8.592J: Solutions for Assignment 7 Spring 2005

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1 8.59J: Solutons for Assgnment 7 Sprng 5 Problem 1 (a) A flament of length l can be created by addton of a monomer to one of length l 1 (at rate a) or removal of a monomer from a flament of length l + 1 (at rate b). Such flaments change ther length through addton or removal at rate (a + b). Hence the probabltes change accordng to dp(l, t) dt = a p(l 1,t) + b p(l + 1,t) (a + b) p(l,t). (1) Snce there are no flaments of less than 1 monomer, the correspondng equaton n ths case s dp(1, t) dt = b p(,t) a p(1,t). () (b) Snce the rates of addton and removal have a rato of a/b, the process wll be staonary f probabltes of successve lengths have ths rato. It s ndeed trval to check that dp(l)/dt = n the above equatons provded that p(l+1) = (a/b)p(l) (a/b) l. After normalzaton, ths leads to the dstrbuton ( ) ( ) b a l p (l) = a 1. (3) b (c) Clearly the above probablty dstrbuton s only vald (postve and normalzable) f b > a,.e. when the rate of removal from the mnus end exceeds that of addtons to the plus end. In such a case, the mean length of flament s gven by l = lp (l) = l=1 b b a. (4) (d) If a > b, on average there are more monomers added to the flament than can be removed, and consequently ts length ncreases lnearly wth tme at a rate of a b. Hence, startng from monomers at tme zero, the mean length at tme t s l(t) = (a b)t. (5) Snce growth proceeds through dscrete probablstc steps, there s also a varance whch adds the contrbutons from the growth and removal steps,.e. l(t) c l(t) l(t) = (a + b)t. (6) The above results can be obtaned n quanttatve manner by Fourer transformng the Master equaton n part (a), to get [ ( ) ( )] ω p(k,ω) = a e k 1 + b e +k 1 p(k,ω). (7) 1

2 Expandng n small k leads to a dsperson relaton ω = (a b)k a + b k + O(k3) = vk Dk +, (8) from whch we can dentfy the drft velocty v = (a b)k, and the dffuson coeffcent D = (a + b)/. The soluton (n real space and tme) of ths drft dffuson equaton s a Gaussan wth the mean and varance calculated earler,.e. p(l,t) = exp Problem [ (l (a b)t) (a + b)t (a) Startng wth the equatons p + t p t ] 1 π(a + b)t. (9) = f + p + + f + p x (v +p + ) + d p + x, (1) = f + p + f + p + x (v p ) + d p x, (11) perform Fourer tranformatons p ± (k,ω) = dx dt e (kx ωt) p ± (x,t), (1) p ± (x,t) = 1 (π) dk dω e (kx ωt) p ± (k,ω). (13) Then to wrte the Fourer transform of equatons 1 and 11, we just need to replace t ω, x k, and p p, to get ω p + = f + p + + f + p + k v + p + d k p +, (14) ω p = f + p + f + p k v p d k p, leadng to (15) p = p + ω + f + k v + + dk f +, (16) p + = p ω + f + + k v + dk f +. (17) Substtutng the last equaton n the prevous one, we get the followng equaton ( f + f + = ω + f + k v + + dk ) ( ω + f + + k v + dk ). (18) Solvng for ω gves ω = 1 [ ((f + + f + ) + k(v v + ) + k d) + (19) ] ± (k(v + + v ) + (f + + f + )) 4f + f +.

3 (b) In order to expand the slowly varyng mode of ω we need to consder only the acoustc mode,.e. the soluton for whch ω(k = ) =. Thus, we expand equaton () to second order n k, to get ( ) v+ f + v f + ω(k) k f + + f + ( d + (f + f + )(v + + v ) (f + + f + ) 3 ) k + O(k 3 ), () from whch we dentfy the drft velocty and dffuson coeffcent (ω vk Dk ) as ( ) v+ f + v f + v =, (1) f + + f + [ D = d + (f + f + )(v + + v ) ] (f + + f + ) 3. () (c) The average poston grows as x = vt, whch the varance n poston s x c = Dt. To get a tme scale beyond whch the dfusson effects are less mportant than the average velocty we need vt Dt,.e. t τ D v. (3) From the numbers provded, we fnd v = m/s, D = m /s, and thus (4) τ 45s. (5) Ths s qute long, suggestng that fluctuatons are very mportant to mcrotubules. Problem 3 (a) We frst look for the steady state wth r =, for whch = vp + D p x We are then left wth an equaton whch s solvable by separaton of varables as dp p (6) = v D, (7) p = Ne x v(x ) D dx. (8) Note that n ths one dmensonal case we are dealng wth a Fokker-Planck equaton whch also descrbes equlbrum n the potental that generates the velocty v. Indeed, the steady state s the correspondng Boltzmann weght. The constant N s determned by normalzaton over the nterval [, ). 3

