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CoQ10 Production: Fed-Batch Fermentation Case Study
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Case study: biotechnological production of CoQ10 by fed-batch fermentation Coenzyme Q10 is a valuable lipidic product synthesized by several organisms. It’s an intracellular metabolite which plays a role in the electron transport chain. Synthesis is oxygen-dependent by R. sphaeroides, and reported yields are up to 770 mg/L (9 mg/g DCW) at 30 °C, controller agitation, pH and aeration in fed-batch culture after ~72 h. Carbon and nitrogen sources vary according to raw-material availability and alternative low-cost sources may be utilized. The product market is expected to grow to $1 billion by 2026 as it’s being widely used as a food and cosmetic supplement due to its bioactive properties. Let the overall bioreaction be: πΆ6 π»12 π6 + πΌπ2 + π½ππ»4 πΆπ → πΎπ + πΏπΆ59 π»90 π4 + ππΆπ2 + ππ»2 π + ππ»πΆπ Species G Glucos e O Oxygen N Ammoniu m chloride X Biomass P Coenzym e Q10 C Carbon dioxide W Water CO2 A Hydroge n chloride HCl Formula C6H12O6 O2 NH4Cl CH1.99O0.5N0.1 C59H90O4 Phase Solid Gas Solid Aqueous Solid Gas Aqueous 26.64 - 863.365 1? 36.46 1.49 - 50 insoluble 44 0.00197 7 1.7 Liquid . 18 1 MW Sp. Gr. 180.16 1.54 53.49 1.519 Tf Tb Solubilit y (water) 148 909 32.0 0.00142 9 -218 -183 0.007 337 383 720 0 100 - H2O 9 Bioreaction πΆ6 π»12 π6 + πΌπ2 + π½ππ»4 πΆπ → πΎπ + πΏπΆ59 π»90 π4 + ππΆπ2 + ππ»2 π + ππ»πΆπ π = πΆπ»1.99 π0.5 π0.19 π π = πΆ 0 0 1 59 1 π» 0 −4 1.99 90 0 π΄ π = −2 0 0.5 4 2 π 0 −1 0.19 0 0 [πΆπ ] [ 0 −1 0 0 0 π =2 πΌ 0 2 1 0 0 πΌ π½ π1 = 6 + (πΌ − π) 0 πΎ π2 = 12 1 0 ∗ πΏ = π3 = 6 + (2πΌ − 2π) π π4 = 0 0 ] π π5 = 0 1 [ ] [ π] Taking RQ = 2; 2α = ε, thus α = 0.5 y ε = 1: π1 = 5 π½ πΆ 0 1 59 0 0 π πΎ π» −4 1.99 90 2 1 2 = 12 π΄ π = 0 0.5 4 1 0 ∗ πΏ = π3 = 3 π π4 = 0 π −1 0.19 0 0 0 [πΆπ ] [−1 0 0 0 1] [ π] [ π5 = 0 ] πΆ6 π»12 π6 + 0.5π2 + 0.18ππ»4 πΆπ → 0.98π + 0.068πΆ59 π»90 π4 + 1πΆπ2 + 2.24π»2 π + 0.18π»πΆπ Hierarchical process synthesis LEVEL 1: DECISIONS ON BATCH vs CONTINUOUS - Technical information: Does any apparatus work in batch mode? Is the process sensitive to upsets & variations? Fed-batch mode suggested, sensitive to variations. - Production rate High or low production rate? Only few days production needed? Few days operational notice? Low production rate, process needs 3 days to complete. - Product lifetime One or two years or longer? Longer. - Value of product… Product value >> manufacturing cost? Yes. LEVEL 2: DECISIONS ON INPUT-OUTPUT STRUCTURE - - Raw materials – Impurities? If the impurities are inert, remove them after reaction, if they’re valuable – Non valuable. If the impurities are inert, present