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SOLUTIONS

In the following sections, solutions are given for some of the exercises.

14.1. SOLUTIONS, CHAPTER 1: VECTOR ANALYSIS

Answer 1a. 11.22

Answer 2. 17

Answer 3. 0.686

Answer 5a. A ·B = 8 − 10 + 2 = 0

Answer 5b. A · B = 0, so one knows the vectors are perpendicular. That’s because A · B =|A||B|cos(θ). With neither |A| and |B| zero, the result can be zero only if cos(θ) = 0,and that will occur only if the vectors are perpendicular (angle of 90 degrees).

Answer 7a. 4

Answer 7b. 0

Answer 9. 42.

Answer 11. The gradient points in the −r direction, i.e., in the “uphill” direction of ψ = 1/r.

Answer 13. The result is the same as the law of cosines.

Answer 15a. No; instead ∇(1/R) = −∇′(1/R). Note the negative sign.

Answer 15b. Yes.

1

2 SOLUTIONS FOR CHAPTER 1: VECTOR ANALYSIS

Answer 17d. Yes.

Answer 19a. With φ = 1, a constant, both the gradient and divergence of φ are zero. Therefore,the first term on the left side of Eq. (1) and the first term on the right side become zero.

Answer 19b. The result is called Gauss’s theorem.

Answer 19c. Withψ = 1 a constant, Gauss’s theorem again results, this time expressed in termsof variable φ rather then ψ.

Answer 20. Yes, it does satisfy Laplace’s equation. See also the text by Stratton where thisequation is used.

Answer 21. The key to these questions is taking advantage of the fact that surface S is a closedsurface. In (a), K · dS is the portion of the surface oriented in the direction of K. In (b),r ·K is a constant, so the result is a multiple of the integral of dS over the whole surface.One may be able to see the general answers by picturing or computing the results for aparticular example, such as a cube.

Answer 22. 2.31

Answer 23. 1.4554

Answer 24. 0.01

Answer 25. 0.001924

Answer 26. –0.2354

BIOELECTRICITY: A QUANTITATIVE APPROACH 3

14.2. SOLUTIONS, CHAPTER 2: SOURCES AND FIELDS

Answer 1a. Because conductivity is the reciprocal of resistivity, the corresponding units willbe 1/(Ωcm) or Siemens/cm.

Answer 1b.σ = 0.01 S/cm

Answer 3. It is given that

φ =Io

4πσr

Consequently,

J = −σ∇φ =−Io4π

∇(1r)

so that evaluating the gradient gives

J =Io

4πr2= 1r

Thus, J has magnitude 1 mA/cm2 and is directed radially.

Answer 5. Take the derivative with respect to x twice to find the divergence. Then find wherethe divergence is zero, which is x = 0.33 cm.

Answer 7.

φ =Ioρ

4πr=

4π504π

√52 + 52

= 7.1uV (microVolts)

Answer 9a. The magnitude is√

22 + 32 + 42 = 5.38 µA.

Answer 9b. The vector area is A = U × V /2. The division by 2 occurs because the crossproduct’s magnitude is that of the parallelogram formed by the vectors, and the triangleis half that. Performing the cross product, A = 0.5az .

Answer 9c. Taking the dot product J ·A = 4 · 0.5 = 2 µA.

Answer 11. Begin with the solution to Exercise 10. Twice differentiate with respect to x (notr). The solution then is

A(x) =Io

4πσ3(x− e)2

r5

.

Answer 13a. ∇Φ = k sech2(x)

Answer 13b. J = −σ∇Φ = −kσ sech2(x)

4 SOLUTIONS FOR CHAPTER 2: SOURCES AND FIELDS

Answer 13c. ∇ · ∇Φ = k sech2(x) tanh(x)

Answer 13d. The answer is the same as that for 13c, since the same question is asked, usingdifferent words.

Answer 15a. The peak of the gradient is at x = 0.

Answer 15b. The maximum slope occurs at the peak of the gradient, so again at x = 0.

Answer 15c. The numerical value of the maximum slope is sech2(x) with x = 0, i.e. 1.

Answer 17a. In equation form, the potential with resistivity 1 from source 1 can be written asφ1

1 = (ρ1/4π)(1/r1) and φ12 = (ρ2/4π)(1/r1). The ratio φ1

2/φ11 = ρ2/ρ1 = 2. The

same argument will apply to potentials from the other source. Therefore, the ratio askedequals 2.

Answer 17b. By definition J = −σ∇φ. Withφ11 = (ρ1/4π)(1/r1), the equation for J includes

no σ (or ρ) term. Therefore, the result is not a function of the resistivity, and the requestedratio is 1.

Answer 17c. Again inspect the equation φ11 = (ρ1/4π)(1/r1). Note that ∇φ has a direction

determined by r, not by ρ, since ρ is a constant insofar as the gradient operation, i.e., ρis not a function of x, y, z. Thus the dot product is 1, since the directions of J does notchange.

Answer 19a. Suitable units for k are µA/cm4.

Answer 19b. At (1, 1, 1) the magnitude of the divergence is 3 µA/cm3.

Answer 19c. At the origin, the divergence is greater than zero. This result is found by using thedefinition of divergence, Eq. (1.23).

Answer 21. Hint: Ignore any interaction between the electrodes. Find the potential on animaginary electrode of the same size and location as a real arising from a point currentsource at an electrode’s center. Make use of linearity to find results for two electrodes asthe sum of results for each one separately.

Answer 23a. k can be mV/cm3

Answer 23b. Electric field E = −∇Φ = −3kx2.

Answer 23c. Sources Iv = ∇ · J = −∇ · σE = 6kx.

Answer 23d. There are sources, proportional to x. There is no source density at x = 0. Sourcedensities are negative when x < 0. Loosely speaking, these sources are associated withcurrents from regions with more positive x to those with less positive x, and analogouslyfor negative x, but no current across the plane x = 0.

Answer 25. Hint: Since any potential field can be considered the superposition of point-sourcefields, choose an arbitrarily located point source and demonstrate the validity of thestatement for this case.


Answer 27. Hint: Think of the two sides of the tank as plates of a parallel plate capacitor, sinceeach one is a conducting region separated by a dielectric. Then use Gauss’s law with avolume enclosing a square centimeter of surface at one interface, or use reference resultsfor capacitance with parallel plates.

Answer 29a. a = 10−3 cm. Thus ri = Ri/a2 = 102/10−6 = 108 Ohms/cm.

Answer 29b. a = 10−3 cm and b = 2 × 10−3 cm. Thus re = Re/(b2 − a2), so re =600/(3 × 10−6 = 2 × 108 Ohms/cm.

Answer 29c. a = 10−3 cm, so rm = Rm/circumference = Rm/4a = 10000/4 × 10−3 =2500 × 103 = 2.5 × 106 Ωcm.

Answer 29d. a = 10−3 cm, so cm = Cm × circumference = Cm × 4a = 1.2 × 4× 10−3 =4.8 × 10−3 microfarads/cm.

Answer 31. The resistance of the total path is the resistance of the top membraneRtop, the resis-tance of the material inside the box, Rinside, and the resistance of the bottom membrane,Rbottom.

Rtop = Rm/Atop. Atop, the area of the top surface, isAtop = a×L = 10×100×10−8 =10−5 cm2. Thus Rtop = 10000/10−5 = 109 Ohms. Rbottom = Rtop.

The resistance of the inside of the box is relatively insignificant, being Rinside = Ri ×a/(L× a) = Ri/L = 100/(100 × 10−4) = 104 Ohms.

Thus the total resistance of the pathRtotal ≈ 2×109, and the steady-state current is only0.1/2 × 109 = 0.05 × 10−9 Amperes, or 0.05 nanoamperes.

Answer 33. Hint: use Gauss’s theorem. Draw an imaginary sphere just outside the inner sphere.

Answer 34. See Smythe, pp. 7–9, on “Lines of Force."

Answer 35. Hint: recall the derivation of the potential from a dipole, as the difference in potentialof two monopoles of opposite sign. Here, think of the quadrupole as two dipoles ofopposite sign with the second moved away from the first by a small amount d. Thenproceed in an analogous fashion.

6 SOLUTIONS FOR CHAPTER 3: BIOELECTRIC POTENTIALS

14.3. SOLUTIONS, CHAPTER 3: BIOELECTRIC POTENTIALS

Answer 1. 148,440 Coulombs

Answer 2. 6.674E11 lbs

Answer 3. 5.092E-8 (m/sec)/(V/m). Use Einstein’s equation.

Answer 4. 24.14 millivolts. The difference arises from the difference in temperature.

Answer 5. 1,074 miles per hour. Balance kinetic and thermal energy. Use

mv2

2=

32RT

No(1)

Answer 8.σ = (DiFZi)ZiCi(F/RT ) (2)

because E = −∇Φ and J = σE

Answer 9A. 9.456E-5 Amperes

Answer 9B. -8.78E-5 Amperes

Answer 10. 4E-7 moles/cm3

Answer 11. 1E-12 moles

Answer 12. 7.448E-6 moles/(cm2/sec)

Answer 13. 2.98E-9 moles/sec

Answer 14. 29.56 cm/sec

Answer 15. 5.37E-5 seconds

Answer 16. 53.1 millivolts

Answer 17. -83.6 millivolts

Answer 18. -60 millivolts

Answer 19. Three ion species are not in equilibrium. (At this transmembrane potential, noindividual ion is in equilibrium.)

