Oxalate (C2O4 2-) is an anion that can contribute to the formation of kidney sto
ID: 300098 • Letter: O
Question
Oxalate (C2O4 2-) is an anion that can contribute to the formation of kidney stones in mammals under certain conditions. Some epithelial cells of the small intestine contain a Cl- / C2O4 2- exchanger which excretes oxalate into the lumen of the small intestine. Consider the following concentrations for an epithelial cell of a human small intestine (temperature 37°C) with a resting membrane potential of -60mV:
Extracellular [Cl-]: 200mM
Intracellular [Cl-]: 15mM
Extracellular [C2O4 2-]: 0.062mM
Intracellular [C2O4 2-]: 0.00025mM
a) Calculate the equilibrium potential (Eion) for each of these ions.
b) If Cl- channels open under these conditions, will Cl- ions move into or out of the cell? Clearly explain why.
c) How much energy (in Joules) would be generated if one mole of Cl- were to move in this direction (the direction you predicted in part b.)? Explain your answer.
d) How many molecules of Cl- must be transported down the electrochemical gradient in order to transport one molecule of oxalate out of the cell? e) Is the exchanger Cl- / C2O4 2- electrogenic? Explain why or why not.
Explanation / Answer
a). The Nernst Equation to calculate the equilibrium potential for chloride at 370C is,
Where,
ECl = equilibrium potential of chloride in millivolts (mV)
Ko = concentration of chloride ions in the exterior of the cell
Ki = concentration of chloride ions in the interior of the cell
Z = valence of the ion (for chloride, it is, -1)
(i). Extracellular [Cl-]: 200mM
Intracellular [Cl-]: 15mM
Thus, ECl = -61 mV log (200 mM / 15 mM) = -68.56 mV
(ii). Extracellular [C2O4 2-]: 0.062mM
Intracellular [C2O4 2-]: 0.00025mM
Thus, EC2o4 2- = -61 mV log (0.062 mM / 0.00025 mM) = -146.034 mV
b). Given that the resting membrane potential is, -60 mV. If the chloride channels open under these conditions, the chloride ions move out of the cell, to make the membrane potential less negative.
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