Biochemical Engineering BIE 5810
Utah State University
Popular in Course
Popular in Biology
verified elite notetaker
Mr. Tessie Labadie
verified elite notetaker
verified elite notetaker
verified elite notetaker
verified elite notetaker
This 36 page Class Notes was uploaded by Mr. Tessie Labadie on Wednesday October 28, 2015. The Class Notes belongs to BIE 5810 at Utah State University taught by Staff in Fall. Since its upload, it has received 92 views. For similar materials see /class/230408/bie-5810-utah-state-university in Biology at Utah State University.
Reviews for Biochemical Engineering
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/28/15
CHAPTER 2 AN OVERVIEW OF BIOLOGICAL BASICS ENGINEER S PERSPECTIVE I 21 ARE ALL CELLS THE SAME a 221 Microbial Diversity p 1112 1 Temperature 7 psychrophiles lt20 C mesophiles 2050 C thermophiles gt50 C Oxygen aerobic anaerobic obligate facultative Nutrients 7 cyanobacteria are photosynthetic and also nitrogen xers convert N2 into NH3 iv Shapes 7 sphericalelliptical coccus cocci cylindricalrod bacillus bacilli spiral spirillum spirilla b 212 Naming Cells p 1214 i Genus species srains substrains eg Eschericia K12 1 genus group of related species 2 species 7 organisms that are substantially alike 3 strainssubstrain 7 variation within species ii Two primary cell types 7 1 prokaryotic and 2 eucaryotic 1 prokaryotes 7 Table 21 22no nuclear membrane a eubacteria and archaebacteria Table22 2 eucaryotes 7 Table 21 22nuclear membrane organelles c 213 7 Viruses p 14 i Need host cell to be functionally active are NOT freeliving ii Size 7 30 to 200 nm nanometers iii DNA or RNA covered by a capsid protein coat iv Lytic cycle 7 host cells lyse to release phages v Lysogenic cycle phase DNA is incorporated into the host DNA and the host continues to multiply in this state vi Phages role in bioprocess technology 1 phage attack on E coli fermentation to make a recombinant protein product can be destructive 2 phages can be used as agents to move genetic material into E coli 3 Killed phage preparation has been used as a vaccine 4 Production of viruslike particles that are empty shells capsid used a vaccines 5 Gene therapy 7 replace virus genetic material with a desired gene capsid acts as a Trojan Horse to protect the gene and to deliver it selectively to a cell type In this case the virus is a biotechnology tool 139 214 Procaryotes p15705 to 3 microns um in radius rapid growth hrs 1 Eubacteria gram stain Gm E subtilis Gm7 L E a Gm better suited to excretion of proteins b Figure 22 p 16 Gm 7 bacterium with outer membrane 3 1 rm g 2 Archae a b c 3 Eucary a F Iquot Mycoplasma 7 Not Gm or Gm 7 clinically important primary atypical pneumonia and common contamination of media used industrially for animal cell culture Acinomycetes longand branched hyphae 7 Antibiotics source Amylolytic and cellulolytic enzymes some for enzymatic hydrolysis of starch and cellulose eg Actinomyces Thermomonospora and Streptomyces Table 23 p 18 7 Bacterial architecture bacteria p 17 7 differ at the molecular level No peptideglycan different lipid composition of cytoplasmic membrane Methanogens thermoacidophiles halobacteria Sources of active enzymes with novel properties otes p 19 7 yeast 5u animal lOu plant 20u Figure 23 p 19 7 Animal and Plant cells Sterols in cytoplasmic membrane strengthen the structure and make membrane less exible Animal cells do NOT have cell wall only cytoplasmic membrane 9 animal cells are shear sensitive fragile and complicates the design of large scale bioreactors for animal cells Reproduction and components 7 p 2022 Fungi l yeasts Fig 26 2 molds Fig 27 i Yeasts 7 Saccharomyces cerevisiae anaerobic alcohol formation and aerobic baker s yeast production Molds 7 antibiotics citric acid Aspergillus m make up a large raction 0f the fermentation industry submerged culture forms pellets 50 u to 1 m can cause nutrient transfer problems oxygen but pellet formation reduced broth Viscosity which can improve bulk oxygen transfer 4 classes 1 phyco 2 asco 3 basidio and 4 deuteromycetes Trichoph on7 athlete s foot Algae p 24 7 diatoms filter aids Chlorella wastewater treatment with single cell protein production gelling agents such as agar and alginic acid II 22 7 CELL CONSTRUCTION p 25 a 221 Introduction 7 living cell structural elements include proteins nucleic acids polysaccharides lipids and the storage materials including fats polyhydroxybutyrate and glycogen b Biological system 7 levels of understanding i 39 39 biology 39 39 39 39 y ii Cellular cell biology microbiology iii Population microbiology ecology Production bioprocess engineering c 222 7 Amino Acids and Proteins p 26 i oc Amino Acids 7 monomers of proteins form peptide bonds 1 Table 24 p 29 Amino acid structure 2 Isoelectric point 7 protein puri cation processes ii Proteins 1 5 categories structural glycoproteins collagen keratin catalytic enzymes gt 2000 known tramport hemoglobin serum albumin regulatory hormones insulin growth hormones protective antibodies thrombin 2 3D structures described at 4 levels of structure p 29 a primary71inear sequence of amino acids ld length b secondaryhydrogen bonding l helixes 2 sheets c tertiaryR group interactions covalent disul de H d quatemarypolypeptide chains interacting 3 Antibodies or immunoglobulins p 31 a immune AbAg reactions diagnostic kits protein separation schemes delivery of anticancer drugs b one of most important products of biotechnology c abzymes impart catalytic activity to antibodies Couple protein engineering to antibodies promises development of speci c catalytic agents 4 Carbohydrates 7 223 p 34 a Modulate some chemical signalinganimalsplants b Monosaccharides 39 Catoms Tale 25 p 35 i Dribose deoxyribose in RNA DNA ii glucose sucrose plant sugar lactose milk and whey c polyscaaharides 3 2 monosaccharides p 37 i biochemical engggt polysaccharide industry dextrins 7 branched