Question

Calculate the membrane potential of a neuron at standard temperature under each of the following conditions: a. The concentration of sodium, potassium and chloride is the same on both the inside and the outside of the membrane. b. The concentration of ions inside and outside the cell are as given below, but the relative permeability is the same for the three ions. c. The concentration of ions inside and outside the cell are as given below, but the relative permeability of K+ is 100 times greater than that of Na+ and Cl-.

(12 Points)

[??+]?=150??; [??+]?=15??; [?+]?=5??; [?+]?=100??; [??−]?=150??; [??−]?=13??;[Na+]o=150mM; [Na+]i=15mM; [K+]o=5mM; [K+]i=100mM; [Cl−]o=150mM; [Cl−]i=13mM;

Answer 1

the membrane potential needs to be calculated for all the three ions at standard temperature for the following conditions :- a)- The concentration of the ions (Na+, K+, Cl-) are same on both the sides of the cell membrane as per the given data then [??+]?=[??+]? , [?+]?= [?+]? and [??−]? =  [??−]? . the basic equation for the calculation of the potential for each of the ions is as follows:-

61.5 X LOG {(ion concentration)outside / (ion concentration)inside } -------(1)

thus plugging the values into eqn 1, we get the membrane potential to be ZERO mV in case (a)

For case b:- where in the given ions have same permeability with the following inside and outside ion concentrations:-

[Na+]o=150mM; [Na+]i=15mM; [K+]o=5mM; [K+]i=100mM; [Cl−]o=150mM; [Cl−]i=13mM

thus plugging the values into eqn 1 we get :-

61.5 X Log (150/15) + 61.5 X Log (5/100) + 61.5 X Log (150/13) = 61.5 X ( 1 + (-1.30) + 1.062) = 61.5 X 0.762 = 46.863 mV , thus the membrane potential is 46.863 mV when the relative permeabilities are same for all the three ions .

for case C :- when the permeability of K+ ions is 100 times greater than that of Na+ and Cl-.

then plugging all the values in the eqn 1 , we get :- 61.5 X ( 1 + 100(-1.30) + 1.062) = - 7868.18 mV

thus the membrane potential decreases in this case C