Na out in Passive cable The voltage gated sodium channel on the left is open and
ID: 3479225 • Letter: N
Question
Na out in Passive cable The voltage gated sodium channel on the left is open and generating inward, depolarizing current. The sodium channel on the right is closed. In order for the channel on the right to open, the membrane potential at that site must depolarize to the threshold for initiating the Hodgkin-Huxley active response. 5a. Ust the S factors that determine the time it takes to reach threshold at the channel on the right. Answer 5b. Write a short essay that explains the role of each factor in determining the rate of impulse conduction along the axon. Include in your explanation definitions for the time and space constants for th network. Answer: 5c. Pick one of the factors and indicate how it must change to slow conduction along the axon-explain the mechanism. AnswerExplanation / Answer
Hi,
This is a model of the neuron action potential based on the cable theory by Lord Kelvin in 1855.
This model tries to elaborate on the action potential and passage of current along the neurons. The impulse travels across the nerve creating a series of depolarising and repolarising events.
5a. This travel is dependent on various factors such as:
1. Lenght of the axon
2. Diameter of the neuron
3. Voltage across the membrane
4. ri intracellular resistence
5. ro outside resistence
6. rm membrane resistence
7. cm capacitance due to electrostatic forces
5b. The time and length of the axon movement changes all other factors. Increase in length of the movement increases with both rm and cm. As the capacitance and resistence increase, charge transfered reduces per unit time and turns the trasnfer slower. It means more time for right channel to depolarise. Also if the ri is lower in one axon than the other, then the conduction velocity increases, channel on the right opens quickly. If the rm increases , it lowers the average leak cross the membrane and increases the conduction velocity.
5c. time = rm x cm
length = square root of rm/rl
The time-scale increases with both the membrane resistance rm and capacitance cm. As the capacitance increases, more charge must be transferred to produce a given transmembrane voltage (by the equation Q = CV); as the resistance increases, less charge is transferred per unit time, making the equilibration slower.
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