solutions to problems in cellular biophysics volume … · solutions to problems in cellular...

SOLUTIONS TO PROBLEMSIN

CELLULAR BIOPHYSICSVOLUME 1: TRANSPORT

Thomas Fischer Weiss

Department of Electrical Engineering and Computer ScienceMassachusetts Institute of Technology

Spring 1997

Date of last modification: October 22, 1997

iii

To Aurice B, Max, Elisa, and Eric

v

PREFACE

The textbooks Cellular Biophysics, Volume 1: Transport (Weiss, 1996a) and CellularBiophysics, Volume 2: Electrical Properties (Weiss, 1996b) contain a collection of exer-cises and problems that have been developed over many years. These problems andexercises allow students to test their comprehension of the material and they also ex-tend the material contained in the textbooks. For learning the material, there is nosubstitute for attempting to solve problems — challenging problems. Solving problemscan reveal which aspects of the material are understood and which are not yet graspedand require further study. This solution book contains solutions to all the exercisesand problems in Volume 1; a companion solution book (Weiss, 1997) contains solutionsto all the exercises and problems in Volume 2. The purpose of making the solutionsavailable is to allow students to check their work. Properly used, these solutions canbe helpful for learning the material. By comparing their solutions with those in the so-lution book, students can obtain an objective evaluation of their comprehension of thesubject material. However, the solutions contained here are more extensive than wouldbe expected for a typical student. Hopefully, these more extensive solutions further ex-plicate the material. There is one caveat concerning a possible misuse of this solutionbook. If a problem is assigned in a subject that uses these textbooks, a student mightbe tempted to just reproduce the solution after only a cursory reading of the problem.This certainly saves time. However, it short-circuits the process of actively strugglingwith the problem. This is where learning takes place and the subject material is assim-ilated. My advice to students is: resist the temptation to use the solution texts in acounterproductive manner. Remember as you struggle to solve a problem, “no pain, nogain.”

The practice of the faculty has been to assign problems for homework one week andthen to issue solutions to students the following week when the work was due. Thesesolutions started out as handwritten and progressed over the years to more elaboratetypeset solutions. Therefore, when I started the project of producing these solutionsbooks, I gathered together all previous solutions — multiple solutions were typicallyavailable for the same problem assigned in different years. These solutions were writtenby various faculty who have taught the subject as well as by graduate student teachingassistants. In reviewing these solutions, I found with some chagrin that several solu-tions were incorrect despite the fact that they had been issued many times (by facultyincluding primarily myself), checked by several teaching assistants, and presumablyread by many hundreds of students. I apologize to students who were led astray byerrant solutions, but these same students should feel guilty for not having caught theerrors. The solutions I have found in error, I have tried to fix. However, my experiencewith compiling past solutions has left me a bit pessimistic about eliminating all errorsfrom these solutions. I invite the reader to communicate with me to point out any re-maining errors. I can be reached via email at [email protected]. I will post errors in thetexts and in the problem solution texts on my homepage on the world wide web whosecurrent address is http://umech.mit.edu:80/weiss/home.html. My homepage canbe reached through the MIT home page which links to the Department of ElectricalEngineering and Computer Science homepage.

As with the textbooks, these solution books were typeset in TEX with LATEX macroson a Macintosh computer using Textures. Spelling was checked with the LATEX spell

vi

checker Excalibur. Theoretical calculations were done with Mathematica and MATLAB.Graphic files were imported to Adobe Illustrator for annotation and saved as encapsu-lated postscript files that were included electronically in the text. Mathematical anno-tations were obtained by typesetting the mathematical expressions with Textures andsaving the typeset version as a file that was read by Adobe Illustrator. The subject istaught with the use of software (Weiss et al., 1992) designed to complement other ped-agogic materials we have used. Some of the problems reflect access to this software,but the software is not required to solve any of the problems.

I wish to acknowledge support from a faculty professorship donated to MIT by Gerdand Tom Perkins. My secretaries (Susan Ross and Janice Balzer) were helpful in compil-ing the solutions from past years. Faculty who have taught the material also developedproblems and solutions over the years. In particular, I wish to acknowledge the contri-butions of my colleagues Denny Freeman and Bill Peake. Finally, my immediate family(Aurice, Max, Elisa, Eric, Kelly, Nico, Sarah, Madison, and Phoebe), which has grown asthe writing has progressed, has continued to support me despite my obsession withwriting texts.

Chapter 1

INTRODUCTION TO MEMBRANES

Exercises

Exercise 1.1 Cellular membranes consist of lipid bilayers. Each bilayer contains phos-pholipids which in turn consist of fatty acids, glycerol, and an alcohol. The fatty acidshave a hydrocarbon chain consisting typically of 16 or 18 carbon atoms. Since a carbon-carbon bond has a bond length of about 1.5 A, the length of a fully stretched out chainof 18 carbon atoms is 25.5 A. Therefore, the portion of the lipid bilayer that is due tothe hydrocarbon chains of fatty acids has a maximum length of about 51 A.

Exercise 1.2 Assume that each position in the protein sequence is equally likely to beany one of the 20 amino acids. Therefore, there are N = 20100 distinct sequences of100 amino acids. This number can be expressed as a power of 10, i.e.,

N = 10100 log10 20 ≈ 10130.

Thus, the number of possible distinct protein sequences consisting of 100 amino acidsgreatly exceeds the number of atoms in the universe. Clearly, not all of the possiblesequences of amino acids are found in nature.

Exercise 1.3

a. The membrane is assumed to be 75 A thick. Therefore, there are (75/5.4)× 3.6 =50 amino acids in a membrane-spanning segment.

b. Figure 1.33 (Weiss, 1996a) shows the opacity of the membrane stained with a heavymetal that combines with the hydrophobic portion of the membrane. The distancebetween the peak opacities is 5.7 nm. Therefore, the hydrophobic domain is lessthan 5.7 nm. If the hydrophobic portion is assumed to span about half of thisdistance, then there are (29/5.4) × 3.6 = 19 amino acids in a segment spanningthe hydrophobic portion of the membrane.

Exercise 1.4

a. The structure is clearly seen if only two amino acids are used — isoleucine (I),which is hydrophobic, and arginine (R), which is hydrophilic. The sequence of 25amino acids is displayed in an α helix in Figure 1.1.

1

2 CHAPTER 1. INTRODUCTION TO MEMBRANES

R25

I24

R23

R22

I21

I20

R19

R18

I17

I16

R15

I14

I13

R

12

R

11I

10

I

9

R

8

R

7

I

6

R

5

R

4

I

3

I

2

R

1

Figure 1.1: An amphipathic α he-lix of 25 amino acids with hydropho-bic amino acids on the left and hy-drophilic amino acids on the right (Ex-ercise 1.4). Each number gives theposition of that amino acid in thesequence. The diagram shows theamino acids in the form of a helicalwheel in which the view is down theaxis of the helix.

b. Twenty-five amino acids yield 25/3.6 = 6.9 turns. Since each turn has a pitch of5.4 A, the axial length of the α helix is 37.5 A.

Exercise 1.5 Both RNA and DNA consist of polymers of nucleotides. Each nucleotideconsists of a pentose linked to an organic base and phosphates. In RNA the pentose isribose; in DNA the pentose is 2-deoxyribose. The bases adenine, guanine, and cytosineare found in both RNA and DNA. The fourth base in RNA is uracil and in DNA is thymine.

Exercise 1.6 Phospholipids are amphipathic with a hydrophilic head and a hydrophobictail. Therefore, they self-assemble into bilayers with the hydrophilic heads facing aque-ous media; cytoplasm on one side and the extracellular solution on the other. Thesebilayers are relatively impermeable to solutes that dissolve in water. Hence, membranesmade of phospholipid bilayers effectively isolate intracellular from extracellular com-partments.

Exercise 1.7 All cells contain a cell membrane consisting of a lipid bilayer which is thepermeability barrier that separates cytoplasm from the extracellular environment. Thecell membrane is highly deformable. Plant cells and some bacteria contain an additionalboundary outside the cell membrane, called the cell wall. The cell wall is relatively rigidand gives mechanical rigidity to the cell. The cell wall is highly permeant to most solutesand to water.

Exercise 1.8 The sequence · · ·TCTAATAGC · · · corresponds to DNA since it containsthymine, and the sequence · · ·UCUAAUAGC · · · corresponds to RNA since it containsuracil.

Exercise 1.9

EXERCISES 3

a. Integral membrane proteins are bound tightly to the membrane and are not read-ily dislodged without the use of detergents that disrupt the membrane. Integralmembrane proteins typically have one or more segments that insert into or spanthe membrane.

b. A procaryote is a single cell without a nucleus or cell organelles. The geneticmaterial and protein synthesis machinery reside in the cytoplasm.

c. Cytosol is the liquid portion of the cytoplasm. The cytosol together with the or-ganelles make up the cytoplasm.

d. In a covalent bond between dissimilar atoms, the electronic cloud is usually dis-posed asymmetrical about the two nuclei. The electronegativity of an atom re-flects its ability to attract the electronic cloud. For example, consider a highlyelectronegative atom such as oxygen bound to a less electronegative atom. Thecenter of gravity of the electronic cloud will be closer to the oxygen nucleus thanto the nucleus of the less electronegative atom.

e. A hydrogen bond is a secondary chemical bond that forms between a hydrogenatom and two electronegative atoms each of which can separately bind hydrogen.The bond links the two electronegative atoms with an interposed hydrogen atom.

f. A pentose is a 5-carbon sugar.

g. The peptide bond is the bond that links two amino acids in a peptide or protein.The peptide bond is formed by a condensation reaction of the carboxyl group ofone amino acid with the amino groups of another.

h. A saturated fatty acid has a hydrocarbon tail that includes the maximum possiblenumber of hydrogen atoms. Each carbon atom in the tail is linked to two hydrogenatoms.

i. An amphipathic molecule has one portion that contains polar chemical groups andanother that contains nonpolar chemical groups.

Exercise 1.10

a. This applies to procaryotes, as well as to both eucaryotic plant and animal cells.

b. This applies to none of these cell types.

c. This applies to eucaryotes.

d. This applies to eucaryotes.

e. This applies to procaryotes, as well as to both eucaryotic plant and animal cells.

Exercise 1.11

a. A hydrophobicity plot is a plot of the hydrophobicity of the amino acids’ side chainsas a function of the amino acid position in the protein.

4 CHAPTER 1. INTRODUCTION TO MEMBRANES

b. Regions of sufficient length (ca. 20 amino acids long) that have a high hydropho-bicity have been found to be regions of the protein that reside in the lipid bilayer.These regions are separated by regions that are hydrophilic that reside either inthe cytoplasmic or extracellular regions.

Chapter 2

INTRODUCTION TO TRANSPORT

Exercises

Exercise 2.1 While elimination of waste products of digestion is an important propertyof the digestive system, the absorption of nutrients derived from the ingested food isequally important.

Exercise 2.2 The surface area of the small intestine is 20 × π(1.5/12) = 7.9 ft2. Thevilli and microvilli have been estimated to increase the surface area by a factor of 600.Hence, the surface area available for absorption is about 4700 ft2 which amounts to asurface area equal to 1.7 tennis courts.

Exercise 2.3 Villi are projections of the intestinal wall into its lumen. They have a lengthof about 1 mm. Microvilli are projection from the surface of an enterocyte and have alength of about 1 µm. Both serve to extend the surface area available for digestionand transport in the small intestine, but their dimensions differ by three orders ofmagnitude.

Exercise 2.4 The influx of glucose on the mucosal side of the enterocyte depends uponthe sodium gradient across the membrane, because the protein carrier that transportsthe glucose transports sodium simultaneously. Since the concentration of sodium inthe lumen of the intestine exceeds that in the cytoplasm of the enterocyte, sodium andglucose flow into the enterocyte on the mucosal surface. Since sodium will tend toaccumulate in the enterocyte due to this mechanism, the role of the sodium/potassiumpump is to remove the sodium from the cytoplasm. This pump maintains the differencein sodium concentration between the lumen of the intestine and the cytoplasm of theenterocyte upon which the influx of glucose into the enterocyte depends.

Exercise 2.5

a. Enterocytes are absorptive epithelial cells that line the wall of the intestine. Theyare responsible for the absorption of the end products of digestion.

b. The glycocalyx is a filamentous structure that coats the microvilli and is attachedto the membrane of the enterocyte. The glycocalyx contains enzymes that are im-portant in the final stages of digestion of nutrients just before they are transportedinto enterocytes.

5

6 CHAPTER 2. INTRODUCTION TO TRANSPORT

c. Tight junctions link the membranes of neighboring epithelial cells. At tight junc-tions the outer leaflets of membranes of apposed cells are fused. These junctionsprovide a barrier to diffusion of substances in a paracellular route (between thecells) in an epithelium.

d. The lumen of the intestine is the region enclosed by the epithelium in which theproducts of digestion and the waste products travel in the course of digestion.

e. Amylase is an enzyme that digests carbohydrates and is secreted by both thesalivary glands and the pancreas. Amylase hydrolyzes the α1,4 linkage commonin carbohydrates such as starch.

Exercise 2.6 The pancreas secretes digestive enzymes into the intestines. These in-clude trypsin, chymotrypsin, carboxypeptidase, lipase, and amylase. The pancreas alsosecretes several hormones, including insulin and glucagon, into the circulatory system.Insulin and glucagon are important in the utilization of the end products of carbohy-drate metabolism.

Exercise 2.7 Tight junctions provide a barrier to diffusion of substances in a paracel-lular route (between the cells) in an epithelium.

Exercise 2.8 In the mouth, the potato is chewed and attacked enzymatically by salivaryamylase which breaks chemical bonds in starch. The partially digested potato enters thestomach and then the small intestine where it is further digested by pancreatic amylaseinto glucose, maltose, and polysaccharide fragments. Maltase and sucrase, located in theglycocalyx of enterocytes, break down maltose into two glucose molecules and sucroseinto glucose plus fructose, respectively. Glucose and fructose, are transported intoenterocytes and then enter the circulatory system. Ultimately, glucose yields carbondioxide and water and produces chemical energy in the phosphate bonds of ATP.

Exercise 2.9 Physiological experiments in which intracellular solutions or the potentialacross the membrane (such as might be important in sodium-linked amino acid trans-port) need to be manipulated while the flux of amino acids are measure are probablymore easily accomplished in the large invertebrate neurons. With presently availabletechniques, the transport would have to be studied on a population of bacteria althoughit could be studied on individual invertebrate neurons. It would be more difficult to dis-cern if there were a population of amino acid transporters that differed subtly in theirtransport properties for experiments on bacteria than on the invertebrate neurons. Iden-tifying the protein carrier may well be equally done on both cell types. However, geneticexperiments in which the portions of the protein carrier are mutated to determine ef-fects on transport are probably done more easily on bacteria which are plentiful, havea short life cycle, and are simpler to manipulate genetically.

Exercise 2.10 Glucose is transported from the lumen of the small intestine into thecirculatory system via membrane transport processes that occur on the apical and ba-solateral portions of enterocytes. In addition, the uptake of glucose by all cells in thebody occurs via membrane resident carriers. Sugar transport into some cell types is

EXERCISES 7

modulated by hormones (e.g., insulin and glucagon) which are released into the circu-latory system by cells in the pancreas. The pancreatic cells also contain membranebound glucose carriers. When the blood glucose concentration changes so does the glu-cose concentration in the interstitial fluid, which changes the uptake of glucose into thepancreatic cells. The change in intracellular glucose concentration leads to modulationof hormonal release. Thus, many of the important stages in carbohydrate utilization inthe body involve membrane processes.

8 CHAPTER 2. INTRODUCTION TO TRANSPORT

Chapter 3

DIFFUSION

Exercises

Exercise 3.1 The permeability of a membrane for a solute equals the ratio of the out-ward solute flux through the membrane to the difference between the inside and outsidesolute concentrations under the assumption that diffusion through the membrane is insteady state.

Exercise 3.2 Given two media 1 and 2 that are in communication with each other andthat contain the solute n. The 1:2 partition coefficient kn is defined as the ratio ofconcentrations of n in liquid 1 to that in liquid 2 when these concentrations have cometo equilibrium.

Exercise 3.3 The continuity equation expresses the conservation of a substance duringthe passage of time. When applied to a volume element, the continuity equation requiresthat the amount of substance that enters the volume element during some time intervalmust equal the increase in that substance contained within the volume during that timeinterval. When applied to a point in space, the continuity equation states that the rateof increase of flux with position equals the rate of decrease of concentration with time.Thus, if the flux is increasing with position at some point, then more flux leaves thatpoint than enters it, and the concentration must be decreasing with time at that point.

Exercise 3.4 Equation 3.1 (Weiss, 1996a) is Fick’s first law which is a macroscopic rela-tion and is independent of any microscopic model of diffusion. Equation 3.18 (Weiss,1996a) is Fick’s first law derived from a particular microscopic model of diffusion —the one-dimensional random walk.

Exercise 3.5 The modern definition of permeability is given by

Pn = φncin − con

.

Pn is the ratio of the flux to the concentration difference across the membrane. Theflux is the rate of transport of solute across a unit area of membrane. Thus, Pn isindependent of the dimensions of cells for cells that have membranes made of identicalmaterial. As shown in Section 3.7 (Weiss, 1996a), diffusion through the membrane of

9

10 CHAPTER 3. DIFFUSION

a cell can be represented as diffusion between two compartments. Conservation ofparticles yields

−dcin(t)dt

= APnVc(cin(t)− con(t)

).

Therefore,

P ′n =APnVc .

Since Pn is by definition independent of the dimensions of a cell, P ′n must depend oncell dimensions. Thus, cells with identical membranes (except for dimensions) havedifferent values of P ′n but the same values of Pn. Early measurements of properties ofmembranes based on estimates of P ′n showed great scatter across different cells. Partof this scatter is attributable to the different dimensions of cells and not to intrinsicdifferences in their membranes.

Exercise 3.6 In steady-state diffusion, the flux and concentration of particles are con-stant in time. In equilibrium, the flux and concentration of particles are constant in timeand, in addition, the flux is zero.

Exercise 3.7 The steady-state time constant (τss ) in the membrane is a measure of thetime required for the membrane to reach steady state. In steady state, the solute concen-tration depends linearly on position in the membrane. The equilibrium time constant(τeq) for the two compartments is a measure of the time required for the baths to reachdiffusive equilibrium. At diffusive equilibrium the concentrations of solute in bothbaths are equal.

Exercise 3.8 Solutes that obey the dissolve-diffuse theory first dissolve in the membraneand then diffuse through it.

Exercise 3.9 Collander plots relate the permeability of solutes through cellular mem-branes to the organic-solvent/water partition coefficient. Because of the large range ofboth variables, the data are normally plotted in logarithmic coordinates. Collander plotsdetermine the relation between the permeability of a cellular membrane for a solute andthe ratio of the solubility of that solute in an organic solvent to that in water. Collanderplots have been used to test the dissolve-diffuse theory of diffusion through cellularmembranes.

Exercise 3.10 Figure 3.1 shows the concentration profile at the time that the membranehas just reached steady-state. The concentration in bath 1 is C and that in bath 2 isnearly zero (a small amount of solute will actually enter bath 2 while the membraneis approaching steady state, so the concentration in bath 2 is not exactly zero). Theconcentration in the membrane decreases linearly from 1.5C at x = 0 to near zero atx = d. The value at x = 0 exceeds the value in bath 1 because the partition coefficientis 1.5.

Exercise 3.11 Bath 2 will come to equilibrium with Bath 1 when c2n(∞) = C and the

concentration of bath 2 will change exponentially from c2n(0) = 0 to c2

n(∞) = C , i.e.,

c2n(t) = C(1− e−t/τeq) for t ≥ 0,

where τeq = V/(APn). The results are shown in Figure 3.2.

EXERCISES 11

0 d x

C

1.5 C

Figure 3.1: Concentration versus distance (Exer-cise 3.10).

0

c2n(t)

t

C

τeq

Figure 3.2: Concentration as a function of time(Exercise 3.11).

Exercise 3.12 A crude approximation can be obtained by computing the time (t1/2)taken for 1/2 the particles released at a point to move at least 200 A, which is

t1/2 =x2

1/2

D= (200× 10−8)2

5× 10−6 = 45× 10−6 ≈ 1 µs.

This calculation gives the order of magnitude of the time it takes for the transmitterto diffuse. This crude estimate does not take the geometry of the synaptic cleft intoaccount.

Exercise 3.13 The equilibrium distribution of solute n results from two physical mech-anisms — gravity which tends to pull the solute to the bottom of the cylinder and diffu-sion which tends to disperse the solute in the solution. The mass of jupiter is 1.9×1027

kg and the mass of the earth is 5.98 × 1024 kg. Therefore, the force of gravity on thesurface is 318 times larger on jupiter than on earth. Therefore, we expect intuitively thatthe solute will be located closer to the bottom of the cylinder on jupiter than on earthas indicated in Figure 3.3. A more quantitative analysis, which is required to obtain amore precise result, follows. The flux of particles due to diffusion and gravity-induced

y

00

cEn (y)

y

00

cJn (y)

Figure 3.3: Spatial dependence of solute concentra-tion in a cylinder on earth cEn(y) and on jupiter cJn(y)(Exercise 3.13). The spatial dependence is exponentialboth on earth and on jupiter, but the space constantis compressed on jupiter. The concentration scale isnormalized.


convection in the y direction is

φ(y) = −Ddcn(y)dy︸︷︷︸

diffusion

+ucn(y)f︸︷︷︸convection

The molar force f = −mg where m is the mass of a mole of solute particles and g isthe acceleration of gravity. The Einstein relation relates the diffusion coefficient to themolar mechanical mobility, i.e., D = uRT . Combining all these relations and noting thatat equilibrium φ(y) = 0 yields the differential equation

dcn(y)dy

+ cn(y)λ

= 0,

where λ = RT/(mg). The solution to this equation is

cn(y) = Ke−y/λ,where K is a constant to be determined by the boundary conditions. Suppose the cylin-der on earth and on jupiter contained the same number of moles of solute. Then

AK∫∞

0e−y/λ dy = Nn,

where A is the cross-sectional area of the cylinder and Nn is the total number of molesof solute n. Evaluation of the integral yields K = Nn/(Aλ) so that

cn(y) = NnAλe−y/λ.

Thus, the spatial distribution of concentration is exponential in y with space constantλ. The space constant is inversely related to g. Hence, the space constant is 318 timessmaller on jupiter than on earth and the concentration at the bottom of the cylinderis 318 times larger. From an earth based frame of reference, it appears that all theparticles lie at the bottom of the cylinder on jupiter.

Exercise 3.14 The impulse response cn(x, t) at xb is larger and reaches its maximumvalue more rapidly than at xa. Therefore, xb is closer to the source than is xa. Hence,xa > xb.

Exercise 3.15 Cell a equilibrates more rapidly than cell b. Therefore, the equilibriumtime constant for cell a is smaller than for cell b, i.e., τeqa < τeqb. Since τeq = Ve/(APn),Pan > Pbn .

Exercise 3.16 Consider the concentrations at the left-hand membrane interface. Asshown in Figure 3.52 (Weiss, 1996a),

ca(0)c1a= ka > 1 and

cb(0)c1b= kb < 1,

where ca(0) and cb(0) are the concentrations of solutes a and b in the membrane at theleft interface, and c1

a and c1b are the concentrations of solutes a and b in bath 1. There-

fore, ka > kb. The same conclusion can be reached by examining the concentrations atthe right membrane interface.

PROBLEMS 13

(mm

ol/L

)5 mm

x

x

0

0

10

20

c(x,t

0)

φ(x,t

0)

Figure 3.4: Flux that results from a sinusoidal con-centration (Exercise 3.18).

Exercise 3.17 Since the concentration in compartment a changes less than that in com-partment b, compartment a has the larger volume. More formally, Figure 3.53 (Weiss,1996a) shows that can(0)− can(∞) < cbn(∞)− cbn(0). But can(∞) = cbn(∞) and cbn(0) = 0.Combining these relations yields can(0) < 2can(∞). Let nan be the number of moles of nin compartment a. Then,

nanVa < 2

nanVa +Vb ,

from which it follows that

Va +Vb < 2Va which implies that Vb < Va.

Exercise 3.18 Based on Figure 3.54 (Weiss, 1996a), the concentration can be expressedas

c(x, t0) = (10+ 10 cos(4πx))× 10−3 (mol/cm3),

where x is expressed in cm. From Fick’s first law

φ(x, t0) = −D∂c(x, t0)dx= 4πD × 10−2 sin(4πx) (mol/(cm2·s)).

The concentration and flux are shown in Figure 3.4.

Problems

Problem 3.1 This problem explores the diffusion of a substance in a cellular processwhen that substance is used up in the cytoplasm.

a. The continuity equation is∂φn∂z

= −∂cn∂t− αnA,

and Fick’s first law is

φn = −D∂cn∂z .These two equations can be combined by taking ∂/∂z of Fick’s first law to yield

D∂2cn(z, t)∂z2 = ∂cn

∂t+ αnA.


0.2 0.4 0.6 0.8 1

−0.5

0.5

0

1

cn(z)

Co

z/l

0.5

1

2

3

Figure 3.5: Plot of the concentration ver-sus position along a cellular process (Prob-lem 3.1). The parameter for each curve is(αnl2)/(DACo). The concentration is posi-tive for 0 ≤ (αnl2)/(DACo) ≤ 2.

b. In the steady state ∂cn/∂t = 0. Therefore,

Dd2cn(z)dz2 = αn

A.

Integrating twice yields

cn(z) = αn2DA

z2 + aoz + bo.

The constants ao and bo are evaluated from the boundary conditions. At z = 0this evaluation gives

cn(0) = Co = b0.

At z = l, φn(l) = 0. Therefore, from Fick’s first law

φn(l) = −D(∂cn∂z

)z=l= −αn

Al−Dao = 0.

Therefore,

a0 = −αnlDA.

c. The concentration can be expressed as

cn(z) = αnDAz(z2− l)+ Co for 0 < z < l.

This relation can be written as

cn(z)Co

= αnl2

DACozl

(z2l− l)+ 1 for 0 < z < l,

which is plotted in Figure 3.5. The concentration must be positive at all positionsalong the process. Since the concentration has its smallest value at z = l,

cn(l) = Co − αn2DA

l2 > 0.

The condition that the concentration is positive at its smallest value places anupper limit on the length of the process lmax which is

lmax =√

2DACoαn

.

PROBLEMS 15

This result shows that lmax increases if: the diffusion coefficient increases, thecross-sectional area of the process increases, the concentration at z = 0 increases,and the rate αn decreases. This makes intuitive sense. For example, we mightexpect that as the rate at which the substance n is removed increases, the lengthof the process over which steady state can be maintained decreases.

To illustrate the numerical consequences of this problem, we examine the expressionfor lmax . The quantity ACo/αn has the units of time, since it is the ratio of the concen-tration per unit length to the rate of change of concentration per unit length. ACo/αnrepresents the time it takes to use up all of the metabolite. Suppose this time is 10 sand D = 10−5 cm2/s (Section 3.3). Then,

lmax =√

2× 10−5 cm2/s× 10s = 0.014 cm

which is very short compared to the axons of many neurons, but is comparable to theflagellum of a sperm cell.

Problem 3.2 Diffusion in the presence of both convection and a chemical reaction is de-scribed by Fick’s first law, modified by the presence of convection, and by the continuityequation, modified by the effect of the chemical reaction,

φ = −D ∂c∂x+ vc,

∂φ∂x

= −∂c∂t−αc.

If these equations are combined they yield a modified diffusion equation that incorpo-rates both convection and the effect of the chemical reaction,

∂c∂t= D ∂

2c∂x2 − v

∂c∂x−αc.

a. If there is no chemical reaction to eliminate particles α = 0.

i. At equilibrium, φ = 0 and ∂c/∂t = 0. Therefore, Fick’s first law becomes

−D dcdx+ vc = 0.

To determine the general form of the solution, try the solution c = Aepxwhich yields

−DApepx + vAepx = 0,

which can be divided by Aepx to yield the characteristic equation

−Dp + v = 0.

The natural frequency is the root of the characteristic polynomial and is

p = v/D.Therefore, the general solution at equilibrium has the form

c(x) = Aevx/D.In the presence of convection, the equilibrium distribution of concentrationis exponential in space.


ii. In the steady state, the flux need not be zero. However, ∂c/∂t = 0. Understeady-state conditions, the diffusion equation becomes

0 = D d2cdx2 − v

dcdx.

To determine the form of the solution, try a solution of the form c = Aepx ,which yields

DAp2epx − vApepx = 0.

Since Aepx 6= 0, the equation can be divided by this factor to yield the char-acteristic equation

Dp2 − vp = 0.

The characteristic polynomial has two roots p = 0 and p = v/D. Therefore,the general form of the solution is

c(x) = A1 +A2evx/D.

The constants A1 and A2 are found by matching the boundary conditionsc(0) = 0 and c(1) = 1 mol/cm3. Application of the boundary condition atx = 0 yields c(0) = 0 = A1 + A2 which implies that A1 = A2. Therefore, thesolution has the form

c(x) = A(1− evx/D

).

Application of the boundary condition at x = 1 yields

c(1) = 1 = A(1− ev/D

),

so that

A = 11− ev/D .

Therefore, the solution is

c(x) = 1− evx/D1− ev/D .

The solution is shown in Figure 3.6 for a diffusion coefficient ofD = 10−5 cm2/sand for convection velocities of±1 cm/s. Note that c(0) = 0 and c(1) = 1 andthe change in c(x) occurs within a narrow range of x because the exponent isvx/D = ±105x. Mathematically, for v > 0 c(x) changes rapidly near x = 1,whereas for v < 0 c(x) changes rapidly near x = 0. Physically, if the velocityis positive then particles are swept from the left to the right, whereas if thevelocity is negative then the particles are swept from the right to the left.

b. Now assume that a chemical reaction removes particles at a rate α ≠ 0. At equi-librium the flux is zero and the diffusive variables are time independent. Note thecontinuity relation for these conditions reduces to

0 = −αc.Therefore, the only solution is c(x) = 0. At equilibrium all of the particles areconsumed.

PROBLEMS 17

0 0.00005

0.5

1

0.99995 1

0.5

1

v

v

c(x)

c(x)

x

x

Figure 3.6: Plots of concentration versusdistance for Problem 3.2.

Problem 3.3 Substitute the solution to the modified diffusion equation g(x, t) into themodified diffusion equation and differentiate by parts to obtain

∂f(x, t)∂t

e−αt −αe−αtf (x, t) = D∂2f(x, t)∂x2 e−αt −αe−αtf (x, t).

Cancellation of common terms on the two sides of the equation and division of bothsides of the equation by e−αt yields

∂f(x, t)∂t

= D∂2f(x, t)∂x2

which shows that since g(x, t) satisfies the modified diffusion equation, f(x, t) satisfiesthe diffusion equation. The method developed in this problem yields a transformationthrough which the modified diffusion equation can be transformed into the diffusionequation. Thus, a knowledge of the solution to one of these equations gives the solutionto the other.

Problem 3.4 The chain rule of partial differentiation is used to relate the partial deriva-tives of g and f . Since there are now 3 variables, a subscript is used to indicate whichvariables are held constant. The partial derivatives with respect to time are evaluatedfirst. (

∂f(z, t)∂t

)z=

(∂g(x, t)∂t

)x+(∂g(x, t)∂x

)t

(∂x∂z

)z,(

∂f(z, t)∂t

)z=

(∂g(x, t)∂t

)x+ v

(∂g(x, t)∂x

)t.

The first partial derivatives with respect to space is evaluated as follows.(∂f(z, t)∂z

)t=

(∂g(x, t)∂x

)t

(∂x∂z

)t,(

∂f(z, t)∂z

)t=

(∂g(x, t)∂x

)t.


−2 −1 0 1 2 3 4 5

100

200

300

400

500

x (mm)

c(x,t

)(m

ol/m

3)

1.3

Figure 3.7: Estimation of peak concen-tration, mean concentration and spatialwidth of the concentration (Problem 3.5).

Similarly, (∂2f(z, t)∂z2

)t=(∂2g(x, t)∂x2

)t.

Now these relations are substituted into the diffusion equation.

∂f(z, t)∂t

= D∂2f(z, t)∂z2 ,

∂g(x, t)∂t

+ v ∂g(x, t)∂x

= D∂2g(x, t)∂x2 .

Therefore,∂g(x, t)∂t

= D∂2g(x, t)∂x2 − v ∂g(x, t)

∂x.

The method developed in this problem yields a transformation through which the mod-ified diffusion equation can be transformed into the diffusion equation. Thus, a knowl-edge of the solution to one of these equations gives the solution to the other.

Problem 3.5

a. Diffusion from a point source in the absence of convection has the solution

c(x, t) = 1√4πDt

e−x2/4Dt, for t > 0.

As shown in Problem 3.4, the solution with convection at velocity v is

c(x, t) = 1√4πDt

e−(x−vt)2/4Dt, for t > 0,

where v = uf , u is the molar mechanical mobility, and f is the molar force.

b. The diffusion coefficient and the mobility can be estimated from the measure-ments shown in Figure 3.7. The peak concentration is about 450 mol/m3 andoccurs at x = 1.6 mm. The half-width of the spatial distribution measured wherethe concentration is down to 1/e = 0.3679 of its peak value is about 1.3 mm.Note that the concentration is down 1/e when (x − vt)2/4Dt = 1. Therefore,(1.3× 10−3)2 = 4D × 200 which implies that D ≈ 2.1× 10−9 m2/s.

PROBLEMS 19

c. The Einstein relation is D = uRT . Therefore, the mobility is

u = 2.1× 10−9 m2/s(8.314 J/(mol· K))(300K)

= 8.47× 10−13 (m/s)/(N/mol).

d. To obtain the force on a mole of particles, use the relation v = uf . The con-vection velocity is determined from the displacement of the peak value of theconcentration, i.e., v · 200 = 1.6 × 10−3 so that v = 0.8 × 10−5 m/s. Therefore,f = v/u = (0.8× 10−5)/(8.47× 10−13) = 9.45× 106 N/mol.

Problem 3.6 The flux through both membranes is the same and steady state applies toboth membranes. Since each membrane obeys Fick’s law for membranes, φn = P1(c1

n−con) = P2(con − c2

n) from which

φnP1+ con = c1

n andφnP2− con = −c2

n.

Summing these two equations yields

φn(

1P1+ 1P2

)= c1

n − c2n,

which can be written as

φn = P1P2

P1 + P2(c1n − c2

n).

Therefore,

P = P1P2

P1 + P2,

which demonstrates that the permeability of the two membranes in series is smallerthan the smaller of the two permeabilities P1 and P2. By taking the reciprocal of thisexpression,

1P= 1P1+ 1P2,

which shows that the reciprocal of the permeabilities add for two membranes in series(two membranes through which the same flux flows).

Problem 3.7 Because steady state applies in the membrane, the flux through patch 1is φ1

n = P1(c1n − c2

n), and that through patch 2 is φ2n = P2(c1

n − c2n). The total flux is

the total quantity of solute flowing through both patches per unit area per unit time.Therefore,

φ = P1A1(c1n − c2

n)+ P2A2(c1n − c2

n)A1 +A2

= P1A1 + P2A2

A1 +A2(c1n − c2

n).

Thus,

P = P1

(A1

A1 +A2

)+ P2

(A2

A1 +A2

).

The permeability of the membrane is a weighted sum of the permeabilities of the twotypes of membrane. Each weighting factor equals the fraction of the area occupied bythat membrane type.


Problem 3.8

a. Let us call the concentration at the membrane surface on side 1, Cs1, and on side2, Cs2. Then in the steady state, the flux can be expressed as

φ = Pm(C1 − C2) = P(Cs1 − Cs2),and the problem is to determine the relation between P and Pm which can bedetermined from Cs1 and Cs2. Steady-state diffusion is assumed to occur in eachunstirred layer. Therefore, in each of these regions, the flux and concentration areindependent of time and satisfy Fick’s first law of diffusion, i.e.,

φ = −Ddc(x)dx

.

Integrating this relation yields∫ cbcadc = −φ

D

∫ badx.

Hence,

cb − ca = −φD(b − a).Application of this result to the two unstirred layers yields

Cs1 − C1 = − φD1d1,

C2 − Cs2 = − φD2d2.

Addition of these equations yields

C1 − C2 = (Cs1 − Cs2)+φ(d1

D1+ d2

D2

).

Division of this equation by φ and identification of terms yields

1Pm= 1P+ d1D2 + d2D1

D1D2= 1P+ 1Pl,

where Pl is the equivalent permeability of the unstirred layer

Pl = D1D2

d1D2 + d2D1.

The measured permeability can also be expressed as

Pm = PPlP + Pl .

Define the reciprocal of the permeability as the diffusive resistance R. Then themeasured diffusive resistance is the sum of the diffusive resistance of the mem-brane and the diffusive resistance of the unstirred layer Rm = R+Rl.

PROBLEMS 21

b. Note that when the permeability of the membrane is large compared to the per-meability of the unstirred layer (P Pl), the measured permeability approachesthe permeability of the unstirred layer (Pm ≈ Pl) and not the permeability of themembrane. If the permeability of the membrane is much lower than that of the un-stirred layer, then the measured permeability approaches that of the membrane.Therefore, measurements of the permeability of highly permeant solutes are af-fected more by unstirred layers than are measurements of the permeability ofpoorly permeant solutes.

