Problems in Biochemical EngineeringJason 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 An enzyme is embedded uniformly within spherical particles (0. and this time V0/S0 = 1. maintaining the same composition and total mass of particles in each case. the particles are loaded with 3 times the amount of active enzyme. In a second preparation using the same particles.138 10 0.00 s-1. 6.127 100 0. b. η(E0 = 20 μM)/η(E0 = 10 μM). or intermediate? Explain the basis for your answer. and the initial reaction rate V0 is measured. strong. 5. and the results are as follows: d (μm) V0/S0 (s-1) 3 0. Calculate the ratio of the two effectiveness factors.-1 For this system.20 min. φ(E0 = 20 μM)/φ(E0 = 10 μM).1 μM. it is found that the rate is proportional to substrate concentration for the conditions tested. The experiment is repeated for a series of average particle diameters d (d = 2R). The reaction rate is proportional to S0 for all of the conditions tested.137 30 0.1 mm diameter) at a concentration E0 = 10 μM. Calculate the ratio of the two Thiele moduli.65 s-1. An industrially important enzyme is embedded uniformly within permeable. with V0/S0 = 0.080 300 0. When these particles are mixed with various substrate concentrations S0. and the product formation rate V0 (moles reacted/pellet volume/time) is measured for each.033 . a. it is determined that there is double the amount of active enzyme per particle (E0 = 20 μM). yielding V0/S0 = 3.In a second preparation. spherical particles with effective concentration E0 = 0. The particles are incubated with different substrate concentrations S0. would you classify the intraparticle diffusion resistance as minimal. 4 x 105 M-1s-1 for the cleavage of a standard polypeptide.4 μM. a. This experiment is repeated for different enzyme loadings.4 0. Neglecting external mass transfer resistance.8 0.010 a. 1000 0.1 0. at a concentration of 10 μM (particle volume basis) with average volume/surface area ratio of 30 μm. The membrane is incubated with different substrate concentrations S0. Estimate the actual catalytic efficiency of the immobilized enzyme. and the product formation rate V0 is measured in each case.2 0. .5 x 10-6 cm2/s. and estimate the catalytic efficiency kcat/KM (μM-1s-1) for the embedded enzyme. estimate the kcat/KM ratio (μM-1s-1) and substrate diffusion coefficient Dp (μm2/s) in the membrane. b. this enzyme exhibits a catalytic efficiency of 5. When the protease is embedded within permeable particles. An immobilized protease is to be used to break down proteins in an industrial process.2 0. Give two possible explanations for the apparent reduction in catalytic efficiency. 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. 8.6 0. while it is subject to strong intraparticle diffusion resistance when E0 = 6. Based on the information in part a and neglecting external mass transfer resistance.0012 0. In solution. The reaction rate is proportional to S0 for all of the conditions tested.0226 a. b. and the results are as follows: E0 (μM) V0/S0 (s-1) 0.4 0. An enzyme is embedded uniformly within a permeable membrane (thickness 2L = 180 μm).0023 0.0158 6.1 μM. the observed catalytic efficiency (at "low" substrate concentrations) is determined to be 1.3 x 105 M-1s-1.0069 1. b.0041 0. Show that the reaction rate reflects the intrinsic enzyme kinetics when E0 = 0. show that the enzyme exhibits intrinsic (reaction- limited) kinetics when d < 10 μm. The effective substrate diffusivity is Dp = 2.0108 3. 19 0. Using quantitative reasoning.5 mm diameter).23 0. at a concentration of 10 μM based on particle volume. What two distinct effects might contribute to the lower observed rate? c.35 0. with the following results: Average reaction rates in (nmol/cm2/min.097 0.37 0. The wells are loaded with different amounts of enzyme E0 and incubated with different substrate concentrations S0. You decide that the two data points collected with E0 = 0.062 0.076 0. .9. a. Under these conditions. You also decide that the data point collected with E0 = 0.36 0.05 nmol/cm2 at the highest agitation speed accurately reflect the intrinsic kinetics of the enzyme. 10. 