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Thermodynamics of Transport Processes: Membrane Transport Sedimentation Biol/Chem 5312 Physical Biochemistry 6 April 2004 lecture 20 1 Transport of material across membranes is one of the most important biological processes. The regulation of this transport is closely linked to the control of metabolism. Transport across biological membranes is usually characterized as belonging to one of three classes: Passive Facilitated Active Transport The ﬁrst two are just examples of diffusion. They follow an approach to an equilibrium state in which the chemical potential of a substance becomes equal on both sides of a membrane. 2 Active transport is quite different. In that case, a substance is pumped across the membrane against its concentration gradient. There must be another source of free energy to drive the transport, such as the free energy released by the hydrolysis of ATP. First, let’s look at Passive Transport. The two chambers are separated by a permeable membrane. In the approach to equilibrium, the chemical potential of the solute (2) ∆x will become equal in both chambers. α β The solute can pass through the membrane. It is important to keep the chambers stirred. 3 Initially, the concentrations on each side of the membrane are not equal, a b C2 ≠ C2 The membrane has a thickness of ∆x. A concentration gradient will develop across this membrane. In steady state, the concentration within the membrane will be constant, and so ∂C2 ∆x =0 ∂t From the continuity equation: α β ∂2C2 a b =0 C2 C2 ∂x 2 4 If the solute is equally soluble inside the membrane as in each chamber, then the concentration gradient will look as shown below. ∂2C2 ∂C2 Since 2 =0 in the membrane, then =K ∂x ∂x where K is a constant Integrating C2 = Kx + B in the membrane b At x = 0, (the β side) B = C2 b ∆x At the other side a C2 = KDx +C2 a b C2 −C2 since K= Dx α β For any x in the membrane: a b a b C2 −C2 C2 C2 C2 (x) = b x +C2 Dx 5 a b C2 −C2 b C2 (x) = x +C2 Dx The red line within the membrane is a straight line from one surface to the other. At the interfaces, the concentrations inside the membrane equal the concentrations in solution. Within the membrane, the ﬂow should be constant: a b ∂C2 C2 −C2 J2 = −D2 = −D2 ∂x Dx D2 is the diffusion coefﬁcient for the ∆x solute in the membrane. This is usually smaller than the value in solution. α β a b C2 C2 6 For a more accurate picture, consider the solubility of the solute in the membrane. In general, it will be less in the membrane phase, than in solution, differing by the partition coefﬁcient f2. mem C2 f2 = sol C2 Then, replace a a C2 by f2C2 and b C2 by f2C2 b b So, a C2 −C2 J2 = −D2 f2 Dx In biological membranes, J can often ∆x be measured, but not D2 f2 or Dx . So permeability is deﬁned: α β D2 f2 P= a b Dx C2 C2 7 a b So, C2 −C2 J2 = −D2 f2 Dx b becomes, J2 = −P2 C2 −C2 a Permeabilities of various solutes are given in Table 14.2 Table 14.2 MEMBRANE PERMEABILITIES P (cm/sec) Membrane K+ Na+ Cl- Glucose Synthetic (PS) <9 x 10-13 <1.6 x 10-13 1.5 x 10-11 4 x 10-10 Erythrocyte 2.4 x 10-10 1 x 10-10 1.4 x 10-4 2 x 10-5 Squid axon 5.6 x 10-8 1.5 x 10-8 1 x 10-8 ND (resting) Squid axon 1.7 x 10-4 5 x 10-6 1 x 10-8 ND (excited) Since all membranes have about the same thickness (∆x), the differences in P must be due to differences in f and D. 8 Another parameter of interest is the transit time, τ. It is deﬁned as the average time required for a particle to cross the distance ∆x. The transit time for one-dimensional diffusion across ∆x: (Dx)2 t= 2D D2 f2 Using P2 = Dx f2 Dx We can write: t= 2 P2 The transit time is inversely proportional to P. 9 Facilitated Transport As seen in Table 14.2, the permeability of some solutes is much higher than others, even though they have a similar chemical nature. This can be due to a mediator of transport. There are two general categories: carriers channels or pores They tend to be rather speciﬁc for particular molecules. The direction of transport is determined thermodynamically, by the concentrations on the two sides of the membrane. 10 Consider the transport of one mole of component i from α to β. DG = µi − µa ∼ RT ln ci − RT ln ca b b i = i b ci DG = RT ln a ci So, the transport of component i from α to β , is favorable, i.e., ∆G<0, only if b ca > ci i If this process continues until the concentrations become equal on both sides, i.e., ∆G = 0, then the system will be at equilibrium, and no further net transport will occur. This is true, whether the transport is passive or facilitated. 