2. stoichiometric aspects of metabolism

2. Stoichiometric Aspects 2. Stoichiometric Aspects of Metabolism of Metabolism

Hans V. WesterhoffFrank BruggemanThanks to Ferda Mavituna

Vangelis Simeonidis

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AIMSAIMS Introduction to elemental balances for Introduction to elemental balances for

biological reactionsbiological reactions Introduction to the concept and use of Introduction to the concept and use of

degree of reductiondegree of reduction

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Lecture contentsLecture contents

2.1 Cell stoichiometry2.2 Elemental balances2.3 Degree of reduction,

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2.1 Cell stoichiometry2.1 Cell stoichiometry

Protein, RNA, DNA, Lipids, Protein, RNA, DNA, Lipids, Lipopolysacchardes, Peptidoglycan, Lipopolysacchardes, Peptidoglycan, and Glycogen. and Glycogen.

While cell composition may vary with cell‑type While cell composition may vary with cell‑type and physiological/environmental conditions, a and physiological/environmental conditions, a typical cell can be assumed to contain:typical cell can be assumed to contain:

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E. coliE. coli

70% water70% water 15% protein 15% protein 7% nucleic acids 7% nucleic acids 3% polysaccharides 3% polysaccharides 3% lipids 3% lipids 1% inorganic ions 1% inorganic ions 0.2% metabolites 0.2% metabolites

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Macromolecular composition of Macromolecular composition of a typical cella typical cellSpeciesSpecies Content (g/g cell)Content (g/g cell) Protein 0.55RNA 0.20DNA 0.03Lipids 0.09Lipopolysaccharides

0.03

Peptidoglycan 0.03Glycogen 0.03Building blocks 0.04TOTAL 1.00

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2.1.1 Proteins2.1.1 Proteins They are the most abundant They are the most abundant

organic molecules within the cell. organic molecules within the cell. All proteins contain All proteins contain C (50%), H C (50%), H

(7%), O (23%) and N (16%).(7%), O (23%) and N (16%). They They also contain also contain sulphur (up to 3%)sulphur (up to 3%) for formation of S‑S bonds. for formation of S‑S bonds.

Molecular weights range from Molecular weights range from 6000 to over 1 million6000 to over 1 million..

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Proteins may serve a Proteins may serve a number of functions:number of functions:

• enzymes (biological catalysts)enzymes (biological catalysts)• regulatory proteins (e.g. insulin) regulatory proteins (e.g. insulin) • transport proteins (e.g. haemoglobin) transport proteins (e.g. haemoglobin) • protective proteins in blood (e.g. antibodies) protective proteins in blood (e.g. antibodies) • toxins (e.g. proteins from toxins (e.g. proteins from Clostridium Clostridium

botulinum) botulinum) • storage proteins (e.g. casein)storage proteins (e.g. casein)• contractile proteins (e.g. flagella) contractile proteins (e.g. flagella) • structural proteins (e.g. collagen)structural proteins (e.g. collagen)

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2.1.2 RNA/DNA2.1.2 RNA/DNA Biological information is stored in Biological information is stored in DNADNA (MW: 2 x l0(MW: 2 x l099) and ) and RNARNA (MW: 2.3 x 10 (MW: 2.3 x 1044 to 1.1 x 10 to 1.1 x 1066). ). The various RNAs which participate in normal cell The various RNAs which participate in normal cell

function serve the purpose of reading and function serve the purpose of reading and implementing the genetic instructions of DNA. implementing the genetic instructions of DNA. Messenger RNAMessenger RNA molecules carry messages from molecules carry messages from DNA to other parts of the cell. These messages DNA to other parts of the cell. These messages are read in the ribosome with the help of are read in the ribosome with the help of ribosomalribosomal RNA RNA. Finally . Finally transfer transfer RNARNA assists in the assists in the translation of the genetic code at the ribosome.translation of the genetic code at the ribosome.

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2.1.3 Other macromolecular 2.1.3 Other macromolecular components of the cellcomponents of the cell

The relative insolubility of The relative insolubility of lipidslipids in water in water leads to their presence predominantly in leads to their presence predominantly in the non‑aqueous biological phases such the non‑aqueous biological phases such as the plasma and organelle as the plasma and organelle membranes. membranes.

FatsFats serve as polymeric biological fuel serve as polymeric biological fuel storage. In addition, lipids constitute storage. In addition, lipids constitute portions of more complex molecules, portions of more complex molecules, such as lipopolysaccharides. such as lipopolysaccharides.

