MitoSeminar II:
Some calculations in
bioenergetics
MUDr. Jan Pláteník, PhD.
Ústav lékařské biochemie 1.LF UK
Helpful comments of Prof. MUDr. Jiří Kraml, DrSc., are acknowledged.
1
Respiratory chain and
oxidative phosphorylation:
Summary
* Transfer of electrons via electron carriers (respiratory
chain) to oxygen in the inner membrane
* Proton pumping from the matrix to the cytosolic site of
the inner mitochondrial membrane (proton gradient).
Protonmotive force = pH gradient + membrane potential
* Protons flow through ATP synthase and power the
synthesis of ATP.
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1
Stoichiometry of oxidative phosphorylation
• If F0 complex has 10 c subunits: one complete turn
means flow of 10 protons.
• The F1 part has three ATP-producing sites: one complete
turn of the shaft means synthesis of three ATP from ADP
+ Pi
• At least 3 H+ have to be pumped out of the matrix for
production of 1 ATP
• One more H+ is consumed for import of phosphate
• Transport of 2 electrons from NADH to oxygen
(complexes I, III, IV) pumps 10 protons, from FADH2 to
oxygen (complexes III, IV) 6 protons .
• Oxidation of 1 NADH produces 2.5 ATP
• Oxidation of 1 FADH2 produces 1.5 ATP
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∆G = ∆H — T ∆S
The chemical reaction can
occur only if the ∆G is negative.
It means: when the products have
less free energy than the reactants
have.
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2
∆G and chemical equilibrium
A + B
C + D
[Ceq][Deq]
Keq =
[C][D]
∆G = – RT ln Keq + RT ln
[Aeq][Beq]
A + B
A + B
→
←
[A][B]
C + D
C + D
A + B ←→ C + D
∆G negative
∆G positive (negative
for reverse reaction)
∆G = 0 (equilibrium)
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Standard free enthalpy
∆Go’ = – RT lnKeq’
1.0M, 25 °C
pH 7.0
R ... universal gas constant 8.3143 J mol-1K-1
T ... absolute temperature in Kelvins (298.15 K = 25oC)
[C][D]
∆G = – RT ln Keq + RT ln
[A][B]
[C][D]
∆G =
∆Go’
+ RT ln
[A][B]
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3
7
Redox potential
X– + Y ↔ X + Y–
X–↔ X + e–
Y + e– ↔ Y–
Difference in affinities to
electrons between two
redox couples, in volts.
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4
Redox potential
Eo …standard redox
potential,
potential, 1.0 M, pH 0
Eo’ … standard redox
potential,
potential, 1.0 M, pH 7
Reference hydrogen electrode:
Eo = 0.0 V Eo’ = –0.42 V
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(from Harper‘s Illustrated Biochemistry, 27th edition, McGraw-Hill Co. 2006)
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5
Relationship between free
enthalpy and redox potential
∆G = – nF ∆E
n … number of transfered electrons
F ... Faraday constant = 96 485 C /mol
C (Coulomb) = J/ V
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The Nernst equation
Tells voltage of galvanic cell, or redox potential, for various
concentrations of components.
∆G =
∆Go’
[C][D]
+ RT ln ————
[A][B]
∆E =
∆ Eo ’
+
∆G = –nF ∆E
RT
[OX]
OX]
+ —— ln ————
nF
[RED]
RED]
Walther Hermann Nernst (1864-1941): Nobel Prize 1920
Equation also called Nernst-Peters
(Peters applied Nernst equation for redox processes)
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Example I: Transfer of electrons from NADH to
oxygen:
oxygen:
NADH + H+ + 1/2O2 → NAD+ + H2O
Redox couples:
couples:
NADH + H+ → NAD+ + 2H+ + 2e–
1/2O2 + 2H+ + 2e– → H2O
Eo’ = – 0.320V
Eo’ = + 0.816V
For the whole reaction:
reaction:
o
∆E ’ = 0.816V – (– 0.320V )= 1.136V
∆Go’ = –nF ∆Eo’
∆Go’ = –2(96.5 kJ V–1 mol –1)(1.136V)
)(1.136V) = – 219.25 kJ mol –1
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Example II: Succinate dehydrogenase
Redox couples:
couples:
succinate → fumarate + 2H+ + 2e–
FAD + 2H+ + 2e– → FADH
FADH2
Eo’ = + 0.0
0.03V
Eo’ = – 0.12
V
0.12V
∆Go’ = –2(96.5 kJ V–1 mol –1)(–
)(–0.15
.15V) = + 28.95 kJ mol –1
Reaction cannot proceed ???
