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

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 141 mM and that inside the cell ([Na+]in) is 23.0 mM, and the cell potential is -49.0 mV (inside negative), calculate the maximum ratio of [glucose]m to [glucose]out that could theoretically be produced if the energy coupling were 100% efficient. Assume the temperature is 37 degreeC

Explanation / Answer

G chem =- RT ln[Na+ in / Na+ out] and G elec = ZF

G = G chem + G elec =- 8.314 x310 ln[23/141]+1x96.5x10^3x- 49x10^-3 =-4673.49-47285 =- 51958.49

we multiply delta G by 2 because the question says that 2 Na+ are needed.

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

-103916.98/-8.314x310=ln[glucose in/glucose out] = 40.32x2.303=log[glucose in/glucose out]

log[glucose in/glucose out]=92.85

taking antilog both side

[glucose in/glucose out]-7.07x10^92

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.