Re-examination Cultivation Technology BB1300 & BB1120 Wednesday, 12 April 2017, 08:00-13:00.

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1 Re-examination Cultivation Technology BB1300 & BB1120 Wednesday, 12 April 2017, 08:00-13:00. Start each main question (1-4) on a separate paper. Write your name (clearly) on each page, number each question and sort the answers in numerical order. The examination maximum is 100 points. A total of 60 points is required to pass this examination. Each successfully completed KS counts for 6 points towards E (not beyond). - Your answers have to be written in English. - Use (correct) units and present full calculations for full credit. - The use of a calculator (empty memory) is allowed. Question 1 (a-e; max. 30 points) a. (max. 6 points) Give the name, correct units and a (brief!) description of the following two dimensionless numbers: N RE and N P. N Re The Reynolds number. Dimensionless numbers by definition have no units. This number is the ratio between the momentum forces ( characteristic length) and the viscous forces that are affecting a fluid. This number can be used to predict laminar or turbulent flow patterns. Not required: N Re = (m v L)/η. N P The Power number. Dimensionless numbers by definition have no units. This number is the ratio between the impeller drag force and the inertial forces affecting the system. If the Power number is known this can for instance be used to estimate the (ungassed) power requirements for mixing in a bioreactor. Not required: N P = P O /(N 3 D i5 ρ). b. (max. 6 points) Give a brief description of the historical development of Biotechnology based on the four different eras (max 10 lines per era). Include period, state of technology and/or technological developments, and give an example of a product that is typical for the developments in each era. Ancient biotechnology (1 st era in the book): Spontaneous fermentation of food- and beverages. Important for flavour, nutritional value (vitamins) and preservation. From ancient history until the discoveries of Pasteur in the 19 th century. E.g. beer. After the discovery of microorganisms as the cause for fermentation, natural microorganisms started to be used for the production of solvents, acids etc. (2 nd era in the book). 19 th century and early 20 th century. E.g. citric acid or butanol. The invention of the microscope by van Leeuwenhoek (or Hooke if you so prefer ) was important. However, while I am proud of this early scientist from Delft, Van Leeuwenhoek didn t really know yet what he was looking at yet. The actual discovery of Pasteur that it involved living microorganism and the development of pure cultures based on Koch s pioneering work was required to really use microorganisms efficiently.

2 Classical strain improvement by mutagenesis and selection and process development (3 rd era in the book) typical for the mid 20 th century. E.g. penicillin. Late 20 th and early 21 st century biotechnology ( 4 th era in book ): Genetic engineering (and synthetic biology) of microbial cell factories for production of pharmaceuticals, fuels, chemicals. E.g. insulin, human growth hormone. c. (max. 6 points) This question concerns wave-bag bioreactors: What is a wave-bag bioreactor? Give a schematic representation of a wave bag bioreactor. Give a short description of the main use for wave-bag bioreactors. Give 2 benefits of this type of bioreactor. A wave-bag bioreactor is essentially an inflated plastic bag containing liquid media (often also containing gas-in and -outlets and sensors) that is rocked in see-saw motion to create waves that simultaneously mix and aerate. These bags are disposable, which makes this relatively expensive. Wave bag are mainly used for cultivation of mammalian cells, such as for instance for pharmaceutical production. In some cases for production or alternatively to prepare inocula. The simplicity and single-use aspect of the wave bag decreases the risk for contamination, which is very important for slow growing mammalian cells on very rich media. Additionally, it enables standardization of operation which is often required for government approval of pharmaceutical production (such as the U.S. FDA). Their lack of cell wall makes mammalian cells extremely fragile. Shear forces from mixing, such as from Rushton impellers, could easily destroy the cells. The gentle nature of the mixing in a wave-bag bioreactor avoids this. d. (max. 6 points) Operation of cultivation processes in bioreactors requires the measurement and control of various parameters regardless of whether it concerns industrial- or lab fermentations. You can take the E. coli fed-batch cultivation performed during the lab-part of this course as an example. Give 6 continuously measured and controlled parameters and describe for each how they are measured and controlled. Examples of parameters that are generally feedback-controlled to a set-point initially set by the process engineer or operator are: Temperature, measured online with a thermometer (often a resistance thermometer) and controlled by computer-regulated flow of cooling water

