Consider the inside and outside of a cell, where the 8 nanometer thick cell memb
ID: 1444500 • Letter: C
Question
Consider the inside and outside of a cell, where the 8 nanometer thick cell membrane has a large number of open Na+ Ion channels. The major contributor to these ions is dissolved NaCI. Lets analyze a toy model of the cells resting potential in which the Na+ ion channels are the only ion channels, they are permanently open, and NaCI is the only contributor of Na + ions. Assume the outside of the cell has a concentration of 150 mM (10^3 Molar) NaCI and a significantly lower concentration inside the cell. Which side of the membrane would have the higher potential? O Inside O Outside Explain why in this situation a potential difference (Nernst potential for Na +) would develop across the membrane. If the Nernst potential for the sodium ions in our toy model is 60 mV, what would the concentration of NaCI be on the inside of the cell? Explain how you got your answer. Now let us refine our model by noting that there is a second source of Na+ ions in the cell: Nal. Suppose the outside of the cell has a concentration of Nal of 0.04 mM and the inside has a Nal concentration of 4 mM. How will the presence of these ions change the Na+ Nernst potential across the membrane? (Assume the Nal is fully ionized in solution.) New Nernst potential = mV. In the refined toy model can you calculate the Nernst potential for Cl- ions? If you can, calculate it and show your work. If you can't, explain why not.Explanation / Answer
A) Sodium (Na+), which is at a high concentration inside and a low concentration outside the membrane.
Sodium (Na+) and chloride (Cl) ions are at high concentrations in the extracellular region, and low concentrations in the intracellular regions. These concentration gradients provide the potential energy to drive the formation of the membrane potential.
B) Nernst equation is given by -
E = Eo (0.0592V/n) * logQ
=> 0.059 = 0.06 - (0.0592V/1) * logQ
=> (0.0592V/1) * logQ = 10-3
=> Q = 1.0396
=> the concentration of NaCl = 19.415 mM
C) Nernst equation is given by -
E = Eo (0.0592V/n) * logQ
=> E = 0.06 - (0.0592V/1) * log(4.04/3.96)
=> E = 59.485 mV ---------------> Nernst Potential
D) E = Eo (0.0592V/n) * logQ
=> E = 0.065 - (0.0592V/1) * log(4.04/3.96)
=> E = - 64.485 mV -------------------> Nernst potential for Cl- ions