Polymerization and force generation

Transcription

1 Polymerization and force generation by Eric Cytrynbaum April 8, 2008 Equilibrium polymer in a box An equilibrium polymer is a polymer has no source of extraneous energy available to it. This does not mean that an equilibrium polymer is always at equilibrium but it cannot move itself away from equilibrium on it own. In the toy problem described here, the initial configuration is away from equilibrium. Actin filaments and microtubules are not equilibrium polymers because they are capable of hydrolyzing ATP and GTP, respectively, which means they are capable of more clever tricks than the polymer described here. Consider the experiment in which a small box of volume V, measured in µm 3, is filled with a solution containing a concentration A 0 of a protein that is capable of polymerizing into a twostranded polymer. At one edge of the box, a couple monomers are glued to the wall to provide a site on which growth can occur (see Figure 1. A k off Figure 1: A polymer in box. Monomers attach to the tip of the polymer at a rate A(t and fall off at a rate k off which is in units of µm per second. The equation describing the change in concentration A(t is da dt = A + k off (1 This can be solved by a number of techniques. I ll use separation of variables. First, separate function of A from functions of t by division: 1 da A + k off dt = 1, 1

2 then integrate with respect to t: 1 da A + k off dt dt = dt, Using a substitution on the left side, u = A(t, du = da dt dt (and immediately replacing the letter u with A, gives: 1 da = dt = t + C. A + k off The LHS can be integrated to get which can be rearranged into 1 log A + k off = t + C A + k off = exp( (t + C = Be kont where B = exp( C, which is necessarily positive. Because B is an unknown positive value, we can simultaneously remove the and allow B to be any value (positive or negative: or, equivalently, A + k off = Be kont A(t = B 2 e kont where the constant B just absorbed a to become B 2 and = k off /. Applying the initial condition, A(0 = A 0, we find that the concentration is A(t = + (A 0 e kont. Note that lim x A(t =. This means that if the concentration starts above, A(t exponentially decays down to. As this decay is occurring, the polymer is growing from zero length to some asymptotic equilibrium length. Unfortunately, when A 0 <, the solution makes it seem as if A(t rises to. However, any rise in concentration has to come at the expense of a shrinking polymer. Because the initial length of the polymer is assumed to be zero, the equations are inappropriate in this case. The intuitive explanation for what happens at these lower concentrations is that no net assembly of polymer is possible so that the concentration simply remains at A 0 indefinitely. To see explicitly what is happening to the length of the polymer, we must recognize that monomers are conserved and hence must appear either in solution or in the polymer. If the length is denoted by l(t, conservation implies that αv δa(t + l(t = αv δa tot (2 whereα is a conversion factor required only because of the units in which quantities are measured (see Table 1, V is the volume of the container, 2δ is the size of a monomer (the 2 is introduced to account for the fact that in a two-stranded polymer, addition of one monomer will increase the polymer length by δ this simplifies several expressions later on and A tot is what the concentration would be if all monomers were in solution. In this example, because the initial length was assumed to be zero, A tot = A 0. Solving equation (2 for l(t, it becomes clear that if A(t were to initially increase from A 0 (as would be the case, apparently, if A 0 <, l(t would have to become negative. This is clearly impossible so we must think of the model in a piecewise manner. Figure 2 shows a few solutions to the problem, taking this piecewise nature into account. 2

3 A(t (A 0 > (A 0 < Figure 2: Concentration as a function of time in the polymer in a box experiment. For A tot >, concentration drops exponentially to. For A tot <, no polymerization can occur (the ostensible growth rate of the polymer is negative so the concentration remains at the initial level. t One way of summarizing the results is to consider the equilibrium state of the system as a function of the total concentration (or, equivalently in this case, the initial concentration. If we define A eq (A tot to be the asymptotic (t concentration and l eq (A tot to be the asymptotic length then { Atot for A A eq (A tot = tot < for A tot > and Figure 3 shows these relationships. { 0 for Atot < K l eq (A tot = D αv δ(a tot for A tot > Table 1: Typical values for parameters Quantity Value V 10 µm 3 A tot 100 µm k off α δ 600 monomers/µm 3 /µm 3 nm 3

4 l eq A eq A tot Figure 3: Asymptotic concentration and polymer length as a function of initial concentration. For initial concentrations below, the concentration is constant so the asymptotic concentration is the same as the initial concentration. For initial concentrations above, the concentration always drops to. All excess monomers above end up in the polymer. A polymer and a spring at equilibrium Now imagine that a spring is placed in the box in such a way that if the polymer grows relatively long, it must compress the spring. Figure 4 shows the geometry of the problem. If l gap is less k sp l gap Figure 4: A polymer and a spring. than the l eq calculated in the Polymer in a Box problem above then the polymer will be able to compress the spring as it grows (see Figure 5. How do we calculate the new equilibrium length, l eq, to which the polymer grows? At equilibrium, the chemical force driving polymerization must exactly oppose the force generated by the compressed spring. In terms of energies, the chemical potential must match the work done by the spring when adding a single monomer. This translates 4

