Transcript T05 oxs med 2013c

Electron Flow in Biological proceses
This topic is covered largely by the computer learning
activity - Oxidation states (Oxs). Slides in this section are
mostly complimentory
Successful study of this topic enables the student to
understand oxidation and reduction half reactions and to
stoichiometrically match electron donors to electron
acceptors in redox reactions.
Example: How much oxygen is needed for the complete
oxidation of 1 mole of glucose to CO2?
Electron carriers
Why is focusing on Electron Flow Important?
Energy is conserved (ATP) from redox reactions
Redox reactions are characterised by a flow of electrons from
e- doner to e- acceptor.
Thus energy metabolism = electron metabolism
The electron flow characterises the bioprocess:
endproducts formed
oxygen requirements
ATP producuced  biomass yield formed
Exception: hydrolysis reactions (e.g. cellulose to sugar)
Overview of Energy Metabolism
simplifying FAD and ATP genration in TCA
glucose
glucolysis
TCA
cycle
glucose 
12 NADH + 2 ATP
Keywords to look up:
Electron carriers
Proton gradient
electron motive force
Hydrogenation = Reduction
Dehydrogenation = Oxidatioin
Cell
ATP
synthase
3H+  1ATP
ETC
NADH 
9 H+
38 ATP
1. Electron Flow: Which direction? Thermodynamics
e- donor + e- acceptor  oxidized donor + reduced acceptor
e.g.
sugar
+
oxygen 
CO2
H2O
for other reactions it is less obvious what is e- donor and
acceptor (e.g. sugar  CO2 + ethanol)
2. How powerful is the transfer? Thermodynamics
Driving force of the reaction – Is it downhill (spontaneous
important to estimate the heat generated and
the potential biomass produced
3. How rapidly is the electron flow? Kinetics
 Substrate oxidation rate  Product formation rate 
 Growth rate  Productivity (e.g. umax, kS)
Interesting demonstration of electron flow Microbial Fuel Cell:
Driving force (voltage) * Speed (amps)  Power (watts)
4. How many electrons?
Stoichiometry
depends on number of molecules reacting and
number of electrons transferred per molecule
allows to quantify the substrates and products of the reaction
allows to establish the fermentation balance for :
electrons
carbon
oxygen
hydrogen
charges
Initially it requires to learn to assign oxidation states to atoms
Example of the use of electron balance to examine
bioprocesses:
Case: Aerobic degradation of organic industrial wastewater
Claim: 80% of organic waste degraded to CO2. Clean water
and CO2 are the only endproducts
Evidence: Only 20% of pollutants left in outflow
Investigation showed: Aeration
capacity (kLa) allowed only 25%
degradation of waste
Question raised: Could it be that
bacteria used the oxygen in the
water molecule to oxidise the
organic pollutants?
Organic
Waste
•
•
•
• • • •
•
•
•
CO2
Clean
Water
O2
Final result: Bacterial biomass (25%) plus flocculated
waste (30%) accumulating in the reactor accounted for the
missing organic carbon.
Conclusion: Electron balance was essential in
developing a clear understanding of the process
Calculation of oxidation state and available electrons
Methanol CH3OH
1. Non charged molecule  Sum of oxidation states is zero
2. Oxygen state of ligands: +IV-II = +II
3.  O.S. of C must be -II
The number of electrons must be calculated by considering
that a carbon of O.S. +IV (CO2) has zero electrons
available. Every more negative O.S. carries a corresponding
number of electrons (+III has 1 e-, -III has 7 e-).
4.  C in methanol (-II): 6 electrons available
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
MSE 2011
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 kLa 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
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 = 12.5 mM/h 
LUR = 4.17 mM/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
NUR= 4/5 OUR (molar)
OUR= 80mg/L/h
2 mmol/L/h
requires 4 e-
/ 32 mg/mmol = 2.5 mmol/L/h  NUR =
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.
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.
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 4.5
CH3-CO-CH3
4
HOOC-COOH 0.5
List the four growth constants with their units. State in
one short sentence what this growth constant means
by referring to its units.