Problem 3.9 Figure 3.36 (Weiss, 1996a) shows a plot of extracellular propanol concen-tration which according to the two-compartment diffusion theory should have the form

ln( con(t)− con(∞)con(0)− con(∞)

)= −t/τeq,

where τeq = Ve/(APp). Ve is the equivalent volume of the two compartments whichapproximately equals the volume of the erythrocyte since the extracellular volume ismuch greater than that of an erythrocyte. A is the surface area of the erythrocyte andPp is the permeability of the membrane to propanol. Because the initial extracellularconcentration is zero, the expression simplifies to

ln(

1− con(t)con(∞)

)= −t/τeq.

Thus, the two-compartment diffusion model predicts that the concentration plotted inthese normalized, semi-logarithmic coordinates should be a straight line as a functionof time. The measurements are consistent with this theory. Therefore, τeq must bedetermined from the measurements, and the permeability computed from τeq as wellas the known surface area and volume of erythrocytes. The measurements indicate thatin 15 ms, the normalized concentration has value 0.1. Therefore, ln 0.1 = −15/τeq,where τeq is in ms. Evaluation of this expression yields τeq = 6.5 ms. The permeabilityis Pp = Ve/(Aτeq) which can be evaluated from

Pp = 104× (10−4)3

137× (10−4)2 × 6.5× 10−3 = 1.2× 10−2 cm/s.

Problem 3.10 This problem involves one-dimensional diffusion from a point source.Hence, the concentration is a Gaussian function of space and time,

c(x, t) = no√4πDt

e−x2/4Dt for t > 0.

a. To find the maximum value of c(x, t) as a function of time for any fixed point inspace, xp, set the partial derivative of c(xp, t) with respect to t to zero.

∂c(xp, t)∂t

= no√4πD

e−x2p/4Dt

(− 1

2t3/2+ (−x

2p/4D)(−1/t2)t1/2

)= 0.


Therefore,1

2t3/2m= x2

p

4Dt5/2m,

which gives

tm =x2p

2D.

b. For x = 1 cm, D = 0.5 × 10−5 cm2/s, tm = 105 s ≈ 1 day. Therefore, if you wantyour coffee to be sweet this morning, stir it!

Problem 3.11

a. For steady-state conditions in the membrane,

φX = PX(ciX − coX).From conservation of X

d(ciXV )dt

= −AφX,and since V is constant,

dciXdt

= − AV φX.Therefore

dciXdt

= − AV PX(ciX − coX),

which can be written asdciXdt

+ APXV ciX =APXV coX.

This is a first-order differential equation with constant coefficients. Hence, thesolution is exponential of the form

ciX(t) = ciX(∞)+ (ciX(o)− ciX(∞))e−t/τeq .The parameters of the exponential equation are obtained from the initial condi-tions and the differential equation and are

ciX(0) = 0, ciX(∞) = coX, τeq =VAPX

.

Therefore, the solution is

ciX(t) = coX(1− e−t/τeq

).

The measurement shows that the time constant for equilibration is 100 s. There-fore,

τeq = VAPX

= 100 =43π(30× 10−4)3

4π(30× 10−4)2PX.

Solving for PX yields

PX = 30× 10−4

3× 10−2 = 10−5 cm/sec.

PROBLEMS 23

b. For steady-state diffusion in the membrane, the permeability is

PX = DXkXd ,

which can be solved for DX to yield

DX = PXdkX = (10−5)10−6

0.2= 5× 10−11 cm2/s.

c. The time constant for the membrane to reach steady state τss is

τss = d2

π2D= (10−6)2

π2 × 5× 10−11 ≈ 2 ms.

d. Since τss τeq, the membrane will be equilibrated instantaneously relative to thetime required for ciX to change appreciably. Therefore, the steady-state assump-tion is justified.

Problem 3.12

a. Since c1(x, t) and c2(x, t) are solutions,

D∂2c1(x, t)∂x2 = ∂c1(x, t)

∂tand D

∂2c2(x, t)∂x2 = ∂c2(x, t)

∂t.

The equations can be scaled by a and b to obtain

D∂2ac1(x, t)

∂x2 = ∂ac1(x, t)∂t

and D∂2bc2(x, t)

∂x2 = ∂bc2(x, t)∂t

.

Addition of these two equations yields

D∂2(ac1(x, t)+ bc2(x, t))

∂x2 = ∂(ac1(x, t)+ bc2(x, t))∂t

,

which shows that ac1(x, t)+ bc2(x, t) is a solution to the diffusion equation.

b. To determine if c(x, t) = C0+γ(t) cos(2πx/λ) is a solution to the diffusion equa-tion, substitute this expression into the diffusion equation which requires evalu-ating the partial derivatives of this function.

∂c(x, t)∂t

= dγ(t)dt

cos(2πx/λ)

∂2c(x, t)∂x2 = −γ(t)

(2πλ

)2

cos(2πx/λ).

Therefore,dγ(t)dt

cos(2πx/λ) = −Dγ(t)(

2πλ

)2

cos(2πx/λ).

This equation can be simplified to give

dγ(t)dt

+D(

2πλ

)2

γ(t) = 0,


whose solution isγ(t) = C1e−t/τ ,

where

τ = 1D

(λ

2π

)2

.

Therefore, the total solution has the form

c(x, t) = C0 + C1e−t/τ cos(2πx/λ).

Substitution of t = 0 indicates that the initial boundary condition is

c(x,0) = C0 + C1 cos(2πx/λ).

Thus, the total solution satisfies the diffusion equation and matches the initialconcentration.

The time constant τ is a measure of the time it takes for the initial sinusoidalspatial distribution of concentration to reach diffusive equilibrium. This timeconstant varies inversely with the diffusion coefficient D. That is, for solute par-ticles that diffuse more rapidly, i.e., have a larger diffusion coefficient, the initialspatial concentration will equilibrate more rapidly. The time constant also variesdirectly as the square of the wavelength of the concentration λ. Thus, the timeconstant increases as the wavelength increases. This makes intuitive sense sincewe expect that rapid variations of concentration in space, which have small wave-lengths, will be smoothed out more rapidly than slow variations in concentration.Thus, high-frequency periodic initial concentrations decay more rapidly than dolow-frequency initial concentrations.

c. In part a it was shown that the diffusion equation is a linear partial differentialequation and the solution for an initial sinusoidal spatial distribution of concen-tration was found in part b. These results can be combined with the use of theFourier series to determine the concentration that results from an arbitrary, peri-odic initial spatial distribution.

A periodic function of x, such as c(x,0), can be expanded in a Fourier series ofthe form

c(x,0) = c0 +∞∑n=1

cn cos(2πnx/λ)+∞∑n=1

dn sin(2πnx/λ).

The initial rectangular wave of concentration is an even function of x. Hence, thecoefficients of the sine terms in the Fourier series must be zero. Furthermore, theconstant component is C0 so that the Fourier series has the form

c(x,0) = C0 +∞∑n=1

cn cos(2πnx/λ).

However cn needs to be determined. These coefficients can be found by multiply-ing both sides by cos(2πnx/λ) and integrating the product over one wavelength.

PROBLEMS 25

Therefore,

1λ

∫ λ/2−λ/2

c(x,0) cos(2πnx/λ)dx = 1λ

∫ λ/2−λ/2

C0 cos(2πnx/λ)dx

+∞∑m=1

cm1λ

∫ λ/2−λ/2

cos(2πmx/λ) cos(2πnx/λ)dx.

The first term on the right-hand side of the equation is the integral over an integernumber of wavelengths of a cosine wave and is zero. The second term can bewritten as two cosinusoids, one at the sum of the spatial frequencies and theother at the difference of the spatial frequencies. The second term is also zeroexcept when m = n. Under this condition, the integrand is cos2(2πmx/λ) =(1/2)(1+ cos(4πmx/λ). Therefore, the second integral equals 1/2 and

cn = 2λ

∫ λ/2−λ/2

c(x,0) cos(2πnx/λ)dx.

The integral can be evaluated over any period of the cosine wave. Because of thesymmetry of the rectangular wave,

cn = 4λ

∫ λ/4−λ/4

C1 cos(2πnx/λ)dx.

Therefore,

cn =(

4C1

λ

)(λ

2πn

)sin(2πnx/λ)

∣∣∣∣λ/4−λ/4 = 2C1

(sin(nπ/2)nπ/2

).

Note that sin(nπ/2) = 0 if n is even. Hence, only the odd terms are non-zero.

The concentration as a function of time for an initial sinusoidal concentrationprofile was found in part b. Now it has been shown that the initial concentration,a rectangular wave, can be expressed as a sum of sinusoidal profiles. Since, thediffusion equation is linear the concentration for each sinusoidal component ofthe initial rectangular wave can be determined and the components then summed.The result is

c(x, t) = C0 +∞∑n=1

2C1

(sin(nπ/2)nπ/2

)e−t/τn cos(2πnx/λ).

where

τn = 1D

(λ

2πn

)2

.

The dependence of the fundamental frequency component (n = 1) on x and t isshown in Figure 3.8. The graph shows a plot of(

4π

)e−t/τ1 cos(2πx/λ)

as a function of the normalized space and time variables, x/λ and (D/λ2)t. Theinitial sinusoidal spatial distribution of concentration remains sinusoidal but its


0

0.05

0.1

−1

0

1

−0.5

0

0.5x/λ

(D/λ2)t

n = 1

Figure 3.8: A plot of the fundamentalcomponent of the solution for an initialrectangular wave of concentration (Prob-lem 3.12).

0

0.05

0.1

−0.30

0.3

−0.5

0

0.5x/λ

(D/λ2)t

n = 3

Figure 3.9: A plot of the third harmoniccomponent of the solution for an initialrectangular wave of concentration (Prob-lem 3.12).

PROBLEMS 27

0

0.05

0.1

−1

0

1

−0.5

0

0.5x/λ

(D/λ2)t

Figure 3.10: A plot of the partial sum ofFourier series components of the solutionfor an initial rectangular wave of concen-tration (Problem 3.12).

amplitude decays as time elapses. Figure 3.9 shows a similar plot for the thirdharmonic of the solution (n = 3). The graph shows a plot of

−(

43π

)e−t/τ3 cos(2π3x/λ)

as a function of the normalized space and time variables, x/λ and (D/λ2)t. Notethat the third harmonic has an amplitude of 1/3 of the fundamental at t = 0 anddecays more rapidly than the fundamental. Figure 3.10 shows a plot of the partialsum of the Fourier series. The graph shows a plot of

15∑n=1

2(

sin(nπ/2)nπ/2

)e−t/τn cos(2πnx/λ)

as a function of the normalized space and time variables, x/λ and (D/λ2)t. Thisis the first 15 terms in the solution for a square wave. At t = 0, the 15-termapproximation to the square wave shows the characteristic Gibbs’ phenomenon.As t increases the high-frequency components decay more rapidly than do thelow-frequency components.

Problem 3.13

a. Consider the cell as one compartment and a vial as the other compartment. Thenconservation of X implies

− 1AdniX(t)dt

= PX(ciX(t)− coX

).

Because the volume of the cell is much smaller than the volume of the vial, it isreasonable to assume that the vial concentration is negligible compared to the cellconcentration, i.e., coX ≈ 0, so that

dniX(t)dt

+ PXAV niX(t) = 0,


where V and A are the volume and surface area of the cell. The solution to thisdifferential equation with initial value of niX(0) = NX is

niX(t) = NXe−t/τ ,

where τ = V/(PXA).b. In the interval ((k− 1)T , kT) the number of moles leaving the cell must equal the

number of moles in the kth vial.

nX(k) = NX(e−(k−1)T/τ − e−kT/τ

)= NX

(eT/τ − 1

)e−kT/τ.

c. To estimate the parameters from the data, take the logarithm of nX(k) to obtain

lnnX(k) = ln(NX(eT/τ − 1)

)− kTτ.

The data indicate that the slope of the line of lnnX(k) plotted versus k is −1/2 sothat T/τ = 1/2. Since T = 10 minutes, τ = 20 minutes. But

PX = VAτ

= (4/3)πr3

4πr2τ= r

3τ= 72× 10−4

3 · 20 · 60= 2× 10−6 cm/s.

The intercept of the line is −23.5 so that

ln(NX(eT/τ − 1)

)= −23.5

which implies that

NX = e−23.5

e0.5 − 1= 6.2× 10−11

0.65= 96 pmol.

Problem 3.14

a. Assume that steady state occurs in the membrane and that diffusion between theinside and outside of the cell can be modeled as two-compartment diffusion. Thenthe concentration has the form

ci(t) = ci(∞)+ (ci(0)− ci(∞))e−t/τeq for t ≥ 0.

The initial concentration of urea in the cell ci(0) = 0. Since Vo Vi, the finalconcentration in the cell will equal that in the bath, ci(∞) = C . Therefore, thesolution will have the form

ci(t) = C(1− e−t/τeq).

The time it takes for the concentration to reach 63% of its final value is the timeconstant

τeq = VeAP

PROBLEMS 29

where

Ve = ViVoVi +Vo ,

and since since Vo ViVe ≈ Vi.

Therefore,

τeq =43πa

3

4πa2P= a

3P.

For a spherical cell with a radius of a = 10 µm

τeq = 10−3 cm(3)(3× 10−4 cm/s)

≈ 1.1 s.

b. The time constant for equilibration of a spherical volume in a solution can beobtained from Figure 3.18 (Weiss, 1996a). For the concentration to reach 63% ofits final value, the unequilibrated fraction is e(t) = 0.37. This value of e(t) for asphere corresponds to

Dτda2 ≈ 0.05.

Therefore,

τd = 0.05a2

D= 0.05(10−3 cm)2

1.4× 10−5 cm2/s≈ 3.6 ms.

c. The ratio of time constants is

τdτeq

≈ 0.05a2/Da/3P

= 0.15PDa.

Therefore, τd/τeq ∝ a.

d. The following conclusions result

• For a 6.7(D/P), τd τeq and diffusion across the plasma membranelimits the rate of urea transport into the cell.

• For a 6.7(D/P), τd τeq and diffusion through the cytoplasm limits therate of urea transport into the cell.

• For a = 10 µm and the parameters given in the problem, diffusion across theplasma membrane is rate limiting.

Problem 3.15

a. If the concentration of sugar in the membrane has reached steady state at timet = 0, the flux of sugar through the membrane must be a constant and the con-centration of sugar must be a linear function of x. Since the partition coefficientk = 1, c(0,0) = c1(0), and c(W,0) = c2(0). Therefore

c(x,0) = c1(0)+ c2(0)− c1(0)W

x for 0 < x < W.


From this equation and Fick’s first law, the flux is

φ(x,0) = −D ∂c(x, t)∂x

∣∣∣∣t=0= −Dc2(0)− c1(0)

W

= 10−5 cm2/s10−4 cm

× 1mol

L× L

103 cm3 = 10−4 molcm2 · s

for 0 < x < W.

b. Because no sugar can enter or exit the two-compartment system, the total amountof sugar in the two compartments must remain constant with time. In particular,the amount at time t = 0, which is equal to c1(0)AL1 + c2(0)AL2 must equal theamount at t → ∞, which is c1(∞)AL1 + c2(∞)AL2. As t → ∞, the concentrationsin the two sides will equilibrate, and c1(∞) will equal c2(∞). Therefore,

c1(∞)AL1 + c1(∞)AL2 = c1(0)AL1 + c2(0)AL2.

Solving for c1(∞) yields

c1(∞) = c1(0)L1 + c2(0)L2

L1 + L2= 1 mol/L× 50 cm+ 0 mol/L× 10 cm

60 cm= 5

6mol/L.

c. Doubling Dsugar should half τeq. The idea is that equilibration occurs becausedifferences between c1(t) and c2(t) cause a flux of sugar through the membrane.By Fick’s first law, the flux is proportional to Dsugar , so doubling Dsugar willdouble the flux. Doubling the flux, doubles the time rate of change of c1(t) andc2(t), and thus halves the time to reach any criterion concentration (e.g., τeq, thetime for c1(t) to come to within a factor of e of its final value).

Problem 3.16

a. The steady-state time constant is

τss = d2

2D.

Therefore, the steady-state time constant for the thin membrane is

τss = (10−4)2

2 · 10−5 = 0.0005 s,

and for the thick membrane is

τss = 12 · 10−5 = 50,000 s.

b. The equilibrium time constant is

τeq = VeAP .

Since the two volumes are identical, Ve = V/2 and since P = Dk/d

τeq = Vd2ADk

.

PROBLEMS 31

Therefore, the equilibrium time constant for the thin membrane is

τeq = 1 · 10−4

2 · 1 · 10−5 · 2= 2.5 s.

Therefore, the equilibrium time constant for the thick membrane is

τeq = 1 · 12 · 1 · 10−5 · 2

= 25,000 s.

c. Note that for the thin membrane τss τeq, i.e., the time it takes for the membraneconcentration profile to reach steady state is much less than the time it takes thetwo volumes to equilibrate. This is the just what is required for Fick’s first law to bevalid at each instant in time in the membrane. For the thick membrane, τss ≈ τeqand the membrane concentration profile is not in steady state. Therefore, Fick’slaw for membranes does not apply at each instant in time.

Problem 3.17 In parts a through f and in part j,φ1 = φ2 = 0 implies that in steady statethe flux through the membrane is zero. Therefore, the concentration in the membranemust be constant for all these parts. In parts g and h, φ1 > φ2 implies that no steadystate is possible since there is a net flux into the system consisting of V1, membrane,and V2. In part i, φ1 = φ2 is the flux through the membrane in steady state. Hence, theconcentration profile in the membrane is linear.

a. D = 0 implies that P = 0 which implies that the flux through the membrane iszero which implies that the concentration is constant. The answer is 9.

b. k = 0 implies that P = 0 which implies that the flux through the membrane is zerowhich implies that the concentration is constant. The answer is 9.

c. D > 0, k > 1 implies there will be a flux of particles through the membrane until V1

and V2 equilibrate. The final value of the concentration is (20V1+10V2)/(V1+V2)which is between 20 and 10 mM/l. The answer is 4 and 5 (if V1 V2).

d. This situation is the same as for part c except that k < 1 which implies that theanswer is 6 or 7. If k = 0, the answer is 9.

e. This is similar to the case considered in part c except that V1 V2. The answeris 5.

f. This is similar to the case considered in part c except that V1 V2 and k = 1. Theanswer is 2.

g. From the discussion given above, the answer is NONE.

h. From the discussion given above, the answer is NONE.

i. With constant flux through the membrane, the concentration in the membrane insteady state must be linear with position. Since k > 1, the answer is 12.

j. There is no flux through the membrane, so there is no change in concentration ofV1 and V2. If k < 1, the answer is 7. If k = 0, the answer is 2.


0

50

100

150

0 0.05 0.1

c(x,0)

c(x,0

) (m

mol

/cm

3 )

MembraneCompartment 1 Compartment 2

x (cm)

c1(0)

c2(0)

c(x,∞) c2(∞)c1(∞)

Figure 3.11: The final (dashed line) and the initial concentration (solid line) of solute in thetwo compartments and the membrane are shown (Problem 3.18). The figure illustrates theconcentration in the membrane and in only a portion of each compartment. The concentrationsin the compartments are uniform in space.

Problem 3.18 An important caveat in this problem is that, where it is so stated, theparameters are to be estimated from the measurements presented in the Figures.

a. Since the membrane:solution partition coefficient is 1, at equilibrium the concen-tration in the two compartments and in the membrane will be the same. Thus,the total quantity of solute at t = 0 can be found and then distributed uni-formly over the total volume of the system. The initial quantity of solute is100 mmol/cm3 · 10 · A. This is distributed over the total volume of the systemwhich is (10+ 1+ 0.1) ·A cm3. Therefore, the final concentration is

1000A11.1A

= 90.09 mmol/cm3,

which is shown in Figure 3.11.

b. This section applies for t = 100 seconds.

i. The concentration profile in the membrane is not a linear function of positionnor is the flux constant. Hence, diffusion in the membrane is not in steadystate.

ii. Because the diffusion regime is not in steady state in the membrane, it is notvalid to use results that depend upon steady-state conditions, such as Fick’slaw for membranes. However, the diffusion coefficient for the solute in themembrane can be obtained from more fundamental considerations by usingFick’s first law evaluated at t = 100

φ(x,100) = −D∂c(x,100)∂x

.

PROBLEMS 33

0

50

100

150

0

5

10

15

0 0.05 0.1

c(x,100)

φ(x,100)

c(x,1

00)

(mm

ol/c

m3 )

φ(x

,100) (µm

ol/(cm2·s))

MembraneCompartment 1 Compartment 2

x (cm)

c1(100)

c2(100)

Figure 3.12: The concentration of solute in the two compartments and the membrane (solid line)are shown at t = 100 (Problem 3.18). The figure illustrates the concentration in the membraneand in a portion of each compartment only. The concentrations in the compartments areuniform in space. The flux of solute is shown (dashed line) in the membrane only. A linetangent to c(x,100) at x = 0 is also shown (dotted).

As indicated in Figure 3.12, a convenient place to measure the slope of c(x,100)is at x = 0+, i.e., just inside the membrane. From the figure,(

∂c(x,100)∂x

)x=0+

≈ −100× 10−3

0.04= −2.5 mol/cm4.

At x = 0+, φ(0+,100) ≈ 12.8× 10−6 mol/(cm2 · s). Therefore,

D ≈ −12.8× 10−6

−2.5= 5.1× 10−6 cm2/s.

c. This section applies for t = 1000 seconds.

i. The concentration profile in the membrane is linear. Hence, the concentrationin the membrane is in steady state.

ii. Note that c2(t) shows a delay and then a portion that has a linear increase inconcentration over a time duration when c(0.04, t) has saturated. Thus, dur-ing this time interval, the concentration in the membrane has reached steadystate and, therefore, the thin-membrane approximation is a good approxima-tion. Hence, during this time interval, c2(t) rises exponentially and has theform C(1 − e−t/τeq). But the time is clearly much less than the equilibriumtime, i.e., t τeq. Therefore, C(1 − e−t/τeq) ≈ C(1 − (1 − t/τeq)) = Ct/τeqwhere C is the final value or equilibrium value of the concentration whichequals 90.09 mmol/cm3. Thus, slope of the concentration must be foundfrom Figure 3.13 during a time interval when steady state has been reached


0

50

100

0

1

2

3

4

5

0 500 1000t (s)

c(0.

04,t)

(mm

ol/c

m3 ) c

2(t) (mm

ol/cm3)

c(0.04,t)

c2(t)

Figure 3.13: The dependence of concentration on time is shown for the time interval (0,1000)for a location in the membrane, c(0.04, t) and in compartment 2, c2(t) (Problem 3.18). Theslope of the line tangent to c2(t) is shown with a dotted line. A line tangent to c(0.04, t)beginning at t = 500 s is also shown.

in the membrane and after the initial transient, i.e., t > 500 ms. The slopeis approximately 3.1× 10−3/700 = 4.4× 10−6 mol/cm3 · s. Therefore, τeq =90.09× 10−3/(4.4× 10−6) = 2.1× 104 s.

iii. The time course for establishing steady state in the membrane consists of asum of exponentials the slowest of which is the steady-state time constantτss . In fact, c(0.04, t) does have a fast transition at the onset and then aslower tail in the response. It is the time constant of the slow tail of theresponse that needs to be estimated. A line starting at t = 500 and tangentto c(0.04, t) crosses the maximum value about 240 s later. Hence, τss ≈ 240s.

iv. Method #1. At t = 1000, the concentration in the membrane is approximatelyat steady state and ∂c(x,1000)/∂x ≈ −100/0.1 = −1000 mmol/cm4. Thus,φ(x,1000) = −D∂c(x,1000)/∂x ≈ 1000D. But by continuity, the flux intocompartment 2 must equal the rate of increase of solute in compartment 2.Therefore, φ(x,1000) = 1000D = (dc2(t)/dt)t=1000 · 1. The derivative canbe evaluated directly from the lower figure as follows

(dc2(t)/dt)t=1000 ≈ 3.1/0.7× 10−3 mmol/cm3 · s.

Combining these relations yields D ≈ 4.5× 10−6 cm2/s.Method #2. A second method involves estimatingD from the equilibrium timeconstant. The equilibrium time constant can be estimated fairly accuratelybecause the thin-membrane approximation holds for times much larger thanthe steady-state time constant. Therefore, τeq = Ve/(PA) for the equilibriumtime constant for two-component diffusion through a thin membrane. Inthis equation, Ve is the equivalent volume of the two compartments, A is thesurface area of the membrane, P = D/d is the permeability of the membrane

PROBLEMS 35

0

50

100

0 50000 100000

Con

centr

atio

n (

mm

ol/c

m3 )

c(0.04,t)

c2(t)

t (s)

Figure 3.14: The dependence of concentration on time is shown for the time interval (0,1000)for a location in the membrane, c(0.04, t) and in compartment 2, c2(t) (Problem 3.18). Theslope of the line tangent to c2(t) is shown with a dotted line. A dashed line tangent to c(0.04, t)beginning at t = 500 s is also shown.

where d is the membrane thickness and D is the diffusion coefficient in themembrane. Therefore,

τeq = V1V2d(V1 + V2)DA

= L1AL2Ad(L1A+ L2A)DA

= L1L2d(L1 + L2)D

= 10 · 1 · 0.111D

= 0.09D

s,

so that D = 0.09/2.1 × 104 = 4.3 × 10−6 cm2/s. Note that this estimate iswithin 16% of the estimated value of D found in part b.

v. The integral is simply the total number of moles of solute that flow throughthe plane at x = 0.1 per unit area in the time interval (0,1000). Thus, sincethe number of moles of solutes in compartment 2 is 0 at t = 0, the numberof moles of solute in this compartment at t = 1000 must equal the quantitytransported into the compartment. Therefore, A

∫ 10000 φ(0.1, t)dt = 1 · A ·

c2(1000) which can be divided by A to yield∫ 10000 φ(0.1, t)dt = c2(1000).

From Figure 3.13 it is apparent that c2(1000) = 3.2 mmol/cm3. Therefore,∫ 10000 φ(0.1, t)dt = 3.2 mmol/cm2.

vi. In a time interval from t = 0 to t = 5τeq, both the concentration in themembrane and in both compartments will be close to equilibrium. There-fore, both will approach a concentration of 90.09 mmol/cm3 with the equi-librium time constant which is 2.1× 104 s as shown in Figure 3.14. Note thatin Figure 3.13 c(0.04, t) appears to saturate at a steady-state value which isabout 60 mmol/cm3 in under 1000 seconds which represents about 4× τss .However, this time is short compared to τeq. Thus, the steady-state value isnot really saturated but changes slowly as the two volumes equilibrate. Fig-ure 3.14 shows that the apparent saturation of c(0.04, t) shown in Figure 3.13is a mirage.


x

c(x)

x

c(x)

r=1 mm r=1 nmFigure 3.15: Distribution of particles in a beakerunder the influence of diffusion and gravity forboth 1 mm particles and 1 nm particles (Prob-lem 3.19). The particles are not drawn to scale.The concentration of particles as a function ofheight is shown beside each beaker.

Problem 3.19 The problem states that the total flux in the positive-x direction (a direc-tion opposite to that of the force of gravity) is φ = φD +φG where φD is the flux dueto diffusion and φG is the flux due to gravity. These fluxes are

φD = −D∂c(x, t)∂xand φG = −Dλ c(x, t).

a. The flux due to diffusion is Fick’s law. The flux due to gravity requires somecomment. Flux due to a body force on the particles has the form φG = upcfpwhere the particle mobility up = D/kT , and the gravitation force on a particle inwater is fp = −meffg. Combination of these terms yields

φG = upcfp = −DmeffgkT

c(x, t)

from which λ = kT/(meffg).

b. At equilibrium φ = 0 and c(x, t) is a function of x only so that

−Ddc(x)dx

− Dλc(x) = 0

which, after canceling some common factors, yields

dc(x)dx

+ c(x)λ= 0

which has the solutionc(x) = c(0)e−x/λ.

c. The space constant λ = kT/(meffg) wheremeff = (4/3)πr3(ρp −ρw). Therefore,

λ = (kT/g)(4/3)πr3(ρp − ρw).

Which has the value

λ = 4.2× 10−17

(4/3)πr3(1)= 10−17

r3 cm.

For r = 1 mm, λ = 10−17/10−3 = 10−14 cm and for r = 1 nm, λ = 10−17/10−21 =104 cm. A 50 ml = 50 cm3 beaker will have a height of a few cm (see Figure 3.15).On the scale of the beaker dimensions, the 1 mm particles will all lie on the bottomof the beaker (within a layer whose dimensions are about 10−14 cm which equals10−4 pm), while the 1 nm particles will be uniformly dispersed in the beaker (witha space constant of 104 cm which is 102 m).

PROBLEMS 37

Microscopeobjective

Microscopecondenser

Emulsion

Focusup anddown

Gravityx

c(x)

Light

Figure 3.16: Schematic diagram of apparatus usedby Perrin to make observations of the concentrationof mastic particles in an emulsion as function ofdepth (Problem 3.19). The structures are not drawnto scale.

d. In the method used by Perrin (Perrin, 1909), both the particles and the spatial distri-bution of particle concentration must be resolved with a light microscope whoselimit of resolution is the wavelength of light, ca. 0.2 µm. This resolution limitplaces both an upper and a lower bound on the particle dimensions that are prac-tical. For particles with a radius of 1 mm, the spatial distribution of concentrationchanges very rapidly, i.e., λ = 10−14 cm = 10−10 µm which is more than 9 ordersof magnitude below the limits of resolution of a light microscope. For 1 nm radiusparticles the spatial distribution is very broad and the difference would have tobe measured over 104 cm = 102 m which is impractical with a light microscope. Inaddition, these particles are too small to be resolved with a light microscope. Sothe strategy is to pick the smallest particles that can be seen. Particles with r = 0.2µm are close to the smallest that can be resolved in a light microscope. These giveλ = 10−17/(0.2 × 10−4)3 cm = 12.5 µm. Thus, the particles can be counted atdifferent depths and the depths needed are in the range of a few to 10’s of µmwhich can be easily obtained with a light microscope with the arrangement shownin Figure 3.16. Note that with 1 µm particles, λ = 10−17/(10−4)3 = 10−5 cm = 0.1µm so the spatial distribution changes appreciable in a few tenths of a µm makingmeasurements versus depth very difficult. Thus, there is a rather narrow rangeof particles that can be used. Perrin used r = 0.14 µm to r = 0.52 µm. Mostmeasurements were obtained for r = 0.212 µm.

Problem 3.20 As shown in the Figure 3.41 (Weiss, 1996a), a regression fit to Collander’smeasurements in Nitella has the following expression

log(PnM2.2) = 1.20 logkn − 0.96.

Thus, the logarithm of the permeability is given by

log Pn = 1.20 logkn − 0.96− 2.2 logM.

Therefore, an estimate of the permeability Pn can be computed from this relation toobtain the results shown in Table 3.1. Both a large partition coefficient and a small


Solute M k Pn Rank(daltons) (cm/s)

A 12 0.1 2.9× 10−5 2B 30 1.0 6.2× 10−5 1C 50 0.02 1.8× 10−7 4D 400 3.0 7.7× 10−7 3

Table 3.1: Characteristics of so-lutes 1-4 (Problem 3.20). M is themolecular weight in Daltons and kis the ether:water partition coeffi-cient, Pn is the estimated perme-ability, and the final column givesthe rank order of the permeabili-ties of these 4 solutes in order ofdecreasing permeability.

molecular weight are important to achieve a high permeability. For example, althoughsolute D partitions best in the organic solvent, it has a large molecular weight, andhence, a low permeability in Nitella membrane.

Problem 3.21 Table 3.6 (Weiss, 1996a) indicates that the permeability of membranesof several types of cells for ethanol is at least one order of magnitude larger than forformamide. From Figure 3.36 (Weiss, 1996a) it is apparent that the permeability of ery-throcyte membranes for methanol is larger than for ethanol. Furthermore, Figure 3.37(Weiss, 1996a) indicates that the permeability of the plant Chara ceratophylla is veryhigh. In general, the permeability of cellular membranes to alcohols is relatively high.Therefore, the permeability of the membrane for methanol should be much higher thanfor formamide. This difference is probably due to the large difference in the solubil-ity of these two substances in organic solvents: methanol is highly soluble in organicsubstances while formamide is not. The difference in solubility is due to the chemicalgroups that these molecules contain. Methanol contains the nonpolar methyl group andthe polar hydroxyl group whereas formamide contains two polar (aldehyde and amino)groups.

Chapter 4

SOLVENT TRANSPORT

Exercises

Exercise 4.1 In a hypertonic extracellular solution, water flows out of the cell throughboth the cell wall and the cell membrane. As water leaves the cell, the cell volume de-creases and the cell membrane shrinks away from the relatively rigid cell wall. Thisleaves a compartment between the cell wall and cell membrane. Because the cell mem-brane is highly selectively permeable and the cell wall is less so, the composition of thiscompartment resembles extracellular rather than intracellular solution.

Exercise 4.2 A property of a solution that depends only on the number of particlesin solution and not on their chemical nature is called a colligative property. Colliga-tive properties of solutions include: osmotic pressure, vapor pressure, freezing pointdepression, and boiling point elevation.

Exercise 4.3 Two solutions are isosmotic if they have the same osmotic pressure.

Exercise 4.4 Two solutions that are isosmotic have the same osmotic pressure. Twoextracellular solutions that are isotonic produce the same cell tonicity. That is, if oneisotonic solution is replaced by another then the cell volume does not change. Oftenthe two terms are used interchangeably, but they are distinct in that isosmoticity can bedetermined by physical chemical means, i.e. with an osmometer. However, establishingisotonicity requires measurements on cells.

Exercise 4.5 The ideal gas law is

pV = RTn or p = RTc,

where p is the pressure that the gas exerts on the walls of a container of volume V , nis the number of moles of gas in the container, and c = n/V is the molar concentrationof the gas in the container. Van’t Hoff’s law is

π = RTCΣ,

where π is osmotic pressure and CΣ is the total concentration of solutes in solution,where the ionization of salts into multiple solutes is taken into account. The pressure

39

40 CHAPTER 4. SOLVENT TRANSPORT

that the gas exerts on the walls is measurable directly. The osmotic pressure of a so-lution is measured with a semi-permeable membrane that allows the solvent but notthe solute to permeate. Alternatively, the osmotic pressure can be estimated with anosmometer under the assumption that it is valid to compute the osmotic pressure fromthe measurement of some other colligative property, e.g., the vapor pressure.

Exercise 4.6 The osmotic pressure of a solution is given by van’t Hoff’s law asπ = RTCΣwhere CΣ is the osmolarity. The osmolarity is the total number of moles of solute insolution divided by the volume of the solution. The osmolarity is usually expressed inmol/L.

Exercise 4.7 The osmotic coefficient χ reflects the deviation of the osmotic pressure ofa real solution π from that predicted by van’t Hoff’s law, i.e.,

χ = πRTCΣ

.

Exercise 4.8 In the left-hand system, osmotic flow of water is from the left to the rightchamber and hydraulic flow of water is from the right to the left chamber. Hence,osmotic equilibrium is possible when the two flows are equal and opposite. In the right-hand system, both osmotic and hydraulic flow of water are from the right to the leftchamber and no osmotic equilibrium is possible.

Exercise 4.9 In the absence of a hydraulic pressure difference across a membrane, theflux of volume is

ΦV = LVRT(C2Σ − C1

Σ

).

When the volume flux is carried entirely by water, the volume flux is proportional to themolar flux of water, i.e., ΦV = φwVw whereφw is the number of moles of water flowingthrough a unit area of membrane per unit time and Vw is the partial molar volume ofwater which is the volume of a mole of water. The osmotic molar flux of water can beexpressed as

φw = Pw(C2Σ − C1

Σ

),

where

Pw = RTLVVw.

Pw is called the osmotic permeability coefficient and is a measure of the flow of water inresponse to a difference in osmotic pressure across a membrane. By using a radioactivetracer for water, the diffusion of water through membranes can be measured in theabsence of an osmotic pressure difference across the membrane. The diffusive flux ofwater can be expressed by Fick’s law for membranes

φw = Pw(c1w − c2

w

),

where Pw is the diffusive permeability (or simply the permeability) of the membrane forwater.

EXERCISES 41

Exercise 4.10 The formal resemblance between the solute flux due to diffusion througha membrane and the osmotic transport of solvent through a membrane is apparent fromthe two flow equations,

φn = Pn(c1n − c2

n

),

ΦV = LVRT(C2Σ − C1

Σ

).

φn has units of mol/(cm2 · s); is the flux of solute n through the membrane; dependson the difference in concentration across the membrane of solute n only. ΦV has unitsof cm/s; is the flux of volume (predominantly water) through the membrane; dependson the difference in total concentration of all solutes across the membrane. In addition,note the difference in the sign that relates the flux to the difference of concentrationacross the membrane.

Exercise 4.11

a. Water transport through a homogeneous, hydrophobic membrane is by the dissolve-diffuse mechanism. Water first dissolves in the hydrophobic membrane and thendiffuses through it. According to this mechanism, water acts as a solute in themembrane. Water transport through a porous membrane is via convection. A dif-ference in osmotic pressure between bath solutions is equivalent to a hydraulicpressure difference across the membrane that acts on the water.

b. A simple criterion is based on estimating Pw , from measurements of water trans-port due to an osmotic pressure difference, and Pw , from measurements of thediffusion of tracer water in the absence of an osmotic pressure difference. Forwater transport through a homogeneous, hydrophobic membrane Pw/Pw = 1,while simple theories of water transport through porous membranes predict thatPw/Pw > 1.

