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 (CH3CH2OH) to acetic acid (CH3COOH) 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 SO42-  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.
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  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
kS
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 (mS) 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 mS*Ymax
- mS*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
CO2
+3
1
-COOH
+2
2
HCOOH, CO, -CO-
• EE = 4-OS
+1
3
-CHO
0
4
-CHOH-
-1
5
-CH2OH
-2
6
-CH2-, CH3OH
-3
7
-CH3
-4
8
CH4
• Note:
Electron equivalent=
Reducing equivalent=
(degree of reduction)
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 O2
OUR = 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 O2
OUR = 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/gS
umax gX/L/h / /gX/L = h-1
ms gS/gX/h = h-1
kS = 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 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 (gX/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 (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
Growth- Overview of Energy Metabolism
simplifying FAD and ATP genration in TCA
CO2
glucose
glucolysis
TCA
cycle
Cell
2 NADH 8 NAD+
8 NADH
ATP
synthase
ETC
O2
2 NADH
1NADH  9 H+
Overall:
36 ATP (+2)
allowing growth
Growth- Simplified Scheme of Energy
preservation as ATP
Important Quantities:
ATP-synthase: 3H+  1 ATP
ETC: 1 NADH  3*3 = 9 H+
1NADH  3 ATP
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
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
reductionCarrier
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
OH
O
OH
• 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+)
• electron buffer
• What are suitable electron
donors and acceptors?
O
Electron carriers exist as
a couple
Working principle of electron carriers
OH
O
OH
• 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+)
• 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
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
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
oxidised form
O
• 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
oxidised form
O
• 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
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
-1
• The electrons in NADH as
the most importanT
electron carrier can also
be visualised
R
H
+1
H
H
H H
N
R
H
-2
R
0
H
N
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+ 0.5O2 +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 reoxidise NADH (fermentations and anerobic respirations)
Energy Metabolism Scheme
simplifying FAD and ATP genration in TCA
CO2
glucose
glucolysis
TCA
cycle
Cell
2 NADH 8 NAD+
8 NADH
ATP
synthase
ETC
O2
2 NADH
1NADH  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 (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-
Lactic Fermentation
- Electron and carbon flow 24 6
ATP
ATP
10 3
20
LDH
12 3
20
10 3
LDH
lactate
12 3
24 6 = glucose (CH2O)6
2 0 = 2 red. equiv.
10 3 = pyruvate (CH3-CO-COOH)
12 3 = hydroxy propanoate =lactate (CH3-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-/H2: Dehydrogenation = oxidation
Add
e-/H2: Hydrogenation = reduction
Pyruvate + 2e-  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(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.
Ethanolic Fermentation
- Electron and carbon flow 24 6 glucose
ATP
ATP
01
10 3
PDC
20
20
10 2
01
10 2
EDH
12 2
10 3
PDC
EDH
ethanol
24 6 = glucose
12 2
20
= 2 red. equiv.
10 3 = pyruvate
Key enzymes:
PDC = pyruvate decarboxylase
EDH = Ethanol dehydrogenase
10 2 = acetaldehyde
12 2 = ethanol
The Entner Doudoroff (KDPG) pathway of ethanolic fermentation
Orgainism: Zymonas mobilis
20
24 6
24 6 = glucose
22 6
24 2 = gluconate
12 3 = GAP
12 3
20
10 3
10 3
10 2
10 2
12 2
12 2
10 3 = pyruvate
ATP
01
01
01
= CO2.
10 2 = acetaldehyde
12 2
= ethanol
Application of Lactic Fermentation
- Silage -
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, pH 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
3H+  1ATP
38 ATP
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 obtained
Ethanol, 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
Ethanolic Fermentation
- Electron and carbon flow OH
O.S.: -1 → 5 electrons
H
C
H
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
24 6 = glucose
20
22 6
24 2 = gluconate
12 3 = GAP
12 3
20
10 3
10 3
10 2
10 2
12 2
12 2
10 3 = pyruvate
ATP
01
01
01
= CO2.
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) O2 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
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
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
20
01
20 5
ATP
24 6 = glucose
20 5 = ribose
2 0 = 2 red. equiv.
10 3 = pyruvate
12 3 = lactate
10 3
20
82
12 2 = ethanol
8 2 =acetate
12 3
12 2
Phosphoketolase pathway = combination of
Pentosephosphate cycle and FBP pathway
0 1 = CO2.
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
20
82
12 2 = ethanol
8 2 =acetate
12 3
12 2
0 1 = CO2.
Presence of oxygen → lactate, acetate and CO2 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%) (2ATP 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 NaCl (2 – 2.5%)
3) Capacity of vessels (concrete, wood) up to 100 tons
4) Incubation (18oC to 20oC) 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 NaCl (2 – 2.5%)
3) Capacity of vessels (concrete, wood) up to
100 tons
4) Incubation (18oC to 20oC) 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% 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, 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)
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
Homogenise
Pasteurise
Add Rennet*
Yougurt (430°)
Heat treatment
(600°)
Kneading
Quark
Fromage frais
(acidic paste)
* Proteolytic enzyme
** Coagulating
Brie
*** Heated stirring
Edamer
Curdling**
Stirring
Settling
Whey
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
12 3
20
PDH
PrDH
20
01
14 3
14 3
ATP
82
12 3
Enzymes: Lactate DH, Pyruvate DH, Propionate DH (PrDH)
Propanoate Formation From Lactate
1. Acryloyl pathway (Clostridium propionicum)
12 3
LDH
12 3
20
PDH
PrDH
The excretion of acetate gains 1
14 3 ATP (acetate kniase),
14 3
20
01
Energetic benefit?
Thus 1/3 ATP/lactate metabolised.
ATP
82
12 3
How to generate ATP from acetate excretion
Phosphate Acetyl transferase:
Acetate~CoA + Pi → Acetyl-P + CoA
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
12 3
12 3
LDH
20
14 3
14 3
PDH
14
20
3
01
Vit B12
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
Propionic Fermentation of Glucose
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