Immobilized Enzymes Coursework

Problems in Biochemical Engineering

Jason Haugh

Department of Chemical & Biomolecular Engineering
North Carolina State University

These problems are provided as a study tool for undergraduate or graduate students taking a course in Biochemical Engineering, and as a "short-cut" for fellow instructors.  More problems will be added as they become available.

Unlike most problems found in textbooks, many of these problems are designed to be done relatively quickly (10-20 minutes), to be used for in-class test or example problems.  Others were designed as homework/final exam problems and therefore take longer; most if not all of those can be shortened as well.  Within each category, problems are listed roughly in order of difficulty.

 

Enzyme and Immobilized Enzyme Kinetics

1. Consider an industrially important enzyme, which catalyzes the conversion of a protein substrate to form a much more valuable product.  The enzyme follows the Briggs-Haldane mechanism:

An initial rate analysis for the reaction in solution, with E0 = 0.10 μM and various substrate concentrations S0, yields the following Michaelis-Menten parameters:

Vmax = 0.60 μM/s; KM = 80 μM.

A different type of experiment indicates that the association rate constant, k1, is

k1 = 2.0 x 106 M-1s-1 (2.0 μM-1s-1).

a. Estimate the values of k2 and k-1.

b. On average, what fraction of enzyme-substrate binding events result in product formation?

 

2. The kinetic properties of the ATPase enzyme, isolated from yeast, which catalyzes the hydrolysis of ATP to form ADP and Pi, are assessed by measuring initial rates in solution, with various ATP concentrations S0 and a total ATPase concentration E0 = 0.60 μM.  From these experiments, it is determined that

Vmax = 1.20 μM/s; KM = 40 μM.

a. Calculate the values of kcat and the catalytic efficiency for ATPase under these conditions.

b. An inhibitor molecule is added at a concentration of 0.1 mM, and the experiments are repeated.  The apparent Vmax and KM are now found to be 0.6 μM/s, and 20 μM, respectively.  Speculate on how this inhibitor works (i.e., specify which species are engaged by the inhibitor).

 

3. The luciferase enzyme in fireflies catalyzes the modification of luciferin, consuming both luciferin and ATP, and producing light (you can probably guess the function of this reaction).  Assuming ATP is in excess, the reaction follows the Briggs-Haldane mechanism, with luciferin as the limiting substrate.

A series of experiments are performed in which 5 μM luciferase enzyme is mixed with various concentrations of substrate S0, and the relative reaction rates are measured in terms of light emission rates, measured using a photomultiplier tube:

S0 (μM)

Relative light units (RLU)

5

3554

10

6262

20

10115

40

14611

80

18786

200

22672

500

24718

1000

25484

a. From the data, estimate the Vmax (in RLU) and KM (in μM).

b. When the substrate concentration is 1000 μM, and the light production is monitored over a period of time, the reaction rate remains relatively constant for approximately 5.0 minutes (300 seconds), after which it rapidly decreases to almost zero.

Estimate the kcat (in s-1) and the catalytic efficiency.

 

4. Consider an enzyme embedded uniformly within particles.  Initial rate experiments are performed using a substrate concentration S0 in the linear range of enzyme saturation, yielding

V0/S0 = 1.85 min.-1

In a second preparation, the particles are loaded with 3 times the amount of active enzyme, yielding

V0/S0 = 3.20 min.-1

For this system, would you classify the intraparticle diffusion resistance as minimal, strong, or intermediate?  Explain the basis for your answer.

 

5. An enzyme is embedded uniformly within spherical particles (0.1 mm diameter) at a concentration E0 = 10 μM.  When these particles are mixed with various substrate concentrations S0, and the initial reaction rate V0 is measured, it is found that the rate is proportional to substrate concentration for the conditions tested, with

V0/S0 = 0.65 s-1.

In a second preparation using the same particles, it is determined that there is double the amount of active enzyme per particle (E0 = 20 μM), and this time

V0/S0 = 1.00 s-1.

a. Calculate the ratio of the two effectiveness factors,

η(E0 = 20 μM)/η(E0 = 10 μM).

b. Calculate the ratio of the two Thiele moduli,

φ(E0 = 20 μM)/φ(E0 = 10 μM).

 

6. An industrially important enzyme is embedded uniformly within permeable, spherical particles with effective concentration E0 = 0.1 μM.  The particles are incubated with different substrate concentrations S0, and the product formation rate V0 (moles reacted/pellet volume/time) is measured for each.  The experiment is repeated for a series of average particle diameters d (d = 2R), maintaining the same composition and total mass of particles in each case.  The reaction rate is proportional to S0 for all of the conditions tested, and the results are as follows:

d (μm)

V0/S0 (s-1)

3

0.138

10

0.137

30

0.127

100

0.080

300

0.033

1000

0.010

a. Neglecting external mass transfer resistance, show that the enzyme exhibits intrinsic (reaction-limited) kinetics when d < 10 μm, and estimate the catalytic efficiency kcat/KM (μM-1s-1) for the embedded enzyme.

b. Show that there is strong intraparticle diffusion resistance when d > 300 μm and estimate the substrate diffusion coefficient in the particle Dp (μm2/s).

