1 A bioreactor with a k La of

1. A bioreactor with a k. La of 25 h-1 with active microbes is aerated resulting in a steady oxygen concentration of 1 mg/L. What is the microbial oxygen uptake rate (in mg/L/h) assuming the oxygen saturation concentration is 8 mg/L?

2. The airflow to a Vinegar producing chemostat running at steady state was interrupted (at 90 sec. below) and oxygen data recorded. a. What is the k. La of the chemostat in h-1 ? b. What was the ethanol (CH 3 CH 2 OH) to acetic acid (CH 3 COOH) conversion rate of the process when it was at steady state?

4. List (in the box next to the molecule) the number of moles of oxygen needed for the complete oxidation to CO 2 of the following compounds CH 3 -CH 2 OH HOOC-COOH CH 3 -CO-CH 3 5. List (in the box next to the molecule) the number of moles of NAHD that can be generated from the complete oxidation to CO 2 of the following compounds: Pentose (CH 2 O)5 CH 3 -COOH H 2 CO 3 6. List (in the box next to the molecule) the number of moles needed for an anaerobic microbe using these substances instead of oxygen as the electron acceptor for the complete oxidation to CO 2 of ethanol (CH 3 -CH 2 OH): NO 3 - N 2 SO 42 - H 2 S Fe 3+ Fe 2+

7. Can microbes use the oxygen atom in the H 2 O molecule as an electron acceptor? Give reasons for your explanation and an example of the end product that would be formed (in the case you think it is feasible).

8. A 10 L chemostat is operated with a flowrate of 0. 6 L/h. An equilibrium is established with a constant oxygen, concentration, p. H, biomass (3 g/L) and substrate concentration. What is the specific growth rate of the microbes in the chemostat and what is the biomass productivity R (g/L/h) of the chemostat? u = D = F/ V = 0. 6 L/h/ 10 L = 0. 06 h-1 Productivity R g/L/h can be calculated from X * D 3 g/L *0. 06 h-1 = 0. 18 g/L/h

9. How would you determine the microbial Yield coefficient from a batch culture and a chemostat culture respectively?

10. Explain the effect of biomass feedback (recycle, retention) on the biomass concentration and productivity R of a chemostat. Use a plot of biomass (X) and productivity (R) versus the dilution rate to illustrate the point.

Steady State Concentration Effect of biomass feedback (here 3 fold): Dotted line no feedback: • Washout occuring early SR X • 3 -fold Feedback approximately: • 3*X 3*R 1/3* S • allows 1/3 reactor size to do same work R S D Dcrit • Feedback essential for pollutant removal. Can be used 100 -fold smaller treatment plant • Note: same assumed feed concentration (SR)And 88

Effects of growth constants on steady state concentrations of biomass and substrate in a chemostat as a function of dilution rate (xaxis) Effect of ms Effect of decrease ks Effect of increased Y Effect of increased μmax And 99

11. Sketch below an example graph of the specific growth rate of a microbe (Y-axis) dependent on the limiting substrate concentration (X-axis). Put a scale (numbers and units) on both axes. Point out in your graph (with an arrow) where 3 of the 4 growth constants can be read from and give their values and units as read from your example graph.

Substrate limitation of microbial growth The two curves are described by two properties: µmax (h-1) The maximum specific growth rate obtained with no substrate limitation (umax (h-1)) µ (h-1) substrate limitation k. S and the half saturation constant (Michaelis Menten constat), giving the substrate concentratation at which half of the maximum u is reached (ks (g/L)). Substrate (g/L) And 1212 Growth- Michaelis Menten model

Effect of Maintenance Coefficient (m. S) on growth Rate The negative specific growth rate (µ) observed in the absence of substrate (when S = 0) (cells are starving, causing loss of biomass over time) µ (h-1) 0 S(g/L) is the decay rate m. S*Ymax - m. S*Ymax And 1313

Relationship between oxidation state and electron equivalents of carbon atoms • The electron equivalents (EE) on a carbon atom is 4 minus the oxidation state (OS) : OS EE Example +4 0 CO 2 +3 1 -COOH +2 2 HCOOH, CO, -CO- • EE = 4 -OS +1 3 -CHO 0 4 -CHOH- -1 5 -CH 2 OH -2 6 -CH 2 -, CH 3 OH -3 7 -CH 3 -4 8 CH 4 • Note: Electron equivalent= Reducing equivalent= (degree of reduction)

