In dog skeletal muscle, extracellularly [Na+] = 150 mM and [K^+] = 2.7 mM, while

Question

In dog skeletal muscle, extracellularly [Na+] = 150 mM and [K^+] = 2.7 mM, while intracellularly [Na^+] = 12 mM and [K^+] = 140 mM. Assume a temperature of 25 degree C and a membrane potential of +60 mV (inside to outside). Calculate the free energy change for one Na^+ being transported from inside to outside the cell in dog skeletal muscle. Calculate the free energy change for one K^+ being transported from outside to inside the cell in dog skeletal muscle. Calculate the free energy change for just the ion transport part of one cycle of a Na^+, K^+, -ATPase in dog skeletal muscle. Based on your calculations, is hydrolysis of a single molecule of ATP enough to drive the ion transport of a Na^+, K^+-ATPase in dog skeletal muscle? Assume that the free energy for hydrolysis of a single molecule of ATP in dug skeletal muscle is -55 kJ.

Explanation / Answer

[Na+] out = 150 mM, [Na+]in = 12mM, [K+]out = 2.7 mM, [K+]in = 140 mM.

(a)Calculate the free energy change for one Na+ being transported from inside to outside in Dog skeleton muscle-

Membrane potential of -60 mV = -60e-3 V

G = q E

= (1 * 1.60e-19 C / ion)(6.022e23 electrons/mole) (0V - (-60e-3 V)) =5781.12J/mole

= 5.78 KJ/mole

b) Calculate the free energy change for one K+ being transported from outside to inside in Dog skeleton muscle

G = q E

       = (1 * 1.60e-19 C / ion)(6.022e23 electrons/mole) ((-60e-3 V) - 0 V) = -5781 J/mole

       = - 5.78 KJ/mole

c) Calculate the free energy change for just the ion transport part of 1 cycle of Na+- K+ ATPase for dog skeletal muscle

During the cycle, the Na+-K+ pump is first phosphorylated and subsequently de-phosphorylated. Binding of intracellular Na+ ions to the pump initiate phosphorylation of the enzyme, while the binding of extracellular K+ ions trigger the de -phosphorylation reaction. After a complete cycle, three Na+ ions have been extruded from and two K+ ions have entered into the cell.

G = qE

= (3 * 1.60e-19 C / ion)(6.022e23 electrons/mole) (0V - (-60e-3 V))

+ (2 * 1.60e-19 C / ion)(6.022e23 electrons/mole) ((-60e-3 V) - 0 V) = 5781 J/mole

= 5.78 KJ/mole

d) Assume that G for hydrolysis of 1 molecule of ATP = -55KJ

Hydrolysis of single molecule of ATP is not enough to drive the ion transport of Na+-K+ ATPase

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