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Growth Kinetics of Cell Cultures

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Bioreaction Engineering Principles

Abstract

In Chaps. 3 and 5, we have discussed how the two important design parameters for cell cultures, yield and productivity, can be derived from experimental data, e.g., from measurements of the substrate consumption and the product formation. Furthermore, we have shown how measured steady-state fluxes in and out of the cell, the exchange fluxes, can be used to calculate the fluxes through the different branches of the metabolic network of a given microorganism.

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Notes

  1. 1.

    An operon contains a cluster of genes that are all under the control of a single regulatory signal or promoter. They are transcribed together onto a single RNA strand and can be translated together. They also found both in prokaryotes and eukaryotes.

  2. 2.

    The true inducer is allolactose, and some small lactose flux – perhaps by passive diffusion into the cell, is permitted.

  3. 3.

    Since the total number of operators of a given type in the cell is very small, it does not make much sense to talk about the fraction of repressor-free operators. However, in a description of enzyme synthesis, one may use (7.47) as an expression for the probability that the operator is repressor free.

  4. 4.

    Conidiphores are modified hyphae on which the asexual spores are formed.

  5. 5.

    The assumption of an infinitely fast reaction is a bit shaky. It means that C has to diffuse at a constant rate up to the boundary z = 0, react instantly with S, and move away toward z = d at the same rate. This is unrealistic as shown by Schultz et al. (1974), unless the following relation is true: Q = kd 2/D >> 1. For a rate constant k = 107 s−1, d = 10 nm, and D = 10−12 m2 s−1, Q = 10 which can be shown to give an overestimation of the flux by about 10%. An error of this magnitude does, however, not invalidate the results obtained above.

  6. 6.

    Benzoic acid as well as citric acid and lactic acid are used to preserve food, exactly because of their inhibition of microbial growth. In acidified food (it could be a soft drink or a fruit juice, or the pH may start to decrease through incipient growth of spoilage bacteria) much of the benzoic acid is on the undissociated form (pK a = 4.1), and the undissociated acid freely diffuses into the cell. In anaerobic yeast cultivation, growth of the yeast almost stops when pH reaches 5, but the ethanol yield increases in a frantic effort of the yeast cell to acquire enough ATP to expulse the protons after the acid has dissociated at the intracellular pH of close to 7. Lactic acid (pK a = 3.88) has the same effect.

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Author information

Authors and Affiliations

  1. Department of Chemical and Biochemical Engineering, Technical University of Denmark (DTU), Lyngby, Denmark

    John Villadsen

  2. Systems Biology, Chalmers University of Technology, Gothenburg, Sweden

    Jens Nielsen

  3. Department of Chemical Engineering, Lund University, Lund, Sweden

    Gunnar Lidén

Authors
  1. John Villadsen
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  2. Jens Nielsen
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Correspondence to John Villadsen .

Problems

Problems

Problem 7.1 Estimation of parameters in the Monod model

From measurements of the residual glucose concentration in a steady-state chemostat at various dilution rates, you can find the following results:

D (h−1)

s (mg L−1)

0.13

11

0.19

14

0.23

18

0.36

38

0.67

85

0.73

513

Calculate by linear regression the parameters in the Monod model. Are any of the data points suspect?

If you want to check the value of μ max determined above, increase the dilution rate in the chemostat to D = 1.1 h−1. This results in a rapid increase of the glucose concentration in the medium. After a while, s >> K s. The result of the change in dilution rate is a decrease in the biomass concentration, and during the wash-out you measure the biomass concentration as a function of time, and obtain the following results:

Time (h)

x (g L−1)

0

5.1

0.5

4.5

1.0

3.7

2.0

2.8

3.0

2.1

4.0

1.4

Determine μ max from this experiment. Discuss the applied method [see also Esener et al. (1981c)].

Problem 7.2Inhibitory effect of lactic acid

Bibal et al. (1988, 1989) studied the inhibition of lactic acid on Streptococcus cremoris, and in this exercise we will analyze their data.