in large amounts, and can be easily separated, remove them before the reactor – Impurities in aqueous solution, inert or taken up for cell metabolism. If the impurities have boiling points lower than the reactants and products, and they’re also inert, recycle them (note: purge unit will be needed!!) – Reaction occurs at liquid state, well below species’ boiling point. If the impurities are also products from the reactor, place feed stream before the unit that will remove the impurities – biomass must be removed, fed-batch is proposed. - Product streams? Recycle unreacted reactants – OK… Recycle intermediate reactants (products) – No Recycle/remove azeotropes with reactants – No Remove (recover) the main product – OK Remove (recover) the valuable by-products – None Remove (as waste)/recycle by-products that are not valuable – OK. - Recycle? Purge Units? For <100% conversion of reactants and reaction in the gas phase. - - Recycle of gases will be necessary – Not necessary. (Check CO2, water) If impurities are present, purge units will be necessary – No. (Check CO2, water). For < 100% conversion and reaction in the liquid phase, Recycle of liquids (reactants) will be necessary – May be possible to recycle carbon source stream. If impurities are present, purge units will be necessary – YES Reversible by-products? Selectivity vs costs? EP2 = Product value – By-products value – Raw-material value Product value (standard): $450 USD/gram*(1000 gram/kg) – 0(no valuable by-products) – $0.6 USD/kg CS - $2.25 USD/kg NS… Approximately $450,000 USD/kg of product. Noting each batch takes around 72 h, reported scale-up volume of up to 150 L: 0.77 g/L*(150 L/3 days)*($450K/kg)*(300 days/year) = $5.2 Million/yr. mm LEVEL 3: DECISIONS ON RECYCLE STRUCTURE & REACTORS - - - Number of reactors? If more than one reactor is needed to get the desired product, more than one reactor will be needed if the conditions are very different 1 reactor? Number of recycle streams? Depends on the number of raw materials and conversion of all reactants… Oxygen comes from air; sugar may be reused (1 stream), find conversion from yield. Take biomass out (centrifuge). Need for compressor/pump? Air is fed at atmospheric pressure. Pump needed for recycle stream. Compressor for CO2 stream. - - Reactor type? Adiabatic or Isothermal? If temperature change is too high or low, heating/cooling will be needed – Isothermal (jacketed reactor). Reaction equilibrium or kinetics? Kinetically controlled or equilibrium reached? - ??? EP3 = EP2 – (reactor cost + pump cost + compressor + heat exchr). LEVEL 4: DECISIONS ON SEPARATION SYSTEMS - Vapour recovery and/or liquid recovery? – Rules If reactor stream is in liquid phase, use liquid recovery If