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LipopolysaccharidesLipopolysaccharides and and peptidoglycanspeptidoglycans participate in the participate in the formation of the cell surface (membranes, formation of the cell surface (membranes, envelopes) and are responsible for the envelopes) and are responsible for the cells' tendency to adhere to each other or cells' tendency to adhere to each other or to walls of reactors, pipes and separators.to walls of reactors, pipes and separators.

They also dictate the cells' resistance to They also dictate the cells' resistance to disruption by physical, enzymatic and disruption by physical, enzymatic and chemical methods. chemical methods.

The number of The number of building blocksbuilding blocks necessary necessary for cellular synthesis varies between for cellular synthesis varies between 75 75 and 100and 100 and these are synthesized from and these are synthesized from 12 precursor metabolites12 precursor metabolites..

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2.2 Elemental balances2.2 Elemental balances Assuming that biomass consists of Assuming that biomass consists of

certain types of macromolecules (e.g. certain types of macromolecules (e.g. protein, RNA), it is possible to protein, RNA), it is possible to calculate calculate an average elemental an average elemental composition for biomasscomposition for biomass from the from the average content of the individual average content of the individual building blocksbuilding blocks. .

The following are typical values:The following are typical values:

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Elemental composition Elemental composition of various cell of various cell componentscomponents

ProteinProtein C H C H 1.581.58 O O 0.310.31 N N 0.270.27 S S 0.0040.004 DNADNAC H C H 1.151.15 O O 0.620.62 N N 0.390.39 P P 0.100.10 RNARNA C H C H 1.231.23 O O 0.750.75 N N 0.380.38 P P 0.110.11 CarbohydratesCarbohydrates C H C H 1.671.67 O O 0.830.83 Phospholipids Phospholipids C H C H 1.911.91 O O 0.230.23N N 0.020.02 P P 0.020.02 Neutral FatNeutral Fat C H C H 1.841.84 O O 0.120.12 BIOMASSBIOMASS C H C H 1.811.81 O O 0.520.52 N N 0.210.21

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Balance equationsBalance equations

XX

PP22

YYSS

PP11

321 2 vvvdtdX

vv22vv11

vv44

vv332x2x

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Balance equationBalance equation For every chemical For every chemical

compoundcompound

For every elementFor every element The balance must The balance must

be closed at all be closed at all times:times:

– For energyFor energy– For redox (electrons)For redox (electrons)

321 2 vvvdtdX

321 20 vvvdtdX

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Example: SCP Example: SCP productionproductiona.a. Write down a stoichiometric Write down a stoichiometric

equation describing equation describing SCP SCP production from methaneproduction from methane. . You may assume that the only You may assume that the only metabolic products are carbon metabolic products are carbon dioxide and waterdioxide and water, and the , and the nitrogen source is ammonianitrogen source is ammonia..

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Example: SCP Example: SCP productionproduction

Single Cell ProteinSingle Cell Protein or or SCPSCP refers to refers to proteinaceous materials which are dried cells proteinaceous materials which are dried cells of micro‑organisms. Example species which of micro‑organisms. Example species which have been cultivated for use in animal or have been cultivated for use in animal or human foods include algae, actinomycetes, human foods include algae, actinomycetes, bacteria, yeasts, molds and higher fungi. bacteria, yeasts, molds and higher fungi. While human consumption of microbial While human consumption of microbial protein is ancient in origin, more recent food protein is ancient in origin, more recent food products involve microbial growth in aerated products involve microbial growth in aerated bioreactors using substrates such as natural bioreactors using substrates such as natural gas and paraffins. gas and paraffins.

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Example: SCP Example: SCP productionproduction

b. How many independent experimental b. How many independent experimental measurements would you need in measurements would you need in order to fully describe the system?order to fully describe the system?

c. Assume that the c. Assume that the oxygen oxygen consumption is 1.35 mol oxygen per consumption is 1.35 mol oxygen per mol methanemol methane. .

Calculate the maximum SCP yield in Calculate the maximum SCP yield in g SCP /g methane.g SCP /g methane.