Answer 20. Two ion species are not in equilibrium. Cl− is in equilibrium.

Answer 21. 143.6 millivolts

Answer 22A. 24.1 Angstroms

Answer 22B. 1.15 microfarads per cm2

Answer 23. 4,000 square micrometers

Answer 24. 4.85E-5 microfarads

Answer 25. 558,036,000 Ohms


Answer 26. 0.515 picocoulombs. Use Q = CV .

Answer 27. 1/10000 Siemens/cm2. Recall that conductivity is the reciprocal of resistivity.

Answer 28. 20 milliseconds

Answer 29. 66 milliseconds. Found from the number of time constants required.

Answer 30A. (Coulombs/sec)/m2

Answer 30B. (moles/sec)/m2

Answer 30C. Coulombs/Volt

Answer 30D. m2/s

Answer 30E. Volts

Answer 30F. (m/sec)/(V/m) = m2/(Vsec)

Answer 31. JK = gK ∗ (Vm − EK)

Answer 32. JNa = gNa ∗ (Vm − ENa)

Answer 33. JL = gL ∗ (Vm − EL)

Answer 34. –67.99 millivolts. Note the negative sign.


Answer 36. 56.04 millivolts. Note the positive sign.


Answer 38. 17.3 microamperes per square centimeter (uA/cm2). Note that the current is out-ward.

Answer 39. –0.960 microamperes per square centimeter (uA/cm2). Note that the current isinward.

Answer 40. 10.38 microamperes per square centimeter (uA/cm2).

Answer 41. –22.27 millivolts per millisecond. Take the time derivative of V = Q/C, andsubstitute Jion to replace dQ/dt.

Answer 42. –7.49 microamperes per square centimeter (uA/cm2). Note that the answer issensitive to precision in the individual current components, because that is followed byan addition of terms that usually have different signs.

Answer 43. –23.69 millivolts

Answer 44. –37.88 millivolts. By definition, a condition that must be met to have a resting(unchanging) potential is Jion = 0.

Answer 45. –5.30 millivolts per millisecond

Answer 50. F , Faraday’s constant

Answer 51. Einstein’s equation shows how the diffusion coefficient can be converted into theconductivity, which is then used in the “electrical” term.

8 SOLUTIONS FOR CHAPTER 4: CHANNELS

14.4. SOLUTIONS, CHAPTER 4: CHANNELS

Answer 1. 741 Ohms

Answer 2. 37,620,000 channels open

Answer 3. The fraction of potassium channels open is 0.0198, or about 2%.

Answer 4. The number of open potassium channels is 4,021.

Answer 5. 5.62 nanoamperes (nA)

Answer 6. Np

Answer 7. Nq

Answer 8. 1

Answer 9.√Npq

Answer 10. M = 10000 q/p

Answer 11. A = 10000 q/Dp

Answer 12. a = 50√q/(πDp)

Answer 13a. 8.69 µm

Answer 14a. In phase 1, the probability p1 that a K channel is open is 0.147

14A Solution in more detail:

1 The clamp voltage V 1m exceeds the resting voltage Vr by vm + 1 = (−40 −

(−60)) = 20 mV.

2 Using v1m in the Hodgkin–Huxley equation for αn gives αn = 0.1582 msec−1.

3 Similarly, βn = 0.09735 msec−1.

4 Thus n, the probability of an individual particle in the channel being in the openposition, is αn/(αn + βn) = 0.619.

5 Thereby, the probability that a K+ channel is open is n4, or 0.147.

Answer 14b. Multiplying cell surface area (600 µm2) by the channel density (40 ch/µm), onefinds the number ofK+ channels to be 24,000. This number times p1 gives 3,525 channelsopen.

Answer 14c. The fluctuation is 4 × 54.8 channels

Answer 14d. 1.41E-9 Amperes, where –9 means ×10−9.

Answer 15a. Probability p2 = 0.642.


Answer 15b. 15,400 channels

Answer 15c. 12.32E-9 Amperes

Answer 17. 4.37

Answer 22. 8.74

Answer 19. 2.82E-9 Amperes.

Answer 20. The ratio is different because phase 2 has a number of open channels that changeswith time, even though the phase 2 voltage is constant. That is, immediately after thetransition (Ex. 19), the number of open channels is the same as the steady state in phase1, because there has been no time for the expected numbers to change. (Even then, thecurrent is not the same as the current in phase 1 at steady state, because the membranevoltage is immediately different.) The number of open channels, and thus the current,evolves as the membrane moves to the steady state for phase 2.

Answer 21a. Probability p1 is 0.025.

Answer 21b. 592 open channels

Answer 21c. The expected fluctuation is 4 × 24 channels.

Answer 21d. 1.48E-10 Amperes

Answer 22a. Probability p2 is 0.894.

Answer 22b. The expected number of open channels is 21,467.


Answer 23. Yes, at steady state.

Answer 24. The ratio is 36.3, i.e., more channels in steady state in phase 2.

Answer 25. The ratio is 196, so a much higher current flows at the steady state of phase 2. It isinteresting to note that this ratio is higher than the ratio of the numbers of open channels.

Answer 26. The ratio is 2.86, much lower than that of the preceding question.

28 Exercise [28] restatement from text: In the model cell, examineNa+ andK+ currentswith V 1

m of -57 mV and V 2m of -20 mV.

a. At t1, what is I1K?


b. At t1, what is I1Na?

c. At t2, what is I2K?

d. At t2, what is I2Na?

Answer 28a. 9.74E-11 AmperesDetail of 28a:

1 The current desired is the product of the K+ current through an open channeltimes the expected number of open K+ channels.

2 (Vm − EK) = 23 mV, so the current through an open channel is Ichan =γ × (Vm − EK) = 2.3E − 13 Amperes.

3 Because vm + 1 = 3 mV, αn is 0.069 and βn is 0.1204, making n = αn/(αn +βn) = 0.364.

4 The expected number of open channels is Nopen = Ntotal × n4 = 24000 ×0.01765 = 424 channels.

5 Thus the current from potassium ions is 424 × 2.3E − 13 = 97.4 picoamperes.

Answer 28b. -5.77E-12 AmperesDetail of 28b:

1 The current desired is the product of theNa+ current through one channel timesthe expected number of open Na+ channels.

2 Vm − ENa = −117 mV, so the current through an open channel is Ichan =γ × (Vm − ENa) = −1.17E − 12 Amperes.

3 Because v1m = 3 mV, αm is 0.274, βm is 3.386, so m is 0.075. Similarly, αh is

0.0603, and βh is 0.063, so h is 0.4889.4 The expected number of open channels is Nopen = Ntotal ×m3h = 5.5 Thus the current due to sodium ion movement is 5 × −1.17E − 12 or –5.8

picoamperes.

Answer 28c. 6.09E-9 Amperes. (Similar to 28A, but different Vm.)Detail of 28c:


2 (Vm − EK) = 40 mV, so the current through an open channel is Ichan =γ × (Vm − EK) = 6.0E − 13 Amperes.

3 Because v1m = 40 mV, αn is 0.3157 and βn is 0.0782, making n = αn/(αn +

βn) = 0.8063.4 The expected number of open channels is Nopen = Ntotal × n4 = 24000 ×

0.4228 = 10147 channels.5 Thus the current from potassium ions is 10147×6.0E−13 = 6088 picoamperes.

Answer 28d. –1.34E-10 Amperes. (Similar to 28B, but different Vm.)Detail of 28d:



2 Vm − ENa = 40 mV, so the current through an open channel is Ichan = γ ×(Vm − ENa) = −0.8 picoamperes.

3 Because v1m = 40 mV, αm is 1.9308, βm is 0.4335, so m is 0.8167. Similarly,

αh is 0.00947, and βh is 0.7310, so h is 0.01279.4 The expected number of open channels is Nopen = Ntotal ×m3h = 167.5 Thus the current due to sodium ion movement is 167 × −0.8 or –133.8 picoam-

peres.

Answer 29a. 7.81E-11 Amperes

Answer 29b. -4.43E-12 Amperes


Answer 29d. -1.61E-10 Amperes

Answer 30. Yes. (Note that these exercises allow comparison of currents at steady-state only.)

32 Question 32, from text. In the model cell, evaluate the requested currents if V 1m is

–55.5 mV, and V 2m is –24.5 mV.

a. At t1, what is I1K?

b. At t1, what is I1Na?

c. At tb, what is IbK?

d. At tb, what is IbNa?

e. At t2, what is I2K?

f. At t2, what is I2Na?



Answer 32c. 5.54E-10 AmperesDetail of c:


2 (Vm − EK) = 55.5 mV, so the current through an open channel is Ichan =γ × (Vm − EK) = 0.555 picoamperes.

3 Because v1m = 35.5 mV, αn is 0.2766 and βn is 0.0802, making n = 0.4515 at

t = ta (using the result of exercise 31).4 The expected number of open channels is Nopen = Ntotal × n4 = 24000 ×

0.0416 = 997 channels.5 Thus the current from potassium ions is 998 × 0.555 = 553 picoamperes.