Exercise 4.12 The solute bombardment theory gives a reasonable explanation for howan additional pressure can develop on the side containing the solute but does not sug-gest a mechanism by which this additional pressure can cause a flux of solvent into theside that has the larger pressure. Furthermore, the solute bombardment theory ignoresthe presence of the solvent to which the momentum incurred by solute bombardmentmust be transferred.

Exercise 4.13

a. At osmotic equilibrium

p1 − p2 = π1 −π2 = RT(C1Σ − C2

Σ).

b. The total volume of water is V1 +V2 = 2 L. Since p1 = p2, at osmotic equilibriumC1Σ(∞) = C2

Σ(∞). Initially, C1Σ(0) = 0.01 mol/L and C2

Σ(0) = 0.02 mol/L becauseNaCl dissociates so that a 0.01 mol/L liter solution has an osmolarity of 0.02 mol/L.


0 0.5 1 1.5

0.5

1

1.5

V1

V2

t/τ

Figure 4.1: Change in volume with time (Exercise 4.13).The volumes are expressed in liters.

The number of moles of glucose is 0.01 mol and of Na+ plus Cl− is 0.02 mol.Therefore at osmotic equilibrium

0.01V1

= 0.02V2

and V1 +V2 = 2.

These two equations are solved by taking the reciprocal of the equation on the leftand substituting the equation on the right to obtain,

100V1 = 50V2 = 50(2− cV1).

Solving these equations yields V1 = 2/3 L and V2 = 4/3 L. The volume can befound as a function of time from the relation

− 1AdV1(t)dt

= LVRT(C2Σ − C1

Σ).

Note that V1(t)+V2(t) = 2 L, N1Σ = 0.01 mol/L and N2

Σ = 0.02 mol/L so that

dV1(t)dt

= 1τ

(1

V1(t)− 2

2−V1(t)

),

where τ = 100/(LVRTA). V2(t) = 2 − V2(T). This differential equation can beintegrated numerically to yield the solution shown in Figure 4.1. The differentialequation for the volume is nonlinear and the solution shows monotonic, but non-exponential changes in volume.

Exercise 4.14 The isotonic volume is the volume of a cell in an isotonic solution. Thus,the isotonic volume is the volume of the cell in its normal environment in the body.

Exercise 4.15

a. The isotonic volume of these cells is the value of the volume when CoΣ = ConΣ whichis when the abscissa equals 1. The isotonic volume is about 2.1× 105 µm3.

b. The fraction of the isotonic volume not due to water is obtained when the soluteconcentration is arbitrarily large. In principle, for such a large solute concentra-tion, the cell shrinks so that no water remains intracellularly and the cell volumeequals that not due to water. The regression analysis in the caption of Figure 4.21(Weiss, 1996a) suggests that when CoΣ → ∞, Vc → 2.03 × 104 µm3. Thus, about10% of the volume of the cell is not due to osmotically active water.

EXERCISES 43

Exercise 4.16 A cell whose shape differs from a sphere can in principle change its vol-ume without changing its surface area. For example, an erythrocyte has the shape of abiconcave disc. The cell can show a large increase in volume without changing its shapesimply by becoming more spherical.

Exercise 4.17 Equation 4.68 (Weiss, 1996a) expresses conservation of water volumebetween the cell interior and the bath solution. The left-hand side of the equation is therate of decrease in the volume of the cell water which equals the right-hand side of theequation which is the flux of water volume that leaves the cell.

Exercise 4.18

a. At equilibrium,

Vc = NiΣCoΣ+V′c ,

whereV′c is the osmotically inactive portion of the total cell volumeVc , and whereit is assumed that V′c Vc . Therefore, the initial and final values of the volumeare

Vc(0) = NiΣCoΣ(0)

and Vc(∞) = NiΣCoΣ(∞)

,

whereNiΣ is constant because the solute is assumed impermeant. Division of thesetwo equations yields

Vc(0)Vc(∞) =

CoΣ(∞)CoΣ(0)

= 2.

Normalizing the concentration yields

c(∞) = 2c(0).

Therefore, the answer is waveform v.

b. The flux of water is related to the change in volume by the conservation relation

− 1A(t)

dV i(t)dt

= ΦV = LVRT(CoΣ(t)− CiΣ(t)

).

If the temperature is decreased, the magnitude of the flux will decrease, and themagnitude of the rate of change of volume will decrease. Hence, the time courseof volume change will be slower. The equilibrium condition that CiΣ = CiΣ does notdepend upon temperature. Therefore, the equilibrium volume will be unchanged.

Exercise 4.19 The time course for equilibration of the volume of a cell with a semi-permeable membrane is given in normalized coordinates by Equation 4.69 (Weiss, 1996a).The time course is controlled by the term αva(t) which is proportional to the ratio ofsurface area to volume of the cell. Hence, for two cells with the same volume, thecell with the larger surface area will show a larger rate of change of volume and willequilibrate faster. This makes intuitive sense since a larger area of contact betweencompartments will allow a greater flow of volume. Therefore, the cylindrical cell withmicrovilli will equilibrate faster.


0 1 2 3

0.0005

0.001

0.0015

0.002

ανt

Cel

lvo

lum

e(m

m3)

Figure 4.2: Time course of cell volume (Exer-cise 4.21).

Exercise 4.20 An experimental test of a theory often involves plotting a measured vari-able against another measured variable and comparing this measured relation to thatpredicted by the theory. When the relation is a straight line, the eye is a good instrumentfor qualitatively checking whether the measured relation and the theoretical relationagree. More quantitative statistical techniques (e.g., linear regression) exist for assess-ing the extent to which a set of data are fit by a straight line. For these two reasons,one commonly used method for analyzing data, is to first find a transformation thatlinearizes the expected relation between variables. Then linear regression techniquesare used to assess goodness of fit.

Exercise 4.21

a. The isotonic volume of a cell is it’s volume in situ, i.e., its volume in its naturalenvironment. If the concentration CoΣ of an impermeant non-ionic solute is equalto ConΣ , the isotonic concentration, then we expect the cell volumeVc to equilibrateto the cell’s isotonic volumeVnc . Thus, if ConΣ /C

oΣ = 1, thenVc = Vnc and the value

of Vnc can be read from the plot.

Vnc ≈ 0.002 mm3

b. Since V i is osmotically active (by assumption), it should shrink as CoΣ increases.In the limit CoΣ → ∞, V i → 0, and Vc → V′c . As CoΣ → ∞, ConΣ /C

oΣ → 0. From the

plot, Vc is about 0.0006 mm3 when ConΣ /CoΣ = 0. Therefore V′c ≈ 0.0006 mm3.

c. If the solution of 200 mmol/L NaCl completely dissociates, then its osmolarity willbe 400 mosm/L. If ConΣ is 200 mosm/L, then ConΣ /C

oΣ will be 1/2. From the plot,

Vc will be approximately 0.0013 mm3.

d. If the cell volume is at equilibrium just before t = 0, then the cell volume willdecrease from an initial value of 0.002 mm3 (part a) to a final value of 0.0013 mm3

(part c). Because the equation for volume change is nonlinear, the time course ofcell volume is not exponential. Figure 4.2 summarizes these facts. The time courseof equilibration will depend on the rate at which water enters the cell. The rate ofwater entry is given by the product of the water flux ΦV (t) and the surface areaof the cell membrane. Assuming that the cell membrane supports no hydraulicpressure difference between the intracellular and extracellular spaces (typical ofanimal cells), the flux of water is equal to the product of the hydraulic conductivityof the cell membrane and the osmotic pressure difference between the intracellular

EXERCISES 45

and extracellular spaces. This osmotic pressure difference is equal to RT timesthe difference in osmolarity between the inside and outside of the cell. Combiningall of these effects, it is apparent that the rate of water entry, and therefore the rateat which the cell volume is changing, depends on: the area of the cell membrane,the hydraulic conductivity of the cell membrane, and the temperature of the cell.This behavior is expressed mathematically as follows

dV i(t)dt

= A(t)LVRT(CiΣ(t)− CoΣ(t)

)which can be expressed in normalized coordinates for cylindrical cells as

dν(t)dt

= αν(ν(t))1/2(

1ν(t)

− c(t)).

The results in Figure 4.2 were obtained by numerically integrating this equationwith c(t) = 2 for t > 0 with the cell volume expressed as

Vc(t) = ν(t)(0.002− 0.0006)+ 0.0006 (mm3).

Exercise 4.22 Water transport in lipid bilayers is not blocked by mercurial compounds,has a ratio of osmotic to diffusive permeability that is near 1, and has a relatively largedependence on temperature. In contrast, water transport through water channels incell membranes is blocked by mercurial compounds, has a ratio of osmotic to diffusivepermeability that exceeds 1, and has a relatively small dependence on temperature.Of course, water transport through a cell membrane that contains water channels is amixture of transport through the lipid bilayer and through the water channels. Whenone or the other of these routes predominates the distinction between the two modesof transport is clearly defined.

Exercise 4.23 These results show that the control oocytes, into which no CHIP RNA wasinjected, show a relatively small osmotically-induced volume increase that is relativelyinsensitive to application of a mercury compound. Thus, these results are consistentwith water transport by diffusion through the lipid bilayer but not through water chan-nels. Addition of CHIP RNA results in a large increase in osmotically-induced watertransport that is reduced by the application of a mercury compound. This latter resultis consistent with water transport via water channels. The experiment demonstratesthat incorporation of CHIP RNA results in the apparent appearance of water channelssuggesting that CHIP RNA codes for the water channel protein.

Exercise 4.24 Vasopressin binds to vasopressin receptors located on the membranes oftarget cells. Via a second messenger, vasopressin binding results in the recruitment ofcytoplasmic water channels and the incorporation of these channels into the membrane.This incorporation increases the osmotic permeability of the cell.


Problems

Problem 4.1 The measurements shown express the pressure and concentration in unitsthat are convenient for making chemical measurements. However, the osmotic pressureis proportional to the molar concentration, i.e., π = RTCΣ where R = 8.3 J/(mol·K) andT = 298K. Therefore, the molar concentration must be expressed in the units used forthe measurements. The data show that a concentration of albumin of 36 g of albuminper kg of water yields an osmotic pressure of 20 cm of water. This pressure can beexpressed in pascals by noting that a column of liquid 20 cm high is equivalent to apressure p = ρgh where ρ is the density of water, g is the acceleration of gravity, andh is the height of the column of water. Therefore,

p = (1 g/cm3) · (980 dyne/g) · (20 cm) = 19600 dynes/cm2 = 1960 Pa.

Note that the concentration is given as the number of grams of albumin per kilogram ofwater. The number of moles of albumin can be obtained from its weight by dividing bythe molecular weight of albumin, M . Dividing the given concentration by the molecularweight of albumin yields the molal concentration of the albumin solution. Assume thatthe concentration is low enough that the molal and molar concentrations are the same.Therefore the molar concentration is (36/M)× 103 moles/m3. Van’t Hoff’s law yields

1960 Pa = 8.3 J/(mol·K)× 298K× (36/(1000 ·M))× 103 moles/m3.

Solving this equation yields a molecular weight of M = 45 kg/mol = 45× 103 g/mol.

Problem 4.2 Assume that all the sucrose dissolves but that only a fraction of the NaCl,α, ionizes. When a salt dissociates each ion contributes to the osmotic pressure

NaCl z Na+ + Cl−.

If α is the fraction of NaCl that is dissociated then 1 − α is the fraction that is notdissociated. Since each dissociated NaCl molecule contributes 2 ions and each undisso-ciated molecule contributes 1 molecule, the osmolarity of the NaCl solution is CNaClΣ =(2α+1−α)CNaCl whereas the osmolarity of the sucrose solution is CsucroseΣ = Csucrose.Therefore,

πNaClπsucrose

= RTCNaClΣRTCsucroseΣ

= (2α+ 1−α)CNaClCsucrose

= 8.754.76

= 1.84.

Therefore, α ≈ 0.84.

Problem 4.3 Sucrose and raffinose have equal osmolarity when they have equal molarconcentrations. Therefore, 59.4 grams per liter of raffinose should have an osmolarityof 0.1 mol/L. Hence, 1 mole of raffinose has a weight of 594 grams which is the molecularweight. Hence, computation of the molecular weight will reveal if one of the sugars hasa molecular weight of 594. The atomic weights are: C, 12; H, 1; O, 16. Therefore, themolecular weights of the sugars are

PROBLEMS 47

Molecular formula Molecular weightC12H22O11 + 3H2O 396 not raffinose,C18H32O16 + 5H2O 594 raffinose,C36H64O32 + 10H2O 1188 not raffinose .

Problem 4.4

a. First, check charge neutrality. Bicarbonate and chloride are the principal anionsand have an osmolarity of 156 mosm/L; sodium and potassium are the principalcations and have an osmolarity of 160 mosm/L. Thus, the cations and anionsare approximately electroneutral and account for an osmolarity of 316 mosm/L.Addition of the osmolarities of the other substances, under the assumption thatall substances are fully dissolved, yields an osmolarity of 324 for the inorganicsubstances and another 157 mosm/L of organic substance to yield 481 mosm/L.

b. Clearly, the sum of the concentrations of all solutes exceeds the measured osmo-larity of tears. Presumably, many of the organic substances are parts of macro-molecules so that it is not valid to simply add the concentrations of individualmoles of nitrogen, for example. Furthermore, some of the salts in tears may notbe fully ionized.

c. If the temperature is 27 then the osmotic pressure in pascals is

π = RTCΣ = (8.314 J/(mol·K))(300K)(320 mol/m3) ≈ 8× 105 Pa,

where the measured osmolarity was used to compute the osmotic pressure.

Problem 4.5

a. From the van’t Hoff Law of osmotic pressure, the osmotic pressure difference mustequal the hydraulic pressure difference.

pfw − psw = πfw −πsw = RT(CfwΣ − CswΣ ),

where fw refers to freshwater and sw refers to seawater. The salt concentrationis greater in the salt water so the osmotic pressure will make ΦV flow from thefresh water to the salt water. To determine psw , determine the osmolarities of thetwo solutions as follows.

CfwΣ = 2

(500g of NaCl106g of water

)(1g of water

cm3

)(1 mole

58 g of NaCl

),

CfwΣ = 1.72× 10−5 moles/cm3,

CswΣ = 2

(40,000g of NaCl

106g of water

)(1 g of water

cm3

)(1 mole

58 g of NaCl

),

CswΣ = 1.38× 10−3 moles/cm3.

Therefore,pfw − psw = RT(CfwΣ − CswΣ ).


Since pfw = 0,

psw = RT(CswΣ − CfwΣ ).

At a temperature of 25C,

psw =(

8.314× 106 Pa

K·(moles/cm3)

)(300K)

(1.36× 10−3moles/cm3

),

psw = 3.39× 109Pa = 34 atmospheres

The pump must apply a hydraulic pressure of at least 34 atm to convert 40,000ppm salt water to 500 ppm salt water.

b. The pressure can also be produced by raising the height of the salt water tank. Oneatmosphere of pressure is produced for each 33.5 feet that the water is raised. Soto produce 34 atmospheres the salt water tank must be raised 1,139 feet withrespect to the fresh water tank.

c. Because you float in the Great Salt Lake and sink at Laguna beach, you realize thatthe density of the salt water is larger in the lake. This means that the concentrationof salt is larger in the lake. The larger salt concentration produces a larger osmoticpressure. Therefore, Utah will need a larger hydraulic pressure to enable reverseosmosis.

Problem 4.6

a. If the beach balls are treated as solutes in a solution, then the side of the poolwith beach balls has a higher osmotic pressure than the side without beach balls.Therefore, water will flow from the side of the pool without beach balls into theside with beach balls.

b. Let the side of the pool with beach balls be denoted by a b subscript and thatwithout by an n subscript. At osmotic equilibrium pb−pn = πb−πn, but πn = 0.Thus, the difference in hydraulic pressure on the two sides is due to the osmoticpressure of the beach balls. If the difference in hydraulic pressure on the twosides is assumed to be due to the increment in the height of the water on the sidewith the beach balls, then

pb − pn = ρgh = RTCbΣ ,where h is the increment in the height of the water in the pool on the side withbeach balls above that on the other side. The molar concentration of beach ballsis

CbΣ =100NA · V = 100

6.022× 1023 · 3 · 6 · 10= 9.23× 10−25mol/m3.

where NA is Avogadro’s number and V is the volume of the side of the pool withthe beach balls. Therefore,

h = RTCbΣ

ρg= (8.314 J/(mol · K)(300K)(9.23× 10−25 mol/m3)

(103 kg/m3)(9.8 N/kg)= 2.3× 10−25 m.

Thus, the height of the water on the side with the beach balls is higher than thaton the side without beach balls by 2.3× 10−25 m.

PROBLEMS 49

Solutes Initial Values Final Values1 2 c1(0) c2(0) x(0) c1(∞) c2(∞) x(∞)

mmol/L mmol/L cm mmol/L mmol/L cm

Glucose Glucose 0 10 5 0 5 0Glucose Glucose 10 10 8 10 10 8Glucose Glucose 30 10 5 20 20 7.5Glucose NaCl 20 10 2 20 10 2Glucose NaCl 80 10 2 32 16 5NaCl NaCl 30 10 5 20 20 7.5

Table 4.1: Table of solute concentrations and partition positions (Problem 4.7).

c. The problem is that the height difference that needs to be measured is very small,i.e., 2.3×10−15 A. This outcome results because the osmotic pressure of 100 beachballs in a pool is very small.

Problem 4.7 At osmotic equilibrium, the difference of hydraulic and osmotic pressureon the two sides of the partition must be the same. Since the partition is free to move,it can sustain no difference of hydraulic pressure. Thus, if osmotic equilibrium isachieved, then the osmotic pressure must be the same on the two sides of the parti-tion which implies that the total solute concentration must be the same.

To solve the problem, the final concentrations and partition positions need to bedetermined. The final concentration on side 1 is

c1(∞) = c1(0)V1(0)V1(∞) = c1(0)x(0)A

x(∞)A = c1(0)x(0)x(∞) ,

and on side 2

c2(∞) = c2(0)V2(0)V2(∞) =

c2(0)(10− x(0)

)A(

10− x(∞))A

=c2(0)

(10− x(0)

)10− x(∞) ,

where A is the cross-sectional area of the compartment. If the concentrations on thetwo sides are equal at equilibrium, then the final concentrations will be the same,

c1(0)x(0)x(∞) =

c2(0)(10− x(0)

)10− x(∞)

which can be solved for x(∞) to yield

x(∞) = 10c1(0)x(0)

c2(0)(10− x(0)

)+ c1(0)x(0)

Once x(∞) is determined, then the final concentrations can be computed from theabove formulas. The answers are shown in Table 4.1. The following are comments onthe different parts:


Figure 4.3: Photographs of the General Sherman giant sequoia and a mangrove tree (reproducedfrom Encarta96 Encyclopedia) (Problem 4.8).

psg πs

po ptg πt

Root

Soil/seawaterΦV

Figure 4.4: Variables involved in water transport in a treeroot (Problem 4.8).

• When c1(0) = 0 water will flow into side 2 pushing the partition to the left. How-ever, osmotic equilibrium cannot be achieved so the partition will move all the wayto the left until x(∞) = 0. The concentrations in side 1, which is now infinitesi-mally thin, remains at zero since the solute does not go through the partition. Theconcentration of side 2 is obtained from the formulas.

• When the initial glucose concentration is the same on both sides, the system isalready in osmotic equilibrium. Hence, nothing changes.

• With a different initial concentration of glucose on the two sides of the membrane,straight forward application of the formulas yields the results. Clearly, with theinitial concentration of glucose on side 1 greater than on side 2, water will flowinto side 1 which will move the partition to the right.

• In the 3 parts that involve NaCl it is important to realize that the total concentrationof solutes is twice the concentration of the salt under the assumption that the saltis fully ionized. Thus, case 4 is already at osmotic equilibrium so there are nochanges in concentration or partition position. Cases 5 and 6 involve using thederived formulas taking total solute concentration into account.

Problem 4.8 Figure 4.3 shows a picture of the General Sherman giant sequoia as wellas a mangrove tree growing with its roots in salt water. Figure 4.4 shows the variablesinvolved in water transport in a tree root.

PROBLEMS 51

Variable Sequoia Mangrove

h (m) 82.9 24.4pg (Pa) 8.1× 105 2.4× 105

CtΣ (mol/m3) 43 43CsΣ (mol/m3) 51 1050πt (Pa) 1.1× 105 1.1× 105

πs (Pa) 1.3× 105 2.6× 106

πt −πs − pg (Pa) −8.3× 105 −2.7× 106

po (Pa) < −8.3× 105 < −2.7× 106

Table 4.2: Summary of osmotic and mechanical factors involved in the rising of sap in the giantsequoia and the red mangrove. The height of the trees is expressed in meters; there are 0.3048meters per foot.

a. At the junction between the roots and the soil/seawater, the flux of water fromthe soil/seawater into the tree root is

ΦV = RTLV ((psg −πs)− (po + ptg −πt)).

where the superscript t is used to denote the tree root and the superscript s todenote either soil or seawater. Let pg = ptg − psg . Therefore, for the sap to rise inthe tree πt −πs − po − pg+ > 0 which implies that po < πt −πs − pg .

b. The pressure due to gravity is pg = ρgh where ρ is the density of water, g is theacceleration of gravity, and h is the height of the tree. This can be expressed asfollows

pg = (103 kg/m3)(9.8 N/kg)h = 9.8× 103h Pa.

The osmotic pressure at 27C is

π = RTCΣ = (8.314 J/(mol · K)(300K)CΣ = 2.49× 103CΣ,

where CΣ is expressed in mol/m3. Note that concentration in mol/L is the sameas in mol/m3. Table 4.2 summarizes the relevant factors. There are number ofinteresting factors involved in the rising of sap that are illustrated in this problem.To interpret results intuitively, recall that atmospheric pressure (which resultsfrom the weight of the atmosphere on the surface of the earth) corresponds to105 pascals also called an atmosphere. Thus, the gravitational pressure due to theheight of the tree must be much larger in the giant sequoia, which is a taller tree,than in the red mangrove. Both pressures correspond to several atmospheres.Both of these pressures tend to make the water flow out of the tree root into thesoil/seawater. Because, the osmolarity of the soil/seawater exceeds that of sapfor both trees, this difference in osmotic pressure also causes water to flow out ofthe tree. However, this effect is much larger in the red mangrove than in the giantsequoia because the osmolarity of seawater greatly exceeds that of the soil. Thus,the capillary forces that make sap rise must be much greater in the red mangrovethan in the giant sequoia.


Problem 4.9 The flux through both membranes is the same

ΦV = RTL1(CoΣ − C1Σ) = RTL2(C2

Σ − CoΣ).CoΣ can be determined as follows

L1(CoΣ − C1Σ) = L2(C2

Σ − CoΣ),(L1 +L2)CoΣ = L1C1

Σ +L2C2Σ.

Therefore,

CoΣ =L1

L1 +L2C1Σ +

L2

L1 +L2C2Σ.

Therefore,

ΦV = RTL1(CoΣ − C1Σ),

= RTL1

( L1

L1 +L2C1Σ +

L2

L1 +L2C2Σ − C1

Σ

),

= RT( L1L2

L1 +L2

)(C2Σ − C1

Σ).

Therefore,

LV = L1L2

L1 +L2.

Problem 4.10 The flux through the composite membrane is the total flow of volumethrough the membrane divided by the membrane area which is

ΦV = Φ1A1 + Φ2A2

A1 +A2,

where Φ1 and Φ2 are the fluxes through patch 1 and 2, respectively,

Φ1 = RTL1

(C2Σ − C1

Σ

)and Φ2 = RTL2

(C2Σ − C1

Σ

).

Combining these two equations yields

ΦV = RT L1A1 +L2A2

A1 +A2

(C2Σ − C1

Σ

),

from which it is clear that

LV = L1A1 +L2A2

A1 +A2.

Problem 4.11

a. Let i and o represent the inside and outside of the tube, respectively. Let ΦVrepresent the flux of water through the membrane with outward flow being takenas positive. The tube contains a volume of solution

V (t) = Ah(t),

PROBLEMS 53

where h(t) is the total height of the solution in the glass tube. Conservation ofwater requires that

dV (t)dt

= −AΦV .A combination of these equations yields

dh(t)dt

= −ΦV .

The volume flux is related to osmotic pressure as follows

ΦV = −LVRT(C1Σ − C2

Σ),

where the hydraulic pressure is assumed negligible, C1Σ = Cs(t), the concentration

of sucrose, and C2Σ = 0. As water flows into the tube, the concentration of sucrose

decreases. However, the total quantity of sucrose remains constant,

Cs(t)V (t) = Cs(0)V (0),so that

Cs(t) = Cs(0)V (0)V (t) = Cs(0)h(0)h(t)

.

A combination of these relations results in the differential equation for h(t),

dh(t)dt

= LVRTCs(0)h(0)h(t).

b. The differential equation for h(t) is separable and can be written as∫ h(t)h(0)

h′(t)h(0)

dh′(t) =∫ t

0RTLVCs(0)dt′,

which upon integration yields

h2(t)− h2(0)2h(0)

= RTLVCs(0)t.

Therefore,

h(t) =√h2(0)+ 2LVRTCs(0)h(0)t.

We evaluate the constants as follows

h(0) = V (0)A

= 20.5

= 4 cm = 0.04 m,

and

2LVRTCs(0)h(0) = 2 · LV · 8.314 J/(mol·K) · 300 K · 102 mol/m3 · 0.04 m,≈ 2× 104LV m2/s.

Therefore,

h(t) =√

0.042 + 2× 104LV t,whose form is shown plotted in Figure 4.5.


0 0.1 0.2 0.3

0.2

0.4

0.6h(t)

αt

Figure 4.5: A plot of h(t) versus αt where α = 2× 104LV(Problem 4.11). The dashed line shows the initial slope ofh(t).

c. If h(t) ≈ h(0) + 0.02t cm initially, then the derivative of height with respect totime is 0.02 cm/s at t = 0. The results from part a show that

dh(t)dt

∣∣∣∣t=0= LVRTCs(0),

so that

LV = dh(t)/dt|t=0

RTCs(0).

The denominator is

RTCs(0) = 8.3 N · m/(mol · K)× 300 K× 100 mol/m3 = 2.49× 105 Pa.

Therefore,

LV ≈ 2× 10−4 m/s2.49× 105 Pa

≈ 8× 10−10 m/(Pa·s).

Alternatively, the initial slope can be obtained from the solution found in part b,and the value of LV determined. For this purpose,

h(t) = h(0)

√1+ 2LVRTCs(0)t

h(0),

h(t) ≈ h(0)(

1+ 12

2LVRTCs(0)th(0)

),

h(t) ≈ h(0)+LVRTCs(0)t.

Hence, the initial slope is LVRTCs(0) which is consistent with the earlier finding.

It is of interest to see if hydraulic pressure due to gravity, which has been ignoredin this problem, makes an appreciable contribution. At an initial height H abovethe water level in the vessel, the hydraulic pressure is

ρgH ≈ 1000 kg/m3 × 9.8 m/s2 ×H m ≈ 104H Pa.

Therefore, for a glass tube with water level of a few centimeters, the hydraulicpressure is negligible compared to the osmotic pressure.

Problem 4.12

PROBLEMS 55

a. The flux of volume is

ΦV = LV((ps − po)− (πs −πo)

).

The hydraulic pressure is due to the head of water and the osmotic pressure dif-ference. The osmotic pressure of the fresh water in the pipe is zero. Therefore,

ΦV = LV((psghs − pogho)− (RTCΣ − 0)

),

= LV(g(pshs − poho)− RTCΣ

).

Note that the difference in hydraulic pressure must overcome the osmotic pressureof seawater in order for the volume flux to be positive.

b. The derivative of the height is obtained as follows

d(Aho(t))dt

= Adho(t)dt

= AΦV .

Substitution for the flux yields(dho(t)dt

)t=0= LV

(ρsghs − ρogho(0)− RTCΣ

).

All the constants are expressed in ISI units. The difference of hydraulic pressureis

ρsghs − ρogho(0) = 9.8ms2

(1.03× 103 kg

m3 × 102 m− 1× 103 kgm3 × 1 m

),

= 106 Pa.

The difference in osmotic pressure is

RTCΣ = 8.314J

mol · K× 300 K× 103 mol

m3 ,

= 2.49× 106 Pa.

Therefore,(dho(t)dt

)t=0= 3× 10−12 m

Pa · s(1− 2.49)× 106 Pa = 4.47 Pa/s.

c. At equilibrium, ΦV = 0. Therefore,

g (ρshs − ρoho(∞))− RTCΣ = 0,

and

ho(∞) = ρsρohs −RTCΣgρo

.

The critical height is one for which ho(∞) > hs so that

ρsρohs − RTCΣgρo

> hs,


and (ρs − ρoρo

)hs >

RTCΣgρo

.

Therefore,

hs > hc = RTCΣg(ρs − ρo).

d. The critical assumption is that the ocean’s composition (osmolarity) is constantwith depth, i.e., that the ocean is well-mixed. If it were then there would be nointrinsic reason that fresh water and energy could not be extracted. On the otherhand, if the ocean were modeled as in equilibrium then the osmolarity would varyexponentially with depth as indicated in Problem 3.19. Thus, as the pipe is lowereddeeper into the ocean, the osmotic pressure would rise exponentially with depth.This increase in osmotic pressure with depth would tend to drive fresh water outof the pipe into the ocean. We investigate the effect of the assumed equilibriumdistribution of concentration. Diffusive equilibrium in the presence of the force ofgravity results in a solute concentration that is an exponential function of depth(see the solution to Problem 3.19), so that

CΣ(hs) = CΣ(0)ehs/λ,where

λ = kTmeffg

.

Combining this result with the earlier inequality, which is required if useful freshwater is to be obtained, yields

hs >RTCΣ(0)g(ρs − ρo)e

hs/λ.

The depth variable can be normalized so that y = hs/λ to give

ay > ey,

where

a = λg(ρs − ρo)RTCΣ(0)

,

=(kTmeffg

)(g(ρs − ρo)RTCΣ(0)

),

= ρs − ρoNAmeffCΣ(0)

.

Note that NAmeff is the effective mass of a mole of salt particles and can be ex-pressed as a molar volume times the effective density. The molar volume timesthe molar concentration is about 1. Thus, a ≈ 1. As shown in Figure 4.6, ay > ey

for values of a > e only. Therefore, this argument suggests that the inequalityis not satisfied and hs > ho(∞) so that no fresh water can be extracted from theocean by this mechanism. The model used here has ignored the change in densityof salt water with depth.

PROBLEMS 57

0.5 1 1.5 2

2

4

6ey

1

2

e3

y

Figure 4.6: A plot of the functions ey and ay versusy for different values of a (Problem 4.12). The func-tions do not intersect for a < e, are tangent for a = eat y = 1, and intersect for a > e.

Side 2MembraneSide 1

0 d x

p1 −RTC1Σ

p2 −RTC2Σ

RTC1Σ

RTC2Σ

p1

p2

ΦV

Figure 4.7: The hydraulic and osmotic pressurein a porous membrane (Problem 4.13). The solidline is p(x) − RTCΣ(x); the dashed line is p(x);the dotted line is RTCΣ(x).

Problem 4.13 The key to this problem is the observation that p−RTCΣ is continuous atthe membrane solution interface. This can be understood as follows. Since the solventis incompressible, all of the water that enters the membrane must also flow out of themembrane. Therefore, the flux of water is continuous through the membrane solutioninterface. It follows from Equation 4.12 (Weiss, 1996a),

ΦV = −κ ∂(p −π)∂x= 0

that p −π must also be continuous through the membrane solution interface.

a. To solve this problem, it is easiest to proceed graphically. Make a scale drawing ofthe membrane pressure variables. RTC1

Σ is the smallest quantity and all variablesare scaled to this quantity. RTC2

Σ = 3RTC1Σ . Inside the membrane RTCΣ = 0 since

no solute can enter the membrane. Hence, RTCΣ can be sketched in the baths andin the membrane as shown in Figure 4.7. Next sketch p1 and p2 in the baths asgiven in the problem. With this accomplished, sketch p − RTCΣ in the baths and,since it must be continuous, sketch it in the membrane. A sketch of both RTCΣand p − RTCΣ in the bath and membrane makes it is easy to sketch p.

b. In the membrane, there is no osmotic pressure gradient so the volume flux is

ΦV = −κdp(x)dx.

Since the slope of p(x) is negative in the membrane, ΦV > 0 as is shown.

c. With convection in the positive x-direction, we expect the concentration profile inthe membrane to resemble the curves such as the ones for γd of 0, 2, or 10 in thefigure. The exact shape depends on the exact parameters.


RTCΣ2

RTCΣ1


RTCΣ(x)

0 d x

p(x)

p1 p2

0

5P

ress

ure

(at

mos

pher

es)

RTCΣ2

RTCΣ1


RTCΣ(x)

0 d x

p(x)p1

p2

(1) (2)

Figure 4.8: Membrane hydraulic and osmotic pressure (Problem 4.14). The long dashes showthe hydraulic pressure, the short dashes show the difference between the hydraulic and osmoticpressure. The latter profile has been offset slightly for clarity. The hydraulic pressure and thedifference between hydraulic and osmotic pressure have profiles that overlap in the membrane.

Problem 4.14 This problem also depends for its solution on understand the continuityof p−π at a membrane solution interface as described in the solution to Problem 4.13.

a. Since there is a discontinuity in the osmotic pressure at each interface, there mustbe an equal discontinuity of hydraulic pressure at each interface as shown in Fig-ure 4.8.

b. Once p is obtained, p − RTc can be sketched as indicated in Figure 4.8.

c. The osmotic pressure on side 1 is RTC1Σ = 2 atmospheres. An atmosphere is 105

Pa, so that

C1Σ =

2× 105

(8.314× 106)(300)mol/cm3 = 80 µmol/cm3

c. The volume flux is from side 2 to side 1 for both cases, but is larger for case (2)since the difference in p − RTCΣ between side 2 and side 1 is 1 for case (1) and 2for case (2). Since the membrane is the same in both cases, the flux is determinedby the difference in p − RTCΣ on the two sides of the membrane only.

Problem 4.15 NaCl solution is an isotonic solution. Assume that NaCl is fully ionizedinto Na+ and Cl−. Therefore, the extracellular osmolarity is CoΣ = 0.3 osmoles/L. Sincethe cell neither shrinks nor swells in this solution, the intracellular osmolarity must alsobe CiΣ = 0.3 osmoles/L.

a. The extracellular osmolarity for the 150 mmol/L glucose solution isCoΣ = 0.15 osmoles/Lso that the extracellular osmolarity is less than the intracellular osmolarity and wa-ter is transported into the cell to achieve osmotic equilibrium. Therefore, the cellswells.

PROBLEMS 59

b. The extracellular osmolarity for the 300 mmol/L glucose solution isCoΣ = 0.3 osmoles/Lso that the extracellular osmolarity equals the intracellular osmolarity. Hence,there is no water transport across the membrane and the cell volume remains thesame.

Problem 4.16

a. i. Because the concentration of solute in Chamber 1 is much larger than thetotal solute concentration in the body fluids, the osmotic pressure π = RTC1

Σis larger than the external osmotic pressure. Because of the properties ofthe piston and drug-delivery orifice, no hydraulic pressure differences exists.

Flux of water into Chamber 1 is ΦV = LVRT(C1Σ − CbfΣ ) ≈ LVRTC1

Σ . As waterflows into Chamber 1, the piston moves to the right and the drug solution isforced out of the orifice at the same volume flow rate.

ii. Chemical energy is stored in the high solute concentration in Chamber 1. Thisstore of energy is dissipated as water flows in and dilutes the solution.

iii. If it is assumed that all the solute in Chamber 1 is in solution initially, thenthe concentration will decrease as water flows in. The concentration will notchange much if ΦV · A · T V1. Thus, for a given volume flow ΦVA, T V1/(ΦVA). Alternatively, if enough undissolved solute is in Chamber 1 tokeep the solution saturated until all the drug is delivered, then the pumpcould operate at a constant rate throughout its life.

b. The volume flux is LVRTC1Σ , and ΦVA = 1 µL/h. Therefore,

LV = ΦVARTC1

ΣA

= 1 µL/h · 10−9 m3/µL · 1/3600 h/s

π · (0.007/2)2 m2 · 8.314 N ·m/(K·mol) · 300 K · 10 mol/L · 103 L/m3

= 2.9× 10−16 m3/(N·s)

The principle of the osmotic pump is the basis of commercially available systems fordrug delivery (e.g., Figure 4.9).

Problem 4.17

a. The equation of water volume conservation is

− 1A(t)

dVc(t)dt

= LVRT(CoΣ −

NiΣVc(t)

).

If the surface areas of the circular ends of the cell are ignored, then the surfacearea of the cell is A(t) = 2πr(t)l and its volume is Vc(t) = πr2(t)l. Substitutionof these relations into the conservation relation yields

− 12πr(t)l

d(πr2(t)l)dt

= LVRT(CoΣ −

NiΣπr2(t)l

),


Figure 4.9: Dimensions (left panel) and schematic diagram (right panel) of commercially avail-able osmotic pumps (Problem 4.16). These Alzet osmotic pumps are made by ALZA Corpora-tion. The osmotic agent causes osmotic water transport into the pump through the semiper-meable membrane. Water entry compresses the impermeable reservoir wall thus forcing drugout of the reservoir through the delivery portal.

