Thermodynamics II: problems Dr Una Fairbrother

Q1. The bacterium Pseudomonas saccharophila contains sucrose phosphorylase an enzyme that catalyses the phosphorolytic cleavage of sucrose Sucrose + Pi glucose 1-phosphate + fructose .................…(1)

a) From the following data, calculate the biochemical standard free energy change for the phosphorolysis of sucrose H2O + sucrose glucose + fructose .................…(2) ΔG BN = -29 kJ mol-1 H2O + glucose 1-phosphate glucose + Pi .................…(3) ΔG BN = -21 kJ mol-1 and b) calculate the equilibrium constant for the phosphorolysis of sucrose

Sucrose + Pi glucose 1-phosphate + fructose ……………..(1) A1: To calculate a)the biochemical standard free energy ( ΔG BN) for the phosphorolysis of sucrose, add the ΔG BN values of the two reactions which combine to yield reaction (1) H2O + sucrose glucose + fructose ΔG BN = -29 kJ mol-1 .............…(2) glucose + Pi H2O + glucose 1-phosphate ΔG BN = +21 kJ mol-1 ..…ie (3) reversed Sucrose + Pi glucose 1-phosphate + fructose ……………..(1) ΔG BN = -8 kJ mol-1 or –8000J

b) to calculate the equilibrium constant (Keq) ΔG BN = -RT lnKeq -ΔG BN = lnKeq RT lnKeq = 8000 J mol-1 = 3.2 (8.315 J K-1mol-1)(298 K) Therefore : Keq = 25

Q2. Could any of the following be used as an energy-rich metabolite for replenishment of cellular ATP from ADP? Metabolite ΔG BN hydrolysis (kJ mol-1 ) Phosphoenolpyruvate -62 1,3 -bisphosphoglycerate -49 Phosphocreatine -43 ATP (to ADP +Pi) -30 Glucose 6-phosphate -14 Glycerol 3-phosphate -9

A2. The approach here is to determine whether the net free energy change for a coupled reaction is negative

eg ΔG BN (kJ mol-1 ) ADP + Pi ATP + H2O +30 Phosphocreatine + H2O creatine + Pi -43 ------------------------------------------------------------- Phosphocreatine + ADP º creatine + ATP -13

Therefore: reaction exergonic and spontaneous, so in this case YES! NOTE: also phosphoenolpyruvate and 1,3 -bisphosphoglycerate when hydrolysed will provide sufficient free energy.

Q3 The standard reduction potential for ubiquinone (Q) is + 0.04V and the standard reduction potential for FAD is -0.22V. Show that the oxidation of FADH2 by Q theoretically liberates enough energy to drive the synthesis of ATP from ADP and Pi under standard conditions.

NOTE: This is a two electron process EO N (V) Q + 2H+ + 2e- QH2 +0.04 - (FAD + 2H+ + 2e- FADH2) -(-0.22) Q + FADH2 QH2 + FAD ΔEO N= (0.04) -(-0.22) V = 0.04 + 0.22 V = 0.26 V

ΔEO N ΔEO N found from (half reaction with more positive EON) - (half reaction with less positive EON) ΔG BN = -nFΔEO N NOTE: F (Faraday’s constant) 96.48kJV-1mol-1 = (-2)(96.48kJV-1mol-1)(0.26V) = -50.17 kJmol-1 NOTE: ADP + Pi ATP + H2O ΔG BN= +30 kJmol-1 Therefore sufficient free energy to support synthesis of ATP!

Q4 In a rat hepatocyte, the concentrations of ATP, ADP + Pi are 3.4mM, 1.3mM and 4.8mM respectively. Calculate the free energy change for the hydrolysis of ATP in this cell (at 25 oC and otherwise standard conditions) How does this compare to the standard free energy change?

ΔG = ΔG BN + RT ln [ADP][Pi] ΔG = ΔG BN + RT ln[C][D] [A][B] For ATP + H2O ADP + Pi then ΔG = ΔG BN + RT ln [ADP][Pi] [ATP][H2O] Thus: ATP + H2O ADP + Pi ΔG BN= -30 kJmol-1

ΔG = -30,000 Jmol-1 + (8.315 JK-1mol1)(298K)ln (1.3 x 10-3)(4.8 x 10-3) (3.4 x 10-3) = -30,000 Jmol-1 + 2480 Jmol-1ln(1.8 x 10-3) -30,000 Jmol-1 - 16,000 Jmol-1 = -46,000 Jmol-1 = -46 kJmol-1 This is about 1½ times the standard free energy change

Summary Calculated: Well done! Standard free energy Equilibrium constant Net free energy change in coupled reactions Reduction potential Free energy change Well done!

Stryer, Biochemistry Segel, Biochemical calculations, Wiley Reading Stryer, Biochemistry Segel, Biochemical calculations, Wiley