Answer 32d. –9.27E-10 AmperesDetail of 32d:



2 Vm − ENa = −84.5 mV, so the current through an open channel is Ichan =γ × (Vm − ENa) = −0.845 picoamperes.

3 Because v2m = 40 mV, αm is 1.615, βm is 0.5565. Thus at t = ta the value of

m is 0.7437. Similarly, αh is 0.011864, and βh is 0.6341, so at t = ta h is0.18365

4 The expected number of open channels is Nopen = Ntotal ×m3h = 1097.

5 Thus the current due to sodium ion movement is 1097 × −0.845 or –927.2picoamperes.

Answer 32e. 4.81E-9 Amperes

Answer 32f. –1.53E-10 Amperes




Answer 33d. –1.92E-9 Amperes

Answer 33e. 7.59E-9 Amperes

Answer 33f. –1.08E-10 Amperes

Answer 34. No, as judged by these exercises, never the same sign. INa and IK have oppositesigns most of the time because (Vm − EK) has a different sign from (Vm − ENa). Forthe sign to be the same, Vm would have to be outside the range EK to ENa.

Answer 35. It is not always true, because |INa| is larger at time tb.


14.5. SOLUTIONS, CHAPTER 5: ACTION POTENTIALS

Answer 1a. gK = gKn4

Answer 1b. gNa = gNam3h

Answer 1c. IK = gK(Vm − EK)

Answer 1d. INa = gNa(Vm − ENa)

Answer 1e. IL = gL(Vm − EL)

Answer 1f. Iion = IK + INa + IL

Answer 1g. Vdot = Is − Iion/Cm

Answer 2a. vm = Vm + 60

Answer 2b. αn = .01(10 − vm)/(exp((10 − vm)/10) − 1)

Answer 2c. βn = 0.125 exp(−vm/80)

Answer 2d. αm = 0.1(25 − vm)/(exp((25 − vm)/10) − 1)

Answer 2e. βm = 4 exp(−vm/18)

Answer 2f. αh = 0.07 exp(−vm/20)

Answer 2g. βh = 1/(exp((30 − vm)/10) + 1)

Answer 2h. dn/dt = αn(1 − n) − nβn

Answer 2i. dm/dt = αm(1 −m) −mβm

Answer 2j. dh/dt = αh(1 − h) − hβh

Answer 3a. ∆Vm = Vdot ∆t

Answer 3b. ∆n = dn/dt ∆t

Answer 3c. ∆m = dm/dt ∆t

Answer 3d. ∆h = dh/dt ∆t

Answer 3e. Vm(t0 + dt) = Vm + ∆Vm

Answer 3f. n(t0 + ∆t) = n+ ∆n

Answer 3g. m(t0 + ∆t) = m+ ∆m

Answer 3h. h(t0 + ∆t) = h+ ∆h

14 SOLUTIONS FOR CHAPTER 5: ACTION POTENTIALS

Answer 4. IfIm = IC + Iion

thenIC = Im − Iion

For IC to be zero, one must have

IC = Im − Iion = 0

ThusIm = Iion

In an isolated patch,Im = Is

so the condition for IC to be zero, and thus no Vm change, is

Is = Iion

Answer 5a. gK 0.73497 mS/cm2

Answer 5b. gNa 4.15057 mS/cm2

Answer 5c. VK [−11.5 − (−72.1)] = +60.6 mV

Answer 5d. VNa [−11.5 − (+52.4]) = −63.9 mV

Answer 5e. IK 44.54 µA/cm2

Answer 5f. INa –265.2 µA/cm2

Answer 5g. IL 11.3 µA/cm2

Answer 5h. Iion –209.4 µA/cm2

Answer 5i. Vm = 209 volts/sec

Answer 6a. vm 48.5 mV

Answer 6b. αn 0.393371 msec−1

Answer 6c. βn 0.068174 msec−1

Answer 6d. αm 2.597745 msec−1

Answer 6e. βm 0.27032 msec−1

Answer 6f. αh 0.006193 msec−1

Answer 6g. βh 0.864127 msec−1


Answer 6h. dn/dt 0.21891 msec−1

Answer 6i. dm/dt 1.40176 msec−1

Answer 6j. dh/dt –0.40895 msec−1

Answer 7a. ∆Vm 10.47 mV

Answer 7b. ∆n 0.010945 no units

Answer 7c. ∆m 0.070088

Answer 7d. ∆h –0.020447

Answer 7e. Vm(t0 + dt) –1.03 mV

Answer 7f. n(t0 + dt) 0.388945 no units

Answer 7g. m(t0 + dt) 0.487088

Answer 7h. h(t0 + dt) 0.456553

Answer 8. Is = −Iion = 209.4 µA/cm2

Answer 9a. gK 11.95 mS/cm2

Answer 9b. gNa 10.87 mS/cm2

Answer 9c. VK [−11.5 − (−72.1)] = +60.6 mV

Answer 9d. VNa [−11.5 − (+52.4)] = −63.9 mV

Answer 9e. IK 724.01 µA/cm2

Answer 9f. INa –694.59 µA/cm2

Answer 9g. IL 11.31 µA/cm2

Answer 9h. Iion 40.731 µA/cm2

Answer 9i. Vm –40.73 mV/msec

Answer 10a. vm 48.5 mV

Answer 10b. αn 0.393371 msec−1

Answer 10c. βn 0.068174 msec−1

Answer 10d. αm 2.597745 msec−1

Answer 10e. βm 0.27032 msec−1


Answer 10f. αh 0.006193 msec−1

Answer 10g. βh 0.864127 msec−1

Answer 10h. dn/dt 0.043058 msec−1

Answer 10i. dm/dt –0.141257 msec−1

Answer 10j. dh/dt –0.08432 msec−1

Answer 11a. ∆Vm –2.037 mV

Answer 11b. ∆n 0.002153

Answer 11c. ∆m –0.007063

Answer 11d. ∆h –0.004216

Answer 11e. Vm(t0 + dt) –13.54 mV

Answer 11f. n(t0 + dt) 0.76115

Answer 11g. m(t0 + dt) 0.94794

Answer 11h. h(t0 + dt) 0.09978

Answer 12. Is = −Iion = −40.73 µA/cm2

Answer 13. Both set A and set B have the same value for Vm and hence IL. With the statevariables of Set A, INa is greater and IK is smaller than with the state variables of Set B.Thus set A leads to a negative total ionic current (and thus a positive ∆Vm. In contrast, setB leads to a positive total ionic current (since potassium dominates) and thus a negative∆Vm.

Answer 14a. Set u = 0.1(10 − vm). Then eu = 1 + u+ u2/2 near u = 0, i.e., near vm = 10.Substituting the expansion into the expression gives αn = 0.1/(1 + u/2) near u = 0.

Answer 14b. At vm = 10 the result of Ex. 14A reduces to αn = 0.1.

Answer 15a. 50 mV/msec

Answer 15b. –50 mV/msec

Answer 15c. 259.4 mV/msec

Answer 15d. 9.27 mV/msec


Answer 16a. A computer procedure that computes every α and β for a given vm, including thecode needed for the special cases is:

// alphab sets values for an, bn, am, bm, ah, bh

// from the value of vm given as the function

parameter

// with consideration of special cases

// of alpha_n and alpha_m.

// Sept 5, 2005

int alphab(double vm)

double u, au;

//n

u = (10-vm)/10;

if (u > 0)au = u; elseau = -u;

if (au > 1.E-7)an=.01*(10.-vm)/(exp((10-vm)/10)-1);

else an = 0.1;

bn=0.125*exp(-vm/80);

//m

u = (25-vm)/10;

if (u > 0)au = u; elseau = -u;

if (au > 1.E-7)am=0.1*(25-vm)/(exp((25-vm)/10)-1);

else am = 1.;

bm=4*exp(-vm/18);

//h

ah=0.07*exp(-vm/20);

bh=1/(exp((30-vm)/10)+1);

return 0;

Answer 16b. Table of α and β values:

i vm an bn am bm ah bh

0 0.000 0.06 0.13 0.22 4.00 0.07 0.05

1 5.000 0.08 0.12 0.31 3.03 0.05 0.08

2 10.000 0.10 0.11 0.43 2.30 0.04 0.12

3 15.000 0.13 0.10 0.58 1.74 0.03 0.18

4 20.000 0.16 0.10 0.77 1.32 0.03 0.27

5 25.000 0.19 0.09 1.00 1.00 0.02 0.38

6 30.000 0.23 0.09 1.27 0.76 0.02 0.50

7 35.000 0.27 0.08 1.58 0.57 0.01 0.62

8 40.000 0.32 0.08 1.93 0.43 0.01 0.73

9 45.000 0.36 0.07 2.31 0.33 0.01 0.82

10 50.000 0.41 0.07 2.72 0.25 0.01 0.88

The α and β values have units of msec−1.


Answer 18: Here are some suggestions to consider for writing code that will be usable for arange of questions:

(1) Keep track of the iteration number (here called the “loop count”) as an integer. (Mostpeople do this instinctively.)

(2) Keep track of elapsed time as an integer. (Most program writers do not do thisinstinctively.) Integer microseconds work well.

(3) Enter and record ∆t as an integer value (µsec). Whenever it is needed for computa-tion, convert it then to a floating point form.

(4) Convert explicitly between the iteration number and the elapsed time, using ∆t.

(5) Keep track of the stimulus start/stop time as integer values.

(6) The goal of items 1 to 5 is to avoid issues of floating point round-off in keeping time,identifying the stimulus starting time, etc.

(7) Keep track of the interval between times that output lines will be created with adistinct integer variable. In general, do not write output for every iteration.

(8) If there is a stimulus, start the first stimulus at t = 0. Begin calculation, however,before that time, so that a baseline can be established.

