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 159 mM and that inside the cell ([Na ]in) is 23.0 mM, and the cell potential is -49.0 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

concnetration of [Na+]in = 23.0 mM

concnetration of [Na+]out = 159 mM

cell potential = - 49.0 mV

Gchem = R T ln ([Na+]in / [Na+]out)

                = 8.314 x 10^-3 x 310 x ln (23 / 159)

Gchem = - 4.983 kJ/mol

G elec = Z F E

               = 1 x 96.5 x - 0.049

               = - 4.7285 kJ/mol

G = 2 x (- 4.7285 - 4.983 )

        = - 19.423 kJmol

G = R T ln ([glucose]in / [glucose]out)

- 19.423 = 8.314 x 310 x ln ([glucose]in / [glucose]out)

([glucose]in / [glucose]out) = 1874