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642:613 Rutgers - Fall 2003 Instructor: Eduardo D. Sontag Sections 2.3-2.5, Membrane Di usion & Transport http://www.math.rutgers.edu/ sontag/613.html Ohm s law for di usion suppose on opposite sides of membrane have chemical at constant concentrations ce and ci ( external and internal ) and membrane has thickness L no6 ux c(0, t) ce ct = D 2 c c(L, t) ci x=0 no? ux x=L (membrane drawn horizontally) assuming no di usion along membrane itself, so model as 1-d problem (same math as adding Neumann conds) could nd time-dependent solution, given initial conds (using separation of variables), but just steady-state here: x c = 0 line; and boundary conds c(x) = ce +(ci ce) L L so J = Dc = D (ce ci) (constant), or ce ci = J D L analogous to Ohm s law V = IR (potential di erence proportional to di erence of charges & ux=current) facilitated di usion: myoglobin (from Protein Data Bank, PDB, http://www.rcsb.org/pdb/molecules/mb3.html) myoglobin is where the science of protein structure really began. . . John Kendrew and his coworkers determined the atomic structure of myoglobin, laying the foundation for an era of biological understanding The iron atom at the center of the heme group holds the oxygen molecule tightly. Compare the two pictures. The rst shows only a set of thin tubes to represent the protein chain, and the oxygen is easily seen. But when all of the atoms in the protein are shown in the second picture, the oxygen disappears, buried inside the protein. So how does the oxygen get in and out, if it is totally surrounded by protein? In reality, myoglobin (and all other proteins) are constantly in motion, performing small exing and breathing motions. Temporary openings constantly appear and disappear, allowing oxygen in and out. The structure in the PDB is merely one snapshot of the protein, caught when it is in a tightly-closed form facilitated di usion when a reaction facilitates di usion; example: oxygen in muscle bers binds to myoglobin oxymyoglobin which results in enhanced transport seems counterintuitive because Mb much larger than O (500 times larger), so di uses slower! model will show enhancement; later, we ll interpret s(0, t) s0 s = O2, e = M b, c = M bO2 s(L, 0) sL s0 x=0 O2 + M b M bO2 k k+ x=L s t e t c t = Ds = De = Dc 2s x2 2e x2 2c x2 + k c k+se + k c k+se k c + k+se (and assume: De = Dc) assume also zero- ux of Mb & MbO2 at boundary i.e., e c = 0 at x = 0, L x x if you have never thought of such boundary conditions, you may wish to interpret them as follows: think of a narrow strip (of very small width ): most particles bounce back far into region, so the net ux at x = L is 0 steady-state analysis: Dssxx + k c k+se = Dcexx + k c k+se = Dccxx k c + k+se = 0 (e + c)xx 0 (linear) & (e + c)x 0 at e + c e0 also, (Dssx + Dccx)x = Dssxx + Dccxx = 0 means that Dssx + Dccx = constant, call it J because it is the sum of the two uxes, i.e. it is the total ux of oxygen (bound or not) integrate: f = J f (0) f (L) = J L, so: J = Ds (s0 sL) + Dc (c0 cL) (where we know s0, sL but not c0, cL) we had: Dssxx + k c k+s(e0 c) = Dccxx k c + k+s(e0 c) = 0 now let = (k+/k )s, u = c/e0, x = Ly, 1 = Ds/(e0k+L2) 10 7, 2 = Dc/(k L2) 10 4 1 yy = (1 u) u = 2uyy so quasi-steady state approx c = e0 s K+s (K=k /k+) now substitute into J = Ds (s0 sL) + Dc (c0 cL) to get: Ds K2 J= (1 + ) (s0 sL) ( = Dce0 , = (s0+K)(sL+K) ) Ds K L expression for ux J Ohm s Law ; factor 1 + enhances e.g. 500 now substitute expression for J , & c as function of s (q-s-s), into Dssx + Dccx = J ; next integrate from 0 to x (any x, not just L), & get soln for s (or ) see book for details) interpretations may also write: Ds Dc sL s0 J= (s0 sL) + e0 L L K + s0 K + sL (second term > 0 since s/(K + s) is increasing) and this exhibits ux as sum of Ohm term plus term that depends on di usion constant Dc one intuition: by binding to Mb, there would be less free O2 