The Na^+-glucose symport system of intestinal epithelial cells couples the \"dow
ID: 221424 • 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 161 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 degree C.Explanation / Answer
First,calculate the free energy.
Delta G=RTln(C in/C out)+ZF Delta V
Delta G=(8.31J/K-mol)(310K)ln(23.0/161 mM)+(+1)96500 J/V mol[-0.049.0V-0V] = -5012 J-4728 J =-9740 J
For 2 moles of Na, this will be =19480 J
Then calclate the ratio of glucose concentrations this free energy can yield
Delta G =(8.31) (310) ln (Glu in/Glu out)
19480=2576.1 ln(Glu in/Glu out)
ln(Ratio of concentrations)=7.561
Ratio=exp(7.561)