MSE 2011 1) A bioreactor with a k. La of 20 h-1 with active microbes is aerated resulting in a steady oxygen concentration of 2 mg/L. What is the microbial oxygen uptake rate (in mg/L/h) assuming the oxygen saturation concentration is 8 mg/L? OUR= 20 h-1 *(8 -2 mg/L) = 120 mg/L/h 2) The airflow to a chemostat running at steady state DO of 5 mg/L (c. S was 8 mg/L) was temporarily interrupted. The oxygen concentration decreased steadily by 0. 05 mg/L every second. What is the k. La of the chemostat in h-1 ? k. LA = 180 mg/L/g / (8 -5 mg/L) = 60 h-1 3) What is the maximum possible rate (in m. M/h) of lactate (CH 3 -CHOH-COOH) oxidation to CO 2 by an aerobic reactor that is limited by an oxygen supply due to a k. La of 50 h-1 assuming an oxygen saturation concentration of 8 mg/L? Lac = 12 e- 1 Lac reacts with 3 O 2 OUR = 50 h-1* 8 mg/L = 400 mg/L/h = 25 m. M/h LUR = 4. 17 m. M/h

List (in the box next to the molecule) the number of moles of oxygen needed for the complete oxidation to CO 2 of the following compounds: CH 3 -CH 2 OH 4. 5 HOOC-COOH 0. 5 CH 3 -CO-CH 3 4 List the four growth constants with their units. State in one short sentence what this growth constant means by referring to its units. Ymax g. X/g. S umax g. X/L/h / /g. X/L = h-1 ms g. S/g. X/h = h-1 k. S = g. S/L

How much NADH can be produced from the complete oxidation to CO 2 of the following compounds: CH 3 -CHOH-CH 2 OH 11 CHOOH 1 benzoate (aromatic ring with a COOH group attached to one of the carbons 15 Can microbes use the oxygen atom in the H 2 O molecule as an electron acceptor? Give reasons for your explanation and an example of the end product that would be formed (in the case you think it is feasible). A chemostat is used to produce microbial biomass for the purpose of recombinant protein production. Lactate (CH 3 -CHOH-COOH) from dairy wastewater is used as the substrate. The yield coefficient of the recombinant strain is 0. 3 g of cells per g of lactate degraded. When interrupting the air flow the oxygen concentration decreased as follows (time is time in sec after interruption): 0 sec: 3 mg/L, 2 sec: 2. 5 mg/L, 4 sec: 2 mg/L, 8 sec: 1 mg/L, 12 sec 0. 2 mg/L. What is the a) lactate oxidation rate, b) the biomass productivity (mg biomass formed/L/h)? OUR = 0. 25 mg/L/s = 900 mg/L/h = 28. 1 mmol/L/h (MW = 32 mg/mmol)/ LUR 9. 38 mmol/L/h 0. 3 g X/ g Lac degraded Needed LUR in mg/L/h LUR (3*12 + 3* 16 +6= 90 mg/mmol)= 844. 5 mg/L/h Productivity = 844. 5 * 0. 3 = 253. 3 mg/L/h

A 20 L chemostat is operated with a flowrate of 0. 6 L/h. An equilibrium is established with a constant oxygen, concentration, p. H, biomass (2 g/L) and substrate concentration. What is the specific growth rate of the microbes in the chemostat and what is the biomass productivity R (g/L/h) of the chemostat? D= 0. 03 h-1 u = 0. 03 h-1 X= 2 g/L R = 0. 06 g. X/L/h In the absence of oxygen, many bacteria can use nitrate (NO 3 -) as electron acceptor and produce N 2 as the endproduct (nitrate respiration or denitrification). What rate of nitrate reduction to N 2 would you expect of a reactor that was switched from aerobic (aerated) conditions to nitrate reducing conditions, if the aerobic reactor had an oxygen uptake rate of 80 mg/L/h? NO 3 - N 2 requires 5 e- while O 2 H 2 O requires 4 e NUR= 4/5 OUR (molar) OUR= 80 mg/L/h / 32 mg/mmol = 2. 5 mmol/L/h NUR = 2 mmol/L/h Contrast batch culture against chemostat culture by pointing out advantages and limitations. Chem +: higher productivity, easier automation, ideal for study Chem-: not for secondary metabolites, prone to cont from outside and backmutations How can you calculate the productivity of a chemostat? Give 3 examples of how the productivity of a chemostat can be approximately doubled by the operator and one statement for each example how this works.

