Binding
Free energy of a solution
Free energy of a solution
The free energy of a “solution” depends on its concentration
It becomes progressively harder and harder (i.e. energetically costly) to force more solute
into a limited space
This is why solutes don’t spontaneously aggregate in one location
We will eventually apply this to the “reaction” A + B ⇆ C, which is shorthand for the actual
reaction nA A + nB B + nC C ⇆ (nA-1) A + (nB-1) B + (nC+1) C
There is an entropic component to solvation (PBoC section 6.2):
Imagine a volume of solvent as comprising N “boxes” in which to put n particles of solute (N = V/h3 if necessary).
There are W = N!/ [n! (N-n)!] ways to do this, so the Boltzmann entropy S = k ln W is
S = kB ln
!
N!
n!(N − n)!
"
In general this expression is complicated, but for n << N we can use the Stirling approximation
(ln N! = N ln N - N) to get
S = −kB [n ln(n/N ) − n]
As expected, this is an extrinsic quantity (it is proportional to system size)
Free energy of a solution
Suppose it costs some solvation energy ε (possibly zero) per solute particle. Then the total
free energy is
G = n! + kB T [n ln(n/N ) − n]
In deriving this, we did not assume anything about the N boxes. This expression applies to a solute being dissolved
in water, but also to a molecules adsorbing onto a surface or to ligands binding to receptors
We do need to have n << N, or else we would have to revisit the approximations on the previous page.
We’ve ignored the cost of excluding other particles (other solutes or water itself) from occupying a site. For water this doesn’t
make much difference, but it can matter under some circumstances.
Since we’re going to minimize free energy, we often don’t care about the amount of free
energy, but only about how free energy changes with n.
We’re more interested in the marginal free energy ∂G/∂n, which is called the chemical potential µ.
µ = ∂G/∂n = " + kB T ln(n/N ) = " + kB T ln(c/c0 )
This form is independent of the unit volume h3 provided it is “small”; equivalently, it is independent of the
concentration c0 which is known as the standard state.
By convention, c0 = 1 M and the chemical potential is
µ = µ0 + kB T ln(c/1M)
Unfortunately, chemists write this as
µ = µ0 + kB T ln c
This frustrates unit analysis and requires you to throw in an extra factor of 1M whenever your units don’t work out.
This is equivalent to setting G=1 in general relativity or ħ=1 in quantum mechanics, which no one does until graduate level in
physics.
Free energy of a solution
Equilibrium between two states
Suppose we’re looking at the simplest possible reaction: A ⇆ B.
This is really nA A + nB B ⇆ (nA-1) A + (nB+1) B, so the free energies on the LH and RH sides
are
GA (nA ) + GB (nB ) ! GA (nA − 1) + GB (nB + 1)
This reaction will proceed until Gtotal = GA + GB is minimized.
That happens when the LHS equals the RHS, ie when
{GA (nA ) − GA (nA − 1)} + {GB (nB ) − GB (nB + 1)} = 0
For large n, this is equivalent to
∂GA (n)
∂GB (n)
· (+1) +
· (−1) = 0
∂n
∂n
which is the same as (by definition of µ)
µA · (−1) + (!A→B + µB ) · (+1) = 0
Free energy of a solution
Substituting in expressions for µ’s of A and B gives
[B]
= e−(µ0B −µ0A )/kB T ≡ e−∆µ/kB T
[A]
Nothing here is specific to solutes being dissolved in water. We can apply it to equally well
to n proteins binding to N sites with binding energy ε.
PBoC does the example of RNA polymerase binding to specific and nonspecific promoter sites (p. 224):
Free energy of a solution
Here εNS and εS are the binding energies associated with nonspecific binding and binding to the particular
promoter site. Presumably εNS > εS so there is an energy penalty associated with nonspecific binding compared to
promoter binding.
The probability of having the promotor occupied (and hence starting transcription) is
pbinding = weightS / (weightNS + weightS) (this is basically just a normalization of the Boltzmann weights).
