1. A bioreactor with a kLa of 25 h-1 with active microbes is aerated resulting in a steady oxygen concentration of 1 mg/L. What is the microbial oxygen

1. A bioreactor with a kLa 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 kLa of the

chemostat in h-1 ?

b. What was the ethanol (CH3-CH2OH) to acetic acid (CH3-

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 CO2 of the following compounds

CH3-CH2-CH2OH HOOC-COOH CH3-CO-CH3

5. List (in the box next to the molecule) the number of moles of NAHD that can be generated from the complete oxidation to CO2 of the following compounds: Pentose (CH2O)5 CH3-COOH H2CO3

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 CO2 of ethanol (CH3-CH2OH):

NO3- N2 SO4

2- H2S Fe3+ Fe2+

7. Can microbes use the oxygen atom in the H2O 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 10L chemostat is operated with a flowrate of 0.6 L/h. An equilibrium is established with a constant oxygen, concentration, pH, 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 = 0.6L/h/ 10L = 0.06 h-1 Productivity R g/L/h can be calculated from X * D

3g/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.

X

S

D

Ste

ady

Sta

te C

once

ntra

tion

Effect of biomass feedback (here 3 fold):Dotted line no feedback:•Washout occuring early

•3-fold Feedback approximately:•3*X 3*R 1/3* S•allows 1/3 reactor size to do same work

•Feedback essential for pollutant removal. Can be used 100-fold 100-fold smaller treatment plant

•Note: same assumed feed concentration (SR)

R

Dcrit

SR

And 88

Effects of growth constants on steady state concentrationsof biomass and substrate in a chemostat as a function of dilution rate (x-axis)

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.

The two curves are described by two properties:

The maximum specific growth rate obtained with no substrate limitation (umax (h-1))

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 limitation of microbial growth

µ(h-1)

Substrate (g/L)

substratelimitation

kS

µmax(h-1)

Growth- Michaelis Menten modelAnd 1212

µ(h-1)

S(g/L)

The negative specific growth rate (µ) observed in the absence of substrate(when S = 0) (cells are starving, causing loss of biomass over time)

is the decay rate mS*Ymax

- mS*Ymax

0

Effect of Maintenance Coefficient (mS) on growth Rate

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) :

• EE = 4-OS

• Note: Electron equivalent=Reducing equivalent=(degree of reduction)

OS EE Example+4 0 CO2+3 1 -COOH+2 2 HCOOH, CO, -

CO-+1 3 -CHO 0 4 -CHOH--1 5 -CH2OH-2 6 -CH2-, CH3OH-3 7 -CH3-4 8 CH4

MSE 2011

1) A bioreactor with a kLa 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= 20h-1 *(8-2 mg/L) = 120 mg/L/h 2) The airflow to a chemostat running at steady state DO of 5 mg/L (cS 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 ?

kLA = 180 mg/L/g / (8-5 mg/L) = 60 h-1 3) What is the maximum possible rate (in mM/h) of lactate (CH3-CHOH-COOH) oxidation to CO2 by an aerobic reactor that is limited by an oxygen supply due to a kLa of 50 h-1 assuming an oxygen saturation concentration of 8 mg/L?

Lac = 12 e- 1 Lac reacts with 3 O2OUR = 50 h-1* 8mg/L = 400 mg/L/h = 25 mM/h LUR = 4.17 mM/h

MSE 2011

1) A bioreactor with a kLa 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= 20h-1 *(8-2 mg/L) = 120 mg/L/h 2) The airflow to a chemostat running at steady state DO of 5 mg/L (cS 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 ?

kLA = 180 mg/L/g / (8-5 mg/L) = 60 h-1 3) What is the maximum possible rate (in mM/h) of lactate (CH3-CHOH-COOH) oxidation to CO2 by an aerobic reactor that is limited by an oxygen supply due to a kLa of 50 h-1 assuming an oxygen saturation concentration of 8 mg/L?

