Answers to Homework 1

15-883 Computational Models of Neural Systems

  1. How closely must inhibition follow excitation to prevent a spike from occurring?

    Answer: For some people, inhibition applied within 1 ms after excitation was enough to block the spike. Other people, with faster computers, were unable to block the spike by hitting S2 after S1. In that case, adjustment of the stimulus parameters is necessary to demonstrate this effect.

  2. Click on S1 once, then again. Why doesn't the neuron spike twice?

    Answer: The neuron has entered its refractory period. The membrane is hyperpolarized for a brief while. In addition, the sodium channels are temporarily inactivated and require time to recover.

  3. What happens after 3 ms of hyperpolarization via S2?

    Answer: This is a demonstration of post-inhibitory rebound. Normally a certain percentage of sodium channels are inactivated; a fraction of the remainder switch stochastically from the closed to the open state. Hyperpolarization undoes the inactivation, so that a large percentage of channels are de-inactivated -- but still closed. When the hyperpolarization is removed, the number of channels that spontaneously switch from the closed to the open state is enough to trigger a spike.

  4. Turn off the fast (voltage-gated) sodium channel. What happens?

    Answer: The cell cannot spike because there are no voltage-gated sodium channels available to depolarize the membrane.

  5. Turn off the delayed rectifier (voltage-gated potassium channel). What happens?

    Answer: Eliminating the voltage-gated potassium channels raises the membrane voltage (because of fraction of these channels are open even when the cell is at rest) and causes the cell to spike. However, without these channels there is nothing to repolarize the membrane, so when the sodium channels inactivate the cell finds a new resting potential close to 0 mv. This is sufficiently positive that the sodium channels' inactivation gates remain closed and the sodium channels do not recover, so further spiking is impossible.

  6. Reduce the external sodium concentration to 100 mM. What is the spike amplitude? Change the external sodium concentration to 550 mM. How does the amplitude change?

    Answer: An external sodium concentration of 100 mM reduces the spike amplitude to around 10 mV. Increasing the concentration to 550 mM raises the spike amplitude to nearly 50 mV. The change in concentration alters the concentration gradient between the inside and the outside of the cell, which in turn alters the equilibrium potential ENa.


Dave Touretzky
Last modified: Mon Jan 22 02:48:49 EST 2001