−dr(t)dt

= LVRT(CoΣ −

NiΣπr2(t)l

),

dr(t)dt

− LVRTNiΣ/(πl)

r2(t)= −LVRTCoΣ .

The last equation has the form

dr(t)dt

+ Ar2(t)

= B.

b. The constants are

A = −LVRTNiΣ

πl,

B = −LVRTCoΣ .

Problem 4.18 As shown in Section 4.7 (Weiss, 1996a), the volume of water in a cellsatisfies the differential equation

dν(t)dt

= ανA(t)(

1ν(t)

− c(t)),

where all the variables are normalized. The shape of the cell is expressed by the shapefactorA(t) which is the cell surface area. To solve the differential equation,A(t)must

PROBLEMS 61

2 4 6 8 10

0.5

1

1.5

2

1.2

1.4

1.6

1.8

2

Constant

Spherical

ανt

surface areaν(t)

c(t)

Cylindrical

Figure 4.10: Comparison of the volume re-sponse of cells to a reduction in the extra-cellular osmolarity by 1/2 for a cell withconstant surface area, a spherical cell, anda cylindrical cell (Problem 4.18).

be expressed in terms of the volume ν(t). This relation depends upon the shape of thecell. For a constant surface area A(t) = 1, for a spherical cell A(t) = ν2/3(t), and fora long cylindrical cell (for which the end caps provide negligible area)A(t) = ν1/2(t).

a. The response to halving the osmolarity is an increase in the cell water volume,but the rate of increase in ν(t) is proportional to A(t). For a constant cell areaA(t) = 1 and for a spherical cell A(t) = ν2/3(t) which is greater than 1 whenthe cell swells as it does when the osmolarity is reduced. Therefore, the rate ofchange is larger for the spherical cell than for the cell with constant surface area.This result is shown in Figure 4.51 (Weiss, 1996a).

b. For a cylindrical cell,A(t) = ν1/2(t) which for ν(t) > 1 lies between the value fora constant surface area cell and a spherical cell. Therefore, the rate of change ofvolume should be larger in a cylindrical cell than in a cell with constant surfacearea and should be smaller than in a spherical cell. This is born out in the solutionsto the three sets of equations shown in Figure 4.10.

Problem 4.19

a. If the salts are assumed to dissociate completely so that each mole of NaCl givesrise to 2 moles of ions and that each mole of CaCl2 gives rise to 3 moles of ions, thenthe osmolarities of the 4 solutions are: NaCl, 300 mosm/L; CaCl2, 600 mosm/L;sucrose, 800 mosm/L; xylose, 150 mosm/L. The order of increasing radius is in theorder of decreasing osmolarity. Therefore, solution 1 must be sucrose, solution 2must be CaCl2, solution 3 must be NaCl, and solution 4 must be xylose. It follows


that the isotonic radius is 80 µm, so that solution 3 must be an isotonic solutionwhich shows that isotonic concentration is 300 mosm/L.

b. At equilibrium,

Vc = 43πr3

c =NiΣCoΣ+V′c ,

where rc is the radius of the cell. Thus, this equation is solved for two of thesolutions as follows,

43π(99.1× 10−4)3 = NiΣ

150× 10−6 +V′c ,43π(60.7× 10−4)3 = NiΣ

800× 10−6 +V′c .

To findNiΣ, subtract these two equations and solve to obtainNiΣ = 5.8×10−10 mol.

c. For an isotonic solution

Vnc =NiΣConΣ

+V′c ,

which can be expressed as follows

V′cVnc = 1− NiΣ

ConΣ Vnc= 1− 5.8× 10−10

(300× 10−6)(4/3)π(80× 10−4)3≈ 1

10.

Problem 4.20

a. The total solute concentration is

CiΣ(t) =NiΣ

Vc(t)−V′c .

Since, NiΣ is constant

CiΣ(0) =NiΣ

Vc(0)−V′c .

Dividing these two equations yields

CiΣ(t)CiΣ(0)

= Vc(0)−V′c

Vc(t)−V′c .

Since the initial extracellular concentration is isotonic, CiΣ(0) = CoΣ(0−).b. The efflux of water from the cell is

ΦV = LVRT(CoΣ(t)− CiΣ(t))

From conservation of water volume,

− 1A(t)

dV i(t)dt

= ΦV (t) = LVRT(CoΣ(t)− CiΣ(t)

).

PROBLEMS 63

J

J

J

J

J

J

J JJ

0 2 4 6 8 10 12Time (minutes)

Arbacia

Vol

um

e (µ

m3 )

2.0 × 105

2.6 × 105

3.2 × 105

Figure 4.11: Measurement of initial slope and finalvalue of volume (Problem 4.20).

But dV i(t)/dt = dVc(t)/dt and for a sphere A = (36π)1/3V 2/3c . These equations

are combined and rearranged to yield

dVc(t)dt

= (36π)1/3V 2/3c (t)LVRT

(CiΣ(0)

Vc(0)−V′cVc(t)−V′c − C

oΣ(t)

).

This is a nonlinear equation for the cell volumeVc(t) given the extracellular soluteconcentration CoΣ(t) and the constants LV and V′c .

c. At t = 0+dVc(t)dt

∣∣∣∣t=0+

= (36π)1/3V 2/3c (0+)LVRT

(CoΣ(0−)− CoΣ(0+)

).

Therefore, LV can be computed from the initial rate of change of cell volume,given the temperature, and the change in osmolarity of the extracellular solution(Figure 4.11). As t →∞, dVc(t)/dt → 0 and

CoΣ(0−)Vc(0)−V′cVc(∞)−V′c − C

oΣ(0+) = 0.

Therefore, V′c can be computed by knowing the initial and final volume of the celland the initial and final value of the extracellular concentration.

Problem 4.21 WThe particular measurements given in the problem are used to estimatethe pore radii using the theoretical results summarized in Figure 4.11 (Weiss, 1996a).All of the parts of the problem involve estimation of the quantity Pw/Pw .

a. First, estimate Pw = RTLV/Vw as follows

Pw = RTLVVw

=

8.314× 106 PaK·(mol/cm3

) · (300 K) ·

(1.4× 10−6 µm/s

)18 mol/cm3 ,

= 194 µm/s.

Therefore, the ratio Pw/Pw = 194/24 = 8.1. Using the theory without hindrance,the pore radius can be estimated from Figure 4.11 (Weiss, 1996a) as r = 1.0 nm.


b. Using the theory with hindrance, the pore radius can be estimated from Figure 4.11(Weiss, 1996a) as r = 0.8 nm.

c. Measurements in erythrocytes suggest that water permeates the membrane bydissolving and then diffusing through the lipid bilayer and via water channels.If mercury compound are assumed to block only the water channels, then boththe osmotic and the diffusive permeabilities through the water channels can beestimated more accurately so that

PwPw =

194− 1224− 12

= 15.2,

which gives a pore radius of r = 1.2 nm.

d. If the results of this problem are assumed to be applicable to erythrocytes, then itfollows that the pore radius is of the order of 1 nm, i.e., of molecular dimensions.Therefore, this radius is clearly large enough to admit water molecules whosediameters are of the order of 0.2 nm. However, there are caveats to be madefor estimates of pore radius that rely on measurements of Pw/Pw as well as ontheories of hindered transport in porous membranes. First, estimates of Pw/Pwbased on measurements vary appreciably across different studies as is made clearby comparing the results cited in this problem with those given in Table 4.2. Inaddition, a variety of theories of hindered transport have been proposed so thatestimating pore radii involves choosing among a number of competing theoriesmost of which are derived from continuum hydrodynamic models whose validityat molecular dimensions is not known for certain. These types of estimates can beput on a firm foundation only if measurements are available for porous membraneswith nm radii pores of known dimensions.

Chapter 5

CONCURRENT SOLUTE AND SOLVENTTRANSPORT

Exercises

Exercise 5.1 For thermodynamic equilibrium, both the solute and water must be inequilibrium. Since the solute is permeant, diffusive equilibrium requires that c1

j = c2j .

Therefore, if c1j 6= c2

j then thermodynamic equilibrium is not possible.

Exercise 5.2 Since the solute is impermeant, the quantity of intracellular solute remainsconstant at its isotonic value. Therefore, the normalized intracellular quantity n(t) = 1.Since the extracellular concentration of solute is doubled during the pulse, the volumeof water in the cell is halved. Therefore, the volume normalized to its isotonic valueν(t) = 0.5.

Exercise 5.3 A small value of α corresponds to solute that is relatively impermeant.That is why there is a relatively small change in n(t) and a relatively large change inν(t). If the solute were impermeant then doubling the extracellular concentration wouldhalve the volume of water in the cell. The case for a solute with α = 0.1 approachesthat of an impermeant solute.

Exercise 5.4 Note that for both ν(t) and n(t), the time course of the onset of the re-sponse differs from the offset. This is most easily seen in ν(t) in which reduction involume in response to the onset of the change in osmolarity is faster than the increasein volume at the offset of the change in osmolarity. These differences in time coursedo not occur for linear differential equations with constant coefficients.

Exercise 5.5 Since the solute is impermeant, the quantity of impermeant solute doesnot change. Therefore, as the volume of the cell decreases, the intracellular concentra-tion of solute increases. At equilibrium, doubling the extracellular osmolarity halvesthe volume and, therefore, doubles the concentration. These trends are apparent in theresults shown in Figure 5.1.

Exercise 5.6 The membrane is more permeable to solute A than to solute B. Therefore,there is a larger transfer of solute A than B and a smaller transfer of volume for solute

65

66 CHAPTER 5. CONCURRENT SOLUTE AND SOLVENT TRANSPORT

0.5

1

1.5

2

−1 1 2 3 4 5 60Normalized time,

2.0

1.0n(t)

2.0

1.0

ανt

ci(t)

co(t)

0.6

0.7

0.8

0.9

-1 1 2 3 4 5 60

0.5

ν(t)

Figure 5.1: Volume, intracellular solute content,and intracellular solute concentration in responseto doubling the extracellular osmolarity (Exer-cise 5.5).

EXERCISES 67

p1 p2

c1j c2j

ΦV

φj

Figure 5.2: Reference directions for fluxes (Exer-cise 5.9).

A than B in order to achieve equilibrium. Therefore, traces 1 and 3 correspond to soluteA and traces 2 and 4 correspond to solute B.

Exercise 5.7 Figures 4.21 and 4.27 (Weiss, 1996a) show that changing solution osmo-larity for the sea water that is extracellular to these invertebrate egg cells results in achange in cell volume that is maintained for a matter of several minutes. Thus, these eggcells are relatively impermeant to the solutes in sea water. However, the volume changesshown in Figure 5.6 (Weiss, 1996a), which result from a change in osmolarity caused bythe addition of ethylene glycol to isotonic sea water, show a transient response thattakes several hundred ms. Putting these two measurements together suggests that themembrane of these eggs cells are much more permeant to ethylene glycol than to thesolutes in sea water.

Exercise 5.8 To convect means to carry. Hence, in solute convection, the solute iscarried by the solvent. Motion of the solvent carries solute along and results in solutetransport. In solute diffusion, solute is transported relative to the solvent due to asolute concentration gradient.

Exercise 5.9

a. The reference direction forΦV must be positive from side 1 to side 2. This directionensures that when the ∆p > 0, ΦV > 0 (Figure 5.2). The solute flux is positive when∆cj > 0. Therefore, the reference direction for φj is the same as for ΦV .

b. To obtain no solute flux requires Pj = 0 and σj = 1. To obtain a non-zero osmotictransport requires LV > 0.

c. Combining equations results in

φj = −Cj(1− σj)σjLVRT∆cj + Pj∆cj = (Pj − Cj(1− σj)σjLVRT)∆cj.

Therefore, a negative slope to the relation between φj and ∆cj can occur if Pj <Cj(1−σj)σjLVRT . Thus, the result is not inconsistent with the Kedem-Katchalskyequations.


Problems

Problem 5.1 The normalized equations of solute and solvent transport of a permeantsolute in the presence of an impermeant solute are

dn(t)dt

= αnA(t)(c(t)− n(t)

ν(t)

),

dν(t)dt

= ανA(t)(n(t)+ n(t)ν(t)

−(c(t)+ c(t)

)).

a. In the presence of solute A only, the cell is in diffusive and osmotic equilibrium.With the substitution of solute B, the cell is initially at osmotic equilibrium, but notat diffusive equilibrium because there is a difference in the concentration of thepermeant solute B across the membrane. Therefore, B diffuses into the cell andincreases the intracellular osmolarity so that water enters and the cell swells. Thiscan be seen from the normalized equations. During the interval 0 < ανt < 25,c(t) = c(t) = 1/2. Because the cell was in an isotonic impermeant solution Afor t < 0, the quantity of impermeant intracellular solute is the isotonic quantity.That is, n(t) = 1 for t < 0. Since this solute is impermeant, n(t) = 1 for t > 0.Therefore, the equations become

dn(t)dt

= αnA(t)(

12− n(t)ν(t)

),

dν(t)dt

= ανA(t)(n(t)+ 1ν(t)

− 1),

with initial values of n(0) = 0 and ν(0) = 1. Therefore, initially dn(t)/dt > 0 andsolute B enters the cell. Furthermore, as n(t) increases dν(t)/dt also increasesso that water enters and the cell swells.

b. & c. At diffusive equilibrium, the concentration of permeant solute must be the sameon the two sides of the membrane. Therefore, since the permeant, normalized ex-tracellular concentration is 1/2 that means the permeant, normalized intracellularconcentration must also be 1/2. At osmotic equilibrium, the total solute concen-trations in the cell must equal that of the extracellular concentration. The total,normalized extracellular concentration is 1 which guarantees that the total intra-cellular concentration is also 1. Thus, the impermeant and permeant intracellularsolutes have concentration 1/2. Since the concentrations are equal, the quantitiesmust be equal. Therefore, since the impermeant quantity is 1, the permeant quan-tity must be 1. Thus, the total normalized quantity of intracellular solute is 2 andthe normalized concentration is 1. Therefore, the normalized volume is 2. Thisresult can also be gotten directly from the equations. At diffusive and osmoticequilibrium, dn(t)/dt = 0 and dν(t)/dt = 0. Therefore, substitution of thesevalues into the normalized equations at equilibrium yields

0 = 12− n(∞)ν(∞) ,

0 = n(∞)+ 1ν(∞) − 1,

from which n(∞) = 1 and ν(∞) = 2.

PROBLEMS 69

Problem 5.2 The normalized equations of solute and solvent transport of a permeantsolute in the presence of an impermeant solute are

dn(t)dt

= αna(t)(c(t)− n(t)

ν(t)

),

dν(t)dt

= ανa(t)(n(t)+ n(t)ν(t)

−(c(t)+ c(t)

)).

a. If osmotic equilibrium is assumed for each instant in time then dν(t)/dt = 0, andthe solution to the second equation is

ν(t) = n(t)+ n(t)c(t)+ c(t) .

b. Substitution of the solution for ν(t) (from part a) into the first of the coupled-flowequations yields, after some rearrangement of terms,

dn(t)dt

= αna(t)(c(t)n(t)− c(t)n(t)

n(t)+ n(t)).

c. Since the cell has constant surface area, a(t) = 1. Thus, the solute diffusionequation becomes

dn(t)dt

= αn(c(t)N − Cn(t)n(t)+ N

),

where c(t) = Cu(t), and where u(t) is the unit step function. Since C is theisotonic extracellular concentration, N/C = 1.

i. At t = 0+, n(0+) = 0. Therefore,

dn(t)dt

∣∣∣∣0+= αn

(CN − C · 0

0+ N

)= αnC,

ii. As t → ∞, dn(t)/dt → 0, and the equation derived in part b applies whichfor the conditions of part c yields

0 =(CN − Cn(∞)n(∞)+ N

),

so that n(∞) = N.

iii. For t < 0, n(t) = 0 and ν(t) = 1. For t > 0

dn(t)dt

= αnC(N −n(t)N +n(t)

),

and

ν(t) = n(t)+ N2C

.

These equations can be rearranged to yield

dn(t)/Ndt

= αn(

1− (n(t)/N)1+ (n(t)/N)

),


0.2

0.4

0.6

0.8

1

2 4 6 8

0.2

0.4

0.6

0.8

1

n(t)/N

ν(t)

αnt

Figure 5.3: The normalized quantity of intracellularpermeant solute (n(t)/N) and of normalized volume(ν(t)) are plotted versus normalized time (αnt) (Prob-lem 5.2).

and

ν(t) = 1+ (n(t)/N)2

.

The solutions can be obtained by noting that the differential equations fordn(t)/dt is separable,∫ (

1+ (n(t)/N)1− (n(t)/N)

)dn(t)/N =

∫αndt.

This equation can be integrated to yield

−(n(t)/N)− 2 ln(1− (n(t)/N)

)= αnt.

The arbitrary constant of integration is clearly zero since this equation issatisfied by the initial condition that n(0) = 0. The solutions for both n(t)/Nand ν(t) are shown plotted versus αnt in Figure 5.3. The solutions show thatdoubling the extracellular osmolarity reduces the cell water volume to 1/2its value. Because the osmolarity was increased with a permeant solute, thesolute diffuses into the cell. An increase in intracellular solute concentrationraises the intracellular osmolarity so that water enters the cell to increase itsvolume. The volume increases until sufficient solute has entered the cell toachieve both diffusive and osmotic equilibrium. The time course of entry ofpermeant solute is the same as the time course of water entry. This is the basisof the method for estimating the solute permeability from measurements ofvolume changes.

d. Using the results of part c.iii, the normalized volume of the cell is

−(2ν(t)− 1)− 2 ln(1− (2ν(t)− 1)

)= αnt.

This solution shows that the time course of the volume is not exponential, butdoes depend upon αn. Thus, αn can be estimated by the following procedure.

PROBLEMS 71

• Measure the volume of the cell as a function of time in response to the changein osmolarity caused by the permeant solute.

• Estimate the volume of the cell that is not water and subtract that from thecell volume to give the water volume. The non-water volume of the cell can beobtained from measurements of the relation between the equilibrium volumeof the cell as a function of the concentration of an impermeant solute.

• Divide the water volume by the isotonic water volume to obtain the normal-ized water volume.

• Compute the quantity −(2ν(t)−1)−2 ln(1− (2ν(t)− 1)

)from the normal-

ized volume.

• Plot this quantity versus time.

• Estimate the slope of the line, e.g., by means of linear regression analysis.The slope of the line equals αn.

• Since αn = PjAn/V in, in order to estimate the permeability of the solute,Pj , from these measurements, the isotonic surface area, An, and the isotonicvolume, V in, need to be measured as well.

Problem 5.3

a. The system is not in diffusive equilibrium, because at t = 0, there is a concentra-tion difference of solute a across the membrane separating compartments 1 and2 and that membrane is permeable to a.

b. The only membrane permeable to water is between compartments 2 and 3. Theosmolarities of the solutions in these two compartments are equal (10 mosm/L)at t = 0. Hence, the system is initially in osmotic equilibrium.

c. Because compartment 1 is so much larger than compartment 2, the composition ofcompartment 1 does not change. Therefore, c1

a(∞) = 10 mmol/L and c1b(∞) = 0.

Since the membrane separating compartment 1 and 2 is permeable to a and sincea is not in equilibrium at t = 0, a will diffuse into compartment 2 until its concen-tration is the same as that of compartment 1, i.e., c2

a(∞) = 10 mmol/L. However,this raises the osmolarity of compartment 2 with respect to compartments 1 and 3.No water flows between compartment 1 and 2, but water does flow from compart-ment 3 to compartment 2 to establish osmotic equilibrium. Osmotic equilibriumoccurs when the osmolarity is the same on the two sides of the membrane thatseparates compartments 2 and 3. The final osmolarity satisfies the equation

10+ (10)(100)V2(∞) = (10)(100)

V3(∞) ,

where the first term on the left-hand side is the concentration of a in compartment2 at equilibrium and the second term is the concentration of b in compartment2. The right-hand term is the concentration of b in compartment 3. In addition,since water is transported only between compartments 2 and 3, water volumes of


those compartments are linked by the equationV2(∞)+V3(∞) = 200. Combiningthese equations to eliminate V3(∞) yields

10+ 1000V2(∞) =

1000200−V2(∞) .

Combining these terms and cross-multiply yields the quadratic equation

(V2(∞))2 − 2000 = 0,

which has one positive solution that equalsV2(∞) = 141 cm3. Therefore,V3(∞) =59 cm3. The concentrations in compartments 2 and 3 can now be computed fromthe volumes. c2

b(∞) = 1000/V2(∞) = 7.1 mmol/L. c3b(∞) = 1000/V3(∞) = 17.1

mmol/L. Since the membrane between compartments 2 and 3 is impermeable toa, c3

a(∞) = 0. Now check if these concentrations are in osmotic equilibrium. Thetotal concentrations in compartments 2 and 3 are: c2

a(∞)+c2b(∞) = 10+7.1 = 17.1

mmol/L and c3a(∞)+c3

b(∞) = 0+17.1 = 17.1 mmol/L. Compartments 2 and 3 arein osmotic equilibrium.

Problem 5.4

a. Answer: C. For an impermeant test solute the cell acts as a perfect osmometerand shrinks in response to an increase in osmolarity of the extracellular solution.Since the solute is impermeant it cannot enter the cell to allow water to flow backin. That is, the volume change is maintained as long as the increase in osmolarityis maintained.

b. Answer: A. This solute equilibrates so rapidly that there is no water flow at all.

c. It is impossible to determine the concentration for A, and difficult for B, but easyfor C. Since the concentration is the same for all three, it is sufficient to computethe concentration for solute C. Since the water volume was halved in response tothe pulse of test solute, cx/C = 1.

d. Answer: The permeability of test solute F is larger than the permeability of testsolute B. Because the volume change is smaller for F than B, less water was ex-changed to achieve equilibrium in F than in B. Hence, more solute must have beenexchanged to achieve equilibrium. Therefore, a smaller volume change for thesame concentration change implies a larger permeability.

e. Answer: E. The larger volume change implies a larger change in osmolarity, andtherefore, a larger value of Cx .

f. The time course of the onset is due to the flow of water through the membraneand is determined by cell geometric factors as well as the hydraulic conductivityof the membrane to water LV which is not varied in this experiment.

Problem 5.5

PROBLEMS 73

a. The total solute flux consists of solute flux by convection and by diffusion,

φj = cj(x)ΦV︸︷︷︸convection

−Ddcj(x)dx︸︷︷︸

diffusion

.

b. Since φj and ΦV are constant, the equation for cj(t) is a first-order, ordinary,linear, differential equations with constant coefficients which can be written as

dcj(x)dx

− ΦVDcj(x) = −

φjD.

The solution can be found from a particular solution plus the homogeneous solu-tion

cj(x) =φjΦV︸︷︷︸

particular

+ Ae(ΦV /D)x︸︷︷︸homogeneous

.

The boundary conditions are

cj(−∞) = C1 = φj/ΦVcj(0) = C0 = C1 +A.

Therefore, A = C0 − C1 and the total solution is

cj(x) = C1 + (C0 − C1)e(ΦV /D)x.

Solute convection and diffusion can be computed from the solute concentration,

cj(x)ΦV = ΦV(C1 + (C0 − C1)e(ΦV /D)x

)−Ddcj(x)

dx= −ΦV (C0 − C1)e(ΦV /D)x.

The results are plotted in Figure 5.4. In the presence of convection, the steady-statespatial distribution of concentration is exponential in space. Note also that while φjand ΦV are constant, both convection of solute cj(x)ΦV and the diffusion of solute−Ddcj(x)/dx are exponential functions of x.

Problem 5.6

a. The right-hand side of Equation 5.43 (Weiss, 1996a) shows that the outward fluxof D2O (considered as the solute) is φj = Pj(cij − coj ). The concentration of D2O

outside the cell exceeds that inside, cij − coj < 0, and Pj > 0, which implies thatφj < 0. That is, Equation 5.43 (Weiss, 1996a) predicts that there will be influx ofD2O. Since, in addition the experiment shows that the cell volume is constant, theleft-hand side of Equation 5.43 (Weiss, 1996a) yields

− 1A

d(cj(t)V i

)dt

= −Vi

Adcj(t)dt

< 0.


Convection

Diffusion

C0

C1

cj(x)

x−D/ΦV

C0ΦV

C1ΦV

−(C0−C1)ΦV

φj(x)Figure 5.4: A plot of the spatial depen-dence of concentration as well as both theconvection and the diffusion componentsof solute flux (Problem 5.5).

Therefore, cij(t) will increase with time.

The right-hand side of Equation 5.44 (Weiss, 1996a) implies that since LV , R, Tand coj − cij are all positive quantities that ΦV > 0, i.e., there is a flow of water outof the cell as a result of the higher osmotic pressure outside the cell due to D2O.But the left-hand side of Equation 5.44 implies that

− 1AdV i(t)dt

= ΦV > 0.

Therefore, V i(t) decreases with time as water flows out of the cell. This conclu-sion, derived from these uncoupled equations, is inconsistent with the experimen-tal findings.

b. & c. The Kedem-Katchalsky equations incorporate a modification of Equation 5.44 (Weiss,1996a) as follows,

− 1AdV i(t)dt

= ΦV = σLVRT(coj − cij).

Therefore, ΦV and consequently dV i(t)/dt can be zero even if coj −cij is non-zeroprovided σ = 0. The reflection coefficient σ is a measure of the ability of themembrane to distinguish solute from solvent. If the solute is indistinguishablefrom the solvent — a reasonable assumption for D2O and H2O — then σ = 0 andno osmotic flow occurs. Since there is no volume flow, ΦV = 0 and the convec-tive term of the Kedem-Katchalsky solute transport equation does not differ fromEquation 5.43 (Weiss, 1996a).

Problem 5.7

PROBLEMS 75

a. Substitute the equation for volume flux into the solute flux equations to yield

φj = Pj∆cj − Cj(1− σj)σjLVRT∆cj.Therefore,

φj∆cj

= Peff = Pj −LVRTCj(1− σj)σj.

b. Since LV , R, T , and Cj are all non-negative quantities and since 0 ≤ σj ≤ 1, theterm LVRTCj(1 − σj)σj is non-negative. Therefore, Peff ≤ Pj . The mechanismfor the difference between the permeability and the effective permeability resultsbecause the transport of solute is due to convection and diffusion. Note that if∆cj > 0, ΦV < 0. Therefore, the diffusive term in the solute transport producesa solute flux that is positive whereas the convective solute flow is negative. Theeffective permeability is the ratio of solute flux to concentration difference andthe solute flux is less than the diffusive solute flux alone.

Problem 5.8 In this problem it is assumed that the solute added extracellularly is dis-tinct from the intracellular solute(s). In addition, it is assumed that the volume of thecell that is not cell water is negligible. Note that solutes 1 and 2 cause a maintainedchange in the volume of the cell. Therefore, these solutes are impermeant (at least onthe time scale shown). Solute 3 causes no change in volume which could occur eitherif it were highly permeant or indistinguishable from water. Solutes 4-6 cause transientchanges in volume as would be the case if these solutes were permeant.

a. The addition of solute 2 results in a decrease in volume of 20% from the isotonicvolume. Therefore, the external concentration solute 2 was increased. At equilib-rium

V i = NiΣCoΣ,

which can be written in normalized form as

ν = 1c,

where ν = V i/V in and c = (CoΣV in)/NinΣ . Therefore, 0.8c = 1 so that c = 1.25.The concentration went from 200 mmol/L to 250 mmol/L which implies that 50mmol/L of solute 2 were added. The solution volume was 10 mL. If the additionof solute 2 is assumed not to change the volume appreciably then the number ofmoles of solute 2 added is 50 mmol/L ×10−2 L = 0.5 mmol.

b. False. When solute 3 is added, there is no change in the volume. This means thatosmotic equilibrium is maintained even though there is an increase in externalosmolarity. There are two possibilities for this to occur. Either solute 3 is highlypermeant and/or solute 3 is indistinguishable from water, i.e., σ3 = 0.

c. False. The return to equilibrium following the initial change in volume is due to theinflux of the permeant solute into the cell. The more permeant solute will enterthe cell more rapidly than the less permeant solute. Since the “time constant”for equilibration is shorter for solute 6 than for solute 4, the membrane is morepermeable to solute 6 than to solute 4 as shown in Figure 5.5.


1 2 3 4 5

1

0

1

0

1

0Solute 4

Solute 5

Solute 6

Normalized time

Nor

mal

ized

vol

um

e

Figure 5.5: Estimation of initial slope (dotted line)of the volume change (Problem 5.8).

d. True. The same reasoning as given in part c. The time constant for solute 5is shorter than for solute 4 as shown in Figure 5.5, so the membrane is morepermeable to solute 5.

e. True — sort of. Solute 3 causes no osmotic effect. Therefore, there are two possi-bilities. Either there is no osmotic effect at all and the solute is indistinguishablefrom water as far as the membrane is concerned. This occurs if σ3 = 0. Anotherpossibility is that the solute is highly permeant and causes a small osmotic effectwhich is too small to be seen on the scale of the figure.

f. True. The initial change in the cell volume is caused by the transport of waterout of the cell in response to the change in osmotic pressure. The membrane isimpermeable to solute 2 because there is no reswelling. The time course of trans-port of solute 5, as can be seen by the return of the volume to its isotonic value,is slow enough that it is reasonable to assume that the initial reduction in volumeoccurs before appreciable amounts of solute 5 have entered the cell. Therefore,the initial change in volume is caused predominantly by water flow without soluteflow. Since the initial change in volume in response to the two solutes is the same,the same change in osmolarity must have caused the responses.

g. True. Since the solute is impermeant no solute is transported across the mem-brane. Therefore, σ1 = 1.

h. True. Solute 4 is transported across the membrane because a change in osmolaritydue to this solute is not maintained. Therefore, P4 ≠ 0.

Problem 5.9

a. Since solute j is impermeant in membrane X and indistinguishable from water inmembrane Y , σx = 1 and σy = 0. Therefore, the volume flux equations for the

PROBLEMS 77

two types of membrane are

ΦVx = Lx(∆p − RT∆cj) since σx = 1,ΦVy = Ly∆p since σy = 0.

The total rate of volume transport is

2AΦV = AΦVx +AΦVy,

where A is the surface area of each type of membrane, so that

ΦV = ΦVx + ΦVy2

,

ΦV = (1/2)(Lx(∆p − RT∆cj)+Ly∆p),ΦV = (1/2)(Lx +Ly)∆p − (1/2)LxRT∆cj,

ΦV = Lx +Ly2

(∆p − Lx

Lx +Ly RT∆cj).

Therefore,

Lm = Lx +Ly2

,

σm = LxLx +Ly .

b. The solute flux equation for the two types of membrane are

φjx = 0 since σx = 1, and Pj = 0,φjy = CjΦV + Py∆cj since σy = 1.

As in part a, the total flux is the average of the two fluxes so that

φj =CjΦV

2+ Py

2∆cj.

Thus, Pm = Py/2.

Problem 5.10

a. Because the dialysate and blood are at the same osmotic pressure under the testconditions, ΦV = LV (pb − pd). Therefore, the slope of Figure 5.24 (Weiss, 1996a)equals LV which is

LV = 6× 10−4 cm/s15× 103 Pa

= 4× 10−8 cm/(s·Pa).

The flux of creatinine is given by the Kedem-Katchalsky equation as

φc = C(1− σ)ΦV + Pc(cb − cd),


where C is the average concentration of creatinine across the membrane. When∆p = pb − pd = 0, ΦV = 0 and φc = Pc(cb − cd). The molar concentrations ofcreatinine are

cb = 0.1 mg/cm3

105 mg/mol= 10−6 mol/cm3 and cd = 4× 10−6 mol/cm3.

Examination of Figure 5.24 (Weiss, 1996a) reveals

Pc = φccb − cd

∣∣∣∣ΦV=0

= −1× 10−10 mol/cm2·s(1− 4)× 10−6 mol/cm3 = 0.33× 10−4 cm/s.

Note that the slope of the relation of φc to ∆p = pb − pd is simply C(1 − σ)LVwhich can be solved for σ to obtain

σ = 1− slopeCLV .

The slope can be estimated from Figure 5.24 as

slope = 10× 10−10 + 1× 10−10 mol/(cm2·s)14× 103 Pa

= 7.9× 10−14 mol/(cm2·s·Pa).

Therefore,

σ = 1− 7.9× 10−14 mol/(cm2·s·Pa)

(2.5× 10−6 mol/cm3)(4× 10−8 cm/(s·Pa))= 0.21.

b. Assume that the same conditions apply as under test condition, then the volumeflux is

ΦV = LV (pb − pd) = (4× 10−8 cm/(s·Pa))(1.24× 104 Pa) = 4.96× 10−4 cm/s,

Thus, the total volume flowing through a 100 cm2 membrane in 6 hours is (4.96×10−4)(100)(60× 60× 6 s) = 1071 cm3 = 1.071 L. The flux of creatinine containsboth a convective and a diffusive term and can be estimated as follows

φc = (0.25× 10−5 mol/cm3)(1− 0.21)(4.96× 10−4 cm/s)+(0.33× 10−4 cm/s)(10−6 − 4× 10−6)(mol/cm3)

= 0.98× 10−9 − 0.1× 10−9 = 0.88× 10−9 mol/(cm2 · s).

Thus, in 6 hours the amount of creatinine removed from the blood is (0.88 ×10−9 mol/(cm2 · s))(100 cm2)(100 g/mol)((60× 60× 6 s) = 0.19 grams.

c. The dialysis machine removes the requisite amount of daily water in 6 hours butnot enough creatinine. A number of changes could be made to improve the per-formance. Pc could be reduced to increase the removal of creatinine but only byabout 10%. σ could be reduced, but this would only increase the solute flux byabout 20%. Increasing either the area of the dialysis membrane or LV by a factor of∼ 5 would increase the water flow by a factor of 5 and also increase the creatinineflow by about the same factor.

Chapter 6

CARRIER-MEDIATED TRANSPORT

Exercises

Exercise 6.1 Properties of carrier-mediated transport mechanisms that are distinctfrom transport by diffusion include:

• flux is facilitated,

• flux is highly structure specific,

• relation between flux and concentration saturates at high concentration,

• flux is inhibited by other solutes,

• flux is regulated by hormones.

Exercise 6.2 The rate of change of extracellular galactose at an initial intracellular galac-tose concentration of 100 mmol/L is the slope of the line for that concentration whichequals (34 mmol/L)/(17 s) = 2 mmol/(L·s). Therefore, the efflux through the membraneof one erythrocyte is

φG = 2× 10−3 mol/(L · s)(1.4× 1013 erythrocytes/L) · (1.37× 10−6 cm2/erythrocyte)

= 0.104 nmol/cm2·s.

Exercise 6.3 If there is a fixed density of carriers in the membrane and each carrier hasa flux that saturates at high solute concentration, then the total flux of solute from apopulation of such carriers will saturate at high solute concentration. The saturatedvalue of the flux will equal the maximum flux per carrier times the number of carriers.However, one can imagine hypothetical mechanisms that would yield saturation of theflux from a population of carriers where the individual carriers did not show saturationof flux. Suppose the number of carriers is not fixed but depends upon solute concen-tration. For example, imagine carriers whose flux/concentration relation follows Fick’slaw for membranes. However, the number of carriers is related inversely to the solute

79

80 CHAPTER 6. CARRIER-MEDIATED TRANSPORT

concentration so that the total flux saturates. In this scheme, the number of carriersis variable, none of the carriers saturates, but the total flux saturates. To be specific,let P(ciS − coS ) be the flux relation for a single carrier and let the number of carriers beN/(K + (ciS − coS )). With this scheme, the maximum flux is PN/K.

Exercise 6.4 This is a simple application of the equilibrium binding of an enzyme andsubstrate. The quantity of bound enzyme is given by

cES(∞) =(

cS(∞)K + cS(∞)

)CET .

Therefore,

cES(∞) =(

95+ 9

)2 = 1.29 mmol/L.

Therefore, the concentration of unbound enzyme is 2− 1.29 = 0.71 mmol/L.

Exercise 6.5 In a second-order binding reaction, an enzyme (E) binds to its substrateto form the bound complex (ES)

S + E z ES.

The dissociation constant for this reaction is

K = cS(∞)cE(∞)cES(∞) ,

where all the concentrations are at their equilibrium values. For a high affinity ligand,more of the enzyme is found bound to substrate which corresponds to a lower dissoci-ation constant (i.e., less of the enzyme is dissociated). Therefore, if KA > KB , then theenzyme has a higher affinity for ligand B.

Exercise 6.6

a. According to Equation 6.37 (Weiss, 1996a), the temperature factor for reactionrates at two different temperatures yields the rate at 20C as

κtα = 3(20−5)/10α = 5.2α,

where α is the rate at 5C. Thus, an increase in temperature from 5C to 20Cincreases the reaction rate by a factor of 5.2.

b. The factorA is proportional to the absolute temperature which changes from 278Kto 293K, a change of 5.4%. Hence, for a small change in temperature, such as15C, the factor A has a much smaller effect on the rate of reaction than does theexponential factor.