  1. (a)

    The influence of lactic acid on the growth of S. cremoris was examined by measuring the maximum specific growth rate during batch growth of the bacterium in media containing various concentrations of lactic acid (p). The results are summarized below:

    p (g L−1)

    μ (h−1)

    0

    0.90

    12.0

    0.68

    39.0

    0.52

    55.0

    0.13

    As discussed in Example 7.11, it is mainly the undissociated form of lactic acid that passes through the cellular membrane, and we will therefore assume that it is only the undissociated acid that has a toxic effect on the cells. Plot the relative specific growth rate, i.e., μ max(p)/μ max(p = 0), versus the concentration of the undissociated acid concentration (in mM). pH = 6.3 was used and pK a for lactic acid is 3.88. Assume that the inhibition model given by (7.19) holds. Find the inhibition constant K i . Plot the model together with the experiments.

  2. (b)

    From the results in (a) you conclude that (7.19) is not well suited for description of the experimental data, since the inhibition by lactic acid seems to be stronger, especially at high values of undissociated lactic acid concentrations (p u). There seems to be a certain maximum concentration of undissociated acid above which growth stops. Next try the model (7.20) to find the influence of p u on μ, and estimate the model parameter. At what concentration of lactic acid will growth stop?

  3. (c)

    Plot the maximum specific growth rate as a function of the pH in a medium containing 1 and 10 g L−1 of lactic acid (total concentration), using the model found in (b).

  4. (d)

    Measuring the yield coefficient on lactose in a steady-state chemostat at different concentrations of lactic acid, Bibal et al. (1988, 1989) found the data below:

    p (g L−1)

    Y sx (g DW g−1)

    0

    0.16

    7.5

    0.16

    13.0

    0.14

    18.5

    0.14

    21.0

    0.14

    32.0

    0.12

    38.5

    0.11

    45.0

    0.10

    48.5

    0.09

    How can you explain the decrease in the yield coefficient with increasing lactic acid concentration?

  5. (e)

    Assume that the maintenance coefficient m s is 0.05 h−1. Calculate the true yield coefficient in (7.27) for p = 0. Using the model derived in (b), calculate the maintenance coefficient as a function of p u. Explain the results.

Problem 7.3Modeling of the lac-operon in E. coli

We will now revisit the model for the lac-operon described in Sect. 7.5.1.

  1. (a)

    The repressor has four binding sites for the inducer (lactose), but in the derivation of (7.47) only the repressor–inducer complex where all four sites are occupied is considered. We now consider binding at all four sites. Specify all the equilibria and the definitions of the association constants. The association constant for formation of X r S i is termed K 1i , and that for formation of X O X r S i is termed K 4i . Binding of the inducer to the repressor operator complex can be neglected (i.e., the equilibrium in (7.39c) is not considered).

    Assume that the affinity for the binding of the repressor to the operator is approximately the same whether no, one, two, or three inducers are bound to the repressor, i.e., K 41 = K 42 = K 43 = K 44. This assumption is reasonable since the repressor probably changes its conformation only when the last inducer is bound to it. With this assumption show that the fraction of repressor-free operators is given by

    $$\hskip 16pt{Q_1} = \frac{{\left[ {{X_{\rm{O}}}} \right]}}{{{{\left[ {{X_{\rm{O}}}} \right]}_{\rm{t}}}}} = \frac{{1 + {K_1}\left[ {{S_{\rm{lac}}}} \right]_{\rm{t}}^4}}{{1 + {K_1}\left[ {{S_{\rm{lac}}}} \right]_{\rm{t}}^4 + {K_2}{{\left[ {{X_{\rm{r}}}} \right]}_{\rm{t}}} + {K_{\rm{S}}}{K_{11}}\left[ {{X_{\rm{r}}}{S_{\rm{lac}}}} \right]{{\left[ {{S_{\rm{lac}}}} \right]}_{\rm{t}}}\left( {1 + {K_{12}}{{\left[ {{S_{\rm{lac}}}} \right]}_{\rm{t}}} + {K_{12}}{K_{13}}\left[ {{S_{\rm{lac}}}} \right]_{\rm{t}}^2} \right)}}. $$
    (1)

    In (1), K 1 = K 41 K 42 K 43 K 44.

    We now assume that the K 11 ≈ K 12 ≈ K 13 ≈ K 14, i.e., the association constant for the fourth inducer is much stronger than the corresponding constants for the first three sites. This assumption follows from our assumption above that the conformation of the repressor changes only when the fourth inducer is bound and when the conformation changes the repressor–inducer complex becomes very stable. What other assumptions are required for reducing (1) to (7.47)?