After some algebra this turns out to be
pbinding
(P/N )e∆!/kB T
=
1 + (P/N )e∆!/kB T
where Δε= εS–εNS is the difference in binding energies between the specific promoter site and nonspecific sites.
As we will see in a few slides, this is exactly the same form as Hill binding.
PBoC curves calculated for
Δε=–2.9 kBT (lac)
Δε=–8.1 kBT (T7)
Free energy of a solution
In the previous derivation, B had only two functions:
nA could be increased or decreased by exchange with B
There was a marginal free energy µB associated with B
We might as well replace B with an external “reservoir” of free energy and of particles. nA
can change by exchange with the reservoir: A ⇆ reservoir.
How many A do we expect at equilibrium?
If the marginal change in GA (aka µA) is different from the marginal change in Greservoir (aka µext), Gtotal = GA +
Greservoir could be decreased by changing nA. At equilibrium we must have µA + µreservoir = 0; ie.
∂GA
− µext = 0
∂nA
We assume that the reservoir is really big, so its µ will never change no matter how much A we pump into or out of
it. That is, µext = constant no matter what.
Then the above equation is equivalent to
∂(GA − µext nA )
=0
∂nA
When a system is free to exchange particles with a large reservoir that has a chemical potential µ, then it is the free
energy G–µN which is minimized rather than simply G.
We found a similar result with our entropic spring was equilibrated against a force. This is a general thermodynamic
phenomenon: when you relax a restriction on Q you add a term fQ Q to the free energy, where fQ is the force conjugate to Q:
basically the thing that converts Q into an energy.
Binding equilibrium
Binding equilibrium
Mass action
Suppose we’re looking at the more complicated reaction A + 2 B ⇆ C.
This is really nA A + nB B + nC C ⇆ (nA-1) A + (nB-2) B + (nC+1) C.
As before, we know the reaction will proceed until the total free energy Gtotal = GA + GB + GC is minimized. The
change in Gtotal for a single rightward reaction is
dGtotal =
dG=0 occurs when
∂GB
∂GC
∂GA
· (−1) +
· (−2) +
· (+1)
∂n
∂n
∂n
−µA − 2µB + (!A+2B→C + µC ) = 0
Putting in our expression for the free energy of A, B and C, we get
[C]
−(!A+2B→C +µ0C −µ0A −2µ0B )/kB T
−∆µ0 /kB T
=
e
=
e
[A][B]2
Since the units don’t make sense, we have to insert (1M)-2 on the RHS.
The RHS is important enough that it has its own name: the equilibrium constant (Keq).
Keq is usually more easily measured than calculated.
Since it has units of 1/(concentration2), we define the dissociation constant Kd so that 1/Kd2 = Keq.
PBoC has a general version of this formula (6.104) for νA A + νB B + ... ⇆ νC C + νD C.
Binding equilibrium
Mass action
A biologically important case is ATP hydrolysis: ATP ⇆ ADP + Pi.
From above, each species has a free energy (per molecule) µi = µ0,i + kBT ln [i].
The change in free energy (products – reactants) for one rightward reaction
∆G = (! + µ0,ADP + µ0,P − µ0,AT P ) + kB T ln
!
[ADP ][Pi ]
[AT P ]
"
≡ ∆G0 + kB T ln
!
[ADP ][Pi ]
[AT P ]
This is the amount of energy made available by cleaving a single phosphate bond. It obviously depends on the
concentrations of ATP, ADP and Pi in the cell.
From PBoC p. 254, ΔG0 ~ -12.5 kBT and in a typical healthy cell [ATP] ~ [ADP] ~ 8 mM while [Pi] ~ 0.4 mM.
These numbers give a free energy harvest of about –20 kBT per ATP hydrolyzed.
By definition, at equilibrium ΔG = 0 and [products] / [reactants ] = Keq, so we have
"
!
