CALIFORNIA INSTITUTE OF TECHNOLOGY

 Division of Biology -- Bi145 Animal Physiology
Some quantitative principles relevant to passive transport processes
Simple diffusion processes and the conduction of electric current by ions in solution are familiar examples of passive transport processes. They can be related to the following theoretical framework:
In considering the transport of a particular species of molecule or ion, it is useful to start by constructing a potential function that expresses the average free energy of these molecules as a function of the relevant measurable variables. The chemical potential ,µ, can be defined as the partial molar free energy for the particular molecules being considered. The average free energy of one of the molecules is then µ/N, where N is Avogadro's number; this can be thought of as the change in the total free energy of the system when one molecule is added to it, but this is rather crude since at any instant the energy of the individual molecules will be distributed around the average value.
The relationship between the chemical potential of molecular species i and the variables that are important in considering transport processes is given by the following expression:
µi = µi0 + RT ln Ci/Cio + ziF(V-Vo) + vi(P-Po) ___________________________(1)
where:
C is, in most cases of biological interest, the concentration in moles/liter. Actually it is the activity which should appear here, and sometimes this distinction is significant.
V is the electrical potential. It seldom appears except as a gradient or a difference in voltage between two regions.
P is the pressure.
µi0 is a standard potential for this species of molecule. Its value depends on the choice of the standard values for Cio, Vo, Po, and on variables, such as the nature of the solvent and the concentration of other solutes, which are not explicitly stated in equation (1). Most commonly, Cio = 1 molar, Vo = 0 and Po = 1 Atm.
zF is the electrical charge per mole for a charged ion. z will often be a positive or negative integer, but in some situations it may be useful to use a fractional value representing a mixture of ionic species, rather than dealing with each species individually. Faraday's constant, F, is 96,500 Joules/volt, or 23.06 kcal/volt.
v is the partial molar volume.
That equation (1) does in fact constitute a potential function can be seen clearly from the fact that the last two terms in equation (1) give the work done per mole when molecules are moved between regions of different voltage or pressure. This may not be so obvious for the term involving concentrations, but this term is familiar from the usual relations between free energy changes and concentrations in chemical reactions.
For comparison with the free energy change values for chemical reactions it is useful to obtain numerical values for µ in units of kcal/mole, in which case equation (1) can be used in the following form (for a temperature of 25 C):
µi = µi0 + 1.37 log(Ci) + 23 ziV + 0.24 v i(P-1) ___________________________(2)
where V is in volts, v i is in liters/mole and P is in atmospheres.

Since µ is a potential function, its first derivatives with respect to spatial coordinates represent forces tending to move the molecules to regions of lower potential, thus minimizing the total free energy of the system. This differentiation requires, of course, that the potential be a continuous function of position, which will normally be true in a homogeneous phase. At a phase boundary, or in other heterogeneous systems, the situation will be qualitatively similar, but other theoretical treatments may be necessary to obtain kinetic parameters. In a homogeneous phase,
Force/mole of the ith species = X i = -grad µi_________________________________ (3)
If the molecules on which this force acts are free to move, they will reach a steady drift velocity at which the accelerating force is balanced by the frictional resistance representing randomization of movement by collisions with surrounding molecules (see Feynman Lectures, Vol. 1, Ch. 43). Experimental work as well as theoretical arguments indicate that this frictional resistance is a linear function of the drift velocity, so that
Drift velocity = -ui grad µi ______________________________________________ (4)
where the constant, ui, is the mobility characteristic of these molecules in a particular solvent. The total movement of molecules in response to the force on them is referred to as the flux , or more accurately the flux density, which is the flux through a unit area of a plane perpendicular to the direction of the flux. This is obtained by multiplying the drift velocity by the number or concentration of molecules per unit volume:

Flux = J i = -Ci ui grad µi ________________________________________________(5)
Since fluxes are normally expressed as moles/sec/cm2 and mobilities in cm/sec/unit gradient, the concentration needs to be expressed as moles/cm3 or as 10-3 x moles/liter. If concentration differences are the only source of forces acting on the molecules:

J i = -uiRTCigrad ln Ci = -uiRTgrad Ci = -Digrad Ci __________________________(6)
This gives us the standard form of Fick's Law which describes the diffusion of a neutral molecule in dilute solution, and a number of other simple diffusion situations. This is an approximation in which both the activity coefficient and the mobility of the diffusing substance are assumed to be independent of concentration, but it is reasonably accurate under the conditions encountered in most biological systems. Di is the diffusion constant .
If voltage differences are the only source of forces acting on the molecules,

