Assignment #5 – Advanced topics in macromolecular linked equilibria

Readings:Schreiber and Fersht, J. Mol. Biol. 248:478-486 (1995).Ames et al J. Biol. Chem. 270:4526-4533 (1995)

Handout on Wyman linkage relations

Problems, due Apr 14

1. The MWC model of allosteric control postulates that the To <==> Ro equilibrium is describedby a single dimensionless number L, a conformational equilibrium constant that must be verylarge to give good cooperativity. Derive for yourself that for a 3-site MWC protein bindingligand x (with microscopic constant k):

<(x) = 3 kx (1+kx)2/[L+(1+kx)3]

Plot this function for L=0, L=1, L=10, L=1000. (Note: In order to get a sense of the shape of thecurves, you'll have to figure out over what range of x you need to make the plots.)

As done in class (for a 4-site system) define Q(x) = L + 1 + 3kx + 3kx2 + x3.Note the statistical factors here. Now, we can calculate:

Now, suppose L is itself controlled by an allosteric inhibitor Y because single Y can bind ONLYto the To state (with binding constant k'). The protein follows the "extended" MWC equilibrium:

k' Lo

Y:To---To---Ro---RX1---RX2---RX3

a. Show that the equation in problem #1 still holds, but that now L=Lo(1+k'y). These equilibria are complicated, so use the Q-method!

With this expanded scheme, Q(x) now includes one additional term:Q(x) = Lo + 1 + 3kx + 3kx2 + x3 + k’y, or, collecting terms:Q(x) = Lo+k’y + (1+kx)3

This is exactly the same form as above, but with L = Lo + k’y.

b. Suppose that Lo=1. (What does this mean?) Show that for this protein to give strongcooperative behavior, Y must be present at concentrations much higher than its dissociationconstant.

To get strong cooperative behavior, the value of L must be large, >100 (see part (a)). SinceLo is only equal to unity, the only way to make L large is to make the term k’y large, i.e., tohave a high conc of Y.

c. Derive an expression for the variation in bound X as the concentration of Y is varied at fixedX concentration. Show that this is the SAME as the variation in bound Y and with x, at fixed y. Why must these be the same?

To do this problem, first take a minute to ponder what I am asking for! What do I meanby “the variation in bound X as the concentration of Y is varied at fixed X concentration”? Think about this – it simply means that you should picture the protein in its environment ofX and Y, with each of these ligands bound to some average extent, given by <x or <y . Thequestion is asking about how does <x change as I change the conc of y (not x!). Translatingthis English-language sentence into mathematical language, it simply asks you to calculate:

, the cross- second derivative.

But this cross-derivative doesn’t depend on the order of differentiation (try it out and see),so it must have the same value as

d. Can you think of any biological situations in which a case like this might operate?

2. Use the Q-method to derive eq (5) in the assigned paper by Ames et al.

3. The tryptophan-repressor protein (TrpR) provides a well-studied example of a DNA-bindingprotein interacting with a specific sequence on double-stranded DNA. Here are some facts.

1. The recognition sequence is palindromic, so that the DNA target presents 2 identicalsites for TrpR.2. TrpR binds to these two identical sites in a highly cooperative fashion, with the secondmolecule binding much more tightly than the first.3. TrpR is able to bind only to double-stranded DNA, and only when its tryptophan-binding site is occupied.

a. What do you think underlies the high co-operativity in the binding of TrpR to DNA?The first TrpR protein to bind interacts only with the DNA, but the second one in additionmakes lateral contact with the already-bound TrpR protein, and this additional interactionmakes this second binding more favorable. (see Branden & Tooze for a discussion of TrpR)

b. Let the dissociation constant for the first TrpR to bind be K1; then, assuming that the secondbinding is of 100-fold higher affinity, derive an expression for the fractional occupancy of theDNA as a function of TrpR concentration. Plot this function, and compare it to an expressionderived by assuming that the second binding is infinitely higher affinity than the first.

This is just a 2-site Adair-binding problem. The fractional occupancy as a function of theTrpR protein (called ligand X here), Y(x), is defined as:

Y(x) = <x(x)/2, since there are two sites at saturation.

But the Q-method easily gives us:

And we have been told that K2=100K1. So:

This is to be compared to the “ideal cooperativity” case in which K2 is infinitely higher thanK1, a case calling for the Hill equation (equivalent to the eq above with the K1x termsneglected):

, where

If you plot these, you’ll see that the S-shaped-ness of the two curves are nearly identical.

