Lecture 26
Energy conversion (continued)
Substrate Oxidation, Generation of
reduced high-energy intermediates and
their oxidation by oxygen
http://www.microscopyu.com/galleries/fluorescence/cells/hela/hela.html
Why oxygen?
Electron re-distribution from less electronegative to more
electronegative atoms occurs with massive energy release:
Electronegativity of common elements:
H < C < S < N < O …Cl < F
Identify the substance and the
reduction state of the first carbon
(red)
Enthalpies of oxidation (combustion)
hydrogen (MW 2)
H2 + ½O2 → H2O
methane (MW 16)
CH4 + 3O2 → CO2 + 2H2O
glucose (MW 180.2)
C6H12O6 + 6O2 → 6CO2 + 6H2O
palmitic acid (MW 256.4)
C16H32O2 + 31O2 – 16CO2 + 16H2O
-286 kJ/mol
(-68.4 kCal/mol)
-891 kJ/mol
(213)
-2840 kJ/mol
(-680)
- 9730 kJ/mol
(-2328)
Compare caloric capacities of the substrates
Oxidation-reduction (Red-ox) reactions usually lead to redistribution of electron densities or complete transfer of
electrons resulting in change of ionization state.
Fe2+ + Cu2+ → Fe3+ + Cu+
or in the form of half-reactions: Fe2+ → Fe3+ + e
Cu2+ + e → Cu+
In biological systems oxidation is often coupled to
dehydrogenation.
1. Direct transfer of electrons
2. As a transfer of H atoms or removal of H atoms coupled to
production of H+
3. As a Hydride ion :H–
4. Through direct combination with oxygen
R-CH3 + (½)O2 → R-CH2-OH
Reduction potentials for mixtures of reductant/oxidant (hlfreaction potentials) are measured using the standard
hydrogen electrode
RT
[electron acceptor]
o
E  E 
ln
nF
[electron donor]
2H+ + 2e → H2
Mg2+ + 2e → Mg (metal)
http://www.chemguide.co.uk/physical/redoxeqia/eomgdiag.gif
Standard Reduction potentials for some half-reactions, Volt
½ O2 + 2H+ + 2e → H2O
+0.816
Fe3+ + e → Fe2+
+0.771
Cytochrome c (Fe3+) + e → Cytochrome c (Fe2+)
+0.254
Fumarate2- +2H+ + 2e → succinate2-
+0.031
2H+ + 2e → H2
0
(standard condition,
condition) pH0)
Pyruvate + 2H+ + 2e → lactate
-0.185
FAD + 2H+ + 2e → FADH2
-0.219
S + 2H+ + 2e → H2S
-0.243
NAD+ + H+ + 2e → NADH
-0.320
NADP+ + H+ + 2e → NADPH
-0.324
α-ketoglutarate + CO2 + 2H+ + 2e → isocytrate
-0.38
2H+ + 2e → H2
-0.414
(pH 7)
NADH + H+ + ½ O2 → NAD+ + H2O
NAD+ + H+ + 2e → NAD
-0.320 V
½ O2 + 2H+ + 2e → H2O
+0.816 V
total difference
DG = -nFEo = -220 kJ/mol
1.14 V
Stage 1
(in the cytosol)
NAD NAD+ + H+ +2e-
NADH
half reaction (-0.32 V)
FAD has smaller
reducing potential
(-0.219 V) so its
reduction can be
coupled to a less
energetically
favorable oxidation
reaction
Beta-Oxidation of
fatty acids
R= CH3-(CH2)n-
R= CH3-(CH2)n-2-
When succinate and oxygen are given to mitochondria, they release H+
Electron transport chain
Complex I
II
III
IV
Energy released in forming water is stored as a PMF
Energy is divided into smaller units ~10-12 protons per water molecule
Iron-sulfur cluster (Fe-S)
Heme
CoQ accepts both electrons and protons
Why add valinomycin?
What conditions are necessary for the experiment to work?
-poorly-buffered medium
-Valinomycin/potassium
-A substrate…NOT NADH…as stated in your book
Is ATP synthesis the reason for the pH returning to the original
value?...
Oxidation rate is slow
Why?
Oxidation rate is fast
Why?