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 (0.1 nmol/cm2. State two reasons why you are so sure. The experiment is repeated for different agitation speeds. Consider an industrially relevant enzyme. c.14 0.13 0.72 1000 0. b.1 nmol/cm2 Agitation speed (rpm) S0 (μM) = 1 10 1 10 10 0.12 0.64 100 0. Show that the observed rate is far lower than one would expect based on the solution kinetics. b.74 a.10 0.) E0 = 0.37 0.05 nmol/cm2 E0 = 0.53 30 0. State at least one reason why you may feel confident about this. S0 = 1 μM at the lowest agitation speed is close to the mass transfer-limited rate. and reaction rates are assessed by accurately measuring accumulation of product. Estimate the value of kcat (min-1) and KM (μM) for this immobilized enzyme. An enzyme is surface-immobilized in microtiter wells.70 300 0. and processes substrate with bulk concentration 100 μM. the initial reaction rate is found to be 5 μM/s (particle volume basis). speculate on how significant each of these effects is in contributing to the lower observed rate.33 0.054 0. 050 cm (500 μm) and an effective substrate diffusion coefficient of 2x10-7 cm2/s. although it may not be necessary). penetrates the particles with an effective diffusivity of 1. a. the enzyme loading is known to be 1. V0. observable modulus may be helpful here. Consider an enzyme embedded uniformly within well-characterized. Estimate the values of Vmax and KM for this immobilized enzyme (the chart of effectiveness factor vs. would you characterize the degree of intraparticle diffusion resistance as minimal.0 30 5.1 μM). At "low" substrate concentrations (< 0. bulk substrate concentrations S(0).5 μM. For each of the three data points above. or strong? b. On a particle volume basis. where the initial rate V0 is expressed on a particle volume basis. the initial rates. is characterized through a series of initial rate experiments (results plotted below). μM V0 (μM/min. intermediate. embedded within spherical particles of 3.) 10 2.0 a. These initial rate data were taken for initial. An immobilized enzyme. permeable particles with a volume/surface area ratio of 0.0 x 10-6 cm2/s. The substrate. does intraparticle diffusion affect the rate significantly? Justify your answer.9 500 33. .11. are expressed on a per-particle volume basis: S(0). 12.0 mm average diameter. added at various concentrations S0. Biotechnol. where d is the height of the channel. The authors claim on p. 13. does intraparticle diffusion affect the rate significantly? Justify your answer. & Bioeng. we assume an expression that is valid for fully-developed. For the purposes of this problem. under the conditions used. Da = Vmax/kLS0. Read the following article: Lee et al. and DS is the diffusion coefficient of H2O2 (a small molecule). Multienzyme catalysis in microfluidic biochips. Estimate the values of kcat (s-1) and KM (μM).6 g/mole) growing aerobically on glucose (C6H12O6). for the surface-immobilized soybean peroxidase (SBP) under maximal enzyme loading conditions (6 μg/cm2) and an initial substrate (H2O2) concentration of 0.. with constant flux (reaction rate) at only the bottom surface: Sh = kLd/DS ≈ 2.125 mM (Figure 5). At "high" substrate concentrations (> 100 μM). You decide to measure the yield coefficients for glucose and oxygen and find that YX/S = 85 g biomass/mole glucose.46N0. . Your goal is to estimate the approximate (order-of-magnitude) value of the Damköhler number. Consider a culture of bacteria with empirical biomass formula CH1. laminar flow between parallel plates.18 (MWB = 23. 83: 20-28 (2003). Do you agree? Cell Metabolism and Growth 1. c.7O0. this is a dimensionless quantity!). estimate the value of Da (watch your units carefully.b. Based on the values given above and others found in the paper. 26 that.692 (a value of ~ 3 suffices as an estimate). the rate of this reaction is not significantly influenced by mass transfer limitations. Estimate the yield coefficient of substrate. b. After accounting for the appropriate liquid concentration of O2 in equilibrium with the headspace.01 g of biomass inoculum and 20 mmol glucose. a. with empirical biomass formula CH1. The possible products are biomass and ethanol (C2H6O). it was found that a total of 15 mmol O2 was consumed.3 g dry cell weight was achieved. Consider a batch culture of bacteria with empirical biomass formula CH1.20 (MWB = 25. What is the maximum possible yield coefficient of biomass (g DCW/mole glucose). a.15 (MWB = 23.6O0. and under what condition is it realized? b.0 g. rigid vessel into which liquid medium containing 10 mmol glucose (C6H12O6) and an excess of ammonium sulfate is added.7O0.2 g/mole). stating all assumptions. This organism secretes no appreciable amounts of product under these conditions. The bioreactor is a closed. b. the headspace gas was analyzed. The total biomass in the culture is calculated to be 1. and under what condition is it realized? 