4. Aequorin (Aq) is a light-emitting protein found in a Pacific Ocean luminescent jellyfish thatworks the night shift. This protein has a useful property: the intensity of emitted light is a

function of Ca++ concentration. This is because the protein has three identical Ca-binding sites,and that somehow, binding of Ca++ is linked to conformational changes leading to an "activated"luminescent form of Aq.

This problem asks you to consider three different ways in which Ca++ binding might drivethe conformational changes. For each of these ways, you must do 3 things: (1) derive anequation for the relative light intensity as a function of Ca++ concentration, by clearly showing theequilibria involved, (2) plot this function on a linear-linear scale, (3) construct a “Hill plot” ofthis function. Don't be over-precise: no more than 2 significant figures.) It’s OK to use computerprograms to make these graphs (this really is NOT a course in arithmetic...)

a. Suppose the 3 Ca++ ions bind to the protein independently, with binding constant k, but thatonly the triply-liganded protein emits light.

Use the Q-method, and remember to use statistical factors to get the macroscopicbinding constants Ki in terms of the intrinsic microscopic binding constant k. In thisproblem, we need to find f PX3(x), which will be proportional to the light intensity:

– look – it’s just a Langmuir cubed!

b. Suppose the 3 Ca++ ions bind independently as above, and that the light emitted by a givenprotein molecule is directly proportional to the number of Ca++ ions bound to that molecule.

Here, the light emitted will just be proportional to the number of Ca++ ions bound.Do you see why?

I/Imax = kx/(1+kx) – Look – a simple Langmuir!

c. Suppose that after the first Ca++ binds with a binding constant k, the protein flips into a light-emitting conformation which increases the affinity of the remaining two sites a zillion-fold.

This is an “ideal cooperativity “ situation calling for the Hill equ:

, where K = (K1K2K3)1/3

After you've done all this, state clearly how the three situations are experimentallydistinguishable.

The graphs of these three equations all have different shapes, which become mostclear if they are plotted on a Hill plot.

5. An ion channel may be blocked from the inside by blocker X and from the outside by blockerY. These blockers bind reversibly to separate sites on the respective sides of the pore and plug itup; thus, only "unblocked" channels conduct ions. Each of these blocking reactions may bestudied separately by applying the blocker, at varying concentrations, on the proper side, anddetermining its own dissociation constant, KX or KY, from the resulting "falling langmuir"inhibition curve.

After failing even to crystallize MgCl2, Chris Catastrophe, a grad student in the Genicklab, decides to do some electrophysiological work, wondering whether the two blockers can plugup the channel simultaneously when added on both sides of the membrane. First, he determinesthat KX = 1 mM and KY = 0.1 mM when measured separately. Then, he varies the concentrationof internally applied X in the presence of a fixed external Y concentration of 0.3 mM. Theexperiment gives a nice falling langmuir inhibition curve, with an "observed dissociationconstant of X," KX

obs.

a. Draw a thermodynamic box describing the channel in its various unblocked and blockedstates, and derive an equation describing the inhibition of channel activity in Chris's experiment.

Kx

C – CX and we are given values for Kx and Ky

Ky | | Kxy

CY – CXY

To get the “observed” or “apparent” Kobs , we have to solve for the fraction ofprotein in the unblocked state as a function of x at fixed y. Remembering that thisproblem is posed in terms of dissociation constants, not the association constantswe’ve been using in class. (But you should be able to make the switch!) The Q-method gives us:

f[C](x) = 1/(1+x/Kx + y/Ky + xy/KxKy ) –

Now get it in Langmuir form! After some algebra, this becomes:

Note that this is a falling langmuir with an amplitude less than unity (i.e., at x=0, thechannel activity is less than 100%, because there is some Y-blocker present. That isevident from the amplitude-factor in the first parenthesis.) Most crucially for theproblem at hand, the “apparent” or “observed” dissociation constant is just givenby the thing that divides x in the demoninator:

This is:

b. What can you conclude, using this equation, if CC finds the values of Kobs below:

Now, we were not told the value of Kxy, but we can find it if we know the value of KX

obs, since these are related by the equation above. Thus, all we have to do is tosolve the above equation for Kxy, using the given values of the K’s:

, where values are in mM units

So now we can just plug in the given values of KXobs :

KXobs Kxy

1 mM 0.1 mM X and Y bind independently and don’t influence each other’s K.3.5 mM 1 mM Y binds to X-occupied channel10-fold more weakly!4 mM infinity! this means that X and Y can NEVER simultaneously bind0.3 mM 0.024 mM Y binds to X-occupied channel4-fold more strongly!

