HOME >HOMEWORK #4 SOLUTION
1. . The absorption of a photon by a rod causes the voltage to decrease by about 2 mV, from -40 to -42 mV for 0.2 seconds. When the cell is at -40 mV, the average rate of release of synaptic vesicles is 1000 per 200 ms. The rate of release of synaptic vesicles decreases 10-fold for every 15 mV decrease in the cell's voltage (in other words, at -55 mV, 100 vesicles are released per 200 ms, at -70 mV, 10 vesicles are released, etc). Given that the actual number of released vesicles is Poisson distributed, would a 2 mV change in voltage produce a significant change in the rate of vesicle release?
(5 points): For a 10-fold reduction in vesicle release for a 15 mV decrease in voltage, the change in the rate of vesicle release with voltage is: Change in release rate = initial rate x (1-10-v/15) where v is the change in the cell's voltage in mV. For a 2 mV change, we get a 265 vesicle decrease (in 200 ms).
The # of vesicles released at a constant voltage is Poisson distributed, so in darkness (at -40 mV), the standard deviation about the mean is 10001/2 = 32 vesicles per 200 ms.
Since 265 vesicles >> 32 vesicles in a 200 ms time window, we will notice this change in voltage.
2. Now imagine a cone cell. The absorption of a single photon by a
cone produces a voltage change of about 0.02 mV for 0.1 seconds. If everything
else is the same as in the rod (i.e. average rate of vesicle release,
voltage dependence of vesicle release, etc., see problem #1), would a
single photon absorbed by a cone produce a significant change in the rate
of vesicle release from the cone?
(4 points): Now we have 500 vesicles released per 100 ms (same rate as rod), but a smaller voltage change in response to an absorbed photon. Plugging in 0.02 mV change gives us: 500 x (1-100.02/15) = 1.5 vesicles per 100 ms decrease in the rate of vesicle release. The standard deviation about the mean due to Poisson fluctuations in vesicle release of 5001/2 = 22. 1.5 << 22 vesicles per 100 ms, so we will not be able to detect a single photon absorbed by a cone.
3. The voltage and current in a bipolar cell is set by the rate of
synaptic vesicle release from the photoreceptors. Would you expect the
magnitude of noise in the bipolar cell (voltage or current fluctuations)
to be larger in complete darkness or bright light? Why?
(3 points): More vesicles are released in the dark than in the light.
Since the fluctuations in the rate of vesicle release scale with the square
root of the mean, the noise will increase as the voltage (and vesicle
release rate) becomes more positive. The vesicles all contain equal numbers
of neurotransmitter, so the noise in the bipolar cell will be larger in
complete darkness, when there is a higher rate of vesicle release and
of fluctuations in vesicle release.
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