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Solution: SCP Solution: SCP productionproduction

The following reaction scheme can be The following reaction scheme can be assumed:assumed:

CHCH44 + +aa O O22 + + bb NH NH33

cc C H C H 1.811.81 O O 0.520.52N N 0.210.21 + + dd CO CO 22 + + ee H H22OO

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CHCH44 + +aa O O22 + + bb NH NH33

cc C H C H 1.811.81 O O 0.520.52N N 0.210.21 + + dd CO CO 22 + + ee HH22OO The following balances can be The following balances can be

written:written: Carbon: Carbon: 1 = c + d1 = c + d Hydrogen: Hydrogen: 4 + 3b = 1.81 c + 2e4 + 3b = 1.81 c + 2e Nitrogen: Nitrogen: b = 0.21cb = 0.21c Oxygen: Oxygen: 2a = 0.52c + 2d + e2a = 0.52c + 2d + e

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Degrees of freedomDegrees of freedom Number of unknowns:Number of unknowns: 55 Number of equations:Number of equations: 44 Degrees of freedom:Degrees of freedom: 11 One equation is still needed to fully One equation is still needed to fully

describe the system. In this particular describe the system. In this particular case, the missing equation is:case, the missing equation is:experimental data:experimental data: a = 1.35a = 1.35

(mol oxygen used per mol methane used)(mol oxygen used per mol methane used)

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Solution of elemental Solution of elemental balancesbalances Solving the above equations Solving the above equations

gives:gives:a = 1.35 (Given)a = 1.35 (Given)b = 0.13b = 0.13c = 0.63c = 0.63d = 0.37d = 0.37e = 1.63e = 1.63

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2.3 Degree of reduction, 2.3 Degree of reduction,

Any solution to the set of N Any solution to the set of N elemental balances is also a solution elemental balances is also a solution to a linear combination of these N to a linear combination of these N elemental balance equations.elemental balance equations.

This fact is used to derive This fact is used to derive a single a single equationequation which may be more convenient which may be more convenient to use in the calculation of stoichiometric to use in the calculation of stoichiometric coefficients (in elemental balances). coefficients (in elemental balances).

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Degree of reduction, Degree of reduction, This equation is theThis equation is the degree of degree of

reduction balancereduction balance which simply which simply states thatstates that the sum of the the sum of the degrees of reduction of degrees of reduction of reactants of a reaction is reactants of a reaction is equal to the sum of the equal to the sum of the degrees of reduction of the degrees of reduction of the products.products.

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2.3.1 Generalised Degree 2.3.1 Generalised Degree of Reduction of Reduction

The The generalised degree of generalised degree of reductionreduction for a compound is the for a compound is the number of electronsnumber of electrons available available for transfer to oxygen on for transfer to oxygen on combustion of the compound.combustion of the compound.

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Generalised Degree of Generalised Degree of ReductionReduction

is the number of electrons is the number of electrons available for transfer to available for transfer to oxygen on combustion of oxygen on combustion of compound compound ii..

does not necessarily have a does not necessarily have a physical meaning.physical meaning.

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Multiplication factors account for Multiplication factors account for the valencies of individual the valencies of individual elements as follows. They elements as follows. They correspond to the number of correspond to the number of electrons on the atoms:electrons on the atoms:

C = +4 (+2 OC = +4 (+2 O2-2- 4e 4e - -+ CO+ CO22))H = +1 (H = +1 ( H H+ + +1e+1e - -))O = - 2 (+2eO = - 2 (+2e-- O O2-2-) ) Now attribute these electrons to the Now attribute these electrons to the

carbon atomscarbon atoms

Generalised Degree of Generalised Degree of Reduction of the Reduction of the CarbonCarbon

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Generalised Degree of Generalised Degree of Reduction; Reduction; carbohydratecarbohydrate

normalised to one carbon normalised to one carbon equivalent. equivalent.

e.g. e.g. CC66HH1212OO66 = CH = CH22OO and and

OCH2κ = 4 + 2 – 2 = 4= 4 + 2 – 2 = 4

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Calculating degrees of reduction: Calculating degrees of reduction: κκ = number of electrons on carbon + = number of electrons on carbon + 44 Methane: CHMethane: CH44; 4H; 4H+ + =4+; C=4+: =4+; C=4+: κκ=8=8 Methanol: CHMethanol: CH33OH 4HOH 4H+ + =4+=4+ ; ; O O2-2-=2-; C=4+: =2-; C=4+: κκ=6=6 Methanal (Formaldehyde): CHMethanal (Formaldehyde): CH22O 2HO 2H+ + =2+=2+ ; O; O2-2-=2-; C=4+: =2-; C=4+: κκ=4=4 Methylic acid (formic acid): COMethylic acid (formic acid): CO22HH22 2H 2H+ + =2+=2+ , 2O, 2O2-2-=4-; C=4+: =4-; C=4+:

κκ=2=2 Carbon dioxyde: COCarbon dioxyde: CO22, 2O, 2O2-2-=4-; C=4+: =4-; C=4+: κκ=0=0 Ethane: CEthane: C22HH66; 6H; 6H+ + =6+: 6/2=3+; C=4+: =6+: 6/2=3+; C=4+: κκ=7=7 Ethene: CEthene: C22HH44; 4H; 4H+ + =4+: 4/2=2+; C=4+: =4+: 4/2=2+; C=4+: κκ=6=6 Ethyne: CEthyne: C22HH22; 2H; 2H+ + =2+: 2/2=1+; C=4+: =2+: 2/2=1+; C=4+: κκ=5=5 Ethanol: CEthanol: C22HH55OH; 6HOH; 6H+ + =6+=6+ , O, O2-2-=2-: 4+/2=2+ C=4+; =2-: 4+/2=2+ C=4+; κκ=6=6 Acetaldehyde: CAcetaldehyde: C22HH44O; 4HO; 4H+ + =4+=4+ , O, O2-2-=2-: 2+/2=1+; C=4+: =2-: 2+/2=1+; C=4+: κκ=5=5 Acetate: CAcetate: C22HH33OOH; 4HOOH; 4H+ + =4+=4+ , 2O, 2O2-2-=4-: 0/2=0; C=4+: =4-: 0/2=0; C=4+: κκ=4=4 Glycerol: CGlycerol: C22HH66OO22 ; 6H ; 6H+ + =6+=6+ , 2O, 2O2-2-=4-: 2+/2=1+; C=4+: =4-: 2+/2=1+; C=4+: κκ=5=5

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Consider aerobic growth of Consider aerobic growth of a microorganisma microorganismBiomass:Biomass:CHCHaxaxOObxbxNNcxcxSSdxdxPPexex Substrates:Substrates: carbon source, carbon source, CHCHasasOObsbs nitrogen source, nitrogen source, CCfNfNHHaNaNOObNbNNNcNcN sulphate, sulphate, HH22SOSO44 and phosphate, and phosphate, HH33POPO44

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Other products of Other products of aerobic metabolismaerobic metabolism In addition to In addition to biomassbiomass, ,

carbon dioxidecarbon dioxide, and , and waterwater;;a a metabolite productmetabolite product CHCHapapOObpbpNNcpcp

is produced.is produced.

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This aerobic growth is This aerobic growth is represented by the following represented by the following stoichiometric equation:stoichiometric equation:

CHCHasasOObsbs + + NN C CfNfNHHaNaNOObNbNNNcNcN + + oo O O22 + +

ss HH22SOSO4 4 + + pp HH33POPO44

CHCHaxaxOObxbxNNcxcxSSdxdxPPexex + + pp CH CHapapOObpbpNNcpcp + +

cc CO CO22 + + ww H H22OO

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Convention for stoichiometric Convention for stoichiometric coefficientscoefficients

ReactantsReactants Stoichiometric Stoichiometric CoeffCoeff

Carbon Substrate:Carbon Substrate: 11Nitrogen Substrate:Nitrogen Substrate: NN

Oxygen:Oxygen: OO

Sulphur Substrate:(SOSulphur Substrate:(SO44)) SS

Phosphorous Substrate:Phosphorous Substrate: PP

Products:Products: Stoichiometric Stoichiometric CoeffCoeff

Biomass:Biomass: Metabolic product:Metabolic product: PP

Carbon dioxide:Carbon dioxide: CC

Water:Water: WW

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Convention for elemental Convention for elemental subscriptssubscripts

CC H :H :aa O :O :bb N :N :cc S :S :dd P :P :ee Carbon Substrate: Carbon Substrate: SS 11 aaSS bbSS ccSS ddSS eeSSNitrogen Subs: Nitrogen Subs: NN fNfN aaNN bbNN ccNN ddNN eeNNOxygen: Oxygen: OOSulphur Subs:(SOSulphur Subs:(SO44))

Phosphorous Subs: Phosphorous Subs: (PO(PO44))

Biomass: Biomass: XX 11 aaXX bbXX ccXX ddXX eeXXProducts: Products: PP 11 aaPP bbPP ccPP ddPP eePP

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The same reaction can The same reaction can also be written asalso be written as



– – CHCHasasOObsbs – – NN C CfNfNHHaNaNOObNbNNNcNcN – – oo O O2 2

––ss H H22SOSO44 – – pp H H33POPO44 == 00

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We forget about N, S and We forget about N, S and P for the momentP for the moment

CHCHaxaxOObxbx + + pp CH CHapapOObpbp+ +


– – CHCHasasOObsbs – – oo O O2 2 == 00

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Aerobic growth Aerobic growth equationequation There are five stoichiometric coefficients and There are five stoichiometric coefficients and

three elemental balances (C, H, O; or C, O, three elemental balances (C, H, O; or C, O, and electrons). and electrons).