sections 0 famylopectin are used as thickeners cellulose chain of Dglucose monomers linked by i l4 glycosidic bonds Resistant to enzymatic hydrolysis Interest in 2 rue99 s converting cellulose wastes into fuels or chemicals 5 Lipids Fats and Steroids 7 222 p 38 a Lipids 7 present in nonaqueous biological phases plasma membranes fatty acid composition i Fats 7 lipids that serve as biological fuel storage ii Phospholipids 7 key membrane component iii Polyhydroxyankanoates PHA eg polyhydroxybutyrate PHB a storage product in some cells p 40 iv Steroids hormones 10398M cholesterol cortisone estrognen progesterone 1 Commercial production depends on microbial conversion because the large number of asymmetric centers makes total synthesis difficult p40 6 Nucleic Acids RNA DNA 7 225 p 40 a Nucleic Acids 7 reproduction of living cells b DNA 7 stores amp preserves genetic information c RNA 7 central role in protein synthesis i mRNA p 46 short half life ii t RNA stable iii r RNA major component of ribosomes Nucleotides 7 i Building blocks of DNA and RNA ii Store Energy 7 ATP ADP iii Reducing power NAD NADP Plasmids 7 circular DNA in cytoplasm nonchromosomal automonous selfreplicating Easily moved in and out of cells and used for genetic engineering 3 1 D Lquot 111 23 CELL NUTRIENTS p 46 a Introduction 7 231 p 46 i Cell composition different from its environment due to semipermeable membrane and energy expenditure to keep it away from thermodynamic equilibrium ii Table 27 p 48 7 80 water iii Macronutrients 7 required at gt10394M CNOHSP Mg K iv Micronutrients 7 required at lt 10394M Mo Zn Cu Mn Ca Na vitamins growth hormones metabolic precursors b Macronutrients 7 232 p 49 see Tables 28 p 49 and 29 p50 i Carbon Source 7 l Heterotrophs use organic compounds 2 Autotrophs use C02 See Table 28 p 49 1 Most common for industrial fermentations 7 molasses sucrose starch glucose dextrin corn syrup and waste sulf1te liquor glucose methanol ethanol and methane are also cheap carbon sources for some fermentations ii Energy Sources 7 l Chemo reduced inorganic chemicals 2 photo light energy c Micronutrients 7 233 p 50 i Iron 7 plays a regulatory role in some fermentation processes 1 deficiency required for excretionof ribo avin by Ashbya os 11 2 conc Regulates penicillin production by g ch sogenum 3 Mn 7 enzyme cofactor plays role in regulation of secondary metabolism and excretion of primary metabolites ii Copper 7in certain respiratorychain components amp enzymes 1 de ciency stimulates penicillin and citric acid production ii39 Cobalt7Vitamin B12 Required in propionic bacteriaamp methanogens iV Mo nitrate reductase and nitrogenase amp for growth on N03 and N2 as sole source of nitrogen V Ca 7 cofactor for amylases and some proteases V1 Vitamins 7 required at concs of 10396 to 103912 M d Growth Media 7 234 p 52 See Table 210 p 52 i De ned media specific amounts of pure chemical compounds with known chemical compositions Complex media 7 contain natural compounds whose chemical composition is not exactly known eg yeast extract peptone molasses or corn steep liquor Usually provides growth factors Vitamins hormones and trace elements often resulting in higher cell yields CHAPTER 4 HOW CELLS WORK 42 The Central Dogma p 105 7 see figure 41 p106 DNA 9 transcription 9 RNA 9 translation 9 proteins 43 DNA Replication p 107 1 Fig 42 7 Initiation of DNA synthesis requires formation ofan RNA primer 2 Fig 43 7 Steps of replication ofbacterial chromosome 44 Transcription p 110 7 mRNA tRNA rRNA 45 Translation Message to Product p113 451 Genetic Code 7 Table 41 p 114 1 Code is degenerate 7 more than one codon specifies a particular amino acid 2 Genetic engineering7code universal 7human protein understood by Ecoli yeast 452 7 Translation7 1 initiation Z elongation and 3 termination 453 7 Posttranslational Processing 7 1 Protein secretion through Cytoplasmic Membrane in prcaryotes 7 p 1 17 a Gm 7 outer membrane blocks release into extracellular compartment b Gm proteins readily pass through cell wall into extracellular compartment 2 Protein secretion in eucaryotes Z pathways by exocytosis A bioprocess engineer must be aware that many proteins are subject to extensive processing after the initial polypeptide chain is made 1 Constitutive exocytosis pathway 7 operates at all times 2 Regulated exocytosis pathway 7 secrete only in response to chemical signals 3 N7linked glycosylation addition of sugars 7 pattern targets protein to a compartment or degradation and removal from the organism if not humanlike Involves both ER and Golgi p117 USE OF PROCARYOTIC CELLS E coli TO SERVE AS HOSTS FOR EXPRESSION OF HUMAN THERAPEUTIC PROTEINS is limited to those proteins where N7linked glycosalyation is NOT present or is UNIMPORTANT Even mammalian cells will show altered patterns of glycosylation when cultured in BIOREACTORS and patterns can shift upon SCALE7UP 46 p 119 Metabolic Regulation HEART OF A LIVING CELL 461 Genetic7level Control Which proteins are synthesized 1 Transcriptional control 7 most common strategy Use Repressor Protein a Feedback Repression 7 by end product Fig 49 p 120 b Induction 7 by substrate Fig 410 p120 2 Definition7 OPERON 7 set on contiguous genes encoding proteins with related functions under the control ofa single promoter7operator 3 Example7 lac operon 7 controls synthesis of 3 proteins in lactose use as CampE source 1 lac z 7 encodes B7galactosidase lactase lactoseglucose galactose Z lac y 7 permease 7 increase rate of uptake of lactose into the cell 3 role of cAMP 7 increases as energy decreases cAMP binds to CAP to form a complex that binds near promoter and enhances RNA polymerase binding to lac prometer 4 Regulon 7 noncontiguous gene products under control of separate promoters can be coordinately expressed in a regulon eg NampF starvation aerobic to anaerobic 1 Constitutive 7 unregulated genes Enzymes involved in glycolysis 2 Example 41 