 

7. An enzyme is embedded uniformly within a permeable membrane (thickness 2L = 180 μm).  The membrane is incubated with different substrate concentrations S0, and the product formation rate V0 is measured in each case.  This experiment is repeated for different enzyme loadings.  The reaction rate is proportional to S0 for all of the conditions tested, and the results are as follows:

E0 (μM)

V0/S0 (s-1)

0.1

0.0012

0.2

0.0023

0.4

0.0041

0.8

0.0069

1.6

0.0108

3.2

0.0158

6.4

0.0226

a. Show that the reaction rate reflects the intrinsic enzyme kinetics when E0 = 0.1 μM, while it is subject to strong intraparticle diffusion resistance when E0 = 6.4 μM.

b. Based on the information in part a and neglecting external mass transfer resistance, estimate the kcat/KM ratio (μM-1s-1) and substrate diffusion coefficient Dp (μm2/s) in the membrane.

 

8. An immobilized protease is to be used to break down proteins in an industrial process.  In solution, this enzyme exhibits a catalytic efficiency of 5.4 x 105 M-1s-1 for the cleavage of a standard polypeptide.  When the protease is embedded within permeable particles, at a concentration of 10 μM (particle volume basis) with average volume/surface area ratio of 30 μm, the observed catalytic efficiency (at "low" substrate concentrations) is determined to be 1.3 x 105 M-1s-1.   The effective substrate diffusivity is Dp = 2.5 x 10-6 cm2/s.

a. Give two possible explanations for the apparent reduction in catalytic efficiency.

b. Estimate the actual catalytic efficiency of the immobilized enzyme.

 

9. Consider an industrially relevant enzyme, which acts upon a small-molecule substrate with the following catalytic properties (measured in solution):

kcat = 5 s-1

KM = 2 μM

The enzyme is embedded uniformly in spherical particles (0.5 mm diameter), at a concentration of 10 μM based on particle volume, and processes substrate with bulk concentration 100 μM.  Under these conditions, the initial reaction rate is found to be 5 μM/s (particle volume basis).

a. Show that the observed rate is far lower than one would expect based on the solution kinetics.

b. What two distinct effects might contribute to the lower observed rate?

c. Using quantitative reasoning, speculate on how significant each of these effects is in contributing to the lower observed rate.

 

10. An enzyme is surface-immobilized in microtiter wells.  The wells are loaded with different amounts of enzyme E0 and incubated with different substrate concentrations S0, and reaction rates are assessed by accurately measuring accumulation of product.  The experiment is repeated for different agitation speeds, with the following results:

  

Average reaction rates in (nmol/cm2/min.)

  

E0 = 0.05 nmol/cm2

E0 = 0.1 nmol/cm2

Agitation speed (rpm)

S0 (μM) =

1

10

1

10

10

 

0.054

0.33

0.062

0.53

30

 

0.076

0.35

0.097

0.64

100

 

0.10

0.36

0.14

0.70

300

 

0.12

0.37

0.19

0.72

1000

 

0.13

0.37

0.23

0.74

a. You decide that the two data points collected with E0 = 0.05 nmol/cm2 at the highest agitation speed accurately reflect the intrinsic kinetics of the enzyme.  State two reasons why you are so sure.

b. Estimate the value of kcat (min-1) and KM (μM) for this immobilized enzyme.

c. You also decide that the data point collected with E0 = 0.1 nmol/cm2, S0 = 1 μM at the lowest agitation speed is close to the mass transfer-limited rate.  State at least one reason why you may feel confident about this.

 

11. Consider an enzyme embedded uniformly within well-characterized, permeable particles with a volume/surface area ratio of 0.050 cm (500 μm) and an effective substrate diffusion coefficient of 2x10-7 cm2/s.  These initial rate data were taken for initial, bulk substrate concentrations S(0); the initial rates, V0, are expressed on a per-particle volume basis:

S(0), μM         V0 (μM/min,)        

10                             2.0

30                             5.9

500                          33.0

a. For each of the three data points above, would you characterize the degree of intraparticle diffusion resistance as minimal, intermediate, or strong?

b. Estimate the values of Vmax and KM for this immobilized enzyme (the chart of effectiveness factor vs. observable modulus may be helpful here, although it may not be necessary).