5 x Figure 5: A polymer and a spring at equilibrium. into the following equation: ( A F δ = k B T ln where F = k sp x eq is the equilibrium spring force (in the general case, called the stall force, k sp is the spring constant, x eq is the distance through which the spring is compressed at equilibrium, k B is Boltzmann s constant, T is the absolute temperature and A = A eq is the monomer concentration at equilibrium for the polymer-spring system. Note that equation (3 is a general expression for the force required to stop a polymer with equilibrium constant from growing when the ambient monomer concentration is A. As before, any concentration can be converted to length using equation (2 so that equation (3 can be written in terms of only one unknown, x eq : ( Atot 1 αv δ k sp x eq δ = k B T ln (l gap + x eq. (4 Unfortunately, this equation is transcendental and thus cannot be solved analytically. However, by drawing curves for the left and right hand sides of the equation, you can get a sense of how the solution changes as a function of the parameters. To solve the equation numerically, Newton s method works well. A growing polymer and a spring So far we have discussed a dynamic polymer approaching equilibrium in the absence of any impinging force and a polymer subject to a spring force at equilibrium. What about the dynamics of a polymer under an opposing force? 1 The derivation of the velocity of growth subject to a force F requires a more advanced set of tools beyond the scope of this set of notes [?]. However, in the case when the monomer on and off rates are slow compared to the time required for a gap at the tip of the polymer to appear by diffusion, the growth rate turns out to be a relatively simple function of the applied force [?]: v = dl ( dt = αv δ Ae F δ k B T k off. (5 1 Add section on mechanisms for generating force - power stroke versus Brownian ratchet (3 5

6 As before, the presence of α and V only serve to convert the given units of concentration to length (see Table 1. Notice that when the polymer eventually stops growing and dl/dt = 0, this expression returns the equilibrium result above (problem 3. Furthermore, in the absence of force (F = 0, the expression reduces to equation (1 after the appropriate conversion between length and concentration (given by equation (2 is made on the left hand side. Relationships like (5 are called force-velocity relationships and are often calculated theoretically, measured experimentally and plotted in various ways (sometime v versus F, sometimes F versus v for all kinds of motors both molecular and otherwise. Technically, equation (5 is only valid when F is a constant but for an example like the polymer and spring, when the monomer size is small compared to x, the spring force opposing growth is roughly constant throughout each monomer addition (increasing only a small amount from k sp x to k sp ( x + δ so it is still a useful relationship for situations in which F changes with time and/or with position of the tip. To give a macroscopic example, think of a person pulling a wagon up a slight incline. When the wagon is empty, the person can step forward at some maximum rate (analogous to ( A k off steps per second. As weight is added to the wagon, the person is only able to move at sub-maximal rates up until the weight gets sufficiently large that no progress is possible (when A exp ( F δ k B T k off = 0. If the weight exceeds the maximum load (analogous to the stall force, the wagon will slide backward. The force-velocity relationship (5 can be interpreted in two ways. First, as written, it tells you the velocity v of polymer growth when a force F is applied to its tip, thus denoted v(f. However, it also describes the force F that a polymer is capable of generating when growing against an obstacle at a rate v relative to the obstacle. This interpretation arises from the fact that equation (5 says that the velocity is constant for a given applied force so that the obstacle applying the force moves without acceleration thus the force applied by the object is exactly matched by the force generated by polymerization (acceleration is zero precisely when the sum of forces is zero. In this way, we can solve for F and write down the force generated by the polymer when it is growing at a rate v, Treadmilling F (v = k BT δ ( ln v AαV δ + A The next step in understanding polymer driven motility is to understand how a polymer can continue to polymerize indefinitely. Our examples so far have involved a polymer attached to the wall of the box and have all reached an equilibrium. The important point of the spring example is that provided the monomer concentration is kept above for the growing tip, the polymer can generate force. Obviously, an equilibrium polymer cannot generate a pushing force indefinitely (i.e. generating work for free so we must consider a polymer that somehow uses energy to keep the concentration above or, in other words, keep the polymer length away from equilibrium. By hydrolyzing ATP after attaching to a polymer but before dissociating from it, actin monomers manage to create a situation in which the dissociation constant at is different at opposite ends of the polymer (see Figure 6. It is worth noting that an equilibrium polymer can have different values of k off and at either end as long as the ratio of the two constants is the same. For generating force in an ongoing manner, we must have different values for K + D = k+ off /k+ on and K D = k off /k on. By convention, biologists have agreed to name the end with the lower the plus end (aka the growing end and the end with the higher the minus end (aka the shrinking end. So from now on, we assume that K + D < K D. And, yes, I got this ordering right if you don t see why, think. 6

