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Intermediate Physics for Medicine and Biology - Russell K. Hobbie & Bradley J. Roth

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9.10 Possible E ects of Weak External Electric and Magnetic Fields

247

conducting medium in which there is an electric field E1 far from the cell. The potential can be determined analytically by solving Poisson’s equation (with zero charge density) in the three regions and matching boundary conditions much as we did to obtain Eq. 9.68. The results, valid for slowly varying applied fields such as a 50or 60-Hz power line field, are shown in Fig. 9.19.10

Only the amplitude of the electric field is shown. Assume the conductivities of the extracellular and intracellular fluids are the same, that a = 10 µm and b = 6 nm, and that σmembrane = 2.4×108σ. The important features of this solution are that the field just outside the cell is roughly the same as the field far away, the field inside the membrane is magnified by a large factor (a/b), and the field inside the cell is multiplied by a very small factor (membrane/bσ). Thus, the cell membrane shields the intracellular space from extracellular electric fields, so these fields are not likely to directly a ect cell organelles and important biomolecules such as DNA. This is reflected in the last line of Table 9.5.

9.10.9 Electrical Interactions and Noise

If an organism is a ected in some way by an external field, then it can be regarded as a detector of that field. The external field can therefore be thought of as a signal. To be detected, the signal must be greater than the noise. The noise can be either thermal (Johnson) noise, shot noise, or noise from the electric currents that normally flow in the body due to nerve conduction and muscle contraction. To have a signal that is not masked by Johnson noise, we must have an electric field E such that

zev

=

zeEd

> 1,

(9.71)

kB T

kB T

 

 

 

where z is the valence of an e ective charge that moves a distance d in the electric field E. Table 9.5 shows the result of a calculation using a field in air of 300 V m1. We use a value z = 10e. For d we use the diameter of the cell, d = 10 µm (though for the membrane perhaps the much smaller thickness of the cell membrane should be used). The values of zeEd/kB T are very small.

One proposal to overcome this signal-to-noise problem is that the biological e ect is due to the averaging of the field over many cells or over time. This was first proposed by Weaver and Astumian (1990), and a specific model has been formulated by Astumian et al.(1995). The model applies the Nernst-Planck equation (Eq. 9.38) and shows that if the concentration of some substance outside the cell is much larger than inside, the response to an oscillating v is “rectification” or a net inward current. This would allow an accumulation of the substance within the cell. The averaging times in their model are 13 h. Weaver and Astumian (1995) review the entire causality problem,

10Calculated using equations in Polk (1995), p. 62.

TABLE 9.5. Comparison of the signal in a cell to thermal

noise for an applied electric field in air of 300 V m1. T = 300 K. z = 10e. d = 105 m.

 

Just outside

In the cell

Inside

 

 

Model

the cell

membrane

the cell

 

 

 

 

 

 

 

 

 

 

 

E (V m1)

1.01

×

105

1.62

×

102

5.40

×

1010

kB T /eE (m)

2.57

3

1.59

 

 

10

7

×

10

 

 

4.79 ×

 

zeEd/kB T

3.9 × 108

6.3 × 105

2.1 × 1012

 

 

 

 

 

 

 

 

 

 

 

including the e ects of shot noise. Adair (2000) reviews many other aspects of the problem.

9.10.10Magnetic Interactions and Noise

The magnetic field is not attenuated at the body surface, as the electric field is. Kirschvink et al. (1992a) reported that the human brain contains several million magnetosomes per gram. Kobayashi et al. (1995) found that contamination with magnetic particles could a ect laboratory experiments with cell cultures, even if the cells being studied do not normally contain magnetosomes. Commercial disposable, presterilized plastic laboratory ware used in tissue culture experiments were found to contain ferromagnetic particles smaller than 100 nm that are readily taken up by white blood cells.

What about the signal-to-noise ratio for magnetic effects? The situation is somewhat more favorable than for the electric case. We saw in Chapter 8 that a single magnetosome has appreciable alignment with the earth’s magnetic field, even in the presence of thermal bombardment. The earth’s field is about 5 × 105 T. For a single magnetosome

mBearth

=

(6.4 × 1017)(5 × 105)

= 0.77. (9.72)

kB T

(1.38 × 1023)(300)

 

 

For a larger magnetosome of radius 100 µm, m = 2 × 1015 A m2 and the energy ratio in the earth’s field is 24. The field due to a typical power line is about 100 times smaller: about 2 × 107 T.

Kirschvink (1992) proposed a model whereby a magnetosome in a field of 104–103 T could rotate to open a membrane channel. As an example of the debate that continues in this area, Adair (1991, 1992, 1993, 1994) argued that a magnetic interaction cannot overcome thermal noise in a 60-Hz field of 5 × 106 T. However, Polk (1994) argues that more biologically realistic parameters, including a large number of magnetosomes in a cell, could allow an interaction at 2 × 106 T.

