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 163 mM and that inside the cell ([Na+]m) is21.0 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

The ratio is 2.06

Now how do we arrive at this ratio

The energetic cost of moving an ion depends on the electrochemical potential, which is the sum of chemical and electrical potential

chemical potential=deltaGchem

delta Gchem=RTln(Na+ in/Na +out)

                  =8.315J/mol.310K ln 21/163

                   =-5.258 KJ/Mol

delta G electric=zF.trans membrane potential

                       =+1.96.5kJ/mol.volt.0.05 V

                       =-4.82kJ

therefore delta G= delta G chem+delta Gelec

                        =-5.258+-4.82

                        =-10.1

there are 2 Na+ transported

so 2 delta G = -RT ln glucose in/glucose out

2.-10.1=-.008315kJ.310.ln glucose in/glucose out

-20.2=-2.57 ln glucose in/glucose out

therefore ln glucose in/glucose out=7.8599

                                                  =2.06

So the maximum ratio is 2.06

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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