Question

5/2400 Resources Give Up Hint Intestinal epithelial cells pump glucose into the cell against its concentration gradient using the Nat-glucose symporter. Recall that the Nat concentration is significantly higher outside the cell than inside the cell. The symporter couples the "downhill" transport of two Nations into the cell to the "uphill" transport of glucose into the cell. If the Na+ concentration outside the cell (INa1.) is 151 mM and that inside the cell ([Na]) is 23.0 mm, and the cell potential is --53.0 mV (inside negative), calculate the maximum energy available for pumping a mole of glucose into the cell. Assume the temperature is 37 °C. AG = What is the maximum ratio of (glucose), to (glucose). that could theoretically be produced if the energy coupling were 100% efficient? O 1.14 O 2300 4.4 x 10-4 O 7.73

Answer 1

Given, [Na⁺]_out = 151 mM

[Na⁺]_in = 23 mM

$\Delta v$ = -53.0 mV

T = 37 oC = 310 K

We Know,

Na+jin AG = RTin + ZFAV [Na+out]

$\Delta G$= 8.314 x 310 x ln(23/151) + 1 x 96500 x -53 x 10^-3

$\Delta G$= -4845.40 - 5114.5

$\Delta G$= 9959.9

Two Na⁺ requires transport of 1 Gglu

$\Delta G$ = -2 x 9959.9

$\Delta G$= -199119.8 J/mol

$\Delta G$ = -19.92 KJ/mol

AG =-RTIN [(Gglu)in [(Gglu)out]

[(Gglu)in] - = -19919.8/(-8.314.-310) "[(Gglu)out]

[(Gglu)in] [(Gglu)out] = 7.73

[(Gglu)in] (Gglu)out] = 2276kJ/mol

[(Gglu)in] [(Gglu)out = AGglu

AGglu = 2276kJ/mol

After round off above value-

AGglu = 2300kJ/mol

Hence the answer is 2300 kJ/mol