How can you calculate the productivity of a chemostat? Give 3 examples of how the productivity of a chemostat can be approximately doubled by the operator and one statement for each example how this works. R (g. X/L/h) = D (h-1) * X (g/L) Can be increased by operator by increasing either D or X D: Double flowrate X: Double SR X: Retain bacteria by recycle or filter to twice the concentration

Growth- Simplified Scheme of Energy preservation as ATP Biological growth requires ATP as the energy source (energy rich phosphate-phosphate bond). ATP is generated mostly during Respiration (Dissimilation) ATP then drives the biomass synthesis (Assimilation) How is it generated ? How much is generated ?

Energy preservation as ATP Four steps for aerobic ATP generation from glucose: 1) Glycolysis : sugar acetate (C 2)) 2) TCA cycle: acetate CO 2 + 4 NADH 3) NADH + O 2 NAD + proton gradient 4) Proton gradient runs a nano-scale “turbine” called ATP synthase

Growth- Overview of Energy Metabolism simplifying FAD and ATP genration in TCA CO 2 2 a cet glucolysis ate glucose TCA cycle Cell 2 NADH 8 NAD+ 8 NADH 1 A TP ATP synthase 3 H+ ETC O 2 2 NADH 1 NADH 9 H+ Overall: 36 ATP (+2) allowing growth

Growth- Simplified Scheme of Energy preservation as ATP Important Quantities: ATP-synthase: 3 H+ 1 ATP ETC: 1 NADH 3*3 = 9 H+ 1 NADH 3 ATP 2 NADH reduce 1 O 2 glycolysis: 1 glucose 12 NADH 1 glucose 12*9 = 108 H+ = 36 ATP + 2 ATP generated from glycolysis via substrate level phosphorylation = 38 ATP

Energy Source for Growth Electron flow: • is critical for the understanding of microbial product formation • allows to understand fermentations • the rate of electron flow determines the metabolic activity • Which direction? Thermodynamics • How powerful ? Thermodynamics • How rapid ? Kinetics • How many ? Stoichiometry, mass balance, fermentation balance

Growth- Simplified Scheme of Energy preservation as ATP How does ATP synthase work? A mechanical turbine that generates a energy rich phopspate bond driven by a proton gradient across the cell membrane See animated clip.

Energy Source for Growth oxidation Electron donor (Reductand) Electron Carrier reduction Electron acceptor (Oxidant) • Microbes catalyse redox reactions (electron transfer reactions) • A redox reaction oxidises one compound while reducing another compound • The electron flow represents the energy source for growth • An energy source must have an electron donor and electron acceptor Electron flow (arrows) electron donor to electron acceptor

Energy Source for Growth oxidation Electron donor (Reductand) Electron flow: • Which direction? Thermodynamics Electron Carrier reduction Electron acceptor (Oxidant) • How powerful ? Thermodynamics • How rapid ? Kinetics • How many ? Stoichiometry, mass balance, fermentation balance Electron flow (arrows) electron donor to electron acceptor

Energy Source for Growth • What are electron carriers? oxidation Electron donor • A redox couple that mediates between donor (Reductand) and acceptor • A redox couple consists of Electro the oxidised and the n reduced form (e. g. NADH reduction. Carrier and NAD+) • acts also as reducing Electron acceptor equivalents buffer (Oxidant) • What are suitable electron donors and acceptors? Electron flow (arrows) electron donor to electron acceptor

Growth- Simplified Scheme of Energy preservation as ATP What do electron carriers look like?