The essential features of all the models are like this. Imagine a particle with magnetic moment m in the

248 9. Electricity and Magnetism at the Cellular Level

B earth

B earth

m

m θ

 

B0

(a)

(b)

FIGURE 9.20. A particle with

magnetic moment m (a)

aligned with the earth’s magnetic field and (b) at an angle θ with the earth’s field because of an applied field B0.

earth’s field. It will tend to align with the field as shown in Fig. 9.20(a). The direction of the magnetic moment with the earth’s field is θ. Apply an alternating field B0 cos ωt at right angles to the earth’s field, as shown in Fig. 9.20(b). There are three torques on the particle. The first is viscous drag, which is proportional to the angular velocity of the particle dθ/dt but in the opposite direction. The second is the torque tending to align m with the earth’s field, −mBearth sin θ. The third tends to align m with the alternating field, mB0 cos ωt cos θ. Assume that the acceleration is so small that the particle is in rotational equilibrium. (This is not necessary, but it simplifies the math.) Then, from Eq. 1.14,

−β dt + mBearth sin θ − mB0 cos ωt cos θ = 0. (9.73)

In order to linearize the equation, assume that θ is small enough so that sin θ ≈ θ and cos θ ≈ 1. The linearized equation is

− mBearthθ = −mB0 cos ωt.

(9.74)

β dt

This is a linear di erential equation with constant coe - cients that can be solved by the techniques of Appendix F. Consider only the particular solution and try a solution of the form

θ = θ1 cos ωt + θ2 sin ωt.

(9.75)

Substitution of this in the equation shows that

θ1 =

 

m2B0Bearth

,

 

 

 

(ωβ)2 + (mBearth)2

 

 

 

θ2 =

ωβmB0

 

,

 

 

(ωβ)2 + (mBearth)2

 

 

and

 

 

mB0

 

 

 

 

θm =

 

 

 

 

,

(9.76)

[(ωβ)2 + (mBearth)2]1/2

where θm is the maximum amplitude: θm2 = θ12 + θ22. We saw in Chap. 4 (Stokes’ law) that the translational viscous drag on a spherical particle is 6πηav. Similarly, the viscous torque on a rotating sphere is 8πηa3(dθ/dt) [Lamb

(1932), pp. 588-589]. The measured values for viscosity inside a cell range from 0.003 to 0.015 N s m2 [Polk (1994)]. Using the average of these, β = 0.009a3. The magnetic moment of a single-domain magnetosome is also proportional to volume: m = 2 × 106a3. This leads to a maximum amplitude that is independent of a:

2 × 106a3B0

θm = [(377)2(0.23)2a6 + (2 × 106)2(5 × 105)2a6]1/2

= 1.5 × 104B0.

(9.77)

Kirschvink originally argued from data about hair-cell deformation that a deflection of 16 or 0.3 radian is needed. This would require B0 = 5 × 105 T. (He had a slightly di erent value because he used a di erent viscosity. He also included the torque due to the force on the channel gate.)

In the absence of the applied field, the thermal fluctuations in angle can be estimated as follows. In the linear approximation, the work required to displace the particle an amount θ from the direction of the earth’s field is

 

 

W =

τ dθ =

mBearthθ dθ = mBearth

θ2

.

(9.78)

2

 

 

 

 

 

 

 

 

 

 

 

Equipartition of energy again gives us

 

 

 

 

 

 

 

thermal =

kB T

=

 

kB T

 

 

=

kB T

.

 

θ2

 

 

 

mBearth

 

× 106)(5 × 105)a3

 

 

 

 

 

(2

 

100a3

 

 

 

 

 

 

 

 

 

 

 

(9.79)

For a 50-nm magnetosome, this gives θrms = 0.58 radian. For a 100-nm magnetosome it is 0.2 radian, comparable to the maximum angles deduced from the model in the preceding paragraph.

Symbols Used in Chapter 9

Symbol

Use

Units

First

 

 

 

 

 

used on

 

 

 

 

 

page

a

Radius

m

232

b

Spacing

m

233

d

Displacement of charge

m

247

e

Electron charge

C

227

f

Force

N

234

f

Fraction of time an ion

 

242

 

 

 

is in a channel

 

 

gK

Potassium conductance

S m2

240

 

 

 

per unit area

 

 

h

Length of cylindrical

m

242

 

 

 

channel

 

 

h

Planck’s constant

J s

244

 

 

 

 

A m2

 

j, j

Electric current density

235

js

Particle current density

m2 s1

234

jv

Volume current density

m s1

235

kB

Boltzmann constant

J K1

227

m

Magnetic moment

A m2

247

n

Number of ions

 

242

pe, pe, ptot

Electric dipole moment

C m

233

 

 

 

 

 

 

 

 

 

 

Problems

249

Symbol

Use

Units

First

0

 

 

Electrical permittivity

C2

N1

229

 

 

 

used on

 

 

 

of vacuum

 

 

m2

 

 

 

 

page

κ

 

 

Dielectric constant

 

 

 

229

p, pc, po

Probability

 

240

η

 

 