reactor stream has 2 phases (separate the 2 phases first) – Liquid phase only Reactor stream is gas (vapour) phase: may need to condense CO2 How to locate & perform separation? EP4 = EP3 – (vapor recovery cost + liquid recovery cost) - Vapour recovery system (VS) – Location No vapour - Liquid recovery system (LS) Decisions on… - No distillation; other types of separations ARE possible. - Cell harvesting? Microfiltration/centrifugation? - Cell disruption: homogenization/bead milling/ultrasound - Cell debris removal: centrifugation/microfiltration. - Extraction by: ORGANIC SOLVENTS (Hexane). It’s not been reported though reversed micelle technology suitable. - Purification: SEC, liquid-liquid extraction, crystallization in vacuum dryer. Mixture analysis: ??? MASS BALANCES BY UNIT U01 + U02 = U11 + U12 U11 = U21 + U22 U21 + U03 = U31 U31 = U41 + U42 U41 + U04 = U51 U51 = U61 + U62 + U63 U61 + U05 = U71 U71 = U81 + U81 BY COMPONENT: REACTOR If the working volume is 2800 L, fed with media containing 50 g/L glucose and 4 g/L NH4Cl, and considering glucose comes as a solution 70%w/v, then the pure water volume would be 2600 L. Additionally, if bioreactor operates at 0.5 vair/vmed/min, that’d make 30 v/v/h. If reactor operates (2800 L/72 h = 38.83 L), then our air volume would be 30 La/Lm/h * 38.83 Lm = 1166.6 Lair/h. By the ideal gas law (1 mol = 22.4 L) and O2 in air is at 21%: 1166.6 Lair/h*(1 gmol/22.4 Lair)*0.21 Lo/Lair = 10.94 mol/h. Thus: UG01 = 10.8 mol/h UN01 = 2.91 mol/h Uo01 = 10.94 mol/h Uw01 = 2000 mol/h. The fixed conversion reactor given by substrate consumption is fixed to 0.9. UG11= UG01(1-0.9) = 0.1 UG01 = 1.08 mol/h UN11= UN01 – 0.18*(0.9 UG01) = 1.2 mol/h Uo12= Uo01 – 0.5*0.9 UG01 = 6.1 mol/h (leaves as gas) Uw11= Uw01+2.24*0.9 UG01 = 2021.8 mol/h UX11 = 0.98*0.9 UG01 = 9.6 mol/h UC12 = 0.9 UG01 = 9.7 mol/ h (leaves as gas) UH11 = 0.18*0.9 UG01 = 1.8 mol/h UP11 = 0.068*0.9 UG01 = 0.66 mol/h IN FIRST DIVIDER (Centrifuge) Mixture contains biomass (X), product (P), water (W), and waste glucose and ammonium chloride (G & N) and produced acid chloride (H, supposed to be neutralized during reactor operation). Also, cell water content is around 70%: Input: π11 = 9.6 πππ 0.66πππ πππ πππ πππ πππ π+ π + 2022 π + 1.08 πΊ + 1.16 π + 1.75 π» β β β β β β Output: π21 = 9.6 πππ πππ πππ π + 0.66 π + 22 π β β β π22 = 2000 πππ πππ πππ π + 1.08 πΊ + 1.16 π β β β IN FIRST MIXER At the first mixer a neutralization reaction occurs, so cell is disrupted. π03 is a solution with 0.25 L of H2SO4 (3%) + 0.4 L of NaOH (14%) are fed are mixed to U21 and heated to 90 °C. Salts are to be removed. π»2 ππ4 + 2ππππ» → ππ2 ππ4 + 2π»2 π Input: UH2SO403 = 0.08 mol/h UNaOH03 = 1.4 mol/h Uw03 = 35 mol/h U21 = 9.6 