Lac = 12 e- 1 Lac reacts with 3 O2OUR = 50 h-1* 8mg/L = 400 mg/L/h = 25 mM/h LUR = 4.17 mM/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: CH3-CH2-CH2OH 4.5 HOOC-COOH 0.5 CH3-CO-CH3 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 gX/gSumax gX/L/h / /gX/L = h-1ms gS/gX/h = h-1kS = gS/L

How much NADH can be produced from the complete oxidation to CO2 of the following compounds:CH3-CHOH-CH2-CH2OH 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 H2O 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 (CH3-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= 90mg/mmol)= 844.5 mg/L/h

Productivity = 844.5 * 0.3 = 253.3 mg/L/h

A 20L chemostat is operated with a flowrate of 0.6 L/h. An equilibrium is established with a constant oxygen, concentration, pH, biomass (2g/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 gX/L/h

In the absence of oxygen, many bacteria can use nitrate (NO3-) as electron acceptor and produce N2 as the endproduct (nitrate respiration or denitrification). What rate of nitrate reduction to N2 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?

NO3- N2 requires 5 e- while O2 H2O requires 4 e-NUR= 4/5 OUR (molar)

OUR= 80mg/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 studyChem-: 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 (gX/L/h) = D (h-1) * X (g/L)

Can be increased by operator by increasing either D or XD: Double flowrate X: Double SRX: Retain bacteria by recycle or filter to twice the concentration

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 ?

Growth- Simplified Scheme of Energy preservation as ATP

Four steps for aerobic ATP generation from glucose:

1) Glycolysis : sugar acetate (C2))

2) TCA cycle: acetate CO2 + 4 NADH

3) NADH + O2 NAD + proton gradient

4) Proton gradient runs a nano-scale “turbine” called ATP

synthase

Energy preservation as ATP

glucose

TCA cycle

ETC

1 ATP 3 H+

glucolysis

8 NADH

1NADH 9 H+

Overall:36 ATP (+2)

allowing growth

Cell

O2

Growth- Overview of Energy Metabolismsimplifying FAD and ATP genration in TCA

CO2

2 NADH2

acet

ate

ATPsynthase

2 NADH

8 NAD+

Important Quantities:

ATP-synthase: 3H+ 1 ATP

ETC: 1 NADH 3*3 = 9 H+

2 NADH reduce 1 O2

glycolysis: 1 glucose 12 NADH

1 glucose 12*9 = 108 H+ = 36 ATP

+ 2 ATP generated from glycolysis via substrate level

phosphorylation = 38 ATP


1NADH 3 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

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.



• 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 donor(Reductand)

oxidation

reduction

ElectronCarrier

Electron acceptor(Oxidant)

Electron flow (arrows) electron donor to electron acceptor


Electron flow:

•Which direction? Thermodynamics

•How powerful ? Thermodynamics

•How rapid ? Kinetics

•How many ? Stoichiometry, mass balance, fermentation balance


oxidation

reduction

ElectronCarrier




• What are electron carriers?• A redox couple that

mediates between donor and acceptor

• A redox couple consists of the oxidised and the reduced form (e.g. NADH and NAD+)

• acts also as reducing equivalents buffer

• What are suitable electron donors and acceptors?


oxidation

reduction

ElectronCarrier



What do electron carriers look like?


Working principle of electron carriers




• electron buffer• What are suitable electron

donors and acceptors?Electron carriers exist asa couple

OH

OH

O

O

Working principle of electron carriers




• electron buffer• What are suitable electron

donors and acceptors?Electron carriers exist asa couple

OH

OH

O

O

Working principle of electron carriers (EC)

• What is the most important difference between the two forms?

• Different number of double bonds

• OH instead of =O

Quinone and hydroquinoneas central pieces of Ubiquinone

OH

OH

O

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?


OH

OH

O

O


• 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.