Working principle of electron carriers • What are electron carriers? • A redox couple that OH mediates between donor and acceptor • A redox couple consists of O the oxidised and the reduced form (e. g. NADH and NAD+) OH • electron buffer • What are suitable electron donors and acceptors? O Electron carriers exist as a couple

Working principle of electron carriers (EC) OH O OH • What is the most important difference between the two forms? • Different number of double bonds • OH instead of =O O Quinone and hydroquinone as central pieces of Ubiquinone

Working principle of electron carriers (EC) OH O • Which form carries electrons? • The reduced form! • Which is the reduced form? • The oxidation states will tell! • Which carbon atoms changed their oxidation state? Quinone and hydroquinone as central pieces of Ubiquinone

Working principle of electron carriers (EC) OH H H OH H O H H H • Which carbon atoms changed their oxidation state? • All carbons that have just one H bonded maintain OS of -1 • The top and bottom C have changed their OS. O Quinone and hydroquinone as central pieces of Ubiquinone

Working principle of electron carriers (EC) • Which carbon atoms changed their oxidation +1 state? H H • All carbons that have just O one H bonded maintain OS H H of -1 +2 +1 H H • The top and bottom C have OH changed their OS. • The reduced form carries H H +2 two more electrons than the O oxidised form • Where are they? Quinone and hydroquinone as central pieces of Ubiquinone OH

Working principle of electron carriers (EC) • Which carbon atoms changed their oxidation state? H H • All carbons that have just O one H bonded maintain OS H H of -1 +1 H H • The top and bottom C have OH changed their OS. • The reduced form carries H H +2 two more electrons than the O oxidised form • Where are they? Quinone and hydroquinone as central pieces of Ubiquinone OH

Working principle of electron carriers (EC) • How many electrons are carried ? • 2 H H • What else is carried? O • a proton H H • Together the electron and +1 H the proton make one H H OH • The reduced electron carrier can also be called a H H hydrogen carrier? +2 • Hydrogenation = adding O hydrogen or electrons to another compound = Quinone and hydroquinone reducing the compound as central pieces of Ubiquinone OH

Working principle of electron carriers (EC) • What can a reduced EC do? • Does a cell also need oxidised EC? OH H H +1 H OH H O H +2 H O Quinone and hydroquinone as central pieces of Ubiquinone

Working principle of electron carriers (EC) H • The electrons in NADH as the most importan. T electron carrier can also be visualised R H H -1 +1 H H H N R H H -2 R N 0 H R NADH/NAD+ as electron carrier • as N is more electronegative than C it is allocated the electrons of C-N bonds (similar to oxygen)

Main advantage of reducing power (NADH) aerobic conditions, NADH = ATP generation: NADH + H+ 0. 5 O 2 +3 ADP + 3 Pi NAD+ +3 ATP +4 H 2 O Respiration balance: combination of ETC and ATP synthase reaction How useful is NADH without O 2 ?

Consequences of O 2 depletion on cells Consequences of O 2 depletion: • No ATP generation • NAHD accumulates and NAD+ is depleted • TCA cycle (requiring NAD+) can’t run • glucose uptake stops NADH (or NADPH) can also be used for anabolism (assimilation) but in addition to reducing power also ATP is needed for assimilation Without O 2 NADH is a problem rather than advantage Anaerobic organisms have developed special metobolic pathways to reoxidise NADH (fermentations and anerobic respirations)

Energy Metabolism Scheme simplifying FAD and ATP genration in TCA CO 2 2 a cet glucolysis ate glucose TCA cycle Cell 2 NADH 8 NAD+ 8 NADH 1 A TP ATP synthase 3 H+ ETC O 2 2 NADH 1 NADH 9 H+ Overall: 36 ATP (+2) allowing growth

Electron flow in fermentations. Anaerobic fermentations (strict sense) make use of internal organic electron acceptors. The electron flow in anaerobic fermentations can be easily demonstrated by documenting the changes in carbon numbers and electron numbers. For example glucose (CH 2 O)6 contains 6 carbons with an oxidation state of zero (4 electrons/carbon). Glucose can be presented as 6 C, 24 e-

Lactic acid fermentation. Anaerobic fermentations (strict sense) make use of internal organic electron acceptors. The electron flow in anaerobic fermentations can be easily demonstrated by documenting the changes in carbon numbers and electron numbers. For example glucose (CH 2 O)6 contains 6 carbons with an oxidation state of zero (4 electrons/carbon). Glucose can be presented as 6 C, 24 e-