3.45N0.0 g DCW/mole) The carbon and nitrogen sources are glucose (C6H12O6) and ammonium salts. Estimate the final amount of glucose in the medium (mmol).YX/O2 = 39 g biomass/mole O2. respectively. Consider an anaerobic fermentation using yeast. A batch culture of this organism initially contains 0. Estimate how much CO2 was produced (mmol).55N0. The headspace contained humidified air initially. YX/S (g DCW/mole glucose) and the final amount of glucose in the medium (mmol). What is the maximum possible yield coefficient of ethanol (moles EtOH/mole glucose). along with carbon dioxide and water. a. Show that the measured values of YX/S and YX/O2 are stoichiometrically consistent with each other. The culture is incubated overnight. during which a net growth of 0. and subsequent optical density measurements suggest that the cells are no longer growing. . After some time. This organism secretes no appreciable amounts of product under these conditions. and speculate on what caused the plateau in biomass amount. 2. evaluate whether or not one can expect significant amounts of secreted product(s). 5. Aerobacter aerogenes is to be cultured aerobically. Considering the case of growth on glucose. aerobic bacterial culture with the following parameters: . The following information was taken from Bailey & Ollis: Empirical biomass formula: CH1. biomass degree of reductance = 4. ------------------------ - Substrate g/g g/mol g/g-C g/g g/mol Glucose 0. a. 6. Give two brief explanations why the answer to part a makes sense (or. and no secreted products are detected. Based on measurements of CO2 concentrations in the headspace and dissolved in the medium. estimate the maximum theoretical yield of biomass per mole of pyruvate (g DCW/mol). The nitrogen source is composed of ammonium salts.49 0. The culture described above is carried out aerobically. The empirical biomass formula is CH1.40 72.5 Pyruvate 0. Estimate the actual biomass yield per mole of pyruvate (g DCW/mol).01 1. Explain the difference between your answers in parts a & b. with glucose (C6H12O6) or pyruvate (C3H4O3) as the growth substrate. c. A bacterial culture is carried out using pyruvate (C3H4O3) as the growth substrate. make an educated guess as to which substrate is more efficient and justify your answer).4 YX/S YX/O2 --------------------------------------------.20 17. you determine that 45 mmol CO2 is liberated for every g DCW biomass produced.9 0. Which growth substrate is more efficient with respect to biosynthesis? b. b.7 1. Consider a batch.4 a.5N0.4. c.17 (MWB = 24.5 g/mol).78N0.24O0.11 35. Compare the actual biomass yields per mole glucose or pyruvate to their respective maximum theoretical yields. Based on this information alone.8O0.2 g DCW/mol).48 15.33 (MWB = 22. 8 % biomass density: 5. 0.01 g DCW/L. Estimate the concentration of glucose remaining in the medium when the data above were taken. (The headspace gas above the culture is allowed to escape into the environment at the same volumetric. expressed as a reduced empirical formula CHpOnNq.9 % mol% CO2 in the headspace: 16. butyrate (C4H8O2). molar.. 7.0 g DCW/L a.2 (MWB = 24. Estimate the maximum specific growth rate. 26: 174-187 (1984)). μmax. H2. flow rate) Sparge gas composition: 20 mol% O2 (8.8 g DCW/mol) Volume of medium in the reactor: 100 L Inoculation density: 0. the cells were still growing exponentially. acetate (C2H4O2). At a time of 10 hours after the lag time. and H2O. and the following data were taken: mol% O2 in the headspace: 4. MW = 180 g/mol) Initial glucose concentration: 20 g/L The bacteria secrete no detectable products other than CO2 and water. as considered in more detail by Papoutsakis (Biotechnol. Bioeng.5 mol% CO2 MO2 (phase equilibrium constant): 40 Carbon source: glucose (C6H12O6.e. considering only the carbon-containing species.Empirical biomass formula: CH2O0. may be simplified and written as follows: . Volumetric flow rate of sparge gas: 40 L/min. b. Other products either do not accumulate or are produced in negligible quantities.5N0. CO2. It is assumed that these bacteria consume glucose and nitrogen-containing salts anaerobically to form: biomass.6 mM). The metabolic pathway. This problem deals with the analysis of butyric acid bacteria fermentation. i. you measure the cell density and substrate concentration: t = 6 h: x = 1. The maximum specific growth rate.2% carbon by weight.24 g/L. b. A microorganism is cultured in a 10 L batch bioreactor. 