7 Diauxic Growth see Figure 411 and Explain 462 7 Enzymatic Level Contro 7 Metabolic Pathway Control p 123 7 inhibition 7 occurs at the enzyme level See Figure 412 Fermentation Specialist 7 tries to disrupt the cell39s control to cause cell to overproduce product of commercial interest 1 lsozymes 7 2 separate enzymes each sensitive to a different end product 2 Concerted Feedback inhibition 7 1 enzyme with 2 allosteric binding sites need high levels of F1 and F2 for full inhibition 3 Sequential Feedback inhibition 7 F1 inhibits E4 F2 inhibits E5 Need both F1 and F2 levels to be high to fully inhibit reaction 4 Cumulative Feedback inhibition 7 effector sites for several end products 47 p124 How the Cell Senses its Extracellular Environment 471 7 Mechanisms to Transport Small Molecules across Cellular Membranes A Energy7lndependent mechanisms 1 Passive Diffusion water oxygen 2 Facilitated Diffusion sugars7eucaryotes glycerol7procaryotes B Energy7Dependent mechanisms 1 Active Transport 2 Group Translocation A1 Fassive Diffusion Concentration gradient JA KP CAB 7 CA1 41 J flux across membrane mol cm2 7 s KP permeability cms CAE extracell conc ofspecies A molcm3 CA1 lntracell conc of species A molcm3 Hydrophobic compounds 7 high diffusivities 10 8 cmZs in cell membranes A2 Facilitated Diffusion protein carrier molecule embedded in membrane binds reversibly with target undergoes conformational change to release A on lntracell side of membrane 1A 1A max C QAI 42 KM CAE KMT CA1 KMT binding affinity of the substrate molcm3 CAE gt CA1 enet flux lNTO cell CAE lt CA1 enet flux OUT of cell 131 Active Transport AGAlNST a concentration gradient proteins in membrane Energy Sources 1 pH gradient of proton7motive force 2 secondary gradients eg Nat or other ions derived from proton7motive force by other transport mechanisms and by hydrolysis of ATP l Proton7motive force See Figure 56 p 142 Tendency of protons to return to the inside of the membrane Hydrogen atoms removed from NADH are carried to outside of membrane 6 are retuned to the cytoplasmic side of the membrane to combine with 02 and H the cytoplasm to form OH on the inside A pH gradient is created across the membrane with CH inside and H outside 2 Molecules transported without coupling to the ion gradients generated by the proton7motive force Hydrolysis ofATP to release phosphate bond energy is used DlRECTLY in transport maltose in E coli JA JA max C KMT CAE B2 Group Translocation Chemical modification of substrate during transport Example 7 Phosphotransferase System 7 uptake of sugars in bacteria with source ofenergy Phosphoenolpyruvate PEP Sugar SE PEP 11 sugar7P 511 pyruvate 131 44 By converting sugar to phosphorylated form sugar is trapped inside the cell Perferable to active transport energy to move sugar into cell then energy to phosphorylate it 472 p 127 Role of Cell receptors in Metabolism and Cell Differentiation Taxis Surface Receptors 7 Response of bacteria involving receptors binding to specific compounds that result in a change in the direction of movement of the flagella Examples chemotaxis aerotaxis phototaxis positively negatively Quorum sensing molecule lntracellular Receptors 7 produced by bacteria whose accumulation is related to cell concentration biofilms eg acylated homoserine lactone Higher cells 7 differentiation Surface Receptors 7 eg Steroids 7 NOT act by themselves but the hormone7receptor complex interacts with gene loci to activate transcription ofa target gene Eg Cell Adhesion can lead to changes in cell morphology that are critical to animal cell growth and physiological function BIECEE 58106810 PROBLEM SET 2 Fall 2003 Date 91603 Date Due 92603 ABET code from svllabus provided for each problem 1 l3be Enzyme kinetics Consider the following enzyme reaction sequence k1 k3 S E ES1 k2 k4 ES2 P E Develop a suitable rate expression for product formation V k5ESZ by using a the equilibrium approach and b the quasiesteadyestate approach 2 l 3 b c e Solve problem 33 page 98 of the textbook a What is the function of fumarase 3 l3bce Solve problem 36 page 99 ofthe textbook 4 l3bce Solve problem 39 page 100 of the textbook 5 l3b c e Solve problem 316 page 102 of the textbook a What type of inhibition is this b Determine the constants Vm Km and K1 c Could you modify the operation ofa biochemical reactor in order to minimize the effect of the inhibitor if so how 6 l 2 3 b e Serratia marcescens is cultured in a minimal medium reactor Oxygen consumption is measured at a cell concentration of 227 gL dry weight ime Oxygen Conc Time Oxygen Conc min mmolL min mmolL 0 025 10 018 2 023 12 016 5 021 15 015 8 a Determine the best kinetic model fit to the data b Determine the rate constant 7 l 2 3 b 3 An enzyme is immobilized on a flat sheet ofpolymer and placed in a stirred reactor The enzyme intrinsic maximum reaction rate is 6x106 molsemg enzyme The amount of enzyme bound to the surface is 1x104 mg enzymecm2 of support The Km value in solution is 2x10393 molL The mass transfer coefficient is 43x10 5 A What is the reaction rate when the bulk concentration of substrate is 4x103 molL B What is the substrate surface concentration C What is the Da value for this system 2 l 2 be Describe simple experiments to determine if the uptake ofa nutrient is by a passive diffusion b facilitated diffusion c active transport or d group translocation Memrbanes Active Transport Problem Given The concentration of chloride ion in blood serum is about 010 M The concentration of chloride ion in urine is about 016 M Find 1 The energy expended by the kidneys in transporting chloride from plasma to urine and 2 how many moles of C1 ions could be transported per mole of ATP hydrolyzed Solution AG 23 RT log CZ 1 23 1987 calmoi 298 log 016 010 7 1362 calmoi 0204 278 calmoi 2 ATP Hydrolysis provides 7700 calmole 7700 calmole ATP 28 C1 ATP 278 calmole C1 The standard free energy change for the movement