Two coefficients are to be determined Two coefficients are to be determined experimentallyexperimentally

These have to do with two degree of These have to do with two degree of freedom: freedom: 1.1. How much product per biomassHow much product per biomass2.2. How much of the substrates is combusted to How much of the substrates is combusted to

deliver the free energy for growth and production deliver the free energy for growth and production

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The generalised degree of The generalised degree of reduction for biomass is given reduction for biomass is given byby

xxx ba 24

CHCHaxaxOObxbx))

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Similarly, Similarly, the degree of the degree of reduction reduction for the carbonfor the carbon source ( source ( CHCHasasOObsbs):):

SSS ba 24

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and the and the degree of degree of reductionreduction for the for the metabolic product metabolic product ( (pp CH CHapapOObpbp))::

PPP ba 24

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Degree of reduction Degree of reduction (number of electrons) (number of electrons) must be conservedmust be conserved

CHCHasasOObsbs + + oo O O2 2 CH CHaxaxOObxbx + + pp CH CHapapOObpbp+ + cc CO CO22 + + ww HH22OO

2

2 2

S O O

biomass p product C CO w H O

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for molecular oxygen: for molecular oxygen:

for water:for water: for nitrogen source:for nitrogen source:

for carbon dioxide:for carbon dioxide:

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O

02

OH

03NH0

3HNO

02CO

Advantage: no need to Advantage: no need to keep track of carbon keep track of carbon dioxide, water, oxygen dioxide, water, oxygen as Oas O2-2-

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Red blood cells and Red blood cells and anaerobic yeastanaerobic yeast Glucose as substrateGlucose as substrate AnaerobicAnaerobic No growthNo growth

Hence product must Hence product must have same degree of have same degree of reduction as substrate: reduction as substrate: lactate: lactate: CHCH33CHOHCOOH=CCHOHCOOH=C33OO33HH6 6 ((κκ=4), =4),

Or (yeast): 2/3alcohol Or (yeast): 2/3alcohol CC22OHOH66 ( (κκ=4+2=6), =4+2=6), plus plus 1/3CO1/3CO2 2 ( (κκ=0): 4=0): 4

productpbiomassOOS 2

CHCH22O + O + O O2 2 CH CHaxaxOObxbx + + pp CH CHapapOObpbp+ + cc CO CO22 + + ww H H22OO

productpbiomass 0404

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Advantage: no need to Advantage: no need to keep track of Oxygen, keep track of Oxygen, waterwater

CHCHasasOObsbs + + oo O O2 2 CH CHaxaxOObxbx + + pp CH CHapapOObpbp+ + cc CO CO22 + + ww H H22OO

productpbiomassOSubstrate 4

5 stoichiometric coefficients; 2 additional element balance equationsHence 2 degrees of freedom

How about with How about with nitrogen?nitrogen?

Has various redox Has various redox statesstates

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Multiplication factors account for Multiplication factors account for the valencies of individual elements the valencies of individual elements as follows: as follows:

C=4 (COC=4 (CO22); H = 1 (H); H = 1 (H++) ) O = - 2 (-O = - 2 (-››OO2-2-) ) N = ‑3 N = ‑3 (NH(NH33)) or or N = 0 N = 0 (N(N22)) or or N = 5 N = 5 (HNO(HNO33))

Generalised Degree of Generalised Degree of Reduction of the Carbon; Reduction of the Carbon; the effect of nitrogenthe effect of nitrogen

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Convention for stoichiometric Convention for stoichiometric coefficientscoefficients

The stoichiometric coefficient for the The stoichiometric coefficient for the main carbon source is taken to be one main carbon source is taken to be one (1).(1).

The nitrogen source is written for The nitrogen source is written for fNfN atoms of carbon. atoms of carbon. If NHIf NH33 is the nitrogen is the nitrogen source, fN=0source, fN=0. .