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. which initially contained 100 g/L growth substrate (considered to be a "high" concentration) and 0. Give the degrees of reductance for all species other than the biomass. Estimate the yield coefficients for acetate and butyrate (mol/mol glucose) under these conditions. . Bioreactors 1. After 6 total hours in culture. with td = 1. b. 2) The degree of reductance of the biomass is generally equal to 4. You are told by a colleague that the doubling time for this culture (in exponential phase) is very reproducible. MWB (g DCW/mol). YX/S (g biomass/g substrate). a. 3) The yield coefficient for biomass is 40.20.0 g DCW/mol glucose. 4) The molar ratio of H2/CO2 produced is found to be 1. Estimate: a. The yield coefficient. S = 73 g/L. μmax (h-1).25 hours. Determine the empirical molecular weight of the biomass.2 g/L biomass. c.29. with a doubling time of 2 h. The total culture time required to reach stationary phase (h). aerobic bacterial culture in a chemostat with sterile feed. 4. μnet. the culture grows exponentially. The results are as follows: D (h-1) x (g/L) S (mM) 0. growing on glucose as the limiting substrate with initial concentration S0 = 10 g/L.1 + S).067 0. a microorganism exhibits the following net specific growth rate.667 5 0 10 a.05 0.5 0. YX/S: μnet (h-1) = 0. Three different dilution rates D are tested for a glucose feed concentration Sf = 10 mM. estimate the maximum specific growth rate μmax (h-1) and the Monod constant KS (mM). d. The total time in culture to reach stationary phase if S0 were 2 g/L. b. The cell density at stationary phase (g/L).248 0. μmax (h-1) b. 3. YX/S (g/g) c. Stationary phase is reached after a total time of 14 h. and the biomass concentration x and glucose concentration S in the exit stream are measured. and yield coefficient. Consider a continuous. tlag (h). Under substrate-limited conditions.c.1 g/L. Consider the growth of a microorganism in batch culture. with S in g/L. . 2. inoculated at a density of 0. Estimate the glucose yield coefficient YX/S (g biomass/mole glucose). e. Estimate: a. assuming that this concentration is also sufficient to support maximal growth. Assuming Monod growth kinetics. After a lag time of approximately 3 h.70 S/(0.208 1. The apparent lag time. 83 0. estimate the maximum cell density achieved.10 2. and the approximate time required to achieve it.52 1. Explain why the cell density can increase as the flow rate is increased. b. b.86 0.0 g DCW of biomass. using the same growth medium as the (sterile) feed.35 2.50 2. the dilution rate D is increased incrementally. At what dilution rate should you operate the chemostat to optimize productivity of a strictly growth-associated product? Explain the basis for your answer.53 0 a.YX/S (g DCW/g) = 0. and the cell density exiting the chemostat is measured for each: D (h-1) x (g/L) 0.74 0.15 2. c. after exponential growth is initiated.32 0. Calculate and compare the overall biomass productivities (g DCW/L/h) of the two scenarios in parts a & b.30 2.17 0.35 h-1.25 2.40 2. 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. In a series of steady state runs.05 1.65 0. a.58 0. . The liquid feed is sterile and contains 50 mM glucose (the limiting growth substrate S).20 2. You decide instead to culture the microorganism in a chemostat.76 0.40 The available growth medium contains 10 g/L substrate.45 2.84 0. Estimate the dilution rate (h-1) at which the chemostat will achieve maximum steady-state productivity of biomass.47 0. as shown here with D < 0.25 0. When a batch bioreactor containing 100 L of the growth medium is inoculated with 1. Is the dilution rate estimated in part a the highest that will support cell growth in the bioreactor? Justify and briefly explain your answer. 7.1 h-1.1S2). estimate the minimum dilution rate at which cell washout would occur. When the dilution rate is set at 0. The specific growth rate of biomass is adequately described by the Monod equation. such that the following constants are known: YX/S = 0. The empirical biomass formula is CH2O0. and no ethanol.45 g/L biomass. Determine the expected concentration of ethanol (mM) in the exit stream. b. and the rate of product formation is given by the Leudeking and Piret equation: rP = (αμ + β)x This system is well characterized.45N0.2 g/L . the exit stream contains 2 mM glucose and 0.4 g/g μmax = 0. modeled according to the phenomenological expression μ (h-1) = 0. 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. b. it is found that the specific growth rate is inhibited at high glucose concentrations S .2 + S + 0. with S in g/L. 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. an excess of ammonium salts. a. Ethanol (C2H5OH) is produced by a microorganism cultured anaerobically in a chemostat. The sterile feed contains a "high" concentration (20 mM) of glucose. 