OH

OH

O

O

H

H

H

H

H

HH

H





• The reduced form carries two more electrons than the oxidised form

• Where are they?Quinone and hydroquinoneas central pieces of Ubiquinone

OH

OH

O

O

H

H

H

H

H

HH

H

+2

+2

+1

+1





• The reduced form carries two more electrons than the oxidised form

• Where are they?Quinone and hydroquinoneas central pieces of Ubiquinone

OH

OH

O

O

H

H

H

H

H

HH

H

+2

+1


• How many electrons are carried ?

• 2• What else is carried?• a proton• Together the electron and

the proton make one H• The reduced electron carrier

can also be called a hydrogen carrier?

• Hydrogenation = adding hydrogen or electrons to another compound = reducing the compoundQuinone and hydroquinone

as central pieces of Ubiquinone

OH

OH

O

O

H

H

H

H

H

HH

H

+2

+1


• What can a reduced EC do?• Does a cell also need

oxidised EC?


OH

OH

O

O

H

H

H

H

H

HH

H

+2

+1


• The electrons in NADH as the most importanT electron carrier can also be visualised

• as N is more electronegative than C it is allocated the electrons of C-N bonds (similar to oxygen)

NADH/NAD+ as electroncarrier

H

RH

R

H

R

H

H

R

HH

H

-2

-1

H

N

N

+1

0

Main advantage of reducing power (NADH)

aerobic conditions, NADH = ATP generation:

NADH + H+ 0.5O2 +3 ADP + 3Pi NAD+ +3 ATP +4 H2O

Respiration balance: combination of ETC and ATP synthase reaction

How useful is NADH without O2 ?

Consequences of O2 depletion on cells

Consequences of O2 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 O2 NADH is a problem rather than advantage

Anaerobic organisms have developed special metobolic pathways to re-oxidise NADH (fermentations and anerobic respirations)

glucose

TCA cycle

ETC

1 ATP 3 H+

glucolysis

8 NADH

1NADH 9 H+

Overall:36 ATP (+2)

allowing growth

Cell

O2

Energy Metabolism Schemesimplifying FAD and ATP genration in TCA

CO2

2 NADH2

acet

ate

ATPsynthase

2 NADH

8 NAD+

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 (CH2O)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 (CH2O)6 contains 6 carbons with an oxidation state of zero (4 electrons/carbon).

Glucose can be presented as 6 C, 24 e-

103 20 103

123 123

20LDH LDH

lactate

ATPATP

246

24620

103123

= glucose (CH2O)6

= 2 red. equiv.

= pyruvate (CH3-CO-COOH)

= hydroxy propanoate =lactate (CH3-CHOH-COOH)

Lactic Fermentation- Electron and carbon flow -

LDH = Lactate dehydrogenase enzyme

Notes on origin of enzyme names

With 2 electrons also 2 protons are transferred electron transfer= hydrogen transfer:

Remove e-/H2: Dehydrogenation = oxidationAdd e-/H2: Hydrogenation = reduction

Pyruvate + 2e- LactatePossible names for the enzyme catalysing the equilibrium (forward and backward reaction):

Lactate dehydrogenaseLactate oxidasePyruvate hydrogenasePyruvate reductase

Quizz: Glucose(6 carbons) is fermented to

2 lactate(CH3-CHOH-COOH) 123

If instead ethanol (CH3-CH2OH) 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 CO2.

103 20 103

102 102

122 122

20

0101PDCPDC

EDH EDH

ethanol

ATPATP

246 glucose

246

20

103

102

122

= glucose

= 2 red. equiv.

= pyruvate

= acetaldehyde

= ethanol

Ethanolic Fermentation- Electron and carbon flow -

Key enzymes:PDC = pyruvate decarboxylaseEDH = Ethanol dehydrogenase

226

20

102

103 103

122

12320

0101

ATP

246 246

01

103

102

122

= glucose

= CO2.