[ADP ][Pi ]
= ∆G0 + kB T ln Keq
0 = ∆G0 + kB T ln
[AT P ]
which gives the relationship
∆G0 = −kB T ln Keq
In practice, ΔG0 is usually deduced from a measurement of Keq.
"
Binding equilibrium
Ligand-receptor binding
An important special case is the reaction L + R ⇆ LR. Proceeding as above, we get
[LR]
= e−(!+µ0LR −µ0L −µ0R )/kB T = (1M )−1 e−∆µ0 /kB T ≡ Keq ≡ Kd−1 .
[L][R]
The fraction of receptors bound is pbound = [LR]/([R]+[LR]). A little algebra yields
pbound =
[L]/Kd
1 + [L]/Kd
Kd is the ligand concentration at which exactly half the receptors are bound.
Once you normalize by Kd, there is only one binding curve for all simple ligand-receptor interactions.
Compare this to the expression for RNA polymerase binding to DNA: P/N plays the role of concentration and
exp(-Δε/kBT) plays the role of Kd.
The lower Kd the tighter the binding
Tight binding: Kd < 10-9 M (Fe(III) to Transferrin)
Medium binding: Kd < 10-6 M (Ca(II) binding to calmodulin)
Loose binding: Kd <10-3 M (ATP binding to hexokinase)
Binding equilibrium
Ligand-receptor binding
When n ligands bind simultaneously, n L + R ⇆ LnR, the fraction of bound receptors is
pbound
([L]/Kd )n
=
1 + ([L]/Kd )n
This family of binding functions are Hill curves and n is the Hill coefficient or cooperativity.
Hill curves
Higher n means steeper curves near Kd.
For the highest sensitivity to [L]
1. Operate near Kd
2. Be highly cooperative (large n).
A little algebra on the Hill function yields
!
"n
c
p
=
1−p
Kd
or, equivalently,
"
!
p
= n ln c − n ln Kd
ln
1−p
ln(p/(1-p)) vs ln c is called a Hill plot.
Its slope is n and intercept is n ln Kd.
Binding equilibrium
Many biological interaction follow Hill
binding curves
A simple interaction (such as O2 binding to
myoglobin, which has only one binding site) has
n=1.
Binding equilibrium
More complicated binding also occurs; it is often highly cooperative.
Everyone’s favorite example is hemoglobin.
Hemoglobin is a tetramer of myoglobin-like subunits, so it has four heme groups and four binding sites.
The binding curve is definitely steeper than a n=1 Hill curve (4 independent binding sites) but not as steep as a n=4 curve (all-ornone binding to 4 sites simultaneously).
This is a general rule: the Hill coefficient is a lower limit on the actual number of binding sites because independently binding sites are
invisible in the Hill plot.
Hemoglobin binding
Binding equilibrium
You can tell something funny is going in with hemoglobin because its Hill plot is not linear:
log (f/(1-f))
Hemoglobin binding
We will return to Hb after we develop the idea of internal states in binding.
Binding equilibrium
Scatchard plots
Suppose there are N binding sites on a protein, each of which binds independently.
The number of bound ligands per protein is ν = N pbound, where pbound is given by the Hill equation,
pbound =
[L]Keq
1 + [L]Keq
Rearranging gives the equation for a Scatchard plot of ν/[L] vs ν:
ν
= N Keq − νKeq
[L]
The slope and intercept are Keq and N.
A nonlinear Scatchard plot means there is cooperativity
Naturally, if you have software to do linear regression you can just plot ν vs [L] and fit
directly to the proper function:
ν=N
[L]Keq
1 + [L]Keq
Internal states:
two-state systems
Internal states
Rationale
To account for identical ligands binding differently, we will introduce the idea of internal
“states” of a receptor.
“States” are biochemically distinct, and therefore usually correspond to distinct conformations.
Initially, internal states were invoked to explain cooperativity, activation, etc.
