The Na+-glucose symport system of intestinal epithelial cells couples the \"down
ID: 43408 • 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 163 mM and that inside the cell ([Na+]m) is21.0 mM, and the cell potential is -53.0 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 degree C.Explanation / Answer
The ratio is 2.06
Now how do we arrive at this ratio
The energetic cost of moving an ion depends on the electrochemical potential, which is the sum of chemical and electrical potential
chemical potential=deltaGchem
delta Gchem=RTln(Na+ in/Na +out)
=8.315J/mol.310K ln 21/163
=-5.258 KJ/Mol
delta G electric=zF.trans membrane potential
=+1.96.5kJ/mol.volt.0.05 V
=-4.82kJ
therefore delta G= delta G chem+delta Gelec
=-5.258+-4.82
=-10.1
there are 2 Na+ transported
so 2 delta G = -RT ln glucose in/glucose out
2.-10.1=-.008315kJ.310.ln glucose in/glucose out
-20.2=-2.57 ln glucose in/glucose out
therefore ln glucose in/glucose out=7.8599
=2.06
So the maximum ratio is 2.06