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:
𝐶6 𝐻12 𝑂6 + 𝛼𝑂2 + 𝛽𝑁𝐻4 𝐶𝑙 → 𝛾𝑋 + 𝛿𝐶59 𝐻90 𝑂4 + 𝜀𝐶𝑂2 + 𝜁𝐻2 𝑂 + 𝜂𝐻𝐶𝑙
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
𝐶6 𝐻12 𝑂6 + 𝛼𝑂2 + 𝛽𝑁𝐻4 𝐶𝑙 → 𝛾𝑋 + 𝛿𝐶59 𝐻90 𝑂4 + 𝜀𝐶𝑂2 + 𝜁𝐻2 𝑂 + 𝜂𝐻𝐶𝑙
𝑋 = 𝐶𝐻1.99 𝑂0.5 𝑁0.19
𝑅𝑄 =
𝐶
0
0
1
59 1
𝐻
0 −4 1.99 90 0
𝐴 𝑂 = −2 0
0.5
4 2
𝑁
0 −1 0.19 0 0
[𝐶𝑙 ] [ 0 −1
0
0 0
𝜀
=2
𝛼
0
2
1
0
0
𝛼
𝛽
𝑐1 = 6 + (𝛼 − 𝜀)
0
𝛾
𝑐2 = 12
1
0 ∗ 𝛿 = 𝑐3 = 6 + (2𝛼 − 2𝜀)
𝜀
𝑐4 = 0
0
]
𝜁
𝑐5 = 0
1
[
]
[ 𝜂]
Taking RQ = 2; 2α = ε, thus α = 0.5 y ε = 1:
𝑐1 = 5
𝛽
𝐶
0
1
59 0 0
𝑐
𝛾
𝐻
−4 1.99 90 2 1
2 = 12
𝐴 𝑂 = 0
0.5
4 1 0 ∗ 𝛿 = 𝑐3 = 3
𝜁
𝑐4 = 0
𝑁
−1 0.19 0 0 0
[𝐶𝑙 ] [−1
0
0 0 1] [ 𝜂] [ 𝑐5 = 0 ]
𝐶6 𝐻12 𝑂6 + 0.5𝑂2 + 0.18𝑁𝐻4 𝐶𝑙 → 0.98𝑋 + 0.068𝐶59 𝐻90 𝑂4 + 1𝐶𝑂2 + 2.24𝐻2 𝑂 + 0.18𝐻𝐶𝑙
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:
𝜇11 = 9.6
𝑚𝑜𝑙
0.66𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑋+
𝑃 + 2022
𝑊 + 1.08
𝐺 + 1.16
𝑁 + 1.75
𝐻
ℎ
ℎ
ℎ
ℎ
ℎ
ℎ
Output:
𝜇21 = 9.6
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑋 + 0.66
𝑃 + 22
𝑊
ℎ
ℎ
ℎ
𝜇22 = 2000
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑊 + 1.08
𝐺 + 1.16
𝑁
ℎ
ℎ
ℎ
IN FIRST MIXER
At the first mixer a neutralization reaction occurs, so cell is disrupted. 𝜇03 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.
𝐻2 𝑆𝑂4 + 2𝑁𝑎𝑂𝐻 → 𝑁𝑎2 𝑆𝑂4 + 2𝐻2 𝑂
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