Question

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1. Two solutions are separated by a semipermeable membrane that is permeable to sodium but not chloride or potassium ions. Solution A on the left contains a solution that is 1 mM NaCl, while Solution B on the right has a concentration of sodium chloride of 100 mM.

There is an electrode in both solutions, and a comparative assessment of voltages between the two compartments is being made. The investigator doing the study decides to call side A ground potential (or, in other words, defines it as 0).

What is the value for the sodium equilibrium potential in this condition? (show how you calculate this) (2)

In one sentence, describe why it has a positive or negative value. (1)

What would be the value of the chloride equilibrium potential? (show how you calculate this) (2)

What is the voltage of side B relative to side A? (2)

Does the chloride equilibrium potential contribute significantly to the voltage actually being measured by the investigator? Why or why not (one sentence)? (2)

2. An investigator is examining the relationship between membrane voltage and ion concentrations in a nerve cell. She experimentally sets the potassium concentration of the inside of the cell to 100 mM.

Draw a graph showing the membrane potential across the cell membrane as the extracellular potassium concentration is varied from 1 mM to 1000 mM. (6)

She then alters the intracellular potassium so that it is 10 mM on the inside of the cell. Draw on the same graph as above the new relation for the membrane potential across the cell as the extracellular potassium is altered from 1 mM to 1000 mM. (3)

3. The mitochondrion has a high concentration of calcium in its interior (~ 1 mM) compared to the cytoplasm (~1 µM).   What would you hypothesize the voltage of the interior of the mitochondrion to be relative to the cytoplasm of the cell if there were a significant permeability to calcium? (6)

Answer 1

Answer:

Q-1):

• To calculate equilibrium potential, we use the Nernst Equation
• 
• Assumming T = 298K
• For Sodium, n = 1
• [X]O = Outside Concentration = 1 mM
• [X]i = Inside Concentration = 100 mM
• So, ENa+ = -118.26 mV
• Side A has a higher concentration of sodium ions compared to side B, hence the flux of sodium ions shall be from side A to side B leading to a negative equilibrium potential.
• The Chloride Equilibrium potential can be calculated using the same equation, except that in this case we need to change the value for valence of ion from +1 to -1.
• Hence, ECl- = -1(-118.26) = 118.26 mV
• If we calculate the voltage of side B, we shall again use the Nernst Equation that we used for equilibrium potential of Side A, with the only change being in inside and outside concentrations which shall now be
• [X]O = Outside Concentration = 100 mM
• [X]i = Inside Concentration = 1 mM
• So, due to this the value of equilibrium potential of side B shall be opposite in sign to that of side A and shall be
• EB = -1(-118.26) = 118.26 mV
• Voltage of side B relative to side A, VBA = EB - EA = 118.26 - (-118.26) = 236.52 mV
• Since Chloride ions cannot pass through the membrane, they do not contribute significantly to the voltage being measured.

Q-2):

Q-3):

• Using Nernst equation,
• Voltage=2.303RTlog(Cainside/Caoutside)/ZF
• = 2.303*8.28*293*log(1000/10)/(2*96000)
• ( since, R= 8.28, T= temperature= 293K, Cainside=1mM=1000uM
• Z=valency=2, F= 96000)
• = 5587.17*2/(2*96000)
• = 0.0582 V
• or 582 mV

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