Assignment 5 Biophys 4322/5322

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1 Assignment 5 Biophys 3/53 Tyler Shendruk April 11, 01 1 Problem 1 Phillips 8.3 Find the mean departure of a 1D walker from it s starting point. We know that the really mean distance from the starting point is (n r n l a 0 (1 since there is no reason to break symmetry (i.e. n r n l N/. But this is so simple that it can t be what he means. To discuss how far the walker departs on average, we should use the MS of. Until the very end, we ll consider. a (n r n l a ( n r n r n l + n l a ( n r n r (N n r + n l a 3n r Nn r + n l a ( 3 n r N nr + n l where we used n l N n r once but kept the n l. We did this because n l n r n. From here on, it doesn t matter which direction is left or right so we drop the subscript r. This means that a ( 3 n N n + n a ( n N n. ( We know that we can use n N/ from either the binomial distribution or just from symmetry. But what about n? There s no simple symmetry argument that I like so we ll resort to using the binomial distribution. We can find the average by summing n weighted by p(n: n n p(n n0 n0 ( n N! 1 n! (N n! N. (3 How to deal with this sum? There is an identity that will help us (I think it s in Phillips and we ll extend it to get what we need: 1

2 If you want you can derive this but I ll just state it for you n0 N! n! (N n! N (a We can use the above identity to get a series of related identities. N! N n n! (N n! N! (n 1! (N n! n0 n0 n1 N N N (N 1! (n 1! (N n! n1 M η1 N M (N 1! (n 1! (N 1 (n 1! M! η! (M η! N N 1 (b Note: n 0 was dropped since it s not allowed in the factorial function. And more importantly for us n0 n N! N n! (N n! nn! (n 1! (N n! n0 N n1 N (N 1 n (N 1! (n 1! (N n! + N n n1 That last identity is exactly what we need to find n n p(n n0 N n0 (N! (n! (N n! (N 1! (n 1! (N n! N (N 1 N + N N 1 (c n N! n! (N n! N [ N (N 1 N + N N 1] N (N 1 + N N N + N N + N N (N + 1. (5

3 and So then MS is a ( n N n a ( N (N + 1 N a ( N + N N a N (6 a N. (7 Problem Phillips 8.5 Find the probability distribution of monomers for a polymer in a cylinder and tethered to one end. We had G n a G x but more generally speaking the diffusion equation for dimensionality d-d is G n a d G. Please notice that in the textbook Phillips used N and I m using n for the length of the polymer. For cylindrical coordinates [ ( G n a 1 r G + 1 ] G 6 r r θ + G z. (8 To solve this we assume separation of variables: G (r, θ, z, n (r Θ(θ Z(z N(n. (9 Now this question is just a painful experience in solving it. We ve all done it before but anyway, the equation becomes [ ( G n a 1 r G + 1 ] G 6 r r θ + G z [ ( 1 N N n a 1 r + 1 Θ 6 r Θr θ + 1 ] Z Z z k where we did the normal thing and noted that there was absolutely no n dependence of the right-hand-side and so recognized it as constant. The solution to that ODE is N(n e kn. (10 3

4 Next we move on to the Z(z function: [ ( k a 1 r 6 r 1 Z Z z 6k a 1 r 6m a µ. + 1 ( r Θ Θr θ + 1 Z + 1 Θ Θr θ ] Z z We defined a new constant/s and so know that the solution to this ODE is Z(z A cos (µz + B cos (µz. There are some boundary conditions. At z 0, G 0 therefore Z 0 which means A 0. Also Z 0 at the other end of the tube z L. In order to ensure this we must demand µl qπ µ qπ L (11 where q [0, 1,...]. This means that the solution is Z B sin ( 1πz. (1 L Let s put off dealing with B since it s just a constant and I m sure it will get absorbed into another constant later. Next up is Θ. We repeat the same thing we ve been doing: The solution to this is 1 Θ Θ θ 6r a p. [ m + k + 1 r ( r ] Θ A cos (pθ + B sin (pθ. (13 We can really say anything about p except p [0, 1,...] (oh and we absorbed the constant from Z here already. Finally, we do the radial component of the function. Let ρ 6 ( k m /a 6 ( k q π /L /a. So then 1 r ( r The solutions to this ODE are Bessel functions: ( ρ r p. (1 (r J p (ρr. (15 We can say something about ρ since we know that G (and therefore must vanish at the wall of the cylinder r ϱ. Let ρϱ α pj

5 be the j th zero of the p th Bessel function. We can rewrite ρ then in terms of α pj ( r (r J p α pj ϱ k α pj a 6ϱ + q π a 6L (16 (17 So then putting all these together (and allowing all values of p, j, q we have G (r, θ, z, n p;j;q n(nz(zθ(θ(r p0 j1 q1 e kn sin ( qπz ( r [A pjq cos (pθ + B pjq sin (pθ] J p α pj L ϱ G p0 j1 q1 ( α e pj a 6ϱ + q π a 6L n sin ( qπz L ( r J p α pj [A pjq cos (pθ + B pjq sin (pθ] ϱ (18 The next thing to do is determine the infinite number of coefficients. I ll update this.pdf for the graduate students but there s not really a lot of biophysics to be learned from this long process so I ll leave it for now. 3 Problem (Again? Phillips Part a Get from Phillips Eq. 8.3 to Eq The probability that two points with N monomers between them in an ideal chain are located r from each other is ( 3/ ] 3 P ( r [ 3 r Na. (19 The probability that a point N monomers from another is a distance r away is ( 3/ ] 3 P (r πna πr exp [ 3r Na. (0 We can show this first off writing r in spherical coordinates and writing r only as it s magnitude: r rˆr + θˆθ + φ ˆφ (1 r r r r. ( 5