= pyruvate

= acetaldehyde

= ethanol

102

122

242 = gluconate

123 = GAP

The Entner Doudoroff (KDPG) pathway of ethanolic fermentationOrgainism: Zymonas mobilis

Application of Lactic Fermentation- Silage -

Silage: Lactic acid fermentation of fodder materialBetter preservation of food energy value than by drying (hay)

Process:

1) Rapid filling of tank (silo)silo with shredded material2) Additves (germination inhibitors, sugars, pH controlers)3) Packing densely and compressing4) Sealing air-tight5) 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 Metabolismsimplifying FAD and ATP genration in TCA

glucose

TCA cycle

ETC

ATPsynthase3H+ 1ATP

glucolysisglucose 12 NADH + 2 ATP

NADH 9 H+

38 ATP

Keywords to look up:Electron carriersProton gradientelectron motive forceHydrogenation = ReductionDehydrogenation = Oxidatioin

Cell

Conclusion:

In the absence of O2 fermentations can be carried out that transfer electrons to internal (synthesised) electron acceptors instead of oxygen.

Useful bioproducts can be obtainedEthanol, organic acids, H2

Lec 5 Overview: Microbial metabolism without O2

• 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

OH

H C H

H C H

H

Ethanolic Fermentation- Electron and carbon flow -

• Energy conserved: 2 ATP from glycolysis (PGK, PK)• Key enzymes: •Pyruvate Decarboxylase, •Ethanol Dehydrogenase

(could also be called ethanol oxidase or acetaldehyde reductase)

O.S.: -1 → 5 electrons


226

20

102

103 103

122

12320

0101

ATP

246 246

01

103

102

122

= glucose

= CO2.

= pyruvate

= acetaldehyde

= ethanol

102

122

242 = gluconate

123 = GAP

The Entner Doudoroff (KDPG) pathway of ethanolic fermentationOrganism: Zymonas mobilis(not examined)

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) hasethical 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 degradation2) Amino acids are not synthesized but taken up from the medium →

faster growth 3) Strains are existing on substrate (e.g. milk, vegetables)4) O2 tolerance of strains5) 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)

103 20 103

123 123

20LDH LDH

lactate

ATPATP

246

246

20

103123

= glucose

= 2 red. equiv.

= pyruvate

= lactate

Homolactic Fermentation- Electron and carbon flow -

LDH = lactate dehydrogenase

O CH

C

H C H

H C H

H

Homo-lactic Fermentation- Electron and carbon flow -

O.S.: 0 → 4 electrons


O.S.: +3 → 1 electron

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). •

205

20

103

123

20

ATP

246 246

01

103

122

= glucose

= CO2.

= pyruvate

=acetate

= ethanol

122

205= ribose

123= lactate

8220

01 20= 2 red. equiv.

82

Heterolactic FermentationPhosphoketolase pathway

Phosphoketolase pathway = combination of Pentosephosphate cycle and FBP pathway

205

20

103

123

20

ATP

246 246

01

103

122

= glucose

= CO2.

= pyruvate

=acetate

= ethanol

122

205= ribose

123= lactate

8220

01 20= 2 red. equiv.

82

Heterolactic FermentationPhosphoketolase pathway

Presence of oxygen → lactate, acetate and CO2 production → 1 additional ATP from acetokinase. No ETP

Heterolactic FermentationOrganisms: 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 FermentationSilage: Lactic acid fermentation of fodder materialProcess:

1) partial drying of fodder2) shredding3) Rapid filling of silo (1 or 2 days)4) packing as densely as possible5) Compressing6) Sealing airtight7) 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%) (2ATP out of 38)

Applications of Lactic FermentationSauerkraut

In principle identical to silage with following modifications:

1) White cabbage as the only plant material2) Cabbage mixed with NaCl (2 – 2.5%)3) Capacity of vessels (concrete, wood) up to 100 tons4) Incubation (18oC to 20oC) for 4 weeks5) Recirculation of brine by pumping for process monitoring (acids)6) About 1.5% lactic acid produced7) 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 FermentationSauerkraut

