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
fructose
glucose 1-phosphate +
.................…(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
• 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
Sucrose + Pi
..…ie (3) reversed
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
(8.315 J K-1mol-1)(298 K)
• Therefore : Keq
= 25
= 3.2
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
Q + 2H+ + 2eQH2
- (FAD + 2H+ + 2e- FADH2)
Q + FADH2
EO N (V)
+0.04
-(-0.22)
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[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:
–
–
–
–
–
Standard free energy
Equilibrium constant
Net free energy change in coupled reactions
Reduction potential
Free energy change
• Well done!
Reading
Stryer, Biochemistry
Segel, Biochemical calculations, Wiley