The resting membrane potential of a cell is largely governed

The resting membrane potential of a cell is largely governed by the face that potassium channels are leaky and can be considered open. Thus, assuming potassium ions are in equilibrium across the membrane, calculate the resting membrane potential. B) How much free energy (per mole) is released or does it take to pump a sodium ion from the inside to the outside of the cell? C) It take one mole of ATP hydrolyzing to ADP + P| to pump three moles of sodium ions across the gradient. What is the intracellular ATP to ADP ratio given that [P_i] = 10 mM and Delta G of ATP hydrolysis is -30 KJ/ml?

Solution

1) Membrane potential (V_m) = RT/zF * ln(Co/ Ci)

Where, R = universal gas constant, 8.3 J/K mol = 1.98 cal / K mol

T = Temperature, 310 K

F = Faraday’s constant, 96500 C/mol = 2.3 * 10⁴ cal / V mol

Z = valence of ion, or potassium it is 1

Co = external concentration of ion (mM)

Ci = internal concentration of ion (mM)

Therefore, V_m = 1.98 * 310 / [1 * (2.3 * 10⁴ )] * ln (2/120)

= (613.8 / 2.3 * 10⁴ ) * (-4.094)

= 0.0266 * (-4.094)

= - 0.1092 V

= -109.2 mV

2) Free energy change deltaG = RT ln (Co/Ci) + zF

= (1.98 * 310) * (2.931) + (2.3 * 10⁴ )

= 1799.16 + (2.3 * 10⁴ )

= 24799.16 cal/mol

= 24.78 kcal/mol

3) ATP à ADP + Pi

The equilibrium constant is calculated as follows :

K’_eq = [ADP][Pi] /[ATP]

According to the given information :

[Pi] = 10 mM = 0.01 M

Therefore, K’_eq = [ADP]/[ATP] * 0.01………(a)

Free energy change is calculated as follows

deltaG = deltaG’ + RT ln K’_eq

24.78 = (-30) + 613.8 ln K’_eq

54.78 = 613.8 ln K’_eq

Therefore,

ln K’_eq = 54.78 / 613.8 = 0.089

2.303 log K’_eq = 0.089

log K’_eq = 0.038

K’_eq = 1.091……….. (b)

Therefore, from a and b,

1.091 = [ADP]/[ATP] * 0.010

[ATP] /[ADP] = 0.010/1.091 = 0.0091 M