Lactic Fermentation - Electron and carbon flow 24 6 ATP 10 3 20 LDH 12 3 20 10 3 LDH lactate 12 3 24 6 = glucose (CH 2 O)6 2 0 = 2 red. equiv. 10 3 = pyruvate (CH 3 -CO-COOH) 12 3 = hydroxy propanoate =lactate (CH 3 -CHOH-COOH) LDH = Lactate dehydrogenase enzyme

Notes on origin of enzyme names With 2 electrons also 2 protons are transferred electron transfer= hydrogen transfer: Remove e-/H 2: Dehydrogenation = oxidation Add e-/H 2: Hydrogenation = reduction Pyruvate + 2 e- Lactate Possible names for the enzyme catalysing the equilibrium (forward and backward reaction): Lactate dehydrogenase Lactate oxidase Pyruvate hydrogenase Pyruvate reductase

Quizz: Glucose(6 carbons) is fermented to 2 lactate(CH 3 -CHOH-COOH) 123 If instead ethanol (CH 3 -CH 2 OH) 122 is the end product, how many can be formed? Carbon balance would suggest 3 (2 carbons)! Electron balance suggests 2 (12 electrons) Electrons are relevant, not carbon. If electrons are balanced any extra carbon must be in the form of CO 2.

Ethanolic Fermentation - Electron and carbon flow 24 6 glucose ATP 01 10 3 PDC 10 2 12 2 20 20 EDH ethanol 10 3 PDC 01 10 2 24 6 = glucose 12 2 Key enzymes: PDC = pyruvate decarboxylase EDH = Ethanol dehydrogenase 20 = 2 red. equiv. 10 3 = pyruvate 10 2 = acetaldehyde 12 2 = ethanol

The Entner Doudoroff (KDPG) pathway of ethanolic fermentation Orgainism: Zymonas mobilis 20 24 6 = glucose 22 6 24 2 = gluconate 12 3 = GAP 12 3 20 10 3 10 2 12 2 ATP 01 01 10 3 = pyruvate 01 = CO 2. 10 2 = acetaldehyde 12 2 = ethanol

Application of Lactic Fermentation - Silage: Lactic acid fermentation of fodder material Better preservation of food energy value than by drying (hay) Process: 1) 2) 3) 4) 5) Rapid filling of tank (silo)silo with shredded material Additves (germination inhibitors, sugars, p. H controlers) Packing densely and compressing Sealing air-tight Avoid contaminatin with decaying material (proteolytic anaerobes such as Clostridia

Silage does not necessarily need a tank: Examples of silage in Australia

Overview of Energy Metabolism simplifying FAD and ATP genration in TCA glucose TCA glucolysis cycle glucose 12 NADH + 2 ATP Keywords to look up: Electron carriers ETC Proton gradient NADH electron motive force Hydrogenation = Reduction 9 H+ Dehydrogenation = Oxidatioin Cell ATP synthase 3 H+ 1 ATP 38 ATP

Conclusion: In the absence of O 2 fermentations can be carried out that transfer electrons to internal (synthesised) electron acceptors instead of oxygen. Useful bioproducts can be obtained Ethanol, organic acids, H 2

Lec 5 Overview: Microbial metabolism without O 2 • Microbial growth is driven by the energy released from the transfer of electrons from donor (reductant, typically organic compounds) to acceptor (oxidant, typically oxygen. • The transfer occurs via mediators (electron carriers) • In the absence of oxygen microbes can ferment sugars by using internal organic mediators (e. g. puruvate, or acetaldehyde) resulting in fermentation products such as ethanol and lactic acid (hydroxy propnanoic acid) • The number of electrons available for reductions (reducing equivalents) on organic substances (including mediators) can be derived from the oxidation states of the carbons

Ethanolic Fermentation - Electron and carbon flow OH O. S. : -1 → 5 electrons H C H H O. S. : -3 → 7 electrons • Energy conserved: 2 ATP from glycolysis (PGK, PK) • Key enzymes: • Pyruvate Decarboxylase, • Ethanol Dehydrogenase (could also be called ethanol oxidase or acetaldehyde reductase)