of an uncharged molecular from one side of a membrane at concentration C1 to the other at concentration C2 under nonequilibrium conditions is AG 23 RT log CZ 1 Transfer of Polycyclic Aromatic Hydrocarbons Between Model Membranes Relation to Carcinogenicity Plant AL H Povvnall and L C Smith ChemxBiol interactions 44 2377246 1983 Key Concepts and Questions Metabolism of PAH by cytochrome P450 is a prerequisite to carcinogenic activity PAHmem 1th on Molecular Volume ofthe PAH is a rateedetermining factor Donor vesicles POPC vvith PAH and Acceptor vesicles POPC vvithout PAH Question Rate limiting step is Thermodynamic parameters determined from a van39t Hoff Arrhenius plot Pig 3 E AG AH AS 7 Table l p242 Question Calculate the Ea value from Fig 3 Question Endoplasmic reticulum what is its role p 240 Solvation of a hydrophobic molecule requires reorganization of water structure Size of cavity formed to accommodate the hydrocarbon molecule is proportional to its molecular area 243 Rate of transfer of lipophilic materials out ofmembranes reflects equilibrium partitioning lf active site of P450 is within the hydrophobic environment of the membrane characteristics of partitioning will determine extent of enzymeesubstrate complex formation Dissociation constant for BaP from microsomes l micromolar Km for hydroxylation ofBaP bby microsomes l micromolar Question What is the likely mechanism for the transfertransport of PAH between membrane and aqueous phase Explain your answer THERMODYNAMICS amp STOICHIOMETRY Estimating Yield amp Electron Acceptor Requirements from Stoichiometry 3C6H1206 802 2NH3 2C5H702N 8C02 14H20 1 3180 832 217 2113 Yglmse A cells 2 21 113 gmole 042 g cells g glucose used A glucose 3180 gmole Calculate Electron Acceptor Equivalent of Electron Donor Express Yield on Electron Acceptor basis CSHIZO6 602 6C02 6H20 2 180 632 e Acceptor Equivalent of glucose is A 1072 6132 gmole 107 g 02 g glucose 3 e39AEglmse A glucose 180 gmole YTEA A cells 2 2113 gmole 039 g cells A glucose as e AE 3180 gmole 107 g 02 g glucose g e AE used Note Actual yield lt YWAE as portion of substrate is oxidized for energy wtime Consider Electron Acceptor Oxygen 1 used for substrate oxidation to CO2 and H20 2 equivalent of biomass C5H702N 502 5C02 NH3 2H20 4 1 13 532 e AE of biomass A 10 5132 gmole 142 g 02g cells A C5H7OZN 113 gmole 3 substrate e donor as e AE n0t transformed to CO2 and H20 Refer to Eqn 1 Glucose is NOT totally oxidized to CO2 amp H20 as some C H and O are incorporated into cells Oxygen actually utilized in Eqn 1 is total glucose oxygen demand e AE of e donor cells formed e AE of biomass e AE utilized e AE cells e AE of oxidized substrate 5 e AE of oxidized substrate 2 e AE utilized e AE cells 6 07 g O2 x 3 moles 1180 g glucose 142 g O2 x 2moles 113 cells g glucose mole glucose g cells mole cells 2 5778 g 02 3209 g 02 2569 g 02 Electron Acceptor Equivalent of Electron Donor is shown as e39AEdonor Now O2 actually used per unit of e AE Eqn 1 used is Oxygen consumed 2569 2 O2 044 2 O1 e AE glucose 3 mole 107 g e AE g glucose 180 gmole g e AE used 044 is less than 1 because of substrate incorporated into cells and not oxidized EXAMPLE 1 Observed Cell YIELD amp Electron Acceptor OXYGEN used Reactor Chemostat p 190 Aerobic CFSTR without recycle Conditions Feedstock e AE 500 gm3 F 1000 m3d e AE not utilized 2 10 gm3 Ef uent single cell protein SCP as X 200 gm3 Find 1 Y obs as g SCP g e AE utilized 2 Oxygen used as g 02 g e AE removed and as g OZday SCP X Q e AEd fee gt 7 Q e AE not utilized Carbon Dioxide Solution 1 Y observed e AEdonm O2 nutrients C5H7OZN CO2 H20 500 gm3 200 gm3 Mass ofX SCPX F a g SCP producedd 200 gm3 1000 m3d 200000 g SCPd b g e AEd removed d 2 500 10 g e AE m3 1000 m3d 2490000 g e AE d c Yobs 200 000 g SCPd 041 g SCP g e39AEd removed 490000 g d e AE 2 Oxygen used as g 02 per d a e AEd mass balance steady state around complex mix no recycle reactor accumulation in out conversion to SCP 0 e AE in e AE out oxygen equivalent of SCP 0 500 10 gm3 1000 m3d 200000 g SCPd 142 g OZg SCP 0 490000 gd 284000 gd 206000 g 02d b Oxygen used in g 02 per g e AEd removed OxygenCOD 1206 000 gd used 042 g 02g e39AEd 490000 g e AEday removed NOTE 1e AE 1 Balance that accounts for cell production amp oxidation is g e AE Cells g e AEd oxidized g e AE removed 041 g SCP g e AEd 142 g 02 g SCP 042 g 02g e AEd 10 g 02g e AEd End of Example 1 Stoichiometry of Biological Reactions R fe Ra fs Rcs Rd 7 R 2 overall balanced reaction fe 2 fraction of electron donor used for energy Ra half reaction for electron acceptor f s 2 fraction of electron donor used for cell synthesis Rcs 2 half reaction for cell synthesis Rd 2 half reaction for electron donor fefs1 8 EXAMPLE 2 Write a Balanced Reaction for the Aerobic Biological Conversion of Acetate To Cells Using NH as the Nutrient Source Given Eqn 7 and half reactions in Table aerobic conditions NH1 fe 041 fs 059 1 Balanced Stoichiometric Equation R feRa fs Rcs Rd feRa 041 1402He 12 H20 2 010302041H041e0205H20 stcs 2 059 15C02120HC03120NH4He 120C5H702N920H20 0118C0200295HCO300295NH4059H050e 00295C5H702N 02655H20 Rd 2 0125CH3COO 0375H20 0125C020125HCO3He R 0125 CH3COO 00295 NH4 0103 02 00295 C5H702N 00955 H20 0095 HCO3 0007 C02 2 Calculate Cell Yield based on acetate Ac and e AEAC Yield 00295 mol cell 1 113 gmol cel 1 333 g cell 2 044 g cells 0125 mol Ac 60 gmol Ac 75 g Ac g Ac Ye AEAC e AEAC 0125 mol Ac 60grnol107 g 02 g AC 2 803 g as e39AE Cal e39AEAC CHCOOH 201 9 2C01 ZHZO OAc 26460 2107 gOZgAc Y e39AEAc 334 g Cells 2 042 g Cellsg e39AEAc 803 g as e AE Nutrient Requirement 00295 mol 14 gmol N 0413 g N 012 g N 00295 mol 113 gmol cell 333 g cells g cells Oxygen Requirement 0103 mol 32 gmol 07 3296 g 07 g O2 00295 mol 113 gmol cell 333 g cells g cells Estimating Biomass Yield from Bioenergetics Key steps in bioenergetics analysis 1 Identify electron donor 2 Calc energy E produced from