6 This means that ( 3/ ] 3 P ( r [ 3 r Na ( 3/ ] 3 [ 3r Na (3 regardless of θ or φ. However, P ( r P (r. To find P (r, let s pause for a moment and imagine that something is in either state A or B. The probability is then p p A + p B. (a This applies for as many states σ as are important to the question: p σ p σ. (b For our particular question, we have a continuous number of states to sum over. Why is this? Because the two angles can be anything; only the distance r matters. So then the sum becomes an integral: p p σ. (c So now back to our question: Only the distance matter so we integrate over the two angles i.e. P (r P ( r 3. Part b θ,φ π π θ0 φ0 π π θ0 φ0 ( 3 πna σ P ( r r sin θdθdφ ( 3/ 3 [ 3r Na 3/ πr exp [ 3r Na ] r sin θdθdφ ]. (5 Get from Phillips Eq to Eq The probability distribution is ( ( 3/ 3 3 r P ( r Na. (6 Let s orient ourselves (or define our axes such that ˆk. This perfectly acceptable choice means there is no ˆφ dependence in the exponent i.e. ( r ( r ( r r + r cos θ. (7 6

7 Ok now we do the same integral as in Part a to find P (r π θ0 π φ0 P ( r r sin θdθdφ ( 3/ 3 π π πna θ0 exp ( 3 r Na ( 3/ 3 π [ 3 ( r + ] π Na r We can do this integral by making the replacement: x cos θ dx sin θdθ c 3r Na r sin θdθ θ0 [ ] 3r cos θ exp Na sin θdθ. (8a (8b (8c which means ( 3/ 3 P (r π [ 3 ( r + ] 1 Na r e cx dx 1 ( 3/ 3 π [ 3 ( r + ] [ ] 1 1 Na r c ecx 1 ( 3/ ( 3 πna r π [ πna 3 exp 3 ( r + ] [e c Na e c] ( [ ( ( ] 1/ 3 r 3 (r 3 (r + Na exp Na. (9 Problem 3 Phillips 8.9 Pick a model for Fig. 8. of Phillips. Compare the model to the data shown. This was all discussed in Section 8.3 of Phillips. Since we re not pulling on the chain with a small force probably the best model would be [ ( fa L Na coth k B T ( fa NaL k B T k BT fa ]. (30 where L( is the Langevin function. From Fig. 8. the stretching seems to go for 5nm before a sudden event occurs (followed by relaxation. The force increase by 00nm in that time. If we were to actually find a fit we would need a value for the Kuhn length and the number of Kuhn segments that make up a titin. Qualitatively, the shape looks pretty similar. 7

8 5 Problem Phillips Part a Write down all the energies, the Boltzman factors and the partition function. a b There are four states: + - These four states lead can each have a different length, direction, energy and Boltzman factor: length direction energy Boltzman factor b + fb e βfb b +fb e βfb a + ɛ fa e β(ɛ fa a ɛ + fa e β(ɛ+fa This generates a partition function Z e βfb + e βfb + e β(ɛ fa + e β(ɛ+fa. (31 5. Part b What s the average length? Let s do this in a more proper way than we often do in this class. partition function gives a free energy The F k B T ln Z (3 From the free energy we can get the average length per monomer by differentiating with respect to force l F f k BT Z T,N Z f k BT [βbe βfb βbe βfb + βae β(ɛ fa βae β(ɛ+fa] Z 1 [ be βfb be βfb + ae β(ɛ fa ae β(ɛ+fa]. (33 Z The average length of the chain is simply 5.3 Part c Plot the average end-to-end distance. The normalized length is L N l (3 Γ L Nb N l l Nb b. (35 8

9 Let a 190nm b 100nm ɛ 300 5k B T ɛβ β.pn nm and plot it. It s quite a sharp transition. 6 Problem 5 Phillips 8.11 Just do it. 7 Problem 6 Harden Use Flory free energy to analyze the conformation of a polymer in a tube (good solvent conditions to derive a scaling relationship for L. The free energy of the chain confined to a tube is F k B T L + τnφ (36 Na where L is the extension of the coil, φ is the volume fraction of monomers and τ T T θ T θ is the temperature distance from the θ-temperature. Since we are concerned with good solvent conditions, τ > 0. It s clear that we find the length of the chain by minimizing the free energy F/ L 0 but the key lies in writing φ as a function of L. The volume of a single monomer is a 3 so the volume of the whole chain is Na 3. The chain is in a tube of cross sectional area πd / D and extends a distance L along the axis. Together, these mean that the volume fraction is and the free energy is φ Na3 LD (37 F k B T L Na + τ N a 3 LD. (38 9

10 Let s minimize the free energy to find the average length of the confined polymer: 1 F k B T L 0 L [ L Na + τ N a 3 ] LD L Na τ N a 3 L D L Na τ N a 3 L D L 3 τ Na N a 3 D L Nτ 1/3 a 5/3 D /3 (39 10