HOME >HOMEWORK #5 SOLUTION

1. Recall from lecture #4 that cells have a capacitance due to their outer membrane being a good insulator.  The specific capacitance of cellular membranes is 1 uF/cm^2.  Consider the case of a spherical cell with a capacitance of 20 pF with an internal K+ and Cl- concentration of 140 mM (millimolar, or millimoles per liter) and an external concentration of 2 mM K+ and Cl-.  The cell's membrane is permeable to K+ only.  Initially (time = 0) the cell is at 0 mV (same voltage inside & outside of the cell).  However, diffusion of K+ across the cell membrane causes the cell very quickly to become negative (because K+ diffuses out of the cell). 

A)     What is the steady state voltage of the cell?

Use the Nernst Equation:

B)     What is the net number of K+ ions that moved out of the cell? 

C)    How much has the K+ concentration changed inside of the cell?

First we must find out the volume of the cell.  We can assume its spherical and find the surface area which we can find the volume from.

Compared with the original concentration of the cell, we see that 1/51851 of the K+ concentration is lost.  This is negligible.

2. Boltzmann's constant, k = 1.38 e-23 Joules/degree Kelvin.  What is kT in SI units at room temperature (T=300 K)?  What is kT in terms of milli-electron volts (energy required to move an electron through a voltage change of 1 millivolt)? 

I'm asking this question not to have you do a trivial unit conversion, but because this number appears EVERYWHERE you're dealing with electronics—the Nernst equation, the voltage dependence of ion channels and semiconductor junctions, Johnson noise, etc.  Why?  Because this is the order of magnitude of the thermal energy of charge carriers in any medium, and if we're in the realm of electronics, we want that energy in terms of volts (times charge).

3. A typical photoreceptor has synapses with many cells.  Retinal cells have a capacitance of about 20 pF.  Suppose that rather than having chemical synapses, photoreceptors were directly electrically coupled to other cells--i.e. all cells change their voltage together. 

This scheme has many problems.  First is that there's no way to make an inverter (an ON bipolar cell).  Another problem comes from the capacitance.  Recall that a single photon produces a voltage change of about 2 mV in 200 ms.  Suppose a rod is coupled to 50 other cells.  How much current must flow through to change the voltage by 2 mV in 200 ms (assume current is constant in time)? 

There are 50 capacative junctions in parallel, so:

This requires a current of:

How does this compare with the magnitude of the single photon current change in the rod?

It’s larger than a single photon current, so it would potentially mask one

4. In honor of Erika's ethanol research, let's look at the (hypothetical) effects of high ethanol concentration on vision.  Turns out thatethanol acts on neurotransmitter receptors:  specifically, it reduces the current that flows in response to glutamate.

Imagine you were monitoring the current flowing into an ON and OFF bipolar cell.  Now imagine that we dump ethanol onto thecells.  Suppose this causes the current flowing through ionotropic receptors in bipolar cells to be reduced.  What effect would this

have on the magnitude of current flowing into the two types of bipolar cells in total darkness?  In bright (saturating) light?   Would the single-photon response be affected?  How?

ON bipolars- don’t have ionotropic receptors, so they aren’t affected.

OFF bipolars- In the light, all current is suppressed, so alcohol won’t affect it.  In the dark, current will be suppressed.  A single photon response will have decreased amplitude and may not be discernable from noise

5. Suppose I build a cell that has a membrane permeable only to sodium ions (Na+).  The Na+ channels (that let Na pass through the membrane) are always open.  I can control the voltage EM of the cell & measure the current IM flowing across the membrane. 

What is the relative ratio of sodium ions outside vs. inside of the cell? 

Observe from the plot that equilibrium is at 50mV, then use the Nernst equation:

Also, give a hand-wavy explanation for why the slopes of the current-voltage relation above are different for extremely positive and negative potentials.  Hint:  think about diffusion. 

The slope is different for extremely positive and negative potentials because negative potential has the electrochemical gradient working with the concentration gradient, and positive potential has the electrochemical gradient working against the concentration gradient.

What do you think the above graph should look like if the internal & external sodium were equal?

It would be a straight line of positive slope passing through the origin.