Generally the program is easiest to manage if elapsed time is converted explicitly fromthe iteration number, because, in this way, the iteration number increases from zero asexpected, while elapsed time can begin with a negative value. Additionally, programdecisions about whether it is time to start or stop the stimulus, or to print output, canbe parametrized on the basis of elapsed time, rather than iteration number. Importantly,interval ∆t can be varied, which often is desirable for testing, independently of thespecification of when the stimulus should occur or output lines should be displayed.

If there is a discrepancy between the output table as printed and your results, consider thefollowing aspects of how the table above is computed and verify their correspondence.Here, line k + 1 is found from line k by:

(1) Using vm for line k to compute α and β values for line k, and ∆n, etc, for the timeof line k. Note that vm for line k = 1 is not used.

(2) Using n, m, h from line k (not line k+1) to get currents and vdot at the time of linek, and from these estimate vm at time k + 1. Note that newer values of n, m, h fortime k + 1 are not used.

(3) Being sure the stimulus does not extend into the period 150 to 200. That is, at time150 the stimulus must stop.

Answer 19. The extended output table for ∆t = 50µsec is:

loopcnt time Is vm vdot n m h

0 -200 0.0 -60.000 -0.0 0.31768 0.05293 0.59612

1 -150 0.0 -60.000 -0.0 0.31768 0.05293 0.59612

2 -100 0.0 -60.000 -0.0 0.31768 0.05293 0.59612

3 -50 0.0 -60.000 -0.0 0.31768 0.05293 0.59612

4 0 200.0 -60.000 200.0 0.31768 0.05293 0.59612


5 50 200.0 -50.000 193.2 0.31768 0.05293 0.59612

6 100 200.0 -40.339 187.5 0.31934 0.06726 0.59343

7 150 0.0 -30.965 -16.1 0.32308 0.09804 0.58617

8 200 0.0 -31.769 -3.2 0.32925 0.14894 0.57257

9 250 0.0 -31.928 17.0 0.33510 0.19253 0.55988

10 300 0.0 -31.077 42.5 0.34081 0.23132 0.54761

11 350 0.0 -28.950 72.6 0.34667 0.26851 0.53503

12 400 0.0 -25.320 108.3 0.35303 0.30769 0.52129

13 450 0.0 -19.905 152.1 0.36032 0.35276 0.50556

14 500 0.0 -12.300 205.7 0.36908 0.40786 0.48727

15 550 0.0 -2.017 264.0 0.37999 0.47705 0.46662

16 600 0.0 11.185 301.8 0.39383 0.56279 0.44473

17 650 0.0 26.277 261.6 0.41141 0.66261 0.42290

18 700 0.0 39.355 114.1 0.43299 0.76511 0.40186

19 750 0.0 45.063 -5.3 0.45755 0.85187 0.38180

20 800 0.0 44.797 -6.4 0.48256 0.91069 0.36273

21 850 0.0 44.475 -26.5 0.50628 0.94580 0.34462

22 900 0.0 43.148 -23.5 0.52874 0.96677 0.32741

23 950 0.0 41.972 -37.1 0.54978 0.97913 0.31106

24 1000 0.0 40.115 -33.6 0.56953 0.98649 0.29554

After time 200 µsec, voltage vm does not continue to fall. Rather, it rises to a substantiallyhigher value in the course of generating an action potential.

Answer 20. ∆t = 2µsec

dTime: 2 microseconds

StimAmplitude: 200

StimDuration: 150

Hh: EK -72.100 ENa 52.400 EL -49.187

Hh: gbarK 36.0 gbarNa 120.0 gL 0.3

Hh: n 0.31768 m 0.05293 h 0.59612

Hh: Cm 1.00 Temper 6.30

Hh: gK 0.367 gNa 0.011 gL 0.300

Hh: Vm -60.000 Vr -60.000

loopcnt time Is Vm vdot n m h

0 -200 0.0 -60.000 -0.0 0.31768 0.05293 0.59612

25 -150 0.0 -60.000 -0.0 0.31768 0.05293 0.59612

50 -100 0.0 -60.000 -0.0 0.31768 0.05293 0.59612

75 -50 0.0 -60.000 -0.0 0.31768 0.05293 0.59612

100 0 200.0 -60.000 200.0 0.31768 0.05293 0.59612

125 50 200.0 -50.155 193.7 0.31842 0.05921 0.59497

150 100 200.0 -40.599 188.7 0.32099 0.07966 0.59039


175 150 0.0 -31.218 -12.5 0.32577 0.11756 0.58045

200 200 0.0 -31.529 1.8 0.33172 0.16300 0.56747

225 250 0.0 -30.967 22.6 0.33761 0.20431 0.55475

250 300 0.0 -29.207 50.0 0.34371 0.24456 0.54155

275 350 0.0 -25.894 85.5 0.35040 0.28738 0.52703

300 400 0.0 -20.557 132.1 0.35820 0.33714 0.51029

325 450 0.0 -12.559 192.6 0.36781 0.39886 0.49086

350 500 0.0 -1.283 260.8 0.38013 0.47712 0.46934

375 550 0.0 12.984 302.2 0.39608 0.57220 0.44718

400 600 0.0 27.507 260.3 0.41601 0.67487 0.42554

425 650 0.0 37.908 146.4 0.43895 0.76850 0.40482

450 700 0.0 42.692 49.7 0.46304 0.84137 0.38509

475 750 0.0 43.894 3.6 0.48679 0.89273 0.36631

500 800 0.0 43.580 -14.3 0.50950 0.92743 0.34845

525 850 0.0 42.646 -22.7 0.53097 0.95056 0.33147

550 900 0.0 41.374 -28.2 0.55117 0.96592 0.31531

575 950 0.0 39.856 -32.6 0.57010 0.97612 0.29995

600 1000 0.0 38.137 -36.3 0.58782 0.98288 0.28533

Answer 20a. vm 49.692 − 39.355 = 3.337 mV

Answer 20b. n 0.46304 − 0.43299 = 0.03005

Answer 20c. m 0.84137 − 0.76511 = 0.07626

Answer 20d. h 0.38509 − 0.40186 = −0.01677

Answer 21a. Yes, linear for t ≤ 100 µsec.

Answer 21b. No, linear for t > 200 µsec.

Answer 22a. For duration 150, just-above-threshold stimulus 45 µA/cm2.

Answer 22b. For duration 150, just-above-threshold stimulus at end of stimulus vm 6.46 mV.

Answer 22c. For duration 300, stimulus 23 µA/cm2.

Answer 22d. For duration 300, vm at end of stimulus 6.38 mV.

The above values depend critically on the fact that the evaluation is for a membrane patch,so that no current is moving longitudinally along intracellular or extracellular path.

Answer 23a. For Is =50, the time to peak is 3100 µsec.

Answer 23b. For Is =200, the time to peak is 750 µsec.

Answer 23c. For Is =500, the time to peak is 450 µsec.


Answer 24. The time duration until stable return within the envelope of the initial conditions is24100 µsec.

Answer 25a. The EL (that corresponds to the new gL) is 264.4 mV.

Answer 25b. Some answers are more fun if you see for yourself. Inspect the return to baselinecarefully, perhaps by plotting vm(t).

Answer 26a. Is =50, 2nd AP at 16800 microseconds from start of first stimulus.

Answer 26b. Is =200, 2nd AP at 8150.

Answer 26c. Is =500, 2nd AP at 5150.

22 SOLUTIONS FOR CHAPTER 6: IMPULSE PROPAGATION

14.6. SOLUTIONS, CHAPTER 6: IMPULSE PROPAGATION

Answer 1a. µF/cm

Answer 1b. Ωcm

Answer 1c. Ω/cm

Answer 1d. Ωcm

Answer 2a. Recall that the HH model has a linear response to transmembrane voltage changesnear the resting potential, so the membrane resistance can be found from the conductancesin the HH model. Use Table 13.5, which gives resting values of HH membrane, to findthe conductances (g values) of HH membrane at rest. Using those values,

Rm = 1/(gK + gNa + gL) = 1, 475 ohm − cm2 (3)

Answer 2b. The resistance R in ohms of a segment of membrane is

R = Rm/As = Ri/(2πaL) = rm/L (4)

Comparing the last two parts of the equation shows that rm = Rm/(2πa). Using theresult forRm found in the answer to the preceding question, one has rm = Rm/(2π0.003)or rm = 78, 248 Ωcm.

Answer 2c. The intracellular resistance per unit length characterizes the volume inside themembrane in a way that takes into account the medium’s resistivity and the fiber’s radius.After ri has been found, one need only make a later choice of a length L (distance alongthe axis) to find resistance R of the segment. Thus resistance R will be

R = riL = RiL/Ai = RiL/(πa2) (5)

To make the last two terms equal, one makes ri = Ri/(πa2). In this fiber, the result isri = 5, 305, 169 ohm/cm.

Answer 2d. The connection of the intracellular resistance per length with the region inside themembrane (in the preceding part) seems obvious, but the statement that “Extracellularcurrents flow to twice the membrane radius” seems much less so. Indeed, in the absenceof a physical boundary of some kind, asserting that currents flow between radius a andradius 2a is at best arbitrary. Even so, the statement recognizes that extracellular currentflows most intensely near the membrane, and it provides a specific basis for computingre, even if it is only an estimate. Note that the cross-sectional area inside a is one-thirdthe cross-section between a and 2a, i.e., assuming 2a as a limiting radius for extracellularcurrent assumes that extracellular current flows through a greater cross-section, but stillwithin a radius of the fiber membrane. With this assumption, then using similar reasoningto that of the previous part, one has

R = ReL/Ae = ReL/[π((2a)2 − a2)] = reL (6)

so that re = Re/[π((2a)2 − a2)]. In this fiber, the result is re = 589, 463 ohm/cm.


