Question

Calculate the Delta G for the following scenarios and state which require energy to happen. a. The internal potassium (K+) concentration is 30x greater than the external concentration. For a cell at 37C with no membrane potential, what is the Delta G? b. The internal potassium (K+) concentration is 30x greater than the external concentration. For a cell at 37C with a membrane potential of -60mV, what is the Delta G? c. The internal chloride ion (CI^-) concentration is 12mM and the external concentration is 145mM. For a cell at 25C with a -90mV membrane potential, what is the Delta G? d. The internal pH is 8.0 and the external pH is 7.0. For a cell at 25C with a -180mV membrane potential, what is the Delta G for the proton gradient?

Answer 1

Delta G = RT ln [X]in/[X]out

= 8.314 x 310 ln 30/1

Delta G = +8766 J/mol

Delta G = RT ln [X]in/[X]out

= 8.314 x 310 ln 30

= +8766 J/mol

delta due to membrane potential

ΔG = (z)(F)(V_m) where

z= the charge on the ion (+1 in this case)

F= 96485

V_m= the membrane potential, about − 60mV

delta G = +1 x 96485 x 0.06

deltaG = -5789 J

So total delta G = 8766 J - 5789 J = +2977 J/mol

delta G = RT ln [X]in/[X]out

= 8.314 x 298 ln 12 x10^-3/145 x 10^-3

=-6173 J/mol

ΔG = (z)(F)(V_m)

delta G = -1 x 96485 x -0.09

delta G = +8683 J

total delta G = 8683 - 6173 = +2509 J/mol

at pH 8; [H+] = 10^-pH = 10^-8 = 1 x 10^-8 M

at pH 7; [H+] = 10^-pH = 10^-7 = 1 x 10^-7 M

delta G = RT ln [X]in/[X]out

= 8.314 x 298 ln 1 x10^-8/1 x 10^-7

=-5704 J/mol

ΔG = (z)(F)(V_m)

delta G = +1 x 96485 x -0.18

delta G = -17367 J

total delta G = -17367 - 5704 = -23072 J/mol

SO only the last process is spontaneous and all the other require energy to happen