Coe cient of viscocity

Pa s

248

q

Charge

C

233

λD

 

 

Debye length

 

m

 

230

r, r

Position

m

229

λ

 

 

Characteristic length

 

m

 

235

r

Radius in cylindrical co-

m

236

ν

 

 

Frequency

 

 

Hz or s1

244

 

ordinates

 

 

ρ, ρext

 

Charge density

 

C m3

229

r

Radius in spherical coor-

m

232

ρ

 

 

Resistivity

 

 

Ω m

235

 

dinates

 

 

σq , σq

 

Charge per unit area

 

C m2

233

t

Time

s

242

σ

 

 

Conductivity

 

S m1

235

u

rv(r)

V m

232

σi

 

 

Standard

deviation

of

A

 

242

u, uo

Energy (normalized to

 

235

 

 

 

current

 

 

 

 

 

v, v

kB T )

 

 

σn

 

 

Standard

deviation

of

 

 

242

Potential

V

227

 

 

 

number of ions

 

 

 

 

vNernst

Nernst potential

V

236

σq

 

 

Standard deviation of

C

 

242

w

Energy

J

240

 

 

 

charge

 

 

 

 

 

x

Position

m

229

σq

 

 

Charge per unit area

 

C m2

246

x

Distance along

m

236

σv

 

 

Standard deviation of

V

 

242

 

cylindrical axis

 

 

 

 

 

voltage

 

 

 

 

 

z

Valence

 

227

τ

 

 

Time constant

 

s

 

235

A, B, A , B

Constants

V

231

τ

 

 

Torque

 

 

N m

248

Bearth

Earth’s magnetic field

T

247

τt

 

 

Tissue time constant

 

s

 

246

B0

Amplitude of applied os-

T

248

θ

 

 

Angle

 

 

 

 

248

 

cillating magnetic field

 

 

φ

 

 

Angle in cylindrical

 

 

 

236

C, C

Concentration

m3

227

 

 

 

coordinates

 

 

 

 

Ci

Concentration of species

m3

230

χ

 

 

Susceptibility

 

N1 s1

234

[Cl] , [Cl ]

i

m3

 

ω, ωs, ω0

Solute permeability

 

237

Chloride concentration

228

ω

 

 

Angular frequency

 

s1

246

C

Capacitance

F

242

ωt

 

 

Characteristic angular

s1

 

D, De , D0

Di usion constant

m2 s1

234

 

 

 

frequency of tissue

 

 

 

 

E, Ex, E0, E1

Electric field

V m1

229

ξ

 

 

Energy in units of kB T

 

 

230

Eext

External electric field

V m1

234

Γ

 

 

Radial concentration

 

 

 

236

Epol

Polarization electric

V m1

233

 

 

 

factor

 

 

 

 

 

 

field

 

 

 

 

 

 

 

 

 

 

 

E

Photon energy

J

244

Problems

 

 

 

 

 

F

Faraday constant

C mol1

227

 

 

 

 

 

F, F

Force

N

234

 

 

 

 

 

 

 

 

 

G

Conductance

S

235

Section 9.1

 

 

 

 

 

J

Current per unit area of

A m2

237

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[K] , [K ]

membrane

m3

 

Problem 1 The chloride ratio between plasma and in-

Potassium

228

terstitial fluid is 0.95. Plasma protein has a valence of

 

concentration

 

 

L

m

235

about

18. In the interstitial fluid, Na = Cl = 155

Separation

 

 

 

 

 

 

 

 

M+ , M+

Concentration of imper-

m3

228

mmol l1. Find the sodium, chloride and protein concen-

M, M

meant cations

m3

 

trations in the plasma and the potential di erence across

Concentration of imper-

228

the capillary wall, assuming Donnan equilibrium.

 

 

meant anions

 

 

 

 

 

 

 

 

 

 

 

[M] , [M ]

Net concentration of im-

m3

228

Problem 2 Suppose that there are

two

compartments

 

permeant ions

 

 

 

 

 

with equal volume V = 1 l, separated by a membrane that

N

Number per unit volume

m3

233

NA

Avogadro’s number

mol1

234

is permeable to K and Cl ions. Impermeant positive ions

[Na] , [Na ]

Sodium concentration

m3

228

have a concentration 0 on the left and M = M+ =10

P

Polarization

C m2

234

mmol l1 on the right. The initial concentration of potas-

R

Gas constant

J K1

227

sium is [K0] = 30 mmol l1 on the left. T = 310 K.

 

 

 

mol1

 

 

 

 

 

(a) Find

the initial concentrations

of potassium

and

R

Resistance

235

chloride on both sides and the potential di erence.

 

Rp

Pore radius

m

236

 

S

Area

m2

229

(b) A fixed amount of potassium chloride (10 mmol) is

T

Temperature

K

227

added on the left. After things have come to equilibrium,

U, W

Energy

J

242

find the new concentrations and potential di erence.