4 (b) On the other hand, for D = we have = v P r x P, (9) x ( x ) rz P = N exp v(z) dz (3) p = N rx ( x v exp rz ) v(z) dz. (31) And agan, N must be determned by normalzaton. (c) For ) v = v, and a ) r =, we are lookng at the soluton we found n the prevous part p = N e v D x. (3) Here, we must mpose the normalzaton, and so we see that we have to choose N = v D, and to requre that v <, otherwse, the probablty dstrbuton we have found has no meanng,.e. p = v D e v x D. (33) In the case b ) wth D =, we have from the soluton n part (a), v rx p = v, (34) πr e whch s meanngful only for v > (see Fg. 1a). For ) v = ax, and a ) r =, agan we get, mposng normalzaton a ax p = e D, (35) πd and ths soluton make sense only for a >. If D =, we fnd p = N ax e rx a, (36) and ths soluton makes sense only for a <. Note n any case that, n order to mpose the normalzaton, whch wll gve us N, we must mpose that the doman of nterest excludes a boundary of the pont (see Fg. 1b). For ) v = v ax, and a ) r =, we get p = N e v D x ax D, (37) and ths soluton make sense only for a >. If D =, we fnd p = N rx ( x v ax exp rz ) v(z) dz (38) = N rx (v ax) v r a 1 e r a x. (39) Ths expresson may have varous meanngful forms dependng on the sgns and magntudes of the parameters r, v and a. For nstance, f a < and v > the functon has an exponental tal, whereas f a > and v >, p s nonzero only on a fnte nterval (see Fg. 1c). 4

5 Problem 4 In ths problem, we have to extend what we dd n class for a sngle state motor to a two states motor. a) Let us label the the probablty of fndng the motor at the ste n th n the frst or n the second state as P(n) and P (n). They are ruled by the followng dfferental equatons: dp(n,t) dt dp (n,t) dt = u P (n 1,t) + w 1 P (n,t) (w + u 1 )P(n,t) (4) = u 1 P(n,t) + w P(n + 1,t) (w 1 + u )P (n,t) b) Defnng the Fourer transform p(ω,k) of P(n,t) as: P(n,t) = dkdω p(ω,k)e (kx ωt) (41) and q(ω,k) for P (t,n), the dfferental equatons become: ω p = u e kd p u 1 p + w 1 q w p (4) ω q = u 1 p u q + w e kd p w 1 q That s: (ω u 1 w ) p + (u e kd + w 1 ) q = (43) (u 1 + w e kd ) p + (ω u w 1 ) q = In order to a non null soluton to exst, we have to requre that the determnant s null, and we get: ω + ωσ + u 1 u (e kd 1) + w 1 w (e kd 1) = (44) where Σ = u 1 + u + w 1 + w. Solvng for ω, and choosng the sgn such that ω(k = ) =, we get: w = 1 ( Σ + ( Σ 4u 1 u (e kd 1) 4w 1 w (e kd 1)) 1 ) (45) c) Taylor expandng up to second order n small k, we get: where ω = vk Dk + o(k 3 ) (46) v = u 1u w 1 w d (47) Σ ( u1 u + w 1 w D = (w 1w u 1 u ) ) Σ Σ 3 d 5

6 v s the drft velocty, and D s the dffuson coeffcent. The Ensten force s gven by: f e = k B T v D = k BT d Σ (u 1 u w 1 w ) (Σ (u 1 u + w 1 w ) (u 1 u w 1 w ) (48) d) Smulatng the effect of a dampng force f d as a suppresson effect on u, we get the new drft velocty: v(f d ) = u 1 u e f d d k BT w 1 w u 1 + u e f d d k BT + w 1 + w d (49) The stallng force f s s the value of f d for whch the drft velofty v(f d ) s zero. We get: f s = k BT d log(w 1w u 1 u ) (5) d) Assumng d = 8.nm, we can verfy how ths model compares to the expermental data. Computng V, D, f s, wth u sec 1, u 5sec 1, w 1 u 1 /1, w u /1, we get: V 4nm/sec (51) D f s 4.7pN 16nm /sec The expmental values are: V 67m/sec (5) D f s 5pN 14nm /sec We can see that the agreement wth the expermental data s reasonable for such a smple model. Problem 5 In the absence of drft, the dffuson equaton predcts fluctuatons n the poston of E. Col, growng wth tme as p( x, t) = D p( x,t), = (53) t x =, and x = 6D t, (54) where D s the dfusson coeffcent, and we have noted that the contrbuton to x comes from the three coordnates. Thus, to calculate the dffuson coeffcent we just 6

7 need to calculate x. The poston vector s the sum of dsplacements from the ndvdual runs,.e. x(t) = vˆn t r, (55) where the tme argument tself s the sum of contrbutons from the runs and tumbles, as t = (t r +t t). Snce the steps are ndependent, we can perform varous averages as Snce, Fnally, x(t) x t r = x = = j = j = j = v ˆn t r = ˆn v t r ˆn j t j r v ˆn ˆn j t r t j r + = v 3 τ r v δ j t r tj r v ˆn t r =, (56) (57) (58) (59) v (t r). (6) 4t 3 + τr e t/τr dt = 3 τ r, we get (61) 1 = v 3 τ r t t r + t t. (6) 4t t r = τr e t/τr dt = τ r, and t t = τ t, gve (63) x = v 3 t τ r, (64) τ r + τ t = D = v τr. (65) 4 τ r + τ t The numbers typcal of E. Col result n D = (µm). (66) s (b) Now, n the presence of a chemcal gradent, we have a preferred moton n the z drecton such that τ r (ˆn) = τ + gˆn.ẑ. We shall assume that the probablty dstrbuton whch regulates the choce of the drecton n whch to move after each tumblng tme s sphercally symmetrc. The only dfference among the drectons s then the avarage duraton of the moton. 7

8 To get the drft velocty we need to calculate the mean poston: x(t) = v ˆn t r = (67) = v ˆn (τ + gˆn.ẑ) = v gˆnˆn.ẑ = (68) = vg n z = 1 3 vg 1 = vg 3 t τ + τ t, leadng to v D = v g. (69) 3ẑ τ + τ t 8

9 8 x p r 4 (a) p D p r.15 (b).1.5 p D x p r, a > and v > 4 3 p D p r a < and v > (c) Fgure 1: Probablty dstrbutons p r and p D. 9

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