If the intracellular concentrations of acetylphosphate (AcP), acetate (Ac), ADP,
ID: 558628 • Letter: I
Question
If the intracellular concentrations of acetylphosphate (AcP), acetate (Ac), ADP, and ATP were 0.1 mM, 3.0 mM, 0.15 mM, and 4.0 mM, respectively, what would be the numerical value of G (in kcal per mol to the nearest hundredth) for the following reaction: AcP + ADP <--> Ac + ATP? Assume a temperature of 25 °C. In the Ch. 12 notes, there is a table that shows the standard free energies (G°') of hydrolysis of various phosphorylated compounds of biological interest. You will need to consult this table for this problem, and be aware of conversions between J and cal.
Standard Free Energies of Phosphate Hydrolysis of Some Compounds of Biological Interest Table 14-4 Compound G"(kJ-mol-1) Phosphoenoipyruvate 1,3-Bisphosphogiycerate ATP ( AMP + PP Acetyi phosphate Phosphocreatine ATP ADP + P) Glucose-1-phosphate -61.9 -49.4 -45.6 -43.1 -43.1 -30.5 -20.9 -19.2 -13.8 -13.8 -9.2 Fructose-6-phosphate Glucose-6-phosphate Glycerol-3-phosphateExplanation / Answer
for the reaction.
AcP --> Ac delta G is -43.1 Kj/mole (1)
ADP--> ATP , deltaG is 30.5 KJ/mole (2)
Eq.1+ Eq.2, AcP+ ADP ----->Ac+ ATP, deltaGo=-43.1+30.5=-12.6 Kj/mole
1KJ= 0.239 KCal,
-12.6 Kj/mole= -12.6*0.239Kcal/mole =-3KCal/mole
Q= [Ac] [ATP]/[AcP] [AdP] = 3*4/(0.1*0.15) = 800
lnQ= 6.7
deltaG= deltaGo+RTlnQ, R= 1.987 Cal/mole.K, T= 25 deg.c= 25+273= 298K
deltaG=-3000 cal/mole+1.987*298*6.7 =967.2Cal/mole