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?

Options:
explanation: 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.
name: +200mV
label: d
answer: no
explanation: 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.
name: +50mV
label: b
answer: no
explanation: 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.
name: -50mV
label: a
answer: yes
explanation: 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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name: -100mV
label: e
answer: no
explanation: 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.
name: -200mV
label: c
answer: no

Correct answer: a