 

V

Particle velocity

m s1

234

 

 

 

 

 

 

 

 

 

 

α

Proportionality

 

240

Problem 3 The extracellular space in cartilage contains

 

constant

 

 

 

 

 

large, immobile, negatively charged molecules called gly-

β

Linear viscous drag coef-

N s m1

234

 

ficient

 

 

coaminoglycans (GAGs). An early sign of osteoarthritis

β

Rotational viscous drag

N s m

248

is the loss of GAGs. The concentration of the GAGs is

 

coe cient

 

 

di cult to

measure directly, but

Shapiro

et al. (2002)

 

 

 

 

250 9. Electricity and Magnetism at the Cellular Level

measured the sodium ion concentration in cartilage using magnetic resonance imaging (see Chapter 18). Assume the interstitial fluid of the body consists of 150 mM of sodium ions and 150 mM of chloride ions, and that both of these ions can move freely between the body fluid and the extracellular space of cartilage. The cartilage sodium ion concentration is measured to be 250 mM. If Donnan equilibrium holds, what is the concentration of the GAGs? For simplicity, assume the GAGs are monovalent.

Section 9.2

Problem 4 Derive the Poisson equation from Gauss’s law in Cartesian coordinates in three dimensions.

Problem 5 Consider ions uniformly dispersed in a solution. Find the average linear separation of the ions for concentrations of 1, 10, 100, and 1,000 mmol l1.

Problem 6 Verify Eq. 9.20.

Problem 7 Verify the parameters presented in Table 9.2. How accurate is the approximation ex 1 + x in this case?

Problem 8 Consider a solution consisting of an equal concentration, C, of monovalent cations and anions.

 

ev

(a) Show that ρext = 2Ce sinh

 

.

kB T

(b) Let ξ = ev/kB T and r = rD , where λD is given by Eq. 9.14. Show that the nonlinear Poisson-Boltzmann equation (Eq. 9.13) becomes 2ξ = sinh ξ.

Problem 9 Analytical solutions to the nonlinear Poisson-Boltzmann equation are rare but not unknown. Consider the case when the potential varies in one dimension (x), the potential goes to zero at large x, and there exists equal concentrations of monovalent cations and anions. Chandler et al. (1965) showed that the solution to the 1-d Poisson-Boltzmann equation, d2ξ/dx 2 = sinh ξ

(see Prob. 8), is ξ(x ) = 4 tanh1 tanh (ξ0/4) e−x , where ξ0 is a constant and 0 < x < ∞.

(a)Verify that this expression for satisfies d2ξ/dx 2 = sinh ξ . (You may need a math handbook with a collection of hyperbolic function identities).

(b)Linearize the Poisson-Boltzmann equation and show that its solution is ξ(x ) = ξ0e−x .

(c)Show that both solutions are equal to ξ0 at x = 0 and equal to 0 at x = ∞.

(d)Compare the solutions for the linear and nonlinear Poisson-Boltzmann equation at x = 0.5 for the cases

ξ0 = 0.1, 1, and 10.

Section 9.3

Problem 10 The value of A used to obtain Eq. 9.30 was determined by saying that as r → 0, the electric field must approach ze/κ4π 0r2. An elaboration of the model would

be to say that the central ion has radius a and that the electric field at r = a must be the same as the field at the surface of the ion, ze/κ4π 0a2. How does this change the expression for v(r)?

Problem 11 Using the method in Sec. 9.3, derive the Poisson–Boltzmann equation in cylindrical coordinates (r, φ, z; see Appendix L) assuming the electric field is radial and does not depend on φ or z. Solutions to the linearized version of this equation are zeroth order modified Bessel functions [see Abramowitz and Stegun (1972)].

Section 9.4

Problem 12 A collection of molecular electric dipoles, each of moment p, are in thermal equilibrium at tem-

perature T . If the dipoles experience

an electric field

of strength E, then determine the

average value of

cos θ, where θ is the angle between the dipole and the electric field. Hint: Assume the dipole orientations follow the Boltzmann distribution, which in this case is

exp (pE cos θ/kB T ), and integrate over all

solid angles

dΩ = 2π sin θdθ. Show that if pE kB T

the average

of cos θ is proportional to E, but if pE kB T the average of cos θ saturates at a value of one. Interpret this physically.

Section 9.5

Problem 13 Find an expression for the slope of the Nernst–Planck constant-field curve in Fig. 9.10 when v is equal to the Nernst potential, v0. Hint: Expand the exponentials in Eq. 9.46 around v0.

Problem 14 Show that when j = 0, Eq. 9.41 gives C(x) = C0e−zev(x)/kB T , as we already know must be true in equilibrium. Hint: Solve for dv/dx.

