Energetics - uni-due.dehc0001/pdf/Microbial Physiology... · 2006. 6. 23. · Energetics....
Transcript of Energetics - uni-due.dehc0001/pdf/Microbial Physiology... · 2006. 6. 23. · Energetics....
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Energetics
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Respiratory Chain
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Respiratory Chain
- Part 1 -
Simplified scheme of the hydrogen and electron transport in a bacterial membrane.
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Respiratory Chain
- Part 2 -
Principles ofATP-synthesis by electron transport phosphorylation
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Respiratory Chain - Part 3 -
Redoxpotentials of the respiratory chain. The chain starts with NAD as the component with the most negative potential and is terminated by Cytochrome-Oxidase as the component with the most positive potential.
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Physiologie Folie 28
Phototrophy
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Physiologie Folie 29a
Chemolithotrophy
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Physiologie Folie 29b
Chemoorganotrophy
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Thermodynamics
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Physiologie Folie 30a
∆G∆H∆S
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Physiologie Folie 30b
Calculation example:
Using ΔGB°–values the ΔG°–value
and the ΔG°´–value for the
formation of lactic acid from
glucose can be calculated.
This reaction is the
metabolic basis for lactic acid bacteria
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Physiologie Folie 30b
Example:glucose 2 lactate + 2 H+
ΔGB° (glucose) = - 917 kJ mol-1ΔGB° (lactate) = - 518 kJ mol-1ΔGB° (H+, pH0) = 0 kJ mol-1ΔGB° (H+, pH7) = - 40 kJ mol-1
for pH = 0:ΔG° = 2 (-518 kJ mol-1) – (- 917 kJ mol-1)
= -119 kJ mol-1
for pH = 7:ΔG° = [2 (-518 kJ mol-1) + 2(-40 kJ mol-1)] – (- 917 kJ mol-1)
= -199 kJ mol-1
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Physiologie Folie 31
Standard formationenergies
for selected important
compounds (∆GB°
-values)
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Physiologie Folie 32
Example :
For the formation of lactic acid from glucose
the ΔG–value can be calculated under
physiological conditions as follows:
ΔG = ΔG° + RT ln [C] [D] / [A] [B]
= ΔG° + RT ln q
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Physiologie Folie 32
Example:
glucose 2 lactate + 2H+
ΔG = ΔG° + RT ln [C] [D] / [A] [B]= ΔG° + RT ln q
physiological concentrations:glucose = 10-2 mol l-1lactate = 10-2 mol l-1protons = 10-7 mol l-1q = 10-16
ΔG = -119 kJ mol-1 + RT ln 10-16
= -210 kJ mol-1
ΔG° = -119 kJ mol-1
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Physiologie Folie 33
Stan-dard redox values E0´
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Physiologie Folie 33a
„Electron tower“
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Physiologie Folie 34
Example :
The oxidation of molecular hydrogen by oxygen is the basis of the energy metabolism of hydrogen bacteria (Knallgasbakterien)
H2 + ½ O2 H2O
ΔG°´ = - 2 x 96,5 kJ V-1 mol-1 x 1,24 V= - 239 kJ mol-1
Using standard values for redox potentials of the reaction partners the free energy of the reaction can be calculated:
2 H+ / H2: E0´ = - 0,42 V½ O2 /H2O : E0´ = 0,82 V
Δ E0´ = 1,24 V
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Physiologie Folie 34a
Depen-dence of redox potential on pH or oxidation/reduc-tion ratio
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Energy conservation
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Physiologie Folie 34b
Energy transfer/storage compounds
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Physiologie Folie 34c
Sub-strate phos-phori-lation
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Physiologie Folie 34d
High energy phosphate bonds
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Physiologie Folie 35
Proton transfer across the membrane
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Physiologie Folie 36
Elec-trontrans-fer phosphori-lation
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Physiologie Folie 36a
F1F0-ATPase
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Respiratory Chain
- Part 1 -
Simplified scheme of the hydrogen and electron transport in a bacterial membrane.
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Respiratory Chain
- Part 2 -
Principles ofATP-synthesis by electron transport phosphorylation
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Physiologie Folie 37
Electrontrans-portchainandtrophiccondi-tion
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Physiologie Folie 37a
Compo-sition ofelectron transport chainversussubstrate type
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Physiologie Folie 38a
Electron transfercom-pounds (1)
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Physiologie Folie 38e
Electron transfer compounds (2)
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Physiologie Folie 38b
Redoxpotential of ETS compounds
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Physiologie Folie 38c
Cytochrome structures (heme and substituents)
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Physiologie Folie 83d
Model of cytochrome c
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Physiologie Folie 39
Top:ETS ofPara-coccusdenitrifi-cansadapts tooxygen
Bottom:ETS andproton trans-location
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Physiologie Folie 40
Photo-syn-thesisand protontrans-loca-tion/ATPase
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Reversed electron transport
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Physiologie Folie 41
Reversedelectrontrans-portfor NADH2formationfrom oxi-dized sub-strates