Biochemical Engineering BIE 5810
Utah State University
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This 25 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 Ronald Sims in Fall. Since its upload, it has received 36 views. For similar materials see /class/230405/bie-5810-utah-state-university in Biology at Utah State University.
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Date Created: 10/28/15
r 675uR 6 3 475R3 032r pp pf 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 5 7 MAJOR METABOLIC PATHWAYS Introduction p133 7 1 Metabolic Engineering remove andor add genes 2 Catabolism 7 process of degrading into smallersimpler products 8L produces energy 3 Anabolism 7 synthesis of more complex compounds 8L requires energy 527Bioenergetics p 134 1 Metabolic reactions in a bacterial cell Fig 51 l Fueling Class I Z Biosynthesis Class H 3 Polymerization Class Ill 2 ATP H20 9 ADP Pi AG 773 kcalmol 51 3 ADP H20 9 AMP Pi AG 773 kcalmol 52 4 Formation of ATP 1 Substrate level phosphorlyation 7 organic compounds as energy source 2 0xidative phosphorylation 7 electron transport 3 Photophosphorlylation 7 electron transport 5 Fig 52 PEP and 137DPG are high energy P cpds 8L P donors while 6 H atoms 7 released in biological redox reactions carried by NAD NADP Fig53 7 p 36 NADH 7 two major functions 1 Reducing Power 7 supply H in biosynthetic reactions eg C02 4 H 9 CH20 H20 autotrophs 53 2 ATP formation in Respiration Metabolism 7 H transferred to 02 Results in formation of up to 3 ATP molecules ATP can be formed if alternate electron acceptor AEA is present N03 53 Glucose Metabolism Glycolysis 8L TCA Cycle Embden7Meyerhof7Pamas pathway 1 p137 Aerobic catabolism of organic compounds 7 3 phases 1 EMF pathway for fermentation of glucose to pyruvate Z Krebs tricarboxylic acid TCA or citric acid cycle for conversion of pyruvate to NADH and C02 3 Respiratory or electron transport chain for formation of ATP by transferring e from NADH to an electron acceptor a aerobic respiration 7 02 as E Z anaerobic respiration 7 N03 8043 Fe2 S0 2 Glycolysis 7 Fig 54 p138 1 Energy APT w in Steps 1 and 3 for phosphorylation 2 Energy ATP gained by substrate7level P 137DP7GA 9 3P7GA PEP 9 pyruvate Question What is the net ATP yield for EMPpathway 3 Pyruvate as key metabolite l anaerobic conditions 7 lactic acid ethanol acetone acetic acid 2 aerobic conditions 7 converted to NADH and C02 through TCA cycle 3 Glycolysis reaction 0ccurs in CYTOPLASM glucose Z ADP Z NAD 2 Pi 9 Z pyruvate 2 ATP 2NADH 14 54 4 Bioenergetics of Glycolysis Efficiency of energy conversion Glycolysis AG a 52000 calmol glucose 2 ADP 2 ATP AG 14 600 calmol glucose AG 7 37400 calmole glucose 1 Efficiency 14 600 cal 28 for lactic acid fermentation 1 52000 cal 2 Efficiency in utilizing total energy potentially available from glucose E 14 600 cal 2 typical of fermentations 686000 cal p 139 TCA cycle main functions 1 provide e NAD H for electron transport chain and biosynthesis 2 supply C skeletons for AMINO ACID synthesis 3 generate energy TCA Cycle occurs in matrix of Mitochondria in Eucaryotes TCA Cycle occurs in membraneebound enzymes in Procaryotes Entry into Krebs Cycle 7 acylation of coenzymeeA by pyruvate Pyruvate NAU COAASH 77777777777777777777 quot9 acetyl CoA NAHD H 55 Pyruvate dehydrogenase Note Acetyl CoA is transferred through the mitochondrial membrane at expense of conversion of 2 NADHs produced in glycolysis to 2 EADH Acetyl CoA key intermediate in metabolism of AMINO ACIDS 8L FATTY ACIDS 6 TCA Cycle 7 Fig 55 p 140 Question Show where Reducing Power is generated and energy is made in TCA Precursors for certain AMINO ACIDS succinate and oceketoglutarate 7 Heterotrophic COz fixation 1 TCA intermediates used for biosynthesis quotshortecircuitquot of TCA cycle 2 Heterotrophic COg fixation by cell to maintain functioning TCA cycle a PCP COz oxalacetate b pyruvate COz oxalacetate c pyruvate COg malate 3 Can be important factor in culturing microbes Growth can be limited by the RATE OF COz FIXATION TO MAINTAIN THE TCA CYCLE 7 a