Similar to silage with following modifications:1) White cabbage as the only plant material2) Cabbage mixed with NaCl (2 – 2.5%)3) Capacity of vessels (concrete, wood) up to

100 tons4) Incubation (18oC to 20oC) for 4 weeks5) Recirculation of brine by pumping for

process monitoring (acids)6) About 1.5% lactic acid produced7) 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

Brin

e R

ecyc

le

Brin

e R

ecyc

le

Applications of Lactic Fermentation

Applications of Lactic FermentationOlives

1) Black (ripe) or green (unripe) olives

2) Pretreatment with 1.5% NaOH saline (reducing bitterness)

3) Washing

4) Place fruit (still alcaline) in brime of 10% NaCl + 3% lactic acid (to neutralise pH)

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 (NaCl)2. Add spices, herbs, dill3. Irradiate surface (UV) and close vessel4. After 3 – 6 weeks 3% lactic acid is produced5. Fermentation pattern like silage

Applications of Lactic FermentationTechnical 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)

CO2 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 Milk

HomogenisePasteurise

Add Rennet*Add starter culture(S. cremoris, S. lactis,L. bulgaricus,S. thermophilusYougurt (430°)

Curdling**StirringSettling

Heat treatment(600°)Kneading Whey

Scolding***CoolingWashingSalting

WheyQuarkFromage frais(acidic paste)

Cottage cheese(granular)

PressuringMaturing

BrieEdamer

Cheddar* Proteolytic enzyme** Coagulating*** Heated stirring

20

123

20ATP

143

123 123

143

82

01

LDH

PDH

Propanoate Formation From Lactate1. Acryloyl pathway (Clostridium propionicum)

The 4 reducing equivalents from lactate oxidation to acetateare merely “dumped” onto two further moles of lactate(dismutation, disproportionation)

Enzymes: Lactate DH, Pyruvate DH, Propionate DH (PrDH)

PrDH

20

123

20ATP

143

123 123

143

82

01

LDH

PDH

Propanoate Formation From Lactate1. Acryloyl pathway (Clostridium propionicum)

PrDH

Energetic benefit?

The excretion of acetate gains 1 ATP (acetate kniase),

Thus 1/3 ATP/lactate metabolised.

How to generate ATP from acetate excretionPhosphate Acetyl transferase:Acetate~CoA + Pi → Acetyl-P + CoAAcetokinase:Acetyl-P + ADP → Acetate + ATP

Propanoate Formation From Lactate

2. Methyl-Malonyl-Pathway (Propionibacteria)

• 2 reducing equivalents from lactate oxidation (exactly: PDHand ferredoxin as e- carrier) are transferred via electrontransport 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 respirationHomoacetogenesis is another example

20

144

20ATP

143

123 123

143

82

01

LDH

PDH

124

104

103

123

ATP

FdETC

Vit B12

01

20

123

103

104

= lactate

= pyruvate

= OAA

143= propionate

124 = fumarate (malate)

144= succinate

Propanoate Formation From Lactate2. Methyl-Malonyl-Pathway (Propionibacteria)

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-CoA, hydrogen gas and CO2.

3) In contrast to the butyric fermentation no acetoacetyl-CoA is formed. Instead two acetyl-CoA 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 O2/L/h• What is the acetone (CH3-CO-CH3)

oxidation rate to CO2.• 16 e- means that 4 O2 accept all el from

acetone• Acetone ox rate is 0.5 mmol/L/h

• OUR is 64 mg/L/h= 2 mmol O2/L/h• What is the nitrate NO3- to N2 reduction

rate• NUR= 2 mmol/L/h * 4/5

Documents

1. A bioreactor with a kLa of 25 h-1 with active microbes is aerated resulting in a steady oxygen concentration of 1 mg/L. What is the microbial oxygen