Exercise 6.7 In steady state, there is a flux of bound solute across the membrane butthe density of carrier in each of the four states is constant. To see how this can occur,consider the density of bound carrier at the inside surface of the membrane. In steadystate, the rate at which carrier binds to solute equals the rate at which bound carrier

EXERCISES 81

Tank

Figure 6.1: Steady state in a hydraulic system (Exer-cise 6.7).

facing the inside surface of the membrane translocates to bound carrier that faces theoutside surface of the membrane. Thus, there is no net change in the density of boundcarrier at the inside of the membrane even though net solute passes through this state.Similar arguments hold for the carrier in each of its four states.

The notion of a steady state is illustrated with a simple hydraulic analogy in Fig-ure 6.1. If the rate of water flow into the tank equals the rate at which water flows outof the tank then the water level in the tank will remain constant. Thus, net water flowsthrough the tank in steady state and the level of water in the tank is constant.

Exercise 6.8

a. Since α is zero, none of the enzyme can translocate to face the extracellular solu-tion. Therefore the densities of outward facing enzymes N

oES and N

oE are zero. The

inward facing densities partition in proportion to the intracellular concentrationof solute and the dissociation constant for the binding reaction. Therefore,

NiES =

ciSciS +K

NET and NiE =

KciS +K

NET .

Since the enzyme cannot translocate, the flux of solute φS is also zero.

b. The case β = 0 is similar to the case α = 0 except that the enzyme can not facethe intracellular solution. Therefore the densities of inward facing enzymes N

iES

and NiE are zero. The outward facing densities partition in proportion to the

extracellular concentration of solute and the dissociation constant for the bindingreaction. Therefore,

NoES =

coScoS +K

NET and NoE =

KcoS +K

NET .

Since the enzyme cannot translocate, the flux of solute φS is also zero.

c. If K = 0, the enzyme cannot dissociate. Therefore, if there is any extracellular orintracellular solute, it will bind to the enzyme and never unbind. Therefore theunbound densities N

iE and N

oE will be zero. The bound densities will partition by

the forward and reverse translocation rate constants, so that

NiES =

βα+ βNET and N

oES =

αα+ βNET .

Since the solute cannot unbind, there will be no transport, φS will be zero.


2 4 6 8 10

NoE/NET

NiE/NET

1

0.8

0.6

0.4

0.2

α/β

Figure 6.2: Density of carrier for a case when thesolute concentration is zero on both sides of themembrane (Exercise 6.9).

1→φS

1

ciS

1

coS

1←φS

Figure 6.3: Plot of sugar influx versus extracellular sugar concentration (right) and of sugarefflux versus intracellular sugar concentration (left) in double reciprocal coordinates (Exer-cise 6.10).

Exercise 6.9 For ciS = coS = 0 there is no carrier bound to enzyme. Therefore, on thisbasis and by inspection of Equations 6.55 and 6.57 (Weiss, 1996a) N

iES = N

oES = 0.

However, from Equations 6.56 and 6.58 (Weiss, 1996a) it follows that

NiE = β

α+ βNET = 1(α/β)+ 1

NET ,

NoE = α

α+ βNET = (α/β)(α/β)+ 1

NET .

These relations are plotted in Figure 6.2. If α/β = 1 then half the carrier is in the insideconfiguration and the other half is in the outside configuration. As α/β is increased,more of the carrier is found in the outside configuration, whereas as α/β is decreased,more of the carrier is found in the inside configuration

Exercise 6.10 For the simple, symmetric, four-state carrier model, the plot of influx ofsugar versus the extracellular concentration of sugar is identical to the plot of efflux ofsugar versus the intracellular concentration of sugar (Figure 6.3).

Exercise 6.11 As can be seen from Equation 6.111 (Weiss, 1996a) for the zero-trans

protocol, the relation between→φS and ciS is a rectangular hyperbola that is a straight

line when plotted in double reciprocal coordinates.

EXERCISES 83

Exercise 6.12 The relation between flux and concentration given by the simple, sym-metric, four-state carrier model is characterized by two macroscopic constants, K, and(φS)max . Hence, measurements of this relation alone can estimate these two parame-ters only. These two parameters are estimated most easily from the intercepts of plotsof flux versus concentration in double reciprocal coordinates. The macroscopic param-eter (φS)max can be expressed in terms of the microscopic parameters as follows,

(φS)max = αβα+ βNET ,

but none of the microscopic parameters can be determined uniquely. Unique determi-nation of the other parameters requires different types of experiments. For example,estimates of the density of carriers in the membrane, NET , can be obtained by bindingradioactive blockers that bind to the carrier and estimating the number of such blockersrequired. The individual rate constants must be obtained from experiments in whichthe state of the carrier can be determined, e.g., photometrically.

Exercise 6.13 Each of the graphs shows the reciprocal of flux plotted versus the recip-rocal of concentration.

a. A competitive inhibitor acts to increase the effective dissociation constant with-out changing the maximum flux. Thus, the only possibility is Graph 1 where theordinate intercept for both lines is the same which implies that the maximum fluxis the same. However, the intercept on the abscissa, which is the reciprocal of thedissociation constant, is larger for curve 2 than for curve 1. This implies that thedissociation constant is smaller in the presence of the competitive inhibitor. Thisis false. Thus, the answer for this part is NONE.

b. A non-competitive inhibitor acts to decrease the effective maximum flux withoutchanging the dissociation constant. Thus, the only possibilities are Graphs 4 and5. Since the ordinate intercept is the reciprocal of the maximum flux, the answeris Graph 4.

c. The simple, symmetric four-state carrier model predicts that the zero-trans effluxas a function of the intracellular concentration is the same as the zero-trans influxas a function of the extracellular concentration. Therefore, Graph 2 is the answer.

d. The general, four-state carrier model predicts that both the effective maximumflux and dissociation constant differ for influx and efflux, but the slope of therelation in reciprocal coordinates is the same. Therefore, Graph 3 is the answer.

Exercise 6.14

a. FALSE. Insulin receptors regulate glucose intake in hepatocytes, adipocytes andmuscle cells only.

b. TRUE. See part a.


c. TRUE. During fasting, the level of glucose in the blood drops and the secretion ofinsulin by β cells in the pancreas is decreased, while the level of glucagon releasedby the α cells in the pancreas is increased. These effects stimulate glycogenolysisin the liver which releases glucose into the circulatory system.

d. FALSE. Muscle cells do not secrete glucose but do secrete other products of gly-colysis, lactate and pyruvate, which can be recycled into glucose by hepatocytes.

e. FALSE. Insulin acts to recruit individual glucose transporters.

f. TRUE. The β cells respond to an increase in blood glucose by releasing insulin.

Exercise 6.15

a. Ans. (2). At equilibrium, a cell that transports the solute by diffusion will have anintracellular concentration of solute that is the same as the extracellular concen-tration.

b. Ans. (2). At equilibrium, a cell that transports the solute by a simple, symmetricfour-state carrier will have an intracellular concentration of solute that is the sameas the extracellular concentration.

c. Ans. (3). A cell that transports the solute by an active transport mechanism canconcentrate the solute intracellularly at a concentration that exceeds the extra-cellular concentration. Thus, (3) is only possible for an active transport system.However, without further specification of the system all three responses (1)-(3) are,in principle, possible for an active transport system.

Exercise 6.16

a. The flux of xylose can be computed directly from

φS = (φS)max

(ciS

ciS +K− coScoS +K

),

= αβα+ βNET

(ciS

ciS +K− coScoS +K

).

Since, ciS = 0

φS = 105/s× 105/s105/s+ 105/s

× 10−10 mol/m2 ×(− 1 mmol/L

1 mmol/L+ 10 mmol/L

)= − 1

22× 10−5 mol

m2 · s.

The rate of change of intracellular xylose can be computed from the flux of xyloseacross the cell membrane by use of the continuity equation,

dcS(t)dt

= −AφSV = −4πr2

43πr3

φS = −3rφS,

EXERCISES 85

where A represents the area of the membrane, V represents the volume of thecell, and r represents the radius of the cell. Therefore

dcS(t)dt

= − 310× 10−6 m

× 122× 10−5 mol

m2 · s= −0.14

molm3 · s

= −0.14mmolL · s

.

b. No.

i. Substituting K = 0 into the general expression for flux of xylose shows that

φS = αβα+ βNET

(ciS

ciS +K− coScoS +K

)

= αβα+ βNET

(ciSciS− c

oScoS

)

= αβα+ βNET (1− 1) = 0.

Thus K = 0 implies no flux of xylose.

ii. The only path for solute transport is to bind enzyme, translocate, and thenunbind on the other side. If the dissociation constant is zero, then the boundenzyme can never unbind. That is, solute will not be released by the boundcomplex. Therefore, if there is no unbinding there can be no transport.

Exercise 6.17

a. The plot shows the flux versus the concentration in double reciprocal coordinates.Thus, small values of concentration coS occur for large values of 1/coS . Maximum

influx←φ occurs when 1/

←φ is a minimum. It follows that the largest influx occurs

for condition 1.

b. Large values of concentration coS occur for small values of 1/coS . Therefore, forhigh concentration the flux is a maximum for condition 2.

c. No. Addition of substance A increases the flux of sucrose when the bath containshigh concentrations of sucrose (i.e. when coS > (1/0.015) mmol/L). Inhibitors(competitive or noncompetitive) always reduce flux. Thus, substance A cannot bea competitive inhibitor.

d. Yes. Addition of a non-competitive inhibitor decreases the equivalent maximumflux without changing the equivalent dissociation constant. Adding substance Bdecreases the maximum flux from (1/0.02) µmol/cm2·s to (1/0.05) µmol/cm2·sand does not change K, which is equal to (1/0.05) mmol/L for both conditions 1and 3. Thus, substance B acts as a non-competitive inhibitor.

Exercise 6.18 GLUT-1, GLUT-2, GLUT-3, etc. refer to distinct glucose transporter pro-teins that have been isolated from cellular membranes and then sequenced. Thesetransporters have distinct amino-acid sequences.


Exercise 6.19 Regions labeled 1-12 in Figure 6.41 (Weiss, 1996a) are all segments ofthe amino acid sequence of the glucose transporter GLUT-1 that are hydrophobic be-cause they have a large hydrophobicity. Each of these segments is 21 amino acids long.Twenty-one amino acids in the form of an α helix have the right length to span the cel-lular membrane. This argument suggests that these segments are the intramembraneportions of this glucose transporter.

Exercise 6.20 Insulin binds to insulin receptors on the surface of the muscle mem-brane. This binding results, through second messengers, in the recruitment of glucosetransporters that are inserted into the membrane to increase transmembrane glucosetransport.

Exercise 6.21 When the blood glucose concentration rises, glucose enters pancreaticβ-cells, and results in the secretion of insulin in the circulatory system. Insulin acts oninsulin receptors of cells and results in the increase in transport of glucose into thosecells thus lowering the blood glucose concentration.

Problems

Problem 6.1

a. The total concentration of binding sites in the membrane is C , so that cb(x, t) +cab(x, t) = C . The dissociation constant is K = ca(x, t)cb(x, t)/cab(x, t), so that(K + ca(x, t))cab(x, t) = ca(x, t)C and

cab(x, t) = ca(x, t)ca(x, t)+KC

b. When K ca(x, t), this equation in part a becomes cab(x, t) ≈ (C/K)ca(x, t), sothat α = C/K. The concentration of binding sites is inversely proportional to thedissociation constant. Therefore, most of the binding sites will be unbound whenthe dissociation constant is large.

c. Conservation of solute a can be expressed in terms of a continuity relation thatrelates the local rate of change of concentration of a to the flux divergence of aas follows,

∂(ca(x, t)+ cab(x, t))∂t

= −∂φa(x, t)∂x

.

The mobile solute obey’s Fick’s first law, so that

φa(x, t) = −D∂ca(x, t)∂x.

Combining the two equations yields

∂(ca(x, t)+ cab(x, t))∂t

= D∂2ca(x, t)∂x2 .

PROBLEMS 87

C

τeq

τss Figure 6.4: Plot of τeq and τss versus C . It has been as-sumed arbitrarily that at C = 0, τeq τss (Problem 6.1).

Using the assumption that ca K, allows substitution of cab(x, t) = (C/K)ca(x, t)which results in

∂ca(x, t)∂t

(1+ C

K

)= D∂

2ca(x, t)∂x2 .

Therefore,

Deff = D1+ (C/K)

The diffusion coefficient is reduced by the presence of binding sites because thebound solute does not diffuse.

d. The concentration in the membrane is assumed to be in the steady state. There-fore, the dependence of concentration on position in the membrane is linear, i.e.,

ca(x, t) = −φa(x, t)Dx + ca(0, t),

and

φa(x, t) = −Dd (ca(0, t)− ca(d, t))

With a membrane:solution partition coefficient of κa, the permeability is

Pa = Dκad .

D appears in the expression for the permeability rather thanDeff because the mem-brane is in steady-state, and therefore, the rate of change in the number of sitesbound to solute is zero. Under these conditions, this problem becomes identical tothe problem for a thin membrane with two compartments that was solved in Sec-tion 3.7 (Weiss, 1996a). Equation 3.59 (Weiss, 1996a) shows that τeq = Ve/(APa)so that

τeq =(V1V2

V1 + V2

)(d

ADka

)The equilibrium time constant is independent of C as shown in Figure 6.4.

e. τss is the time constant for the spatial distribution inside the membrane to reachsteady state. This time constant can be determined from the solution to the dif-fusion equation for the mobile solute in the membrane. As was found in part c,this diffusion equation contains an effective diffusion coefficient that is less thanthe diffusion coefficient in the absence of binding sites. With Deff as the diffusion


coefficient, this problem is similar to finding the time constant τss in Section 3.7(Weiss, 1996a). Therefore,

τss = d2

π2Deff= d2

π2D/(1+ C/K) =d2(K + C)π2DK

So τss is linearly related to C , and as C increases so does τss as shown in Figure 6.4.

f. If τeq τss , then the assumption that the concentration in the membrane is insteady state is valid for computing the time course of equilibration across themembrane. Since τss ∝ C , as C increases, so does τss . Therefore, for given bathdimensions, as C increases the assumption that τeq τss gets poorer and poorer(Figure 6.4).

The conceptually difficult part of this problem is that there are two diffusion coeffi-cients that apply: one is D, the diffusion coefficient of the mobile (unbound) solutewhich appears in Fick’s first law; the other is Deff, the effective diffusion coefficient thatoccurs in Fick’s second law. D applies when the concentration of bound solute is notchanging, i.e., in steady state. Deff applies when there is a time rate of change in the con-centration of bound solute. Therefore, D determines the equilibration time τeq underthe assumption that τeq τss which validates the steady-state assumption. However,during the time the concentration in the membrane is reaching steady state, there is atime rate of change of concentration of solute a governed by Deff.

Problem 6.2 The fluxes across the two barriers are φ1 and φ2,

φ1 = α1N1 − β1Nm,φ2 = α2Nm − β2N2.

a. Under steady-state condition φ1 = φ2 = φ so that

α1N1 − β1Nm = α2Nm − β2N2.

With the relations N1 = c1d and N2 = c2d, the solution for Nm is

Nm = α1N1 + β2N2

α2 + β1= d

(α1c1 + β2c2

α2 + β1

).

The flux is

φ = α1N1 − β1Nm,

φ = α1dc1 − β1d(α1c1 + β2c2

α2 + β1

),

φ = d(α1α2c1 − β1β2c2

α2 + β1

).

b. If the flux must be zero when the concentration is the same on the two sides of themembrane then α1α2 = β1β2. If this condition is satisfied then the flux satisfiesFick’s first law, and

P = d α1α2

α2 + β1= d β1β2

α2 + β1.

PROBLEMS 89

Sugar K (φS)max Sugar

1 20 40 12 10 20 I3 10 40 G4 20 80 B

Table 6.1: Transport parameters (Problem pb:badger).

c. According to the theory of absolute reaction rates

α1 = Ae(0−E1)/kT ,α2 = Ae(Em−E2)/kT ,β1 = Ae(Em−E1)/kT ,β2 = Ae(0−E2)/kT .

The constant A is the same for all rates since all rates are the same when theenergies are all zero. Therefore,

P = d Ae(Em−E1)/kTAe(0−E2)/kT

Ae(Em−E2)/kT +Ae(Em−E1)/kT.

This expression can be simplified to

P = d 1eE1/kT + eE2/kT

.

Therefore, the permeability depends only on E1 and E2 and not on Em. As thebarrier energies E1 and E2 are increased, the permeability is decreased.

Problem 6.3 For a simple, symmetric, four-state carrier mechanism

1←φs= 1(φs)max

(1+ K

cos

).

The intercept at 1/coS = 0 is 1/(φS)max ; the intercept for 1/←φs is−1/K. The parameters

determined from the graph are given in Table 6.1.

Problem 6.4 Conservation of soluteM , under the assumption that the cell volume doesnot change appreciably, yields

φs = −VAdci(t)dt

.

For the conditions that both co K and ci(t) K (the latter follows from the former),the flux is given by

φs ≈ (φs)max(ci(t)K

− co

K

).

These two equations can be combined to yield(VK

A(φs)max

)dci(t)dt

+ ci(t) = co.


The solution to this first-order, ordinary differential equation with constant coefficientsis

ci(t) = co(1− e−t/τ

),

since ci(0) = 0 and where

τ = VKA(φs)max

Therefore,

(φs)max = VKAτ =(10−10 cm3) · (10−4 mol/cm3)

(10−6 cm2) · (102 s)= 10−10 mol/(cm2 · s).

Problem 6.5

a. Both E and ES diffuse across the membrane with the same diffusion coefficient D.As shown in Section 3.5 (Weiss, 1996a), under steady-state conditions the spatialdependence of concentration is linear. Hence,

cE(x) = coE − ciEd

x + ciE,

cES(x) = coES − ciESd

x + ciES,

where x = 0 is the location of the inside and x = d is the location of the outsideinterface of the membrane and solution. These equation are linear in x and havethe correct boundary values. For example, for x = 0, cE(0) = ciE and x = d,cE(d) = coE . In addition, the flux is gotten by Fick’s law and is

φE = −DdcE(x)dx

= −DcoE − ciEd

φES = −DdcES(x)dx

= −DcoES − ciESd

.

Since, φE +φES = 0coE − ciEd

= −coES − ciESd

,

so that the slopes of the linear concentration dependencies have the same magni-tude but opposite signs.

Since N = cd, the equations can be expressed in terms of densities (N) instead ofconcentrations,

NE(x) = NoE −N

iE

dx +N

iE,

NES(x) = NoES −N

iES

dx +N

iES,

NoE −N

iE

d= −N

oES −N

iES

d.

The spatial dependencies of the carrier densities are shown in Figure 6.5.

PROBLEMS 91

0 d x

NiES

NoES

NiE

NoE

NES(x)

NE(x)

slope =N

oE −N

iE

d

slope =N

oES −N

iES

d

Figure 6.5: Spatial distribution ofdensity of bound and unboundcarrier as a function of position inthe membrane (Problem 6.5).

b. The total carrier densities are

NE = 1d

∫ d0

NE(x)dx and NES = 1d

∫ d0

NES(x)dx,

and are most easily obtained graphically from Figure 6.5.

NE = NoEd+ (1/2)(Ni

E −NoE)d

d= N

iE +N

oE

2.

Similarly,

NES = NoESd+ (1/2)(Ni

ES −NoES)d

d= N

iES +N

oES

2.

c. The flux of S is easily derived

φS = φES = −DcoES − ciESd

= Dd2 (N

iES −N

oES) = α(Ni

ES −NoES).

To proceed further, expressions for NiE and N

iES are required. But the total density

of carrier is

NET =NE +NES = NiE +N

oE

2+ N

iES +N

oES

2.

In addition, since the carrier remains in the membrane

φE +φES = 0 = α(NiE −N

oE)+α(Ni

ES −NoES),

which implies thatNiE −N

oE +N

iES −N

oES = 0,

andNiE +N

iES =N

oE +N

oES.

These equations are solved for NiES first. A combination of the two relations de-

rived thus far yields

NET = NiE +N

oE +N

iES +N

oES

2=N

iE +N

iES.


The interfacial reactions yield

coSNoE = KN

oES and ciSN

iE = KN

iES.

A combination of these equations yields

NET = KciSNiES +N

iES =

(KciS+ 1

)NiES,

from which

NiES =

(ciS

ciS +K

).

By a similar argument

NoES =

(coS

coS +K

).

Therefore,

φS = (φS)max(

ciSciS +K

− coScoS +K

),

where

(φS)max = αNET = Dd2 NET .

d. For ciS K and coS K,φS ≈ PS(ciS − coS ),

where

PS = αNET

K.

Problem 6.6

a. SinceφE+φES = 0, α(NiE−N

oE)+α(Ni

ES−NoES) = 0. Hence, No

ES+NoE =N

iES+N

iE .

Therefore,

NET =NoEI +

NiE +N

oE

2+ N

iES +N

oES

2=N

oEI +N

oE +N

oES.

Substituting for the equilibrium reactions at the interfaces yields

NET = coI NoE

KI+N

oE +N

oES,

=(

1+ coIKI

)NoE +N

oES,

=(

1+ coIKI

)NoESKScoS

+NoES,

=((

1+ coIKI

)KScoS+ 1

)NoES.

PROBLEMS 93

0 2 4 6 8 10

0.2

0.4

0.6

0.8

10123456

←φS

(φS)max

coSKS

coIKI

Figure 6.6: (Prob-lem 6.6).

Therefore,

NoES =

coS

coS +K((

1+ coIKI

)NET .

Since ciS = 0, NiES = 0 and

φS = −αNoES = −αNET

coS

coS +K((

1+ coIKI

) ,so that (φS)max = αNET . Hence,

←φS= (φS)max

coS

coS +K((

1+ coIKI

) .

b. A plot of←φS /(φS)max versus coS/KS for several values of coI /KI is shown in Fig-

ure 6.6.

c. An increase in coI decreases φS because I combines with carrier E and therebymakes E unavailable to bind with S and carry it across the membrane. That is, Icompetes with S for E.

Problem 6.7 From conservation of glucose

− 1Ad(VcG(t))

dt= φG(t),

where φG(t) is the outward flux of glucose, A is the surface area of the cell, and V isthe volume of the cell. Assume that the

• volume does not change;

• transport through the membrane is via a simple, symmetric, four-state carriermechanism; and

• the extracellular concentration of glucose is zero.


Therefore,dcG(t)dt

= − AV φM(

cG(t)cG(t)+K

).

Because the cell is spherical, A/V = 4πr2/((4/3)πr3) = 3/r . Hence,

dcG(t)dt

= −3φMr

(cG(t)

cG(t)+K).

• Experiment #1 — For cG K,

dcG(t)dt

≈ −3φMrK

cG(t),

ordcG(t)dt

+ 3φMrK

cG(t) = 0.

Therefore, τ = rK/(3φM) = 10 · 60 = 600 s.

• Experiment #2 — For cG K,

dcG(t)dt

≈ −3φMr= −10−8 mol/(cm3 · s),

Therefore,

φM = r3× 10−8 = 30× 10−4

3× 10−8 = 10−11 mol/(cm2 · s),

K = 600 · 3 ·φMr

= 600 · 3 ·φM30× 10−4 = 6× 10−6 mol/cm3.

Problem 6.8 Assume that the simple, four-state carrier model applies to this cell. There-fore, the efflux of sugar in the presence of no extracellular sugar is

→φS= (φS)max

ciSciS +Ki

,

which for ciS K gives→φS= (φS)maxKi

ciS.

This is similar to Fick’s law for membranes with a permeability PS = (φS)max/Ki. FromChapter 3 (Weiss, 1996a), the equilibration time τeq = Ve/(APS). Therefore,

τeq = VeKiA(φS)max

.

Hence, for larger Ki/(φS)max , τeq will be larger. From the plot it is apparent that thevalues of Ki and (φS)max are: for Sugar 1, 20 mmol/L and 20 µmol/cm2·s, with a ratioof 1; for Sugar 2, 100 mmol/L and 50 µmol/cm2·s with a ratio of 2; for Sugar 3, 20mmol/L and 50 µmol/cm2·s, with a ratio of 0.4. Thus, since D-glucose equilibratesmost rapidly it must be Sugar 3. Since D-xylose equilibrates most slowly, it must beSugar 2. The results are summarized in Table 6.2.

PROBLEMS 95

Sugar Ki (→φs)max

Solute # (mmol/L) (µmol/cm2 · s)

D-glucose 3 20 50D-galactose 1 20 20D-xylose 2 100 50

Table 6.2: Transport parameters and identification of Sugars 1, 2, and 3 (Problem 6.8).

Problem 6.9

a. Conservation of glucose yields

− 1A(t)

d(V (t)cG(t))dt

= φG = φm(

cG(t)cG(t)+K

),

where it is assumed that the external glucose concentration is zero. Because thecell is spherical

− 14πr2(t)

d((4/3)r3(t)cG(t))dt

= φm(

cG(t)cG(t)+K

).

If the change in osmolarity due to the change in glucose concentration in the cell issmall, then the cell radius does not change and the differential equation becomes

−r3dcG(t)dt

= φm(

cG(t)cG(t)+K

).

b. Rearrangement of this differential equation yields

dcG(t))dt

= −3φmr

(cG(t)

cG(t)+K).

Cross multiplication and integration yields∫ cG(t)cG(0)

(1+ K

c′G(t)

)dc′G(t) = −

3φmr

∫ t0dt′,

which results in

cG(t)− cG(0)+K ln(cG(t)cG(0)

)= −3φm

rt.

c. If K cG(t) then

K ln(cG(t)cG(0)

)≈ −3φm

rt,

orcG(t) ≈ cG(0)e−(3φm/(rK))t.

Note that for this condition, the relation between flux and concentration is lin-ear, and therefore, the differential equation is linear with constant coefficientsand gives rise to an exponentially decaying glucose concentration. Intuitively, ifthe rate of efflux of glucose is proportional to the intracellular concentration ofglucose then exponential kinetics result.


Slope Intercept(φS)max

→φS

KSciSKR

= 110

1+ KSciS= 3

(φS)max←φS

KScoSKR

= 120

1+ KScoS= 2

Table 6.3: Transport parameters (Prob-lem 6.10).

d. If K cG(t) then

cG(t)− cG(0) ≈ −3φmrt,

or

cG(t) ≈ cG(0)− 3φmrt.

In this limit, the relation between flux and concentration is a constant, the flux issimply the maximum glucose flux possible for the carrier. Hence, the intracellularglucose concentration drops linearly as a constant quantity of glucose flows outof the cell.

Problem 6.10 From the simple, symmetric, four-state carrier model

(φS)max→φS

= 1+ KSciS

(1+ c

iRKR

),

(φS)max←φS

= 1+ KScoS

(1+ c

oRKR

).

The parameters estimated from the graphs are shown in Table 6.3. We determine ifthese results are consistent with a set of parameters for the simple, symmetric, four-state carrier model. First, examine the implications of the slopes.

KS/(ciSKR)KS/(c0

SKR)= c

oS

ciS= 1/10

1/20= 2.

Then examine the implications of the intercepts,

1+ KSciS

= 3⇒ KSciS= 2,

1+ KScoS

= 2⇒ KScoS= 1.

Therefore, the slopes and intercepts are both consistent with coS/ciS = 2. The above

equations also show that KS = 2ciS and KR = 20. Thus, there is a consistent set of pa-rameters that fit the measurements using a simple, symmetric, four-state carrier model.

Problem 6.11

PROBLEMS 97

r (φG)max K(µm) (µmol/(cm2· s)) (mmol/L)

6 25 509 25 16 2/3

27 50 16 2/3

Table 6.4: Transport parameters (Prob-lem 6.12).

a. For steady-state conditionsφE+φES = 0. But, φE = 0 which implies thatφES = 0.But since the only way the solute crosses the membrane is in a complex with thecarrier, φS = 0.

b. In the steady-state φS = 0. Hence, there is no net flux of S across the membraneand the internal concentration cannot change. There is one caveat. If the concen-tration is changed transiently then there will be a transient change in state of thecarrier until a new steady state is achieved. During this interval a small amountof solute can be transported but this transport ceases as a new steady state isestablished.

c. Although the net flux in steady state is zero, the unidirectional fluxes are notzero although they are equal in magnitude. Therefore, there will be a net flux ofradioactively labeled S. Therefore, while the concentration of total S in the celldoes not change, the proportion of radioactive S will increase until the ratio ofradioactive to non-radioactive S is the same inside and outside the membrane.

d. Because solute can exchange across the membrane without a change in concentra-tion, this transport mechanism is called exchange diffusion.

Problem 6.12

a. At equilibrium ciG = coG = 0.01 mmol/L for all three types of cells.

b. The parameters of the simple, symmetric, four-state carrier mechanism can be es-timated directly from the data given. The intercepts for the abscissa and ordinateare −1/K and 1/(φG)max , respectively. The results are given in Table 6.4. Theflux of glucose is

φG(t) = (φG)max(

ciG(t)ciG(t)+K

− coG(t)coG(t)+K

).

But, for coG = 0.01 mmol/L which is much less than K for all three cell types,

φG ≈ (φG)maxK(ciG(t)− coG(t)).

Conservation of glucose, and the assumption that the change in volume of thecells is small, results in

−Vi

AdciG(t)dt

≈ (φG)maxK

(ciG(t)− coG(t)),



V iKA(φG)max

dciG(t)dt

+ ciG(t) = coG(t).

The solution isciG(t) = C

(1− e−t/τ

)for t ≥ 0,

where C is the bath concentration of glucose, and

τ = V iKA(φG)max

= rK3(φG)max

.

The time constants are

τ6 = (6× 10−4 cm) · (0.05mmol/cm3)3(2.5×mmol/cm2 · s)

= 0.4 ms

τ9 = (9× 10−4 cm) · (0.0167mmol/cm3)3(2.5×mmol/cm2 · s)

= 0.2 ms

τ27 = (27× 10−4 cm) · (0.0167mmol/cm3)3(5×mmol/cm2 · s)

= 0.3 ms

c. The cell with radius 9 µm equilibrates most rapidly and the cell with radius 6 µmequilibrates least rapidly.

Problem 6.13

a. The equivalent flux is the maximum flux of glucose and is obtained from the or-dinate intercept which is 1/φeq ≈ 0.018 L · min/mmol from which φeq ≈ 55mmol/L·min. The slope of the line fit to the points in the absence of cytochalasinis approximately (0.042−0.018)/(0.05−0) = 0.48 min. Since the slope isKeq/φeq,Keq = 26.4 mmol/L.

b. A similar calculation is used in part b as in part a. The ordinate intercept is thesame as in part a. Therefore, φeq ≈ 55 mmol/L·min. Since the slope of the linefit to the points is approximately (0.15−0.018)/(0.042−0) = 3.1 min, Keq = 170mmol/L.

c. The effect of the cytochalasin B is to increase Keq while φeq is the same with andwithout cytochalasin B. Thus, cytochalasin B, when applied extracellularly, acts asa competitive inhibitor to glucose efflux.

Problem 6.14

a. If the concentration of glucose in the enterocyte is constant then the net rate ofinflux of glucose must be zero. Therefore, Aaφa = Abφb.

PROBLEMS 99

b. Since the transport of glucose through the basolateral membrane can be repre-sented by a simple, symmetric, four-state carrier model, the relation of efflux toconcentration is

φb = φmax(

ciscis +K

− coscos +K

).

c. For cos = 0, combining expressions yields

Aaφa = Abφmax(

ciscis +K

).

To solve for cis , it is easiest to divide both sides of the equation by Abφmax , invertboth sides of the equation and solve for 1/cis and then take the reciprocal to obtain

cis = K(Aaφa/Abφmax)

1− (Aaφa/Abφmax).

d. i. Note when φa = 0, cis(∞) = 0. This is apparent from the results of part c andis consistent with the plot shown in the problem. The physical explanation isthat when the apical flux is zero, the basolateral flux is zero. Hence, for zeroglucose flux through the basolateral membrane, the intracellular concentra-tion must equal the extracellular concentration of glucose which is zero.

ii. In steady state, as the total rate of influx of glucose in the apical region Aaφaincreases, the efflux through the basolateral region increases to equal theinflux. A higher efflux in the basolateral membrane occurs at higher intracel-lular concentrations of glucose. As the efflux is driven toward saturation, asmall increment in efflux requires a large increment in intracellular glucoseconcentration. A steady-state solution exists provided the influx through theapical membrane is less than the maximum efflux through the basolateralmembrane Abφmax .

iii. When the influx exceeds the maximum efflux, i.e., whenAaφa > Abφmax , theenterocyte can no longer transport glucose out of the basolateral membranefast enough to match the rate of influx through the apical membrane. Hence,no steady state is possible for φa/φmax > Ab/Aa = 1/24. This accounts forthe unphysical result in this region, which is that the equations are satisfiedonly for negative values of the intracellular concentration.

Problem 6.15

a. The flux of the tracer is

φ∗S (t) = (φS)max(

C + ci∗S (t)C + ci∗S (t)+K

− C + co∗S (t)C + co∗S (t)+K

).

Evaluation of the first term on the right-hand side under the assumption thatci∗S (t) C yields

C + ci∗S (t)C + ci∗S (t)+K

= CC +K

1+ ci∗S (t)C

1+ ci∗S (t)C+K

,


≈ CC +K

(1+ c

i∗S (t)C

)(1− c

i∗S (t)C +K

),

≈ CC +K

(1+ c

i∗S (t)C

− ci∗S (t)C +K

),

≈ CC +K

(1+ K

C(C +K)ci∗S (t)

).

The second term in the flux equation has a similar form as the first,

C + co∗S (t)C + co∗S (t)+K

≈ CC +K

(1+ K

C(C +K)co∗S (t)

).

A combination of these expressions yields

φ∗S (t) ≈ (φS)max(

CC +K

)(1+ K

C(C +K)ci∗S (t)− 1− K

C(C +K)co∗S (t)

),

φ∗S (t) ≈ (φS)max(

K(C +K)2

)(ci∗S (t)− co∗S (t)).

Therefore,

γ = (φS)max(

K(C +K)2

).

b. Conservation of labeled solute yields

−Vi

Adci∗S (t)dt

= γ(ci∗S (t)− co∗S (t)).

The total quantity of labeled solute is co∗S (0)V o = N∗S so that conservation oflabeled solute also requires ci∗S (t)V i + co∗S (t)V o = N∗S so that

dci∗S (t)dt

+ γAV i(

1+ Vi

V o)ci∗S (t) =

γAN∗SV iV o .

Therefore,

α = (φS)maxAV i(

1+ Vi

V o)(

K(C +K)2

)and β = (φS)maxAV iV o

(K

(C +K)2)N∗S

c. The differential equation for ci∗S (t) is linear, ordinary, and has constant coeffi-cients. Hence, the solution has the form

ci∗S (t) = ci∗S (∞)(1− e−αt

),

where

ci∗S (∞) =βN∗Sα

= N∗SV i +V o .

Hence, the steady-state value of the concentration is just the total quantity ofradioactive sugar divided by the total volume of the system.

PROBLEMS 101

d. The steady-state value of the concentration of tracer gives no information aboutthe kinetic parameters but the rate constant α does. If all the parameters asidefrom K and (φS)max are known or can be determined independently, then mea-surements of α for different values of C allows estimation of the kinetic parame-ters.

e. One major advantage of the equilibrium exchange method is that there are no os-motic pressure differences created. Hence, there is no transport of water and thevolume of the cell does not change. In addition, the time dependence of concen-tration is an exponential time function making estimation of parameters simple.In contrast, the zero-trans integrated flux method requires much more complexanalysis methods with the possibilities of errors. One disadvantage of the equilib-rium exchange method is that a series of measurements have to made at differentvalues of C in order to estimate the kinetic parameters.

Problem 6.16

a. Transport of solute by diffusion satisfies the differential equation

− 1Ad(V iciS(t))

dt= φS = PS(ciS − C).

Since the volume of the cell is constant,

V iAPS

d(ciS(t))dt

+ ciS(t) = C.

If the initial concentration ciS(0) = 0, then the solution to the differential equationis

ciS(t) = C1(1− e−t/τ1) for t ≥ 0,

where C1 = C and τ1 = V i/APS = (4/3)πr3/(4πr2PS) = r/(3PS).b. From the binding reaction

K = ciSciM

ciSM,

and from conservation ofM , ciM+ciSM = CT . Combining these two equations yieldsKciSM = ciS(CT − ciSM) which can be solved to give

ciSM =ciS

K + ciSCT .

For K C it follows that K ciS since ciS ≤ C . Therefore,

ciSM ≈CTKciS.

Conservation of S yields

−Vi

Ad(ciS(t)+ ciSM(t))

dt= φS = PS(ciS − C),


System Solute concentration Time constantUnbound Total

No binding C Cr

3PSWith binding C C

(K + CTK

)r

3PS

(K + CTK

)

Table 6.5: The equilibrium value of intracellular solute concentration and the time constant ofequilibration are given for the system without and with binding (Problem 6.16).

where ciS(t)+ ciSM(t) is the total concentration of S in the cell both unbound andbound to M . But, ciS(t)+ ciSM(t) = (1+ (CT/K))ciS(t) Combining these equationsyields

V i(1+ (CT/K))APS

d(ciS(t))dt

+ ciS(t) = C.

