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Homework #3 for Yue section of Systems Bioengineering I, Fall 2014, page 1 of 8

Homework 3: More gating and action-potential generation(100 points in total)

1. (80 points total) ION-CHANNEL GATING. As a star engineer at Frankenstein Inc, you have

been asked to build the first artificial spiking cell. Inspired by your lectures on the Hodgkin-

Huxley model, you decided to use a voltage-gated Na channel (with 3 mactivation gates and1 h inactivation gate) in your cell. Your Na channel has a conductance (gNa) of 32 pS andyou inserted precisely 500 channels (hand-counted and in the correct orientation). The

channel was designed to have a steady-state inactivation curve (h(), black); and steady-

state activation (m3(), blue) as shown in Figure 1.

Figure 1. Steady state probabilities of activation (m3()) and inactivation (h())

For your first experiment, you verified that your Na channel was functional by recording thewhole-cell currents after insertion of only the Na channels. The extracellular and intracellular

solutions you used were as follows:

Table 1. Intracellular and extracellular solutions

Extracellular Intracellular

150 mM NaCl

4 mM KCl

1 mM MgCl20.5 mM CaCl2

10 mM pH buffer

(pH adjusted to 7.4)

15 mM NaCl

140 mM KCl

1.5 mM MgATP10 mM pH buffer

(pH adjusted to 7.4)

probability

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Your first recordings are shown below. In the following subsections, carefully deduce the

gating parameters at Vstep= -10 mV, m, m, h, h. Then, predict and plot the response of

these channels to an entirely new voltage protocol.

Figure 2. Whole-cell Na current from your first artificial cell.

A. (5 pts) Derive the I* current-voltage relationship for the sodium current, assuming that all

channels are open and not inactivated (m = 1, h = 1) and given the experimental conditions

described above. For simplicity, you may assume a battery-resistor model of permeation.

Vreversal= 25*ln(150/15) = 57.6 mV

I* =Nchannel*gNa*(V-Vreversal)

= 500 * 0.032 pA/mV * (V57.6)= 16*(V57.6)

-100 mV

-10

-100

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1B. (15 pts) Using the relation from 1A, convert the trace into a time waveform for the

probability of being open. Plot the waveform carefully, presumably using computer software like

excel or MATLAB.

I* ( -10 mV) = -1081.6 pA; I*(-100 mV) = -2521.6 pA;

Obtained by dividing the whole-cell current (Figure 2) byI*(-10 mV) between 0 and 8 msec and

I*(-100 mV) for t < 0 msec and t > 8 msec

1C. (20 pts) Fit a time waveform proportional to the timecourse of h(probability that hgate is

enabled) for the time period between 0 and 8 msec. Reproduce the open probability trace from1B and show the fit in red, give the parameters and form of the model below, and interpret these

parameters to deduce the values of h and h at V= -10 mV. For t< 0 ms, Assume that thevoltage was held at -100 mV for a long period (long enough to allow m and h gates to reach

steady-state)The exact steady-state values can be obtained from excel spreadsheet.

h(t) = h(, -10mV) + (h(, -100mV) - h(, -10mV))*exp(-t/ where h+h).

h(t) = 0.0074 + (1 - 0.0074)*exp(-t/

h(, -10mV) = hh+h) ~ 0.0074.

Moreover, h(t) = 0.0074 + (1 - 0.0074)*exp(-t/ is a good fit,as shown in red

Thus h = 0.0074/1.1 = 0.0067 ms-1

and h+ h = 1/1.1 => h~ 0.9 ms-1

.

(Full points even if h was deduced to be ~ 0 and hwas approximated to 0.9).

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8

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0

0.5

1

0 2 4 6 8

1D. (20 pts) Use the fit in part B to determine the time waveform for m(probability of mgates

being enabled) between 0 and 8 msec. Deduce the values for mand m

Since the overall probability waveform (part 1B) is given by m3* h, the timecourse of m

3could

be obtained by dividing through the overall probability waveform by the timecourse of h as

determined in part C. This procedure yields the blue line shown in figure. To get m, we take thethird root of the blue line, yielding the thick gray line.

m(t) = m(, -10mV) + (m(, -100mV) - m(, -10mV))*exp(-t/m where m m+m).But from excel sheet, m(, -100mV) = 0.

m(, -10mV) = (.94)^(1/3) = 0.98 = mm+m)So we fit the gray waveform with the function

m(t) = 0.98*(1-exp(-t/0.265) (red fit)

Hence, m+ m= 1/0.265 = 3.76 ms-1

m/ (m+ m) = 0.98.

Solving two equations and two unknowns, m= 0.98*3.76 = 3.7 ms-1

andm= 0.07 ms-1

.

1E. (10 pts) Having deduced the parameters above in parts A through C, see if you can predictthe behavior of the same cell for a different voltage protocol. Here, assume the channel has

reached steady state with respect to a holding potential of +100 mV. Now plot the Na current

between 0 and 8 ms, when the voltage is returned to -10 mV.If we hold at +100 mV for a long time the Na channels will inactivate (h(,+100mV) = 0. When

the voltage is returned to -10 mV, the Na channels will still be inactivated. So the whole-cell

current will be ~ 0 as shown in figure.

