 6.3. Quantifying Growth Kinetics

6.3. Quantifying Growth Kinetics



Unstructured Structured .

Most idealized case

Cell population treated as Multicomponent

one component solute average cell description

single component Multicomponent

heterogeneous description of cell-to-cell

individual cells heterogeneity

Actual case

Nonsegregated Segregated .

Balanced Growth

(approximation)

Balanced Growth

(approximation) Average cell

approximation

Average cell approximation

6.3.2. Using Unstructured Nonsegregated Models to Predict Specific Growth Rate



 Monod equation

m_m : maximum specific growth rate K_s : saturation constant

 Other equations

m = ^m

^S

K

+ S

_____

Other equations

for substrate-limited growth

When more than one substrate is growth rate limiting



m

m

m

m

m



m

m

m

m

m



m

m

m

m

where the lowest value of

m

6.3.2.2. Models with growth inhibitors



 Substrate inhibition

 Product inhibition

 Inhibition by toxic compound

Substrate Inhibition of Cell Growth

Enzyme Inhibition

Noncompetitive inhibition

Competitive inhibition

Product Inhibition of Cell Growth

Cell Growth in Ethanol Fermentation

Inhibition by Toxic Compounds

Enzyme Inhibition

Noncompetitive inhibition Competitive inhibition

Enzyme Inhibition

Uncompetitive inhibition

Inhibition by Toxic Compounds

6.3.2.3. The logistic equation

dX/dt = m ^X

dX/dt = k (1-X/X_∞) X, X(0)=X₀

Sigmoidal

(stationary phase involved)

X = ^X

^e

kt

1- (X

/X

) (1 - e

)

______________

6.3.2.4. Growth model

for filamentous organisms

 In suspension culture : formation of microbial pellet

 On solid medium : colony (long and highly branched cells)

 In the absence of mass transfer limitations

R ∝ t R: radius of a microbial pellet radius of a mold colony dR/dt = k_p = constant Eq(1)

Growth model for filamentous organisms

 M : mass of pellet

dM/dt = r (4pR²) (dR/dt)

= k_p r (4pR²) (by Eq(1))

= k_p (36pr)^1/3 (4/3 pR³r)^2/3

= g M2/3 Eq(2)

where g = k_p (36pr)^1/3 M = 4/3 pR³r

Growth model for filamentous organisms

 Eq(2)

dM/dt = g M2/3 Eq(2)

M^-2/3 dM = g dt 3 [M^1/3] = g t

M = (M₀^1/3 + (g t)/3)³

≈ (g t /3)³

M₀ (the initial biomass) is usually very small compared to M.

M M₀

6.3.3. Chemically Structured Model

 Since it is not practical to write material balances on every cell

component, we must select skillfully the key variables and processes of major interest in a particular application when formulating a

structured kinetic model.

 Mass balance inside the cell

 ^Batch

 V_R : total volume of liquid in the reactor

 C_i : extrinsic concentration of component i

 X : extrinsic concentration of biomass

Intrinsic and Extrinsic Concentrations



 The amount of a compound per unit cell mass (or cell volume)



 The amount of a compound per unit reactor volume

Intrinsic and Extrinsic Concentrations

Intrinsic and Extrinsic Concentrations

Single-cell model for E. coli by Shuler and colleagues





stoichiometric and kinetic parameters, almost all of which can be determined from previous

literature an the biochemistry of E. coli growth.

Single-cell model for E. coli by Shuler

and colleagues

7. Stoichiometry of Microbial Growth and Product Formation





 Growth yield and etc.

 RQ : respiratory quotient

– Defined as the moles of CO₂ produced per mole of oxygen consumed

7.3. Stoichiometric Calculations



One mole of biological materials is defined as the amount containing 1 gram atom of carbon, such as CH_aO_bN_d.

Assumption: No extracellular products other than H₂O and CO₂ are produced.

Elemental Balances

CH_mO_n + a O₂ + b NH₃  c CH_aO_bN_d + d H₂O + e CO₂ Eq(1) where CH_mO_n : 1mole of carbohydrate

CH_aO_bN_d : 1 mole of cellular material Elemental Balances

C: 1 = c + e

H: m + 3b = ca + 2d Eq(2)

O: n + 2a = cb + d + 2e

N: b = cd

RQ = e/a Eq(3)

We have five equations and five unknowns (a, b, c, d, e). With a

measured value of RQ, these equations can be solved to determine the stoichiometric coefficients.

___________

7.3.2. Degree of Reduction



 The concept of degree of reduction has been

developed and used for proton-electron balances in bioreactions.



 Defined as the number of equivalents of available electrons per gram atom C.

Degree of Reduction





 Methane (CH₄): 1x4 + 4x1 = 8

g = 8/1 = 8

 Glucose (C₆H₁₂O₆): 6x4 + 12x1 + 6x(-2)=24 g = 24/6 = 4

 Ethanol (C₂H₅OH): 2x4 + 6x1 + 1x(-2) = 12 g = 12/2 =6

Aerobic production of a single extracellular product

CH_mO_n + a O₂ + b NH₃  substrate (s)

c CH_aO_bN_d + d CH_xO_yN_z + e H₂O + f CO₂ Eq(4) biomass (b) product (p)

g_s = 4 + m - 2n

g_b = 4 + a - 2b - 3d Eq(5) g_p = 4 + x - 2y - 3z

For CO₂, H₂O, and NH₃, the degree of reduction is zero.

___________

Aerobic production of a single extracellular product



• elemental balance on C, H, O, and N

• available electron balance

• energy balance

• total mass balance



 We would typically choose a carbon, a nitrogen, and an available-electron balances.

Balance Equations

C: c + d + f =1 Eq(6)

N: c d + d z = b Eq(7)

elec: c g_b + d g_p = g_s – 4 a Eq(8)

RQ = f / a Eq(9)

Yx/s = c Eq(10)

Yp/s = d Eq(11)