If the initial concentration ciS(0) = 0, then the solution to the differential equationis

ciS(t) = C2(1− e−t/τ2) for t ≥ 0,

where C2 = C and τ2 = V i(1+ (CT/K))/APS == r(1+ (CT/K))/(3PS).c. The final value of the (unbound) intracellular solute concentration and the time

constant for equilibration can be estimated directly from measurements solutetransport for a cell. If only these quantities are obtained for a single known valueof C and if r is determined, it is still not possible to determine which of themechanisms described in parts a and b occur in the cell. For a single value of C , theresults in part b could have been obtained by pure diffusion with no binding, if thepermeability of the cell membrane had the value PS/(1+ (CT/K)). Since PS is notknow a priori, there is no way to tell from a single measurement which mechanismapplies. However, the mechanism could be determined by performing additionalexperiments. Table 6.5 summarizes the results for the two mechanisms. Thesetwo mechanisms can be distinguished by the following methods.

• Determine the effect of CT on τ . When there is no binding of solute to M ,the time constant is independent of the concentration of M in the cell. Con-versely, if there is binding to M , the time constant increases as the concen-tration of M increases because some of the solute that enters the cell by dif-fusion binds to M and is not available to equilibrate the intracellular soluteconcentration.

• Measure the total concentration of S in the cell. In the absence of binding thetotal quantity of S in the cell is

ciS(t) = C(1− e−t/τ1) for t ≥ 0.

At equilibrium, ciS(∞) = C . In contrast, in the presence of binding the totalquantity of S in the cell is the sum of the unbound and bound S

ciS(t)+ ciSM(t) =(

1+ CTK

)ciS(t) = C

(1+ CT

K

)(1− e−t/τ2) for t ≥ 0.

PROBLEMS 103

Therefore, at equilibrium, the total concentration of S in the cell is C(1 +(CT/K)) which exceeds that in the bath. The total concentration of S in thecell can be measured with a variety of methods, e.g., radioactive tracers, flamephotometry or electronprobe analysis.

Problem 6.17

a. True. This is a basic assumption of the simple, symmetric four-state carrier modeland guarantees that the carrier does not leave the membrane.

b. Ans. ii. Note K = ciSniE/niES . With niE = niES (as shown) K = ciS .c. True. States Eo and ESo do not change. The change from state (1) to state (3)

could have been achieved by changing ciS from ciS = K as in (1) to ciS = 0 as in (3).

d. False. Note that niES < noES which implies that niE/n

iES > 1 which implies that

K/ciS > 1 which implies that ciS < K. Also noES > noE , so that coS > K which implies

that coS > ciS which implies that φS < 0.

e. True. niE = niES which implies that ciS = K. noE = noES which implies that coS = K.Therefore, ciS = coS and φS = 0.

f. True. niES = 0 and niE > 0 implies that ciAniE/K = 0 which implies that for K

bounded ciA = 0.

g. False. Changing K changes both ciSniE/n

iES and coSn

oE/n

oES . But noE/n

oES doesn’t

change.

h. Ans. (4). niE = 0 implies that ciS K, and noES = 0 implies that coS = 0. In general,

φS = (φS)max(

ciSciS +K

− coScoS +K

).

The first term is maximized when ciS K and the second term is minimized whencoS = 0. For these conditions, φS = (φS)max .

Problem 6.18 The influxes of glucose into β-cells and erythrocytes are

φβ = (φG)max cGcG + 20

and φe = (φG)max cGcG + 0.5

,

where φβ is the influx of glucose into β-cells, φe is the influx of glucose into erythro-cytes, (φG)max is the maximum flux of the carrier, and cG is the glucose concentrationin the blood in mmol/L.

a. At cG = 5 mmol/L,

φβ = 525(φG)max and φe = 5

5.5(φG)max.


Hence,φβφe= 5/25

5/5.5= 0.22.

At this concentration of glucose, the relation of flux to concentration is near satu-ration for the erythrocyte but well below saturation for the β-cell. Hence, the ratiois much less than 1.

b. For the β-cell the percentage increase in flux is

(φβ)10 − (φβ)5(φβ)5

× 100 = 10/30− 5/255/25

× 100 = 66.7%,

whereas for the erythrocyte

(φe)10 − (φe)5(φe)5

× 100 = 10/10.5− 5/5.55/5.5

× 100 = 4.7%.

Because the flux relation is near saturation for the erythrocyte, doubling the bloodglucose concentration does not change the influx of glucose much for this cell.However, the same change in concentration causes a large change in the influx ofglucose into the β-cell because the flux relation is well below saturation for thiscell.

c. For the normal range of blood glucose, the erythrocyte glucose transporter is nearsaturation so that the influx of glucose into the cell is maximal. The erythrocyteincorporates as much glucose as is possible until glucose accumulates in the cellto reduce the influx. The erythrocyte is an example of cell that utilizes glucose forits metabolic needs. The β-cell glucose transporter is well below saturation at thenormal blood glucose concentration. Hence, a change in blood glucose leads to arelatively large change in glucose influx. This is important for proper functioningof the β-cell as a controller of blood glucose level. The β-cell releases insulin inthe blood stream, by a complex mechanism described elsewhere (Weiss, 1996a), inresponse to an increase in intracellular glucose. Thus, the sensitivity of the β-cellsystem depends upon the slope of the relation of glucose influx to blood glucoseconcentration which is a small when the transport system is near saturation.

Problem 6.19

a. First, consider ciS/→φS plotted versus ciS .

i. To find the relation, form the ratio

ciS→φS= ciS

(φS)maxciSciS+K

= 1(φS)max

(ciS +K)

Therefore a plot of ciS/→φS versus ciS is a straight line with slope 1/(φS)max

and intercepts at −K and K/(φS)max as shown in Figure 6.7.

PROBLEMS 105

ciS→φS

(φS)max

ciS−K

K

(φS)max

1

Figure 6.7: ciS/→φS plotted versus ciS (Prob-

lem 6.19).

ciS→φS

ciS−K

ciS→φS

ciS

(φS)max

1

Noncompetitive inhibition Competitive inhibition

Figure 6.8: Effect of increasing the concentration of an inhibitor is indicated by a dashed line(Problem 6.19).


ii. As the concentration of a noncompetitive inhibitor is increased, (φS)max de-creases (Table 6.2 (Weiss, 1996a)). In these coordinates, a change in concen-tration of a noncompetetive inhibitor corresponds to an increase in the slopeof the lines while the intercept on the abscissa remains fixed (Figure 6.8).

iii. As the concentration of a competitive inhibitor is increased, K increases (Ta-ble 6.2 (Weiss, 1996a)). In these coordinates, a change in concentration of acompetetive inhibitor corresponds to a shift of the lines parallel to the ab-scissa without a change in the slope of the lines (Figure 6.8).

b. Next, consider→φS plotted versus

→φS /ciS .

i. To find the relation, form the ratio

→φS /(φS)max = ciS

ciS +K,

→φS /(φS)max

ciS= 1

ciS +K.

A combination of these two relation yields

→φS /(φS)max = ciS

→φS /(φS)maxciS

.Next express ciS in terms of

→φS /((φS)maxciS), which is

ciS =1

→φS /((φS)maxciS)

−K,

which yields

→φS /(φS)max =

1→φS /((φS)maxciS)

−K→φS /(φS)max

ciS

.Simplifying this result yields

→φS= (φS)max −K

→φSciS

.Hence, this transformation results in a straight line whose slope is −K andwhose ordinate intercept is (φS)max as shown in Figure 6.9.

ii. As the concentration of a noncompetitive inhibitor is increased, (φS)max de-creases (Table 6.2 (Weiss, 1996a)). In these coordinates, a change in con-centration of a noncompetetive inhibitor corresponds to a decrease in theordinate intercept with the slope fixed (Figure 6.10).

iii. As the concentration of a competitive inhibitor is increased, K increases (Ta-ble 6.2 (Weiss, 1996a)). In these coordinates, a change in concentration ofa competetive inhibitor corresponds to an increase in the magnitude of theslope without changing the ordinate intercept (Figure 6.8).

PROBLEMS 107

1

(φS)max

−K

→φS

ciS

→φS

(φS)max

K

Figure 6.9:→φS plotted versus

→φS /ciS

(Problem 6.19).

1

−K

→φS

ciS

→φS

(φS)max

→φS

ciS

→φS

Noncompetitive inhibition Competitive inhibition

Figure 6.10: Effect of increasing the concentration of an inhibitor is indicated by a dashed line(Problem 6.19).


Problem 6.20

a. The total flux of solute is φS =→φS −

←φS where Equations 6.111 and 6.113 (Weiss,

1996a) give expressions for the unidirectional fluxes for the general, four-statecarrier model. At arbitrarily low concentrations, these two equations reduce to

→φS≈

ciSKgK2g/φz

= φzKgciS and

←φS≈

coSKgK2g/φz

= φzKgcoS .

Hence, at arbitrarily low concentrations,

φS ≈ φzKg (ciS − coS ),

and the permeability is

PS = φzKg =aiaoai + ao

(bihiho

aibiho + aigiho + aobihi

)NET .

This expression can be simplified by making use of the relationaobigohi = aibogihoto yield

PS = aobigohi(ai + ao)(bigo + gigo + bogi)NET .

b. Symmetry of the rate constants implies that ai = bi = α, ao = bo = β, go = gi,and ho = hi which results in

PS = βαgihi(α+ β)(αgi + g2

i + βgi)NET ,

= αβhi(α+ β)(α+ β+ gi)NET .

If the interfacial reactions are assumed more rapid than the translocation reactionthen gi α+ β and

PS = αβhigi(α+ β)NET .

If the interfacial binding reactions are so fast that they are assumed to be at equi-librium, then K = gi/hi and

PS = (φS)maxK,

where

(φS)max = αβα+ βNET .

Thus, the permeability of the general, four-state carrier model reduces to thatfor the simple, symmetric, four-state carrier model when the rate constants aresymmetric and the interfacial reactions are assumed to be fast compared to thetranslocation reactions.

Problem 6.21

PROBLEMS 109

a. In the steady state the rate of change of state density is zero and the followingequations arise. The first two equations represent the rate equations for N

iE and

NoE . The third equation represents conservation of carrier. −(ai + hiciS)ai

1

ao−(ao + hocoS )

1

gigo1

N

iE

NoE

NES

= 0

0NET

.The solutions are ratios of sums of products of rate constants

NiE

NET= M

iED,

NoE

NET= M

oED,NES

NET= MES

D,

where

MiE = aogo + aogi + gihocoS ,

MoE = aigi + aigo + gohiciS,

MES = aihocoS + aohiciS + hihociScoS ,D = M

iE +M

oE +MES.

b. Using the same method as used in the text (Weiss, 1996a), assume that some ofthe solute on the inside bath is marked S∗. Then the efflux can be written as

→φS= goNES∗

ciSciS∗.

In the steady state

dNES∗

dt= 0 = hiciS∗Ni

E − (gi + go)NES∗ ,

which has included the explicit assumption that no S∗ binds to Eo. This leads tothe relation

NES∗ = hiciS∗gi + goN

iE.

The efflux of solute is obtained by combining these expressions to give

→φS= gohi

gi + go ciSN

iE.

The influx is found by the replacement i↔ o to obtain

←φS= giho

go + gi coSN

oE.

c. Examination of the equation for→φS reveals that the numerator can be written as

a sum of two terms, one proportional to ciS and another proportional to ciScoS .

The denominator can be written as a sum of four terms, one constant, another


proportional to ciS , another proportional to coS , and another proportional to ciScoS .

Hence, the flux can be written in the form

→φS=

ciS(Kg + coS )(K2g/φz)+ (Kg/φio)ciS + (Kg/φoi)coS + (1/φ∞)ciScoS

.

The form of the relation between the flux and the concentrations for the three-state carrier mechanism is identical to that for the four-state carrier mechanism.Therefore, one cannot distinguish between the three-state and four-state carriermechanisms from measurements of flux versus concentration alone.

d. The macroscopic parameters can be written in terms of the microscopic parame-ters as follows.

Kg = aogogiho

+ goho+ 1,

φz = gohi(aogo + gogi + giho)2giho(gi + go)2(ai + ao) NET ,

φio = go(aogo + gogi + giho)(ao + go)(gi + go) NET ,

φoi = gohi(aogo + gogi + giho)ho(ai + gi)(gi + go) NET ,

φ∞ = gigogi + goNET .

Problem 6.22

a. Because the zero-trans efflux and influx have different parameters, these resultsare inconsistent with the simple, symmetric, four-state carrier model.

b. The rest of the problem deals with the general, four-state carrier model. Table 6.9(Weiss, 1996a) shows φeq and Keq for the different protocols in this problem.

i. By definition φio = 1.98± 0.31 and φoi = 0.53± 0.038 mmol/(L·min).

ii. By definition φ∞ = 7.54± 0.45 mmol/(L·min).

iii. The relation1φz+ 1φ∞

= 1φio

+ 1φoi

,

yields

φz = 1(1/φio)+ (1/φoi)− (1/φ∞) .

Substitution of the mean values for the parameters yields

φz = 0.44 mmol/(L·min).

iv. The value of Kg can be computed as shown in Table 6.6. The standard errorin Kg is computed from the standard error in Keq only. Hence, this estimateof the standard error in Kg is a lower bound because it does not include theuncertainties in the values of other parameters.

PROBLEMS 111

Protocol Value of Kg (mmol/L)

Zero-trans efflux Kg = Keqφz/φio = (0.4± 0.12) · 0.44/1.98 = 0.089± 0.027Zero-trans influx Kg = Keqφz/φoi = (0.073± 0.069) · 0.44/0.53 = 0.061± 0.057Infinite-cis efflux Kg = Keqφio/φ∞ = (0.252± 0.096) · 1.98/7.54 = 0.066± 0.025Infinite-cis influx Kg = Keqφoi/φ∞ = (0.937± 0.226) · 0.53/7.54 = 0.066± 0.016Equilibrium exchange Kg = Keqφz/φ∞ = (1.29± 0.11) · 0.44/7.54 = 0.075± 0.006

Table 6.6: Estimated dissociation constant for different protocols (Problem 6.22).

v. Note that all the estimates of Kg are consistent in that their mean values donot differ from one another by more than one standard error. Thus, thesecomparisons reveal no inconsistency with the predictions of the general, four-state carrier model.

Chapter 7

ION TRANSPORT AND RESTINGPOTENTIAL

Exercises

Exercise 7.1 a. I; b. H; c. X (mol/(cm·s·V)); d. C; e. G; f. E; g. J.

Exercise 7.2

a. Since only potassium permeates the membrane at equilibrium, the membrane po-tential must equal the potassium equilibrium potential. Assuming normal roomtemperature of 25C,

Vm = VK = 59 log10

(c2Kc1K

)= 59 log10(1/10) = −59 mV.

At equilibrium, the flux due to diffusion (which is from volume 1 to volume 2) isbalanced by drift of potassium ions (which must be from volume 2 to volume 1).The drift is in this direction if the potential in volume 1 is less than the potential involume 2 so that the positive potassium ions flow from a higher to a lower electricpotential.

c. Application of a battery to the two volumes completes an electric circuit consist-ing of the battery and the membrane as shown in Figure 7.1. The membrane isrepresented by an equivalent conductance and a battery in series.

Vm

GK VK

+ −

+ −

Im

com

par

tmen

t 1

com

par

tmen

t 2

Figure 7.1: Electric network model of membrane and source(Exercise 7.2).

113

114 CHAPTER 7. ION TRANSPORT AND RESTING POTENTIAL

b. The network shows that Im = GK(Vm − VK). When Vm > VK , Im > 0, i.e., thecurrent flows from the left to the right compartment. But, Vm = −30 mV andVK = −59 mV, so that Vm − VK = −30+ 59 = 29 mV and Im > 0.

Exercise 7.3 An ionic solution obeys electroneutrality if its constituent ions contain nonet charge so that

∑n znFcn = 0. Electroneutrality results from the strong electrostatic

forces between charges. A net charge in a solution will tend to be neutralized over atime scale greater than the relaxation time and over a distance scale greater than theDebye length.

Exercise 7.4 The Nernst equilibrium potential defines the condition for equilibrium ofan ion across a membrane permeable to that ion. The condition is

Vn = RTznF

ln

(concin

),

where Vn is the Nernst equilibrium potential defined as the inside minus the outside po-tential across the membrane, R is the molar gas constant, T is the absolute temperature,zn is the valence of ion n, F is Faraday’s constant, and con and cin are the concentrationsof ion n on the outside and inside of the membrane, respectively. When the potentialacross the membrane equals the Nernst equilibrium potential, the passive flux of thation is zero. The physical bases of the equilibrium arises because ions are transported bydiffusion resulting from a difference of concentration and by drift due to the presenceof an electric field across the membrane. When the fluxes due to drift and diffusion areequal and oppositely directed, the net flux is zero and equilibrium occurs.

Exercise 7.5 In steady state the ionic flux through the membrane, the concentrationof ions in the membrane, and the voltage across the membrane are all constant withrespect to time. Electrodiffusive equilibrium requires all of the conditions for steadystate plus the condition that the ionic flux through the membrane is zero. At equilib-rium, the potential across the membrane equals the Nernst equilibrium potentials ofeach permeant ionic species. Rest requires all of the conditions for steady state plusthe condition that the net current through the membrane (total across ionic species)is zero. Quasi-equilibrium requires all of the conditions for steady state plus that thenet flux of each ionic species (summed across all of the transport mechanisms for thatspecies) is zero.

As an example, suppose external electrodes pass a constant current through themembrane of a cell. For this case, the membrane could come to a steady-state condi-tion. It could be at electrodiffusive equilibrium if the membrane contains active trans-port mechanisms to carry all of the current from the external electrodes through themembrane. By definition, the cell is not at rest. Furthermore, the cell could not be inquasi-equilibrium, since the external current must be carried through the membrane bysome ionic species.

Exercise 7.6 The statement is largely correct except for the parenthetical phrase. Thesolution would not blow up. The excess charges would repel each other and wouldultimately reside on the boundaries of the vessel enclosing the solution.

EXERCISES 115

+ −

++

+

+++

++

+

+++

+

+

+

+

+

+

+

+ +

+

+

−

−−

−

−

− −−

−−−

−−

−−

−−

− −−

− −

−

Vm

Figure 7.2: Two compartments separated by a membranepermeable to the positive ion only (Exercise 7.8) showinganions and cations in the baths.

+

++

+

+

+

+

−−

−

−

−

−−

+ −Vm

Figure 7.3: Two compartments separated by a membrane permeable to thepositive ion only (Exercise 7.8) showing anions and cations near the mem-brane.

Exercise 7.7

a. The answer is iii. There must be cations in the solution to satisfy electroneutrality.

b. Yes, the sodium and chloride concentrations must be equal.

Exercise 7.8

a. At electrodiffusive equilibrium the tendency of the cation to diffuse from right toleft must be balanced by a tendency to drift from left to right. Thus, if Vm > 0the cations will tend to drift from left to right. At electrodiffusive equilibrium thedrift and diffusion just cancel and there is no net flux of the cation.

b. Both solutions must be electrically neutral. If both the anion and the cation areunivalent, then there must be the same number of cations as anions on each sideof the membrane as shown in Figure 7.2.

c. Near the membrane, in a layer of solution whose thickness is of the order of theDebye length, there is net charge on each side of the membrane — net cations onthe side with positive potential and net anions on the side with negative potentialas is shown in Figure 7.3.

Exercise 7.9 The two solutions in (1) and (3) are electrically neutral whereas those in (2)are not. Solution 2 is an unphysical ionic solution — you cannot obtain such a solutionby dissolving salts in water! Therefore, case (2) is not a possible answer to either part aor part b.

a. Both (1) and (3) are possible distributions of ions at equilibrium. In both cases,a potential appears across the membrane so that the left compartment is at ahigher potential than the right compartment in order to prevent diffusion of thepotassium ions from the right to the left compartment. It is interesting to notethat the osmolarity of the solution in the right compartment is larger than that


in the left compartment for (1). Therefore, since the walls and the membraneare rigid, a hydraulic pressure appears across the membrane such that the rightcompartment is at a higher hydraulic pressure than the left compartment.

b. Since both potassium and chloride are permeant, at equilibrium the potentialacross the membrane must equal both the potassium and the chloride equilib-rium potential. But, the valences have different signs, so that

VK = RTF ln

(crKclK

)= VCl = −RTF ln

(crClclCl

),

which implies thatcrKclK= c

lClcrCl,

where clK , crK , clCl, and crCl are the concentration in the left and right compartmentsfor potassium and for chloride. Thus, the concentration ratios of potassium andchloride must be reciprocal. Therefore, only distribution (3) is possible.

Exercise 7.10

a. True or false questions

i. True. By definition the total current flowing out of a cell is zero at rest.

ii. True. In this region, the slope of the line relating the resting membrane poten-tial to the logarithm of the extracellular potassium concentration approachesthat for a potassium electrode. Therefore, Vom ≈ VK .

iii. False. Since the pump is nonelectrogenic,

Vom =GK

GK +GNaVK +GNa

GK +GNaVNa.

But since Vom ≈ VK , GK/(GNa +GK) ≈ 1 which implies that GNaGK .

iv. True. VNa = 59 log10(150/15) = 59 mV. Vom = VK < 0 for the range ofconcentrations shown. Therefore, VNa > VK .

v. True. Note that for coK ≈ 100 mmol/L, Vom = VK = 0. Therefore, since VK =59 log10(c

ok/c

iK), then ciK = 100 mmol/L.

vi. True. Since the pump is nonelectrogenic, no net current is carried by thepump so that IaK + IaNa = 0.

vii. False. At rest, Im = 0 and since the pump is nonelectrogenic, IaK + IaNa = 0.Therefore, IpK + IpNa = 0, and IpK = −IpNa

viii. True. If the pump maintains constant sodium concentration in the cell thenIaNa + IpNa = 0. Therefore, IaNa = −IpNa = −GNa(Vom − VNa)

b. For the nonelectrogenic pump, the equation in part a.iii holds. Note that asGNa/GKincreases the second term in the equation increases and the first term decreases.In particular, the first term determines the slope of the relation of Vom to coK . Thus,a reduction of this term will reduce this slope. This theoretical prediction could bethe basis of the measured change in slope. According to this argument, GK/GNashould be larger at −100 mV than at −125 mV.

EXERCISES 117

Exercise 7.11

a. The Nernst equilibrium potentials are

VK = 60 log10

(15

150

)= −60 mV,

VNa = 60 log10

(15015

)= +60 mV.

b. Note that when Im = 0, Vm = Vom = −40 mV.

c. When Vm = VK , IK = 0. Therefore, Im = INa or INa/Im = 1.

d. The slope of the curve that relates Vm to Im equals the total resistance of themembrane. Hence, the reciprocal is the membrane conductance.

Gm = 0.4× 10−9

40× 10−3 = 10 nS.

e. The resting potential of the cell is related to the ion conductances and Nernstequilibrium potentials by the relation

Vom = GKGK +GNaVK +

GNaGK +GNaVNa,

−40 = −GK10

60+ GNa10

60.

In addition, GK +GNa = 10 nS. Combining these two equations yields

−406= −(10−GNa)+GNa.

Hence, GNa = 5/3 nS and GK = 25/3 nS. These results fit with the fact that the rest-ing potential is much closer to the potassium than to the sodium equilibrium po-tential. Hence, we expect the potassium conductance to greatly exceed the sodiumconductance.

Exercise 7.12 We explore Sharp Wan’s contention by assuming that puncturing themembrane leads to a leakage resistance Rl in parallel with the membrane which canbe represented by the network shown in Figure 7.4. Note that the resting potential is

Vom =Rl

Rl +RmVK =Rl

Rl +Rm59 log10

(coKciK

).

Thus, the slope of Vom versus log10 coK is 59Rl/(Rl+Rm). Therefore, a slope of 59/2 =

29.5 mV/decade is obtained if Rl = Rm.

Exercise 7.13 The direct effect of active transport on the membrane potential of a cellresults when the active transport mechanism transports a net current across the mem-brane. This contributes a component of the membrane potential that is directly at-tributable to active transport in the sense that if the current source is eliminated (e.g.,


Vm

+−

+

−

VK

Rm

Rl

Figure 7.4: Equivalent electric network of the cell mem-brane and a parallel leakage path caused by punctur-ing the membrane with a recording micropipette (Exer-cise 7.12 and Problem 7.12).

by a specific blocker substance), the contribution of the active transport mechanism tothe membrane potential goes rapidly to zero. The indirect effect results because if theactive transport mechanism is blocked, then ions will flow down their electrochemicalpotential gradients to reduce these gradients. Hence, ion concentrations will changewhich will result in changes in the Nernst equilibrium potentials of the permeable ions.The changes in Nernst equilibrium potentials will result in a change in the resting valueof the membrane potential. Because, it takes time for the ion concentrations to changeappreciably, the change in potential that results from the indirect effect takes longer tobe manifested than that caused by the direct effect.

Exercise 7.14 Cell #1 is permeable to K+ only. The resting membrane potential willbe the Nernst equilibrium potential for potassium Vom = VK = 59 log10(10/150) = −69mV. Cell #2 is permeable to Cl− only. The resting membrane potential will be the Nernstequilibrium potential for chloride Vom = VCl = −59 log10(160/160) = 0 mV. Cell#3 ispermeable of both K+ and Cl−. Since the compositions of the solutions are the same asfor Cell#1 and Cell#2, the Nernst potentials are the same as for those cells, i.e., VK = −69mV and VCl = 0 mV. Therefore, both ions cannot be in equilibrium and there will benet transport of each ion down its concentration gradient, and the potential across themembrane will change with time. Cell#4 is also permeable to both K+ and Cl−, but thecompositions of these ions are different. The two Nernst equilibrium potentials areVK = 59 log10(10/150) = −69 mV and VCl = −59 log10(150/10) = −69 mV. Therefore,both Nernst equilibrium potentials are the same and both ions will be at equilibrium atthe potential Vom = −69 mV.

Exercise 7.15

a. Potassium. When V = 0, current flow is due to diffusion alone. If K+ were perme-ant, diffusion of K+ would be upward because the concentration of K+ is greaterin compartment 2 than in compartment 1. Thus, if K+ were permeant, currentwould flow upward through the membrane and then through the wire in the di-rection opposite to the reference direction — i.e., I would be negative. If Na+ werepermeant, diffusion of Na+ would be downward because the concentration of Na+is greater in compartment 1 than in compartment 2. Thus, if Na+ were permeant,current would flow downward through the membrane and through the wire in thereference direction — i.e., I would be positive. Because the concentration of Cl−is the same in each compartment, there would be no net diffusion of Cl− even if

EXERCISES 119

I

V

G =1

59S

VK = 59 mV

+

−+

−

Figure 7.5: Circuit for ionic system in Ex-ercise 7.15

it were permeant, so I would be zero. Thus if I = −1 mA when V = 0, then thepermeant ion must be potassium.

b. The circuit is shown in Figure 7.5. The Nernst equilibrium potential for potassiumis

VK = RTF lnc2Kc1K= 59 mV× log

(1 mmol/L

0.1 mmol/L

)= +59 mV.

The conductance G can be computed from the condition that I = −1 mA whenV = 0. Therefore

G = IV − VK =

−1 mA−59 mV

= 159

S.

c. The current I can be determined directly from the model in part b.

I = G(V − VK) = 159(1− 0.059) = 15.9 mA.

Exercise 7.16

a. The solution is

φpNa φpK φaNa φaK VomI I D I I

where “D” implies decreases, “I” implies increases. When the cell equilibrates inthe control solution, there is active transport of sodium out of the cell (φaNa > 0)and active transport of potassium into the cell (φaK < 0). The ouabain in thetest solution will slow the pump so that φaNa → 0 and φaK → 0. Therefore φaNadecreases andφaK increases. Since the pump is electrogenic and produces positiveoutward current, slowing the pump will cause Vom to increase. Increasing Vom willtend to increase both φpNa and φpK .

b. The solution is

φpNa φpK φaNa φaK VomD D I D D

where “D” implies decreases, “I” implies increases. When the cell equilibratesin the control solution, the pumps are disabled because their action depends onintracellular ATP. Therefore φaNa = φaK = 0. Adding intracellular ATP during thetest interval will start the pumps, making φaNa > 0 and φaK < 0, i.e. φaNa increasesand φaK decreases. Starting the pump causes Vom to decrease, and thereby causesφpNa and φpK to decrease.


Problems

Problem 7.1

a. The conductivity of an electrolyte is related to its composition as follows

σe =∑nunz2

nF2cn =∑n

z2nF2

RTDncn.

We shall compute the conductivity using SI units. Note that concentrations ex-pressed in mmol/L are numerically equal to concentrations expressed in mol/m3,and that the diffusion coefficient needs to be expressed in m2/s. For axoplasm,

σax = F2

RT(DNacNa +DKcK +DClcCl),

σax = (9.65× 104)2

(8.314)(291)(1.33× 10−9 · 50+ 1.96× 10−9 · 400+ 2.03× 10−9 · 100),

σax = 4.06 S/m.

Therefore, the resistivity is ρax = 1/σax = 1/4.06 = 0.2466 Ω·m which equals24.6 Ω·cm. Therefore, the estimated resistivity is smaller than the measured re-sistivity; the ratio is 24.6/30 = 0.82. A number of factors are likely to contributeto a difference between the estimated and measured resistivity. In computing thetheoretical estimate it was assumed that the concentration equaled the activity ofeach ion. Typically, the activity coefficients are less that one. Hence, the estimatedconductivity should overestimate the measured conductivity, and the estimatedresistivity should underestimate the measured resistivity. Thus, this factor couldaccount for some of the difference. In addition, the composition of axoplasm thatis given is clearly incomplete — it is not electroneutral. Hence, some other ionsmust be present. Taking these additional ions into account would increase theconductivity and decrease the resistivity. This factor would make the differencebetween the estimated and measured resistivities larger.

b. The conductivity of the external solution is estimated to be

σe = (9.65× 104)2

(8.314)(291)(1.33× 10−9 · 460+ 1.96× 10−9 · 10+ 2.03× 10−9 · 540),

σe = 6.65 S/m,

so that the resistivity is 15.0 Ω·cm. Hence, the theoretical estimate of the re-sistivity is smaller than the measured value; the ratio of estimated to measuredresistivity is 15/22 = 0.68. Some of the same factors discussed in connection withthe resistivity of axoplasm may affect the difference between the measured andestimated resistivity of the external solution. Thus, without further analysis wecannot really account for the difference in resistivity of axoplasm and the externalsolution. Factors such as those suggested by Katz may make a contribution to thedifference.

PROBLEMS 121

c. The Stokes-Einstein relation can be solved for the radius to yield

a = kT6πDη

= (1.381× 10−23 J/K)(291 K)6πD(0.9× 10−3 Pa · s)

= 2.37× 10−19

D.

Therefore,

aNa = 2.37× 10−19

1.33× 10−9 = 1.78× 10−10 m = 1.78 A,

aK = 2.37× 10−19

1.96× 10−9 = 1.21× 10−10 m = 1.21 A,

aCl = 2.37× 10−19

2.03× 10−9 = 1.17× 10−10 m = 1.17 A.

Thus, the estimates of the radius of the ions are within a factor of 2 of the crystalradii. A number of factors could account for the differences. First, the diffusioncoefficients used for the entire problem are those for infinite dilution. The diffu-sion coefficient at the concentrations in axoplasm and in the external solution willdiffer from those at infinite dilution. Furthermore, the diffusion coefficient givenin the problem were for a temperature of 25C and the temperature assumed inthe problem was 18C. In addition, ions in solution bind to water molecules to givea hydrated ion whose radius is larger than the crystal radius of the ion. Finally,the Stokes-Einstein relation is most accurate for large spherical particles in waterand its validity for small ions may be questionable.

Problem 7.2

a. The relation between the potential and the concentration is obtained from threerelations: steady state yields

J = J+ + J− = 0,

electroneutrality yieldsc+(x) = c−(x) = c(x),

and the Nernst-Planck equation for current density of each ion yields

Jn = −znF(unRT

dcn(x)dx

+unznFcn(x)dψ(x)dx

).

Combination of these 3 relations yields

−F(u+RT

dc(x)dx

+u+Fc(x)dψ(x)dx

)+ F

(u−RT

dc(x)dx

−u−Fc(x)dψ(x)dx

)= 0,

which can be simplified to

(u+ +u−)Fc(x)dψ(x)dx= (u− −u+)RT dc(x)dx

.

Rearranging terms yields

dψ(x)dx

= −(u+ −u−u+ +u−

)RTFc(x)

dc(x)dx

= −(u+ −u−u+ +u−

)RTFd ln c(x)dx

.


b. The potential difference across the membrane can be determined by integratingthe potential across the membrane

Vj =∫ 0

ddψ(x) = −

∫ 0

d

(u+ −u−u+ +u−

)RTFd ln c(x)dx

dx,

= −(u+ −u−u+ +u−

)RTF

ln c(x)∣∣∣∣0

d

=(u+ −u−u+ +u−

)RTF

ln(c(d)c(0)

).

Because, the concentrations are assumed to be continuous at the interfaces,

Vj =(u+ −u−u+ +u−

)RTF

ln

(c2

c1

).

c. The relation between flux and current density is Jn = znFφn so that

φ+ = −u+RT dc(x)dx−u+Fc(x)dψ(x)dx

,

φ− = −u−RT dc(x)dx+u−Fc(x)dψ(x)dx

.

Therefore,

φ = φ+ +φ− = −(u+ +u−)RT dc(x)dx− (u+ −u−)Fc(x)dψ(x)dx

,

and after substitution for dψ(x)/dx

φ = −(u+ +u−)RT dc(x)dx+ (u+ −u−)

2

u+ +u− RTdc(x)dx

,

which after rearrangement of terms yields

φ = −utRT dc(x)dx,

where

ut = 4u+u−u+ +u− .

d. Plots of ut/u+ and Vj/((RT/F) ln(c2/c1)) are shown in Figure 7.6. The liquidjunction potential is zero when the mobilities of the anion and cation are thesame. The normalized potential is positive if the cation mobility exceeds the anionmobility and is negative when the anion mobility exceeds the cation mobility.

e. i. The system is not in equilibrium because there is net flux of both the cationand the anion. At equilibrium, φ+ = φ− = 0. In contrast, in the liquidjunction

φ+ = φ− = φ = utRTd (c2 − c1),

so that φ 6= 0 if c2 6= c1.

PROBLEMS 123

−1

−0.5

0.5

1

10−2 10−1 101 102

Vj/((RT/F ) ln(c2/c1))

u+/u−

ut/(4u+)

Figure 7.6: Plots of the effectivemobility and the junction poten-tial as a function of the ratio ofmobilities of the cation and anion(Problem 7.2).

ii. Further analysis allows the questions to be answered precisely. The equationthat links the flux to the concentration gradient can be integrated since theflux is constant to yield∫ d

0φdx = −

∫ c(d)c(0)

utRT dc(x),

φd = −utRT (c(d)− c(0)) ,φ = utRT

d(c2 − c1),

where it is assumed that the concentration is continuous through the interfacebetween the liquid junction and the bath solutions. Since the concentrationgradient is constant, the concentration profile in the membrane must be linearand can be expressed as

c(x) = c1 − (c1 − c2)xd,

which satisfies the requisite boundary conditions at x = 0 and x = d. Thepotential in the liquid junction can be found from the relation between theconcentration and potential gradient found in part a,

ψ(x)−ψ(0) = −RTF

(u+ −u−u+ +u−

)ln(c(x)c(0)

),

ψ(x)−ψ(0) = −RTF

(u+ −u−u+ +u−

)ln

(c1 − (c1 − c2)xd

c1

).

The expressions for both the concentration and potential in the liquid junc-tion can be used to compute the diffusive and drift components of the fluxwhich for the cation are

(φ+)diffusion = +u+RTd

(c1 − c2

),

(φ+)drift = −u+RTd

(u+ −u−u+ +u−

)(c1 − c2

).

The sum of the fluxes due to diffusion and to drift is

φ+ = (φ+)diffusion + (φ+)drift =2u+u−u+ +u−RT(c

1 − c2).


ψ(x

)−

ψ(0

)

RT/F

c(x)/c1

x/d

x/d

u+/u− = 2

u+/u− = 1/2

0 0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1

0.2 0.4 0.6 0.8 1

−0.2

−0.1

0.1

0.2

Figure 7.7: Dependence of concentration and elec-tric potential versus position in the liquid junction(Problem 7.2). The results are shown for c1/c2 = 2and for both u+/u− = 2 and u+/u− = 1/2.

The anion fluxes are

(φ−)diffusion = +u−RTd

(c1 − c2

),

(φ−)drift = +u−RTd

(u+ −u−u+ +u−

)(c1 − c2

),

and

φ− = (φ−)diffusion + (φ−)drift =2u+u−u+ +u−RT(c

1 − c2).