Many states that were proposed just to fit models have been identified structurally.
For purposes of enumeration, we will often assign a state variable (here σ) to different states.
Initially you should just consider this a label.
Later we will use clever assignments of σ to encode different energies of the state.
Complicated proteins can have many relevant internal states
Hemoglobin has 4 or 6, depending on the model.
We will work our way up from simpler systems that have only two important states.
Two-state systems
Bistable proteins
Many proteins fit well to a binary description:
The assignment which state has σ=0 and σ=1 is completely arbitrary at this point.
Two-state systems
All these systems have no half-way state.
Maybe this isn’t always obvious, but you can sometimes
check whether the protein is binary:
Since there is no halfway state, the only way to respond to
an outside signal is to modulate the on/off probability:
The average conductance of the channel varies continuously
although any individual channel is only fully open or fully
closed.
“Average” means average of a single channel over time or
average over many identical channels at a particular instant.
Two-state systems
In this model, there are two states: open and closed
They have baseline energies ε0,open and ε0,closed
Applying a negative voltage makes opening less likely, so the free energy Gopen = ε0,open - qV for some effective
charge q.
popen =
e−Gopen /kB T
1
=
−G
/k
T
−(∆!
−G
/k
T
0 +qV )/kB T
1+e
e open B + e closed B
with Δε0 = ε0,closed - ε0,open.
This fits the data very well for q=2.5e (e is the charge on the electron).
Two-state systems
The same logic applies to any system where you can guess the energy associated with a state
change based on the physics:
Creating a surface of area A with surface tension τ requires an energy τ A, so
The change in area ΔA of a mechanosensitive channel requires an energy τΔA.
Put this into the free energy: Gopen = ε0,open - τΔA. Eventually
popen =
e−Gopen /kB T
1
=
−G
/k
T
−(∆!
0 +τ ∆A)/kB T
1+e
e open B + e−Gclosed /kB T
which fits the observed p-τ curve very well:
opening probability for the
mechanosensitive channel MscL
Two-state systems
PBoC represents this schematically as a free energy surface
that changes with external load
This really is schematic, since they use essentially the same
figure for phosphorylation of an enzyme:
Since there is no physics rationale behind the free energy change
upon phosphorylation, the free energy changes I1 and I2 are just
parameters in the model.
Phosphorylation changes the relative probabilities of active/
inactive states: the unphosphorylated enzyme is usually inactive
while the phosphorylated enzyme is usually active.
PBoC does this quantitatively based on the changes in the energies
of the active vs inactive states due to phosphorylation.
If we could continuously phosphorylate a molecule, there would
be a sigmoidal curve for % activity vs level of phosphorylation,
just like on the previous two slides.
Two-state systems
Ligand binding
Here we have a two-state protein: ligand bound vs ligand free.
The energy difference between bound and unbound (εb) is fixed
Generate the sigmoidal curve by changing the amount of free ligand [L] in solution
Instead of computing the free energy of [L] explicitly, just treat free L as a very large
reservoir of L’s operating at a chemical potential µ.
The free energy change upon binding is +εb (the actual binding) –µ (removing one
L from the sea of dissolved L’s), hence the weights in the figure.
You can ascribe the –µ to the free energy of solution of L or you can simply stick it onto
the binding energy itself. Either way works.
Using state variables
You can figure out a single expression for the energy of the bound (σ=1) and unbound (σ=0) states:
G(σ) = σ("b − µ)
The Boltzmann factors are
state 0: exp(−βG(0)) = 1
state 1: exp(−βG(1)) = exp(−β("b − µ))
β is the standard abbreviation for (kBT)-1.
