Neuroscience 26 Exam 1 Suggested brief answersSpring 2007
1.a. Golgi method - way of staining a neuron’s soma and dendrites completely; it also stains only a small fraction of the neurons, so each stained neuron’s structure can be seen clearly.
b. Cell birthday - time of last cell division before a neuron becomes post-mitotic.
c. Myelin - membranous wrapping of axons, speeds impulse conduction
d. Filopodia - protrusions at tip of a growing axon, shows amoeboid-like movement
e. IPSP - a change in membrane potential of a postsynaptic neuron, usually hyperpolarizing, makes it more difficult for the neuron to reach threshold for firing an impulse.
2. (1) Peptides are synthesized in the soma and transported down the axon to the synapse; small-molecule transmitters are synthesized in axon terminals by enzymes which had been transported down the axon.
(2) Cell bodies are on the inside, surrounded by white matter, in spinal cord; cell bodies are on the outside, surrounding white matter, in the telencephalon.
(3) Raising external [K+] depolarizes neurons, i.e. makes the resting potential less negative. Raising external [Na +] has negligible effect on the resting potential, but would be expected to increase the amplitude of action potentials, i.e. make the peak more positive.
3. (1) The NT at excitatory synapses does depolarize the dendrite, but that depolarization is an EPSP, not an impulse.
(2) Indeed there is no current or change in membrane potential at a synapse where Erev=Vrest, but that doesn’t mean the synapse has no effect - it stabilizes the membrane near Vrest when other excitatory synapses are active, and thus inhibits the neuron from firing an impulse.
(3) No, those are not the same - activation and inactivation are separate, largely independent processes; a Na channel can be inactivated and also activated at the same time.
4. (a)EHCO3- = (62mV/(-1)) x log10 (20 mM/10 mM) = (-62 mV) * 0.3 = -19 mV.
(b) A synapse that caused an increase in postsynaptic membrane permeability to bicarbonate would have a reversal potential equal to EHCO3- , i.e. -19 mV. This level is much less negative than typical threshold voltages for initiating an impulse, so this synapse would be excitatory. [Note: Many students began by reasoning incorrectly that bicarbonate would enter the neuron if the membrane became permeable to it. However, which direction bicarbonate will flow also very much depends on the membrane potential. When the cell is at typical resting potentials around -65 mV, the strong inside negativity will repel negatively-charged HCO3- and it will flow out, against its concentration gradient. That’s the meaning of the equilibrium potential for bicarbonate - it will flow out at any membrane potential more negative than -19 mV, and it will flow in at any membrane potential less negative than -19 mV.]
5. (A) The two brief stimuli allow the membrane to reach resting potential and even hyperpolarize during the refractory period. This is partly caused by Na inactivation, but during this time inactivation is reduced to its normal value at the resting potential. When a constant depolarizing stimulus is applied, the membrane never gets to return to resting potential, let alone hyperpolarize, so Na inactivation isn’t reduced, so Na channels never open as much during the second mini-impulse. K channels are also kept more open in the constant-stimulus case, which also reduces the amplitude of the 2nd impulse.
(B) The impulses will die out at the collision point, since the membrane on each side of the collision point is refractory (i.e. high Na inactivation, high K permeability).
6. Competition for space in the target tissue and competition for trophic factor released by the target tissue both affect cell death. In C, there is vacant space where inputs from the removed optic nerve would innervate, giving ganglion cells from the intact eye more chance to find a target and therefore survive. In D, lack of any electrical activity might lead to more trophic factor being released by the target tissue innervated by the TTX retina, since that tissue would be “seeking” input, and thus there would be a higher rate of cell survival in both eyes. In E, again the target tissue expecting impulses from the TTX-injected eye might release more trophic factor, leading to more cell survival in the non-TTX eye.
[Problem set was worth 2 points added to the exam score if largely correct; 1 point if handed in but with mistakes on at least 3 questions.]