Neuronal membranes as nature's capacitors
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Neuronal membranes as nature's capacitors
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[post_date] => 2025-01-09 08:15:15
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[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Neurons are specialized to facilitate communication from one part of the body to another, and these signals of communication are achieved through electrical potentials. The cell membrane of a neuron creates an ionic potential by separating various ions across its cell membrane. Any construct that can separate an electric charge can be considered a capacitor, even the lipid bilayer membrane of a neuron! Capacitance (C, measured in Farads) is defined formulaically as the amount of stored charge divided by voltage, as seen in the formula C= Q/V.
Various ion pumps and channels are utilized by neurons to maintain a transmembrane ionic gradient, which produces an electric potential energy (Voltage). The pumps and channels themselves act an electrical conductor, much like a wire in a circuit. Therefore, we can describe the electrical properties of a neuron membrane in the physical terms of capacitors and resistors, and accordingly calculate the voltage and resistance of this membrane.
An action potential is the means by which a neuron can send a signal elsewhere throughout the body. By allowing the transmembrane voltage to increase through ion channels, an electric current can form, which then propagates down the axon of a neuron. As shown in figure 1, a neuron will normally have a resting potential around $-70$ mV, because the inside of a cell is more negative than the outside. When the transmembrane voltage becomes more negative than the resting potential, it is said to hyperpolarize, while when it becomes more positive than the resting potential, it is said to depolarize. Over time, the ionic potential will naturally depolarize. Therefore the negative electric potential must be maintained by Na+/K+ pumps, which exchanges $3 $ Na+ ions out of the cell for every $2$ K+ ions in.
To initiate an action potential a stimulus is required to reach the threshold level of a voltage-gated ion channel. This stimulus opens the channel, allowing more Na+ to enter the cell to propagate the action potential. A careful interplay of Na+ entering and K+ leaving the cell creates depolarization and repolarization, respectively.
Figure 1. Graph of voltage changes during an action potential
Adapted from wikipedia user Chris 73, CC-BY-SA 3.0
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[question] => Ouabain is a toxin which blocks the action of the Na+/K+ pump. Given enough time, what effect will ouabain have on a neuron’s membrane potential?
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[answer] => 2
[description] => Reason for the Correct Answer:
If ouabain is blocking the Na+/K+ pump, then the electric potential of -70 mV cannot be maintained.
Recall that over time, the ionic potential will naturally tends towards depolarization.
If the transmembrane ionic gradient is not maintained by the Na+/K+ pump, then the ionic potential will be slowly lost, resulting in 0 V.
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[answers] => Array
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[0] => Array
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[each_answer] => A. It will result in hyperpolarization, resulting in a V of -60mV
)
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[each_answer] => B. It will result in depolarization, resulting in a V of 0 mV
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[each_answer] => C. It will result in depolarization, resulting in a V of +65 mV
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[each_answer] => D. It will result in hyperpolarization, resulting in a V of -100 mV
)
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[quiz_unique_key] => 3873426850
[question] => During a specific portion of an action potential, Na+ channels are inactivated, and Na+ can not enter the cell. Which portion of the graph would indicate the start of Na+ channel inactivation?
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[answer] => 4
[description] => Reason for the Correct Answer:
Na+ entering the cell would cause depolarization.
The repolarization of a cell is caused by K+ leaving the cell.
The peak of the action potential signifies the closing and inactivation of Na+ channels, because no more Na+ is entering the cell, and therefore it no longer is depolarizing.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Repolarization curve
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[1] => Array
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[each_answer] => B. Refractory period
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[each_answer] => C. Depolarization curve
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[each_answer] => D. Peak of the action potential
)
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => Because membranes act as capacitors, ion channels can be modeled as a parallel circuit. Using the properties of the membrane given, what would be the capacitance across the membrane?
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[answer] => 1
[description] => Reason for the Correct Answer:
Because cell membranes can be represented as a parallel RC circuit, the potential difference across the capacitor is the same as the potential difference across the resistor.
You can use Ohms law and the equation for capacitance (C=Q/V) to calculate the capacitance of the membrane.
Ohm’s law is V = I * R so:
Once we know the voltage across the membrane, we can input this into the capacitance equation and solve:
However, remember that capacitance must be a positive value.! We used -10V to solve for C, but we could have just as easily used 10V, as the direction of current doesn’t matter for determining capacitance. Thus, the final answer should be:
C = 0.3 pF (picoFarads)
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[0] => Array
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[each_answer] => A. 
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[1] => Array
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[each_answer] => B. 
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[2] => Array
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[each_answer] => C. 
)
[3] => Array
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[each_answer] => D. 
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[quiz_unique_key] => 2261298308
[question] => A researcher measures a 10mV decrease in transmembrane voltage of a neuron over 2 milliseconds. The capacitance of the section of membrane studied is known to be 1 μF (1×10-16 F). How much current is flowing across the membrane?
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[description] => Reason for the Correct Answer:
Use the formulas for current 
and 
capacitance tp solve for I.
Since capacitance of a membrane is constant, voltage and charge will undergo a proportional change over time.
We know that the capacitance is 1 x 10-6 F, and the voltage is 10 mV so:
Knowing that 
C means we can solve for current:
)
[answers] => Array
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[0] => Array
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[each_answer] => A. -5 x 10-6 Amps
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[1] => Array
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[each_answer] => B. -5 x 10-5 Amps
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[2] => Array
(
[each_answer] => C. 5 x 10-5 Amps
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[each_answer] => D. 5 x 10-6 Amps
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[quiz_unique_key] => 2377279144
[question] => If no other ions are involved, how would decreasing the concentration of sodium outside the cell affect the graph of action potential?
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[answer] => 3
[description] => Reason for the Correct Answer:
Sodium is responsible for the depolarization phase of an action potential.
If the concentration of sodium is lower outside the cell, then the overall voltage will be decreased, because there is less of a difference in charge being separated across the membrane.
If the concentration of sodium outside the cell is lower than it was previously, and no other ions are involved then the current across the membrane will decrease, because this means the amount of charge is also less. A decrease in current means the action potential will take a longer time.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. The action potential would peak at a higher voltage, over a longer time
)
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[each_answer] => B. The action potential would peak at a higher voltage, over a shorter time
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[each_answer] => C. The action potential would peak at a lower voltage, over a longer time
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[each_answer] => D. The action potential would peak at a lower voltage, over a shorter time
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