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
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
Problem 1P
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Solve 4
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd6957468-6fb1-4659-9083-987b06676d6d%2Ff1e80fdc-ddf7-4685-8581-40c0e49d724c%2Ftwplqh_processed.jpeg&w=3840&q=75)
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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