The physical basis of these results can now be explored. Consider the resultsfor the condition c1 > c2. The concentration and potential in liquid junctionare shown for such a condition in Figure 7.7. For the condition c1 > c2, thenet flux of both cations and anions is φ > 0. Now suppose the mobilityof the cation exceeds that of the anion, i.e., u+/u− > 1, then the potentialgradient is positive. Under these conditions the cation flux due to diffusionis positive and the cation flux due to drift is negative. Thus, the magnitude ofthe cation flux is reduced below that attributed to diffusion alone. In contrast,the diffusive flux and the drift add for the anion. Thus, the potential gradientthat arises in the liquid junction decreases the magnitude of the flux of themore rapidly mobile cations and increases the magnitude of the flux of theslower anions. Thus, the net flux of each ion is the same and electroneutralityis maintained in the liquid junction. If the anions are more mobile, then thesign of the potential gradient is reversed and once again the flux of the fasterion, now the anion, is decreased and that of the slower ion increased. Theresultant cation and anion fluxes are equal so that the net current densitythrough the junction is zero.

f. The arrangement for recording the membrane potential with an intracellular mi-cropipette and the resulting junction potentials are shown in Figure 7.8 The mea-sured potential Vmeas is related to the resting potential across the membrane Vom

PROBLEMS 125

Cell

VKCl/cyt

+

−

Bath

VKCl/bath

+

−+

−V om

+ −Vmeas

Vj1

Vj2

Figure 7.8: Arrangement formeasuring the membrane po-tential and the accompanyingliquid junction potentials (Prob-lem 7.2).

as followsVmeas = Vj1 + Vom − Vj2.

Thus, the net junction potential is Vj = Vj1 − Vj2 which can be estimate as

Vj1 =(u+ −u−u+ +u−

)RTF

ln(

0.152

)−(u+ −u−u+ +u−

)RTF

ln(

0.0052

)=

(u+ −u−u+ +u−

)RTF

ln(

0.150.005

).

This can be evaluated using the mobilities from Table 7.2 (Weiss, 1996a),

Vj1 =(

7.89− 8.217.89+ 8.21

)RTF

ln 30 =(

7.89− 8.217.89+ 8.21

)59 log10 30 = −1.7 mV.

g. One important criterion for choosing a binary electrolyte to fill the micropipettesis minimization of the junction potential between the electrolyte in the pipette andintracellular solution. Junction potentials are minimized when the mobilities ofthe cation and anion are the same. According to Table 7.2 (Weiss, 1996a) the bestcombination for minimization of the junction potential is CsI. However, filling themicropipette with CsI would yield a junction with CsI on the micropipette side andKCl on the intracellular side. The mobilities of potassium and chloride are alsoquite close. Since intracellular solutions resemble KCl in ion composition, KCl isthe best choice.

Problem 7.3

a. Using Kirchhoff’s voltage law

Vi = VAg/AgCl−p+VAgCl/KCl−p+VKCl/cyt+Vom−VKCl/bath−VAgCl/KCl−r−VAg/AgCl−r .

b. If the potential across the junction between the silver and silver chloride is con-stant and the same at both electrodes, that term will drop out of the equation toyield

Vi = VAgCl/KCl−p + VKCl/cyt + Vom − VKCl/bath − VAgCl/KCl−r .c. Note that VAgCl/KCl = VoAgCl/KCl − (RT/F) ln cCl where VAgCl/KCl ≈ 222 mV for a

1 mol/L solution of KCl at 25C.


i. Since the chloride concentrations is 2 mol/L in both the micropipette and inthe salt bridge, the potentials at the AgCl/KCl boundary is the same for bothelectrodes. Therefore, these two terms cancel to yield

Vi = VKCl/cyt + Vom − VKCl/bath.

ii. If the salt bridge at the reference electrode is removed then the contributionfrom the two AgCl/KCl is

(222− (RT/F) ln cpCl)− (222− (RT/F) ln cbCl) = (RT/F) ln

(cbClcpCl

).

where cpCl and cbCl are the concentrations in the micropipette and bath, re-spectively. Without the salt bridge, the junction potentials at the AgCl/KClboundaries make a net contribution to the measured potential that dependsupon the chloride concentration. This argument emphasizes the importanceof using salt bridges to measure cellular potentials.

Problem 7.4 The derivation given here is based on one given elsewhere (Patlak, 1960).

a. As stated in the problem,→φn= Pincin and

←φn= Poncon where Pin and Pon are indepen-

dent of concentration. Therefore,

→φn←φn

= PincinPoncon

.

Let con be the outside concentration of ion n when it is in electrodiffusive equilib-rium across the membrane in the presence of a membrane potential Vm. Becausethe efflux and influx of ion n are equal at electrodiffusive equilibrium,

1 = PincinPoncon

.

Therefore, since Vm equals the Nernst potential when the external concentrationis con,

PinPon= c

on

cin= e(znF/RT)Vm.

Substitution of this result into the ratio of unidirectional fluxes yields

→φn←φn

= cincone(znF/RT)Vm = e−(znF/RT)Vne(znF/RT)Vm = e(znF/RT)(Vm−Vn).

b. The outward current density due to ion n is Jn = znF(→φn −

←φn). Under resting

conditions,∑n Jn = 0 and Vm = Vom. Since the ions are all univalent, the valence

can be dropped and the summation separated into two summations: one for thecations alone and the other for the anions,∑

+nJn +

∑−nJn = 0,

PROBLEMS 127

where the lower limit of the summation is used in a non-standard manner; +nmeans a summation on all cations, and −nmeans a summation on all anions. Thefluxes can be substituted for the current densities and Faraday’s constant can befactored out to yield ∑

+n(→φn −

←φn)−

∑−n(→φn −

←φn) = 0.

This relation can be rewritten as

∑+n

←φn

→φn←φn− 1

+∑−n

→φn

←φn→φn− 1

= 0,

∑+nPoncon

(cincone(F/RT)Vm − 1

)+∑−nPincin

(concine(F/RT)Vm − 1

)= 0,

eF/(RT)Vm(∑+nPoncin +

∑−nPincon

)−(∑+nPoncon +

∑−nPincin

)= 0.

Therefore,

eF/(RT)Vm =∑+n Poncon +

∑−n Pincin∑

+n Poncin +∑−n Pincon

,

from which

Vom =RTF

ln

(∑+n Poncon +

∑−n Pincin∑

+n Poncin +∑−n Pincon

).

Note that for the special case Pin = Pon = Pn, the expression becomes

Vom =RTF

ln

(∑+n Pncon +

∑−n Pncin∑

+n Pncin +∑−n Pncon

),

which is the same as the potential derived from the Goldman theory.

Problem 7.5

a. Because, the membrane is thin compared to the radius of the cell, the capacitancecan be approximated by that of a parallel plate capacitance,

C = κmεoAd

,

where C is the capacitance, κm is the dielectric constant, εo is the permittivity offree space, A is the area of each plate, and d is the distance between the plates.Hence,

κm = CAdεo= Cm dεo = (10−2 F/m2)

70× 10−10 m8.854× 10−12 F/m

= 7.9

Dielectric constants of oils and waxes are typically in the range 2-5 and those ofglasses are in the range 3-10. Water has a dielectric constant of 81. Thus, thedielectric constant estimated above is closer to that of glasses than of oils andwaxes. However, there is some uncertainty in our estimate. For example, the


thickness of the membrane was chosen as 70 A. The hydrophobic region of thelipid bilayer may be somewhat smaller than that. If the hydrophobic portion of themembrane is taken to have a thickness of 50 A, then the estimate of the dielectricconstant is 5.6 which is near the range of oils and waxes.

b. The magnitude of the electric field in the membrane is |Vom|/d = 70× 10−3/70×10−10 = 107 V/m. This is two orders of magnitude greater than the dielectricstrength of air. Oils have dielectric strengths of about 107 V/m and glasses havedielectric strengths of about 108 V/m.

c. The surface charge per unit area on the inside of the membrane is

Q = CmVom = −10−6 F/cm2 · 70× 10−3 V = −70 nC/cm2.

Therefore, the total charge on the inside surface of the membrane is

qs = QA = −70 nC/cm2 × 4π(25× 10−4)2 cm2) = −5.5 pC.

This surface charge can be compared to the total quantity of charge due to potas-sium ions in the cell.

qiK = ρiKV = FciKV= (9.65× 104 C/mol)(0.2× 10−3 mol/cm3)((4/3)π(25× 10−4)3 cm3))= 1.26 µC.

Therefore, the ratio qs/qiK = 5.5×10−12/1.26×10−6 = 4.4×10−6, i.e., the surfacecharge is a very small fraction of the total charge available in the cell.

d. The resistance of the membrane is

Rm = ρmdA so that ρm = RmAd = Rmd.

Hence,

ρm = (103 Ω · cm2)/(70× 10−8 cm) = 1.4× 109 Ω · cm.

The resistivity in Ω ·cm is about 10−6 for copper, > 1013 for insulators and about103-105 for semi-conductors.

e. The Nernst equilibrium potential for potassium is

VK = 59 log10

(coKciK

)= 59 log10

(0.0020.2

)= −118 mV.

f. At rest JC = Cm(dVm(t)/dt) = 0. Therefore, JpK + JaK = 0. Hence,

JaK = −JpK = −GK(Vm − VK) = −10−3(−70+ 118)× 10−3 = −48 µA/cm2.

PROBLEMS 129

x0

Axon

cn(x, t)

xp

E

Figure 7.9: The direction of the electric field is shownfor the spatial dependence of concentration shown(Problem 7.6).

g. If JaK = 0 then the membrane potential changes from −70 mV to the potassiumequilibrium potential −118 mV. This change in potential accompanies a change inthe surface charge on the membrane from−5.5 pC to−5.5(−118/−70) = −9.3 pC.Since the membrane is permeable to potassium only, a net amount of potassiumof 9.3 − (5.5) = 3.8 pC must have crossed the membrane from the inside to theoutside to account for the change in potential. This represents a fractional changeof potassium charges of −3.8 × 10−12/1.26 × 10−6 = −3 × 10−6. Therefore, thepercentage change in potassium concentration is −0.0003%.

Problem 7.6

a. The drift velocity isv = uKzKFE. Hence, whenv > 0,E > 0 as shown in Figure 7.9.

b. Equation 7.58 (Weiss, 1996a) shows that at any time, the concentration has a max-imum at x − vt = 0 and the amplitude is reduced by 1/e for (xw − vt)2 = 4DKt.Call xw−vt = ∆xw(t). Therefore, xw is the position at which the concentration is1/e of its maximum value, and ∆xw(t) is the width of the concentration from thelocation of the maximum to the location xw at time t. Since (∆xw(t))2 = 4DKt,the change in width between two times t0 and t1 is (∆xw(t1))2 = (∆xw(t0))2 +4DK(t1 − t0). The data in Figure 7.10 shows the count rate versus position (datapoints) and a Gaussian curve fit to the data points. Because the curve averagesover the data points, the diffusion coefficient will be estimated from the curve. Asindicated in Figure 7.10, the widths can be estimated for the two times so that

DK = (∆xw(t1))2 − (∆xw(t0))24(t1 − t0)

= (1.43× 10−2)2 − (0.87× 10−2)2

4 · 445 · 60= 1.2× 10−9 m2/s.

c. Both the diffusion coefficient and the mobility can be estimated from the mea-surements in the presence of an electric field. The diffusion coefficient can beestimated as in part b, although because these measurements are taken over ashorter time, the estimate based on the figure alone will be quite inaccurate. Theinaccuracy occurs because the difference of the squares of the widths of the con-centrations must be determined and these widths are difficult to read accurately


J J

J

J

J

J

J

J

JJ

E

E

E

E

E

E

E

E

E

EE0

100

200

300

400

500

600

700

0 1 2 3 4 5 6Distance (cm)

Cou

nts

/min

ute

1.43

0.87

Figure 7.10: Construction used to estimate thediffusion coefficient from the measurements(Problem 7.6).

J

J

J

J

J

J

J

J

JE

E

EE

E

E

EE

E

E0

50

100

150

200

250

0 1 2 3 4 5

Distance (cm)

Cou

nts

/min

ute

0.66

1.00.84

Figure 7.11: Construction used to estimatethe diffusion coefficient and the mobilityfrom the measurements (Problem 7.6).

from the figure. The estimate of the diffusion coefficient is

DK = (∆xw(t1))2 − (∆xw(t0))24(t1 − t0)

= (1.0× 10−2)2 − (0.84× 10−2)2

4 · 37 · 60= 3.3× 10−9 m2/s.

The mobility can be estimated from the change in location of the peak value ofthe concentration. As shown in Figure 7.11, the peak moves from near 1.9 cm tonear 2.6 cm, a distance estimated to be about 0.66 cm, in 37 minutes. Therefore,the drift velocity is

v = 0.66× 10−2

37× 60= 3.0× 10−6m/s.

From the relation that v = uKzKFE, the mobility is estimated as

uK = vFE =

3.0× 10−6 m/s(9.65× 104 C/mol) · (54.8) V/m

= 5.7× 10−13 (m/s)/(N/mol)

PROBLEMS 131

d. The diffusion coefficient is related to the mobility by the Einstein relation. There-fore, the diffusion coefficient can be calculated from the estimated mobility asfollows

D = uRT = (5.7× 10−13)(8.314)(300) = 1.4× 10−9 m2/s.

This value agrees within 15% with the estimate of the diffusion coefficient in partb but differs appreciably from the more inaccurate estimate of the diffusion coef-ficient obtained in part c.

Problem 7.7

a. The network with passive transport only consists of the capacitance in parallelwith the two conductances. The time course of the response of this network isgiven by the passive time constant

τp = CGNa +GK =

10−9 F(5× 10−9 + 5× 10−7) S

≈ 2 ms.

This time constant is much faster than the time course of the response to the injec-tion which has a time constant of 5 minutes. Hence, for the measured membranepotential shown, the rate of change of membrane potential is so small that thecurrent through the capacitance is negligible compared to that through the otherbranches. Hence, the membrane capacitance can be ignored for the remainder ofthis problem. The measured time constant cannot be accounted for by the passivenetwork.

Can we account for the magnitude of the change in membrane potential by passivetransport alone? The membrane potential is

Vm =(

GNaGNa +GK

)VNa +

(GK

GNa +GK

)VK − IaNa + IaK

GNa +GK .

Since IaNa, IaK , and VK are constant during the injection of sodium,

Vm(t)− Vm(0) = ∆Vm(t) =(

GNaGNa +GK

)∆VNa(t) ≈ 10−2∆VNa(t).

Therefore, the change in the sodium equilibrium potential is 100 times larger thanthe change in membrane potential. Since the change in membrane potential is16 mV, the sodium equilibrium potential would have to change by 1.6 V! Is thisconsistent with the experiment?

∆VNa(t) = RTF

ln

(coNaciNa(t)

)− RTF

ln

(coNaciNa(0)

)= −RT

Fln

(ciNa(t)ciNa(0)

),

= −RTF

ln

(∆ciNa(t)+ ciNa(0)

ciNa(0)

)= −RT

Fln

(∆ciNa(t)ciNa(0)

+ 1

),

≈ −60 log10

(∆ciNa(t)ciNa(0)

+ 1

)mV.


The quantity ∆ciNa(t)/ciNa(0) can be computed for t = 0.5 min = 30 s from the

data given. ciNa(0) = 6× 10−6 mol/cm3 and

∆ciNa(30) = (48× 10−12 mol/min)(0.5 min)4× 10−6 cm3 = 6× 10−6 mol/cm3 = ciNa(0).

Therefore,

∆VNa(30) ≈ −60 log10

(6× 10−6

6× 10−6 + 1

)= −60 log10 2 ≈ −18 mV.

Hence, the change in VNa is much too small to account for the measured change inmembrane potential. Therefore, passive transport alone cannot account for eitherthe time constant or the magnitude of the response to the injection of sodium.

b. In part a it was shown that

Vm = 1GNa +GK

(GNaVNa +GKVK − (IaNa + IaK)) ,but that a change in VNa cannot account for the change in Vm. The point of thispart is to determine whether a change in IaNa can account for the change in Vm.Conservation of sodium requires that a change in sodium concentration must bedue to a change in sodium current

Vc d∆ciNa(t)dt

= −∆IaNa(t)F

= −Vcτa∆ciNa(t).

This equation reduces to

d∆ciNa(t)dt

+ ∆ciNa(t)τa

= 0,

which is a first-order, linear, ordinary differential equation whose homogeneoussolution is exponential with time constant τa,

∆ciNa(t) = ciNa(30)e(t−30)/τa for t > 30 s.

The change in active sodium current can be found from the change in concentra-tion,

∆IaNa(t) =FVcτa

∆ciNa(t) =FVcτaciNa(30)e(t−30)/τa for t > 30 s.

The membrane potential change is related to the change in active sodium currentso that

∆Vm = − ∆IaNa(t)

GNa +GK =FVc

τa(GNa +GK)ciNa(30)e(t−30)/τa for t > 30 s.

If we choose τa = 5 · 60 s,

∆Vm ≈ −(105 C/mol)(4× 10−6 cm3)(6× 10−6 mol/cm3)(5 · 60 s)(0.5× 10−6 S)

e(t−30)/τa for t > 30 s,

≈ −16e(t−30)/300 mV, for t > 30 s.

Thus, the model with active sodium transport can account for the measurements.

PROBLEMS 133

+

−

+−VK

G

G G

V om

Figure 7.12: Equivalent circuit model of a cell with two mem-brane patches (Problem 7.8).

Problem 7.8

a. Patch 1 contains two conductances each of value G and each equilibrium potentialis zero because solution A has the same composition as intracellular solution.Patch 2 contains a potassium conductance whose value is G and an equilibriumpotential that has the value

VK = 59 log10

(cBKcoK

)mV = 59 log10

(15

150

)= −59 mV.

The circuit is shown in Figure 7.12.

b. Thus, the resting potential is the potassium equilibrium potential divided by theresistive divider network,

Vom =1/(2G)

1/(2G)+ 1/GVK =VK3= −19.7 mV.

Problem 7.9

a. The cell has passive and active transport of both sodium and potassium. Hence,an appropriate model for the resting potential of the cell is the one given in Fig-ure 7.13. The resting potential can be derived by superposition of the contribu-tions of each source with the other sources set to zero which yields

Vom =GNa

GNa +GK VNa +GK

GNa +GK VK −1

GNa +GK (IaNa + IaK)︸︷︷︸

Direct contribution

,

with the direct contribution of active transport to the resting membrane potentialidentified. The active transport currents are

IaNa = zNaFφNa = 1 · (9.65× 104) · (3× 10−17) ≈ 3 pA,IaK = zKFφK = 1 · (9.65× 104) · (−2× 10−17) ≈ −2 pA.


Vm

+

−

GNa GK

VNa VK+−

+−

IpNa

IaNa

IpK

IaK

Figure 7.13: Network model for iontransport in Problem 7.9.

Therefore, the active component of the membrane potential is

Voma = −(3− 2)× 10−12

10−10 = −10 mV.

b. Because the cell is at equilibrium, IpNa + IaNa = 0 and IpK + IaK = 0. Therefore,

IpNa = −3× 10−12 = GNa(Vom − VNa),IpK = 2× 10−12 = GK(Vom − VK).

The Nernst equilibrium potentials can be computed

VNa = 59 log1010615

≈ 50 mV,

VK = 59 log103

150≈ −100 mV.

Combining these results yields

GNa = −3× 10−12

(Vom − 50)× 10−3 ,

GK = 2× 10−12

(Vom + 100)× 10−3 ,

where Vom is expressed in mV. In addition,

GNa +GK = 10−10 S.

Combining these equations to eliminate GNa and GK yields

−3× 10−12

(Vom − 50)× 10−3 +2× 10−12

(Vom + 100)× 10−3 = 10−10,

which can be manipulated to give

(Vom)2 + 60Vom − 1000 = 0,

and has two solutions Vom = −30(1 ± 1.45) mV. The two values are −73.5 and+13.5 mV. Since GK > GNa, the resting potential must be nearer to VK than to VNaso that the resting potential must be Vom = −73.5 mV.

PROBLEMS 135

c. The conductances can be computed as follows

GNa = −3× 10−12

(−73.5− 50)× 10−3 = 24.3 pS,

GK = 2× 10−12

(−73.5+ 100)× 10−3 = 75.5 pS.

Problem 7.10 The direction of passive transport of each ion can be determined usingthe relations

Jn = Gn(Vom − Vn) where Vn = RTznF

ln

(concin

).

The Nernst equilibrium potential for each ion is

VNa = 59 log100.115= −128 mV,

VK = 59 log101

28.7= −86 mV,

VCl = 59 log101.338= 86 mV.

Hence,

JNa = GNa(−86+ 128)× 10−3 = GNa(42× 10−3),JK = GK(−86+ 86)× 10−3 = 0,JCl = GCl(−86− 86)× 10−3 = −GCl(172× 10−3).

Therefore, the passive current density carried by sodium is outward, which implies thatthe passive flux of current is outward. Hence, to maintain sodium concentration in sap,sodium must be transported inward by active transport mechanisms. The passive cur-rent density carried by potassium is zero. Hence, no active transport of potassium isneeded. The passive current density carried by chloride is inward. The inward currentcarried by a negative ion implies an outward flux of chloride. Hence, chloride must betransported inward by an active transport mechanism to maintain chloride concentra-tion in the sap.

Problem 7.11

The results from the two cells are analyzed separately for parts a and b.

Cell 1 From the graphs it is apparent that Cell 1 acts as a perfect potassium elec-trode, i.e., the resting potential equals the potassium equilibrium potentialand is independent of external sodium concentration. Therefore, GNa = 0and GK = Gm = 10 nS. Also, when Vom = −60 mV (the normal resting poten-tial), conK = 20 mmol/L. The value of ciK can be computed either by notingthat Vom = VK = 0 when conK = ciK or by computing −60 = 60 log10(20/ciK).Either way, ciK = 200 mmol/L. Since Vom is independent of coNa, conNa cannot becomputed from these results. The results are shown in Table 7.1.


Cell 1 Cell 2

GK (nS) 10 8GNa (nS) 0 2conK (mmol/L) 20 10ciK (mmol/L) 200 356conNa (mmol/L) × 200ciNa (mmol/L) 20 20

Table 7.1: Conductances and concentrations forCells 1 & 2 (Problem 7.11).

Cell 2 Since the resting potential depends on both coNa and coK , the membraneis permeable to both ions. Since the pumps are non-electrogenic, the directeffect on the membrane potential can be ignored. Therefore,

Vom =GK

GK +GNaVK +GNa

GK +GNaVNa,


Vom ≈GK

GK +GNa60 log10

(coKciK

)+ GNaGK +GNa60 log10

(coNaciNa

).

From the graphs

GKGK +GNa60 = 48 mV/dec and

GNaGK +GNa60 = 12 mV/dec,

which can be solved to yield

GKGK +GNa =

4860= 0.8 and

GNaGK +GNa =

1260= 0.2.

Therefore, GK/GNa = 4 satisfies both equations (which are not independent).Since GK + GNa = 10 nS, 4GNa + GNa = 5GNa = 10 nS and GNa = 2 nS andGK = 8 nS. Using these values

Vom = 48 log10

(coKciK

)+ 12 log10

(coNaciNa

).

But ciNa = 20 mmol/L. Also from the graphs at −60 mV, conK = 10 mmol/L,conNa = 200 mmol/L. Thus, only ciK must be determined, and it can be deter-mined from results at any membrane potential, i.e.,

0 = 48 log10

(200

ciK

)+ 12 log10

(20020

),

from which

log10

(200

ciK

)= −12

48= −1

4,

which implies that ciK = 356 mmol/L. The parameters are given in Table 7.1.

PROBLEMS 137

Vm

+−

+−

+

−

10

+−

Jm

Cell 1 Cell 2

8 2

V 1K V 2

K V 2Na

Figure 7.14: Network models for Problem 7.11.V1K = 60 log10(c

oK/200), V2

K = 60 log10(coK/356), and

V2Na = 60 log10(c

oNa/20). Units are: nS for conduc-

tances, mmol/L for concentrations, mV for potentials.

c. The networks are shown in Figure 7.14.

Problem 7.12

a. Because the cell membrane acts as a perfect potassium electrode, the resting po-tential of the cell must equal the potassium equilibrium potential,

Vom = VK =RTF

ln

(coKciK

)≈ 59 log10

(15

150

)= −59 mV.

b. The membrane and the leakage path are in parallel as shown in Figure 7.4.

c. The membrane resistance is Rm = 1/(GmA) where Gm is the specific resistanceof the membrane and A is the surface area of the cell. The membrane resistanceis

Rm = 1

(10−3 S/cm2)(4π(15× 10−4)2 cm2= 35.4 MΩ.

Since Rl = 100 MΩ, the resting potential of the cell with the shunt resistancecaused by the leakage path is

Vm = 100100+ 35.4

(−60) = −44.3 mV.

d. The relation of the membrane potential to the Nernst equilibrium potential is

Vm = 100100+ 35.4

VK = 0.74× 59 log10

(coK

150

)= 43.6 log10 c

oK − 94.8 mV,

where coK is in mmol/L.

e. The effect of the shunt resistance is to reduce the slope of the relation of mem-brane potential to the logarithm of the external potassium concentration from 59to 43.6 mV/decade of concentration. The exact value of this reduction dependsupon the resistance of the leakage path. Note that the data for the frog skeletalmuscle for concentration above 3 mmol/L shows a linear dependence of resting


Vm

+−

+−

+

−

GK

VK VL

GLFigure 7.15: Network model to interpret the measurements of Lingand Gerard (Problem 7.13).

potential on the logarithm of the external concentration of potassium with a slopeless than that of a potassium electrode. This raises the question of whether theleakage path caused by the penetrating micropipette is in part responsible forthese results.

Problem 7.13

a. The following apply to Adrian’s measurements.

i. For large values of coK the measurements approach that of a perfect potassiumelectrode which implies that the membrane is permeant to potassium only.Therefore, the conductances for all other ions is zero. Thus, GK = Gm =2× 10−6 S.

ii. Since, the membrane potential acts as a perfect potassium electrode at highextracellular concentration,

Vom = VK =RTF

ln

(coKciK

),

so that Vom = 0 when the inside and outside concentrations are the same.Therefore, reading the extrapolation of the measurements to a potential of 0mV yields ciK ≈ 120 mmol/L.

b. The following apply to Ling and Gerard’s measurements.

i. Clearly the data of Ling and Gerard do not conform to the predictions ofa perfect potassium electrode even at high potassium concentration. Thus,ions in addition to potassium must flow between the inside and the outside ofthe cell. The additional flow of ions is represented by the network shown inFigure 7.15. In the absence of the micropipette, the measurements of Adriansuggest that for large values of coK only potassium ions permeate the mem-brane. This pathway is represented by the potassium conductance in serieswith the Nernst equilibrium potential for potassium. It is hypothesized thatinsertion of the micropipette caused another pathway, a leakage pathway, forcurrent to flow between the inside and the outside of the cell. Therefore, the

PROBLEMS 139

two currents, one due to potassium flowing through the membrane and theother due to ions flowing through the leak, are in parallel. An equivalent cir-cuit for the leakage path consists of a conductance in series with a battery —a Thevenin’s equivalent network.

ii. The resting potential for the network model in Figure 7.15 is found by writ-ing Kirchhoff’s current law, setting the total membrane current to zero, andsolving for the membrane potential to obtain

Vom =GK

GK +GLVK +GL

GK +GLVL.

This can be expressed in terms of the potassium concentration

Vom =GK

GK +GL58 log10

(coKciK

)+ GLGK +GLVL mV.

Because the slope of Vom versus coK is 44 mV/dec,

GKGK +GL =

4458= 0.76,

which can be solved to determine that GL/GK = 0.32. The total membraneconductance of the frog sartorius muscle fiber is given, and in part a it wasdetermined that the total conductance of the muscle fiber is due to potassium.Thus, GK = 2 × 10−6 and GL = 0.64 × 10−6 S. Note that the value of coK forwhich Vom = 0 is approximately the same for both sets of measurements. Atthis concentration, the measurements of Adrian show that VK = 0. Therefore,VL = 0. This is a satisfying result, since the model for the hole caused by thepuncture of the membrane is simply that of a shunt resistance with VL = 0.Finally, VK = 58 log10(c

oK/c

iK).

iii. Clearly the calculations support the hypothesis that the Ling and Gerard mea-surements at high potassium concentration behave as if they are compro-mised by the shunt resistance caused by the insertion of the micropipette.Although consistent with this hypothesis, the results do not prove the hy-pothesis.

iv. The model shown in Figure 7.15 cannot account for the saturation of theresting potential at low extracellular potassium concentration.

Problem 7.14

a. If the model involves only the passive transport of sodium and potassium thenthe network diagram is as shown in Figure 7.16. The circuit can be solved to yield

VEP =(

GNaGNa +GK

)VNa +

(GK

GNa +GK

)VK.

This shows that VEP will be closest to the Nernst equilibrium potential of the morepermeant ion. The Nernst equilibrium potential for each ion is required for a


−

VEP

+

+−

+−

IpNa IpK

GNa GK

VNa VK

Figure 7.16: Model for the endolymphatic potentialincluding passive transport of sodium and potassium(Problem 7.14).

quantitative analysis. At 38C,

VNa = RTF

ln

(cpNaceNa

)= 61 log

(1402

)= 113 mV,

VK = RTF

ln

(cpKceK

)= 61 log

(5

150

)= −90 mV.

Thus, VEP = 100 mV is close to VNa = 113 mV, which implies that GNa GK . Aquantitative estimate of γ = GNa/GK can be obtained from the equation for VEPwhich can be rewritten as

100 =(γ

γ + 1

)113−

(1

γ + 1

)90 mV,

which can be solved for γ = 14.6. Thus, the passive model can account for thevalue of VEP provided that the conductance for sodium is 14.6 times that for potas-sium.

b. If passive transport of sodium and potassium accounts for VEP with GNa/GK =14.6 then

VEP =(

14.614.6+ 1

)VNa +

(1

14.6+ 1

)VK mV,

=(

14.614.6+ 1

)61 log

(cpNaceNa

)+(

114.6+ 1

)61 log

(cpKceK

)mV,

= 57 log

(cpNaceNa

)+ 3.9 log

(cpKceK

)mV.

Therefore, the model predicts that a change in concentration of sodium in peri-lymph from 140 to 40 mmol/L should result in a change in VEP of

∆VEP = 57 log(

402

)− 57 log

(1402

)= 57 log

(40

140

)= −31 mV.

Also a change in concentration of potassium from 5 to 1 mmol/L should result ina change in VEP of

∆VEP = 3.9 log(

1150

)− 3.9 log

(5

150

)= 3.9 log

(15

)= −2.7 mV.

PROBLEMS 141

−

VEP

+−

IpK

IaK

GK

VK

+

Figure 7.17: Model for the endolymphatic potentialincluding active potassium transport (Problem 7.14).

These predictions of the passive model are not consistent with the results shown.Hence, the passive model does not account for the effect of a change in ion con-centration on VEP .

c. Part b shows that a model with purely passive transport does not account forthe results. Furthermore, since a change in the sodium concentration results inonly a small change in VEP , and a change in potassium concentration results in alarge change in VEP , a model that includes potassium transport is required. Thesimplest model that involves both active and passive potassium transport is shownin Figure 7.17. According to this model, at rest IpK + IaK = 0 so that GK(VEP −VK)+IaK = 0 which can be solved for VEP = VK − IaK/GK . Therefore, according to thismodel

VEP = 61 log

(cpKceK

)− I

aKGK mV.

Now how does this model compare with the measurements? Note that for cpK = 5mmol/L, VK = 61 log(5/150) = −90 mV. This implies that for VEP = 100 mV,IaK/GK = VK−VEP = −90−100 = −190 mV. Then, when the solution is changed sothat cpK = 1 mmol/L, VK = 61 log(1/150) = −133 mV, VEP = −133+190 = 57 mV.This prediction of the model with both active and passive potassium transport fitsapproximately with the experiments that are given.

Problem 7.15

a. Because the pump transports 3 sodium molecules out and 2 potassium moleculesinto the cell for each molecule of ATP split, the pump transfers net charge. Hence,this is inherently an electrogenic pump. However, the pump does not necessarilymake a large contribution to the normal resting potential of the cell.

b. An electric network model for the cell is shown in Figure 7.18. The resting potentialof the cell is obtained from the relation

Vom =GK

GK +GNaVK +GNa

GK +GNaVNa −JaK + JaNaGK +GNa .


Vm

+

−

VNa VK+−

+−

GNa

JpNa

JaNa

JpK

JaK

GK

Figure 7.18: Network model for iontransport in Problem 7.15.

Note that becauseGK GNa, the contribution of the sodium equilibrium potentialto the resting potential is small. The contribution of the the potassium equilib-rium potential to the resting potential is easily computed. Because, GK GNa,GK/(GK +GNa) ≈ 1 and

VK = 59 log10

(coKciK

)= 59 log10

(15

156

)= −60 mV.

Therefore, Vom ≈ VK , and the pump makes no contribution to the normal restingpotential.

c. The results are interpreted as follows: injection of sodium stimulates the pumpand this causes a contribution of −20 mV to the resting potential. Hence, if α isthe utilization of ATP by the pump in a unit area of membrane per second,

∆Vom = − JaK + JaNaGK +GNa ,

= −F φaK +φaNaGK +GNa ,

= −F−2α+ 3αGK +GNa = −

αFGK +GNa ≈ −

αFGK.

Therefore,

α ≈ −∆VomGKF

= (20× 10−3)(1× 10−3)9.65× 104 = 2.1× 10−10 mol/(cm2 · s).

Problem 7.16

a. If there is no net hydrogen ion transport at rest, the hydrogen ion concentrationratio must have a Nernst equilibrium potential of −70 mV,

−70 = 59 log10coHciH= 59 log10 c

oH − 59 log10 c

iH.

The pH of the extracellular and intracellular solutions are expressed in terms ofthe concentrations as

pHo = 7.3 = − log10 coH and pHi = − log10 c

iH

PROBLEMS 143

Distance (mm)0 2 4 6 8 10 12 14

0

5

10

15

20

Cou

nts

/min

0.3

Figure 7.19: Construction used to es-timate the mobility from the mea-surements (Problem 7.18). The longdashed line marks the peak of the fitof the count rate by a Gaussian func-tion. The short dashed line marks themid-point of the horizontal line thatdenotes the spatial extent of the ini-tial injection of radioactive calcium.

where the concentration are expressed in mol/L. A combination of these equationsyields

−7059= −1.19 = pHi − pHo.

Therefore, pHi = −1.19 + 7.3 = 6.11. Thus, the intracellular pH is acidic underthese conditions.

b. Since hydrogen is not at equilibrium at rest and since the pH of cells is constant,there must be additional hydrogen transporting mechanisms in cellular mem-branes.

Problem 7.17 To determine the validity of the suggestion, the work required to movesodium out of the cell against the electrical and chemical gradients must be computed.Equation 7.92 (Weiss, 1996a) gives a formula for the free energy required.

∆GNa = −νNaFVom + νNaRT ln

(coNaciNa

),

= −4(9.65× 104)(−70× 10−3)+ 4(8.314)(300) ln 8,= 2.7× 104 + 2.1× 104 J/mol = 48 kJ/mol.

Hence, the energy per mole required to pump out sodium exceeds that available fromhydrolysis of ATP. Dr. Tropsnart’s model is not energetically feasible.

Problem 7.18

a. As indicated in the solution to Problem 7.6, the mobility can be determined fromthe drift velocity.

i. Figure 7.19 shows that the center of the count rate distribution has movedslightly to the left of the center of the initial injection region. The displace-ment is at most 0.3 mm.

ii. Therefore, an upper bound on the drift velocity is

v = 3× 10−4 m(159 · 60) s

= 3.1× 10−8 m/s,


and an upper bound on the molar mechanical mobility is

uCa = vFE =

3.1× 10−8 m/s(9.65× 104 C/mol) · (51) V/m

= 6.3× 10−15 (m/s)/(N/mol).

iii. According to Table 7.2, the molar mechanical mobility for calcium in water(at infinite dilution and at 25C) is 1.60×10−13 (m/s)/(N/mol). Therefore, theestimate based on the data in Figure 7.19 shows that the mobility of calciumin axoplasm is less than 4% that in water. More extensive measurements(Hodgkin and Keynes, 1957) suggest that the mobility of calcium in axoplasmis less than 2% of that in water.

b. The diffusion coefficient can be obtained from the width of the count rate distri-bution. At time 159 minutes after injection, the curve that fits the measurementshas the parameter 2

√Dt = 0.262 cm. Therefore,

D = (0.131× 10−2)2

159 · 60= 1.8× 10−10 m2/s.

According to Table 7.2, the diffusion coefficient for calcium in water (at infinitedilution and at 25C) is 0.4 × 10−9 m2/s. Therefore, the estimate based on thedata in Figure 7.19 shows that the diffusion coefficient of calcium in axoplasm isless than 45% that in water. More extensive measurements (Hodgkin and Keynes,1957) suggest that the diffusion coefficient of calcium in axoplasm is about 10%of that in water.

c. Both the diffusion coefficient and the mobility of calcium in axoplasm are lowerthan in water. This likely results because calcium binds to a number of differentintracellular components and is not free to diffuse.

Problem 7.19

a. The answer is vi. Since the concentration of chloride in the cell is constant, thereis no flux of chloride across the cell membrane. Since there is no flux of chloride,there is no chloride current, and the resting potential is independent of the chlo-ride conductance GCl and the Nernst equilibrium potential for chloride, VCl. Themembrane potential is determined entirely by the flow of sodium and potassiumions. If that membrane potential did not equal the Nernst equilibrium potentialfor chloride, then chloride ions would flow passively across the membrane and al-ter the internal concentration of chloride until the chloride equilibrium potentialequalled the membrane potential.

b. If the concentration of sodium and potassium in the cell are constant, then therecan be no net flux of either ion. Therefore,

GNa(Vom − VNa)+ IaNa = 0,GK(Vom − VK)+ IaK = 0.

PROBLEMS 145

Combining these equations yields

GNa(Vom − VNa)GK(Vom − VK) = I

aNaIaK.