Answer 3a. rm = 63, 700 Ωcm

Answer 3b. cm = 0.0377 µF/cm

Answer 3c. ri = 1, 273, 000 ohm/cm

Answer 3d. re = 169, 000 ohm/cm.

Answers 4A–C are taken from material in the text of Chapter 6.

Answer 4a. Intracellular current is

Ii =−1

(ri + re)[∂Vm

∂x− Ire] (7)

When I = 0 because there is no stimulus, the 2nd term drops out.

Answer 4b. Because I = Ii + Ie, the extracellular current Ie = I − Ii, and thus,

Ie =1

(ri + re)[∂Vm

∂x+ Ire] (8)

When I = 0 (no stimulus) the second term again drops out.

Answer 4c. The transmembrane current

Im =1

(2πa)(ri + re)(∂2Vm

∂x2 − reip) (9)

where ip = 0 because there is no stimulus. As the question asks for Im, i.e., current percm2, the Im equation requires that dimensions (e.g., of a) are in cm.

Answer 5a. Vm(x) = av tanh(x) has negative Potentials for x < 0, positive for x > 0. Theaction potential is moving left, toward the more negative region.

Answer 5b. Ii(x) = −ai sech2(x). Intracellular current is flowing left.

Answer 5c. Ie(x) = +ae sech2(x). Extracellular current is flowing right.

Answer 5d. Im(x) = −2am sech2(x) tanh(x). Membrane current is outward left, inward right.

Answers 5a–d. The expressions given depend on undefined constants av , ai, ae, and am, whichare proportionality constants that include such factors as the radius and the axial resis-tances. The proportionality constants are not defined in detail here, so that this exercisecan focus on the curve shapes rather than magnitudes.

Answer 6a. (v0 − 2v1 + v2)/(R+ r) milliamperes

Answer 6b. (v1 − 2v2 + v3)/(R+ r) milliamperes

Answer 6c. (v2 − 2v3 + v4)/(R+ r) milliamperes

24 SOLUTIONS FOR CHAPTER 6: IMPULSE PROPAGATION

Answer 7. The difference in Question 7 (relative to Exercise 6) is that in Ex. 7 it is asked thatmembrane current be found “per unit area.” Dividing the earlier result by the geometricfactors of one segment, one has

I3m =

1(2πa∆x)

(v2 − 2v3 + v4)(R+ r)

mA/cm2 (10)

Answer 8. The transmembrane current at crossing 0, at the left end, is the same as the intra-cellular axial current between crossings 0 and 1. Thus I0 = (v1 − v0 + rS)/(R + r)milliamperes.

Answer 9. The expression must take into account the different manner for computing the deriva-tive, and the different area of the segments at the fiber’s ends. Thus

I0m =

2(πa∆x)

(v1 − v0 + rS)(R+ r)

(11)

Answer 10a. I1m = +1.085 mA/cm2

Answers 10A-b. Note that the computation can be simplified by substituting λ algebraicallyinto the equation for Im.

Answer 11a. I0m = −6.508 mA/cm2

Answer 11b. I1m = +6.508 mA/cm2

Answers 11a–b. These answers can be obtained by finding the solution again from scratch, oras a variation on the answers to exercise 10, looking at the relationships of the givenvoltages.

Answer 12. From Chapter 6, the rate of change of Vm with time t is

∂Vm(xo, t)∂t

=[Im(xo, t) − Iion(xo, t)]

Cm(12)

Answer 13a. 6.18 milliseconds

Answer 13b. 240.5 mV/msec

Answer 13c. 10.3 millivolts

Answer 13d. –7.84 microamperes/cm2

Answer 13e. –248.4 microamperes/cm2


Answer 14a. 2.5 msec

Answer 14b. 280.2 mV/msec

Answer 14c. 12.4 millivolts

Answer 14d. –3.57 µA/cm2

Answer 14e. –284 µA/cm2

Answer 15a. The mesh ratio is 3.01.

Answer 15b. The mesh ratio indicates instability. The bad outlook is tempered by the HHmembrane having fairly low Rm, and the resistance present along the extracellular aswell as intracellular axial direction, which is not taken into account in the mesh ratiocalculation.

Answer 15c. The manifestation of instability will be that, at some time step, values of Vm willbegin to oscillate with increasing amplitude, and after a few time steps will be out of thecomputable range.

Answer 15d. Stability can be markedly improved by reducing ∆t to a lower value. For example,making ∆t = 1 µsec would reduce the mesh ratio to 0.30, a change in mesh ratio thatlikely would correspond to a stable calculation.

Answer 15e. Try it and see.

26 SOLUTIONS FOR CHAPTER 7: ELECTRICAL STIMULATION OF EXCITABLE TISSUE

14.7. SOLUTIONS, CHAPTER 7: ELECTRICAL STIMULATION OF EXCITABLETISSUE

Answer 1. rheobase

Answer 2. threshold

Answer 3. chronaxie

Answer 4a. seconds

Answer 4b. Volts

Answer 4c. Volts

Answer 5. a/(1 − exp[−500/(RC)]

)Answer 6. W

Answer 7. 1000U/R

Answer 8. 1000W/a

Answer 9. W

Answer 10. Begin by rearranging the Weiss–Lapicque equation so that e−T/τ is the only termon the left. Rewrite the rearranged equation twice, once for each condition. Divide thefirst equation by the second, and combine terms so that there is only one exponential termon the left. Take the natural log, and rearrange the result to solve for τ . The result is:

τ = (D − d)/ log

(I

i

(a− i)(a− I)

)

where log is the natural log.

Answer 11. 20 millivolts

Answer 12. 2,000 Ωcm2

Answer 13. 125.07 µA/cm2

Answer 14. 2.0 msec.

Answer 17. 0.115214 cm

Answer 18. λ = 0.212 cm

Answer 18. λ = 0.212 cm

Answer 18. λ = 0.212 cm


Answer 19. τ = 1.8 msec

Answer 21a. P (x) = S/√

(h2 + (x− e)2)

Answer 21b. A(x) = S(−1/r3 + 3(x − e)2/r5). Note the marked difference between theanswer 21A (which naively seems to be the effect of the stimulus on the fiber), andAnswer 21B. Although A(x) is not an equation for Vm produced by the stimulus, it doesat least provide the initial direction of change for Vm.

Answer 21c. A(x) = −S(1/r3 − 3e2/r5)

Answer 22a. The distance where A(x) > 0 is 8.6 cm.

Answer 22b. The distance where A(x) > 0 is 1.4 cm.

Answer 22c. No. In stimulation, extent of change is important, but amplitude is also important,and often more so. A significant point of interest is, however, that both source and sinkstimuli produce regions of positiveA(x), as this result indicates that either polarity mightserve as a stimulus, but would stimulate to different regions of the fiber.

Answer 23a. A(x) = 3S/(4h2)

Answer 23b. A(x) = S(1/h2 − 1/(h2 + d2)(3/2) + d2/(h2 + d2)(5/2))

Answer 24. A(x) > 0 is a region where, at least initially, Vm is growing more positive, andthus is more likely to reach threshold and initiate an action potential.

Answer 25. 3 extrema

Answer 26. 4

Answer 27. The ratio L2/L1 = 2. The answer shows that the portion of the fiber being depo-larized by the stimulus is affected markedly by the location of the electrode relative tothe fiber.

Answer 28. The ratio L2/L1 = 2. Note that a ratio other than 1 occurs because of a changein the separation of the electrodes, not a change in orientation of the electrode pair. Thechanged ratio indicates that more of the fiber is being affected, i.e., L2 > L1.

Answer 29. A(x = 0.3) = 28.5 mV/cm2

Answer 30a. 8,100 Joules

Answer 30b. 2,160 Joules

Answer 31a. About 3.2E9 stimuli

Answer 31b. About 5.7E9 stimuli

Answer 32a. 175 mV/msec

Answer 32b. 150 mV/msec

28 SOLUTIONS FOR CHAPTER 7: ELECTRICAL STIMULATION OF EXCITABLE TISSUE

Answer 33a. 1.33 Volts

Answer 33b. 0.995 Volts

Answer 34a. 1.19 milliamperes

Answer 34b. 2.50 milliamperes

Answer 35. The design meets both requirements.

Answer 36. Design fails requirement 1 but meets requirement 2.

Answer 37. Design meets requirement 1 but fails requirement 2.

Answer 38. Design meets both requirements.

Answer 39. The authors know the answer to this question, but have chosen not to include ithere, so as not to take away the reader’s enjoyment of finding the solution.


14.8. SOLUTIONS, CHAPTER 8: EXTRACELLULAR FIELDS

Answer 1. At t = 10 milliseconds. On the graph, one sees the midpoint of the upstroke at about32 mm. Dividing by the velocity (4 mm/msec), one finds that this action potential occursabout 8 msec after the one at x = 0. Additionally, parameter t1 adds a 2 msec time delayto the upstroke.

Answer 2. See Figure 14.1. The graphs here were made with Matlab r©. Matlab is the registeredtrademark of The Mathworks Inc.

