Transcript Degree of reduction

2. Stoichiometric Aspects
of Metabolism
Hans V. Westerhoff
Frank Bruggeman
Thanks to Ferda Mavituna
Vangelis Simeonidis
AIMS
Introduction to elemental balances for
biological reactions
 Introduction to the concept and use of
degree of reduction

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Lecture contents
2.1 Cell stoichiometry
2.2 Elemental balances
2.3 Degree of reduction, K
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2.1 Cell stoichiometry

While cell composition may vary with
cell-type and physiological/environmental
conditions, a typical cell can be assumed to
contain:
Protein, RNA, DNA, Lipids,
Lipopolysacchardes, Peptidoglycan, and
Glycogen.
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E. coli
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70% water
15% protein
7% nucleic acids
3% polysaccharides
3% lipids
1% inorganic ions
0.2% metabolites
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Macromolecular composition of
a typical cell
Species
Protein
RNA
DNA
Lipids
Lipopolysaccharides
Peptidoglycan
Glycogen
Building blocks
TOTAL
Content (g/g cell)
0.55
0.20
0.03
0.09
0.03
0.03
0.03
0.04
1.00
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2.1.1 Proteins
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They are the most abundant organic
molecules within the cell.
All proteins contain C (50%), H (7%),
O (23%) and N (16%). They also
contain sulphur (up to 3%) for
formation of S-S bonds.
Molecular weights range from 6000 to
over 1 million.
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Proteins may serve a
number of functions:
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enzymes (biological catalysts)
regulatory proteins (e.g. insulin)
transport proteins (e.g. haemoglobin)
protective proteins in blood (e.g. antibodies)
toxins (e.g. proteins from Clostridium botulinum)
storage proteins (e.g. casein)
contractile proteins (e.g. flagella)
structural proteins (e.g. collagen)
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2.1.2 RNA/DNA
Biological information is stored in DNA
(MW: 2 x l09) and RNA (MW: 2.3 x 104 to 1.1 x 106).
 The various RNAs which participate in normal cell
function serve the purpose of reading and
implementing the genetic instructions of DNA.
Messenger RNA molecules carry messages from
DNA to other parts of the cell. These messages are
read in the ribosome with the help of ribosomal
RNA. Finally transfer RNA assists in the translation
of the genetic code at the ribosome.
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2.1.3 Other macromolecular
components of the cell
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The relative insolubility of lipids in water leads
to their presence predominantly in the
non-aqueous biological phases such as the
plasma and organelle membranes.
Fats serve as polymeric biological fuel
storage. In addition, lipids constitute portions
of more complex molecules, such as
lipopolysaccharides.
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Lipopolysaccharides and peptidoglycans
participate in the formation of the cell surface
(membranes, envelopes) and are responsible for
the cells' tendency to adhere to each other or to
walls of reactors, pipes and separators.
They also dictate the cells' resistance to
disruption by physical, enzymatic and chemical
methods.
The number of building blocks necessary for
cellular synthesis varies between 75 and 100 and
these are synthesized from 12 precursor
metabolites.
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2.2 Elemental balances