But if you wanted to be more precise,m(t) = m(,-10mV) + (m(,+100mV) - m(,-10mV))*exp(-t/m)m(t) = 0.98 + (0.02)*exp(-t/0.265).

h(t) = h(,-10mV) + (h(,+100mV) - h(,-10mV))*exp(-t/h)h(t) = 0.00740.0074*exp(-t/1.1).

I(t) = m(t)^3*h(t)*gNa*Nchannels*(V-Vreversal).

This would result in a very small current (steady-state ~ 8 pA).

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1F. (10 pts) Though satisfied with the success of your initial experiments, you tried to repeat

them again as any good scientist would. However, this time around, your fellow engineers had

poor quality control resulting in 20% of your Na channels (i.e. 100/500) being defective. Thesedefective channels lacked an inactivation gate (h) but their activation gate (m) was normal. Plot

the whole cell currents in response to the voltage waveform shown below. Assume your

recording solutions are the same as before.

INa= 0.8*Nchannels*gNa*m3(t)*h(t) (V - Vreversal) +0.2*Nchannels*gNa*m

3(t)(V - Vreversal)

where m(t) and h(t) were determined in 1C and 1D.

m(t) = m(,-10mV) + (m(,-100mV) - m(,-10mV))*exp(-t/m)m(t) = 0.98 * (1exp(-t/0.265))

h(t) = h(,-10mV) + (h(,-100mV) - h(,-10mV))*exp(-t/h)

h(t) = h(t) = 0.0074 + (1 - 0.0074)*exp(-t/

So then,INa= 12.8*m

3(t)*h(t) (V57.6) +3.2*m

3(t)(V57.6)

-100 mV

-10

-100

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2. (20 points total) ACTION-POTENTIAL GENESIS

The defective Na channels in your artificial cell were a premonition, but it takes a lot more than amere 20% non-inactivating (persistent) Na channels to deter you from achieving your goal for

creating a spiking artificial cell. So you inserted 50 potassium (K) channels (always open) with a

conductance of 40 pS into your cell. The intracellular and extracellular solutions are as describedin Table 1 (page 1). You have an electrode in the cell to record and manipulate voltage.

2A. (2 pts) Precisely write the equation for the whole cell K current in your artificial cell. You

may assume a battery-resistor model for permeation.

IK=Nchannel,K*gK*(V-Vreversal,K)

Vreversal,K= 25mV*ln(4/140) = -88.9 mV

IK= 50*0.04 pA/mV*(V+88.9)

2B. (5 pts) On the graph below, carefully plot the quasi-instantaneous current-voltage relations

for IK, INa and IK + INa before the Na channel inactivation sets in (h = 1) with quantitativeprecision. Mark all fixed points of the system and label them as stable or unstable. What is the

voltage threshold for activation (Vth)? If you depolarize the cell to -20 mV by a stimulus what

voltage will the system stabilize at?

You may assume that Na channel activation (m) gates equilibrate with changes in voltage faster

than the inactivation gate (h). Also remember that 20% of the Na channels lack an hgate but

have normal mgates.INa= 0.8*Nchannel,Na*gNa*m

3(,V)*h(V)*(V-Vreversal) + 0.2*Nchannel,Na*gNa*m

3(,V)* (V-Vreversal)

But h= 1. So,INa= 16*m3(,V)*(V-57.6). Red =IK, black =INa, blue =INa+IK. Vth~ -35 mV,

so if you depolarize to -20 mV, the voltage will stabilize near the stable fixed point at +40mV.

INa

IK

INa+IK

Stable fixed point

Unstable fixed point

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2C. (5 pts) Now suppose you let the system evolve after the stimulus and Na channels with an

inactivation gate (h) are fully inactivated (h= 0). On the graph below, carefully plot the quasi-

instantaneous current-voltage relations for IK, INaandIK+INawith quantitative precision. Markall fixed points of the system and label them as stable or unstable. What voltage does the system

stabilize at?

INa= 0.2*Nchannel,Na*gNa*m3(,V)* (V-Vreversal)

= 3.2 * m3(,V)* (V-57.6)

Here the non-inactivating fraction of Na current is sufficiently large that there are still 3 fixedpoints, so the system is not able to relax to the resting potential (stable point at -90 mV). Thus

the system would stabilize at ~ 0 mV.

INa

IKINa+IK

Stable fixed point

Unstable fixed point

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2D. (4 pts) Following part 2C, you realized that your artificial cell is stuck midstream in a

perpetual action potential. Devise a strategy to allow the cell to repolarize, using methods from

which the company Medtronic would profit. You are not allowed to change the channelcomposition in your cells or the bath solutions.

To repolarize the cell, you simply need withdraw current to bring the voltage to be just below theunstable fixed point at ~ -25 mV. Now the system can repolarize back to the resting potential(stable fixed point at -90 mV).

2E. (4 pts) Assuming 20% of your Na channels continue to be defective (lack h gate) as aconsequence of the manufacturing process, how would you redesign your system to allow it to

spike reliably.

One way would be to increase the number of K channels (in this case to 70 channels). Now

again, the K current is large enough to overwhelm the non-inactivating Na current resulting in a

single stable fixed point at -90 mV as shown below.

You could also decrease the number of Na channels to about 350, then again the system will

evolve to have only a single stable fixed point at -90 mV.

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