Figure 1. Vm(t) at position x = 24 mm.

Answer 3. In an action potential the activation phase always comes first. Therefore, on the timewaveform, the activation phase (fast upstroke) is on the left (lower time). On an actionpotential waveform traveling toward higher x, the activation phase is on the right (higherx), since activation is moving to the right. Otherwise the action potentials appear largelythe same, though real action potentials may show more subtle differences.

Answer 4. See Figure 14.2. Here V maxm = 95.5 mV/ms. For comparison, measured transmem-

brane potentials from healthy tissue usually have V maxm between 100 and 200 mV/ms, as

abnormalities inevitably leads to a lower value. This (unnatural) AP also is on the slowside.

Answer 5a. Amax = 23.8708, xa = 32.0070, va = −14.9099, i = 640. Here xa is the xcoordinate of the point of maximum slope, va is the value of Vm at that point, and Amaxis the magnitude of the slope at that point.

Answer 5b. Vm = −Amax(x− xa) + va

Answer 5c. Bmax = 6.2494, xr = 19.9950, vr = −10.0319, i = 400

Answer 5d. Vm = Bmax(x− xr) + vr

30 SOLUTIONS FOR CHAPTER 8: EXTRACELLULAR FIELDS

Figure 2. Vm(t) at position x = 24 mm. V maxm = 95.5 mV/ms.

Answer 5e. xcross = 29.35, Vmcross = 48.4

Answer 5f. Activation crossing at x = 33.9 mm. Recovery crossing at x = 12.0 mm.

Answer 6b. With the velocity doubled, the AP waveform of Vm(x) occupies twice the distancealong x. Also, the leading edge is about twice the distance from the origin.

Answer 6d. With velocity doubled, the AP wave shape of Vm(t) remains the same, except thatthe time of onset is shorter.

Answer 7. The axial current is plotted in Figure 14.3.

Figure 3. Ii(x) at time t = 10 msec. The vertical scale extends to 2.5E-4 mA.


Answer 8. The transmembrane current is plotted in Figure 14.4.

Figure 4. Transmembrane current im(x)) at time t = 10 msec.

Answer 17. The transmembrane current is plotted in Figure 14.5. In the figure, each panelshows the potential that arises from the lumped monopolar sources (the solid line) and thepotential that arises from the continuous sources (the dashed line). Panels A, B, and C arefor distances h of 0.1, 1.0, and 10.0 millimeters, as marked. All plots are for time t = 10msec. When h is small, as in panel A, the approximation of the continuous by lumpedsources is poor, so lumpiness is striking. Conversely, as h increases to 1 millimeter, thepotentials as generated by the lumped sources are smoother. At 10 millimeters, the lumpedsolution becomes similar to those from the continuous sources, even though distance hstill is less than the distance between source M1 and source M3.

Answer 27. V = 7.162 × 10−5 V. This value was found using Eq. (8.71). The potentials are tobe found at points far enough from the cell that a simplification is possible, because thesolid angle is virtually a constant and so can be factored from the equation.

32 SOLUTIONS FOR CHAPTER 8: EXTRACELLULAR FIELDS

Figure 5. Extracellular potential Φe(x) from Lumped Monopolar and Continuous sources.


14.9. SOLUTIONS, CHAPTER 9: CARDIAC ELECTROPHYSIOLOGY

Answer 1. Cardiac electrophysiology, compared to that of nerve. Correct statements are num-bers 1, 3, and 5.

Answer 2. Cardiac structure. Correct choices are 1, 2, 3, and 4.

Answer 3. AV node propagation: choice 2

Answer 4. Correct choices are 1, 2, 3, 4, 5, 9

Answer 5. 10

Answer 6. 4.1E9 ohms

Answer 7. Multiply milliseconds by 0.001 to convert the values to seconds.

Answer 8. 22 ± 3 ms

Answer 9. 44 ± 4 ms

Answer 10. Purkinje strand. The correct responses are 2 and 3.

Answer 11. 300E6 ohms

Answer 12. 100 ± 20 mV

Answer 13. 0.3 ± 0.05 m/s

Answer 14. 0.0015 ± 6E-4 m

Answer 15. 0.036 ± 0.005 V. Some variability depends on exactly how the edges are defined.

Answer 16. 40 ± 15 mV Making an enlarged figure is helpful. Note the calibration bar.

Answer 17. 15 ± 5 mV Making an enlarged figure is helpful. Note the calibration bar.

Answer 18. Constants in the differential equation are a = 1/(cqR) and b = u/(cq).

34 SOLUTIONS FOR CHAPTER 9: CARDIAC ELECTROPHYSIOLOGY

Answer 19. In the solution for v, a = uR and b = cqR.

Answer 20. About Purkinje fibers. Choices are 3 and 5.

Answer 21. Connexon selectivity. Choices are 1, 2, and 3.

Answer 22. Choices are all the statements.

Answer 23. 80% of 8E10 cells.

Answer 24. 4.2E-4 s

Answer 25. vm rises 0.04 V

Answer 26. 2.451 megaohms

Answer 27. Cell 2 has transmembrane voltage of 0.019 V relative to rest.

Answer 28. 0.021 V

Answer 29. Time required to reach threshold is 2.3541E-4 s.

Answer 30. Nominal velocity 0.288 m/s

Answer 31a. 0.143 m/s

Answer 31b. 0.282 m/s

Answer 31c. 0.144 m/s

Answer 31d. 0.659 m/s

Answer 32. Increasing the velocity. Choose 1 and 4. Choices other than 4 are evaluated in thepreceding question, while 4 is evaluated in an earlier section.

Answer 33. Select choices 1, 2, 6, 7, and 8.

Answer 34. 1, 2, 4, and 5


Answer 35. 0.002618 S/cm

Answer 36. 0.523599

Answer 37. 923.99784 1/cm−1

Answer 38. 209 Ωcm

Answer 39. 0.083 cm

Answer 40. 0.1617 cm

Answer 41. 0.0265 V

Answer 42. 0.0707 V

Answer 43. –0.0900 V

Answer 44. –0.0242 V

Answer 45. 0.0604 V

Answer 46. 0.0465 V

Answer 47. –0.097 V

Answer 48. Select 2, 4, 5, 6, 7

Answer 49. Select 2 and 3. The design fails because requirement 1 is not satisfied.

Answer 50. Select 1 and 3. The design fails because requirement 2 is not satisfied.

Answer 51. 0.8

Answer 52. –0.12 V. Note that ∆vm has a negative sign. It is the value across the excitationwave, i.e., the value in the tissue ahead of the excitation wave minus the value in theregion the excitation wave has passed.

Answer 53. –0.6389


Answer 54. 250 Ωcm

Answer 55. 300 Ωcm

Answer 56. 0.0042E-4 V

Answer 57. φLL = 5.04E − 5 V

Answer 58. VII = 9.62E − 5 V

Answer 59. VII = −1.07E − 4 V

Answer 75a.

∫v

−σφ∇2(

1r

)dV =

∫H

σM∇φH · dSHin

r−∫

H

σMφH∇(

1r

)· dSHin

−∫

L

σMφL∇(

1r

)· dSLin −

∫L

σLφL∇(

1r

)· dSL

−∫

B

σMφB∇(

1r

)· dsB (13)

In Eq. (??), vector surface element dSHin points into the heart region H, vector surfaceelement dSB points out of the body surface, vector surface element dSLin points intothe lung region L, and vector surface element dSL points out of the lung region L. Thelung conductivity is designated σL, while the remaining torso conductivity is describedby σM .

In getting Eq. (??) from Eq. (13.18), several particular aspects of current flow in the torsovolume conductor were used. First, ∇2φ is zero at every point within the volume, sincethe volume has been drawn in such a way as to exclude the active tissue in the heart and toexclude regions where σ makes a transition (i.e., the site of secondary sources). Second,current flow across the lung boundary is continuous. Finally, the normal component ofelectric field, ∇φ · n, approaches zero as a point approaches the body surface, since theexterior of the body surface is a nonconductor (hence no current can leave the volumeconductor).

Answer 75b. ∫V

−σφ∇2(

1r

)dV = 4πδ(r)σBφb (14)

where φb and σB are the potential and conductivity at the point b, where r = 0. Thispoint is chosen arbitrarily and located just inside the body surface. This is the point forwhich potentials will be determined.


Answer 75c.

4πσBφb = σM

∫H

∇φH · dSHin

r− σM

∫H

φH∇(

1r

)· dSHin

−(σL − σM )∫

L

φL∇(

1r

)· dSL − σM

∫B

ΦB∇(

1r

)· dSB (15)

Answer 75d. As a first step we get:

φb = − σM

4πσB

∫H

∇φH cdotdSH

r− σM

4πσB

∫H

φHarH

· nH

r2HdSH

+(σL − σM )

4πσB

∫L

φLarL

· nL

r2LdSL +

σM

4πσB

∫B

φBarB

· nB

r2BdSB (16)

where unit normal vector nH points out of heart region H,nL points out of lung regionL, and nB points out of the body region B. Vectors arH

, arL, and arB

are all of unitmagnitude and their directions are associated with the radius vectors (rH , rL, rB) fromb to the heart, lung, and body surface, respectively.