Problem 15 The discussion surrounding Eqs. 9.35–9.42 was for a model of ions in a pore with constant electric field. It is also possible to write an integral version of the Nernst–Planck equation. Consider a single channel in which the current is the same for all values of x, the distance along the channel. If the di usion constant and cross-sectional area of the channel are allowed to vary, and with the usual substitution u(x) = zev(x)/kB T , Eq. 9.42 becomes

dC

+ C(x)

du

i = j(x)S(x) = −zeD(x)S(x)

 

 

.

dx

dx

(a) Show that if each term is multiplied by eu, this can be written as

 

ieu(x)

u(x) dC

 

u(x) du

 

 

= −ze e

 

+ C(x)e

 

 

.

D(x)S(x)

dx

dx

(b) Show that if the integration is carried from x1 to x2, then the current in the channel is

i =

ze C(x2)eu(x2) − C(x1)eu(x1)

 

I

 

,

where the integral

 

 

 

 

I = x2

eu(x) dx

,

 

 

S(x)D(x)

 

 

x1

 

 

 

 

 

 

contains all the information about the channel.

Problem 16 Cardiac cells have a potassium channel, called “IK1,”which shows inward rectification (larger current for potentials more negative than the potassium Nernst potential, VK , than for potentials more positive than VK ). This channel sometimes is said to show “anomalous rectification,” Why is it anomalous? [The mechanism of anomalous rectification is described by Nichols et al. (1996)].

Section 9.6

Problem 17 Consider a channel that is 100 times more permeable to potassium than to sodium (ignore all other ions).

(a)Write an equation for the reversal potential as a function of the intracellular and extracellular sodium and potassium ion concentrations.

(b)Assume [Ki] = 150, [Nai] = 50, and [Nae] = 150 mM. Plot vr versus [Ke] using semilog paper. On the same plot, draw the potassium Nernst potential as a function

of [Ke].

Problem 18 Calculation of the permeability ratios from measurement of the reversal potential is di cult because the concentrations inside the axon are not known. One can overcome this by measuring how the reversal potential (Eq. 9.57) changes as outside concentrations are varied. Obtain an equation for the shift of reversal potential if two measurements are made: one in which the concentration Ca1 = 0, the other with Cb1 = 0.

Section 9.7

Problem 19 A patch-clamp experiment shows that the conductance of a single Ca2+ channel is G = 25 pS. The membrane thickness is b = 6 nm. Use v = 50 mV.

(a)Assuming that the resistivity of the fluid in the channel is ρ = 0.5 Ω m, find an expression and numerical value for the channel radius a.

(b)If the conductance per unit area is 1,200 S m2, find the number of pores per unit area.

(c)The current is i = Gv, where v is the applied voltage. Find an expression for n, the number of calcium ions per second passing through the channel, in terms of whichever of parameters G, v, b, and a are necessary.

(d)How many calcium ions are in the channel at one time, if the calcium concentration is C mmol l1?

Problems 251

Problem 20 A potassium channel might have a radius of 0.2 nm and a length of 6 nm. If it contained potassium at a concentration of 150 mmol l1, how many potassium ions on average would be in the channel?

Problem 21 How long does it take for a sodium ion to drift in the electric field (assumed constant) through a membrane of thickness L and applied potential v? How long does it take to move by pure di usion? Find numerical values when the membrane is 6 nm thick and potential di erence is 70 mV.

Problem 22 Suppose that a sodium pore when open passes 10 pA and jNa = 0.2 A m2. Calculate the number of open pores per unit area and the average linear spacing between them.

Problem 23 Calculate the current density of sodium ions in a region of length 6 nm due to (a) pure di usion when there is no potential di erence and the concentrations are 145 and 15 mmol l1, (b) pure drift when the concentration is 145 mmol l1 and the potential di erence is 70 mV, and (c) both di usion and drift if the electric field is constant.

Problem 24 Patch-clamp recording is done with a micropipette of radius 1 µm.

(a)If the pipette encircles a single channel with conductance 20 pS, what is the channel current when the channel is open and the voltage across the membrane is 20 mV away from the Nernst potential for the ion in question? Make a simple estimate using Ohm’s law.

(b)Assuming a capacitance of 0.01 F m2, what current charges the capacitance of the membrane patch under the micropipette if a 20-mV change occurs linearly in 5 µs?

Problem 25 The following circuit illustrates the e ects that must be considered when an electrode is used to measure the properties of a patch of membrane. R1 is the resistance of the electrode. R2 and C are properties of the membrane. The applied voltage v0(t) is a step at t = 0. The electrode current is i(t). The voltage across the membrane patch is v(t).

 

 

 

i(t)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+

 

 

 

 

 

 

 

 

 

 

+

 

 

R

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

v 0(t )

 

 

 

 

 

R

 

 

 

 

v(t )

 

 

 

 

C

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(a) Show that

v0(t) = R1C dv + R1 + R2 v(t). dt R2

(b) Show that the time constant is τ = R1R2C/(R1 + R2) and that τ → R1C if R1 R2, τ → R2C if R1 R2.

2529. Electricity and Magnetism at the Cellular Level

(c)If v0(t) is a step of height v0 at t = 0, show that

v(t) = v0

 

R2

 

1 − e

t/τ

R1 + R2

 

 

and

v0

 

 

 

 

R2

 

i(t) =

1 +

 

 

 

 

 

e−t/τ .

