Biochemistry part a/b question

Question

5. (a) What is the free energy change if 1 mole of Na+ is transported across a cell membrane from the inside of the cell where the concentration is 1.0 µM to the outside of the cell where it is 100 mM? Assume the temperature is 37˚C and the membrane potential is 70 mV (inside negative).

(b) Will the hydrolysis of 1 mole of ATP be sufficient to drive the transport of 1 mole of Na+ (at pH 7.4 and 37˚C), assuming the concentrations of ATP, Pi, and ADP are 4.60 mM, 5.10 mM and 310 µM, respectively [ATP + H2O à ADP + Pi + H+ ∆G˚’ = -31 kJ/mol].

Answer 1

You should start by finding the equation for the bioenergetics of transport:

delta G = nRTln(C2/C1) + nFZdelta psi (membrane voltage change)

The constants are F= 96485 J/V* M or J/ C

R= 8.314 J/ mol * K

We are told that the concentrations are 1.0 uM and 100 M that the temperature is 310 K and membrane potential is at 70 mV. We know that it is more energetically unfavorable to move something against its' concentration gradient so this 1st part should be a positive delta G because (-G = favorable). Likewise we are moving something positive (Na+) to a more positive charge environment which is also unfavorable. So, both parts should be positive G. Be sure to check and convert units to J, M, K then just plug it in.

a) You should get ^G= + 30.5 KJ .

b) Since this is (SLIGHTLY) lower than the energy we are told the hydrolysis of ATP (-31kJ)