8.7 h-1 KS = 0. Consider a culture of bacteria that secrete a product in a chemostat operated at steady state. It is safe to assume that ethanol is the only secreted product. If the glucose concentration in the feed is 10 g/L. In bench-scale experiments.6S/(0. a. with sterile feed. Consider cell growth in a chemostat at steady state. 6.19 under these conditions.c. Estimate the factor by which power imparted by the impeller changes.3 g/g-h The liquid feed to the chemostat is sterile and contains 10 g/L of the limiting growth substrate. a. c. Consider the growth of a microorganism in batch culture.2 g/g β = 0. Determine the value of the maintenance coefficient (g substrate/g DCW-h). a. Estimate the value of the saturation constant.3 g DCW/g. b. maintenance (10%). 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. . the cell density doubles every 0. and you have data suggesting that dissolved oxygen is unacceptably heterogeneous within the reactor. When the cell density reaches its peak value.75 h. The batch reactor is inoculated with 0. the observed substrate yield coefficient is 0. KS (g/L). and substrate consumption is allocated towards biosynthesis (60%). Oxygen Mass Transfer/Bioreactor Scale-up 1. It is a large-volume process. stirred-tank bioreactor.05 g/L. 9. What dilution rate will optimize the productivity of the chemostat (g product/h)? An approximate answer is sufficient.α = 0. as well as product formation (30%). Estimate the factor by which the value of kLa' changes.01 g DCW/L and 10 g/L substrate. The product formation is strictly growth-associated. Give two good reasons why this change may be detrimental. a. Consider an air-sparged. b. With this in mind. c. When the substrate concentration is high. Estimate the maximum cell density and the time (after lag phase) required to achieve it. the substrate concentration is measured and found to be 0. Consider that a high product concentration (g/L) is also desirable. In an attempt to shorten the blend time. b. you increase the stirring speed of the impeller by a factor of 2. YX/O2. and the cell biomass doubles every 2 hours under fully aerated conditions. estimate the ratio Ni (8000L)/Ni (20L). and you are told that the mass transfer rate constant. 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.000 L. where II refers to the 10. For partial credit. From the ideal gas law.II/Ni. . An aerobic culture is to be carried out in a 100 L stirred-tank batch reactor. Calculate the maximum cell density (g DCW/L) that will be achieved before oxygen becomes limiting for growth. The yield coefficient based on oxygen consumption. The reactor is sparged with enriched air [Cg(O2) = 20 mM] at a flow rate Fg = 20 L/min. Estimate the minimum value of kLa' (s-1) required to achieve maximal growth for the entire duration of the culture.30 mM.2. The medium contains sufficient glucose to yield maximal growth. You now wish to scale up the process to 10.I (P/V)II/(P/V)I Rei. and the modified Henry's Law constant for O2 is M = 40.02 s-1. If mixing is in the turbulent regime. Estimate the following ratios.I Ni. The modified Henry's law expression describing liquid-vapor equilibrium for oxygen is Cg = MCL*.000 liters..II/Rei. measured under process conditions. b. a.000 L scale and I refers to the 100 L scale: dt. The desired concentration of biomass at the end of the run is 1 g/L. b. The same culture is to be scaled up to 8.II/dt. keeping the same kLa' and Fg/V.0 = 8. with M = 40 for the culture conditions described here. Aeration is achieved by sparging air through the tank with agitation. Consider a 20 liter aerobic batch culture. The microorganisms consume 20 mmol O2/g DCW/h. is kLa' = 0.I 3. a. is known to be 30 g biomass/mole O2. by what factor does the impeller speed change? In other words. at least state whether the required impeller speed increases or decreases. the entering oxygen concentration is calculated to be Cg. assuming the same gas flow rate per unit fluid volume (Fg/V) at both scales. Cg = 8. Rei?" Him: "Uh… yeah. d. At a certain time during exponential growth.500 L and scale I is 50 L. Just maintain the same Reynolds number. After an initial lag phase. After graduation you accept a position in a biotech company. the absolute growth rate of the cells (rx) will drop. If you are not able to answer one or more parts. b. who received his degree from {insert name of rival school here}.20 mM).500 L process scale. and soon after." You decide to perform an analysis of this strategy. You discuss the problem with another recently hired chemical engineer. State two reasons why scaling up in this manner is a really bad idea. An air-sparged. Calculate the following ratios. 