The Na^+-glucose symport system of intestinal epithelial cells couples the \"dow

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^+]m) is 19.0 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

Apply:

dG1 = RT*ln([Na+]in / [Na+] out)

and also

dG2 = Z*F*dY

fod

dG totla = dG1 + dG2

so

dG1 = RT*ln([Na+]in / [Na+]out)

T = 37°C= 310 K

dG1 = 8.314*310*ln(19/161) = -5507.686 J/mol

for

dG2 = Z*F*dY

dG2 = 1*96500*(-53*10^-3) = -5114.5 J/mol

so

dG total = -5507.686 -5114.5 = -10622.186 J/mol

for Na+ transport...

n = 2

so

2dG = -RT*ln (glucose in / glucose out)

-2*10622.186 = -8.314*310 * ln(glucose in/ gulocse out)

so

ln(glucose in/ gulocse out) = (-2*10622.186) / ( -8.314*3109) = 8.24

(glucose in/ gulocse out) = exp(8.24) = 3789.540

(glucose in/ gulocse out) = 3789.540

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