 

12. An immobilized enzyme, embedded within spherical particles of 3.0 mm average diameter, is characterized through a series of initial rate experiments (results plotted below), where the initial rate V0 is expressed on a particle volume basis.  The substrate, added at various concentrations S0, penetrates the particles with an effective diffusivity of 1.0 x 10-6 cm2/s.  On a particle volume basis, the enzyme loading is known to be 1.5 μM.

a. At "low" substrate concentrations (< 0.1 μM), does intraparticle diffusion affect the rate significantly?  Justify your answer.

b. At "high" substrate concentrations (> 100 μM), does intraparticle diffusion affect the rate significantly?  Justify your answer.

c. Estimate the values of kcat (s-1) and KM (μM).

 

13. Read the following article:
Lee et al.  Multienzyme catalysis in microfluidic biochips.  Biotechnol. & Bioeng., 83: 20-28 (2003).

Your goal is to estimate the approximate (order-of-magnitude) value of the Damköhler number,

Da = Vmax/kLS0,

for the surface-immobilized soybean peroxidase (SBP) under maximal enzyme loading conditions (6 μg/cm2) and an initial substrate (H2O2) concentration of 0.125 mM (Figure 5).

For the purposes of this problem, we assume an expression that is valid for fully-developed, laminar flow between parallel plates, with constant flux (reaction rate) at only the bottom surface:

Sh = kLd/DS ≈ 2.692  (a value of ~ 3 suffices as an estimate);

where d is the height of the channel, and DS is the diffusion coefficient of H2O2 (a small molecule).

Based on the values given above and others found in the paper, estimate the value of Da (watch your units carefully; this is a dimensionless quantity!).

The authors claim on p. 26 that, under the conditions used, the rate of this reaction is not significantly influenced by mass transfer limitations.  Do you agree?

 

Cell Metabolism and Growth

1. Consider a culture of bacteria with empirical biomass formula

CH1.7O0.46N0.18    (MWB = 23.6 g/mole)

growing aerobically on glucose (C6H12O6).

You decide to measure the yield coefficients for glucose and oxygen and find that

YX/S = 85 g biomass/mole glucose;

YX/O2 = 39 g biomass/mole O2.

This organism secretes no appreciable amounts of product under these conditions.

a. Show that the measured values of YX/S and YX/O2 are stoichiometrically consistent with each other, stating all assumptions.

b. A batch culture of this organism initially contains 0.01 g of biomass inoculum and 20 mmol glucose.  The culture is incubated overnight, and subsequent optical density measurements suggest that the cells are no longer growing.  The total biomass in the culture is calculated to be 1.0 g.  Estimate the final amount of glucose in the medium (mmol), and speculate on what caused the plateau in biomass amount.

 

2. Consider an anaerobic fermentation using yeast, with empirical biomass formula

CH1.7O0.45N0.15     (MWB = 23.0 g DCW/mole)

The carbon and nitrogen sources are glucose (C6H12O6) and ammonium salts, respectively.  The possible products are biomass and ethanol (C2H6O), along with carbon dioxide and water.

a. What is the maximum possible yield coefficient of biomass (g DCW/mole glucose), and under what condition is it realized?

b. What is the maximum possible yield coefficient of ethanol (moles EtOH/mole glucose), and under what condition is it realized?

 

3. Consider a batch culture of bacteria with empirical biomass formula

CH1.6O0.55N0.20     (MWB = 25.2 g/mole).

The bioreactor is a closed, rigid vessel into which liquid medium containing 10 mmol glucose (C6H12O6) and an excess of ammonium sulfate is added.  The headspace contained humidified air initially.  After some time, during which a net growth of 0.3 g dry cell weight was achieved, the headspace gas was analyzed.  After accounting for the appropriate liquid concentration of O2 in equilibrium with the headspace, it was found that a total of 15 mmol O2 was consumed.

This organism secretes no appreciable amounts of product under these conditions.

a. Estimate the yield coefficient of substrate, YX/S (g DCW/mole glucose) and the final amount of glucose in the medium (mmol).

b. Estimate how much CO2 was produced (mmol).

 

4. A bacterial culture is carried out using pyruvate (C3H4O3) as the growth substrate.  The nitrogen source is composed of ammonium salts.  The empirical biomass formula is CH1.8O0.5N0.17 (MWB = 24.2 g DCW/mol).

a. Based on this information alone, estimate the maximum theoretical yield of biomass per mole of pyruvate (g DCW/mol).

The culture described above is carried out aerobically, and no secreted products are detected.  Based on measurements of CO2 concentrations in the headspace and dissolved in the medium, you determine that 45 mmol CO2 is liberated for every g DCW biomass produced.

b. Estimate the actual biomass yield per mole of pyruvate (g DCW/mol).

c. Explain the difference between your answers in parts a & b.