4. The conditions are such that dissolved oxygen will ultimately limit cell growth. the cell density will remain constant. where scale II is 1. stirred-tank bioreactor is operated in batch mode. the gas flow is turned off for 60 seconds. (kLa')II/(kLa')I. 5.50 min-1. but cell death and maintenance under oxygen-limited conditions are negligible." You: "You mean the impeller Reynolds number. NII/NI.c. at least state whether the ratio will be greater than. CL* = 0. Consider a batch culture sparged with air (Fg/V = 0. Power imparted per volume fluid. and the dissolved oxygen concentration (CL) is monitored as a function of time: . or equal to one. 6. True or False (you must justify your answer): When the dissolved oxygen concentration drops below critical levels.0 mM. (P/V)II/(P/V)I. Stirring speed. Discuss two potential drawbacks of the scale-up strategy described in part b. less than. a. where your first assignment is to scale up an aerobic microbial culture from the well-characterized 50 L pilot scale to the 1. the cells in the reactor grow exponentially. Him: "That's easy. c. b. Assuming that the same Fg and V are to be maintained.1 mM in 30 seconds. You decide to maintain the same circulation time (determined by Ni) and kLa' values. Estimate the cell density at which the culture will cease to grow exponentially. Estimate the value of kLa' (s-1). the dissolved oxygen concentration is fit well by CL (mM) = 0. M = 40). a. the dissolved oxygen concentration is CL = 0.2/Ni. is increased by a factor of 3.50 g DCW/L. when the sparging is turned off briefly. after which they stop doing so.1) that yields the same kLa' value? .After the air is turned back on. and by approximately what factor? Why is this not such a good strategy? 7. Estimate the specific growth rate μ (h-1) during this period. Impeller #1 has 1/2 the diameter of the tank (di/dt = 1/2) and a power number of 1 (aerated). where t is time in seconds. Ni.2 mM. At a certain point during exponential phase. b. At the time of this experiment.083t). An agitated batch bioreactor is aerated with O2-containing gas (Cg = 10 mM. The stirring speed.173 . c. From the time when the measurement was taken. By what factor will the kLa' change? Will the maximum oxygen transfer rate change by approximately the same factor? Justify your answer.133exp(-0. the cell density is determined to be 0. The process is to be scaled up by a factor of 10 in terms of volume. c. a. the cells continue to grow exponentially for 4. You wish to choose between two different impellers for this bioreactor. Impeller #2 has 1/3 the diameter of the tank and a power number of 2 (aerated). CL drops to 0. what is the ratio of stirring speeds for the two impellers (Ni. What process variable will need to be adjusted.0.0 hours. and you wish to keep the kLa' and Fg/V constant (as is typically done for mass transfer considerations).App = 10 pM (1. but the cells' oxygen demand is low. the subunits are precoupled. with a measured value of KD. such that the assumption of 1:1 ligand/receptor stoichiometry is reasonable in this case.85 a. The HT-2 cell line is a line of mouse T cells that proliferates in response to the cytokine. After graduation. The concentration of IL-2 required to elicit half-maximal steady-state binding to cells is extremely low. through a multi-subunit receptor expressed on the T cells. estimate the rate of ligand depletion. Comment on whether daily re-feeding is warranted. in (pM IL-2 consumed)/day. typically) are much lower than those typically used in microbial culture. c. which is in charge of process design for large-scale animal cell culture. you go to work for a medium-sized pharmaceutical company in their Cell Culture Manufacturing group. 3. 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). and the medium initially . you discover that the maximum oxygen transfer rate (OTR) for the process scales much differently than for "traditional" fermentation processes. Based on the typical values cited above.Animal Cell Culture/Cell Signaling 1. with a maximum value of 1. Interleukin-2 (IL-2) stimulates T cells during the immune response. is known to be 0. Accordingly.000 molecules/cell. A flask of cells was inoculated at 3x104 cells/mL. ke . You wish to determine how often you will need to change the medium in order to maintain the IL-2 concentration.1 min-1. and so the gas flow rates (Fg/V ~ 0. is oxygen depletion from the bubbles significant? b. We assume here that all internalized IL-2 molecules are degraded.0 x 10-11 M) The steady state number of IL-2/receptor complexes is somewhat lower than for most growth factors. 