R1 + R2

R1

(d)Plot v(t) and i(t).

(e)The case R1 R2 is called voltage-clamped. Find expressions for v(t) and i(t) in that case and plot them. Where does the transient current flow? For fixed R2, what is the time constant?

(f ) In the current-clamped case, R1 R2 and i0 = v0/R1. Find expressions for v(t) and i(t) and plot them. For fixed R2, what is the time constant?

(g) Make numerical plots of v(t) and i(t) when v0 = 150 mV, R1 = 106 , C = 5 pF, and R2 = 1011 .

Problem 26 A patch-clamp experiment is done with a micropipette having a resistance of 106 . When 150 mV is applied across the membrane, the current is 0 when the pores are closed and 1 pA when one channel is open. The membrane capacitance is 4 ×103 F m2. The microelectrode tip has an inner radius of 20 µm. What is the time constant for voltage changes? Does it depend on whether the channel has opened or closed?

Section 9.8

Problem 27 Weaver and Astumian (1990) derived Eq. 9.63a for the thermal noise of the transmembrane potential using a di erent method than in Sec. 9.8. A resistor has a voltage noise spectral density, σe2(f ) (in units of V2 Hz1), such that σe2(f ) = 4kB T R, where f is the frequency. It corresponds to voltage e in the figure. Weaver and Astumian represented the membrane as a parallel combination of resistance R and capacitance C. The voltage across the capacitor, v, is the transmembrane potential.

 

R

 

 

+

i

 

+

e

 

 

v

Ð

 

C

Ð

 

 

 

(a)For a particular frequency f , derive a relationship between the spectral density of the voltage fluctuations of the transmembrane potential, σv2(f ), and σe2(f ). (Hint: Derive an equation governing the voltage in an RC circuit, and then solve it using the methods described in Appendix F.)

(b)Integrate σv2(f ) over all frequencies to get the volt-

age fluctuations σv2.

#

(c) Estimate σv2 for a spherical cell of radius 10 mm,

having a membrane capacitance per unit area of 0.01 F m2.

Section 9.9

Problem 28 In some nerve membranes a region of “negative resistance” is found, in which the current decreases as the voltage is increased.

(a)Where have we seen this behavior before?

(b)To see why it happens, consider two cases. The current through the membrane is given by j = g(v)(v − v0), where g(v) is a property of the membrane, and the Nernst potential v0 depends on the ion concentration on either

side of the membrane. For this problem let v0 = +50 mV. Calculate j as a function of v for two cases: (a) g(v) = 1; (b) the conductance increases rapidly with voltage: g(v) = (5.6 × 107)e0.288v (v in mV).

(c) Negative resistance increases the sensitivity of the ampullae of Lorenzini, as measured by Lu and Fishman. To see why, calculate the output voltage in a tworesistance voltage divider network (as in Fig. 6.23) and discuss what happens if R2 is negative.

Section 9.10

Problem 29 Estimate the transmembrane potential that corresponds to the threshold for electroporation. Compare it to the normal cell resting potential.

Problem 30 Here is one way that signal-to-noise ratio can be improved. Suppose that there are N receptors, connected in the nervous system in such a way that an output response requires a logical AND between all N receptors. Whether or not there is a response is sampled every T seconds. If the signal exists, all N receptors respond. If the signal does not exist, each receptor responds to thermal noise with a probability p (which might be p = e−U/kB T , where U is an activation energy). Assume that p is the same for each receptor, and that whether a receptor has responded to thermal noise is independent of the response of all other receptors and also independent of its response at any other time.

(a)What is the signal-to-noise (S/N ) ratio as a function of N ? Suppose that N = 8. Plot S/N as a function of p.

(b) Find U/kB T vs N for S/N = 4.

Problem 31 Here is another way to look at the signal- to-noise ratio.

(a)Show that the energy of a charged parallel-plate capacitor can be written as κ 0E2V /2, where V = Sd is the volume between the plates. This is a special case of a general relationship that the energy per unit volume associated with an electric field is κ 0E2/2.

(b)Use the information about the magnitude of the electric field in the cell membrane from Fig. 9.19 to calculate the total electrostatic energy in the membrane.

(c)Compare the ratio of the total electrostatic energy to kB T when the air field is 300 V m1. This overestimates the ratio, because the energy is spread over the entire membrane and is not available to interact in one place.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

References

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Problem 32 Obtain Eq. 9.79 from the expression U =

Adair, R. K. (2000). Static and low-frequency magnetic

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63: 415–454.

 

Problem 33 Electric fields in the body caused by expo-

Adair, R. K., R. D. Astumian, and J. C. Weaver (1998).

Detection of weak electric fields by sharks, rays

and

sure to power lines are produced by two mechanisms: di-

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rect coupling to the power line electric field, and Fara-

 

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this endogenous field using the data in Fig. 7.23).

