The Na –glucose symport system of intestinal epithelial cells couples the \\\"do
ID: 192703 • Letter: T
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.
Explanation / Answer
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