In considering active transport by Na+-K+-ATPase, 3 Na+ are pumped out of the ce

Question

In considering active transport by Na+-K+-ATPase, 3 Na+ are pumped out of the cell and 2 K+ are pumped in for each ATP that is hydrolyzed to ADP + Pi. Given a G for ATP hydrolysis of -10 kcal/mol, and that V is -60 mV (more electronegative inside), and that the pump typically maintains the internal Na+ at 10mM, external Na+ at 145 mM, internal K+ at 140 mM and external K+ at 5mM, what is the efficiency of the pump (i.e., what fraction of the energy available from ATP hydrolysis is used to drive transport)?

Explanation / Answer

The ion concentrations are Internal Na+ = 10 mM, External Na+ 145 mM; Internal K+ = 140 mM and External K+ = 5 mm ; G for ATP hydrolysis of -10 kcal/mol and Membrane potential = -60

The efficiency of the pump = Energy required to move ions / Energy released by ATP hydrolysis.

Energy required to move Na+ = RT ln(Conc inside/ Outside)

R = 1.98 Cal /mol. K and Temperature - 37 = 310 K

By substituting

G for Na+ = 1.98 X 310 ln(10/140)

= 1619.8 cal / mol.

G for K+ = 1.98 X 310 ln(140/5)

= 2045 cal/mol.

The G for the electrical component = zF; Where Z- membrane potential, F- Faraday constant and is membrane potential.

By substituting F= 23,062 cal mol-1 V-1 and 1 for charge of ion

G = 23.062 X -60 X 1

= 1383 cal /mol

So Na + G = 1620 +1383 = 3003 cal/mol

K + G = 2045 - 1383 = 662 cal / mol.

Energy from 1 mole of ATP is used to move 3 Na+ and 2 K+

Hence Total energy for moving ions = (3 X 3003) + (2 X 662) = 10333 cal. or 10.1 Cal.

Energy released from hydrolysis of 1 mol of ATP is given as 10.1 kcal.

Thus efficiency of the pump = 100 X (10.1/10) = 100%

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