The Entner Doudoroff (KDPG) pathway of ethanolic fermentation Organism: Zymonas mobilis (not examined) 24 6 = glucose 20 22 6 24 2 = gluconate 12 3 = GAP 12 3 20 10 3 10 2 12 2 ATP 01 01 10 3 = pyruvate 01 = CO 2. 10 2 = acetaldehyde 12 2 = ethanol

Special features of Entner Doudoroff pathway • 1 NADH, 1 NADPH • Only 1 ATP (less biomass as byproduct) • Only one pyruvate through GAP (bottleneck) → faster? Special features of Zymomoanas • Higher glucose tolerance • Higher product yield (less ATP → less biomass) (100 g ethanol / 250 g glucose) = 78% molar conv. eff • Not higher ethanol tolerance

Special features of Entner Doudoroff pathway (not examined) • 1 NADH, 1 NADPH • Only 1 ATP (less biomass as byproduct) • Only one pyruvate through GAP (bottleneck) → faster? Special features of Zymomoanas • Higher glucose tolerance • Higher product yield (less ATP → less biomass) (100 g ethanol / 250 g glucose) = 78% molar conv. eff • Not higher ethanol tolerance

Bio-ethanol from sugar cane as fuel (Brasil) • Distillation costs more energy than ethanol fuel value • Separation costs higher than fermentation costs Research (1990’s) • Thermophilic strains (Clostridium using cellulose) • Finding more ethanol resistant strains Controversial topic: Bioethanol from sugar (first generation bio-ethanol) has ethical problems. Current research: Bio-ethanol from cellulosic waste (straw, wood, paper) Requires enzymes. (e. g. Simultaneous saccharification/ fermentation)

Lactic Fermentation - Occurrence If plant or animal material containing sugars and complex nitrogen sources is left in the absence of oxygen → lactic acid bacteria take over Selective enrichment Natural fermentation (since prehistoric times) Why do lactic acid bacteria take over sugar conversion on rich media? : 1) Simple metabolism → fast degradation 2) Amino acids are not synthesized but taken up from the medium → faster growth 3) Strains are existing on substrate (e. g. milk, vegetables) 4) O 2 tolerance of strains 5) Production of inhibitory acid (ph <5) Examples: Milk, whole meal flour, vegetables,

Lactic Fermentation - Organisms Lactic acid bacteria (Lactobateriacease) • gram positive • non motile • obligate anaerobics • no spores • aerotolerant • no cytochromes and catalase • fermentation of lactose • no growth on minimal glucose media • requirement of nutritional supplements (vitamins, amino acids, etc. ) • when supplied with porphyrins → they form cytochromes !? ! (indicating that they were originally aerobic organisms that have lost the capacity of respiration, metabolic cripples)

Homolactic Fermentation - Electron and carbon flow 24 6 ATP 10 3 20 LDH 12 3 20 10 3 LDH lactate 12 3 24 6 = glucose LDH = lactate dehydrogenase 20 = 2 red. equiv. 10 3 = pyruvate 12 3 = lactate

Homo-lactic Fermentation - Electron and carbon flow O CH C O. S. : +3 → 1 electron H C H H O. S. : 0 → 4 electrons O. S. : -3 → 7 electrons Strategy: 1) Aerotolerant → can ferment with strict anaerobes are still inhibited by oxygen 2) Simple quick metabolism and usage of carbohydrates 3) Production of acid, inhibiting competitors

Significance: Why do lactic acid bacteria not spoil food but preserve it? • Only ferment sugars (24 e-) to lactate (2* 12 e-) nutritional value not significantly altered • Don’t degrade proteins • Don’t degrade fats • Acidity suppresses growth of food spoiling organisms (eg. Clostridia) • enhances nutritional value of organic material (example sauerkraut, Vit. C, scurvy) • Complex flavour development (diacetyl) • Examples: • Yogurt, sauerkraut, buttermilk, soy sauce, sour cream, cheese, pickled vegetables, • technical lactic acid for the production of bio-plastic (hydroxy acids allow chain linkages via ester bonds between hydroxy and carboxy group). •

Heterolactic Fermentation Phosphoketolase pathway 24 6 20 01 20 5 ATP 20 24 6 = glucose 20 5 = ribose 2 0 = 2 red. equiv. 10 3 = pyruvate 12 3 = lactate 10 3 12 3 20 82 12 2 = ethanol 12 2 Phosphoketolase pathway = combination of Pentosephosphate cycle and FBP pathway 8 2 =acetate 0 1 = CO 2.