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Assuming that biomass consists of certain
types of macromolecules (e.g. protein,
RNA), it is possible to calculate an average
elemental composition for biomass from
the average content of the individual
building blocks.
The following are typical values:
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Elemental composition of
various cell components
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Protein
DNA
RNA
Carbohydrates
Phospholipids
Neutral Fat
BIOMASS
C H 1.58 O 0.31 N 0.27 S 0.004
C H 1.15 O 0.62 N 0.39 P 0.10
C H 1.23 O 0.75 N 0.38 P 0.11
C H 1.67 O 0.83
C H 1.91 O 0.23N 0.02 P 0.02
C H 1.84 O 0.12
C H 1.81 O 0.52 N 0.21
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Balance equations
S
v1
dX
 v1  v2  2  v3
dt
v2
X
2x
Y
v4
P1
v3
P2
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Balance equation
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For every chemical
compound
For every element
The balance must be
closed at all times:
– For energy
– For redox (electrons)
dX
 v1  v2  2  v3
dt
dX
0
 v1  v2  2  v3
dt
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Example: SCP production
a.
Write down a stoichiometric
equation describing SCP
production from methane.
You may assume that the only
metabolic products are carbon
dioxide and water, and the
nitrogen source is ammonia.
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Example: SCP production
Single Cell Protein or SCP refers to proteinaceous
materials which are dried cells of
micro-organisms. Example species which have
been cultivated for use in animal or human foods
include algae, actinomycetes, bacteria, yeasts,
molds and higher fungi. While human
consumption of microbial protein is ancient in
origin, more recent food products involve
microbial growth in aerated bioreactors using
substrates such as natural gas and paraffins.
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Example: SCP production
b. How many independent experimental
measurements would you need in order
to fully describe the system?
c. Assume that the oxygen consumption
is 1.35 mol oxygen per mol methane.
Calculate the maximum SCP yield in
g SCP /g methane.
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Solution: SCP production
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The following reaction scheme can be
assumed:
CH4 +a O2 + b NH3
cCH
1.81
O
0.52N 0.21
+ d CO 2 + e H2O
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CH4 +a O2 + b NH3
c C H 1.81 O 0.52N 0.21 + d CO 2 + e H2O
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The following balances can be written:
Carbon:
Hydrogen:
Nitrogen:
Oxygen:
1=c+d
4 + 3b = 1.81 c + 2e
b = 0.21c
2a = 0.52c + 2d + e
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Degrees of freedom
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Number of unknowns: 5
Number of equations: 4
Degrees of freedom:
1
One equation is still needed to fully
describe the system. In this particular
case, the missing equation is:
experimental data:
a = 1.35
(mol oxygen used per mol methane used)
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Solution of elemental
balances
Solving the above equations gives:
a = 1.35 (Given)
b = 0.13
c = 0.63
d = 0.37
e = 1.63
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2.3 Degree of reduction, K
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Any solution to the set of N
elemental balances is also a solution
to a linear combination of these N
elemental balance equations.
This fact is used to derive a single
equation which may be more convenient
to use in the calculation of stoichiometric
coefficients (in elemental balances). 23
Degree of reduction, K

This equation is the degree of
reduction balance which simply
states that the sum of the degrees
of reduction of reactants of a
reaction is equal to the sum of the
degrees of reduction of the
products.
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2.3.1 Generalised Degree of
Reduction
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The generalised degree of reduction
for a compound is the number of
electrons available for transfer to
oxygen on combustion of the
compound.
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Generalised Degree of Reduction
k
is the number of electrons
available for transfer to oxygen
on combustion of compound i.
k
does not necessarily have a
physical meaning.
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Generalised Degree of
Reduction of the Carbon
Multiplication factors account for the
valencies of individual elements as
follows. They correspond to the
number of electrons on the atoms:
C = +4 (+2 O2-  4e -+ CO2)
H = +1 ( H+ +1e -)
O = - 2 (+2e- O2-)
Now attribute these electrons to the
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carbon atoms