 

5. Aerobacter aerogenes is to be cultured aerobically, with glucose (C6H12O6) or pyruvate (C3H4O3) as the growth substrate.  The following information was taken from Bailey & Ollis:

Empirical biomass formula: CH1.78N0.24O0.33 (MWB = 22.5 g/mol); biomass degree of reductance = 4.4

                                                            YX/S                                                                YX/O2
                                    ---------------------------------------------                            -------------------------
Substrate                      g/g                  g/mol               g/g-C                            g/g                  g/mol

Glucose                       0.40                  72.7                  1.01                           1.11                  35.5
Pyruvate                      0.20                  17.9                  0.49                           0.48                  15.4

a. Compare the actual biomass yields per mole glucose or pyruvate to their respective maximum theoretical yields.  Which growth substrate is more efficient with respect to biosynthesis?

b. Give two brief explanations why the answer to part a makes sense (or, make an educated guess as to which substrate is more efficient and justify your answer).

c. Considering the case of growth on glucose, evaluate whether or not one can expect significant amounts of secreted product(s).

 

6. Consider a batch, aerobic bacterial culture with the following parameters:

Empirical biomass formula: CH2O0.5N0.2 (MWB = 24.8 g DCW/mol)

Volume of medium in the reactor: 100 L

Inoculation density: 0.01 g DCW/L.

Volumetric flow rate of sparge gas: 40 L/min.

(The headspace gas above the culture is allowed to escape into the environment at the same volumetric, i.e. molar, flow rate)

Sparge gas composition: 20 mol% O2 (8.6 mM), 0.5 mol% CO2

MO2 (phase equilibrium constant): 40

Carbon source: glucose (C6H12O6; MW = 180 g/mol)

Initial glucose concentration: 20 g/L

The bacteria secrete no detectable products other than CO2 and water.

At a time of 10 hours after the lag time, the cells were still growing exponentially, and the following data were taken:

mol% O2 in the headspace: 4.9 %

mol% CO2 in the headspace: 16.8 %

biomass density: 5.0 g DCW/L

a. Estimate the maximum specific growth rate, μmax.

b. Estimate the concentration of glucose remaining in the medium when the data above were taken.

 

7. This problem deals with the analysis of butyric acid bacteria fermentation, as considered in more detail by Papoutsakis (Biotechnol. Bioeng., 26: 174-187 (1984)).  It is assumed that these bacteria consume glucose and nitrogen-containing salts anaerobically to form: biomass, expressed as a reduced empirical formula CHpOnNq, butyrate (C4H8O2), acetate (C2H4O2), CO2, H2, and H2O.  Other products either do not accumulate or are produced in negligible quantities.

The metabolic pathway, considering only the carbon-containing species, may be simplified and written as follows:

C6H12O6  --->  6 CHpOnNq

C6H12O6  --->  2 AcCoA + 2 CO2

AcCoA  --->  C2H4O2

AcCoA + C2H4O2  --->  C4H8O2

Further analysis is predicated on the following observations:

            1) The biomass is generally 46.2% carbon by weight.

            2) The degree of reductance of the biomass is generally equal to 4.29.

            3) The yield coefficient for biomass is 40.0 g DCW/mol glucose.

            4) The molar ratio of H2/CO2 produced is found to be 1.20.

a. Determine the empirical molecular weight of the biomass, MWB (g DCW/mol).

b. Give the degrees of reductance for all species other than the biomass.

c. Estimate the yield coefficients for acetate and butyrate (mol/mol glucose) under these conditions.

 

Bioreactors

1. A microorganism is cultured in a 10 L batch bioreactor, which initially contained 100 g/L growth substrate (considered to be a "high" concentration) and 0.2 g/L biomass.  You are told by a colleague that the doubling time for this culture (in exponential phase) is very reproducible, with

td = 1.25 hours.

After 6 total hours in culture, you measure the cell density and substrate concentration:

t = 6 h: x = 1.24 g/L; S = 73 g/L.

Estimate:

a. The maximum specific growth rate, μmax (h-1).

b. The yield coefficient, YX/S (g biomass/g substrate).

c. The cell density at stationary phase (g/L).

d. The total culture time required to reach stationary phase (h).

e. The apparent lag time, tlag (h).

 

2. Consider the growth of a microorganism in batch culture, inoculated at a density of 0.1 g/L, growing on glucose as the limiting substrate with initial concentration S0 = 10 g/L.  After a lag time of approximately 3 h, the culture grows exponentially, with a doubling time of 2 h.  Stationary phase is reached after a total time of 14 h.

