s ]T

[L ]K d[L]

Thus from the dissociation phase, one can calculate the koff

and from the above equation one can determine K

d and hence k

on.

e.g. of the use of surface plasmon resonance

Calpastatin binding to calpain:

Each of domains 1-4 of calpastatin can bind to one calpain molecule.


Kinetic constants for the binding of calpastatin to calpain were measured with the Biacore instrument:

Domain kon

koff

Kd (k

off/k

on)

1 2.0x107 9.0x10-5 4.5x10-12

2 3.0x106 1.1x10-2 3.8x10-9

3 5.0x106 3.0x10-3 6.0x10-10

4 6.5x106 3.0x10-4 4.8x10-11

Note that the largest and smallest kon

values differ by less than 10-fold, while the largest and smallest k

off values differ by a factor

of more than 120, resulting in a more than 1000-fold range of Kd

values.


Domains 1 and 4 were chosen for crystallization trials and a structure of the complex of domain 4 with calpain was obtained:

Measurement of “signals”

In an equilibrium,

where “s” represents sites, often s (or L) can be distinguished from sL by some spectroscopic method such as fluorescence, absorbance, NMR, ESR etc. The total signal (S) is some linear combination of the signals from s (or L) and sL.

s + L sL

S=S f [s ]Sb [sL]

where Sf and S

b are coefficients of signals from free s (or L) and

bound sL, respectively.

Sig

nal

S0

Sb > S

f

Sb < S

f

[L]


Sig

nal

S0

Sb > S

f

Sb < S

f

[L]

The points where [L] = 0 and [L]→∞ can eliminate the terms Sb

and Sf, respectively:

S0=S f [s ]T & S∞=Sb[s ]T

These can be substituted into the previous expression to give:

S=S0 f S∞

where θ is the fraction of sites bound and f is the fraction of sites free (i.e. 1-θ).

S=S f [s ]Sb [sL]


If ΔS = S – S0 and ΔS

max = S

∞-S

0, the fraction of sites bound is:

=S

Smax

Sig

nal

S0

[L]

ΔSmax

ΔS

For example, if the signal is 0.6 in the absence of L and 1 at saturating [L], then when the signal is 0.8:

θ = (0.8-0.6)/(1-0.6) = 0.5

By determining θ at several values of [L] one can calculate Kd.

e.g. of measurement of fluorescence

Calpain is activated by the binding of Ca2+ and in the process a single tryptophan side chain, undergoes a rearrangement from being exposed to solvent (Ca2+-free) to being partially buried (Ca2+-bound). This results in an increase in fluorescence when Ca2+-binding triggers the conformational change.

Deviations from ideal behaviour

If a Scatchard plot (θ/[L] vs [L]) has the following shape, what might you conclude?

θ/[L]

θ

Recall that the slope is -1/Kd. At low θ the slope is steep, while

at high θ it is less steep.


Here are two curves plotted directly:

Both curves represent binding to two binding sites. The red curve, b(x) represent simple binding (K

d = 10 nM) while the

green curve, c(x), contains two binding sites, one with Kd = 1

nM and the other 10nM.

Binding to different independent sites

While it may not be obvious from the shapes of the curves that two different binding sites are present, the Scatchard plot would make it obvious:

θ/[L]

θ

The slope at low θ is steep, representing binding to the tighter of the two binding sites (it saturates first), while the slope at high θ represents the binding to the weaker binding site.

Binding to different independent sites

For two different, but independent binding sites, the direct plot of θ vs [L] shows a curve that results from the linear combination of the equations for each binding site:

= 1 2=n1 [L]

K d1[L]

n2[L]K d2[L]

Although each class of binding site would conform to the Scatchard model, their sum does not. In the event that the K

d

values for the two sites are very different, the direct plot may make the presence of two sites fairly obvious. In that case the resulting curve may appear to be the superposition of the two binding curves.

Direct plots for 2 different sites

The two graphs below represent binding to two different binding sites, with a 100-fold difference in K

d on the left and a 1000-fold

difference in Kd on the right:

Kd1

= 1 nM, Kd2

= 100 nM Kd1

= 0.1 nM, Kd2

= 100 nM

Scatchard plot for 2 different sites

While the linear combination of the Scatchard equations is straightforward for the direct plot, it not possible to rearrange that equation to a get a linear equation in the form of θ/[L] vs θ. One can often deconvolute the curve into two ideal Scatchard plots

θ/[L]

θ

intercept=n1

K d1

n2

K d2

intercept=n1n2

[L]=

n1

K d1

−

K d1

[L]=

n2

K d2

−

K d2

Scatchard plot for 2 different sites

The Kd1

values can be determined by the initial slope at low θ, while the slope at high θ can allow estimation of K

d2 or via

determination of n2 from the x-intercept and hence K

d2.

θ/[L]

θ

slope=−1K d1

slope=−1K d2

intercept = n1 intercept = n

1 + n

2


If a Scatchard plot (θ/[L] vs [L]) has the following shape, what might you conclude?

θ/[L]

θ

Recall that the slope is -1/Kd. At low θ the slope is shallow

while at high θ it is much steeper.


θ/[L]

θ

At low θ the shallow slope indicates weaker binding, while at high θ the steeper slope indicates stronger binding. But wouldn't the tighter binding sites be filled first?

Of course this is an indication of positive cooperativity.

Derivation of Hill equation

Hill developed an empirical model of binding to account for this.

A number (h) of ligands, bind in a concerted fashion to n sites on the macromolecule. If the binding of one ligand alters the remaining sites and thereby influences the binding of other ligands, then the binding is said to be cooperative. If h > 1, then the binding of one ligand enhances the binding of others (positive cooperativity), while if h < 1, the binding of one ligand hinders the binding of others (negative cooperativity).

M + hL MLh

The average number of bound and free sites per macromolecule are given by:

K d=[M ][L]h

[MLh]

=[sL ][M ]t

=h [MLh]

[M ]Tn−=

[s ][M ]T

=n[M ]

[M ]T

Derivation of Hill equation

Dividing the first equation by the second gives:

n−=

h [MLh]

n [M ]

Rearrangement of to give:K d=[M ][L]h

[MLh][MLh]=

[M ][L]h

K d

and substitution into the equation above gives:

n−=

h [M ][L]h

K d

n [M ]=

hn[M ][L]h

K d

If we define Kd' = nK

d/h

or 1/K

d' = h/nK

d and take the logarithm

of both sides we get the Hill equation:

log

n−=h log[L]−logK d

'

Hill plot

Plotting log(θ/(n-θ) versus log[L] should give a straight line:

log

n−

log[L]

slope = h (Hill coefficient

y-intercept = - logK'

0

Population distribution with multiple...

Documents

Transcript of Population distribution with multiple...