2. and the endocytic rate constant. Actually. For a cell density of 105 cells/mL (108 cells/L) and an initial IL-2 concentration of 100 pM. with OTRmax scaling as (P/V)0.03 min-1) and kLa' values (~ 20 h-1.25us0. The process is a stirred-tank with air sparging. Evaluate whether or not the strategy described above is sound. interleukin 2 (IL-2). The process volume is to be scaled up by a factor of 100. and it is assumed that all internalized ligand molecules are degraded. Estimate how much time it will take if the centrifuge is operated at 2. Estimate the value of μmax (h-1) under these conditions. steady-state number of IL-2/IL-2 receptor complexes per cell for this cell line.): 5 10 a. A chromatographic separation produces two peaks. Estimate the maximum. 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. estimate the new resolution value if the length of the column is extended to 10 cm. At various times. Rs. Product Recovery 1. Solids are effectively pelleted from a sample in 5 minutes using a centrifuge operated at 1. Calculate the resolution value. For the chromatographic separation described in part a.contained a saturating dose (100 pM) of IL-2. It may also be presumed that cell growth is limited by IL-2 concentration.000 rpm. a. while keeping the same flow rate. 2.1 min-1).): 15 40 Standard deviation (min. . b. an aliquot of the cell suspension was taken out and the cell density was quantified. of these two peaks.000 rpm. the peaks exhibit the following characteristics: Peak 1 Peak 2 Mean residence time (min. b. For a 5 cm long column. and comment on the ability to achieve both high yield and high purity. The spherical beads to be used in the column are impermeable.0 Peak width (min. A chromatographic separation produces two peaks with the following characteristics: tr.0 2. The following results are obtained: Unconjugated Ligand-conjugated Protein X Protein Y Protein X Protein Y Average residence time (min. A contaminating protein. pulses of Protein X or Protein Y are added to the column. One can increase the yield of Protein X by collecting for a longer period of time. An affinity column is used to separate a specific protein called Protein X from spent culture medium. a. meaning that neither Protein X nor Protein Y can penetrate the particles.0 5.0 6. What causes the broadening of the Protein Y peaks? What additional factor(s). avg (min. Protein Y. c. For the column loaded with ligand-conjugated beads. Estimate the % yield of Protein X in the product.) 5.) Peak 1 20 5 Peak 2 45 10 .0 5. b.3. contribute to the broadening of the Protein X peak in the ligand-conjugated column? 4.0 (Note that the peak width w = 4 σt assuming Gaussian peaks).0 12. Protein X is collected between 10 and 15 minutes. is also found in the medium.4 2. In pilot experiments. if any.) 2. 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).) σt (min. List two potential drawbacks associated with this strategy. You wish to decide between keeping just the first fraction (which will have the higher concentration of B. Rs.In an effort to collect the protein in Peak 1. Chromatograms are acquired at two different flow rates: Q = 2 mL/min. An affinity column (100 mL total volume) is used to separate a specific protein called protein B from spent culture medium. suppose that you collect two fractions of the flow-through: from 35 to 40 minutes. is also found in the medium.4 Peak width (min. In the case of Q = 2 mL/min. x plot). or pooling the two fractions together (as if you collected from 35-45 minutes). x plot). a.2 1. 5. protein A. Using quantitative reasoning. A contaminating protein.) 22 37 4. and from 40 to 45 minutes.. Protein A Protein B Protein A Protein B Average residence time (min.6 6. .6 11. b. It is not possible to accurately estimate the product purity without knowing the peak concentrations. Calculate the values of the resolution.4 7. a. CB). for each flow rate. (see above for erf(x) vs.) 4. speculate on whether high purity will be achieved.1 (Note that the peak width w = 4 σt assuming Gaussian peaks). b. Estimate the fractional recovery (yield) of the protein in peak 1. Estimate the following ratios: YieldB(35-45')/YieldB(35-40') CB(35-45')/CB(35-40') (see above for erf(x) vs. Q = 10 mL/min. Comment on the ability to achieve both high yield and high purity in each case. you collect material coming off the column between 15 and 22 minutes.