5mM -1.5 mM – impossible, a negative value! The constraints of a thermodynamicbox make this value impossible. It wouyld violate thermodynamics!

c. In each case above, calculate the apparent standard-state free energy of binding of X tothe channel already blocked by Y. Compare these to the free energy of binding of X to theunblocked channel, to get an idea of the nature of the "interaction" between the two blockers onthe same channel.

Just “-RT-log” these K values and see how they satisfy the thermobox.

d. Do problem #c for KXobs = 5 mM, and show that this is an "impossible" situation. Explain why

it could never happen.

6. Calbindin is a small globular protein that is a homodimer, and that can undergo aconformational. Each of the identical subunits has a binding site for Ca2+. In the "T" state, theCa2+-binding sites are disrupted so that effectively they don't exist, but the transition to the "R"state creates the two sites. The "intrinsic" binding constant to the individual, isolated site is ko.

However, the 2 Ca2+ ions do not bind independently of one another in the R- state. Thisis because Ca2+ is charged, and the two sites are actually quite close together -- only 7 Å apart -- in the dimer. Thus, there is an electrostatic repulsion between the Ca2+ ions when they are bothon their sites. This energy of electrostatic repulsion has a value , (kcal/mol).

a. Draw out a scheme for the linked equilibria involved here.

To – Ro – RX1 – RX2 – just a simple 2-site linked scheme. Lo K1 K2

We must also express the macroscopic K’s in terms of the intrinsic microscopicbinding constant for the site.

K1 = 2ko (note statistical factor of 2 for first binding)

(note stat factor of ½ and repulsion factor)

b. Suppose that the only form of the protein that has "activity" is the form with 2 Ca2+ ionsbound. Derive an expression for how the protein's activity will vary with Ca2+ concentration, interms of Lo, ko and ,.

Use the Q-method to solve for f[RX2]:

c. Assuming that Lo = 100, give a rough sketch of the activity as a function of [Ca2+], andcompare this to what this would look like if there were no electrostatic repulsion. Use a value of, that you calculate explicitly from coulombic repulsion.

The point here is that the T<=>R equilibrium gives you a positive cooperativity,while the electrostatic repulsion tends to give negative cooperativity. You have toplot out this equation to see which effect wins out. The electrostatic repulsion canbe calculated from the distance between Ca ions from a coulombic potential to be~2.4 kcal/mol. Thus, the repulsion factor comes out to be ~0.02.

7. Human growth hormone receptor (R) is an integral membrane protein that acts as the initiaionpoint in the elaborate, complicated cellular proliferation response to growth hormone. Itsstructure is known, as is its mode of action, which is a "dimerization" mechanism. Monomeric Ris randomly distributed and mobile in the plasma membrane. A single growth hormone molecule-- let's call it "X" for ligand -- binds with high affinity to two R molecules, but with low affinityto only one. The ternary complex R2X can be considered a "sandwich" structure in which thetwo R proteins are the slices of bread glued together by the growth hormone jam. Whendimerized in this way, the ternary complex becomes "activated" by autophosphorylation, andthen an elaborate cascade of events occurs leading to cell growth. The binary complex, RX, istotally inactive in leading to these downstream responses.

a. Set up the appropriate equilibria to solve for the concentration of ternary complex, C, as afunction of total receptor concentration in the membrane, RT, and free circulating growthhormone, in the blood X. You may assume that the concentration of binary complex, RX, isnegligibly low.

The reaction to consider is the binding of a single ligand to two receptor molecules:

2R + X – R2X

Define equilibrium const: K = R2X/(R*x)Also remember that the total conc of receptor in the membrane, T, is a constant, given by:

T = R + 2R2X so R = T-2R2X

These two relations lead to a quadratic eq for C as a function of T:

(R2X)2 + C(1+1/(4Kx )) + T2/4 = 0, which solves to:f(x) = 2R2X/T = frac of protein in active state =

f(x)

Hold on! That is sure no langmuir, or even a rational function! But it’s the correct answer. It’s arising function of x that saturates at a value of f=1.

b. It is found that the response of a given cell to growth hormone is very sensitive to theexpression level of receptor in that cell. Why is this, both intuitively, and according to youranswer in (a)?

Hint: Do not try to use the Q-method here – it won’t work! Just set up the equilibriainvolved, take a deep breath, and plow ahead.

This depends on the concentration of total receptor (T), by the equation above. Intuitively, that’sbecause the receptor has to dimerize to form complex, and the equilibrium is driven to the rightas R increases.