when culture is initiated at low density with little accumulation of inctracellular COz or b when a gas sparge rate into a fermentation tank is high 54 Respiration 141 Question How many max ATPs are produced through EMP and TCA cycle with 02 as terminal electron acceptor Show calculations using spreadsheet and connect biochemical reaction to the production of reducing power See Table 51 p141 l Oxidative phosphorylation r forming ATP from electron transport chain a 3 ATPs for each NAHD H b 2 ATPs for each FADHZ in eucaryotes 2 Figure 56 7 Respiratory cytochrome chain a regenerate NADs for glycolysis b generate ATPs for synthesis 3 Overall reaction of glucose metabolism in eucaryotes C6H1206 36 Pi 36 ADP 6 OZ 6 C02 6 H20 36 ATP 57 PO 36 ATP 12 O 3 ATPO for NADH PO 4 ATP 2 O 2 ATPO for FADHZ Energy Balance for Efficiency 36 ATPs X 73 kcalmole ATP 263 kcal captured AGO 686 kcal for oxidation of glucose according to equation 57 Efficiency 263 kcal captured 686 kcal released 38 Now consider the relationship between standard free energy change AG and the equilibrium constant for a chemical reaction Consider aA bB cC dD abcd no moles of ABCD 571 Free Energy Change constant Temp and Press f standard free energy concentration of reactants and products of the reaction AG AGquot RTln CC Dd 572 ANSb So the standard free energy change for glucose oxidation is 686 kcalmole AG AGO only when all reactants and products are present at 10 M Now how about the concentrations of reactants and products in a human cell Human erythrocyte has been evaluation of steadyrstate concentrations of intermediates in glycolysis When these concentrations of reactants and products are considered the energy recovery efficiency is about 53 Glucose 79 lactate 2 ATP AGO 47 kcalmol Efficiency 146 kcal captured 52 kcal released 28 In muscle tissue for glycolysis efficiency A the actual efficiency is higher than 53 close to 60 Back to p 141 For Procaryotes 7 ATPNADH H lt 2 ATPFADHS l 55 p142 CONTROL SITES IN AEROBIC GLUCOSE METABOLISM I Glycolysis ATP 1 phosphofructokinase active phosphofructokinase inactive ADP Phosphofructokinase is ALLOSTERIC enayme activated by ADP 8 Pi and inactivated by ATP 2 Oz concentration 7 regulatory effect called Pasteur Effect Glycolysis Rate higher under anaerobic conditions Wny does aerobic conditions slow down rate of glycolysis Explain 3 High NADHNAD reduces the rate of glycolysis II Krebs Cycle High ATPADP and NADHNAD ratios reduce the processing rate of the TCA cycle III Electron Transport Chain inhibition 1 Block e39 transport in span between NADH and ubiquinone act on NADH dihydrogenase Rotenone plant substance used by South American Indians as fish poison Amytal barbituate drug Piercidin antibiotic that resembles ubiquinone and can compete With it 2 Block e transport in span between cytochrome b to c Antimycin A isolated from Streptomyces griseusl 3 Block e39 transport from cytochrome aa3 to oxygen Hydrogen cyanide HCN Hydrogen sulfide HZS Carbon monoxide CO 56 p 143 METABOLISM OF NITROGENOUS COMPOUDS Used by the cell as carbon nitrogen and energy sources 1 Amino Acids converted to organic acids by deamination R7CH7COOH HZO NAD 9 R7C7COOH NH3 NADH H 510 NHZ O a Organic acids 7 oxidized for energy production ATP b NH3 used in protein and nucleic acid synthesis 2 Transamination7change amino acids to organic acids 8 other amino acids amino group exchanged for keto group of 27keto acid 4 Nucleic Acids7purinesyrimidines9 urea acetic acid 9 NH3 COz 57 p 144 NITROGEN FIXATION N26H6e 9 2NH3 514 nitrogenase 7 Need reducing conditions no oxygen 7 Rhizobium and Azospirillum 7 used for agricultural purposes 7 x N and provide ammonium to plants They are bioprocess products 58 p 144145 METABOLISM OF HYDROCARBONS 1 Aliphatic HCs 7 Bioremediation 7 requires oxygen Pseudomonas Mycobacteria a Low water solubilig is a barrier to rapid metabolism Refer to article quotTransfer of Polycyclic Aromatic Hydrocarbons Between Model Membranes Relation to Carcinogenicityquot Authors Plant AL HJ Pownall and L C Smith Journal