The Na –glucose symport system of intestinal epithelial cells couples the \\\"do
ID: 79651 • 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 143 mM and that inside the cell ([Na ]in) is 21.0 mM, and the cell potential is -54.0 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
Maximum energy available from transport of sodium ions into the cell consists of two parts, that available from the transport down the chemical gradient and that available from transport down electrical gradient, where R is gas constant, T is temperature in Kelvin, Z is charge on the ion (+1), F is Faraday constant (96.5 kJ/mol *V)and delta phi is membrane potential which is -54mV
Gchem = RT ln (Na+ in / Na+out) and Gelec = Z F
Minimum energy needed to pump glucose against a concentration gradient is given by:
G = RT ln [glucose in/ glucose out]
G chem. = -4.181 and G elec = -5.3075
G = 9.7255
There are 2Na ions transported: 2* G = -RT*ln [glucose in/glucose out]
G = RT ln [Na]out/[Na]in +ZF delta phi
=2 (273+37) ln(143/21)
glucose in/glucose out = 2000