11 Active Transport How do cells accumulate solutes against a considerable concentration gradient, across a biological membrane? In some cases, the solute is modiﬁed after transport, such as the phosphorylation of sugars. Or, sometimes there are very high afﬁnity binding sites inside the cell. In these cases, the effective concentration of the solute remains low inside the cell. These cases are not examples of active transport. In active transport, solutes are moved across a membrane against a concentration gradient due to its coupling to a favorable chemical reaction. Typically, the hydrolysis of ATP drives unfavorable transport. b Ci DGi = RT ln a + DGR Ci 12 Example, Na+-K+- ATPase [Na+]in = 0.01 M ATP [K+]in = 0.10 M ADP +Pi 3Na+ ↓ ↑ 2K+ ∆ψ = - 0.07 V (negative inside) Calculate Na+ and [Na+]out = 0.14 M K+ separately. [K+]out = 0.05 M 13 ATP ADP +Pi [Na+]in = 0.01 M [K+]in = 0.10 M 3Na+ ↓ ↑ 2K+ ∆ψ = - 0.07 V (negative inside) [Na+]out = 0.14 M [K+]out = 0.05 M in CNa DGNa = RT ln out + ZNa FDyin→out CNa 140 DGNa = (8.31J/K · mol)(310) ln + (1)(96, 480J/V · mol)(0.07V ) = 13.55kJ/mol 10 out CK DGK = RT ln in + ZK FDyout→in CK 100 DGK = (8.31J/K − mol)(310K) ln = 0.97 kJ/mol + (1)(96, 480J/V − mol)(−0.07V ) = 0.97kJ/mol 5 14 If 3 Na+ and 2 K+ are transported per cycle of the transporter, DGtotal = 3 × 13.55kJ/mol + 2 × 0.97kJ/mol = 42.52kJ/mol Inside a cell, the free energy of ATP hydrolysis is estimated to be about -50 kJ/mol. So, hydrolysis of one ATP is sufﬁcient for the transport of 3 Na+ and 2 K+. The transport of the Na+ is much more costly, because it is against the membrane potential. 15 The Goldman equation, shown below, shows how the membrane potential is derived from all ions that are present, and where they each might have different permeabilities. RT Â+ Pi+ [B+ ]a + Â− Pi− [Xi− ]b Dy = ln i F Â+ Pi+ [B+ ]b + Â− Pi− [Xi− ]a i Each sum is over all cations (B) or all anions (X), multiplied by the appropriate permeabilities (P). Note that if all P are zero (or nealy zero) except for one ion, then the Goldman equation reduces to this form: RT [B]a Dy = ln b F [B] The Goldman equation is useful for calculating changes in the membrane potential when ions change permeability, as occurs during nerve impulses. 16 Sedimentation We will now take one more look at sedimentation, using the formalism of irreversible thermodynamics. In that case we will think of sedimentation as a process that is moving towards an equilibrium state, one in which the total potential will be uniform throughout the system. In a 2-component system, this will be met when: µ ∂˜ 2 ∂µ2 = − M2 w2 r = 0 ∂r ∂r ↑ ↑ ↑ total chemical centrifugal potential potential potential The Flow equation: µ ∂˜ 2 ∂µ2 J2 = −L2 = −L2 − M2 w2 r ∂r ∂r 17 µ ∂˜ 2 ∂µ2 J2 = −L2 = −L2 − M2 w2 r ∂r ∂r ∂µ2 Previously we have analyzed ∂r ∂µ2 RT dC2 = M2 n2 rw2 r + ¯ ∂r C2 dr C2 Substituting for L L2 = where f2 is the frictional coefﬁcient NA f2 C2 RT dC2 J2 = − M2 n2 w2 r + ¯ − M2 w2 r NA f2 C2 dr Reorganizing, M2 (1 − n2 r) 2 ¯ RT ∂C2 J2 = w rC2 − NA f2 NA f2 ∂r 18 M2 (1 − n2 r) 2 ¯ RT ∂C2 J2 = w rC2 − NA f2 NA f2 ∂r but, M2 (1 − n2 r) ¯ S= NA f2 RT and D= NA f2 ∂C2 so J2 = S2 w rC2 − D2 2 ∂r Except for minor corrections, e.g., slight pressure dependence of n2 and r , ¯ this description of ﬂow processes during sedimentation is complete. 19 ∂C Initially, the concentration is uniform =0 ∂r When the centrifugation begins, only the centrifugal potential is important. As a boundary forms, concentration becomes less uniform, and both terms become important. If the centrifugation is ended (ω=0), then the concentration gradient will eventually go away. How to monitor the moving boundary? ∂C Previously, we used the maximum peak, i.e. ∂r for calculation of the rate of boundary movement. 20 In the plateau region (below the boundary) ∂C =0 ∂r so, J = sCp w2 r There must be a point that moves with velocity v’, such that J = Cpv’ and then v s= 2 wr This point, rb, occurs at the second moment of the concentration gradient. R ¯ ∂C r2 dr This is deﬁned as rb = R 2 ∂r ∂C ∂r dr See Fig. 14.9 21 In a sharp boundary, the second moment will move at approximately the same rate as the peak of the concentration gradient. The previous has come from application of Fick’s First Law. What about the Second Law? ∂C2 ∂2C2 = D2 2 ∂t ∂x To apply the continuity equation to centrifugation ∂C2 ∂J2 =− ∂t ∂x we must notice that in centrifugation the cross-section is proportional to r. Previously, in one-dimensional ﬂow, the cross-section was independent of x. 22 Since the force is directed radially during centrifugation, ∂C2 1 ∂rJ2 =− ∂t r r ∂r t Substitute for J ∂C2 J2 = s2 w rC2 − D2 2 ∂r yields, ∂C2 1 ∂ 2 ∂C2 =− 2 r s2 w C2 − rD2 ∂t r ∂r ∂r Perhaps not surprisingly, this is a difﬁcult equation to work with, but approximate solutions have been found for several particular situations. One thing it explains is the “radial dilution” effect. Fig. 14.10 23 In moving boundary sedimentation, the concentration of the plateau is not constant, but instead decreases during the experiment. In the plateau, the concentration at any time is equal to a constant, Cp. ∂C2 1 ∂ 2 ∂C2 So, in the plateau region, =− 2 r s2 w C2 − rD2 ∂t r ∂r ∂r ∂Cp 1 ∂ 2 becomes =− r s2 w2 Cp ∂t r ∂r ∂Cp and = −2s2 w2 Cp ∂t o −2s2 w2t or integrated: Cp = C e The plateau concentration drops exponentially with time. 24 25