Chem7Biol Interactions 44 2377246 1983 b Metabolism7 rst step7oxygenation by oxygenases alcohol aldehyde acid 2 Aromatic Hydrocarbons 7Metabolism 7 aerobic oxygenases See example for Benzene p 145 Benzene 9 catechol 9 cis7cis7muconate 9 B7keto adipate 9 acetyl CoA succinate Anaerobic metabolism 7 alternate electron acceptors NOg39 SO42 Mn4 Fe3 59 p 145 OVERVIEW OF BIOSYNTHESIS l Pentose7phosphate pathway Hexose7monophosphate pathway Fig 57 a Reducing Power produced NADPH for anabolism b Carbon skeletons 3 4 5 8 7 C for biosynthesis using NADPH l ribose purines coenzymes aromatic amino acids 2 amino acid 7 commercial products 7 families 7 Fig 58 p 147 c Fatty Acid synthesis 7 stepwise buildup of acetyl7CoA l acetyl7CoA C02 9 malomyl7CoA intermediate of FA synthesis 2 C02 requirement 7 can lengthen start7up phase for commercial prod supply key lipids eg oleic acid d Polysaccharide synthesis 7 from hezose eg glucose is common If lt 6 carbons EMP pathway can operate in reverse to produce glucose GLUCONEOGENESIS a start with pyruvate but pyruvate7PEP is irreversible in glycolysis EMP b pyruvate COZ ATP H20 9 oxalacetate ADP Pi ZHT oxalacetate ATP 9 PEP ADP C02 517 c after PEP glycolysis reversible up to fructose7l67diphosphate need two enzymes NOT in EMP l fructose7l 67diphosphatase 2 glucose767phosphatase 510 p 148 ANAEROBIC METABOLISM l Anaerobic Resppiration Section 52 7 a same pathways as aerobic metabolism b different terminal electron acceptor Alternate Electron Acceptor eg Denitrification Dissimilatory N05 9 N2 c Fermentation7generation of energy without electron transport chain 1 Fig 59 7 lactic acid 8 alcoholic fermentations 2 Fig 5107 Fermentation of sugars from lactic acid d Fig 511 7 Entner7Doudoroff Pathway 7 l fermentation of glucose by Zymomonas 2 production of 1 mole ATP per mole glucose not 2 a low energy yield forces more glucose into ethanol and less into cell mass than for yeast which uses glycolysis to produce pyruvate yields 2 moles ATPmole glucose 1 Fix COZ by enzyme ribulose bisphosphate carboxylase a Key step in Calvin Cycle Ribulose715rdiphosphate C02 9 2 glyceric acid737phosphate b See Fig 512 p 151 1 C02 xation 2 Reduction of xed C02 3 Regeneration of C02 acceptor 2 Photosynthesis 7 6 C02 H20 9 CeleOe 6 02 519 light 1 Phase 1 Light Phase H20 NADP Pi ADP 9 oxygen NADPH H ATP 520 light 2 Phase 2 Dark Phase COZ NADPH H ATP 9 16 glucose NADPT ADP Pi 3 Photophosphorylation 7 ATP generation in photosynthesis whereby light absored by chlorophyll results in electron excitation and return to normal state by emitting light quanta uorescence and electron that travels through electron transport chain 512 p 152 SUMMARY 7 class read on you own End Chapter 5 CHAPTER 1 WHAT IS A BIOPROCESS ENGINEER III Example Applications a Medicines Organ growth in reactors Foods Computers based on biological molecules rather an silicon Superorganisms to degrade pollutants Consumer products Industrial processes IOH1999quot Biological Engineering a Biological systems obey rules of chemistry and physics and are subject to engineering analysis b Living cells are predictable and can be modeled and quanti ed c Living processes can be constructed on commercial scales d Combine the skills of the engineering with the skills of the biologist Descriptions and Characterizations a Biotechnology 7 use or development of methods of direct genetic manipulation for a socially desirable goal eg a chemical a plant gene therapy an organism to degrade a waste b Bioengineering 7 Engineers working with biotechnology including work on medical and agricultural systems c Biological Engineering 7 Emphasizes applications to plants and animals d Biochemical Engineering 7 Extension of chemical engineering to systems using a biological catalyst to bring about desired chemical transformations e Biomedical Engineering 7 getting closer to biochemical engineering especially in the areas of cell surface receptors and animal cell culture f Biomolecular Engineering 7 de ned by National Institutes of Health as research at the