bacterial redox reaction 3 Calc E needed for converting growth carbon source to cell matter 4 Calc Cell Yield based on balance between E prod and E needed for synthesis Amount of E needed for cell synthesis 2 f carbon source nitrogen source Pyruvate is used an the intermediate for cell synthesis in this analysis Energy for Cell Synthesis AGS 2 AG AGC AGH AG 2 Free Energy 9 Km K Where AGs 2 convert 1 eeq of carbon source to cell matter AGp 2 convert 1 eeq of carbon source to pyruvate calc using AG half reactions to convert carbon source to pyruvate K 2 fraction of energy captured remainder as heatentropy assume 06 m 1 if AGp gt0 endergonic 1 if AGp lt0 exergonic AGc to convert 1 eeq of pyruvate to 1 eeq of cell 3141 kJeeq cell AGN per eeq of cells to reduce nitrogen used to ammonia kleeq 1746 for N03 1361 for N02 1585 for N2 0 for NH4 Electron Donor fraction used for energy fe fraction used for cell synthesis fs Equation relating energy made available left side 2 energy for synthesis right side K AGR fefs AGS where fe fs 1 10 8 K 2 fraction of energy captured assume 06 AGR 2 energy released from redox reactions kJmole e transferred fe e mole fraction of substrate used for energy fs e mole fraction of substrate used for synthesis AGS 2 energy used for cell synthesis growth kJmole e transfer for cell growth EXAMPLE 3 Estimate Biomass Yield Using Bioenergetics Calculate growth on acetate with O2 as electron acceptor Given Cell YIELD for acetate as carbon source e donor heterotrophs aerobic NH4 nutrient 1 g cells as e AE AC g e AEAC utilized 2 g cells g e AEAC utilized 1 Calc K AGR with 02 as electron acceptor K AGR fefs AGS where fe fs 1 10 8 Solve for acetate redox reaction with oxygen using Table kJ mole e Eqn 18 18CH3COO 38H20 9 18CO2 18HCO3 H e 2768 Eqn 4 140 H e 9 12 H O 78 14 18CH3COO 1402 9 18CO 18HCO3 18H20 10582 AGR Calc energy produced amp captured by cell 2 KAGR 060 10582 6342 kJmole e A Calc energy needed per e39 mole of cell AGs AGs 2 1 AGc x 9 Km K AGN 0 NH3 as nitrogen source AGc 3141 kJmole eeq cell pyruvate to cell AGp acetate to pyruvate reaction eq 18 and eq 15 kJmole e eq 18 18 CH3COO 38 H20 9 18 CO2 18 HCO3 e H 2766 eq 15 15CO2 110HCO3 H e 9 110CH3COCOO 25H20 3578 AGp 812 Because AGpgt0 gt m 1 energy is required Calc AGs 812 3141 0 4494 kJmole e 06 1 1 Determine fe amp fs with eq 10 fefs AGS K AGR 4494 6342 0707 Since fe fs 1 gt fe1 fe 0707 gt fe 0707 1707 041 g energy e39AEAcg e39AEAc utilized thus fs 059 lg cell 1e39AEAg 1e39AEc utilizedl 1 Determine yield based on e39AEAc g cells g e AE utilized For cells C5H7OZN 1 g cells 2 142 g e39AEcells Yield 21059 g cell ge AEg ge AEL utilized 2 042 g Cellsg ge AEL utilized 142 g cell e AE g Cells EXAMPLE 4 Estimate Biomass Yield Using Bioenergetics Calculate growth on acetate with CO2 as electron acceptor Given Cell YIELD for acetate as e donor heterotrophs methanogenic NH4 nutrient 1 g cells as e AE AC g e AEAC utilized 2 g cells g e AEAC utilized l Calc K AGR with CO2 as electron acceptor K AGR fefs AGS where fe fs 1 10 8 Solve for acetate redox reaction with CO2 using Table kJmole e Eqn 18 18CH3COO 38H2O 9 18CO2 18HCO3 H e 2768 Eqn 8 18CO2 H e 9 18CH4 18HCO3 2411 18CH3COO 38H2O 9 18CH4 14HCO3 AGR 357 Calc Energy captured by cell 2 K AGR 060 357 2142 kJmole e g Calc energy needed per e39 mole of cell AGS AGs 2 AG AGc AGiN 9 Km K AGp 2 same as for acetateO2 812 kJ mole e AGc 3141 kJmole e cells AGN 0 AGs 1812 3141 0 4494 kJmole e synthesis same as for aerobic 06 10 1 Determine fe and Is with eq 10 fefs AGS K AGR 4494 2142 210 Since fe fs 1 22gt fe 1 fe 210 22gt fe 21 1 fe 0954 g energy e39AEAcg e39AEAc utilized thus fs 1 0954 0046 g cell e39AEAcg e39AEAc utilized 4 Determine yield based on e39AEAc g cells g e AE utilized For cells C5H7OZN 1 g cells 2 142 g e AECells Yield 2 0046 g cell e AEg e AE utilized 2 0032 g Cells g e AE utilized 142 g cell e AE g Cells 5 Compare Yields for acetate oxidation as a function of electron acceptor Electron acceptor Yield g Cell g e AEAC Product 02 042 CO2 H2O CO2 0032 CH4 l Examples of halfreactions for construction of a balanced equation for an overall bacterial reaction Overall Reaction RRdf5Rastc Electron donor half reactions Rd Generic organic 16 CaHchNd 23619 Hzo COZ 9 NH3 11 c wheree4ab2c 3d Inorganic NH4 31120 3 N0339 H e39 L L L l 2 12 16 HZS16HS 2 H20 8 304 16H e Fe2 Fe3 e39 Electron acceptor half reactions Ra Aerobic 39139 02 H e39 H20 Anaerobic NO339 W g e39 Tg N2 15 H20 denitri cation l 12 1 1 L 8 304216He 16 H23 16 HS 2H20 sulfate I respiration C02He39CH4HZO methane fermentation Cell synthesis half reactions RC ammonia N source co2 516 NH3 H e39 516 C5H702N H20 39 5 1 22 1 LL nitrate N source 28 C02 28 6 28 H e 28 C5H702N 28 H20 N03 I Halt reactions ior biological systems Reaction number Half reaction AG Wb kJ per electron equivalent Reactions for bacterial cell synthesis R j Ammonia as nitrogen source C022ioHC05 5NHJ H e39 gc5H702NggH2o Nitrate as nitrogen source 2 ENOg T cog w equot 2i8c5H702NgH20 Reactions for electron acceptors Ra Nitrite 3 gNog 3w e g N g H20 4323 Oxygen 4 02 H equot H20 78lt Nitrate 5 9405 EH39 e em gHYO 7t67 Sullite I 6 gsogwine HQS HS HO 1360 Suliate 7 50 H equot HQSfEHSf Hzo 2127 Carbon dioxide methane lermentation 8 gco2 H e gcm EH20 2411 Reactions for electron donors Rd Organic donors heterotrophic reactions Domestic wostewater 9 co2 sic NH HCOg H39 equot 5 coH903N 2 quot H20 3180 Protein amino acids proteins nitrogenous organics to 3 5 co2 3 23 NH1H e i cmHuosN 3 H20 3222 Formate it ch05 H e g HCOO H20 4807 Reaction Enumber Half reaction Glucose l 12 gm H e 536an gHzo Carbohydrate cellulose starch sugars I 13 C02H equot CHZOgHzo Methanol 14 gcozH e39 gCH30HgHzo Pyruvate 15 gco2 Hco3 H e cmcocooquot gHzo Ethanol 16 gC02H e39 CHJCH20H H2O Propionote l7 gco2 i HCOj H equot 4 CH3CH2COO Ti H20 Acetate 18 gco2 g HCOg H e CH3COO39 3 H20 Grease lots and oils 19 gcoHH39 e ggceH1 O gHzo