3 (industrial scale; cooling coils or jacket) or heating (lab scale; jacket or heating finger ). At laboratory scale bioreactors can also be heated by heating blankets. ph, measured online with a ph electrode, controlled by computer-regulated addition of base and/or acid. Choice of ph can depend on the cell factory used, the type of process, chemical properties of substrate- or product. Liquid inflow and/or outflow, measured and controlled with mass flow meters and valves. Control depends on type of fermentation (batch, fedbatch, continuous) and other required additions or outflows. Gas flow, often simultaneously measured and controlled with a mass flow controller, but others options are available. Set-point are often chosen for optimal aeration. Liquid level inside bioreactor, e.g. conductivity sensor, acoustic level sensor or balance depending on scale. Depending on the fermentation type the level of the liquid is only measured (batch) or both measured and controlled (continuous or optionally fed-batch) through control of liquid in- and outflow. Level measurements can also be important in view of foam control. RPM of the stirrer, various measurement options available such as for instance stroboscopic measurement. This in turn affects the impeller rotational speed and mixing. Controlled via the power input or gear box of the stirrer engine. Important for proper mixing as well as gas-transfer properties of the bioreactor. Other equally correct answers are possible. e. (max. 6 points) This question concerns the concept of viscosity: Give a definition of viscosity. Explain the concept of Newtonian- and non-newtonian fluids and give an example of each of these two categories. Why is viscosity important for cultivation technology? Viscosity is a property of the fluid that dictates how fast, i.e. at which shear rate, a liquid will move in relation to the applied shear force. The dynamic vicosity (η; Pa*s), which includes the mass, is most commonly used. Newtonian fluids have a linear correlation between the shear rate and the shear force, which means that their dynamic viscosity is constant with shear rate. Water is an example of this. For non-newtonian fluids the correlation between the shear rate and shear force is not linear, meaning their dynamic viscosity is not constant. An example of this would be a high concentration mixture of starch and water. Viscosity is a crucial parameter to determine the power input required for the desired degree of mixing, but also for aeration with influence on bubble size, flow patterns etc. The type of fluid (Newtonian or non-newtonian fluids) has a huge input on the type of bioreactor and impellers to use in the design phase of a new process.

4 Question 2 (a-d; max. 28 points) a. (max. 7 points) Continuous cultivation in CSTR-bioreactors is often used for academic research of microorganisms under highly defined conditions. Derive the mass balances of biomass (X) and substrate (S) for this type of cultivation using biomass specific consumption/production rates. Do not make assumptions, with the exception of 100% viability. You do not have to consider product formation. Motivate all steps and explain/define parameters. Mass balance: Accumulation = In - Out + Production Consumption (+/- Transfer) 100% viability means no consumption. No transfer of biomass to other phases ( I don t even consider that an assumption). Biomass mass balance: d(c x V L )/dt = F in C x,in - F out C x,out + µ C x V L C x = biomass concentration in the bioreactor. V L = volume of the liquid phase. d(c x V L )/dt = change of total biomass amount in bioreactor in time (accumulation). F in = Inflow of liquid. F out = Outflow of liquid (common to be equal to zero). C x,in = biomass concentration in the ingoing liquid flow. C x,out = biomass concentration in the outgoing liquid flow. µ = specific growth rate. Substrate mass balance: d(c s V L )/dt = F in C s,in - F out C s,out - q s C x V L C s = substrate concentration in the bioreactor. C s,in = substrate concentration in the ingoing liquid flow. C s,out = substrate concentration in the outgoing liquid flow. See above for others. b. (max. 7 points) Two questions concerning the concept of steady state: Explain the concept of steady state. If you make a change in the ingoing feed in a chemostat how long do you have to wait until a new steady state? Motivate your answer mathematically. Steady state is a concept that indicates that no changes occur in time. In the context of mass balances over a bioreactor that means d(c i V L )/dt = 0 and that all the positive- and negative terms on the right side of the mass balance cancel each other out. Interestingly, the properties of a microbial cultivation in a CSTR-based continuous culture automatically result in establishment of such a steady state. When, after initial changes, the steady state is approached close enough (this process is asymptotic) the system remains stable. Under these conditions the specific growth

5 rate of the microorganisms is (assuming ideal mixing and constant volume) identical to the dilution rate (F out /V L ). This greatly facilitates study of the physiology of different microorganisms under highly defined conditions. With a computer you can of course show numerically at what time the new steady state is approached close enough. This is often defined as within 2% of the asymptotic value. That is obviously not available at the exam or when a quick decision has to be made in for instance a lab. In general the rule of thumb for recovery of continuous systems from disturbances is: response = e -φ. Where in this context φ would be number of times the volume of the bioreactor has been changed. This tells us that after approximately 4 volume changes the deviation from the eventual steady state is <2%. The inverse of the dilution rate is the mean residence time (mean τ i ) of the liquid in the bioreactor (i.e. the time it takes for one volume change). This is (V L /F) and has the units of time. The time you have to wait can be approximated by 4 V L /F. c. (max. 7 points) What is a wash-out curve with respect to continuous cultivation and describe a common cause for this phenomenon? Give a graphical representation of curves for: C x (or X),C s (or S), the specific growth rate (µ), the specific substrate uptake rate (q s ) and the volumetric biomass productivity (r x ). Motivate your answers. Dear students, your teacher failed to notice that in the field of cultivation technology different people use the term wash-out curve for two different things (see below). To make it more confusing both of those things are highly relevant and I unknowingly again asked for the first of these two options (without using the name wash-out curve ) in question 4c. I should have noticed that earlier and for that I apologise. Sorry! Now to business. My mistake can of course never go at the expense of your grades. I kept the highest of your scores for either 2c or 4c, discarded the lowest and normalized everything back to 100 by dividing by This positively influenced everyone s point totals. This is the reason why your total score can be higher than the sum of your points. Option 1 (See top panel below for plot; this is what the book calls a wash-out curve ). Scientists can use a plot of these 5 parameters from steady-state chemostat cultures against the dilution rate to decide what would for instance be an optimal dilution rate for a process. Such a plot could be derived through modelling (see session Prof. Enfors), or through manual calculations or in extreme cases even through real measurements (each set of points for one dilution rate would be an independent steady state!!!). The cause for this would be the desire of scientists to have such a plot