mol/h X + 0.66 mol/h P + 22 mol/h W Output: UNa2SO431 = 0.08 mol/h UNaOH31 =1.24 mol/h UX31 = 9.6 mol/h UP31 = 0.66 mol/h Uw31 = 58.4 mol/h IN CLARIFIER Assuming total fatty acid content in dry cells is about 10%. We’ll assume a fraction of water content boiled at previous step and crystals are left dry. Input: U31 = 69.9 mol h Output: UP41 = 0.66 mol/h UX41 = 1 mol/h (lipid crystals, biomass base) UX42 = 8.6 mol/h (as cell debris) UNa2SO442 = 0.08 mol/h UW42 = 58.4 mol/h UNaOH42 = 1 mol/h IN SOLVENTIZER This stage will assume all product mass is going to be dissolved (absorbed) by the organic solvent (hexane). 0.5 L/h is added = 3.8 mol/h. INPUT U41 = 0.66 mol/h P + 1 mol/h lipid crystals = 1.66 mol/h U04 = 3.8 mol/h Output: U51 = 5.46 mol/h IN VAPORIZER All solvent is vaporized and leaves as gas. Cell debris fatty crystals are removed, and we may consider some product leaves attached to debris. Input U51 = 5.46 mol/h Output UP61 = 0.59 mol/h UX61 =0.1 mol/h U62 = 3.8 mol/h UP63 = 0.07 mol/h UX63 = 0.9 mol/h PREPARATION FOR PURIFICATION Raw product stream U61 is diluted in a hexane/isopropanol mixture (10:2) as mobile phase for chromatography. Input UH05 = 2 mol/h UI05 = 0.69 mol/h U61 = 0.69 mol/h Output U71 = 3.38 mol/h CHROMATOGRAPHY UNIT Assume continuous chromatographer. All product is recovered? Assuming pure product stream: Input U71 = 3.38 mol/h Output UP81 = 0.59 mol/h UH81 = 0.026 mol/h UI81 = 0.005 mol/h U81 = 0.621 mol/h UH82 = 1.974 UI82 = 0.685 mol/h UX82 = 0.1 mol/h U82 = 2.759 mol/h %MASS BALANCE FOR GLUCOSE %GLOBAL BALANCE %Coefficients Y_G=-1 Y_N=-0.18 Y_O=-0.5 Y_X=0.98 Y_P=0.068 Y_C=1 Y_W=2.24 Y_A=0.18 ETA_G=0.95 ETA_S=1 Y2_OH=-2 Y2_S=-1 Y2_F=1 Y2_W=2 %SPLIT FRACTIONS Z1_G=1 Z1_N=1 Z1_A=1 Z1_X=1 Z1_P=1 Z1_O=0 Z1_W=0.01 Z2_X=0.1 Z2_P=0.95 Z2_W=1 Z2_F=1 Z2_OH=1 Z3_X=0.05 Z3_P=0.95 Z3_H=1 Z4_P=0.99 Z4_X=0.001 Z4_H=0.001 Z4_I=0.001 x1=0.5 x2=5 x3=6 %Across reactor: M01_G=0.05*M01 M11_G=M01_G+Y_G*ETA_G*M01_G M21_G=0 M22_G=Z1_G*M11_G %MASS BALANCE FOR NITROGEN SOURCE M01_N=0.004*M01 M11_N=M01_N+(Y_N)*(ETA_G)*M01_G M21_N=0 M22_N=Z1_N*M11_N %MASS BALANCE OXYGEN: M02_O=0.5*M01_G M12_O=M02_O+Y_O*ETA_G*M01_G M22_O=M12_O %MASS BALANCE CARBON DIOXIDE M12_C=Y_C*ETA_G*M01_G %MASS BALANCE HYDROCHLORIC ACID M11_A=Y_A*ETA_G*M01_G M22_A=Z1_A*M11_A M21_A=0 %MASS BALANCE BIOMASS M11_X=Y_X*ETA_G*M01_G M21_X=Z1_X*M11_X M22_X=(1-Z1_X)*M11_X M31_X=M21_X M41_X=Z2_X*M31_X M42_X=(1-Z2_X)*M31_X M51_X=M41_X M61_X=Z3*M51_X M63_X=(1-Z3_X)*M51_X M71_X=M61_X M81_X=Z4_X*M71_X M82_X=(1-Z4_X)*M71_X %MASS BALANCE PRODUCT M11_P=(Y_P)*(ETA_G)*M01_G M21_P=Z1_P*M11_P M22_P=(1-Z1_P)*M11_P M31_P=M21_P M41_P=Z2_P*M31_P M42_P=(1-Z2_P)*M31_P M51_P=M41_P M61_P=Z3*M51_P M63_P=(1-Z3_P)*M51_P M71_P=M61_P M81_P=Z4_P*M71_P