Generalised Degree of
Reduction; carbohydrate
k
normalised to one carbon
equivalent.
 e.g.
C6H12O6 = CH2O
κ CH 2 O
and
= 4+2–2=4
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Calculating degrees of reduction:
κ = number of electrons on carbon + 4
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Methane: CH4; 4H+ =4+; C=4+: κ=8
Methanol: CH3OH 4H+ =4+ ; O2-=2-; C=4+: κ=6
Methanal (Formaldehyde): CH2O 2H+ =2+ ; O2-=2-; C=4+: κ=4
Methylic acid (formic acid): CO2H2 2H+ =2+ , 2O2-=4-; C=4+: κ=2
Carbon dioxyde: CO2, 2O2-=4-; C=4+: κ=0
Ethane: C2H6; 6H+ =6+: 6/2=3+; C=4+: κ=7
Ethene: C2H4; 4H+ =4+: 4/2=2+; C=4+: κ=6
Ethyne: C2H2; 2H+ =2+: 2/2=1+; C=4+: κ=5
Ethanol: C2H5OH; 6H+ =6+ , O2-=2-: 4+/2=2+ C=4+; κ=6
Acetaldehyde: C2H4O; 4H+ =4+ , O2-=2-: 2+/2=1+; C=4+: κ=5
Acetate: C2H3OOH; 4H+ =4+ , 2O2-=4-: 0/2=0; C=4+: κ=4
Glycerol: C2H6O2 ; 6H+ =6+ , 2O2-=4-: 2+/2=1+; C=4+: κ=5
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Consider aerobic growth of a
microorganism
Biomass:
CHaxObxNcxSdxPex
Substrates:
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carbon source, CHasObs
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nitrogen source, CfNHaNObNNcN
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sulphate, H2SO4
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and phosphate, H3PO4
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Other products of aerobic
metabolism
 In
addition to biomass,
carbon dioxide, and water;
 a metabolite product
CHapObpNcp
is produced.
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This aerobic growth is
represented by the following
stoichiometric equation:
CHasObs + aN CfNHaNObNNcN + ao O2 +
as H2SO4 + ap H3PO4
g CHaxObxNcxSdxPex + bp CHapObpNcp +
bc CO2 + bw H2O
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Convention for stoichiometric coefficients
Reactants
Carbon Substrate:
Nitrogen Substrate:
Oxygen:
Sulphur Substrate:(SO4)
Phosphorous Substrate:
Products:
Biomass:
Metabolic product:
Carbon dioxide:
Water:
Stoichiometric Coeff
1
aN
aO
aS
aP
Stoichiometric Coeff
g
bP
bC
bW
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Convention for elemental subscripts
Carbon Substrate:
Nitrogen Subs:
Oxygen:
S
N
C
1
fN
H :a O :b N :c
aS
bS
cS
aN bN cN
S :d P :e
dS
eS
dN eN
1
1
aX
aP
dX
dP
O
Sulphur Subs:(SO4)
Phosphorous Subs: (PO4)
Biomass:
X
Products:
P
bX
bP
cX
cP
eX
eP
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The same reaction can
also be written as
g CHaxObxNcxSdxPex + bp CHapObpNcp +
bc CO2 + bw H2O
– CHasObs – aN CfNHaNObNNcN – ao O2
– as H2SO4 – ap H3PO4 = 0
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We forget about N, S and P
for the moment
g CHaxObx + bp CHapObp+
bc CO2 + bw H2O
– CHasObs – ao O2 = 0
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Aerobic growth equation
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There are five stoichiometric coefficients and three
elemental balances (C, H, O; or C, O, and
electrons).
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Two coefficients are to be determined
experimentally
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These have to do with two degree of freedom:
1. How much product per biomass
2. How much of the substrates is combusted to deliver the
free energy for growth and production
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The generalised degree of
reduction for biomass is given by
(g CHaxObx)
k x  4  ax  2bx
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Similarly, the degree of
reduction for the carbon
source ( CHasObs):
k S  4  a S  2bS
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and the degree of reduction
for the metabolic product
(bp CHapObp):
k P  4  aP  2bP
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Degree of reduction
(number of electrons) must
be conserved
CHasObs + ao O2 g CHaxObx + bp CHapObp+ bc CO2 + bw H2O
k S  aO kO 
2
 g  k biomass  b p  k product  b C  k CO2  b w  k H 2O
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Advantage: no need to keep
track of carbon dioxide,
water, oxygen as O2
for molecular oxygen:
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for water:
k O  4
2
kH O  0
2
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for nitrogen source:
k HNO  0 k NH  0
3
3
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for carbon dioxide:
k CO  0
2
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Red blood cells and
anaerobic yeast
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Glucose as substrate
Anaerobic
No growth
CH2O + 0 O2 0 CHaxObx + bp CHapObp+ bc CO2 + bw H2O
k S  a O  k O  g  k biomass  b p  k product
4  0Hence
 4 product
 0  kmust
 b p  k product
biomass
have
2
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same degree of reduction
as substrate: lactate:
CH3CHOHCOOH=C3O3H6
(κ=4),
Or (yeast): 2/3alcohol
C2OH6 (κ=4+2=6), plus
1/3CO2 (κ=0): 4
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Advantage: no need to keep
track of Oxygen, water
CHasObs + ao O2 g CHaxObx + bp CHapObp+ bc CO2 + bw H2O
k Substrate  a O  4  g  k biomass  b p  k product
5 stoichiometric coefficients; 2
additional element balance equations
Hence 2 degrees of freedom
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How about with nitrogen?
Has various redox states
Generalised Degree of
Reduction of the Carbon;
the effect of nitrogen
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Multiplication factors account for the
valencies of individual elements as
follows:
C=4 (CO2); H = 1 (H+)
N = -3
(NH3)
or
N=0
(N2)
N=5
(HNO3)
O = - 2 (-›O2-)
or
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Convention for stoichiometric coefficients