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Example II: Succinate dehydrogenase
Redox couples:
couples:
succinate → fumarate + 2H+ + 2e–
FAD + 2H+ + 2e– → FADH
FADH2
Eo’ = + 0.0
0.03V
Eo’ = – 0.12
V
0.12V
But,
But, if fumarate is consumed in the next reaction,
reaction, and FAD
reoxidized,
reoxidized, the actual ratios succinate/
succinate/fumarate and
FAD/FADH2 would not be 1:1
If succinate : fumarate is 500:1, redox potential of the system
using the Nernst equation would be:
be:
RT
[OX]
OX]
o
’
∆E = ∆E + —— ln ———— = 0.03 + (–
(–0.08) = – 0.05 V
nF
[RED]
RED]
And for FAD/FADH2 500:1 ∆E = –0.12 + 0.08 = – 0.04 V
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Likewise malate dehydrogenase reaction:
Redox couples:
couples:
malate → oxaloacetate + 2H+ + 2e–
NAD+ + 2H+ + 2e– → NADH + H+
Eo’ = – 0.17
V
0.17V
o
E ’ = – 0.32
V
0.32V
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8
Ionic gradient on a membrane can also do work
Na Na
Na
Na
Na Na
Na
Na
Na
Na
Na
Na
Na
Na
c1
∆G = – RT ln ————
c2
c(Na+)1 = 120 mmol/l
c(Na+)2 = 12 mmol/l
= –(8.3143 × 298.15) × ln 10 =
= – 5.7 kJ/mol
Equation is valid if c1 > c2 (diffusion to c2); for 37 °C the coefficient is – 5.9
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9
Protonmotive force
∆ p = ∆ pH + ∆ ψ
Difference in
concentration of
protons
- about 1 pH unit
…from Nernst
equation is 60 mV
Electric potential
on the inner
membrane.
Must be measured,
value e.g. 160 mV
∆ p = 60 mV + 160 mV = 220 mV
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Efficiency of mitochondrial production of ATP
Oxidation of 1 mole of NADH leads to pumping of 10 mol protons
and production of 2.5 mol ATP, protonmotive force is 220 mV
Oxidation of 1 mol NADH: ∆Go’ = – 219.25 kJ/mol
kJ/mol
+
Protonmotive force ∆p = 220 mV corresponds to: ∆µH
∆G = ∆µH+ = – F · ∆p = – 21.2 kJ/mol
kJ/mol
is electrochemical
proton gradient
For pumping 10 mol protons: ∆G = 10 x (21.2) = 212 kJ
…Most of the energy from oxidation is converted
to proton gradient
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10
Efficiency of mitochondrial production of ATP
Oxidation of 1 mole of NADH leads to pumping of 10 mol protons
and production of 2.5 mol ATP, protonmotive force is 220 mV
Oxidation of 1 mol NADH: ∆Go’ = – 219.25 kJ/mol
kJ/mol
+
Protonmotive force ∆p = 220 mV corresponds to: ∆µH
∆G = ∆µH+ = – F · ∆p = – 21.2 kJ/mol
kJ/mol
is electrochemical
proton gradient
For pumping 10 mol protons: ∆G = 10 x (21.2) = 212 kJ
For 1 ATP is needed: ∆Go’ = 30.5 kJ/mol,
kJ/mol,
In the cell really (excess of ATP): ∆G’= asi 50 kJ/mol
4 protons do work 84.9 kJ/mol,
kJ/mol, which
is certainly enough for 1 ATP
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Efficiency of mitochondrial production of ATP
Oxidation of 1 mole of NADH leads to pumping of 10 mol protons
and production of 2.5 mol ATP, protonmotive force is 220 mV
Oxidation of 1 mol NADH: ∆Go’ = – 219.25 kJ/mol
kJ/mol
+
Protonmotive force ∆p = 220 mV corresponds to: ∆µH
∆G = ∆µH+ = – F · ∆p = – 21.2 kJ/mol
kJ/mol
is electrochemical
proton gradient
For pumping 10 mol protons: ∆G = 10 x (21.2) = 212 kJ
For 1 ATP is needed: ∆Go’ = 30.5 kJ/mol,
kJ/mol,
In the cell really (excess of ATP): ∆G’= asi 50 kJ/mol
Oxidation of 1 mol NADH gives theoretically energy for
making 4.4 mol ATP, practically 2.5 mol is produced
… efficiency cca 57 % (for
(for standard conditions cca 35 %)
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