The Na+-glucose symport system of intestinal epithelial cells couples the \"down
ID: 984868 • 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 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 37degreeC.Explanation / Answer
Delta G = RTln x [glu]in/[glu]out + 1(96485)(0.49)
= 8.314 j/mol (310K) ln(159/23) + 47277.65
= 52259.64 j/mol this is for one Na+
= 104519.29 x 2 = 9963.99 j/mol
= 104.519 kj/mole
Now we can find the ratio-
Delta G = RTln x [glu]in/[glu]out
104519.29 /8.314 (310K) = ln [glu]in/[glu]out
ln X ln [glu]in/[glu]out = 15.17
x = 3897158
Its very huge.