J i = -Ci ui ziF grad V ___________________________________________________(7)
which is also familiar, although it is common to use a mobility constant which has units of cm/sec//volt/cm, in which case the factor ziF doesn't appear explicitly.
The purpose of these arguments is to show that the basic flux equation, equation (5), is a reasonable generalization which includes the most familiar transport phenomena, so that it can be used as a definition of passive transport. Passive transport is transport which is driven by the gradient of chemical potential of the molecular species which is being transported. It is always in the direction of chemical equilibrium for this species; that is, it tends to level out any differences in chemical potential. It is always associated with a decrease in the total free energy of that species in the system under consideration.
Obviously, any transport that does not use only this source of energy to overcome the frictional resistances to movement must utilize some other source of energy, and if the transport occurs in opposition to a gradient of chemical potential, additional energy is required to increase the potential of the transported molecules. Such transport, brought about by energy derived from the metabolism of an organism, is referred to as active transport .
This distinction between active and passive transport is not completely unambiguous, because it has neglected the possibility of interactions between a flux of one kind of molecule and the other molecules in the solution, which may induce fluxes of these molecules which would not be expected on the basis of their electrochemical potential distributions. The appearance of such fluxes can be described by the term codiffusion . Codiffusion phenomena occuring at membranes are more commonly referred to as coupled transport . A general theoretical treatment of codiffusion processes has been developed as part of the theory of the thermodynamics of irreversible processes (see Katchalshy & Curran, Nonequilibrium Thermodynamics in Biophysics , Harvard, 1965, Chapter 8). It involves the following type of analysis: Suppose we have a system in which there are two diffusing molecular species. This can be represented formally by

J i = LiiX i + LijX j

J j = LjiX i + LjjX j_____________________________________________(8)

X i and X j are the forces defined much as in equation (3), and the L's are transport coefficients. Lii and Ljj can often be obtained by arguments similar to those we have already used, but in general, the coupling coefficients Lij and Lji are more difficult to determine. However, an important general result of the theory of irreversible thermodynamics is that the forces and fluxes can be defined in such a way that Lij = Lji. The matrices set up to deal with systems with a larger number of components are also symmetrical if the terms are properly defined. Although there are some situations where this theoretical framework is useful, we will not consider it in any more detail.

Some important equilibrium situations
 Equation (1) is also a convenient point of departure for considering a number of important equilibrium situations that might arise in a two-compartment system in which some components can pass freely from one compartment to another. A two-compartment system might consist of two immiscible solvents, or of two aqueous solutions separated by a selectively permeable membrane.
At equilibrium there must be no net transport of components between the two compartments, and the electrochemical potential of any component that is free to move between the two compartments must be the same in both compartments.

Case 1
 Consider a two phase system, with both phases at the same pressure and voltage, but
µ10 not equal to µ20. The ratio between the solute concentrations in the two phases is given by

C1/C2 = exp(( µ20 - µ10)/RT) ____________________________________(9)

The ratio C1/C2 will be independent of concentration, and will equal the partition coefficient characteristic of the particular combination of components. This situation generally arises when a solute is distributed between two immiscible solvents, and we then speak about, say, the ether-water partition coefficient for urea, etc.

 Case 2
 Both compartments are at the same pressure and µ10 = µ20. If the solute molecules are neutral, equation (1) requires that the concentrations be equal at equilibrium. However, with charged molecules, it is possible for the concentrations in the two compartments to be unequal at equilibrium if there is an appropriate voltage difference between the two compartments. This voltage difference is then obtained from the familiar Nernst equation , which can be obtained from equation (1):

V1 - V2 = (RT/zF) ln C1/C2 _________________________________(10)

Or, from equation (2), the following form can be obtained which gives values in volts, for a temperature of 25 C.

V1 - V2 = (1/z) 0.059 log C1/C2 ________________________(11)

The activity coefficient of the solute ion is assumed to be the same in both phases.

Case 3
 Osmotic equilibrium with neutral molecules and µ10 = µ20.
A concentration difference between the two compartments may be maintained at equilibrium if there is a pressure difference:

P1 - P2 = (RT/v ) ln C1/C2 (12)

In most cases of biological interest, movement of solute molecules between the two compartments is restricted, but the solvent, water, can diffuse freely across the interface between the two compartments and must therefore satisfy the condition for equilibrium. Under these conditions the activity of water cannot be accurately expressed by its concentration in moles/liter, but must instead be expressed in terms of its mole fraction in the solution. Usually it is more convenient to have an expression for the pressure difference in terms of the solute concentrations, rather than the mole fraction of water. If the solute concentration is represented by c, the number of moles of water in one liter of solution will be given by
(1/v ) - kc
where kc represents the number of moles of water which must be missing from one liter of solution in order to accommodate c moles of solute. The mole fraction of water in the solution is
((1/v)-kc)/ ((1/v)-kc+c) = 1/( 1 + cv/(1-kcv))

In most situations of biological interest, where the solutions are relatively dilute, cv and kcv will both be small compared to one, since v for water is approximately 0.018 liters/mole. Thus approximately, the mole fraction is given by 1/(1+cv ) and its logarithm is approximately equal to -cv . Substituting this into equation (12) leads to the familiar expression for the osmotic pressure:
P1 - P2 = RT(c1 - c2), ______________________________(13)
The symbol P is often used for osmotic pressure, to distinguish it from hydrostatic pressure.
The pressure must be greater on the side with the greater solute concentration.

