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 153 mM and that inside the cell ([Na ]in) is 19.0 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

Delta Gchem = RTln(Na+ in / Na+ out)

R= gas constant= 8.314 J/mol*K= 0.008314KJ/K*mol

Temperature T= 37⁰C

Temoerature in K= 273+37=310K

Sodium in= 19mM

Sodium out= 149mM

Delta Gchem = (0.008314 KJ/K * mol)(310 K)(19mM / 149 mM) = 2.577 * ln( 0.1275)

                    = 2.577* (-2.06)= -5.31 KJ/mol

Faradays constant=96.5 KJ/V, Z= +1

Cell potential= Delta psi= -49mV= -0.049V

Delta Gelec = ZF(cell potential)

Z=1 as sodium has 1 charge

Delta Gelec= (+1)(96.5 KJ/V)(-0.049 V) = -4.7285 KJ/mol

There are 2 sodium ions transported per glucose molecule

Delta G = 2 * (Gchem + Gelec) = 2 * (-5.31KJ/mol + -4.7285 KJ/mol) = -20.08KJ/mol

The energy allows the transport of glucose against a concentration gradient. For every 2 mol of Na+ transported, one mol of glucose is transported.

Hence, delta G= 20.08 KJ/mol (otherwise its unfavourable)

DeltaG=RTln([glucose in]/[glucose out])

ln[Glu]in/[Glu]out= Delta G/RT= 20.08/310* 0.08314=7.78

([glucose in]/[glucose out]= e^7.78=2392.27