1. Bioenergetics of Anaerobic Microbial Reactions Traditional and new processes

2. Methanogenesis Archaebacteria All methane producing bacteria (MPB) are archeabacteria Archeabacteria are different from all other life forms (3rd primary kingdom) ArcheabacteriaProkaryotes Eukaryotes Methanogens Halobacterium etc. Eubacteria Cyanobacteria etc. Plants Animals Fungi etc. This new taxonomic division of lifeforms into three kingdoms is based on phylogenetically conservative features: 1. Ether linked membrane lipids rather than ester links 2. No murein in cell wall (→ penicillin resistant) 3. Different protein synthesis (→ streptomycin resistant)

3. 4. More subunits in RNA polymerase 5. Unique coenzymes (e.g. F 420) Special features in archaebacteria: High temperature tolerance (115°C) Acid resistance (pH 1, at 90°C) Pyrodiction Sulfolobus Unique use of light ( no photosystem)Halobacterium Archaebacteria are likely to be the earliest life forms (Symbiont hypothesis)

4. Methanogenesis 0 1 CO2 MFCOH MPCH2 MPCH3 CoMCH3 CH4 H2 MF MP MF CoM MP CoM 2 1 4 1 6 1 8 1 Formyl - C Methylene - C Methyl - C Methane ATP Only the last step allows the generation of 1ATP Pathway of CO2 Reduction

5. Methanogenesis Electron transport chain CH3 - CoM + F420 CH4 + F420 + CoM 2 H+ ADP + Pi ATP ATP - ase Membrane Principle mechanism of ATP generation: Methyl - Respiration 2 H+

6. Methanogenesis 6 1 Acetate and Methanol Metabolism During methanol conversion to methane only 1 ATP is generated via methyl – respiration. 4 CH3OH → 3 CH4 + CO2 + 2 H2O 8 1 0 1 In the presence of hydrogen methanol is only used as electron acceptor → methyl – respiration. 6 1 H2 + CH3OH → CH4 + H2O 8 1 2 0

7. Acetoclastic methanogens split acetate into a methyl group and a carboxy group. The two electrons of the carboxy group are transferred to the methyl group → methyl – respiration. 6 1 CH3 - COOH → CH4 + CO2 8 1 2 1 0 1 8 2 Methyl - respiration is the only way of ATP generation in methanogenesis

8. Overall the energy conserving reaction in methanogenesis is from proton translocation via electron transfer to the methyl group generated during metabolism. Methanogenesis = anaerobic methyl respiration

9. Revision: Gibbs Free Energy Change The Gibbs Free Energy (G) of substrates and products can be calculated. For exergonic (=spontaneous = downhill) reactions the G of the substrates is higher than that of the products. Hence the change in G (=ΔG) for exergonic reactions is negative. This ΔG is the driving force of the reaction. As the reaction proceeds, the diminishing substrate concentration and increasing product concentration cause the difference in G to become smaller and smaller until an equilibrium is reached at which ΔG = zero.

10. Revision: Gibbs Free Energy Change Handbooks provide values of the Gibbs Free Energy Change as: ΔG o assumes that all reactants and products are a unity (concentrations are 1 M and gas partial pressures are 100 kPa) ΔG o’ assumes that all reactants and products are at unity, however the proton concentration is not 1 M (  pH 0) but 10-7 M (  pH 7). This value makes more sense for most biological systems ΔG is the actual Free Energy Change under experimental conditions and changes any moment as the reaction proceeds

11. 11 Methanobacillus omelianskii, observed features 1. degrades ethanol to acetate and methane gas: CH3-CH2OH + CO2 --> CH3-COOH + CH4 2. Also grows with H2 as e-donor and CO2 as e- acceptor: 4H2 + CO2 --> CH4 + 2H2O Suspicions about the purity of culture were raised because: 1. H2 inhibited ethanol degradation but not CH4 production (Preference for H2?) 2. No growth on ethanol after prolonged cultivation on H2 + CO2 (Mutagenic loss of ethanol dehydrogenease?) 3. Bubbling inert N2 gas through ethanol degrading culture --> CH4 production stopped but ethanol oxidation to acetate continued.

