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Selected Solutions to End of Chapter 13 Problem The second reaction in Glycolysis (an “isomerase”) is written backwards in this problem (which makes the real answers different than that in the textbook), it should be: Glucose-6-phosphate Fuctose-6-phosphateKeq = 0.508 This is analdose ketose. Think about this reaction: one ring structure going to another. a. What is the ΔGo’ ? Well… ΔGo’ = -RT ln Keq ΔGo’ = -(2.48 kJ/mole) ln 0.51 ΔGo’ = 1.7 kJ/mole b. If the concentration of glucose-6-phosphate is 0.5M and fructose-6-phosphate is 1.5M, then what is the ΔG ? Well… ΔG= ΔGo’ +RT lnQ So, ΔG= 1.7 kJ/mole + RT ln (1.5 M / 0.5M ) ΔG= 1.7 kJ/mole + (2.48 kJ/mole x ln 3) ΔG= 4.4 kJ/mole Note: this is different from the answer in the text which was calculated from the reverse of the glycolytic pathway. c. Why areΔGo’ and ΔG different? One is under standard conditions, the other, ΔG is the “actual’ ΔGin the conditions that either are artificial (in this case) or natural in the cell with the actual concentrations of metabolites in the cell. This is what we will look at to see the actual thermodynamics of whole metabolic pathways. ΔGo’ for coupled reactions: glucose-1-phosphate  glucose-6-phosphate ΔGo’ = -7.3 kJ/mole glucose-6- phosphate  fructose-6-phosphate ΔGo’ = 1.7 kJ/mole So, the overall ΔGo’ is -5.6 kJ/mole now, what’s the Keq? From: ΔGo’ = -RT ln Keq lnKeq = - ΔGo’ / RT = - (5.6 kJ/mole) / (2.48 kJ/mole) = 2.25 lnKeq = 2.25 so Keq = 9.5 Note that this answer differs from that in the textbook, this is because they rounded out too much before getting to the calculated answer. 12. Some more coupled reactions: a. Phosphocreatine + H2O Creatine + PiΔGo’ = -43.0 kJ/mole ADP + Pi ATP + H2O ΔGo’ = 30.5 kJ/mole Overall: Phosphocreatine + ADP Creatine + ATP ΔGo’ = -12.5 kJ/mole b. ATP + H2O ADP + Pi ΔGo’ = -30.5 kJ/mole Fructose + Pi Fructose-6-phosphate + H2O ΔGo’ = 15.9 kJmole Overall: ATP + Fructose ADP + Fructose-6-phosphate ΔGo’ = -14.6 kJ/mole. 14. This problem is just like problem 6 above. Do it and get the answers which is for the ΔG in the cell: -13 kJ/mole. You need to use the RT at 37oC which is 2.58 kJ/mole and not that much different from that at 25oC, but use the right one anyhow. 19. Daily use of ATP by Human Adults. First of all, the calculation of the ΔG of ATP synthesis from known concentrations of the substrates and products in a liver cell at 37oC. Then we can calculate the weight of ATP used each day in a 150 pound human (68 kg). a. The ΔG. Data: ΔGo’ of ATP hydrolysis is -30.5 kJ/mole and just the opposite sign for synthesis. In liver cell: [ATP] = 3.5 mM, [ADP] = 1.5 mM, and [Pi] = 5.0 mM. So for the synthesis: ADP + Pi ATP + H2O ΔG = ΔGo’ + RT lnQ hints: RT has to be at 37oC (310o K), put concentrations in M. so, RT = (8.315 J/mole K)(310 K) = 2.58 kJ/mole ΔG = 30.5 kJ/mole + (2.58 kJ/mole) ln ( ([0.0035 M])/([0.0015 M][0.005 M])) ΔG = 30.5 kJ/mole + (2.58 kJ/mole)(6.15) = 46.4 kJ/mole What is the take home message? Thermodynamics in the cell is the real energy changes that take place or need to be supported. The standard ΔG shows the tendency to produce or require energy. b. Data: the average human (68 kg) needs the standard 2,000 diet calories (one diet calorie is really 1000 cal = kcal) which in kJ is 8,360 kJ / 24 hours. Assume half of food energy is used to make ATP. What is the weight of ATP your body makes/day. So the strategy is to get the moles of ATP this represents and use the MW to get the weight. The MW of ATP = 507 g / mole or 0.507 kg / mole Half of the energy = 8,360 kJ / 2 = 4,180 kJ. 4,180 kJ / (46.4 kJ / mole) = 90 moles 90 moles x (0.507 kg / mole) = 45.6 kg WOW that is 2/3’s your weight ! c. That is certainly not t