Using the equation for the solid angle in the equation above (??) gives

φb = − σM

4πσB

∫H

φHdΩH − σM

4πσB

∫H

∇φH · dSH

r

+(σL − σM )

4πσB

∫L

φLdΩL +σM

4πσB

∫B

φBdΩB (17)

This expression gives the potential at one point on the body surface in terms of several in-tegrals. The integrals thereby can be interpreted as identifying the major factors affectingthe potential at a body surface point.

The first two integrals take into account the potentials and the gradients of potential (i.e.,the currents) that exist on the cardiac surface. Although shown here as two separateintegrals, it is useful to note that the potentials and gradients on the cardiac surface arenot independent of each other.

The third integral takes into account the lack of uniformity in the volume conductor.Specifically, in this derivation the integral takes into account the different conductivityassigned to the lung volume. In problems with other inhomogeneities, additional integralswould be present to take into account their effects. For example, the skeletal muscle isanother inhomogeneity (which also is anisotropic) that is thought to play a significantrole in determining the body surface potentials.

Finally, the fourth integral shows that the potential at a point on the body surface alsodepends on the potentials at other points on the body surface. Why is this so? The bodysurface defines the interface between the conducting torso and the nonconducting air,locating an obvious discontinuity in conductivity. Accordingly, secondary sources ariseat this surface and the potential of any surface point will depend on the field generatedby all such secondary sources. In fact, the secondary source distribution will be suchthat the field it generates when added to the field from all other sources (i.e., appliedsources) must satisfy the aforementioned boundary condition, namely, ∇φ · an = 0 on


B. In general, the body surface sources cause the potentials measured there to be higherthan they otherwise would be, by approximately a factor of 2 or 3.

While the integrals in (14.17) arise from field theoretic considerations, note that thecontribution from the lung and body surface contains the secondary source double layersdescribed by (8.64). In the sense of (14.17) the first two terms could be thought torepresent primary sources, namely, a double layer (with strength σMφH , as shown inthe first integral) and a single layer (with strength σM∇φH

· an, as shown in the secondintegral). However, such sources are actually equivalent sources, and (14.17) can only beused to calculate fields outside the heart.

The integrals in the expression above make unequal contributions to the body surfacepotentials. A first approximation to the resulting body surface potential’s variation frompoint to point can often be obtained after ignoring the last two integrals altogether. Fur-ther, because of their different r dependence, the contributions from the first and secondintegrals may be quite unequal.

Answer 75e. The equation is the same except that the lung integral drops out.


14.10. SOLUTIONS, CHAPTER 12: FUNCTIONAL ELECTRICAL STIMULATION

Answer 24. The exercises ask for portions of computer code that handle the Frankenhauser–Huxley (FH) membrane model. The responses below are taken from a report by ChadR. Johnson, which includes a complete program listing. (Johnson CR. 2005. A mathe-matical model of peripheral nerve stimulation in regional anesthesia, pp. 359–367. PhDdissertation, Department of Biomedical Engineering, Duke University.)

Answer 24A. Code to initialize the necessary FH variables.

function [] = iFH(vm)

% IFH Initialize FH membrane model.

% iFH initializes the parameters and gating

% variables for the FH membrane model.

%

% INPUT vm: Transmembrane potential vector.

% -- PREAMBLE ------------------------------------------

global pFH;

global CONST;

global DATA;

global UDATA;

% -- pFH STRUCTURE -------------------------------------

% Set FH parameters in structure.

pFH = struct( ...

’Vrest’, -70.0, ... % mV

’Nai’, 13.74, ... % mM

’Nao’, 114.5, ... % mM

’Ki’, 120.0, ... % mM

’Ko’, 2.5, ... % mM

’El’, 0.026, ... % mV

’PNa’, 8.0e-3, ... % cm/sec

’PK’, 1.2e-3, ... % cm/sec

’PP’, 0.54e-3, ... % cm/sec

’gbarl’, 30.3, ... % mS/cmˆ2

’gmyel’, DATA.gmyel, ... % mS/cmˆ2

’R’, 8314.4, ... % mJ/(K mole)

’T’, UDATA.temp, ... % K

’Q10’, 0, ...

’F’, 96485.0, ... % coulombs/mole

’D’, 0, ... % Constant, defined below.

’Rv’, 0, ... % Indices for active nodes.

’M’, 0, ... % Indices for myelination.

’malpha’, 0, ... % Current rate constants...

40 SOLUTIONS FOR CHAPTER 12: FUNCTIONAL ELECTRICAL STIMULATION

’mbeta’, 0, ...

’halpha’, 0, ...

’hbeta’, 0, ...

’nalpha’, 0, ...

’nbeta’, 0, ...

’palpha’, 0, ...

’pbeta’, 0, ...

’gates’, 0, ... % Gating variables.

’I’, 0); % Ionic currents.

% -- INITIALIZE FH MODEL ------------------------------

% Find indices for nodes and non-nodes.

pFH.Rv = find(DATA.ldx);

pFH.M = find(1-DATA.ldx);

% Define constant D (RT/F).

pFH.D = pFH.R.*pFH.T./pFH.F;

% Recalculate Nernst potentials for given temperature.

pFH.El = (pFH.T./293.15).*pFH.El;

% Define Q10’s (from Rattay review, 1993).

pFH.Q10.malpha = 1.8.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.mbeta = 1.7.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.halpha = 2.8.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.hbeta = 2.9.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.nalpha = 3.2.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.nbeta = 2.8.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.palpha = 3.0ˆ((pFH.T-293.15)./10.0);

pFH.Q10.pbeta = 3.0.ˆ((pFH.T-293.15)./10.0);

% Q10’s for PNa and PK

% from Frankenhaeuser and Moore:

% Q10.PNa = 1.3; Q10.PK = 1.2.

pFH.Q10.PNa = 1.3.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.PK = 1.2.ˆ((pFH.T-293.15)./10.0);

% Q10 for leakage current: added by CRJ.

pFH.Q10.gbarl = 1.2.ˆ((pFH.T-293.15)./10.0);

% Initialize rate constant vectors, msec-1.

pFH.halpha = pFH.Q10.halpha.*0.1.* ...

(-10.0-vm(pFH.Rv))./ ...

(1.0-exp((vm(pFH.Rv)+10.0)./6.0));

pFH.hbeta = pFH.Q10.hbeta.*4.5./ ...

(1.0+exp((45.0-vm(pFH.Rv))./10.0));

pFH.malpha = pFH.Q10.malpha.*0.36.* ...


(vm(pFH.Rv)-22.0)./ ...

(1.0-exp((22.0-vm(pFH.Rv))./3.0));

pFH.mbeta = pFH.Q10.mbeta.*0.4.* ...

(13.0-vm(pFH.Rv))./ ...

(1.0-exp((vm(pFH.Rv)-13.0)./20.0));

pFH.nalpha = pFH.Q10.nalpha.*0.02.* ...

(vm(pFH.Rv)-35.0)./ ...

(1.0-exp((35.0-vm(pFH.Rv))./10.0));

pFH.nbeta = pFH.Q10.nbeta.*0.05.* ...

(10.0-vm(pFH.Rv))./ ...

(1.0-exp((vm(pFH.Rv)-10.0)./10.0));

pFH.palpha = pFH.Q10.palpha.*0.006.* ...

(vm(pFH.Rv)-40.0)./ ...

(1.0-exp((40.0-vm(pFH.Rv))./10.0));

pFH.pbeta = pFH.Q10.pbeta.*0.09.* ...

(-25.0-vm(pFH.Rv))./ ...

(1.0-exp((vm(pFH.Rv)+25.0)./20.0));

% Initialize gating variable vectors.

pFH.gates.m = pFH.malpha./(pFH.malpha+pFH.mbeta);

pFH.gates.h = pFH.halpha./(pFH.halpha+pFH.hbeta);

pFH.gates.n = pFH.nalpha./(pFH.nalpha+pFH.nbeta);

pFH.gates.p = pFH.palpha./(pFH.palpha+pFH.pbeta);

% Adjust pFH.Vrest to retain equilibrium.

INa = pFH.Q10.PNa.*pFH.PNa.*(pFH.gates.m(1).ˆ2.0).* ...

pFH.gates.h(1).*(pFH.Vrest.*pFH.F./pFH.D).* ...

((pFH.Nao-pFH.Nai.*exp(pFH.Vrest./pFH.D))./ ...

(1.0-exp(pFH.Vrest./pFH.D)));

IK = pFH.Q10.PK.*pFH.PK.*(pFH.gates.n(1).ˆ2.0).* ...

(pFH.Vrest.*pFH.F./pFH.D).* ...

((pFH.Ko-pFH.Ki.*exp(pFH.Vrest./pFH.D))./ ...


IP = pFH.Q10.PNa.*pFH.PP.*(pFH.gates.p(1).ˆ2.0).* ...




Il = pFH.Q10.gbarl.*pFH.gbarl.*(-pFH.El);

Iion = INa+IK+IP+Il;

while (abs(Iion)>CONST.myzero)
















pFH.Vrest = pFH.Vrest - ...

((UDATA.dt.*1.0e-3)./(mean(DATA.Ctot))).*Iion;

end

% Adjust El to get Iion=0 at rest.

pFH.El = (INa+IK+IP)./( pFH.Q10.gbarl.*pFH.gbarl);

% Initialize current vectors: include all nodes.

pFH.I.Na = zeros(size(vm));

pFH.I.K = zeros(size(vm));

pFH.I.P = zeros(size(vm));

pFH.I.l = zeros(size(vm));

pFH.I.myel = zeros(size(vm));

Answer 24B. Code to compute the membrane currents for the current membrane state:

function [] = iFH(vm)

% IFH Initialize FH membrane model.