12. 12 Hypothesis of Marvin P. Bryant: Interspecies Hydrogen Transfer, Claims: 1. Methanobacillus consists of an association of two microbes: 2. Pelobacter depends on H2 removal by MPB or N2 flushing. 3. MPB needs needs Pelobacter to provide H2 as e-donor. 4. Both depend on each other (syntrophy) Theory of Bryant on Interspecies Hydrogen Transfer Ethanol Acetate H2 CH4 CO2

13. 13 Lack of driving force of reaction 1: CH3-CH2OH + H2O --> CH3-COO- + H+ + 2 H2 DGo'= + 9.6 kJ/mol Bryant's argument: reaction 2 has excess driving force 4 H2 + HCO3- + H+ --> CH4 + 3 H2O DGo'= - 135.6 kJ/mol Total reaction: DGo'= - 116.4 kJ/mol Question: Energy sharing possible (pulley analogy)? Bryant’s hypothesis was not readily accepted: Ethanol Acetate H2 CH4 CO2

14. 14 Bryant’s hypothesis was not readily accepted: Ethanol Acetate H2 CH4 CO2

15. 15 Energy Sharing How can two different microbes share the common energetic potential? The actual free energy change of the reaction DG depends on the product to substrate ratio (P/S) and can be calculated from the standard DG= DGo + 5.69 kJ * log (P/S) DGo'= + 9.6 kJ/mol- 135.6 kJ/mol Ethanol Acetate H2 CH4 CO2 For high substrate concentrations (P/S less than 1) the DG will be more favourable (negative). Interspecies hydrogen transfer was postulated to operate at extremely low H2 partial pressures o 1- 10 Pa (compared to 100,000 Pa for standard conditions)

16. 16 Interspecies hydrogen transfer was postulated to operate at extremely low H2 partial pressures o 1- 10 Pa (compared to 100,000 Pa for standard conditions. Low H2 increases the driving force of reaction 1 while it decreases the driving force of reaction 2. DGo'= + 9.6 kJ/mol- 135.6 kJ/mol Ethanol Acetate H2 CH4 CO2 A sharing in driving force is possible by lowering H2 until reaction 1 becomes feasible (exergonic) but not so low that reaction 2 becomes endergonic. Reaction1 Reaction 2 Energy Sharing

17. 17 Eventually Bryan’s hypothesis was not only confirmed but found to be a general feature of anaerobic systems (e.g. rumen, anaerobic digesters, sediments) The interspecies hydrogn can be intercepted by chemical and biological means (more powerful H2 users than methanogens. DGo'= + 9.6 kJ/mol- 135.6 kJ/mol Ethanol Acetate H2 CH4 CO2 Other substrates found to degrade via H2 transfer include all fatty acids, alcohols, many amino acids 30 % of electron flow in anaerobic digesters passes via H2. Use of energetic calculations critical to understand (exploit, control, optimise) many biological reactions) Reaction1 Reaction 2 Significance of Interspecies Hydrogen Transfer

18. 18 How can the energetic values be determined ?: The example of Methanobacillus omelianskii and interspecies hydrogen transfer has shown the importance of DG calculations: Example problem: Establish the standard DGo for the anaerobic conversion of ethanol (12 e-) to acetate (8 e-) by Pelobacter (syntrophic partner in M. omlianskii) 1. Establish proper equation: (4 electrons transferred) CH3-CH2OH + H2O CH3-COO- + H+ + 2 H2 (-181.75) + (-237.18) (-369.4) + 0 + 0 2. Look up the standard Gibbs free energy of formation (Gfo) 3. Subtract the Gfo of substrates from Gfo of products --> DGo= + 49.53 kJ/mol 4. This gives the energetic situation (negative = spontaneous = exergonic = downhill) for standard conditions: Room temperature, partial pressure of all gases = 100 kPa, all concentrations 1 mol/L. Why is the established value not the one calculated by Bryant for the same reaction?