Estimate:

a. μmax (h-1)

b. YX/S (g/g)

c. The total time in culture to reach stationary phase if S0 were 2 g/L, assuming that this concentration is also sufficient to support maximal growth.

 

3. Consider a continuous, aerobic bacterial culture in a chemostat with sterile feed.  Three different dilution rates D are tested for a glucose feed concentration Sf = 10 mM, and the biomass concentration x and glucose concentration S in the exit stream are measured.  The results are as follows:

D (h-1)

x (g/L)

S (mM)

0.05

0.248

0.067

0.5

0.208

1.667

5

0

10

a. Estimate the glucose yield coefficient YX/S (g biomass/mole glucose).

b. Assuming Monod growth kinetics, estimate the maximum specific growth rate μmax (h-1) and the Monod constant KS (mM).

 

4. Under substrate-limited conditions, a microorganism exhibits the following net specific growth rate, μnet, and yield coefficient, YX/S:

μnet (h-1) = 0.70 S/(0.1 + S), with S in g/L;

YX/S (g DCW/g) = 0.40

The available growth medium contains 10 g/L substrate.

a. When a batch bioreactor containing 100 L of the growth medium is inoculated with 1.0 g DCW of biomass, estimate the maximum cell density achieved, and the approximate time required to achieve it, after exponential growth is initiated.

b. You decide instead to culture the microorganism in a chemostat, using the same growth medium as the (sterile) feed.  Estimate the dilution rate (h-1) at which the chemostat will achieve maximum steady-state productivity of biomass.

c. Calculate and compare the overall biomass productivities (g DCW/L/h) of the two scenarios in parts a & b.  What other consideration will make the batch process even less productive compared to the chemostat?

 

5. Consider a culture of bacteria making a valuable product in a chemostat operated at steady state.  The liquid feed is sterile and contains 50 mM glucose (the limiting growth substrate S).  In a series of steady state runs, the dilution rate D is increased incrementally, and the cell density exiting the chemostat is measured for each:

D (h-1)

x (g/L)

0.05

1.58

0.10

2.17

0.15

2.47

0.20

2.65

0.25

2.76

0.30

2.83

0.35

2.86

0.40

2.84

0.45

2.74

0.50

2.25

0.52

1.32

0.53

0

a. Explain why the cell density can increase as the flow rate is increased, as shown here with D < 0.35 h-1.

b. At what dilution rate should you operate the chemostat to optimize productivity of a strictly growth-associated product?  Explain the basis for your answer.

c. At what dilution rate should you operate the chemostat to optimize productivity of a strictly non-growth-associated product?  Explain the basis for your answer.

 

6. Consider cell growth in a chemostat at steady state, with sterile feed.  In bench-scale experiments, it is found that the specific growth rate is inhibited at high glucose concentrations S , modeled according to the phenomenological expression

μ (h-1) = 0.6S/(0.2 + S + 0.1S2), with S in g/L.

a. If the glucose concentration in the feed is 10 g/L, estimate the minimum dilution rate at which cell washout would occur.

b. Is the dilution rate estimated in part a the highest that will support cell growth in the bioreactor?  Justify and briefly explain your answer.

 

7. Ethanol (C2H5OH) is produced by a microorganism cultured anaerobically in a chemostat.  The empirical biomass formula is CH2O0.45N0.19 under these conditions.  The sterile feed contains a "high" concentration (20 mM) of glucose, an excess of ammonium salts, and no ethanol.  When the dilution rate is set at 0.1 h-1, the exit stream contains 2 mM glucose and 0.45 g/L biomass.  It is safe to assume that ethanol is the only secreted product.

a. Determine the expected concentration of ethanol (mM) in the exit stream.

b. The dilution rate chosen probably does not optimize biomass productivity.  Briefly explain how you know this to be the case and speculate on why this dilution rate may have been chosen.

 

8. Consider a culture of bacteria that secrete a product in a chemostat operated at steady state.  The specific growth rate of biomass is adequately described by the Monod equation, and the rate of product formation is given by the Leudeking and Piret equation:

rP = (αμ + β)x

This system is well characterized, such that the following constants are known:

YX/S = 0.4 g/g

μmax = 0.7 h-1

KS = 0.2 g/L

α = 0.2 g/g

β = 0.3 g/g-h

The liquid feed to the chemostat is sterile and contains 10 g/L of the limiting growth substrate.

a. What dilution rate will optimize the productivity of the chemostat (g product/h)?  An approximate answer is sufficient.

b. Consider that a high product concentration (g/L) is also desirable.  With this in mind, how might you adjust the dilution rate from the value given in part a?  Choose a new dilution rate and give the reasoning behind your answer.