To turn the Boltzmann factors into probabilities we need to normalize them using the sum of all Boltzmann factors:
this is called the grand partition function
or, doing the sum out
Z = Σσ e−βG(σ) = Σσ e−βσ(#b −µ)
Z = 1 + e−β("b −µ)
Two-state systems
The average occupancy is !σ" = 0 · p(0) + 1 · p(1) = Σσ σp(σ)
You can compute this by hand, or by noticing that
∂ ln Z/∂µ = Σσ βσp(σ)
!σ" = kB T
∂ ln Z
∂µ
This case is simple enough that it’s easier to write down the probabilities by hand than to construct the partition function. When
the system gets more complicated it will be easier to calculate Z and take derivatives.
The µ makes this a grand canonical ensemble (appropriate for equilibrium N in contact with a particle reservoir) as opposed to a
canonical ensemble (fixed N).
In the end,
e−β("b −µ)
(c/c0 )e−β("b −µ0 )
!σ" =
=
1 + e−β("b −µ)
1 + (c/c0 )e−β("b −µ0 )
where the expression for the chemical potential of a dilute gas was used in the last step.
Comparing this to the Hill function, we see that
Kd = c0 eβ("b −µ0 ) = 1M eβ∆µ0
Internal states:
multistate systems
Multistate systems
Hemoglobin
Hb has four subunits, each of which has one binding site (a heme group).
Hb looks roughly like 4 Mb grouped together, except that unlike Mb it does not have a simple Hill-like binding
curve.
Two models for the cooperativity of Hb: MWC and Pauling
Monod-Wyman-Changeaux (MWC) model (PBoC p. 272)
For each subunit,
There is an internal state variable (T or R).
T is more probable than R when no O2 is bound
With no O2, ET=0 and ER=ε.
O2 binding tips the balance the other way, so R is favored:
With bound O2, ET = εT and ER = ε + εR, with εT > ε + εR
This is just like the hypothetical phosphorylation energy landscape.
To make it cooperative, assume that either all subunits switch from T to R or none switches (“symmetric” model)
This is interesting because the actual O2
sites are far from each other.
Binding in one subunit changes the energy
of binding in another subunit, far away. This
is called allostery.
PBoC illustrates this for a 2-subunit “Hb”:
Without the all-or-none requirement, there
would be 16 possible states instead of 8.
Multistate systems
The MWC energy is E(σm , σ1 , σ2 , σ3 , σ4 ) = (1 − σm )"T
σm=0 for the T state and σm=1 for the R state
4
!
σi + σm
i=1
"
" + "R
4
!
i=1
σi
#
There is only one σm for all four subunits (unlike the σi’s) because the entire tetramer is in the T or R state.
This notation is what makes the system cooperative. Very subtle.
If we had four σm’s (σm,1 .. σm,4) this would not be a cooperative system. That is, there would be only one Kd regardless of how
many O2 were bound.
Within each state T or R, oxygens bind completely independently, so the partition function is
Z = (1 + x)4 + e−β" (1 + y)4
x = e−β("T −µ) =
c −β("T −µ0 )
e
c0
y = e−β("R −µ) =
c −β("R −µ0 )
e
c0
This form of the partition function Z is sometimes called the binding polynomial.
N = kBT ∂(ln Z)/∂µ. After some algebra,
best fit of Imai data to MWC model
4
(1 + x)3 x + e−β" (1 + y)3 y
!N " = 4
(1 + x)4 + e−β" (1 + y)4
3
2.5
number bound
The best fit (green) to binding data (blue) is much better than
a non-cooperative model (red).
3.5
2
1.5
1
0.5
0
−1
10
0
10
1
10
ppO2
2
10
3
10
Multistate systems
Pauling model (PBoC p. 274)
The sites are not independent, so the binding energy depends on whether adjacent sites are already bound
(“sequential model”)
The model assumes that Hb is a pyramid, so any two sites are adjacent. (Which is not obvious from the PBoC illustration).
The Pauling energy is
4
!
4
J !!
E(σ1 , σ2 , σ3 , σ4 ) = "
σi +
σi σj
2
i=1
i=1
j!=i
The cross term (J term) contains the cooperative interaction.