Heterolactic Fermentation Phosphoketolase pathway 24 6 20 01 20 5 ATP 20 24 6 = glucose 20 5 = ribose 2 0 = 2 red. equiv. 10 3 = pyruvate 12 3 = lactate 10 3 12 3 20 82 12 2 = ethanol 8 2 =acetate 0 1 = CO 2. Presence of oxygen → lactate, acetate and CO 2 production → 1 additional ATP from acetokinase. No ETP

Heterolactic Fermentation Organisms: E. g. Leuconostoc spp. Lactobacillus brevis Strategy: • Use of parts of the pentose phosphate cycle which is designed for synthesis of pentose (DNA, RNA). → • Aerotolerant, simple pathway, quick metabolism, suited for substrate saturation. Application: Sourdough bread, Silage, Kefir, Sauerkraut, Gauda cheese (eyes) In the presence of oxygen, reducing equivalents from glucose oxidation are transferred to oxygen, allowing the gain of an additional ATP via acetate excretion Key enzymes of FBP pathway missing (Aldolase, Triosephosphate isomerase).

Application of Lactic Fermentation Silage: Lactic acid fermentation of fodder material Process: 1) partial drying of fodder 2) shredding 3) Rapid filling of silo (1 or 2 days) 4) packing as densely as possible 5) Compressing 6) Sealing airtight 7) Additives (germination inhibitors, sugars, organic acids) 8) Avoid contamination with decaying fodder (Clostridia, proteolytic bacteria) Nutrient loss: 1. drying of fodder hay (25%), 2. ensilaging (10%) (2 ATP out of 38)

Applications of Lactic Fermentation Sauerkraut In principle identical to silage with following modifications: 1) White cabbage as the only plant material 2) Cabbage mixed with Na. Cl (2 – 2. 5%) 3) Capacity of vessels (concrete, wood) up to 100 tons 4) Incubation (18 o. C to 20 o. C) for 4 weeks 5) Recirculation of brine by pumping for process monitoring (acids) 6) About 1. 5% lactic acid produced 7) Sterilisation of product to have cooked sauerkraut (German). Raw (fresh sauerkraut used in salads) 8) Problem: 1 to 15 tons of highly polluted effluent per ton of cabbage

Applications of Lactic Fermentation Similar to silage with following modifications: 1) White cabbage as the only plant material 2) Cabbage mixed with Na. Cl (2 – 2. 5%) 3) Capacity of vessels (concrete, wood) up to 100 tons 4) Incubation (18 o. C to 20 o. C) for 4 weeks 5) Recirculation of brine by pumping for process monitoring (acids) 6) About 1. 5% lactic acid produced 7) Sterilisation of product to have cooked sauerkraut (German). Raw (fresh sauerkraut used in salads) 8) Problem: 1 to 15 tons of highly polluted effluent per ton of cabbage Brine Recycle Sauerkraut

Brine Recycle Applications of Lactic Fermentation

Applications of Lactic Fermentation Olives 1) Black (ripe) or green (unripe) olives 2) Pretreatment with 1. 5% Na. OH saline (reducing bitterness) 3) Washing 4) Place fruit (still alcaline) in brime of 10% Na. Cl + 3% lactic acid (to neutralise p. H) 5) Sugar addition to accelerate fermentation (Lactobacillus plantarum) 6) Incubate for several months until lactic acid >0. 5% 7) Wooden barrels or plastic tanks

Pickled Gherkins 1. Cover gherkins in 3% salt brine (Na. Cl) 2. Add spices, herbs, dill 3. Irradiate surface (UV) and close vessel 4. After 3 – 6 weeks 3% lactic acid is produced 5. Fermentation pattern like silage

Applications of Lactic Fermentation Technical lactic acid Use: Leather – Textile – and Pharmaceutical Industry Bioplastics (Polylactic acid, biodegradable) Food acid (flavourless, non volatile) e. g. in sausages Product yield: 900 g per g of sugar Substrate: whey, cornsteep liquor, malt extract, ideally: sugars (15% cane or beets) Strains: Lactobacillus bulgaricus, Lactobacillus delbrueckii Duration: 5 days batch culture

