The Na –glucose symport system of intestinal epithelial cells couples the \\\"do

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 151 mM and that inside the cell ([Na ]in) is 23.0 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 °C.

Explanation / Answer

The energy available from the transport of Na ions into the cell consists of two parts, from the electrical gradient and the chemical gradient

G chem = RT ln[Na+ in / Na+ out]

G elec = ZF

R: gas constant

T:temperature in Kelvin

Z :charge on the ion (+1)

F:Faraday constant (96.5 kJ/(mol·V)

:membrane potential

For the G chem = RT ln[Na+ in / Na+ out] = (0.00831KJ/molK)(310.15K)ln(0.023/0.151) = -4.85 KJ/mol

For the G elec = ZF = (1)(96.5KJ/molV)(-0.053V)=-5.1145 KJ/mol

Total G would be G chem + G elec, however, because two Na ions are needed to transport glucose

2*G = -RT*ln[glucose in/glucose out]

2*[(-4.85 KJ/mol)+(-5.1145 KJ/mol)]= - (0.00831KJ/molK)(310.15K)ln[glucose in/glucose out]

ln[glucose in/glucose out] = 2*[9.96]/[0.00831*310.15]

ln[glucose in/glucose out] =7.72

glucose in/glucose out=2253

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