% iFH initializes the parameters and gating

% variables for the FH membrane model.

%

% INPUT vm: Transmembrane potential vector.

% -- PREAMBLE -------------------------------------

global pFH;

global CONST;

global DATA;

global UDATA;

% -- pFH STRUCTURE --------------------------------


% Set FH parameters in structure.

pFH = struct( ...

’Vrest’, -70.0, ... % mV

’Nai’, 13.74, ... % mM

’Nao’, 114.5, ... % mM

’Ki’, 120.0, ... % mM

’Ko’, 2.5, ... % mM

’El’, 0.026, ... % mV

’PNa’, 8.0e-3, ... % cm/sec

’PK’, 1.2e-3, ... % cm/sec

’PP’, 0.54e-3, ... % cm/sec

’gbarl’, 30.3, ... % mS/cmˆ2

’gmyel’, DATA.gmyel, ... % mS/cmˆ2

’R’, 8314.4, ... % mJ/(K mole)

’T’, UDATA.temp, ... % K

’Q10’, 0, ...

’F’, 96485.0, ... % coulombs/mole

’D’, 0, ... % Constant, defined below.

’Rv’, 0, ... % Indices for active nodes.

’M’, 0, ... % Indices for myelination.

’malpha’, 0, ... % Current rate constants...

’mbeta’, 0, ...

’halpha’, 0, ...

’hbeta’, 0, ...

’nalpha’, 0, ...

’nbeta’, 0, ...

’palpha’, 0, ...

’pbeta’, 0, ...

’gates’, 0, ... % Gating variables.

’I’, 0); % Ionic currents.

% -- INITIALIZE FH MODEL ----------------------------

% Find indices for nodes and non-nodes.

pFH.Rv = find(DATA.ldx);

pFH.M = find(1-DATA.ldx);

% Define constant D (RT/F).

pFH.D = pFH.R.*pFH.T./pFH.F;

% Recalculate Nernst potentials for given temperature.

pFH.El = (pFH.T./293.15).*pFH.El;

% Define Q10’s (from Rattay review, 1993).

pFH.Q10.malpha = 1.8.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.mbeta = 1.7.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.halpha = 2.8.ˆ((pFH.T-293.15)./10.0);


pFH.Q10.hbeta = 2.9.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.nalpha = 3.2.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.nbeta = 2.8.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.palpha = 3.0ˆ((pFH.T-293.15)./10.0);

pFH.Q10.pbeta = 3.0.ˆ((pFH.T-293.15)./10.0);

% Q10’s for PNa and PK

% from Frankenhaeuser and Moore:

% Q10.PNa = 1.3; Q10.PK = 1.2.

pFH.Q10.PNa = 1.3.ˆ((pFH.T-293.15)./10.0);

pFH.Q10.PK = 1.2.ˆ((pFH.T-293.15)./10.0);

% Q10 for leakage current: added by CRJ.

pFH.Q10.gbarl = 1.2.ˆ((pFH.T-293.15)./10.0);

% Initialize rate constant vectors, msec-1.

pFH.halpha = pFH.Q10.halpha.*0.1.* ...

(-10.0-vm(pFH.Rv))./ ...

(1.0-exp((vm(pFH.Rv)+10.0)./6.0));

pFH.hbeta = pFH.Q10.hbeta.*4.5./ ...

(1.0+exp((45.0-vm(pFH.Rv))./10.0));

pFH.malpha = pFH.Q10.malpha.*0.36.* ...

(vm(pFH.Rv)-22.0)./ ...

(1.0-exp((22.0-vm(pFH.Rv))./3.0));

pFH.mbeta = pFH.Q10.mbeta.*0.4.* ...

(13.0-vm(pFH.Rv))./ ...

(1.0-exp((vm(pFH.Rv)-13.0)./20.0));

pFH.nalpha = pFH.Q10.nalpha.*0.02.* ...

(vm(pFH.Rv)-35.0)./ ...

(1.0-exp((35.0-vm(pFH.Rv))./10.0));

pFH.nbeta = pFH.Q10.nbeta.*0.05.* ...

(10.0-vm(pFH.Rv))./ ...

(1.0-exp((vm(pFH.Rv)-10.0)./10.0));

pFH.palpha = pFH.Q10.palpha.*0.006.* ...

(vm(pFH.Rv)-40.0)./ ...

(1.0-exp((40.0-vm(pFH.Rv))./10.0));

pFH.pbeta = pFH.Q10.pbeta.*0.09.* ...

(-25.0-vm(pFH.Rv))./ ...

(1.0-exp((vm(pFH.Rv)+25.0)./20.0));

% Initialize gating variable vectors.

pFH.gates.m = pFH.malpha./(pFH.malpha+pFH.mbeta);

pFH.gates.h = pFH.halpha./(pFH.halpha+pFH.hbeta);

pFH.gates.n = pFH.nalpha./(pFH.nalpha+pFH.nbeta);

pFH.gates.p = pFH.palpha./(pFH.palpha+pFH.pbeta);

% Adjust pFH.Vrest to retain equilibrium.
















while (abs(Iion)>CONST.myzero)















pFH.Vrest = pFH.Vrest - ...

((UDATA.dt.*1.0e-3)./(mean(DATA.Ctot))).*Iion;

end

% Adjust El to get Iion=0 at rest.

pFH.El = (INa+IK+IP)./( pFH.Q10.gbarl.*pFH.gbarl);

% Initialize current vectors: include all nodes.

pFH.I.Na = zeros(size(vm));

pFH.I.K = zeros(size(vm));


pFH.I.P = zeros(size(vm));

pFH.I.l = zeros(size(vm));

pFH.I.myel = zeros(size(vm));

Answer 24C. Code to advance the values of the gating variables:

function [] = gatesFH(vmNew,dt);

% GATESFH Update FH rate constants and gating variables.

% GATESFH(vmNew,dt) updates the membrane channel

rate

% constants using the vector of updated nerve fiber

% transmembrane potentials vmNew, then finds the gating

% variables for the next iteration given the integration

% time step dt.

% -- PREAMBLE ------------------------------------------

global pFH;

% -- RATE CONSTANTS --------------------------

% Calculate rate constant vectors, in msec-1.

halphaNew = pFH.Q10.halpha.*0.1.* ...

(-10.0-vmNew(pFH.Rv))./ ...

(1.0-exp((vmNew(pFH.Rv)+10.0)./6.0));

hbetaNew = pFH.Q10.hbeta.*4.5./ ...

(1.0+exp((45.0-vmNew(pFH.Rv))./10.0));

malphaNew = pFH.Q10.malpha.*0.36.* ...

(vmNew(pFH.Rv)-22.0)./ ...

(1.0-exp((22.0-vmNew(pFH.Rv))./3.0));

mbetaNew = pFH.Q10.mbeta.*0.4.* ...

(13.0-vmNew(pFH.Rv))./ ...

(1.0-exp((vmNew(pFH.Rv)-13.0)./20.0));

nalphaNew = pFH.Q10.nalpha.*0.02.* ...

(vmNew(pFH.Rv)-35.0)./ ...

(1.0-exp((35.0-vmNew(pFH.Rv))./10.0));

nbetaNew = pFH.Q10.nbeta.*0.05.*...

(10.0-vmNew(pFH.Rv))./ ...

(1.0-exp((vmNew(pFH.Rv)-10.0)./10.0));

palphaNew = pFH.Q10.palpha.*0.006.* ...

(vmNew(pFH.Rv)-40.0)./ ...

(1.0-exp((40.0-vmNew(pFH.Rv))./10.0));

pbetaNew = pFH.Q10.pbeta.*0.09.* ...

(-25.0-vmNew(pFH.Rv))./ ...

(1.0-exp((vmNew(pFH.Rv)+25.0)./20.0));

% -- GATING VARIABLES -------------------------------


% Update gating variables’ vectors for next iteration.

pFH.gates.m = (pFH.gates.m + (dt./2.0).* ...

(pFH.malpha + malphaNew - pFH.gates.m.* ...

(pFH.malpha + pFH.mbeta)))./ ...

(1 + (dt./2.0).*(malphaNew + mbetaNew));

pFH.gates.h = (pFH.gates.h + (dt./2.0).* ...

(pFH.halpha + halphaNew - pFH.gates.h.* ...

(pFH.halpha + pFH.hbeta)))./ ...

(1 + (dt./2.0).*(halphaNew + hbetaNew));

pFH.gates.n = (pFH.gates.n + (dt./2.0).* ...

(pFH.nalpha + nalphaNew - pFH.gates.n.* ...

(pFH.nalpha + pFH.nbeta)))./ ...

(1 + (dt./2.0).*(nalphaNew + nbetaNew));

pFH.gates.p = (pFH.gates.p + (dt./2.0).* ...

(pFH.palpha + palphaNew - pFH.gates.p.* ...

(pFH.palpha + pFH.pbeta)))./ ...

(1 + (dt./2.0).*(palphaNew + pbetaNew));

% Store current rates for next iteration.

pFH.malpha = malphaNew;

pFH.mbeta = mbetaNew;

pFH.halpha = halphaNew;

pFH.hbeta = hbetaNew;

pFH.nalpha = nalphaNew;

pFH.nbeta = nbetaNew;

pFH.palpha = palphaNew;

pFH.pbeta = pbetaNew;

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