Question

HW 11-1

Recall that the Na + concentration is significantly higher outside the cell than inside the cell. The symporter couples the "downhill" transport of two Na + ions into the cell to the "uphill" transport of glucose into the cell. If the Na + concentration outside the cell ( [ Na + ] out ) is 149 mM and that inside the cell ( [ Na + ] in ) is 19.0 mM, and the cell potential is − 55.0 mV (inside negative), calculate the maximum energy available for pumping a mole of glucose into the cell. Assume the temperature is 37 °C.

Δ G gluc = kJ/mol

2. What is the maximum ratio of [glucose]in to [glucose]out that could theoretically be produced if the energy coupling were 100% efficient?

a. 1.131.13

b. 38003800

c. 8.248.24

d. 2.6×10−4

Answer 1

Given, [Na+]out = 149 mM

[Na+]in = 19 mM

$\Delta v$ = -55.0 mV

T = 37 oC = 310 K

We Know,

$\Delta G$= 8.314 x 310 x ln(19/149) + 1 x 96500 x -55 x 10^-3

$\Delta G$= 10615.55

Two Na+ requires transport of 1 Gglu

$\Delta G$ = -2 x 9959.9

$\Delta G$= -21231.10 J/mol

$\Delta G$ = -21.23 KJ/mol

[(Gglu)in] AGglu = -RTln [(Gglu)out = -21.23/(-8.314 x 310)

AGglu = 8.24kJ/mol

Option c. 8.24 kJ/mol