H
C
H
C
CH
H
O
H
O
O
C
C
H
H
H
H
C
H
H
H
C
H
O
C
O H
O H
C
H
O O
Stoichiometric equations
Correct Stoichiometry of biochemical Redox Reactions:
How much ethanol acid can be formed from the fermentation of 1
mole of glucose?
(CH2O)6
 x CH3-CH2OH
1. 24 e-
 12 e- (2 ethanol will be produced)
(CH2O)6
 2 CH3-CH2OH
2. 6C
 4 C (2 bicarbonate are produced)
(CH2O)6
 2 CH3-CH2OH + 2 HCO3-
3. 6O
 8O (2 water are consumed)
(can’t be 3 ethanol !)
(CH2O)6 + 2 H2O  2 CH3-CH2OH + 2 HCO34. 16H
 14H (2 protons are produced)
(CH2O)6 + 2 H2O  2 CH3-CH2OH + 2 HCO3- + 2H+
5. By balancing hydrogen, the charges should be also balanced
 The reaction will decrease the pH (however some acidity will
escape as CO2 (as HCO3- + H+  CO2 + H2O)
Background carbonate equlibrium
The Carbonate equilibrium:
Essential Inorganic Chemistry Background
CO2 in water:
acidic condtions
CO2 + H2O  H2CO3
at neutral pH (taking protons away):
H2CO3

HCO3- + H+
at basic pH (taking more protons away):
HCO3-

CO3-- + H+
Overall: the production of CO2 has the capacity to supply protons.
For neutral pH use HCO3- rather than CO2.
Alternatively the number of electrons available can be
determined from the number of reduced bonds (C-C, C-H).
Each reduced bond corresponds to 2 electrons available.
Because of its high electro negativity oxygen atoms “posses the electron
couple bonding to the carbon atom.
OH
H
C
H
H
Use the oxidation state of compounds when the molecule is
very large or the structure formula is not known.
1. Glucose formula: (CH2O)6
2. Oxidation state of H and O atoms: 0
3. O.S. of each C atom: 0
4. Electrons available per C atom: 4
5. Total number of electrons available 24
6. Electrons required to reduce O2 (to H2O): 4
7. The oxidation of one glucose to 6 CO2 requires (24/4) 6 O2
Since the electrons available per molecule is critical to know
for fermentation balances organic molecules can be
characterised by the number of electrons and carbons. For
glucose this is 24 6.
The oxidation of glucose to gluconic acid removes 2
electrons. Thus gluconic acid C6H12O7 can be
characterised as 22 6.
Gluconic acid, C6H12O7
1. Temporarily assign 4 electrons to each carbon by
neglecting all ligands (this would be correct if all bonds
were C-C (e.g. or when total O.S. of all ligands were zero)
2. Calculate O.S. of all ligands: -14+12= -2
3. Add number of O.S. of ligands to number of electrons
assigned under 1): 24+ (-2)= 22
Succinate, COOH-CH2-CH2-CHOOH, C4H6 O4
1. 4*4 = 16 electr., 2. O.S. ligands: -8+6= -2,
3. Assigned electr. (16)+ O.S. of ligands: (-2) = 14
Fumarate, COOH-CH=CH-COOH, C4H4O4
1. 4*4 = 16 electr., 2. O.S. ligands: -8+4= -4,
3. 16+(-4) = 12
Malate,
COOH-CH2OH-CH2OH-COOH, C4H8O6
1. 4*4 = 16 electr., 2. O.S. ligands: -12+8= -4,
3. 16+(-4) = 12
Ethanol,
CH3-CH2OH, C2H6O
1. 2*4 = 8 electr., 2. O.S. ligands: -2+6= +4,
3. 8 + 4 = 12
Glucose
HCO
HOCH
HOCH
HOCH
HOCH
HOCH
H
Gluconate
OOCH
HOCH
HOCH
HOCH
HOCH
HOCH
H
Rote learning approach to linking electron
numbers (reducing equivalents) with
H
oxidation -2
states of organics
0
-1
+1
NADH/NAD+ as electron
carrier
• The electrons in
NADH as the most
importan electron
carrier can also be
visualised
How do electrons get from donor to acceptor?