Inorganic donors autotrophic reactions 20 Fe3 e Fez 21 12N03 H39e39 gN05gHzo 22 N05 H39e gNHggH20 23 gNO H39e39 NH3 H20 24 iSOZ39 H e gs Hzo 25 503 H39e HZSl15HS39HQO 26 3303 H39e S2O HZO 27 nggHwe NH 28 H e e 39 H2 29 1550339 H39 e39 503 H20 AGquot Wllb kJ per electron equivalent 4196 4184 3751 3578 3179 2791 2768 2761 7AAO AO15 3450 3262 1948 2128 2130 2747 4046 4433 Adapted from McCarty 1975 and Sawyer et al 1994 bReactonts and products or unit activity except 10quot7 se m standard halfcell potentials of biological interest V Half Cell n E39o voltsbC acetate 002 2Hpyruvate H20 2 070 succinate 002 2Hozketoglutarate H20 2 067 acetate 2Hacetaldehyde H20 2 O6O chlorophyll Pmquot 1 O6 3Pglycerate 2Hglyceraldehyde3P H20 2 O55 ferredoxin oxred 1 O43 2HH2 2 042 H COgformate 2 O42 acetyl CoA 2Hacetaldehyde CoASH 2 041 C02 pyruvate 2Hmalate H20 2 033 NADP 2HNADPH W 2 o32 Iipoate0x 2Hlipoatered 2 029 13diPglycerate 2HVglyceraldehyde 3P 2 O29 acetoacetate 2llBhydroxybutyrate 2 O27 s 2Hst 2 023 chlorophyll P P f 1 02 acetaldehyde 2Hethanol 2 O20 pyruvate 2Hllactate 2 019 FAD flavin adenine dinucleotide 2HFADH2free 2 018g oxaloacetate 2Hmalate 2 017 aketoglutarate NH 2Hglutamate H20 2 014 pyruvate NH 2Halanine H20 2 013 standard hydrogen half cell 2HH2 2 E0 000 methylene blue oxred 2 001 fumarate 2Hlsuccinate 2 003 cytochrome b Fe3Fe2 1 006 ubiquinone oxred 2 010 Cu2Cu 1 015 hemoglobin Fe3Fe2 1 017 cytochrome c Fe3Fe2 1 022 cytochrome a Fe3Fe2 1 029 211 02H202 2 030 chlorophyll l fVP IJ 1 04 N05 2HNo H20 2 042 30239 2Hso H20 2 048 Fe3Fe2 1 077 211 12 021120 2 082 chlorophyll P PIOI 1 09 an is the number of transferred electrons bFor a 2equot reaction AE of 010 volt corresponds to A06 of 46 kcalmole cDetermined at pH 7 and 25 C relative to the standard hydrogen half cell pH 0 dChlorophyll at the photoreactive center of photosystem I in photosynthesis P PT and P represent the excited the electrondeficient and the ground state forms respectively of the pigment molecule See Chapter 12 ePi represents inorganic phosphate HPOE fChlorophyll at the photoreactive center of photosystem II in eucaryotic photosynthesis P P and P91 represent the excited the electrondeficient and the groundstate forms respectively of the pigment See Chapter 12 EThis value is for the free coenzyme Different flavoproteins carrying bound FAD Chapter 11 vary in Ed from approximately 00 to 03 volt 0 Both chemotrophs and phototrophs obtain energy from redox reactions Chemotrophs require an environmental source of high energy electrons eg an organic molecule and an acceptor to which Thermodynamics amp Bioenergetics 102103 Consider the relationship between standard free energy change AG and the equilibrium constant for a chemical reaction aA bB cC dD abcd no moles of ABCD 571 Kleq pH7 C C D d l 1 Ala Blb AG AGO RT ln CC Dd 572 A1303b Free energy change standrad free energy change fconstituent concs If we allow reaction 571 to proceed toward equilibrium free energy change will decrease during the drive to equilibrium In the drive toward equilibrium the reaction will do work at constant temp and pressure When the system reaches equilibrium no further net chemical change will occur and AG 0 At equilibrium AG 0 and Y 0 AGO RTaneq 13 AGYo Standard Free Energy temp 25 C 298 K pressure 1 atrn concentration of reactants and products l molal concentration use 1 Molar R 198 calK 7 mole In addition biochemists use two non standard conditions 1 pH7 and 2 water concentration assumed constant By convention both H1 ion concentration and water concentration are left out of the equilibrium expression 1 l and are included in the value of Keq for any reaction in which they appear Then AG0 is shown as AGYo At equilibrium AGY and expression becomes AG 0 7 RT ln Keq 14 AG AGYo only when all reactants and products are present at 10 M Now how about the concentrations of reactants and products in a human cell Sample calculation Given Enzyme phosphoglucomutase catalyzes reaction Glucoser 1 phosphate 79 glucose767phosphare Start with 0020 M 1713 add the enzyme and allow reaction to proceed the nal equilibrium mixture will have 0001 M Crer and 0019 M GrBrP at 25C and pH7 Calculate equilibrium constant and Standard Free Energy change Keq G671D 0019 19 Crer 0001 The Standard Free Energy change would be AG O 7 RT ln 7 1987 calmol X 298 K In 19 AG o 1743 calmo Standard Free Energy change of a chemical reaction is the Difference between sum of free energies of products and sum of free energies of reactants all in standard state ie conc of l Molar temp 25C pressure 1 atm Each chemical compound has a characteristic intrinsic free energy by Virtue of its molecular structure Standard free energy change AG 0 can be expressed as AG O 2 Goproducts 2 Goreactants 15 For reaction 571 AG O CGOC dGOD aGOA bGOB 16 Example Enzyme Fumarase r Fumarate to Malate Fumarate H20 Malate AG 0 Z Aij Omalate Ofumarate 0 H20 720198 70144417 5669 7088 kcalmol Chapter 6 7 HOW CELLS GROW 61 p 155 Introduction 7 1 Substrates Cells Extracellular Products More Cells 61 ZS X 2P n 3 Net Speci c Growth Rate Mm 1 62a X it um g 7 kd 62b 5 Cell growth in terms of cell number concentration MR l m MR net specific replication rate 63 N dt M R indicates that kd 0 cell death unimportant or ignored 62 p 156 Batch Growth 621 Quantifying Cell Concentration 6211 Determine Cell Number Density 1 Petroff7Hausser slide or hemocytometer 2 Petri dishes agar 7 viable cells 7 plate countquot method 3 Particle counter Fig 61 6212 Determine Cell Mass Concentration 1 Direct methods 7 11 Dry weight 12 Packed cell volume centrifuge fermentation broth 13 Optical density turbidity using spectrometer for light transmission 2 Indirect methods 7 fermentation process molds S or P 21 lntracellular components RNA DNA protein Fig 62 211 DNA 8L protein concs remain fairly constant 212 Cell Protein 7 total amino acids Biuret Lowry Kjeldahl 22 lntracellular ATP