Solution:
Stoichiometric Coefficients: relate relative molar quantities of different chemical species in a reaction
Yield Coefficients: same as above (but commonly expressed in relative mass quantities, rather than molar quantities)
Solution:
Lag, exponential, deceleration, stationary, and death/decline phases
Solution:
Primary Metabolites: Cell growth related products;
rate of formation is proportional to μ*x
Secondary Metabolites: Non-growth related products;
rate of formation is proportional to x.
Solution:
Unsegregated Models: Cells are all considered to respond in a homogeneous manner
Segregated Models: Cells are divided into different subgroups, each responding differently to environment
(e.g., different specific growth rate, etc.)
Reaction Stoichiometry:
a C6H12O6 (glucose) + ... ---------> C6H10NO3 (yeast) + ...
Solution:
Any one of the chemical species can act as the basis. Here we assign unity to biomass (yeast).
Reaction Stoichiometry:
a C6H12O6 + b O2 + c NH3 ---------> C6H10NO3 + d H2O + e CO2
(glucose) (yeast)
Solution:
Elemental Balance:
Glucose Oxygen ammonia biomass water CO2
----------------------------------------------
C: 6*a = 6 + 1*e ... eqn(1)
H: 12*a + 3*c = 10 + 2*d ... eqn(2)
N: 1*c = 1 ... eqn(3)
O: 6*a + 2*b = 3 + 1*d + 2*e ... eqn(4)
We have four equations (eqns (1)-(4))
--> Cannot solve for the 5 stoichiometric coefficients (a, b, c, d, e).
Need one more equation from some sort of measurements.
|Acidic|Neutral|Basic|Acidic then basic|Basic then acidic|No predictable pattern|
Solution:
As cells take up ammonia (NH3), sulfuric acid is
left behind in the broth.
(NH4)2SO4 --> 2NH3 + H2SO4
The broth becomes increasingly acidic in the absence of pH
control. Had the nitrogen source been nitrate, the opposite
would be true. Note that only ammonia NH3 is taken up. Had
ammonium ion NH4+ been taken up, the cells
will quickly become positively charged, which does not happen in
practice.
Solution:
We add NH4OH to counterbalance pH shift due to
NH3 consumption. Thus, by following NH4OH
addition in pH control, we measure NH3 uptake rate
(NUR). Combined with another measurement (say CER), we have an
additional equation needed to solve the stoichiometry.
measurements: CER/NUR=e/c ... eqn(5)
We have five equations (eqns (1)-(5))
--> Solve for the 5 stoichiometric coefficients (a, b, c, d, e).
Solution: RQ measurements: RQ=CER/OUR=e/a ... eqn(5) We have five equations (eqns (1)-(5)) --> Solve for the 5 stoichiometric coefficients (a, b, c, d, e).
t OUR CER x
(min) (g/L-min) (g/L)
-----------------------
0 : : 1
1 : : ?
2 : : ?
3 : : ?
: : : ?
Solution:
First, material balance coupled with the RQ measurements gives
the stoichiometry coefficients, which are closely related to
yield coefficients. Secondly, from the yield coefficients, we
can numerically integrate the rate expression (say with the
Euler's Method) to find biomass concentration x.
Finally, integrate the rate expression given OUR in mole oxygen per time per fermentor volume.
dx/dt=OUR(t)*b(t)*MWyeast
Approximate dx/dt numerically as: dx/dt=(xi+1-xi)/(ti+1-ti)
xi+1=xi+OUR(ti)/b(ti)*MWyeast*(ti+1-ti)
We can also go through yield coefficients:
Species Formula MW
-------------------------------
Glucose C6H12O6 180
Yeast C6H10NO3 144
Ammonium Sulfate (NH4)2SO4 132
glucose-to-biomass yield: Yx=MWyeast/a/MWglucose=144/180/a
oxygen-to-biomass yield: YO=MWyeast/b/MWoxygen=144/32/b
:
(Similarly, calculate other yield coefficients.)
Finally, integrate the rate expression.
dx/dt=OUR(t)*MWoxygen*YO(t)
xi+1=xi+OUR(ti)*MWoxygen*YO*(ti+1-ti)
Solution:
When μ is a function of all variables (μ=μ(x,s)), we must
integrate the coupled set of ODEs simultaneously.
Biomass: dx/dt = μ*x (plus endogenous decay term -mx*x or other applicable terms)
Substrate: ds/dt = - μ*x/Yx (plus maintenance term -ms*x or other applicable terms)
Initial conditions: x(0), s(0)
Solution:
x, s [=] g/L
t [=] hr
μ [=] hr-1
Y [=] g cell/g substrate
Solution:
Many forms exist. One empirical way is to include a substrate
or biomass term in the denominator.
Substrate inhibition: μ = μm*s/(K+s+Ki*s2) ... Andrew's model of substrate inhibition
μ = μm*s/(K+s)/(Ki+s)
Biomass inhibition: μ = μm*s/(K+s+Kx*x)
μ = μm*s/(K+s)/(Kx+x)
μ = μm*s/(K*x+s) ... Contois model of cell inhibition
Solution:
Same as in batch, but add the transport terms due to feed
(D*sf) and out-flow (-D*s).
Usually, there is no x in the feed; xf=0.
Biomass: dx/dt = D*(xf-x) + μ*x
Substrate: ds/dt = D*(sf-s) - μ*x/Y
Initial conditions: x(0), s(0)
where D = F/V
Solution:
Steady-state means no change with time, i.e. d?/dt=0.
Biomass: dx/dt = 0 = D*(xf-x) + μ*x
Substrate: ds/dt = 0 = D*(sf-s) - μ*x/Y
Steady states are:
washout steady-state (stable if D>μ)
x=0 & s=sf
non-washout steady-state
μ(x,s)=D
sf-s=x/Y(x,s)
(Solve the above two equations for two unknowns x & s.; multiple solutions may exist.)
Stability: examine eigenvalues arising from linearlized stability analysis around the steady-states.
Solution:
Operating parameters: D=F/V & sf.
maximize profit = D*x - α*D*sf
where α is the price of substrate relative to cell biomass
Solution to d(profit)/dD = 0 gives D that yields maximum profit.
|Batch|Fed-Batch|CSTR|
Solution:
CSTR gives a higher cell productivity than batch or fed-batch.
Solution:
Yes, cell recycle allows operation at higher cell concentration, higher
dilution rate, and higher cell productivity.
Solution:
No, cell immobilization is suitable for non-growth related products.
Cell biomass is a product directly related to cell growth.
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