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CHCHasasOObsbs + + NN C CfNfNHHaNaNOObNbNNNcNcN + + oo O O22 + +

ss HH22SOSO4 4 + + pp HH33POPO44



Full stoichiometric Full stoichiometric equationequation

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The same reaction can The same reaction can also be written asalso be written as




––ss H H22SOSO44 – – pp H H33POPO44 == 00

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Neglect S and PNeglect S and P

CHCHaxaxOObxbxNNcxcx + + pp CH CHapapOObpbpNNcpcp + +



== 00

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If we neglect the S and P If we neglect the S and P content of the biomass, then content of the biomass, then the the generalised degree of generalised degree of reductionreduction for biomass is: for biomass is:

NsourceN

xxxx c

cba 24

CHCHaxaxOObxbxNNcxcx + + pp CH CHapapOObpbpNNcpcp + +


– – CHCHasasOObsbs –C –CfNfNHHaNaNOObNbNNNcNcN – – oo O O2 2

== 00

NNNNsource baf 24

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Aerobic growth Aerobic growth equationequation There are eight There are eight

stoichiometric coefficients stoichiometric coefficients and six elemental balances. and six elemental balances.

Two coefficients are to be Two coefficients are to be determined experimentally.determined experimentally.

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The generalised degree of The generalised degree of reduction for biomass is given reduction for biomass is given byby

xx

xN

NNNxxx

ed

cc

bafba

56

2424

CHCHaxaxOObxbxNNcxcxSSdxdxPPexex))

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Similarly, Similarly, the degree of the degree of reduction reduction for the carbonfor the carbon source ( source ( CHCHasasOObsbs):):

SSS ba 24

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and the and the degree of degree of reductionreduction for the for the metabolic product metabolic product ( (pp CH CHapapOObpbpNNcpcp ) )::

PN

NNNPPP c

cbafba 2424

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The valency of nitrogen is usually taken The valency of nitrogen is usually taken as – 3 (as – 3 (NHNH33). ).

However, this can be different for other However, this can be different for other nitrogen sources.nitrogen sources.

(OK for as long as one makes the same (OK for as long as one makes the same choice for the entire reaction/system)choice for the entire reaction/system)

Generalised Degree of Generalised Degree of ReductionReduction

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Degree of reduction for Degree of reduction for ATPATP

In all reactions where ATP is In all reactions where ATP is involved, it simply acts as a involved, it simply acts as a carrier of free energy.carrier of free energy.

It does not get involved with It does not get involved with electron transfer directly.electron transfer directly.

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Degree of reduction Degree of reduction for metabolic redox for metabolic redox carrierscarriersNADH NADNADH NAD++ + H + H++ + 2e + 2e--

NADPHNADPH NADPNADP++ + H + H++ + 2e + 2e--

FADHFADH22 FAD + 2 HFAD + 2 H++ + 2e + 2e--

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Degree of reduction, Degree of reduction, , for , for various compounds various compounds

CompoundCompound Methane (CHMethane (CH44) ) 88Glucose Glucose 44Glucose-6-Glucose-6-phosphate phosphate

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PyruvatePyruvate 3.333.33ATP ATP 00NADH NADH 22NADPH NADPH 22FADHFADH22 22HH22 22

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Appreciate how stoichiometry can be applied Appreciate how stoichiometry can be applied for the quantitative description of metabolism.for the quantitative description of metabolism.

Express a microbial process as a simplified, Express a microbial process as a simplified, stoichiometrically balanced reaction.stoichiometrically balanced reaction.

Appreciate the information needed for this Appreciate the information needed for this (from the literature/experiments).(from the literature/experiments).

Understand the concept of degree of Understand the concept of degree of reduction and its use in the elemental reduction and its use in the elemental balances to find the stoichiometric balances to find the stoichiometric coefficients for biological reactions.coefficients for biological reactions.

Learning OutcomesLearning OutcomesYou should be able toYou should be able to

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construct ATP balances for anaerobic construct ATP balances for anaerobic and aerobic metabolism.and aerobic metabolism.

use stoichiometry, yield coefficients and use stoichiometry, yield coefficients and yield factors, to comment on and yield factors, to comment on and compare efficiencies of biological compare efficiencies of biological processes with different strains, processes with different strains, microorganisms, different substrates, microorganisms, different substrates, different products, in aerobic and different products, in aerobic and anaerobic processes. anaerobic processes.

Learning OutcomesLearning Outcomes You should be able toYou should be able to

2. stoichiometric aspects of metabolism

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