concentration mg ATPmg cells 7 constant luciferin 02 777777777 779 light 64 luciferase ATP conc of a typical bacterial cell 1 mg ATP g dry7weight cell 23 Nutrients used for cell mass production N P S O 2 carbon source Note This relates to the stoichiometry material covered previously 24 Products of cell metabolism 7 241 7 anaerobic conditions 7 ethanol lactic acid 242 7 aerobic conditions 7 C02 243 7 changes in pH eg nitrification denitrification 244 7 viscosity change 7 due to extracellular polysaccharide 62 160 Growth Patterns and Kinetics in Batch Culture 1 Batch growth curve7Eigure 63 phases include lag log deceleration stationary death 2 Lag 7 adaptation of cells to a new environment Note 1 Many commercial fermentation plants rely on batch culture 2 To obtain high productivit the lag phase must be as short as possible 21 Low concs of some nutrients and growth factors can extend the lag phase 22 Example 7 Enterobacter aerogenes 7 lag phase increases as Mg 2 is decreased Mg2 is a activator of the enzyme phosphatase 23 Example 7 heterotrophic cells require C02 fixation and excessive sparging removes metabolically generated COz too rapidy 24 Age of inoculum culture how long maintained in a batch condition minimize 7 cells are adapted to growth mediumconditions before inoculation 7 cells should be exponential phase 7 inoculum size should be large 5 to 10 by volume 25 Multiple lag phases 7 diauxic growth see Example 41 3 Log or exponential phase 7 balanced growth 7 all cells grow at same rate average composition remains constant 7 31 Net Specific Growth Rate remains the same during this phase 32 Exponential growth rate is First Order with respect to X g um X XX0 at t0 65 dt lntegration yields 11 X mt l or XX0 umtt 66 0 Where X and X0 are cell cons at time t and t0 33 Doubling time time required to double microbial mass or number td ln 2 0693 67 Hnet Hnet t d ln 2 68 H R 4 Deceleration phase 7 1 Depletion of one or more essential nutrients or 2 Accumulation of toxic byproducts of growth 3 Changes occur over a very short period of time log to deceleration phase 4 Cell physiology under conditions of nutrient limitations are studies in continuous culture 5 Stationary phase 7 net growth rate 0 1 Growth rate death rate 2 Metabolites 7 primary growth7related 7 secondary nongrowth7related 3 Phenomena occurring during Stationary phase7 1 Total cell mass constant but number of viable cells decrease 2 Cell lysis mass decrease 2 growth on lysis products 3 Secondary metabolites 7 produced as results of deregulation 4 Endogenous metabolismcatabolism7cell reserves for building blocks 8L energy 1 g 7kdX or XXsoe kdt 69 it XSD cell mass conc at beginning of Stationary phase Because S 0 ug 0 in Stationary phase Z If inhibitory product is produce and accumulates ethanol7yeast manage by 1 Dilution oftoxified medium 2 Add unmetabolizable chemical compound to complex the toxin 3 Simultaneous removal of toxin 6 Death phase 7 rate of death usually follows first7order kinetics g 7de or N Nse dk W 610 it where NS is the concentration of cells at the end of the Stationary phase and k d is the first7order death rate constant 7 Yield Coefficient Growth Yield in a fermentation is YXs i M 611 Apparent Growth Yield at end of a fermentation batch growth period AS ASassimilation ASassimilated into ASgrowth ASmaintenance 612 into biomass extracell prod energy energy Other yield coefficients YxO2 AX YpS Q 613 614 A 02 AS Table 61 p 167 lists values of YXg and YXO2 for substrates 8L organisms Maintenance Coefficient specific rate of substrate uptake for cell maintenance M7 dSdtm 615 X 1 During Stationary phase 7 endogenous metabolism is used for quotMquot 2 Maintenance 7 repair damage components transfer nutrients inout of cells motility adjust osmolarity of cell interior 8 Microbial products are classified into three major categories Fig 66 p 168 1 Growth7associated products are produced simultaneously with microbial growth 01p l YPx Hg 616 X it eg production of a constitutive enzyme 2 Nongrowth7associated product formation occurs during Stationary Phase Specific rate of Product formation is constantquot qp B constant 617 eg Secondary metabolites 7 antibiotics eg penicillin 3 Mixed7growth7associated product formation occurs during slow growth 8L Stationary phases Specific rate of product formation qp ocug 3 618 eg lactic acid fermentation xanthan gum 8L some secondary metabolites 9 Example 61 7 Growth rate amp Yield 7 Solution 1 Max Specific New Growth Rate net lfl 7 ll 01 hi1 l2 l1 2 Apparent Growth Yield Y M 41 7 04 g cellg subtrate AS 063 7 100 3 Maximum cell concentration if 150 g glucose used with same inoculum size Xmax X0 YSO g cellsL 623 p 169 7 Now Environmental Conditions affect Growth Kinetics 1 Temperature above optimal level 7 specific replication rate Fig 67 m M R715d N 619 1 Both 12 and k d vary with temperature Arrhenius Eq H R Ae7EaRT 7 k d Aie7EdRT 63920 Ea activation energy for thermal growth 10720 kcalmol Ed activation energy for thermal death 60780 kcalmol 2 Temp 7 affects Yield coefficient Yxs eg single 7cell protein production 1 Affects Maintenance coefficient m Em 15720 kcalmol 3 Temp 7 affects rate 7limiting step in fermentation process 1 Rate of bioreaction gt rate of diffusion immobilized cell system 2 Ea for molecular diff 6 kcalmol Ea for bioreactions gt 10 kcalmol 2 pH 7 bacteria 378 animal cells 65775 non7optimum gt quotmquot increase NOg39 utilization causes increase in pH Organic acid production decreases pH C02 production affects pH 3 Dissolved Oxygen 7 consumption rate gt supply rate at high cell concentration 1 Growth rate can be proportion to Oz conc MichaeliseMenten relation 2 Fig 69 172 7 growth rate as fOz while mass is determined by S 3 Critical 02 as DOW 571 0 bacteria yeast 1050 mold gtlt 4 Rate of Oz transfer NO2 mg Ozlehr kLaC CL 62l kL oxygen transfer coefficient cmh a gaseliquid interfacial area cmZcms