This is a finite-size version of the most famous system in statistical mechanics: the Ising model.
From this energy, the partition function is (PBoC 7.44)
Z = 1 + 4e−β("−µ) + 6e−2β("−µ)−βJ + 4e−3β("−µ)−3βJ + e−4β("−µ)−6βJ
and the mean occupancy is N = kBT ∂(ln Z)/∂µ:
4x + 12x2 j + 12x3 j 3 + 4x4 j 6
!N " =
1 + 4x + 6x2 j + 4x3 j 3 + x4 j 6
−βJ
x ≡ e−β("−µ) = ce−β("−µ0 ) j ≡ e
Actually, there’s a more involved Adair model that includes 3-body interactions (energy K) and 4-body interactions (energy L).
See PBoC p. 277-278 for details). This doesn’t fit the oxygen binding data any better, but can account for competitive binding
with, for example, CO.
Multistate systems
The end result (with physiological parameters) is a suppression of the intermediate states that have 2 or 3 O2
bound:
Qualitatively, you can bind one O2, but as soon as you bind the second the third and fourth immediately bind as well, giving a
Hill coefficient of about 3.
As you suppress intermediate states you approach the all-or-none behavior of the Hill function (as soon as one ligand binds, they
all bind)
You can interpret this as Kd’s that decrease with the number of oxygen bound, so binding the first O2 makes it easier to bind the
next one.
ΔE1
ΔE1
ΔE1
Multistate systems
All these models can be parameterized either by energies or by Keq (or, equivalently, Kd).
Chemists tend to use Keq, which are more easily measured.
Physicists tend to use energy, which make the relationship between different states more transparent.
The various Keq are not independent; this seems more obvious to me in the energy picture.
ΔE1
ΔE3
ΔE2
ΔE4
[TAB ]
= KT A = e−β(ET +A −ET ) = e−β(∆E1 )
[TB ][A]
What is the equilibrium for the diagonal?
From the energy picture, it is
[RAB ]
= e−β(∆E1 +∆E2 ) = e−β(∆E3 +∆E4 )
[TB ][A]
From the binding equilibrium picture it is
[RAB ]
= KT A YAB = YB KRA
[TB ][A]
Multistate systems
If the Keq are interpreted as microscopic equilibrium constants, then we need to account for
the different number of available binding sites in each intermediate state
For n identical binding sites with identical microscopic binding equilibria KT and KR for the T and R states, the
macroscopic equilibiums constant for 1-binding, 2-binding, etc. are all different:
Multistate systems
Flagellar rotary motor
The motor of the bacterial flagellum drives it full speed CW or CCW.
This looks very much like a two-state system.
The probability of going CW or CCW is modulated by the binding of a signaling protein.
There are many (perhaps 34) binding sites for CheY on the motor
The motor is like a 34-subunit Hb whose allosteric state you can observe
from outside
Response
regulatorthe
output in bacterial
cell.
Hill function with n=2.5,
Kd=18.8 µm
From Alon et al, “Response regulator
output in bacterial chemotaxis”,
EMBO J 17:4238
Fig. 1. Fraction of time tethered cells spend turning CW (fCW). Cells
induced with various amounts of IPTG were tethered and their motion
was analyzed using video tracking. Each point is an average of data
Fig. 3. Distribution of CW bias between different individ
Plotted is the fraction of time spent CW, fCW, divided by
fCW over all cells with the same CheY induction level, "
Multistate systems
Flagellar motor
Make a motor model analogous
to the MWC model
U.Alon et al.
re-constitute a full bundle. A crude estimate of the typical
time for such hydrodynamic interactions is the bacterium
N length (a few micrometers)
N divided by its velocity (10–
0
20 µm/s), yielding a tumble duration of the order of tenths
of seconds.