near left but end boundary conditions say that one has s0 there so more must ow in (di usion tends to equalize!) while at other end, opposite happens, and more ows out a related example: muscle respiration consider problem of getting O2 to center of muscle muscle ber thought of as cylinder 0 r a radial di usion, ignoring longitudinal di usion now recall Laplacian in polar coordinates, 2-dim case: write f (r, , t) = c(r cos , r sin , t); then (some calculus): +2 r r r 2 (all terms on the RHS evaluated at r, , t) so Laplacian for radially symmetric c (write f as c ): D c r r r r r 2 (for spherically symmetric c in 3-d, 1 r 2 r c r 2 r ) ( 2c)(r cos , r sin , t) = 2f + 1 f 1 2f so steady-state equation (same variables as before): Ds s r + k c k+se g = 0 r r r Dc c r k c + k+se = 0 r r r where g = constant consumption of oxygen in muscle we impose boundary conditions: dc ds dc s(a) = sa, (a) = 0, (0) = 0, (0) = 0 dr dr dr (at r = 0: if continuing in same direction, even function, so derivative zero at that point, or think of small circle around zero: balance of ows in/out is zero) with 1 = Ds/(e0k+a2), 2 = Dc/(k a2), = g/k , = (k+/k )s, u = c/e0, r = ay, y ( 1 + 2u)y y = y so integrate: y( 1 + 2u)y = A + y 2/2 ( 1 + 2u)y = A/y + y/2, and integrate again: 1 + 2u = A ln y + y 2/4 + B but soln not well-de ned at y = 0 unless A = 0, so A = 0 at boundary y = 1: 1 (1) + 2u(1) = /4 + B what (1) = 1 is so that s(0)=u(0)=0 ( just enough O2)? evaluate above at zero B = 0 too, so, using also 1 1 y y y = (1 u) u y and q-s-s approx, substitute u as function of to obtain: 1 1 + = 1 + 1 4 1 where = 2/ 1, so can solve for 1( ) ( consumption g) note that it is an increasing function Mb helps diminish the minimal concentration 1 needed to prevent oxygen debt (concentration at center falls to 0) 1 2 so plot of 1 vs 4 1 = Ds/(e0k+a2) small if k+ (binding constant) large, = Dc/(k a2) large if k small (unbinding), large if high a nity Uniport, e.g. Glucose Transport Si (Se) = glucose inside (external), Ci (Ce) = carrier protein, open to inside (external), Pi (Pe) = complex, open to inside (external) Si + Ci Pi k k+ Se + Ce Pe k k+ Pi Pe Ci Ce k k k k assume also that Se and Si 0 (constant supply rate outside & removal rate inside) si = k pi k+sici Ji se = k pe k+sece + Je pi = kpe kpi + k+sici k pi pe = kpi kpe + k+sece k pe ci = kce kci k+sici + k pi ce = kci kce k+sece + k pe at steady state: 0 = d/dt(si + se + pi + pe) = Je Ji, so Ji = Je = J note that pi + pe + ci + ce = total amount of receptors c0 use rhs s for se, pi, pe, ci plus pi + pe + ci + ce = c0 to solve for J in terms of se, si (seen as parameters when solving for rest) se si J = ac0 (si + K1)(se + K2) K3 for suitable constants (see book) J1 J2 horiz axis se, vertical J , di erent si s (non-dimensionalized) variations: symport in which two go in simultaneously sodium outside cell is high compared to inside (due to Na/K pump) so gradient provides energy for transport (but no additonal energy required) or antiport, where exchange inside/outside e.g. symport: when both sodium and glucose are attached, conformational change in protein molecule happens, and both are transported to inside cell http://jimswan.com/237/channels/channel graphics.htm (skip equations - similar to uniport, but exponents appear, as in cooperativity) active transport: e.g.: electrogenic Na+-K+ pump three sites for Na+ attachment on inside surface of carrier, and two for K+ outside ATPase on intracellular surface hydrolyzes ATP, releasing energy that causes carrier conformational change this pumps the 3 Na+ ions out and then potassium attaches and 2 K+ ions are pumped in http://jimswan.com/237/channels/channel graphics.htm on balance, more + s pumped out, but on the other hand negative (other) ions are not permeable - this creates a polarization accross the membrane see book for details (skipping - no time. . . )

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