19. 19 Considering pH for thermodynamic calculations DG= DGo + 5.69 kJ * log (P/S) Standard conditions imply that also protons are at a concentration of 1 mol/L (pH 0 !). How to convert reaction energetics to consider pH 7 rather than 0: 1. Use formula : DG= DGo + 5.69 kJ * log (P/S) DG= 49.53 kJ + 5.69 kJ * log (P/S) 2. Replace P by the concentration desired: DG= 49.53 kJ + 5.69 kJ * log (0.0000001 /S) = 49.53 kJ - 39.83 kJ = +9.7 kJ 3. Conclusion: Reaction requires less energy to be run but is still not spontaneous as DG is positive. What about the other products and substrates ?

20. 20 How to calculate actual reaction energetics by considering all P and S concentrations CH3-CH2OH + H2O --> CH3-COO- + H+ + 2 H2 DG= DGo + 5.69 kJ * log ((P 1 *P 2 *P 3 ) / (S 1 *S 2 *S 3 )) Example : Does the reaction become favourable given that P 1 = aceate =1 mM, = 0.001 of std. cond. P 2 = protons = 10-7 M P 3 =H2 = 10 Pa = 0.0001 of std. cond. S 1 = ethanol = 10 mM = 0.01 of std. cond. 1. Use formula : DG= 49.53 kJ + 5.69 kJ * log (P 1 *P 2 *P 3 /(S 1 *S 2 *S 3 )) 2. Replace P and S by the concentrations desired: DG= 49.53 kJ + 5.69 kJ * log (0.001*10-7*(0.0001) 2 /(0.01*1))= DG= 49.53 kJ - 5.69 kJ * log 10-16= 49.53 kJ - 5.69 kJ * 16 = -41.51 kJ 3. Conclusion: Reaction is now thermodynamically possible and can support growth.

21. 21 How to calculate actual reaction energetics by considering all P and S concentrations CH3-CH2OH + H2O --> CH3-COO- + H+ + 2 H2 DG= DGo + 5.69 kJ * log ((P 1 *P 2 *P 3 ) / (S 1 *S 2 *S 3 )) Example : Does the reaction become favourable given that P 1 = aceate =1 mM, = 0.001 of std. cond. P2=P2=

22. 22 Anaerobic digestion, a 3 stage process In contrast to aerobic degradation, AD requires different groups of specialised bacteria: Hydrolysis and fermentation Acetogenesis via hydrogen production Methanogenesis of acetate and H2 Fermentative Bacteria (e.g. Clostridia) Volatile fatty acids (VFA) alcohols, amino acids OHPA MPB CH4 H2 CO2 Acetate CO2 Complex organic matter (starch, cellulose, fats, protein)

23. 23 Anaerobic digestion, role of H2 concentration 1/3 of the electron flow proceeds via H2 H2 concentration is very low (1/ 40,000 of atmosphere = 40 ppm) but flux is high If H2 increases higher than 100 ppm to 1000 ppm the OHPA can not operate any more (DG is positive) Extremely low H2 level is critical Digester overloading with easily fermetable organics (sugars)  fast H2 production  H2 production > H2 consumption  H2 accumulation  OHPA don’t convert organic acids  fermenting bacteria produce more VFA than H2 acetate  drop in pH  killing MPB  Overloading of anaerobic digesters with excess substrate results in failure due to H2 buildup  Bottleneck similar to Crabtree effect, acidification Methanogenesis of acetate and H2

24. 24 What is the expected CH4 concentration in biogas for different organics Compound ElectonsE/COS%CH4 content Carbons Glucose24 6 4050 Ethanol12 2 6-275 Lactate12 3 4050 Propionate14 3 4.7-0.759 Butyrate20 4 5-162.5 Butanol24 6 6-275 Oxalate 2 2 1312.5 Formate 2 1 2225 Methanol 616-275 Methane 8 1 8-4100 CO2 0 1 0+40