  2. (b)

    Lee and Bailey (1984d) also modeled the lac-operon, but they included binding of the repressor to a nonspecific binding site in the chromosome (X d). Again we neglect binding of inducer to the repressor–operator complex, and the equilibria are therefore

    $$ {X_{\rm{r}}} + n{S_{\rm{lac}}}\;\mathop { \leftrightarrow }\limits^{{K_1}} \;{X_{\rm{r}}}n{S_{\rm{lac}}}, $$
    (2a)
    $$ {X_{\rm{o}}} + {X_{\rm{r}}}\mathop {{\; \leftrightarrow }}\limits^{{K_2}} \;{X_{\rm{o}}}{X_{\rm{r}}}, $$
    (2b)
    $$ {X_{\rm{o}}} + {X_{\rm{r}}}{S_{\rm{lac}}}\;\mathop {{ \leftrightarrow \;}}\limits^{{K_4}} {X_{\rm{o}}}{X_{\rm{r}}}{S_{\rm{lac}}}, $$
    (2c)
    $$ {X_{\rm{d}}} + {X_{\rm{r}}}\;\mathop {{ \leftrightarrow \;}}\limits^{{K_5}} {X_{\rm{d}}}{X_{\rm{r}}}, $$
    (2d)
    $$ {X_{\rm{d}}} + {X_{\rm{r}}}{S_{\rm{lac}}}\;\mathop { \leftrightarrow }\limits^{{K_6}} \;{X_{\rm{d}}}{X_{\rm{r}}}{S_{\rm{lac}}}. $$
    (2e)

    By assuming that [X d]t ≈ [X d], show that

    $$ {Q_1} = \frac{{\left[ {{X_{\rm{O}}}} \right]}}{{{{\left[ {{X_{\rm{O}}}} \right]}_{\rm{t}}}}} = \frac{{1 + {K_5}{{\left[ {{X_{\rm{d}}}} \right]}_{\rm{t}}} + {K_1}\left[ {{S_{\rm{lac}}}} \right]_{\rm{t}}^n\left( {1 + {K_6}{{\left[ {{X_{\rm{d}}}} \right]}_{\rm{t}}}} \right)}}{{1 + {K_5}{{\left[ {{X_{\rm{d}}}} \right]}_{\rm{t}}} + {K_1}\left[ {{S_{\rm{lac}}}} \right]_{\rm{t}}^n\left( {1 + {K_6}{{\left[ {{X_{\rm{d}}}} \right]}_{\rm{t}}}} \right) + {K_2}{{\left[ {{X_{\rm{r}}}} \right]}_{\rm{t}}}}}, $$
    (3)
  3. (c)

    Lee and Bailey (1984d) specified the parameters in (3) to be

    $$ \eqalign{ {K_1} = {1}{0^7}\;{{\hbox{M}}^{ - {1 }}},\quad {K_2} = {2}\; \times \;{1}{0^{{12}}}\;{{\hbox{M}}^{ - {1}}}{,}\quad {K_4} = {2}\; \times \;{1}{0^9}\;{{\hbox{M}}^{ - {1}}} \hfill \\{K_5} = {1}{0^3}\;{{\hbox{M}}^{ - {1}}},\quad {K_6} = {1}.{5}\; \times \;{1}{0^9}\;{{\hbox{M}}^{ - {1}}}. \hfill \\}<!endgathered> $$

    Furthermore, they state that

    $$ {\left[ {{X_{\rm{d}}}} \right]_{\rm{t }}} = { 4}\; \times \;{1}{0^{ - {2}}}\;{\hbox{M}}\ {\hbox{and}}\ { }{\left[ {{X_{\rm{r}}}} \right]_{\rm{t }}} = { 2}\; \times \;{1}{0^{ - {8}}}\;{\hbox{M}}{.} $$

    Plot the value of Q 1 using both (3) and (7.47) as a function of the inducer concentration [S lac]t. Comment on the result.

    The parameters given by Lee and Bailey are for IPTG (isopropyl-β-d-thiogalactosidase), a frequently used inducer in studies of the lac-operon. Assume that the parameters are the same for lactose as inducer and calculate the concentration of lactose (in mg L−1) to give Q 1 = 0.5. Discuss why even this low concentration of lactose leads to induction of the lac-operon.