By electron carriers
Examples: NAHD, CoQ, FADH, Ferredoxin
Electron carriers are present in the cell in its reduced and
oxidised form. The ratio of reduced to oxidised e- carrier
reflects the energy situation (“starving” to “overfed”)
Reduced e- carriers are also called reducing equivalents.
Definition of reducing equivalents:
1 reducing equivalent =
1 electron or one electron equivalent in form of a hydrogen
atom
Significance of reducing equivalents for the microbial cell.
Advantage or disadvantage?
Reducing equivalents must be produced and consumed
during microbial metabolism.
Consumption is by using other compounds as electron
acceptors.
Electron acceptors can be either:
providing the conservation of ATP (electron transport
phosphorylation, respiration)
or just a dumping ground (fermentations using internal
electron acceptors
Electron carriers
Energy Source for Growth
• Microbes catalyse redox
Electron donor
reactions (electron transfer
(Reductand)
reactions)
• A redox reaction oxidises
one compound while
Electron
Carrier
reducing another compound
reduction
• The electron flow represents
the energy source for
Electron acceptor
growth
(Oxidant)
• An energy source must
have an electron donor and
Electron flow (arrows) electron electron acceptor
oxidation
donor to electron acceptor
Electron carriers
Energy Source for Growth
oxidation
Electron donor
(Reductand)
Electron
Carrier
reduction
Electron acceptor
(Oxidant)
• Electron flow :
• is critical for the
understanding of microbial
product formation
• allows to understand
fermentations
• the rate of electron flow
determines the metabolic
activity
Electron flow (arrows) electron
donor to electron acceptor
Electron carriers
Energy Source for Growth
oxidation
Electron donor
(Reductand)
Electron
Carrier
reduction
Electron acceptor
(Oxidant)
• Electron flow:
• Which direction? 
Thermodynamics
• How powerful ?
Thermodynamics
• How rapid ?  Kinetics
• How many ? 
Stoichiometry, mass
balance, fermentation
balance
Electron flow (arrows) electron
donor to electron acceptor
Electron carriers
What is a redox reaction?
The most significant biological reactions relevant to bioprocesses are “redox reactions”. Redox reactions are characterised
by an electron transfer from an electron donor to an electron acceptor. The electron donor is oxidised by loosing electrons
while the electron acceptor is reduced when it receives the electrons. As often together with the electrons also a proton is
transferred, biochemists refer to an oxidation as dehydrogenation (loss of e- and H+ meaning loss of H) and to reduction as
hydrogenation. Consequently enzyme names can be confusingly called dehydrogenases and hydrogenases instead of
oxidases and reductases, respectively.
Strictly speaking an oxidation by itself will not occur as it is only an electrochemical half reaction. Half reactions are
characterised by either showing electrons as a reactant or a product (e.g. Fe 2+  Fe3+ + e-). The reaction can only exist in
the real world if it is coupled with a suitable opposing half reaction: The oxidising half reaction needs to be coupled with a
reducing half reaction to become a full redox reaction.
How can we decide whether a compound is an electron donor or acceptor?
Depending on the half reaction it is coupled with, many compounds can be an electron donor in one moment and an electron
acceptor at a different moment. For example pyruvate can accept electrons to be reduced to lactate (lactate dehyrogenase
in lactic fermentation) or it can release electrons (and CO2) to be oxidised to acetate. To know whether a compound is likely
to accept or give electrons one would need to know the other player in the overall redox reaction. If it is a strong oxidising
agent (tending pull electrons away from other compounds) it will turn the first compound into an electron donor and if it is a
strong reducing agent it may force electrons onto the first compound turning it into an electron acceptor. It can be exactly
prodicted under which conditions a compound will be an acceptor or donor of electrons. This area of chemistry is
thermodynamics. The redox potential and the Gibbs Free Energy change (Delta G) are the useful measures to predict what
will happen in the reaction.
Enzyme names.