 

9. Consider the growth of a microorganism in batch culture.  When the substrate concentration is high, the cell density doubles every 0.75 h, the observed substrate yield coefficient is 0.3 g DCW/g, and substrate consumption is allocated towards biosynthesis (60%), maintenance (10%), as well as product formation (30%).   The product formation is strictly growth-associated.

The batch reactor is inoculated with 0.01 g DCW/L and 10 g/L substrate.

a. Estimate the maximum cell density and the time (after lag phase) required to achieve it.

b. Determine the value of the maintenance coefficient (g substrate/g DCW-h).

c. When the cell density reaches its peak value, the substrate concentration is measured and found to be 0.05 g/L.  Estimate the value of the saturation constant, KS (g/L).

 

Oxygen Mass Transfer/Bioreactor Scale-up

1. Consider an air-sparged, stirred-tank bioreactor.  It is a large-volume process, and you have data suggesting that dissolved oxygen is unacceptably heterogeneous within the reactor.  In an attempt to shorten the blend time, you increase the stirring speed of the impeller by a factor of 2.

a. Estimate the factor by which power imparted by the impeller changes.

b. Estimate the factor by which the value of kLa' changes.

c. Give two good reasons why this change may be detrimental.

 

2. An aerobic culture is to be carried out in a 100 L stirred-tank batch reactor.  The reactor is sparged with enriched air [Cg(O2) = 20 mM] at a flow rate Fg = 20 L/min., and the modified Henry's Law constant for O2 is M = 40.  The microorganisms consume 20 mmol O2/g DCW/h, and you are told that the mass transfer rate constant, measured under process conditions, is kLa' = 0.02 s-1.

a. Calculate the maximum cell density (g DCW/L) that will be achieved before oxygen becomes limiting for growth.

b. You now wish to scale up the process to 10,000 L, keeping the same kLa' and Fg/V.  Estimate the following ratios, where II refers to the 10,000 L scale and I refers to the 100 L scale:

dt,II/dt,I

Ni,II/Ni,I

(P/V)II/(P/V)I

Rei,II/Rei,I

 

3. Consider a 20 liter aerobic batch culture.  The yield coefficient based on oxygen consumption, YX/O2, is known to be 30 g biomass/mole O2.  The medium contains sufficient glucose to yield maximal growth, and the cell biomass doubles every 2 hours under fully aerated conditions.  The desired concentration of biomass at the end of the run is 1 g/L.

Aeration is achieved by sparging air through the tank with agitation.  From the ideal gas law, the entering oxygen concentration is calculated to be

Cg,0 = 8.30 mM.

The modified Henry's law expression describing liquid-vapor equilibrium for oxygen is

Cg = MCL*,

with M = 40 for the culture conditions described here.

a. Estimate the minimum value of kLa' (s-1) required to achieve maximal growth for the entire duration of the culture.

b. The same culture is to be scaled up to 8,000 liters, and in order to maintain consistent mass transfer rates you design the system such that the impeller power/volume (P/V) ratio is the same at both scales.  If mixing is in the turbulent regime, by what factor does the impeller speed change?  In other words, estimate the ratio Ni (8000L)/Ni (20L).  For partial credit, at least state whether the required impeller speed increases or decreases.

c. Discuss two potential drawbacks of the scale-up strategy described in part b.

 

4. After graduation you accept a position in a biotech company, where your first assignment is to scale up an aerobic microbial culture from the well-characterized 50 L pilot scale to the 1,500 L process scale.  You discuss the problem with another recently hired chemical engineer, who received his degree from {insert name of rival school here}.

Him: "That's easy.  Just maintain the same Reynolds number."

You: "You mean the impeller Reynolds number, Rei?"

Him: "Uh… yeah."

You decide to perform an analysis of this strategy.  Calculate the following ratios, where scale II is 1,500 L and scale I is 50 L.  If you are not able to answer one or more parts, at least state whether the ratio will be greater than, less than, or equal to one.

a. Stirring speed, NII/NI.

b. Power imparted per volume fluid, (P/V)II/(P/V)I.

c. (kLa')II/(kLa')I, assuming the same gas flow rate per unit fluid volume (Fg/V) at both scales.

d. State two reasons why scaling up in this manner is a really bad idea.

 

5. An air-sparged, stirred-tank bioreactor is operated in batch mode.  After an initial lag phase, the cells in the reactor grow exponentially.  The conditions are such that dissolved oxygen will ultimately limit cell growth, but cell death and maintenance under oxygen-limited conditions are negligible.

True or False (you must justify your answer):
When the dissolved oxygen concentration drops below critical levels, the absolute growth rate of the cells (rx) will drop, and soon after, the cell density will remain constant.