  4. (d)

    Show that for an antiinducer (neglect binding of the repressor to nonspecific sites)

    $$ {Q_1} = \frac{{\left[ {{X_{\rm{O}}}} \right]}}{{{{\left[ {{X_{\rm{O}}}} \right]}_{\rm{t}}}}} = \frac{{1 + {K_1}\left[ {{S_{\rm{lac}}}} \right]_{\rm{t}}^{\rm{m}}}}{{1 + {K_1}\left[ S \right]_{\rm{t}}^{\rm{n}} + {K_1}{K_4}\left[ S \right]_{\rm{t}}^{\rm{n}}{{\left[ {{X_{\rm{r}}}} \right]}_{\rm{t}}}}}. $$
    (4)

Problem 7.4Facilitated transport through membranes

  1. (a)

    In Example 7.10, (2) [s a >> s b] and negligible contribution from the non-carrier- associated transport, the total volumetric flux v through the membrane area A is

    $$ JA{ } = \frac{{DA}}{d}\frac{{{c_{\rm{t}}}{s_{\rm{a}}}}}{{{K_{\rm{m}}} + {s_{\rm{a}}}}} = \frac{{{v_{\max }}{s_{\rm{a}}}}}{{{K_{\rm{m}}} + {s_{\rm{a}}}}}{\hbox{where}}\,\,{K_{\rm{m}}} = \frac{{{{K\prime}_{\rm{eq}}}}}{K}{\hbox{and}}\,{v_{\max }} = \frac{{DA{c_{\rm{t}}}}}{d}. $$
    (1)

    For uptake of glucose in human erythrocytes (red blood cells), the following data are found experimentally:

    Glucose concentration (mmol L−1)

    Glucose flux (mmol min−1)

    1.0

    0.09

    1.5

    0.12

    2.0

    0.14

    3.0

    0.20

    4.3

    0.25

    5.0

    0.28

    Make a double reciprocal plot of (JA)−1 = v −1 versus s a and determine the parameters v max and K m.

  2. (b)

    In the example, an infinitely fast equilibrium for the reversible reaction \( S{ } + { }C\overset{{{r_{\rm{m}}}}}{\longleftrightarrow} SC \) is postulated.

    Consider the other extreme: an infinitely fast diffusion of C and SC across the membrane, and a slow reaction.

    Now c and sc are both constant and have the values \( \bar{c} \) and \( \bar{s}\bar{c} \) while the solute concentration must vary across the membrane according to

    $$ D\frac{{{{\text{d}}^2}s}}{{{\hbox{d}}{z^2}}} - {k_1}s\bar{c} - {k_2}\bar{s}\bar{c} = 0. $$
    (2)

    with boundary conditions given by (5) of the example.

    Solve the differential equation (2), and determine the values of the two constant carrier concentrations in terms of c t, the two rate constants, and \( \bar{s} = K\displaystyle\frac{{{s_{\rm{a}}} + {s_{\rm{b}}}}}{2}. \)

    [You may consult Note 6.2 since the mathematics involved in this problem is essentially the same as that used in the note, except that plane-parallel symmetry is used here and spherical symmetry in the note]

Problem 7.5Phosphotransferase-based membrane transport systems

Many interesting physiological phenomena can be observed due to the complexity of sugar transport systems.

  1. (a)

    In Benthin et al. (1993a), stable oscillations are observed in both μ and r p because fructose can be taken up by two PTS systems.

    Describe the oscillatory phenomenon, and discuss the mechanism proposed by the authors to explain the complex behavior of the cultivation.

  2. (b)

    In Benthin et al. (1993b), the uptake of glucose and mannose through two specific sites on the transmembrane Mannose-PTS transporter is discussed. It turns out that the transporter is able to distinguish between α glucose and β glucose, an astonishing specificity of the protein.

What is the difference between α and β glucose?

Describe the set of experiments used by the authors to arrive at the conclusion that there must be two different and specific sites on the medium side for uptake of nearly identical sugars.

Discuss the mathematical model used to explain the widely different uptake profiles for α and β glucose after a pulse addition of a mixture of two different sugars.

Finally, discuss the adaptation of the transport system for a chemostat cultivation with constant dilution rate when the medium is changed from mannose limitation to glucose limitation.

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Villadsen, J., Nielsen, J., Lidén, G. (2011). Growth Kinetics of Cell Cultures. In: Bioreaction Engineering Principles. Springer, Boston, MA. https://doi.org/10.1007/978-1-4419-9688-6_7

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