This equation can be expressed in terms of the dimensionless ratios γ = GNa/GKand α = −IaNa/IaK as

GNa(Vom − VNa)GK(Vom − VK) = −α.

Cross multiplying and combining terms yields the solution

Vom =γVNa +αVKγ +α .

This result demonstrates the validity of the argument given in part a, namely thatVom is independent of GCl and VCl. Using the numerical values provided yields

Vom =0.1 · 68+ 1.5 · (−68)

0.1+ 1.5= −59.5 mV.

c. Reduction of the external concentration of chloride from 150 to 50 mmol/L changesthe Nernst equilibrium potential for chloride as follows

∆VCl = −RTF ln

(50

ciCl(0+)

)− −RT

Fln

(150

ciCl(0−)

)= RTF

ln(

15050

),

if it is assumed that the intracellular concentration of chloride does not changeinstantaneously, i.e., ciCl(0+) = ciCl(0−). The temperature is not given, but theconcentrations and the Nernst equilibrium potentials for sodium and potassiumare given. Therefore,

68 = RTF log10 e

log(

14010

),

which shows that the RT/(F log10 e) ≈ 59 mV. Therefore,

∆VCl = 59 log(

15050

)≈ 28 mV.

i. The change in Vom that results from a sudden change in the chloride equilib-rium potential of 28 mV can be determined easily by superposition. The con-tribution of the chloride equilibrium potential to Vom is simply (GCl/Gm)VCl.Therefore, the change in Vom that results from a change in VCl is

∆Vom =GClGm∆VCl =

11+ 1+ 0.1

28 = 13.3 mV.

Therefore, Vom(0+) = −59.5+ 13.3 = −46.2 mV.

ii. Since the equilibrium value of the resting potential is independent of chlorideconcentration, it must be that Vom(∞) = −59.5 mV.


iii. After the cell has equilibrated, the chloride equilibrium potential adjusts itselfto the resting potential. Therefore,

−59.5 = −59 log

(50

ciCl(∞)

),

so that ciCl(∞) ≈ 5 mmol/L.

iv. The fundamental idea is that since chloride is transported only passivelyacross the membrane, at equilibrium the Nernst equilibrium potential forchloride must equal the resting potential. The resting potential is determinedby the passive and active transport of sodium and potassium only. Therefore,any change in chloride concentration causes a transient change in the mem-brane potential which is restored to its original value by a net flow of sodiumand potassium ions. The resulting change in potential drives a flow of chlo-ride ions so that chloride will once again be in equilibrium with the restingpotential.

Further discussion of problemTo analyze this problem more completely, let us determine the intracellular chlorideconcentration when the extracellular concentration is 150 mmol/L which is

−59.5 = −59 log

(150

ciCl(0+)

),

so that ciCl(0+) ≈ 15 mmol/L. With 150 mmol/L of external chloride the cell is in equi-librium with an intracellular concentration of chloride of 15 mmol/L. With a suddenreduction in extracellular chloride concentration, the system is taken out of equilib-rium and the chloride equilibrium potential is increased by 28 mV which results in achange in the potential across the membrane of 13.3 mV. Thus, chloride is now outof equilibrium across the membrane. The direction that chloride ions flow at t = 0+can be determined by noting that IpCl(0+) = GCl(Vm(0+) − VCl(0+)) and that the signof the current is determined only by the sign of the difference in potential. This dif-ference in potential is (−59.5 + 13.3) − (−59.5 + 28) = −14.7 mV. Therefore, the netchloride current is inward and the chloride flux is outward. However, the equilibriumpotential is determined by sodium and potassium concentrations and conductances andis independent of the chloride conductance and concentration. Therefore, the equilib-rium membrane potential is the same for both values of external chloride concentration.Therefore, chloride moves out of the cell to reduce the chloride equilibrium potentialback to −59.5 mV. In the process the intracellular concentration of chloride is reducedfrom 15 to 5 mmol/L.

The existence of a net chloride transport out of the cell implies that the assumptionsmade in the problem statement — i.e., that the intracellular concentrations of sodiumand potassium and the volume of the cell were constant — cannot be correct. Theefflux of chloride ions alone violates intracellular electroneutrality. Thus, if 10 mmol/Lof chloride are transported out of the cell, then 10 mmol/L of cation must also betransported out of the cell. Therefore, the intracellular osmolarity decreases by 20mosm/L and water must be transported out of the cell to reduce the cell volume and torestore osmotic equilibrium.

Chapter 8

CELLULAR HOMEOSTASIS

Exercises

Exercise 8.1

a. The ions are all univalent, and the sum of the cation concentration equals the sumof the anion concentrations for both solutions at t = 0. Therefore, both solu-tions satisfy electroneutrality. The osmolarity of both solutions is 200 mosm/L.Therefore, the solutions are in osmotic equilibrium. Ionic equilibrium can be as-sessed from the Nernst equilibrium potentials for the permeant ions. In general,Vn = 59 log10(c2

n/c1n) mV at normal room temperature. Therefore, for the cation,

VB = 59 log10(50/100), and for the anion, VC = −59 log10(100/50). Thus, bothNernst equilibrium potentials have the same value which is Vom = VB = VC . There-fore, the system is also in ionic equilibrium at t = 0.

b. From the analysis in part a, Vom = VB = VC = 59 log10(50/100) = −18 mV. The neg-ative membrane potential produces a drift of ions that just cancels the diffusionof both B and C.

c. The network model displayed in Figure 8.1 shows that IB = GB(Vm − VB) andIC = GC(Vm − VC). Since Vm = −40 mV and VB = VC = −18 mV, IB = −22GBand IC = −22GC in mA if G has units of siemens. Thus, there will be a net currentfrom side 2 to side 1 for both the cation and the anion. That implies that the cationwill flow from side 2 to side 1 to increase c1

B and decrease c2B . The anion will flow

Vm

GB

VB+−

IB

side 1

side 2

+−

+−

VC

IC

GC

Figure 8.1: Network model for Exercise 8.1.

147

148 CHAPTER 8. CELLULAR HOMEOSTASIS

from side 1 to side 2 to increase c2C and decrease c1

C . The flow will stop when theconcentrations have changed sufficiently that the Nernst equilibrium potentials ofboth B and C are VB = VC = −40 mV.

d. If an uncharged, impermeant protein is added to side 1 to give a concentration ofthis protein of 10 mmol/L = 10 mol/m3, electroneutrality is unaffected, ionic equi-librium is unaffected. However, side 1 will have an increase of osmotic pressure of10 mosm/L. Because the membrane and the walls of the volume are rigid, no watercan flow. Therefore, a hydraulic pressure will develop so that p1 − p2 = π1 −π2.Therefore, p1−p2 = RT10 ≈ 25 kPa so that volume 1 will be at a higher hydraulicpressure than volume 2.

Exercise 8.2 a. 13; b. 12; c. 8; d. 16; e. 6; f. 2; g. 15; h. 3; i. 10; j. 1; k. 7.

Exercise 8.3

a. Water.

b. Cells with vasopressin receptors respond to an increase in systemic vasopressinwith an increase in water flux in response to a difference in osmotic pressureacross the membrane.

c. Glucose.

d. Cells with insulin receptors respond to an increase in systemic insulin with anincrease in glucose flux in response to a difference in glucose concentration acrossthe membrane.

e. Both vasopressin and insulin act by binding to their respective receptors on themembrane. Via second messengers, vasopressin and insulin result in the recruit-ment of cytoplasmic water channels and glucose carriers, respectively, and theirincorporation into the membrane.

Problems

Problem 8.1 There are 5 concentrations in this problem and the values of two of themare given. Therefore, 3 independent equations are needed to specify the remainingconcentrations. Electro-neutrality of the two solutions provides 2 equations; osmoticequilibrium across the membrane provides the third. Since, the membrane is permeableto chloride, the potential across the membrane at equilibrium must be the chlorideequilibrium potential. Electroneutrality of the two solutions yields

c1Cl + c1

A = c1K and c2

Cl = c2K,

which becomesc1Cl + 135 = c1

K and 150 = c2K,

PROBLEMS 149

so that c2K = 150 mmol/L. Osmotic equilibrium yields

c1Cl + c1

A + c1K = c2

Cl + c2K,

which givesc1Cl + 135+ c1

K = 300.

Substitution of electroneutrality of side 1 into osmotic equilibrium yields

2c1Cl + 270 = 300,

so that c1Cl = 15 and c1

K = 150 mmol/L. For equilibrium of chloride

Vm = VCl = RTzClF

ln

(c2Cl

c1Cl

)= −59 log10

(15015

)= −59 mV.

Problem 8.2 At equilibrium several conditions must hold simultaneously.

• The flux of each ion is zero. Since there are only two permeant ions in this problem,the flux of each must be zero, i.e.,

φNa = φCl = 0.

If transport is by passive means only, then the potential across the capillary wallmust equal the Nernst equilibrium potential of both sodium and chloride. If thereference direction for potential is defined as positive when the inside of the cap-illary (the plasma) is at a positive potential with respect to the interstitial fluidsthen Vom = VNa = VCl, which implies that

RTF

lnciNacpNa

= −RTF

lnciClcpCl,

which yields the Donnan ratiociNacpNa

= cpCl

ciCl,

where the superscripts i and p refer to interstitial fluids and plasma, respectively.

• The flux of volume is zero at equilibrium. Therefore, the difference in hydraulicpressure across the capillary wall must equal the difference of osmotic pressure,i.e., pp − pi = πp −πi.

• Both the plasma and the interstitial fluids obey electroneutrality. The interstitialfluids are defined as 155 mmol/L NaCl. Hence, ciNa = ciCl, which obeys electroneu-trality. For electroneutrality of plasma,

∑znc

pn = 0. Therefore, cpNa−cpCl+zpcpP =

0, where zP and cpP are the valence and the concentration of protein in plasma.

a. For this part zP = 0.


i. Electroneutrality implies that since zP = 0, cpNa = cpCl. The Donnan ratiotogether with electroneutrality of the interstitial fluids and plasma impliesthat (ciNa)2 = (cpNa)2. Therefore, ciNa = cpNa. Therefore, cpNa = cpCl = 155mmol/L.

ii. Since all the concentrations are the same on both sides of the capillary wall,the Nernst equilibrium potentials for sodium and chloride are zero and thepotential across the membrane is zero.

iii. Because of the presence of the uncharged protein in plasma, the plasma is athigher osmotic pressure than the interstitial fluids, so that 27C

πp −πi = RT(311− 310)× 10−6

=(

8.314× 106 Pa

K · (mol/cm3)

)(300 K)(10−6 mol/cm3)

= 2.5 kPa.

iv. Therefore, pp − pi = 2.5 kPa. The plasma is at a higher hydraulic pressurethan the interstitial fluids.

b. For this part zP = −50.

i. Electroneutrality of plasma requires that cpNa = cpCl+50cP . It is convenient toexpress all concentration in mmol/L in this problem. With these units cP = 1.The Donnan ratio shows that since ciCl = ciNa, therefore,

155

cpNa= c

pNa − 50

155,

which results in the following quadratic equation

(cpNa)2 − 50cpNa − (155)2 = 0.

A similar quadratic equation is satisfied by chloride

(cpCl)2 + 50cpCl − (155)2 = 0.

Solutions for positive concentrations are cpNa = 182 and cpCl = 132 mmol/L.

ii. The potential across the membrane is simply the Nernst equilibrium potentialfor the ions which must be the same.

Vom = 59 log155182

= −59 log155132

= −4 mV.

iii. The osmotic pressure difference is

πp −πi = RT(182+ 132+ 1− 310)× 10−6

=(

8.314× 106 Pa

K · (mol/cm3)

)(300 K)(5× 10−6 mol/cm3)

= 12.5 kPa.

PROBLEMS 151

iv. Therefore, pp − pi = 12.5 kPa. The plasma is at a higher hydraulic pressurethan the interstitial fluids.

Note that the presence of a small concentration of highly-charged impermeant solutehas a large effect on the equilibrium values of the concentrations, causes a potential toexist across the capillary wall, and greatly increases the hydraulic pressure differenceacross the wall.

Problem 8.3 At t = 0 the osmolarity of the extracellular solution is increased from300 mosm/L to 360 mosm/L. Therefore, water will leave the cell to establish osmoticequilibrium, the cell will shrink, and the concentrations of impermeant solutes in thecell will increase.

a. At equilibrium

CoΣ =NiΣV i .

Therefore, if there are two different extracellular osmolarities then

Co1Σ = NiΣ

V i1 and Co2Σ = NiΣ

V i2 .

The quantity of intracellular solute does not change since the solutes are all im-permeant. The ratio of equations yields

Co1Σ

Co2Σ= V

i2

V i1 .

Therefore,

V i2 = Co1Σ

Co2ΣV i1 = 300

360100 = 83.3 µm3.

b. The intracellular concentration of any component is the ratio of the quantity tothe volume. Therefore,

ninV i1 = c

i1n and

ninV i2 = c

i2n .

Taking the ratio of these equations yields

ci2n =V i1V i2 c

i1n .

This result can be used to find the concentrations of intracellular solutes at equi-librium as follows,

ci2 = 10083.3

180 = 216 µm3,

ci5 = 10083.3

120 = 144 µm3.

The concentrations of all other intracellular solutes is zero.


Problem 8.4

a. Electroneutrality of the intracellular and extracellular solutions implies that

z+ci+ + z−ci− = 0 and z+co+ + z−co− = 0.

b. At osmotic equilibrium, the osmolarity of the solutions on the two sides of themembrane are equal

ci+ + ci− = co+ + co−.

c. The independent variables are iii, iv, vi, vii, and viii. The dependent variables arei, ii, v, ix, and x.

d. The equations for electroneutrality yield

co− = −z+z−co+ and ci− = −

z+z−ci+.

Substitution into the equation of osmotic equilibrium yields

ci+(

1− z+z−

)= co+

(1− z+

z−

),

which implies that ci+ = co+ and ci− = co−.

e. From the relation ci− = co−ni−V = co− so that V = n

i−co−.

f. The resting membrane potential is

Vom = V+ =RTz+F

ln

(co+ci+

)= RTz+F

ln(1) = 0.

g. The results for the binary and z+-z− electrolytes are very similar. In both cases,both the cation and the anion have the same intracellular and extracellular con-centrations, and the potential across the membrane is zero. In each case, the cellact as a perfect osmometer. The only differences are due to electroneutrality. Inthe binary electrolyte ci− = ci+, in the z+-z− electrolyte ci− = −(z+/z−)ci+.

Problem 8.5

a. Electroneutrality of the intracellular solution implies that

ci+ + z−ci− = 0.

b. At osmotic equilibrium, the osmolarity of the solutions on the two sides of themembrane are equal

ci+ + ci− = 2C.

PROBLEMS 153

−5 −4 −3 −2 −1 0

0.2

0

0.4

0.6

0.8

1

z−

ci+/2C

ci−/2CFigure 8.2: Dependence of cation and anion concentra-tions on the valence of the anion (Problem 8.5).

c. The independent variables are iii, iv, vi, vii, and viii. The dependent variables arei, ii, v, ix, and x.

d. The equation for electroneutrality yields

ci− = −1z−ci+.

Substitution into the equation of osmotic equilibrium yields

ci+(

1− 1z−

)= 2C,

which implies that

ci+ =(z−

z− − 1

)2C and ci− =

(1

1− z−)

2C.

The dependence of the concentrations on the valence of the anion is shown inFigure 8.2. To satisfy electroneutrality, z− < 0. Hence, we plot the concentrationsonly over this range. Although a single anion has a valence that is an integer, amixture of anions could have an equivalent non-integer valence. Hence, it is validto plot the concentrations as a function of negative, real values of the valence. Asa consequence of electroneutrality of the intracellular solution, as the valence ofthe anion is made more negative, the cation concentration increases and the anionconcentration decreases. At a valence of z− = −1, the concentrations of the cationand the anion are equal, and equal their extracellular concentrations.

e. From the relation

ci− =ni−V =

(1

1− z−)

2C,

it follows that

V = ni−(1− z−)

2C.

The volume is plotted versus the anion valence in Figure 8.3. As the valence be-comes more negative, the volume increases because electroneutrality requires alarger concentration of cation to neutralize the anion. Hence, the osmotic pressureincreases intracellularly and water enters to increase the cell volume and establishosmotic equilibrium.


−5 −4 −3 −2 −1 0

2

0

4

6

z−

V2C/ni

− Figure 8.3: Dependence of volume on the valenceof the anion (Problem 8.5).

V om

RT/F

z−

−5 −4 −3 −2 −1

−1

1

2

3

4

Figure 8.4: Dependence of resting potential on thevalence of the anion (Problem 8.5).

f. The resting membrane potential is

Vom = V+ = RTz+F

ln

(co+ci+

)

= RTF

ln

C(z−z−−1

)2C

= RT

Fln(z− − 1

2z−

).

The dependence of the resting potential on the valence of the anion is shown inFigure 8.4. When the valence has a small magnitude (−1 < zi < 0), the concen-tration of the cation required to satisfy electroneutrality is small which makes theresting potential positive (to prevent influx of cation). At a valence of z− = −1, theconcentrations of anion and cation are the same and each equal C , the extracellu-lar concentrations of cation and anion. Hence, the resting potential is zero. As thevalence is further decreased, the intracellular concentration of cation increasesabove its extracellular concentration and the resting potential becomes negative(to prevent efflux of cation).

g. In contrast to the binary electrolyte, the concentrations of both the anion andcation are not equal for the electrolyte in this problem. Furthermore, there is apotential across the membrane for this electrolyte. However, the cell acts as anosmometer in both cases although the osmotic behavior for the electrolyte in thisproblem depends on the intracellular anion valence.

Problem 8.6

a. Electroneutrality implies that

ci+ − ci− + zicii = 0.

PROBLEMS 155

b. At electrodiffusive equilibrium the potential across the membrane must equal theNernst equilibrium potential of both the cation and the anion, i.e.,

Vom = V+ = V− =RTF

ln

(Cci+

)= −RT

Fln

(Cci−

),

which implies thatCci+= c

i−C,

and that ci+ci− = C2.

c. A combination of the two equations yields

ci+ −C2

ci++ zicii = 0,

which yields(ci+)2 + (zicii)ci+ − C2 = 0.

This quadratic equation can be factored to yield

ci+ =−zicii ±

√(zicii)2 + 4C2

2.

Since the concentration must be a positive quantity, the positive root is chosen onphysical grounds,

ci+ =√(zicii)2 + 4C2 − zicii

2.

But ci− = C2/ci+. Rationalization to remove the radical in the denominator yields

ci− =√(zicii)2 + 4C2 + zicii

2.

The concentrations can be expressed more conveniently as

ci+C

=√√√√(zicii

2C

)2

+ 1− zicii

2C,

ci−C

=√√√√(zicii

2C

)2

+ 1+ zicii

2C.

These normalized concentrations are plotted in Figure 8.5. When zicii/(2C) = 0,which occurs if zi = 0 and/or if cii = 0, the intracellular concentrations of cationsand anions are equal to their extracellular concentrations. The total concentra-tion of cations and anions, or the osmolarity, is twice that value. If zi > 0, theconcentration of cations decreases and the concentration of anions increases inorder to satisfy electroneutrality. With a positively charged impermeant ion, theconcentration of intracellular anion concentration exceeds the extracellular anionconcentration. If zi < 0, the concentration of cations increases and the concentra-tion of anions decreases in order to satisfy electroneutrality.


−3 −2 −1 1 2 3

1

2

3

4

5

6

ci+/C ci−/C

zicii

2C

Figure 8.5: Dependence of concentrations of cation andanion on the concentration of impermeant solute (Prob-lem 8.6).

0.5 1 1.5 2

2

4

6

cii2C

∆π

2RTC 1

2

3

Figure 8.6: Dependence of the osmotic pressuredifference on the concentration of impermeant so-lute (Problem 8.6). The parameter with each traceis the value of |zi|.

d. The osmotic pressure difference is

πi −πo = RT(ci+ + ci− + cii − 2C),

= RT

√(zicii)2 + 4C2 − zicii

2+√(zicii)2 + 4C2 + zicii

2+ cii − 2C

,= RT

(√(zicii)2 + 4C2 + cii − 2C

).

Let ∆π = πi −πo, then

∆π2RTC

=√√√√(zicii

2C

)2

+ 1+ cii2C− 1,

which is plotted in Figure 8.6. The osmotic pressure is an even function of zi.Hence, we consider plots of the dependence of the osmotic pressure difference onthe concentration of impermeant ion for different values of |zi|. When the con-centration of impermeant ion is zero, the osmotic pressure intracellularly equalsthat extracellularly. Increasing the concentration of impermeant ion, for eitheran anion or a cation, increases the osmotic pressure difference above the osmoticpressure of the extracellular solution.

e. The osmotic pressure difference is zero when√(zicii)2 + 4C2 = 2C − cii .

Notice that √(zicii)2 + 4C2 ≥

√4C2 = 2C ≥ 2C − cii .

PROBLEMS 157

Therefore osmotic pressure will be zero only if the two “≥” are in fact “=”. Thesecond “≥” is satisfied with equality only if cii = 0, which also guarantees that thefirst “≥” is satisfied with equality. Therefore the osmotic pressure is zero if andonly if cii = 0.

Problem 8.7

a. Figure 8.8 (Weiss, 1996a) deals with a cell model in which there is a binary elec-trolyte on both sides of the membrane and an additional impermeant intracellularion of valence zi. The membrane is permeant only to the cation. The results inFigure 8.23 (Weiss, 1996a) are for zi = 1, and indicate that c− = 1, c+ = 1 − ni,and ci = ni. Therefore, the intracellular osmolarity is

c− + c+ + ci = 1− (1− ni)+ ni = 2,

which equals the normalized extracellular osmolarity.

b. The concentrations of all intracellular charges are added as follows

c+ − c− + ci = 1− ni − 1+ ni = 0.

Therefore, the intracellular solution satisfies electroneutrality.

c. Physically plausible solutions occur only if nizi ≤ 1. Since zi = 1 in this case,physically plausible solutions occur only if ni ≤ 1. In this range, all concentrationsand quantities are non-negative and the volume is a non-negative quantity.

Problem 8.8

a. i. Electroneutrality of the outside solution is∑n zncon = 0 which yields

coNa = coCl.

ii. Electroneutrality of the inside solution is∑n zncin = 0 which yields

ciNa = ciCl.

iii. Since there is no hydraulic pressure difference across the tubule, osmoticequilibrium across the membrane implies that CoΣ = CiΣ which yields

coNa + coCl = ciNa + ciCl +nMV .

iv. For electrodiffusive equilibrium of chloride ions

Vom = VCl = −RTF

ln

(coClciCl

).


+

−

VNa+−

+−

JaNa

VCl

GCl GNa

V om

Figure 8.7: Equivalent circuit for current flow be-tween the inside and outside of the tubule (Prob-lem 8.8).

v. For quasi-equilibrium of sodium, the net flux of sodium is zero. Therefore,the current density carried by sodium must be zero

JaNa +GNa(Vom −

RTF

ln

(coNaciNa

))= 0.

b. Current between the inside and outside of the tubule consists of a passive chloridecurrent and both a passive and an active sodium current as indicated in Figure 8.7.

c. A combination of quasi-equilibrium of sodium and electrodiffusive equilibrium ofchloride yields

Vom = −RTF

ln

(coClciCl

)= − J

aNaGNa

+ RTF

ln

(coNaciNa

),

which results in

JaNaGNa

= RTF

ln

(coClciCl

)+ RTF

ln

(coNaciNa

),

JaNaGNa

= RTF

ln

(coClc

oNa

ciClciNa

).

Since, ciCl = ciNa and coCl = coNaJaNaGNa

= 2RTF

ln

(coNaciNa

).

Therefore,

ciNa = coNae−FJaNa/(2RTGNa),

ciCl = coCle−FJaNa/(2RTGNa).

The equations for Vom and ciCl can be combined to yield

Vom = −JaNa

2GNa.

From osmotic equilibrium

V = nMcoNa + coCl − (ciNa + ciCl)

= nM2coNa − 2ciNa

= nM2coNa

(1− e−FJaNa/(2RTGNa)

) .

PROBLEMS 159

0 1 2 3

1

2

3

4

5

V(2coNa/nM )

JaNa/(2RTGNa/F )

Figure 8.8: Normalized volume of the tubule as afunction of the normalized current density due toactive sodium transport (Problem 8.8).

d. Examination of the expressions derived in part c reveals that as JaNa → 0, ciNa →coNa, ciCl → coCl, V

om → 0, and V → ∞ (Figure 8.8). That is, if active transport

of sodium is reduced to zero then the concentrations of sodium and chloride inthe tubule approach the concentrations outside the tubule, the potential betweenthe inside and the outside of the tubule approaches zero, and the tubule swellswithout bound. Only an infinite volume can satisfy osmotic equilibrium whenthe concentrations of sodium and potassium inside and outside are the same andthere is a fixed quantity of impermeant solute inside. An infinite volume makesthe osmolarity of this component zero. Examination of the expressions derivedin part c also reveals that as JaNa → ∞, ciNa → 0, ciCl → 0, Vom → −∞, and V →nM/(coNa). That is, if the active transport of sodium is increased without limit, thenthe concentrations of sodium and chloride in the tubule approach zero. Sodiumconcentration approaches zero because sodium is pumped out faster than it leaksinto the tubule. Chloride must follow sodium out to maintain electroneutrality.Therefore, the internal osmotic pressure is due only to M and the osmolarity isnM/V which must equal the outside osmolarity which is 2coNa. For an arbitrarilylarge active sodium current, the potential between the inside and outside of thetubule gets arbitrarily negative. That is, the tubule is hyperpolarized by the activesodium current as can be seen from Figure 8.7.

Problem 8.9

a. i. Solute A is impermeant so that the number of moles of A on the two sides ofthe partition remains the same. Water flows from side 2 to side 1 to equili-brate the concentration. The initial values of variables are as follows,

n1A(0) = c1

A(0)AL2

and n2A(0) = c2

A(0)AL2.

Therefore, the total number of moles of A are

NA =(c1A(0)+ c2

A(0)) AL

2.

Therefore, the final concentrations are

c1A(∞) = c2

A(∞) =(c1A(0)+ c2

A(0))AL2

AL,


c1A(∞) = c2

A(∞) =c1A(0)+ c2

A(0)2

= 10+ 52

= 7.5 mmol/L.

To find the location of the partition, note that the final concentration of A onside 1 is

c1A(∞) = c1

A(0)(AL/2)Ax(∞) ,

7.5 = 10(L/2)x(∞) .

Therefore, x(∞) = 2L/3. This result can be checked by considering c2A,

c2A(∞) = c2

A(0)(AL/2)A(L− x(∞)) ,

7.5 = 10(L/2)x(∞) .

Solving this equation also yields x(∞) = 2L/3.

ii. The equilibrium in concentration will be the same as for solute A discussedin part a.i. c1

B(∞) = c2B(∞) = 7.5 mmol/L, except that now both water and

solute B will flow through the partition.

iii. Several conditions must hold for the two solutions, the position of the par-tition, and the potential across the partition. The solutions must obey elec-troneutrality, the two baths must be in osmotic equilibrium, and the two bathsmust be in electrodiffusive equilibrium. These relations are checked in turn.

• Electroneutrality is checked first. Both side 1 and side 2 contain 10mmol/L of cation and 10 mmol/L of anion. Hence, both solutions sat-isfy electroneutrality.

• The osmolarity is 20 mmol/L on both sides of the partition. Hence, thereis osmotic equilibrium at t = 0 and there is no transport of water acrossthe partition.

• Electrodiffusive equilibrium needs to be checked for ions E and F onlysince only these are permeant. At equilibrium, the potential across themembrane must equal the Nernst equilibrium potentials of both ions.Thus, both of these equilibrium potentials needs to be checked.

VE = RTF

ln

(c2E(0)c1E(0)

)= 26 ln

(5

10

)= −18 mV,

VF = RTF

ln

(c2F(0)c1F(0)

)= −26 ln

(105

)= −18 mV,

Therefore, V(∞) = −18 mV and there is no net transport of solutes E andF. The system is at equilibrium at t = 0 and remains at equilibrium.

PROBLEMS 161

b. Conservation of solute B requires that

− 1Adn1

B(t)dt

= φB = PB(c1B(t)− c2

B(t)),

− 1Adn1

B(t)dt

= PB

(n1B(t)Ax(t)

− NB −n1B(t)

A(L− x(t))

),

dn1B(t)dt

= −PB(Ln1

B(t)−NBx(t)x(t)(L− x(t))

).

Therefore, αn = −PB , βn = L, and γn = NB .

c. Conservation of water volume requires that

− 1Ad(Ax(t))dt

= ΦV = LVRT(c2B(t)− c1

B(t)),

dx(t)dt

= LVRT(c1B(t)− c2

B(t)),

dx(t)dt

= LVRT(n1B(t)Ax(t)

− NB −n1B(t)

A(L− x(t))

),

dx(t)dt

= LVRTA

(Ln1


).

Therefore, αx = LVRT/A, βx = L, and γx = NB .

d. The partition is permeable to solute B so that both solute B and water will flowacross the partition in response to a solute concentration difference. The param-eter δ is a measure of the relative permeability of the partition to solute and towater. Intuitively, if δwere arbitrarily large then solute B would cross the partitionto equilibrate the compartment concentrations before any water was transported.Thus, n would change a great deal and y would not change much. Therefore,the correct answers are n3 and y3. Alternatively, for arbitrarily small values of δ,very little solute would be transported before the water transport had establishedosmotic equilibrium across the partition. Thus, for small values of δ, n would notchange much and y would change appreciably. This case corresponds to n1 andy1

The following addendum explains how the results shown in Figure 8.26 (Weiss, 1996a)were obtained. From parts b and c

dn1B(t)dt

= −PB(Ln1


),

dx(t)dt

= LVRTA

(Ln1


).

These equation are normalized by dividing by NB and L as follows

dn1B(t)/NBdt

= −PBL

(n1B(t)/NB − x(t)/L

(x(t)/L)(1− x(t)/L)

),

dx(t)/Ldt

= LVRTNBAL2

(n1B(t)/NB − x(t)/L

(x(t)/L)(1− x(t)/L)

).


The normalized variables are defined as n = n1B(t)/NB , y = x/L, and normalized time

as

τ =(LVRTNB

AL2

)t,

to obtain

dn(τ)dτ

= −δ(n(τ)−y(τ)y(τ)(1−y(τ))

),

dy(τ)dτ

=(n(τ)−y(τ)y(τ)(1−y(τ))

).

where

δ = ALPBNBRTLV .

The two normalized coupled equations have similar right-hand sides. Hence,

dn(τ)dτ

= −δdy(τ)dτ

,

which can be integrated to yield

n(τ) = no − δy(τ).

Therefore, both n and y have the same time dependence. The solutions were obtainedby numerical integration of the coupled differential equations.

Problem 8.10 Since the intracellular solutions are the same for all the cell models, cer-tain relations are common to all the cell models. Electroneutrality of the intracellularsolution requires

ciNa + ciK − ciCl + zAciA = 0,

where zA is the valence of anion A. Electroneutrality of the extracellular solution re-quires that

coNa + coK − coCl = 0.

Osmotic equilibrium across the membrane requires that

ciNa + ciK + ciCl + ciA = coNa + coK + coCl.

a. For Model 1, electrodiffusive equilibrium for potassium requires that the flux ofpotassium is zero,

φK = GKF (Vom − VK) = 0.

Therefore, Vom = VK .

i. False. The membrane is impermeable to chloride. Hence, there is no fluxof chloride no matter what the values of the chloride concentrations or themembrane potential.

ii. True.

PROBLEMS 163

iii. False. Suppose the membrane were impermeant to all ions. Then a change incomposition of the extracellular solution would result in a change in cell vol-ume to satisfy osmotic equilibrium. If the membrane is permeable to potas-sium, then an infinitesimal flow of potassium across the membrane estab-lishes electrodiffusive equilibrium so that the potential across the membraneis the Nernst equilibrium potential for potassium. Thus, potassium is in equi-librium. Since the quantity of charge required to establish the potential is sosmall, no appreciable change in concentration occurs and there is not furtherchange in cell volume.

iv. False. Since the cell water volume is bounded, there is no necessity that ciNa =0.

b. For Model 2, electrodiffusive equilibrium requires that both φK = 0 and φCl = 0.Therefore,

φK = GKF (Vom − VK) = 0,

and

φCl = GClF (Vom − VCl) = 0.

Therefore, Vom = VK = VCl.i. True.

ii. True.

iii. False. There is no sodium-potassium pump in this model.

c. For Model 3, at quasi-equilibrium for chloride, the net chloride flux must be zero,

φCl = −GClF (Vom − VCl)+ 2αco = 0.

At quasi-equilibrium for potassium, the net potassium flux must be zero,

φK = GKF (Vom − VK)+αco − νKαATP = 0.

At quasi-equilibrium for sodium, the net sodium flux must be zero,

φNa = αco + νNaαATP = 0.

In addition, at rest the total current through the membrane must be zero,

Jm = GK(Vom − VK)+GCl(Vom − VCl)+ (νNa − νK)FαATP = 0.

The cotransporter makes no direct contribution to the resting potential becauseit passes no net current. The solution for Vom in terms of the other variables is

Vom =(

GKGK +GCl

)VK +

(GCl

GK +GCl)VCl −

((νNa − νK)FGK +GCl

)αATP .

i. False. It can be seen from above that the resting potential does not equal thechloride equilibrium potential in general.


ii. False. It can be seen from above that the resting potential does not equal thepotassium equilibrium potential in general.

iii. False. As can be seen from the equation for quasi-equilibrium of sodium, thetwo rates αco and αATP are linked but neither is constrained to be zero.

iv. False. A change in αATP results in a direct change in Vom which changes thepotential (Vom − VCl) that drives chloride through the membrane.

d. For Model 4, at electrodiffusive equilibrium for chloride, the net chloride flux mustbe zero,

φCl = −GClF (Vom − VCl) = 0.

Therefore, Vom = VCl. At quasi-equilibrium for potassium, the net potassium fluxmust be zero,

φK = GKF (Vom − VK)− νKαATP = 0.

At equilibrium for sodium, the net sodium flux must be zero,

φNa = νNaαATP = 0.

Therefore, αATP = 0 which implies that (GK/F)(Vom − VK) = 0 so that Vom = VK .

i. True.

ii. True.

iii. True.

e. For Model 5, at electrodiffusive equilibrium for chloride, the net chloride flux mustbe zero,

φCl = −GClF (Vom − VCl) = 0.

Therefore, Vom = VCl. At quasi-equilibrium for potassium, the net potassium fluxmust be zero,



φNa = GNaF (Vom − VNa)+ νNaαATP = 0.

Elimination of αATP between the last two equations yields

αATP = φpKνK= −φ

pNaνNa

,

where

φpK =GKF(Vom − VK) and φpNa =

GNaF(Vom − VNa).

Therefore,φpKνK+ φ

pNaνNa

= 0.

PROBLEMS 165


Jm = GK(Vom − VK)+GNa(Vom − VNa)+ (νNa − νK)FαATP = 0,

where the chloride current is zero and has been omitted.

i. True.

ii. False. This is true only if αATP = 0 which is not required in general.

iii. False. This is true only if φpK = φpNa = 0 which is not required in general.

iv. True.

f. For Model 6, at quasi-equilibrium for potassium, the net potassium flux must bezero,



φNa = GNaF (Vom − VNa)+ νNaαATP = 0.


Jm = GK(Vom − VK)+GNa(Vom − VNa)+ (νNa − νK)FαATP = 0.

A solution for αATP is obtained from one of the flux relations, e.g.,

αATP = GKFνK

(Vom − VK),

which is substituted into the resting condition equation to yield

GK(Vom − VK)+GNa(Vom − VNa)+ (νNa − νK)GKνK(Vom − VK) = 0.

Rearranging terms yields(GK +GNa +GK

(νNa − νKνK

))Vom = GKVK +GNaVNa + (νNa − νK)

GKνKVK,

which, after collecting terms yields

Vom =(

GK/νKGK/νK +GNa/νNa

)VK +

(GNa/νNa

GK/νK +GNa/νNa)VNa

i. False. Chloride is not permeant.

ii. False. This is true only if αATP = 0 which is not required in general.

iii. True.

Bibliography

Hodgkin, A. L. and Keynes, R. D. (1957). Movements of labelled calcium in squid giantaxons. J. Physiol., 138:253–281.

Patlak, C. S. (1960). Derivation of an equation for the diffusion potential. Nature,188:944–945.

Perrin, J. (1909). Movement Brownien et realite moleculaire. Ann. Chim. Phys., 58:5–114.

Weiss, T. F. (1996a). Cellular Biophysics. Volume 1: Transport. MIT Press, Cambridge,MA.

Weiss, T. F. (1996b). Cellular Biophysics. Volume 2: Electrical Properties. MIT Press,Cambridge, MA.

Weiss, T. F. (1997). Solutions to Problems in Cellular Biophysics. Volume 2: ElectricalProperties. MIT Press, Cambridge, MA.

Weiss, T. F., Trevisan, G., Doering, E. B., Shah, D. M., Huang, D., and Berkenblit, S. I.(1992). Software for teaching physiology and biophysics. J. Sci. Ed. Tech., 1:259–274.

167

solutions to problems in cellular biophysics volume … · solutions to problems in cellular...

Documents