 

6. Consider a batch culture sparged with air (Fg/V = 0.50 min-1; Cg = 8.0 mM; CL* = 0.20 mM).  At a certain time during exponential growth, the gas flow is turned off for 60 seconds, and the dissolved oxygen concentration (CL) is monitored as a function of time:

After the air is turned back on, the dissolved oxygen concentration is fit well by

CL (mM) = 0.173 - 0.133exp(-0.083t), where t is time in seconds.

a. At the time of this experiment, the cell density is determined to be 0.50 g DCW/L.  Estimate the cell density at which the culture will cease to grow exponentially.

b. The stirring speed, Ni, is increased by a factor of 3.  By what factor will the kLa' change?  Will the maximum oxygen transfer rate change by approximately the same factor?  Justify your answer.

c. The process is to be scaled up by a factor of 10 in terms of volume.  You decide to maintain the same circulation time (determined by Ni) and kLa' values.  What process variable will need to be adjusted, and by approximately what factor?  Why is this not such a good strategy?

 

7. An agitated batch bioreactor is aerated with O2-containing gas (Cg = 10 mM, M = 40).  At a certain point during exponential phase, the dissolved oxygen concentration is CL = 0.2 mM; when the sparging is turned off briefly, CL drops to 0.1 mM in 30 seconds.

a. Estimate the value of kLa' (s-1).

b. From the time when the measurement was taken, the cells continue to grow exponentially for 4.0 hours, after which they stop doing so.  Estimate the specific growth rate μ (h-1) during this period.

c. You wish to choose between two different impellers for this bioreactor.  Impeller #1 has 1/2 the diameter of the tank (di/dt = 1/2) and a power number of 1 (aerated).  Impeller #2 has 1/3 the diameter of the tank and a power number of 2 (aerated).   Assuming that the same Fg and V are to be maintained, what is the ratio of stirring speeds for the two impellers (Ni,2/Ni,1) that yields the same kLa' value?

 

Animal Cell Culture/Cell Signaling

1. After graduation, you go to work for a medium-sized pharmaceutical company in their Cell Culture Manufacturing group, which is in charge of process design for large-scale animal cell culture.  The process is a stirred-tank with air sparging, but the cells' oxygen demand is low, and so the gas flow rates (Fg/V ~ 0.03 min-1) and kLa' values (~ 20 h-1, typically) are much lower than those typically used in microbial culture.  Accordingly, you discover that the maximum oxygen transfer rate (OTR) for the process scales much differently than for "traditional" fermentation processes, with OTRmax scaling as

(P/V)0.25us0.85

a. Based on the typical values cited above, is oxygen depletion from the bubbles significant?

b. The process volume is to be scaled up by a factor of 100, and you wish to keep the kLa' and Fg/V constant (as is typically done for mass transfer considerations).   What process variable will have to be adjusted and by what factor?   (Partial credit for identifying the variable and whether it must increase or decrease).

c. Evaluate whether or not the strategy described above is sound.

 

2. Interleukin-2 (IL-2) stimulates T cells during the immune response, through a multi-subunit receptor expressed on the T cells.  Actually, the subunits are precoupled, such that the assumption of 1:1 ligand/receptor stoichiometry is reasonable in this case.  The concentration of IL-2 required to elicit half-maximal steady-state binding to cells is extremely low, with a measured value of

KD,App = 10 pM (1.0 x 10-11 M)

The steady state number of IL-2/receptor complexes is somewhat lower than for most growth factors, with a maximum value of 1,000 molecules/cell, and the endocytic rate constant, ke , is known to be 0.1 min-1.  We assume here that all internalized IL-2 molecules are degraded.

You wish to determine how often you will need to change the medium in order to maintain the IL-2 concentration.

For a cell density of 105 cells/mL (108 cells/L) and an initial IL-2 concentration of 100 pM, estimate the rate of ligand depletion, in (pM IL-2 consumed)/day.  Comment on whether daily re-feeding is warranted.

 

3. The HT-2 cell line is a line of mouse T cells that proliferates in response to the cytokine, interleukin 2 (IL-2).  A flask of cells was inoculated at 3x104 cells/mL, and the medium initially contained a saturating dose (100 pM) of IL-2.  At various times, an aliquot of the cell suspension was taken out and the cell density was quantified.  The time course data are displayed below on linear and semi-log plots:

                       

The internalization rate constant of IL-2/IL-2 receptor complexes has been measured (0.1 min-1), and it is assumed that all internalized ligand molecules are degraded.   It may also be presumed that cell growth is limited by IL-2 concentration.

a. Estimate the value of μmax (h-1) under these conditions.

b. Estimate the maximum, steady-state number of IL-2/IL-2 receptor complexes per cell for this cell line.