The stoichiometric coefficient for the main
carbon source is taken to be one (1).

The nitrogen source is written for fN atoms
of carbon. If NH3 is the nitrogen source,
fN=0.
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Full stoichiometric
equation
CHasObs + aN CfNHaNObNNcN + ao O2 +
as H2SO4 + ap H3PO4
g CHaxObxNcxSdxPex + bp CHapObpNcp +
bc CO2 + bw H2O
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The same reaction can
also be written as
g CHaxObxNcxSdxPex + bp CHapObpNcp +
bc CO2 + bw H2O
– CHasObs – aN CfNHaNObNNcN – ao O2
– as H2SO4 – ap H3PO4 = 0
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Neglect S and P
g CHaxObxNcx + bp CHapObpNcp +
bc CO2 + bw H2O
– CHasObs – aN CfNHaNObNNcN – ao O2
=0
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If we neglect the S and P content of
the biomass, then the generalised
degree of reduction for biomass is:
g CHaxObxNcx + bp CHapObpNcp +
bc CO2 + bw H2O
– CHasObs –CfNHaNObNNcN – ao O2
=0
k x  4  a x  2bx 
cx
k Nsource
cN
k Nsource  4 f N  aN  2bN
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Aerobic growth equation
 There
are eight stoichiometric
coefficients and six elemental
balances.
 Two
coefficients are to be
determined experimentally.
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The generalised degree of
reduction for biomass is given by
(g CHaxObxNcxSdxPex)
4 f N  aN  2bN
k x  4  ax  2bx 
cx
cN
 6d x  5ex
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Similarly, the degree of
reduction for the carbon
source ( CHasObs):
k S  4  a S  2bS
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and the degree of reduction
for the metabolic product
(bp CHapObpNcp ):
4 f N  a N  2bN
k P  4  aP  2bP 
cP
cN
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Generalised Degree of
Reduction

The valency of nitrogen is usually taken as –
3 (NH3).
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However, this can be different for other
nitrogen sources.
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(OK for as long as one makes the same
choice for the entire reaction/system)
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Degree of reduction for ATP

In all reactions where ATP is involved,
it simply acts as a carrier of free
energy.

It does not get involved with electron
transfer directly.
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Degree of reduction for
metabolic redox carriers
NADH
NAD+ + H+ + 2e-
NADPH
NADP+ + H+ + 2e-
FADH2
FAD + 2 H+ + 2e-
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Degree of reduction, K, for various
compounds
Compound
K
Methane (CH4)
Glucose
Glucose-6-phosphate
Pyruvate
ATP
NADH
NADPH
FADH2
H2
8
4
4
3.33
0
2
2
2
2
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Learning Outcomes
You should be able to
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Appreciate how stoichiometry can be applied for the
quantitative description of metabolism.
Express a microbial process as a simplified,
stoichiometrically balanced reaction.
Appreciate the information needed for this (from the
literature/experiments).
Understand the concept of degree of reduction and
its use in the elemental balances to find the
stoichiometric coefficients for biological reactions.
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Learning Outcomes
You should be able to
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construct ATP balances for anaerobic and
aerobic metabolism.
use stoichiometry, yield coefficients and yield
factors, to comment on and compare
efficiencies of biological processes with
different strains, microorganisms, different
substrates, different products, in aerobic and
anaerobic processes.
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