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HOME >HOMEWORK #6 SOLUTION

1.  Recall the calcium current in the photoreceptor that controls the synaptic vesicle release.  Assume that the calcium current is small compared to other currents in the cell and does NOT affect the membrane voltage.  Shown below is the current-voltage relationship for an OPEN calcium channel in the photoreceptor:

What is shown in the plot is merely for OPEN channels.  The actual current-voltage relationship is due to the fact that the probability of a given sodium channel being open increases as voltage increases, and goes to zero at something like –70 mV.    (The overall current is suppressed for more negative voltage because more of the channels are closed.)

Negative currents represent current (i.e. Ca+ ions moving) INTO the cell, ground is outside the cell.  Explain what the calcium channels must do in order to decrease the magnitude of the calcium current as the voltage becomes more negative (i.e. explain qualitatively the voltage dependence of the Ca+ channels).  If you wish, you may draw a plot showing the open probability vs. voltage for your Ca+ channel.

3A.  NOISE:  An ion channel is either closed where it has conductance g = 0, or open where it has conductance g = 1 (units arbitrary).  As discussed in class, K+ channels are constantly switching between closed and open.  If we look over a time window that is long compared to the time scale of channel openings and closings, we can calculate the mean conductance x and the variance (conductance noise) V2.

The mean conductance x has a value of 1 in the extreme case that the channel is always open, and 0 in the case that the channel is always closed; in both these cases the variance (noise, or amount the conductance is flickering) V2 = 0.  When the channel is open half the time, x = 0.5, and the variance V2 = .25.

The explicit dependence of the variance (noise) on x (fraction of time channel spends open) is: V2 = x - x2.  A plot of the standard deviation, V, is therefore a semicircle of radius 1/2 centered at x = .5.

Give a hand-wavy explanation for why the variance depends on the mean the way it does.  Explain why you would expect no noise (due to the channel) if the channel is always open or always closed, and why the noise is at a maximum when the channel is open half the time.

Hand-Wavy:  If the mean is 0 or 1, we are given that the variance is 0. This is because the channel is always either open or closed, and there is no flicker in the current magnitude. The other extreme case is when the mean = 0.5. The variance is greatest when the channel spends half the time open because the channel is making the greatest # of transitions between open closed.

Or, if you can, show explicitly that V2 = x - x2 (you must if you're a senior or graduate student in physics).

3B.  The ion channels in the photoreceptors are open less than 10% of the time even in complete darkness.  In light, they open even more rarely.  When these channels are flickering between open & closed, they give rise to current fluctuations (and hence noise).  Does keeping the open probability low at all times help reduce the effects of noise in the photoreceptor?

Yes, it does.  Because of the semicircular shape, the farther from 0.5 you are, the lower your variance is.

Would this noise (noise of the ion channels) be larger or smaller in bright light compared to darkness?

In the light, (nearly) all channels are closed, so there will be less noise than in the dark

4.  Consider a Ca2+ ion in solution.  It has a greater atomic mass than Na+ or K+, twice the electric charge, and a Pauling radius between that of Na+ and K+.  How should the ionic mobility of the Ca2+ ion compare with Na+ and with K+?  Why?

According to the equipartition theory, we would expect all of them to have the same kinetic energy, 3/2 kT.  Since this is equal to 1/2 m v2, the more massive the object is, the slower it would move.  But, this completely ignores the effects of hydration.  We know that the degree of hydration (amount of water molecules associated/attached to an ion) increases with [1] decreased radius (accessibility issues) and [2] higher charge (ionic attraction).  So, the calcium ion will definitely be slower than the potassium ion because of calcium’s smaller radius and higher charge.  From what we know, it is ambiguous as to which wins out with respect to sodium, which has a smaller radius than calcium, but less charge.  (In fact, the charge wins out, and calcium diffuses more slowly in water than does sodium)

5.  Imagine a cell with a high K+ concentration inside (150 mM) and low K+ concentration outside (2 mM).  The Na+ concentration inside is 5 mM, and is 150 mM outside.  Normally, there are a few K+ channels open, and the cell is at -65 mV or so.  If the cell membrane becomes more positive, an action potential is generated.  Based on diffusion, electricity, and the properties of ion channels, explain as much as you can about the action potential (shown below).

Initial depolarization causes sodium channel opening [1], and sodium influx causing a rise in potential towards the sodium equilibrium potential [2].  However, the sodium channels open only transiently.  The delayed opening of potassium channels [3] allow potassium to diffuse back out of the cell [4], restoring equilibrium potential.

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