Enzyme names can be confusing. It is not easy to give enzymes a proper name that describes its metabolic activity. This is
because enzymes typically catalyse both the forward and the backward reaction. For example the enzyme that catalyses the
reduction of pyruvate to lactate could either be called pyruvate reductase or lactate oxidase. As a proton is transferred
together with the electron, it is actually called lactate dehydrogenase instead of pyruvate reductase. By realising the
reversibility of enzyme reactions and that hydrogenases are reductases it should not be too difficult to derive the role of
certain enzymes from their names.
Organic acids dissociation
Lactic acid  Lactate- + H+
Electron carriers
Energy Source for Growth
oxidation
Electron donor
(Reductand)
Electron
Carrier
reduction
Electron acceptor
(Oxidant)
• 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?
Electron flow (arrows) electron
donor to electron acceptor
Electron carriers
Working principle of electron carriers
OH
O
OH
O
Electron carriers exist as
a couple
• 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?
Working principle of electron carriers (EC)
OH
O
OH
O
Quinone and hydroquinone
as central pieces of Ubiquinone
• What is the most important
difference between the two
forms?
• Different number of double
bonds
• OH instead of =O
Working principle of electron carriers (EC)
OH
O
OH
O
Quinone and hydroquinone
as central pieces of Ubiquinone
• Which form carries
electrons?
• The reduced form!
• Which is the reduced
form?
• The oxidation states will
tell!
• Which carbon atoms
changed their oxidation
state?
Electron carriers
Working principle of electron carriers (EC)
OH
H
H
O
H
H
OH H
H
H
H
O
Quinone and hydroquinone
as central pieces of Ubiquinone
• 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.
Electron carriers
Working principle of electron carriers (EC)
OH
+1
H
H
O
H
H
+1
OH H
+2
H
+2
H
H
O
Quinone and hydroquinone
as central pieces of Ubiquinone
• 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.
• The reduced form carries
two more electrons than the
oxidised form
• Where are they?
Working principle of electron carriers (EC)
OH
H
H
O
H
H
+1
OH H
H
H
+2
H
O
Quinone and hydroquinone
as central pieces of Ubiquinone
• 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.
• The reduced form carries
two more electrons than the
oxidised form
• Where are they?
Electron carriers
Working principle of electron carriers (EC)
OH
H
H
O
H
H
+1
OH H
H
H
+2
H
O
Quinone and hydroquinone
as central pieces of Ubiquinone
• 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 compound
Electron carriers
Working principle of electron carriers (EC)
• What can a reduced EC do?
• Does a cell also need
oxidised EC?
OH
H
H
O
H
H
+1
OH H
H
H
+2
H
O
Quinone and hydroquinone
as central pieces of Ubiquinone
Electron carriers
Working principle of electron carriers (EC)
H
-1
• The electrons in
NADH as the most
importan electron
carrier can also be
visualised
H
R
H
N
R
+1
H
H
H
H
H
-2
R
N
R
NADH/NAD+ as electron
carrier
H
0
Ubiquinone as electron carrier
Oxidised
O
+II
C
O
C
C
C
O
C
C
C
C
2 electrons
2 protons
+II
Reduced
OH
+I
O
C
O
C
C
C
O
C
C
C
C
+I
OH
NADH as electron carrier
Oxidised
H
-I
C
H
C
C
H
C
H
C
+I
+N
R
2 electrons
1 protons
Reduced
H
H
-II
C
R
H
C
C
H
C
H
C
±0
N
R
R
Stoichiometric equations
Correct Stoichiometry of biochemical Redox Reactions:
How much lactic acid can be formed from the fermentation of 1 mole
of glucose and what is the effect on the pH?
(CH2O)6
 CH3-CHOH-COO-
1. 24 e-
 12 e- (two lactate will be produced)
(CH2O)6
 2 CH3-CHOH-COO-
2. 6C
 6C (no CO2 is produced)
(CH2O)6
 2 CH3-CHOH-COO-
3. 6O
 6O (water is neither consumed or produced)
(CH2O)6
 2 CH3-CHOH-COO-
4. 12H
 10H (2 protons are produced)
(CH2O)6
 2 CH3-CHOH-COO- + 2H+
2 protons are produced, hence the reaction will lower the pH.