interface of biology and chemical engineering and is focused at the molecular level g Bioprocess vs Biochemical Engg 7 Bioprocess Engg Is broader than biochemical engg and includes the work of mechanical electrical and industrial engineers to apply their disciplines to processes based on living calls or subcomponents of cells Includes equipment design sensor development control algorithms manufacturing strategies etc Focus of this course Application of chemical engineering principles to systems containing biological catalysts with an emphasis on systems making use of biotechnology The cell itself is now a designable component of the overall process V VI Perspectives of Biologist and Engineers a Fundamental training i Biology 7 qualitative improvements in experimental laboratory tools descriptive models interpretation of laboratory data from complex systems Formulation of testable hypotheses experimental design and data interpretation Engineering 7 emphasis on physical and mathematical sciences mathematical formulations quantitative models and approaches to complex systems Generally unfamiliar with experimental techniques and strategies used by life scientists Life Science Engineer 7 needs a solid understanding of biology microbiology biochemistry cell biology and its experimental tools to apply engineering principles to biological systems Story of Penicillin How biologists and engineers work together a Fquot 0 3 1 D Iquot 1928 Alexander Fleming St mary s Hospital in London Staphylococcus aureus Observations 7 lack of bacterial growth near contaminated particle Fleming s genius was to realize that this observation was meaningful and not a failed experiment Killing organism was Penicillium notatum Norman Heatley 7 biochemist play the role of bioprocess engineer He developed an assay to monitor amount of penicillin made to determine the kinetics of the fermentation During WWII Howard Florey of Oxford approached US companies 7 Merck P zer Squibb and the USDA I Illinois In 1940 fermentation for the production of a pha1meceutical was an unproved approach versus chemical synthesis which had more confidence and penicillin is a fragile and unstable product USDA Northern Reginal research Lab in Peoria IL made major contributions including development of a medium corn steep liquor lactose based medium to increase productivity about 10 fold New strain Penicillium ch sogenum is superior and was used Manufacturing challenges 39 One method was growth on the surface of moist bran Difficulties in temperature control sterilization and equipment size Surface method 7 milk bottle containers bottle plants Difficulties 7 long growing cycle and labor intensive Submerged tank process engineers 7 Problems of oxygen supply and heat removal are current problem in antibiotic fermentor design Pfizer completed the first plant for commercial production of penicillin by submerged fermemtation iv Other hurdles in product recovery and purification v Merck 7 realized that people who understood both engineering and biology were NOT available They assigned an chemical engineer and a microbiologist to work together on each aspect Progress has involved better understanding of L ii39 L V mold physiology metabolic pathways penicillin structure methods of mutation and selection process control reactor design The Penicillin process established a paradigm for bioprocess development and biochemical engineering that sill guides the 9959 S profession viii No similar paradigm has emerged from our experience with quot quotJ 39 J cells 39 39 DNA VII Bioprocess Regulatory Constraints a US Food and Drug Administration 7 Three phases for drug approval takes 15 years from discoverythroughprocess approval i Phase I 7 safety ii Phase II 7 efficacy amp side effects iii Phase III 7 large scale testing 1000 to 3000 patients b GMP 7 good manufacturing practice also GLP 7 good laboratory practices and SOP 7 standard operating procedures c FDA approval is for the product AND the process together
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