# Thermodynamics of Transport Processes Membrane Transport by cgg10267

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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.

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
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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

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

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

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.
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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
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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

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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.
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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
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

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.
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