 

 

 

Bastian, J. (1994). Electrosensory organisms. Physics

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 34 Derive the equations for the electric field

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Bockris, J. O’M., and A. K. N. Reddy (1970). Modern

shown

in Fig.

9.19. Use

the following

method.

Let

Electrochemistry. New York, Plenum, Vol. 1.

 

the potentials be voutside

=

A cos θ/r

2

− E1r cos θ and

 

 

Bren, S. P. A. (1995). 60 Hz EMF health e ects—a

vintracellular = Br cos θ, where A and B are unknown con-

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Carstensen, E. L. (1995). Magnetic fields and cancer.

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Ohm’s law:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Chandler, W. K., A. L. Hodgkin, and H. Meves (1965).

 

 

 

 

∂voutside

 

 

 

 

 

 

σ

 

 

 

 

 

 

 

The e ect of changing the internal solution on sodium

 

outside

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

∂r

 

 

 

 

 

 

 

 

 

 

inactivation and related phenomena in giant axons. J.

 

 

 

 

 

 

 

 

r=a

 

 

 

 

 

 

= σ

 

 

 

 

∂vintracellular

 

 

 

 

Physiol. 180: 821–836.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Crouzy, S. C., and F. J. Sigworth (1993). Fluctuations

 

intracellular

 

 

 

 

 

 

 

 

 

 

 

 

 

 

∂r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r=a

 

 

 

in ion channel gating currents: Analysis of nonstationary

 

 

 

 

 

 

 

 

 

 

 

σmembrane

 

 

= (voutside

vintracellular)

 

 

 

 

 

 

shot noise. Biophys. J. 64: 68–76.

 

 

 

b

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r=a

 

DeFelice, L. J. (1981). Introduction to Membrane

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(a)

Verify

that

the

expressions

for

voutside

and

Noise. New York, Plenum.

 

vintracellular obey Laplace’s equation and behave properly

Denk, W., and W. W. Webb (1989). Thermal-noise-

at r = 0 and r = ∞.

 

 

 

 

 

 

 

 

 

 

 

 

limited transduction observed in mechanosensory recep-

(b) Use the boundary condition to determine A and B.

tors of the inner ear. Phys. Rev. Lett. 63(2): 207–

(c) Use your expressions for the potential to determine

210.

 

the electric fields given in Fig. 9.19.

 

 

 

 

 

Doyle, D. A., J. M. Cabral, R. A. Pfuetzner, A. Kuo, J.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M. Gulbis, S. L. Cohen, B. T. Chait, and R. MacKinnon

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1998). The structure of the potassium channel: Molecu-

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10

Feedback and Control

We now turn to the way in which the body regulates such things as temperature, oxygen concentration in the blood, cardiac output, number of red or white blood cells, and blood concentrations of substances like calcium, sodium, potassium and glucose. Each of these is regulated by a feedback loop. A feedback loop exists if variable x determines the value of variable y, and variable y in turn determines the value of variable x.

Suppose that x is the deviation of a bullet from its desired path. A bullet has no feedback; after it has left the gun, its deviation from the desired path is determined by the initial aim of the gun, fall due to gravity, drift caused by the wind, and air turbulence. An accuracy of one part in 104 (about a tenth of an inch in 50 ft) is quite good. A car, on the other hand, is steered by the driver. If deviation from the center of the lane x becomes appreciable, the driver changes y, the position of the steering wheel. The value of y determines x through the steering mechanism and the tires. It is possible to have a car deviate less than 1 ft from the desired position within a lane after driving 3,000 miles, an accuracy of one part in 107. This is an example of negative feedback. If x gets too large, the factors in the feedback loop tend to reduce it.

Negative-feedback systems can generate oscillations of their variables. We see oscillations in physiological systems on many di erent time scales, from the rhythmic activity of the heart, to changes in the rate of breathing, to daily variations in body temperature, blood pressure, and hormone levels, to monthly variations such as the menstrual cycle, to annual variations such as hibernation, coloring, fur growth, and reproduction.

It is also possible to have positive feedback. Two bickering children can goad each other to new heights of anger. Positive feedback initiates the action potential described in Chapter 6: depolarization of the axon leads to increased sodium permeability, which further speeds depolarization. Blood pressure is regulated in part by sen-

sors in the kidney. A patient with high blood pressure may su er damage to the blood vessels, including those feeding the kidneys, which reduces the blood pressure at the sensors. The sensors then ask for still higher blood pressure, which accelerates the damage, which leads to still higher blood pressure, and so on.

The simplest feedback loop consists of two processes: one in which y depends on x and another in which x depends on y. The loop can have many more variables. Steering the car, in addition to the variables of lane position and steering wheel position, involves vision, neuromuscular processes, all of the variables in the automobile’s steering mechanism, and the Newtonian mechanics of the car’s motion—with external variables such as the behavior of other drivers continually bombarding the system.

Sections 10.1–10.3 deal with the relationships between the feedback variables when the system is in equilibrium or in the steady state, and none of the variables are changing with time. The techniques for determining the operating point—the steady-state values of the variables—are graphical and can be applied to any system if the relationship among the variables is known.