M82_P=(1-Z4_P)*M71_P %MASS BALANCE SULFURIC ACID M03_S=x1 M31_S=(Y2_S)*(ETA_S)*M03_S %MASS BALANCE SODIUM HYDROXIDE M03_OH=18*x1 M31_OH=M03_OH+(Y2_OH)*(ETA_S)*(M03_S) M42_OH=Z2_OH*M31_OH M41_OH=(1-Z2_OH)*M31_OH %MASS BALANCE SODIUM SULPHATE M31_F=(Y2_F)*(ETA_S)*(M03_S) M42_F=M31_F %MASS BALANCE HEXANE M04_H=x2 %FEED TO EVAPORIZE, SOLVATIZER M51_H=Z3_H*M04_H M62_H=M51_H M05_H=x3 M71_H=M05_H M82_H=(1-Z4_H)*M71_H M81_H=Z4_H*M71_H %MASS BALANCE ISOPROPANOL M05_I=(1/6)*x3 M71_I=M05_I M82_I=(1 - Z4)*M71_I M81_I=Z4_I*M71_I %MASS BALANCE WATER M01_W=M01-M01_G-M01_N M11_W=M01_W+(Y_W)*(ETA_G)*(M01_G) M21_W=Z1_W*M11_W M22_W=(1-Z1_W)*M11_W M03_W=(100/97)*M03_S+(100/86)*M03_N M31_W=M03_W+(Y2_W)*(ETA_S)*(M03_S) M41_W=0 M42_W=(Z2_W)*M31_W %streams M01=M01_G+M01_N+M01_W M02=M02_O M11=M11_G+M11_N+M11_X+M11_P+M11_A+M11_W M12=M12_O+M12_C M21=M21_P+M21_X+M21_W M22=M22_G+M22_N+M22_A+M22_W M03=M03_S+M03_OH+M03_W M31=M31_OH+M31_F+M31_W+M31_P+M31_X M41=M41_P+M41_X M42=M42_X+M42_W+M42_OH+M42_F M04=M04_H M51=M51_P+M51_X+M51_H M61=M61_P+M61_X M63=M63_P+M63_X M63=M63_H M05=M05_H+M05_I M71=M71_X+M71_P+M71_H+M71_I M81=M81_P+M81_X+M81_H+M81_I M82=M82_P+M82_X+M82_H+M82_I μ01 μ02 μ03 μ04 μ05 μ11 μ12 μ21 k G O N X P C A W H S B F I Streams μ22 μ31 μ41 μ42 μ51 μ61 μ62 μ63 μ71 μ81 μ82 k G O N X P C A W H S B F I SUM 2013.71 0.00536324 0 0.00144509 0 0 0 0 0.99319167 0 0 0 0 0 1 μ01 μ02 36.48 10.94 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0.95942982 0 0 0 0.00219298 0 0.03837719 0 0 0 0 1 1 μ03 μ04 2.69 3.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0.74349442 0 0 0 0 0 0 0 0.25650558 1 1 μ05 μ12 μ21 32.26 15.8 2036.14 0 0 0.00053042 0 0 0.38607595 0 0 0.00058935 0 0.29758215 0.0047148 0 0.02045877 0.00032414 0 0 0.61392405 0 0 0.00088403 0 0.68195908 0.99295726 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 μ11 Streams μ62 μ61 μ51 μ42 μ41 μ31 0.69 5.46 68.32 1.66 69.98 2004.04 0 0 0 0 0 0.00053891 0 0 0 0 0 0.00057883 0 0 0 0 0 0 0 0.13718205 0.60240964 0.12587822 0.18315018 0.14492754 0 0.12087912 0.85507246 0 0.00943127 0.39759036 0 0 0 0 0 0 0 0 0 0 0 0.00089819 0 0 0 0.85480094 0.99798407 0.83452415 0 0 0.6959707 0 0 0 0 0 0 0 0 0 0 0 0 0.01814988 0 0.01771935 0 0 0 0.00117096 0 0.00114318 0 0 0 0 0 0 1 1 1 1 1 1 μ22 μ82 μ81 μ71 μ63 2.759 0.621 3.38 0.97 3.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.03624502 0 0.92783505 0.0295858 0 0 0.07216495 0.17455621 0.95008052 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0.59171598 0.04186795 0.71547662 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.20414201 0.00805153 0.24827836 0 1 1 1 1 1 E = I1 * R1 1/R1 = 1/Ra + 1/Rb1 E = I2 * R2 1/R2 = 1/Ra + 1/Rb2 Rb1 = 1 Rb2 = 10 E = 12
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