Question

The Na –glucose symport system of intestinal epithelial cells couples the \"downhill\" transport of two Na ions into the cell to the \"uphill\" transport of glucose, pumping glucose into the cell against its concentration gradient. If the Na concentration outside the cell ([Na ]out) is 163 mM and that inside the cell ([Na ]in) is 23.0 mM, and the cell potential is -51.0 mV (inside negative), calculate the maximum ratio of [glucose]in to [glucose]out that could theoretically be produced if the energy coupling were 100% efficient. Assume the temperature is 37 °C.

Answer 1

The maximum energy (or work) available from the transport of Na ions into the cell consists of two parts, that available from the transport down the chemical gradient, and that available from the transport down the electrical gradient, where R is the gas constant (8.3145 J/mol/K), T is the temperature in Kelvin, Z is the charge on the ion (+1), F is the Faraday constant (96485 J/(mol·V), and Δψ is the membrane potential (-51 X 10^-3 V).

ΔG chem = RT ln[Na+ in / Na+ out] and ΔG elec = ZFΔψ

The minimum energy (or work) needed to pump glucose against a concentration gradient is given by:

ΔG = RT ln[glucose in/glucose out]

Remember there are 2 Na+ transported:
2*ΔG = -RT*ln[glucose in/glucose out]

ΔG chem = RT ln[Na+ in / Na+ out] = (8.314 X 310.15) ln [(23/163)] = -5049.533     ( in joules)

ΔG elec = ZFΔψ = 1 X 96485 X -51 X10^-3 = -4920.735                                           ( in joules)

ΔG = ΔG chem + ΔG elec = -5049.533 -4920.735 = -9970.268 J = -9.970268 kJ

2*ΔG = -RT*ln[glucose in/glucose out]

ln[glucose in/glucose out] = -2*ΔG/RT = (-2 X -9.970268) / 0.008314 X 310.15 = 7.7331   ( in kilo joules)

[glucose in/glucose out] = e^7.7331 = 2282.6675

The maximum ratio of [glucose]in to [glucose]out that could theoretically be produced if the energy coupling were 100% efficient = 2282.6675