E = 61/z *log(xo/xi) You made an artificial cell that has a membrane completely
ID: 820211 • Letter: E
Question
E = 61/z *log(xo/xi)
You made an artificial cell that has a membrane completely impermeable to all ions. There are no ion channels or pumps
in the membrane. You fill the cell with 100 mM KCl, and place it in an extracellular solution containing 10 mM KCl.
(1) What is the electrical membrane potential measured between the inside and the outside of the cell? Why?
You insert into the membrane of this cell ion channels that are selective only for K+ ions (note: these are the so-called
"leak" channels that are open all the time, regardless of membrane potential).
(2) What is the membrane potential now? Why?
You insert even more K+ channels into the membrane, which makes it even more permeable to K+ ions.
(3) What is the membrane potential now?
You add 5 mM NaCl inside the cell, and 100 mM NaCl to the extracellular solution.
(4) Will the membrane potential change, and if so, will it become more negative or more positive? (Don
Explanation / Answer
E = 61/z *log(xo/xi)
1)
xo = 10mM
xi = 100mM
for KCl, z = +1
E = 61*log(10/100) = -61 mV
2)
leak channels allow the K+ ions to flow from the cell, since the volume inside cell is very less compared to the outside volume the concentration inside the cell tend to become equal to concentration outside
the new potential would be 0
3)
same thing happens
4) since outside concentration of NaCl is more than inside concentration, the term log(xo/xi) will be positive
therefore the potential should become more positive
5)
membrane potential will change when leak channels specific to NaCl are introduced, the concentration of NaCl inside will tend to increase and get equal to outside potential.
therefore the potential should decrease and eventually it should reach zero
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