N One may view the flagella interactions as an
additional level of information processing in the chemotaxis response. The result of this processing is that whereas
CW
N individual motors displayN
prolonged CW intervals, swim0
ming cells keep their tumble duration short and independent of P-CheY. This behavior of swimming cells is close
to the expected optimal behavior. Since it is only during
runs that the bacteria sample their environment and reach
7
Assuming that L0=10 and N=30
gives
K=7.6
2.0should
µM beand
favorable
regions,
each±
tumble
kept as brief as is
necessary to randomize the heading of the subsequent run.
Ψ=2.0 ± 0.2.
An additional difference between tethered motors and
free-swimming cells is observed at very high P-CheY
The CheY-P affinity betweenlevels.
CW Whereas
and CCW
differs
by only
ΔG
CW rotation
increases
monotonically
(Figure 1), swimming cells become less tumbly at P-CheY
= kBT ln Ψ = 1.4 kBT: small. levels higher than ~50 µM (Figures 4 and 5). This is
probably caused by the runs associated with CW rotating
The increased affinity has
to be
because
we needin aflagella
bundles
thatsmall
have previously
been documented
switch
mutants
that
lock
the
motor
in an
extreme CW
Hill coefficient of 2.5 though we have 30 binding
sites.
state (Khan et al., 1978; Wolfe and Berg, 1989). The
average run speed drops abruptly at high P-CheY levels
to ~50% of the wild-type run speed (Figure 6), in agreement with the observation that speed generated by CW
bundles is about half the speed generated by CCW bundles
(Khan et al., 1978). It should be noted that CW runs only
predominate at P-CheY levels that exceed the total CheY
content of wild-type cells. At all levels of P-CheY that
could possibly be attained in wild-type cells (i.e. 0 to
~50 µM), the tumbling frequency increases with increasing
P-CheY. It is only at P-CheY levels exceeding the maximal
physiological level that the tumbling frequency begins to
decrease with increasing P-CheY. This latter behavior can
be detrimental since it leads to reverse chemotaxis where
cells move towards repellants and away from attractants
(Khan et al., 1978). Strikingly, the cell appears to have
tuned the signal transduction pathway (kinase activity and
CheY concentration) to give the highest possible maximal
tumbling response that does not carry over to the reverse
chemotaxis regime.
Z = L (1 + c/K) + (1 + Ψc/K)
=
(1 + Ψc/K)
L (1 + c/K) + (1 + Ψc/K)
microscopic Kd’s
p
Fig. 9. Allosteric model for the motor switch. The motor is assumed
to be in one of two states, CCW or CW. It displays spontaneous
stochastic transitions between the states, with a (large) equilibrium
constant L0. The motor is assumed to have N independent, identical
binding sites (represented here by open circles) for P-CheY (stars).
Both of the rotation states can display multiple P-CheY occupancy
states, with between 0 and N sites bound by P-CheY proteins. The
probability of P-CheY binding is not the same for the two states. The
equilibrium dissociation constants for P-CheY binding are K for
binding to the CCW state and K/! for binding to the CW state
(vertical arrows). Transitions between CW and CCW states with m
binding sites occupied by P-CheY occur with equilibrium constants Lm
(horizontal arrows). In the model, P-CheY binds preferentially to the
CW state (! "1), and thus increasing cytoplasmic concentrations of
P-CheY shift the equilibrium towards a higher fraction of time spent
Multistate systems
RNA hairpins
Some systems that look like they should be multistate are so cooperative that they are
effectively two-state.
Pull on a RNA hairpin with an optical trap, measuring the extension as a function of force.
Multistate systems
RNA hairpins
An experimental trace looks like
Multistate systems
RNA hairpins
This fits beautifully to a two-state model of RNA unzipping.
Note that there is only one free parameter in this model (ΔF0). Once you pick ΔF0 to give
the proper midpoint of the folding curve, the rest of the curve is mandated
Because, just like we saw for simple ligand binding, there is only one family of curves for two-state binding. Once
the pick the Keq (here it’s called Feq = ΔF0), there is no freedom to change the shape of the curve any further.