Question

The Na glucose symporter transports glucose from the lumen of the small intestine into cells lining the lumen. Transport of 1 glucose molecule is directly coupled to the transport of 2 Nations into the cell. 2 Na*(out) + 1 glucose(out) - 2 Nat(in) + 1 glucose(in) Assume the following conditions at 37 °C: [Nalin - 12 mm, [Na lout = 145 mm, (glucoseJout = 28 pm, and Ay =-72 mV (inside negative). Faraday constant F = 96,485 C/mol, ideal gas constant R = 8.314 J/mol K. a. What is AG for transport of 2 Nat from outside to inside under these conditions? b. What is the upper limit for (glucose)in under these conditions? Note: must show calculations. Answers without calculating procedures receive no credit. Apical surface Basal surface Blood Intestinal lumen Microvilli Epithelial cell 3 Na 2 Na* Na K

Answer 1

we have,

a.

Na+ (out) ===> Na+ (in)

$\Delta$G = RT ln (Na+ (in) / (Na+ (out) + $\Delta \Psi$ z(Na+) F

$\Delta$G = 8.314 J/K-mol*310 K ln (12 mM / 145 mM) + (-72 mV) *(+1)* 96485 C/mol

$\Delta$G = - 6423 J/mol + (-6947 J/mol)     ( 1 CV = 1J )

$\Delta$G = - 13.37 kJ/mol

b.

The free energy for Na+ transport can be consumed for transporting Glucose inside against conc. gradient :

thus, thus $\Delta$G =   + ve for this transport , this is balanced by -ve free energy of Na+ transport.

thus,

$\Delta$G = RT ln (Glu (in) / (Glu (out)   <~ 13370 J/mol

max. 13370 J/mol

thus,   RT ln (Glu (in) / (Glu (out)   = 13370 J/mol

8.314 J/K-mol*310 K ln ( (Glu (in) / 28 uM) = 13370 J/mol

ln ( (Glu (in) / 28 uM) =   5.19

(Glu (in)) max = ~ 5012 uM     or 5 mM