Case 4 -- Donnan Equilibrium
 Osmotic equilibrium involving charged molecules needs to be considered when we are dealing with cells and some other biological systems. A cell will contain proteins and other macromolecules that will probably carry a net negative charge at typical intracellular pH. Consider a case where the concentration of macromolecules, P, is 5 mM, with an average charge of 5 (-) per molecule. The proteins therefore have an ionic contribution of 25 milliequivalents (meq), and 25 meq of positive ions will be required in the cell to balance these charges. Let's assume for simplicity that these positive charges are supplied by Na+ ions, at a concentration of 25 mM.
The total concentration of osmotically active solute in the cell is then 5 + 25 mM = 30 mM. (This is sometimes referred to as 30 milliosmoles.) If the cell is surrounded by distilled water, there will be a strong driving force for water to enter the cell to dilute the solutes inside the cell. If there is a strong cell wall around the cell that can withstand this osmotic pressure, the cell will come to equilibrium with the pressure given by equation (13).
For a flexible animal cell, we will try to solve the problem by putting salt in the environment around the cell. Suppose the solution on the outside of the cell is a 15 mM solution of NaCl. The concentration of osmotically active solute will be 15 mM Na+ + 15 mM Cl- = 30 mM (or 30 mosmoles). Under these conditions, the water concentrations inside and outside of the cell will be at equilibrium. However, the ion concentrations inside and outside the cell are not at equilibrium. This will defeat our efforts to solve the problem of osmotic equilibrium if there is any leakage of NaCl across the cell membrane .
If the cell membrane is permeable to Na+ and Cl- ions, there will be a strong inward concentration gradient for Cl- ions, and Cl- ions will move into the cell. There will be an outward concentration gradient for Na+ ions, and Na+ ions will move out of the cell. Both of these ion flows will cause a movement of electrical current out of the cell. This charge separation can only occur until the membrane capacitance is charged up to a voltage that will prevent any further current flow. This will happen almost instantaneously, as a result of small ion flows that are insignificant in comparison to the total concentrations of ions in the system.
We now have a situation with a voltage across the membrane, with the outside + relative to the inside. This voltage difference will slow down the influx of Cl- ions, and will reverse the outward flow of Na+ ions. There will be equal influxes of Na+ and Cl- ions into the cell, without any flow of electric current across the membrane. This process will come to a stop when the concentrations of Na+ and Cl- ions inside the cell have reached values such that both the Na+ ion concentration difference across the membrane and the Cl- concentration difference across the membrane are in equilibrium with the membrane voltage, according to equation (10). In this situation:
[Na+]out/ [Na+]in = [Cl-]in/ [Cl-]out

We also have the equation for charge balance inside the cell:

[Na+]in = [Cl-]in + 5[P-5].

From these equations, the concentrations of [Na+] = 32 mM and [Cl-] = 7 mM inside the cell can be calculated. When these equations are satisfied, the system is said to be in Donnan equilibrium. However, it is not a true equilibrium situation because there is still an osmotic imbalance, with total concentrations inside the cell of 32 + 7 + 5 = 44 milliosmoles, compared to 30 milliosmoles outside. So this does not solve the osmotic problem, although it does reduce it.
A more typical biological situation would be one where the external salt concentration is higher, say 0.15 M, as in typical vertebrate body fluids. In this case, the same procedure leads to the result that [Na+] = 163 mM and [Cl-] = 138 mM inside the cell. The osmotic imbalance is now reduced to 6 milliosmoles, but it still hasn't been eliminated.
However, we can now see that this cell can achieve osmotic equilibrium if it can actively pump Na+ ions out of the cell, so that the internal Na+ concentration is lowered to 160 mM. This will not require much energy, since the desired concentration is very close to the equilibrium concentration. The Cl- ion concentration must fall to 135 mM, to maintain charge balance. The total solute concentration inside the cell will then be 160 + 135 + 5 = 300 mM, which is equal to the 300 mM concentration resulting from 150 mM NaCl outside of the cell.
The rate at which the active transport mechanism is required to pump Na+ ions out of the cell is determined by the rate at which these ions leak back into the cell through the cell membrane. The Na+ ion permeability of most cell membranes is normally low, so this energy requirement is small. It does not depend on the permeability of the membrane to water or to Cl- ions. Both of these can be optimized to meet other cell requirements, without affecting the energy needed to maintain the cell at a constant volume.

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