Applications of Lactic Fermentation Sourdough bread Biological raising agent (homo- and heterolactic fermentation) CO 2 produced from heterolactic bacteria Necessary for rye bread to increase digestibility Health bread (lipid, proteins unchanged, vitamins produced) Pre-acidified (stomach friendly) Complex flavour development Increased shelf life

Cheese Production Add Rennet* Yougurt (430°) Heat treatment (600°) Kneading Quark Fromage frais (acidic paste) * Proteolytic enzyme ** Coagulating Brie *** Heated stirring Edamer Milk Homogenise Pasteurise Curdling** Stirring Settling Whey Pressuring Maturing Add starter culture (S. cremoris, S. lactis, L. bulgaricus, S. thermophilus Scolding*** Cooling Washing Salting Cottage cheese (granular) Cheddar

Propanoate Formation From Lactate 1. Acryloyl pathway (Clostridium propionicum) The 4 reducing equivalents from lactate oxidation to acetate are merely “dumped” onto two further moles of lactate (dismutation, disproportionation) 12 3 LDH 20 PDH 01 12 3 ATP 82 Pr. DH 20 14 3 12 3 Enzymes: Lactate DH, Pyruvate DH, Propionate DH (Pr. DH)

Propanoate Formation From Lactate 1. Acryloyl pathway (Clostridium propionicum) 12 3 LDH 20 PDH 01 12 3 ATP 82 Pr. DH Energetic benefit? The excretion of acetate gains 1 14 3 ATP (acetate kniase), 14 3 20 Thus 1/3 ATP/lactate metabolised. 12 3 How to generate ATP from acetate excretion Phosphate Acetyl transferase: Acetate~Co. A + Pi → Acetyl-P + Co. A Acetokinase: Acetyl-P + ADP → Acetate + ATP

Propanoate Formation From Lactate 2. Methyl-Malonyl-Pathway (Propionibacteria) • 2 reducing equivalents from lactate oxidation (exactly: PDH and ferredoxin as e- carrier) are transferred via electron transport phosphorylation to fumarate (fumarate respiration) resulting in one extra ATP (2/3 ATP/lactate metabolised). • Reverse TCA cycle. Fumarate reduction is an example of anaerobic respiration Homoacetogenesis is another example

Propanoate Formation From Lactate 2. Methyl-Malonyl-Pathway (Propionibacteria) 12 3 LDH 20 14 3 PDH 14 20 3 01 Vit B 12 Fd ATP 14 4 82 ETC 14 4 0 1 12 4 ATP 12 4 20 10 4 10 3 12 3 = lactate = propionate = succinate = fumarate (malate) 10 4 = OAA 10 3 = pyruvate

Propionic Fermentation of Glucose

Butyric Fermentation

Acetone Butanol fermentation

Homoacetogenesis The homoacetogenesis starts like the butyric acid fermentation: 1) Use of the fructose bisphosphate pathway (FBP) leading to 2 puruvate and 2 NADH. 2) Oxidative decarboxylation of pyruvate to acetyl-Co. A, hydrogen gas and CO 2. 3) In contrast to the butyric fermentation no acetoacetyl-Co. A is formed. Instead two acetyl-Co. A are intermediate products.

Homoacetogenesis

• Specific growth rate u in chemostat culture • Get the D from F/V • D=u

• E- acceptor from NADH in fermentations • For example acetaldehyde in ethanolic ferm

• Effect of growth constants on productivity R in a chemostat • R depends on X and D • Increased umax allows higher D • Increased Ymax gives higher X • Ms not much diff

• OUR is 64 mg/L/h= 2 mmol O 2/L/h • What is the acetone (CH 3 -CO-CH 3) oxidation rate to CO 2. • 16 e- means that 4 O 2 accept all el from acetone • Acetone ox rate is 0. 5 mmol/L/h

• OUR is 64 mg/L/h= 2 mmol O 2/L/h • What is the nitrate NO 3 - to N 2 reduction rate • NUR= 2 mmol/L/h * 4/5