kLa volumetric oxygen transfer coefficient h l CSat CLDo conc 5 Oxygen Uptake Rate OUR mg Ozh qo2X ugx 622 YXO2 6 OTR OUR g X kLaCrCL 623 YXO2 Or 7 YXo2 kLaC CL 624 dt p 173 8 Redox Potential Eh E O M log 1302 23 log H 625 4F F 9 lonic Strength 7 l affects transport of nutrients in and out of cells and 2 solubility of Oz 3 l l 2 CiZiZ 626 eg NaCl inhibitory at gt 40 gL osmotic pressure max conc glucose 100 gL ethanol 50 gL 624 Heat Generation by Microbial Growth you read 63 p 175 Quantifying Growth Kinetics 631 Introduction Structured model 7 divides cells mass into components Unstructured model 7 assumes fixed cell mass composition balanced growth Valid in single7stage steady7state continuous culture 8L log phase of batch culture Fails during any transient condition Segregation 7 divide culture into individual units cells7 differ from each other 632 Unstructured Nonsegregated Models to Predict Specific Growth Rate 6321 Substrate7limited growth p176 uy 11ES Fig 611 MONOD EQN 630 Ks S 7 Michaelis 7 Menten kinetics for enzymes applied to whole cells for growth 7 If endogenous metabolism is not important then mt MY p 177 For rapidly growing dense cultures W umS 631 K50 SO S Where So is the initial substrate concentration and K30 is dimensionless Skip equations 633 through 638 6322 Models with growth inhibitors High S P and with inhibitory substances growth rate depends on inhibitor concentration 1 Substrate inhibition 7 skip 2 Product lnhibition 7 a Competitive My 2 lm S 642 Ks1 PKp S b Noncompetitive MY 2 um i 643 l KgS l PKp Example Ethanol fermentation from glucose ETOH is inhibitor at concentrations above 5 3 Inhibition by toxic compounds 7 same as enzyme kinetic expressions a Competitive Mg lm S 646 Ksl lK1 S b Noncompetitive Mg gm 647 1 KgS 1 lKI c Uncompetitive Mg d include death term in rate expression Hg ims kyd KS S J33 K S l lK1 HIKr j 648 649 6323 The Logistic Equation To describe growth curve in Fig 63 combine growth equation 62 with Monod equation 630 and assume no endogenous metabolism l lnetx dt l lnetllms KS 8 To get J S X Clt KS S Relationship between microbial yield Y and substrate consumption is Y Q dX y d8 d8 lntegrate over X and S to get X 7 X0 sz SO 7 S See text for Equations 652 and 653 Logistic equations are a set of equations that characterize GROWTH in terms of 62a 630 650 651 CARRYING CAPACITY CC Specific growth is related to amount of unused CC pg k17XXoo x00cc Therefore M dx 9 RX 1 XXoo dt X dt Xo e 17 x9 16 Xoo Equ 656 is represented by the growth curve in Fig 612 182 X Study 7 Example 62 Logistic Equationz pages 1817182 654 655 656 6324 p 183 Growth models for filamentous organisms or submerged microbial pellet SKlP 633 p183 Models for Transient Behavior 7 shift in environmental or cultural conditions 6332 Chemically Structured Models Fig 614 1 All reaction should be expressed in terms of intrinsic concentrations amount of a compound per unit cell mass or cell volume Extrinsic concentration amount of a compound per unit reactor volume 7 cannot be used in kinetic expressions 2 The dilution of intrinsic concentration by growth must be considered 321 VRX r 660 dt 634 p 189 Cybernetic Models 7 Process is goal seeking eg maximization of growth rate 1 Initially motivated by desire to predict response of a microbial culture to growth on a set of substitutable carbon sources 2 Recently 7 identify regulatory structure of a complex biochemical reaction network eg cellular metabolism 3 Newest use 7 metabolic engineering 7 relating information on DNA sequences in an organism to physiologic function See Chapter 8 64 p 189 How Cells Grow in Continuous Culture 641 Introduction 7 1 Constant environmental conditions for growth product formation 2 Determine response of cells to the environment 642 Specific Devices for Continuous Culture 1 Chemostat Constant chemical environment 7 constant nutrient cell product 2 Turbidostat 7 cell concentration maintained constant Study environmental stress select cell variants or mutants with desirable properties 3 Plug Plow Reactor PPR 643 ldeal Chemostat CPSTR with pH and DO control units Fig 618 Material balance on cell concentration around the chemostat FXO 7 FX tth VR 7 kdX VR VR 664 dt Solve for g 97 g HEX VR 7 th VB 664a Dl VR VR VR VR F D dilution rate l Where 6 retention time in chemostat V From 664 FXO 7 FX ng VR 7 kdX VR g VR dt 0 7 FX HEX VR 7 0 0 steady state Therefore FX ng VR Divide by X F pg VR Solve for pg pg F l D 666 VR 6 Therefore setting the Dilution D rate sets the Growth Rate 1 g NOW substitute Monod equation pg D Emil S 667 Ks S Find max Kg 7 from plot of lug versus lS 1 K S Ks 5 K 11 Mg pm 8 MM S MM 8 HM S um Y In X b Use 667 to solve for relate S as a function of D D KS T max S D Ks DS umax S DKS l lmaxSTDSSlJ39maxTD Therefore S D K 668 max 7 Material balance on limiting substrate around the chemostat FSO HEX VR 1 quVRl ClS VR 669 YMWS Yps dt FSO 7 FS 7 ng VR 1 r 0 extracell prod 0 steady state Y XS F 80 i HEX VR 1 Y XS F Q 7 S D SO 7 S HEX 670 VR YMws Since Hg D at steady state if kd 0 then DSOrS DX Y gt03 Therefore X YMXS SO S 671 Using Equation 668 the steadyrstate cell concentration can be expressed as X YMXS SO D Kg 672 max 7 Now lets consider endogenous metabolism Equ 666 D pg becomes D pg 7 kd um 673a Or pg D kd 673b Substitute 673b into steady7state substrate balance and extracell 130 then Eqn 669 becomes DSO 7 S 7 D l5dX 0 673c Y gt03 Where YMWS denotes the maximum yield coef cient no maintenance or endogenous respiration Eqn 673c can be rearranged to DQQQ 7 1 2 ki 0 674 X Y XS Dp 7D ikd 70 67520 Y XS WW8 Y MS Divide by D and solve for lYAPXS 1 1ltd 675b YAPXs YMm YMws D l ms 676 YAPXs W D Y b m X Where ms kg 677 YMws While YMWS is a constant YAPXS varies with growth conditions if kd gt 0 Find YMX S and mg plot 1 YA versus lD slope ms intercept 1 Y XS r 675uR 6 3 475R3 032r pp pf
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'