 

Product Recovery

1. Solids are effectively pelleted from a sample in 5 minutes using a centrifuge operated at 1,000 rpm.  Estimate how much time it will take if the centrifuge is operated at 2,000 rpm.

 

2. A chromatographic separation produces two peaks.  For a 5 cm long column, the peaks exhibit the following characteristics:

 

Peak 1

Peak 2

Mean residence time (min.):

15

40

Standard deviation (min.):

5

10

a. Calculate the resolution value, Rs, of these two peaks, and comment on the ability to achieve both high yield and high purity.

b.  For the chromatographic separation described in part a, estimate the new resolution value if the length of the column is extended to 10 cm, while keeping the same flow rate.

 

3. An affinity column is used to separate a specific protein called Protein X from spent culture medium.  A contaminating protein, Protein Y, is also found in the medium.  The spherical beads to be used in the column are impermeable, meaning that neither Protein X nor Protein Y can penetrate the particles.  In pilot experiments, pulses of Protein X or Protein Y are added to the column, which is packed with beads that are either unconjugated (no binding to Protein X or Protein Y) or ligand-conjugated (bind specifically to Protein X).  The following results are obtained:

 

Unconjugated

Ligand-conjugated

 

Protein X

Protein Y

Protein X

Protein Y

Average residence time (min.)

5.0

5.0

12.0

5.0

Peak width (min.)

2.0

2.0

6.4

2.0

(Note that the peak width w = 4 σt assuming Gaussian peaks).

a. For the column loaded with ligand-conjugated beads, Protein X is collected between 10 and 15 minutes.  Estimate the % yield of Protein X in the product.

b. One can increase the yield of Protein X by collecting for a longer period of time.  List two potential drawbacks associated with this strategy.

c. What causes the broadening of the Protein Y peaks?  What additional factor(s), if any, contribute to the broadening of the Protein X peak in the ligand-conjugated column?

 

4. A chromatographic separation produces two peaks with the following characteristics:

                      tr, avg (min.)         σt (min.)

Peak 1                  20                      5

Peak 2                  45                     10

In an effort to collect the protein in Peak 1, you collect material coming off the column between 15 and 22 minutes.

a. Estimate the fractional recovery (yield) of the protein in peak 1.

b. It is not possible to accurately estimate the product purity without knowing the peak concentrations.  Using quantitative reasoning, speculate on whether high purity will be achieved.

(see above for erf(x) vs. x plot).

 

5. An affinity column (100 mL total volume) is used to separate a specific protein called protein B from spent culture medium.  A contaminating protein, protein A, is also found in the medium.  Chromatograms are acquired at two different flow rates:

 

Q = 2 mL/min.

Q = 10 mL/min.

 

Protein A

Protein B

Protein A

Protein B

Average residence time (min.)

22

37

4.4

7.4

Peak width (min.)

4.6

11.2

1.6

6.1

(Note that the peak width w = 4 σt assuming Gaussian peaks).

a. Calculate the values of the resolution, Rs, for each flow rate.  Comment on the ability to achieve both high yield and high purity in each case.

b. In the case of Q = 2 mL/min., suppose that you collect two fractions of the flow-through: from 35 to 40 minutes, and from 40 to 45 minutes.  You wish to decide between keeping just the first fraction (which will have the higher concentration of B, CB), or pooling the two fractions together (as if you collected from 35-45 minutes).  Estimate the following ratios:

YieldB(35-45')/YieldB(35-40')

CB(35-45')/CB(35-40')

(see above for erf(x) vs. x plot).

 

  • Hi,
    I did an experiment in school last week investigating the effect of temperature on both free and immobilized lipase. We used temperatures of 40 and 80 degrees C. At 40, both were active but at 80 only the immobilized lipase was active. However, its activity was greatly reduced to that at 40. I understand that the immobilisation process makes the enzymes more stable but why is this and why, if it is more stable, is the activity reduced at all at a higher temperature? In other words, what is the explanation for the decrease in activity in immobilised lipase between 40 and 80 degrees C when with free lipase the enzyme is simply denatured and activity is completely halted?
    Any ideas would be greatly appreciated!! Thanks

  • The immobilisation process makes the enzyme MORE stable, not completely stable. So, the activity of the immobilized enzyme will be greater than that of the free enzyme. Basically, just because its more stable it does not mean that it wont have reduced activity at certain temperatures.

  • The alginate beads (I presume) stabilise the bonds within the tertiary structure of the immobilised lipase, thus makes them more stable. However, SOME of the immobilised lipase will still denature, and so there is less activity at 80 degrees. The lipase which does not denature is still active and so can still hydrolyse whatever substrate it may be, so there is still some activity.

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