When the system is not at equilibrium, it returns to the equilibrium point if the system is stable. Although the equations describing this return to equilibrium are usually not linear, Secs. 10.4–10.6 discuss how linear systems behave when they are not at the operating point. A linear system may “decay” exponentially to the steadystate values, or it may exhibit oscillations.

Most systems are not linear. Section 10.7 discusses systems described by nonlinear equations in one or two dimensions, introducing some of the vocabulary and graphical techniques of nonlinear systems analysis. It closes with an example of resetting the phase of a biological oscillator. Section 10.8 introduces the ideas of period doubling and chaotic behavior through di erence equations and

15.47p

256 10. Feedback and Control

FIGURE 10.1. Schematic curves of the relationship between thyroid hormone (T3) and thyroid stimulating hormone (TSH) in the thyroid gland and in the pituitary.

 

80

 

 

 

 

 

 

 

 

70

20

60

p =O2 demand, mmole min-1

 

, torr

60

40

80

 

 

 

 

 

 

 

B

 

 

 

 

 

 

 

 

 

 

CO2

50

 

 

 

 

 

 

 

P

40

 

 

 

A

 

 

 

= alveolar

 

 

 

 

 

 

 

 

 

 

 

 

 

30

 

 

 

C

 

 

 

20

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10

 

 

 

 

 

 

 

 

0

 

 

 

 

60

 

100

 

0

 

 

20

40

80

 

 

 

 

y = ventilation rate, l min-1

 

the logistic map. It then describes a linear map that appears to be chaotic but is not. Section 10.9 shows how a linear di erential equation that depends on a fixed delay in the variable can exhibit either damped or continuous oscillations. Section 10.10 summarizes the earlier sections, and Sec. 10.11 gives several biological examples.

A great deal of work was done on modeling physiological feedback systems between 1950 and 1975. Books from that era include Riggs (1970) and Stark (1968). A contemporary book that describes physiological models in detail, including computer modeling, is Khoo (2000).

FIGURE 10.2. The pressure of CO2 in the alveoli of the lungs decreases as the ventilation rate is increased. The di erent curves correspond to di erent total metabolic rates.

Figure 10.2 shows this relationship when the pH and PO2 of the blood are fixed. As ventilation rate rises, PCO2 falls. We are ignoring several other feedback loops [Riggs (1970, pp. 401–418)]. If the metabolic rate rises, PCO2 also rises. Experiments show that ventilation rate y and alveolar PCO2 (which we will call x) are related by (Riggs, op. cit.)

10.1Steady-State Relationships Among Variables

Any feedback loop can be broken down, conceptually at least, into separate processes that relate a dependent variable to an independent variable and possibly to some other parameters. Figure 10.1 shows an example. In the first process the thyroid gland, in response to thyroid-stimulating hormone (TSH) from the pituitary, produces the thyroid hormones thyroxine (T4) and triiodothyronine (T3). An increase of TSH increases production of T3 and T4. These processes depend on other parameters, such as the amount of iodine available in the body to incorporate into the T3 and T4. In the second process, the pituitary increases the production of TSH if the concentration of T3 in the blood falls. It may also respond to T4 and other variables as well. (This is an oversimplification. The pituitary actually responds to hormones secreted by the hypothalamus. The hypothalamus is responding to the levels of T3 and T4.)

For a quantitative example, consider a simple model relating the amount of carbon dioxide in the alveoli (air sacs of the lung) and the rate of breathing (ventilation rate). If the body is producing CO2 at a constant rate, a given ventilation rate corresponds to a definite value of PCO2 , the partial pressure of carbon dioxide in the alveoli. In the steady state the amount of CO2 exhaled (the volume of gas leaving the lungs per minute times PCO2 ) is just equal to the amount produced in the body.

x =

 

.

(10.1)

y − 2.07

In these equations y (the independent variable) is measured in l min1, x (the dependent variable) is in torr, and parameter p is the body’s oxygen consumption in mmol min1. A typical resting person requires p = 15 mmol min1.

Equation 10.1 can be derived using a simple model for respiration. Let the metabolic rate of the body be described by o, the rate of oxygen consumption in mol s1. The respiratory quotient F relates o to the rate of CO2 production, so

Rate of CO2 production = F o.

(10.2)

A typical value of F is 0.8.

Carbon dioxide is removed from the body by breathing. If the rate at which air flows through the alveoli is1 (dV /dt)alveoli in m3 s1, then the rate of removal is obtained from the ideal-gas law:

Rate of CO2 removal = x(dV /dt)alveoli .

RT

The rate (dV /dt)alveoli is less than the ventilation rate y because air in the trachea and bronchi does not exchange

1Strictly speaking, (dV /dt)alveoli is not the derivative of a function V . (It always has a positive value, and the lungs are not ex-

panding without limit!) We use the notation to remind ourselves that it is the rate of air exchange in the alveoli.