7 - A + A - k off + k off y(t x(t Figure 6: A polymer with both ends free to polymerize and depolimerize. We define x(t to be the position of the plus end and y(t to be the position of the minus end. about what happens when A > = k off / and A < in equation (1. The treadmilling equation for the concentration of monomers is da dt = k+ ona + k + off k ona + k off (6 and the equations for the positions of the plus (x(t and minus (y(t ends of the polymer are dx dt dy dt = δαv (A + k + off = δαv k+ on(a K + D, (7 = δαv (kona k off = δαv k on(a K D. (8 Pop quiz: why is there a minus sign in front of the y equation? As before, there is a conservation law for this system. In particular, what quantity is conserved? In the first section, we wrote down the conservation law based on intuition for the problem but we could have done it mathematically instead - we ll do that here. This time, we ve written equations for all variables in the system, x, y and A. Notice that a particular combination of the equations above gives a zero sum; in particular, δαv times the A equation added to the difference between the x and y equations: da dt + dx dt dy dt = 0. Integrating this up, we find that δαv A(t+x(t y(t = C. Clearly, the constant must be δαv A tot. Physically, this conservation law says that the total amount of monomer, converted into a length (δαv A(t plus the length of the polymer (x y is constant. That is, monomers are neither created nor destroyed and must either be in monomer or polymer form. Notice that the A equation is expressed entirely in terms of A and the rate constants so it can be solved without considering the other variables. The solution (to be verified as an exercise is A(t = + (A tot e kont where = (k + off + k off /(k+ on + kon and = + + kon. One important feature of this equation is that it immediately tells us that the steady state concentration is. The location of this constant relative to K + D and K D is important because this determines the sign of both dx/dt and dy/dt at steady state (see equations (7,(8. Problem 7 below asks you to examine this in detail. 7

8 At steady state, you should find (after working through the problems below that the plus end grows continually because lim t A(t = > K + D. As we saw in the previous section, a polymer tip that is growing under such a condition (ambient concentration greater than the dissociation constant has the capacity to generate a force. Problem set 1. Sketch the length of the polymer as a function of time for the polymer in a box example for three different values of A tot : one value less than and two different values greater than. 2. Sketch the force-velocity curve (v versus F given by equation (5. Label (i the point at which the applied force is zero - the velocity here is called the free velocity, (ii the point at which the motor stalls - the force at this point is called the stall force, (iii the interval of forces for which the motor moves forward and (iv the interval of forces for which the motor is overwhelmed by the applied force and moves backward. 3. Assume that the applied force, F, is constant (i.e. not generated by a spring but instead by an invisible hand. Convert equation (5 into an equation for A(t by replacing all appearances of l(t using (2. Show that as t, the solution to the resulting equation, A(t, approaches A eq, the solution to equation (3, by showing that A eq is a steady state and, using graphical (phase line arguments, that it is a stable steady state. 4. Assume that the applied force, F, is generated by the compressed spring whereby F = k sp (l(t l gap. There is now the added complication that l(t also appears in the exponent on the right hand side of equation (5. Proceed as in the previous problem - derive an equation for the steady state and show using graphical (phase line arguments that there is only one steady state and that it is stable. 5. Draw the steady state length l eq = l gap + x eq as a function of A tot for the polymer and spring example. As equation (4 cannot be solved explicitly for l eq as a function of A tot, you can solve for A tot as a function of l eq and plot the desired function l eq (A tot by plotting A tot (l eq with the l eq axis vertically and the A tot axis horizontally 2. Keep in mind that until A tot is large enough that the polymer touches the spring, this problem is indistinguishable from the case shown in Figure Fill in each table entry with either grows or shrinks : A < K + D K + D < A < K D K D < A plus end... minus end Show that the value of A(t as t (i.e. is greater than K + D and less than K D using the assumption that K + D < K D. Use these inequalities to show that the plus end grows and the minus end shrinks. Furthermore, show that the growth rate and shrink rate are equal at steady state this is precisely treadmilling! 2 Suggested by Derek Inman thanks, Derek. 8

9 8. Calculate the solutions A(t, x(t and y(t to equations (6, (7 and (8. A(t is plotted below. Plot x(t and y(t on the axes provided assuming that x(0 = y(0 = 0. A(t A x(t, y(t t t 9