QuestionSBA
QuestionChoice
The equilibrium potential (V) of an ion across a membrane is equal to the natural logarithm of the concentration of the ion on the outside of the membrane divided by the concentration of the ion within the membrane (ln([X]o/[X]i)), multiplied by the gas constant (R) and the temperature in kelvin (T), and divided by the ion's valence (z) and Faraday's constant (F). This is described by Nernst's equation: V = (RT/zF)ln([X]o/[X]i). By multiplying z by -2, the V is multiplied by -1/2. If the original equilibrium potential is +100mV, the new equilibrium potential is -50mV. To get +200mV, the relative concentrations would need to be changed.
10001164
d
+200mV
QuestionChoice
The equilibrium potential (V) of an ion across a membrane is equal to the natural logarithm of the concentration of the ion on the outside of the membrane divided by the concentration of the ion within the membrane (ln([X]o/[X]i)), multiplied by the gas constant (R) and the temperature in kelvin (T), and divided by the ion's valence (z) and Faraday's constant (F). This is described by Nernst's equation: V = (RT/zF)ln([X]o/[X]i). By multiplying z by -2, the V is multiplied by -1/2. If the original equilibrium potential is +100mV, the new equilibrium potential is -50mV. To get +50mV, the ion would have to be substituted for a divalent cation.
10001162
b
+50mV
QuestionChoice
The equilibrium potential (V) of an ion across a membrane is equal to the natural logarithm of the concentration of the ion on the outside of the membrane divided by the concentration of the ion within the membrane (ln([X]o/[X]i)), multiplied by the gas constant (R) and the temperature in kelvin (T), and divided by the ion's valence (z) and Faraday's constant (F). This is described by Nernst's equation: V = (RT/zF)ln([X]o/[X]i). By multiplying z by -2, the V is multiplied by -1/2. If the original equilibrium potential is +100mV, the new equilibrium potential is -50mV.
10001161
a
-50mV
QuestionChoice
The equilibrium potential (V) of an ion across a membrane is equal to the natural logarithm of the concentration of the ion on the outside of the membrane divided by the concentration of the ion within the membrane (ln([X]o/[X]i)), multiplied by the gas constant (R) and the temperature in kelvin (T), and divided by the ion's valence (z) and Faraday's constant (F). This is described by Nernst's equation: V = (RT/zF)ln([X]o/[X]i). By multiplying z by -2, the V is multiplied by -1/2. If the original equilibrium potential is +100mV, the new equilibrium potential is -50mV. To get -100mV, the ion would have to be substituted for a monovalent anion.

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10001165
e
-100mV
QuestionChoice
The equilibrium potential (V) of an ion across a membrane is equal to the natural logarithm of the concentration of the ion on the outside of the membrane divided by the concentration of the ion within the membrane (ln([X]o/[X]i)), multiplied by the gas constant (R) and the temperature in kelvin (T), and divided by the ion's valence (z) and Faraday's constant (F). This is described by Nernst's equation: V = (RT/zF)ln([X]o/[X]i). By multiplying z by -2, the V is multiplied by -1/2. If the original equilibrium potential is +100mV, the new equilibrium potential is -50mV. To get -200mV, the relative concentrations would need to be changed.
10001163
c
-200mV
Concept
Chapter
2693
3822
Membrane Potential
Topic
152
Neuroscience
12779
If the equilibrium potential of a monovalent cation across a membrane is +100mV, what would its equilibrium potential be at the same relative concentrations if it were a divalent anion?
a