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4. During the propagation of an action potential, there is a transient reversal of the local membrane potential from ~-70 mV to ~+50 mV, a process called depolarization. This depolarization is due to the stimulated inward movement of Nations into the cell through Na+ channels. The depolarization in turn provides a voltage signal that quickly (within milliseconds) activates local K+ ion channels to permit the flow of K+ ions back out of the cell during a process called repolarization. On the plot below, use a solid line to draw the change in membrane potential observed during an action potential and label the depolarization and repolarization phases. Membrane potential (mv) + 60 + 40 answer. + 20 0 - 20 - 40 60 - 80 Time (ms) Since depolarization during the action potential is due to the movement of Na* ions into the cell, this means that the A4 for Nat is transiently changing from the value you calculated in question #1a. Specifically, A calculated from the ratio of [Na*]outside/[Na] inside, will now become negative when using the Nernst equation. If we assume A4 for Na* to be -50 mV at the peak of the action potential, calculate the ratio of [Na]outside/[Na] inside under this condition. Show your work below and compare this ratio to the [Na]outside/[Na+Jinside ratio in the resting state. You can also use drawings to explain your

4. During the propagation of an action potential, there is a transient reversal of the local membrane potential from ~-70 mV to ~+50 mV, a process called depolarization. This depolarization is due to the stimulated inward movement of Nations into the cell through Na+ channels. The depolarization in turn provides a voltage signal that quickly (within milliseconds) activates local K+ ion channels to permit the flow of K+ ions back out of the cell during a process called repolarization. On the plot below, use a solid line to draw the change in membrane potential observed during an action potential and label the depolarization and repolarization phases. Membrane potential (mv) + 60 + 40 answer. + 20 0 - 20 - 40 60 - 80 Time (ms) Since depolarization during the action potential is due to the movement of Na* ions into the cell, this means that the A4 for Nat is transiently changing from the value you calculated in question #1a. Specifically, A calculated from the ratio of [Na*]outside/[Na] inside, will now become negative when using the Nernst equation. If we assume A4 for Na* to be -50 mV at the peak of the action potential, calculate the ratio of [Na]outside/[Na] inside under this condition. Show your work below and compare this ratio to the [Na]outside/[Na+Jinside ratio in the resting state. You can also use drawings to explain your

Biology: The Dynamic Science (MindTap Course List)
Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN:9781305389892
Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan
Chapter39: Information Flow And The Neuron
Section: Chapter Questions
Problem 5TYK: The major role of the Na+/K+ pump is to: a. cause a rapid firing of the action potential so the...
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4. During the propagation of an action potential, there is a transient reversal of the local membrane potential from ~-70 mV to ~+50 mV, a process called depolarization. This depolarization is due to the stimulated inward movement of Nations into the cell through Na+ channels. The depolarization in turn provides a voltage signal that quickly (within milliseconds) activates local K+ ion channels to permit the flow of K+ ions back out of the cell during a process called repolarization. On the plot below, use a solid line to draw the change in membrane potential observed during an action potential and label the depolarization and repolarization phases. Membrane potential (mv) + 60 +40 + 20 0 - 20 - 40 answer. - 60 - 80 Time (ms) Since depolarization during the action potential is due to the movement of Na* ions into the cell, this means that the Au for Nat is transiently changing from the value you calculated in question #1a. Specifically, A calculated from the ratio of [Na*]outside/[Na+] inside, will now become negative when using the Nernst equation. If we assume Au for Nat to be -50 mV at the peak of the action potential, calculate the ratio of [Na]outside/[Na*]inside under this condition. Show your work below and compare this ratio to the [Na]outside/[Na+Jinside ratio in the resting state. You can also use drawings to explain your
Transcribed Image Text:4. During the propagation of an action potential, there is a transient reversal of the local membrane potential from ~-70 mV to ~+50 mV, a process called depolarization. This depolarization is due to the stimulated inward movement of Nations into the cell through Na+ channels. The depolarization in turn provides a voltage signal that quickly (within milliseconds) activates local K+ ion channels to permit the flow of K+ ions back out of the cell during a process called repolarization. On the plot below, use a solid line to draw the change in membrane potential observed during an action potential and label the depolarization and repolarization phases. Membrane potential (mv) + 60 +40 + 20 0 - 20 - 40 answer. - 60 - 80 Time (ms) Since depolarization during the action potential is due to the movement of Na* ions into the cell, this means that the Au for Nat is transiently changing from the value you calculated in question #1a. Specifically, A calculated from the ratio of [Na*]outside/[Na+] inside, will now become negative when using the Nernst equation. If we assume Au for Nat to be -50 mV at the peak of the action potential, calculate the ratio of [Na]outside/[Na*]inside under this condition. Show your work below and compare this ratio to the [Na]outside/[Na+Jinside ratio in the resting state. You can also use drawings to explain your
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