6 Motivation of figure: This is a combination of the mass balances for biomass, substrate with the Pirt and Monod equations. At steady state µ = D and hence the black line. Above D crit this becomes funny since in the actual mathematical steady state there will be no cells (C x = 0). The biomass concentration follows from the combination of the equations where the decrease on the left is mostly caused by maintenance (Pirt) and the decrease on the right by affinity (Monod). The biomass mass balance gives r x. q s comes from the Pirt equation. C s,in follow from Monod (via Pirt to get q s ). Option 2 (See bottom panel below for plot; this is the wash-out curve I was thinking about when I made the exam). We start with a normal steady-state chemostat culture of a microorganism (left of the dashed line). At some point something happens that causes problems for the microorganisms. This could for instance be a change to a carbon source it cannot use, a ph or temperature that is too high or low for growth, or as illustrated here a change in dilution rate. In this case the dilution rate is changes from a value <µ max to a value > µ max. The biomass mass balance tells us that the new steady state will then have a C s of zero. This type of wash-out curve describes the time dependent transition from the original steady state to the new steady state. On the right side of the dashed lines the microorganisms are trying to grow and consume substrate (hence the flat lines for µ and q s ) as fast as possible, but they simply cannot keep up. The biomass mass balance results in the profile for c x. r x (µ c x ) would initially go up (since µ goes up) but since c x decreases it will keep decreasing in time until it also reaches zero. C s follows from the substrate mass balance.

7 Figure top panel: Each asked parameters plotted on the y-axis as a function of the dilution rate (x-axis).this is the wash-out curve as defined in the course book (page 129). For each point on the x-axis, all 5 values will have to be determined (or predicted) for separate steady states. Figure bottom panel: Wash-out curve describing transition from a dilution rate < µ max, to a dilution rate > µ max. This is a plot describing what happens in time before and after that switch (dashed vertical line). This is what Ton really wanted to ask.

8 d. (max. 7 points) Derive an equation to describe exponential growth of microorganisms in a batch process. Describe the parameters used in this equation. Mass balance on biomass without in-out flow: d(c x V L )/dt = + µ C x V L C x = biomass concentration V L = Volume µ = specific growth rate t = time C x V L = M x = total amount of biomass in system Integration from t=0 to t=t with M x (0) = M x,0 and M x (t) = M x gives: ln(m x ) - ln(m x,0 ) = ln(m x /M x,0 ) = µ t M x = M x,0 e (µ t) Assuming constant volume the following is also correct C x = C x,0 e (µ t)

9 Question 3 (a-c; max. 21 points) The fungus Penicillium chrysogenum is grown in a bioreactor which is known to have both an ideally mixed gas- as well as liquid phase. The liquid volume in the bioreactor is 5 m 3 and the pressure in the gas phase is 2 bar. Under the conditions prevailing in the bioreactor the Henry-constant for oxygen is 1.37 mol O 2 per m 3 liquid per bar. Measurements with an oxygen electrode show that the dissolved oxygen concentration in the liquid phase (C O2 ) is equal to 0.20 mol m -3. The in-going gas stream is 6250 mol h -1 and contains 21% oxygen. The out-going gas stream is 6640 mol h -1 and contains 15% oxygen. The characteristic time of the gas phase in the bioreactor is much smaller than the characteristic time of the liquid phase. The influence of the in- and out-going liquid streams on the overall oxygen balance is negligible. a. (max. 7 points) Calculate the equilibrium concentration of oxygen in the liquid phase (C O2 *; mol m -3 ) under the conditions prevailing in the bioreactor. C O2 * = α 0 P y O2 The y O2 in the bioreactor is (ideal mixing) equal to y O2,out = C O2 * = = mol m -3. b. (max. 7 points) Calculate the overall oxygen consumption rate (R O2 ) for the bioreactor in mol h -1. O 2 balance of gas phase: dn O2 /dt = F N,in y O2,in - F N,out y O2,out T N,O2 Pseudo-steady state 0 = T N,02 T N,O2 = mol h -1 O 2 balance on liquid phase: dm O2 /dt = F in c O2,in F out c O2,out R O2 + T N,O2 Pseudo-steady state & negligible solubility (for the balance) R O2 = T N,O2 = mol h -1. c. (max. 7 points) Calculate the K L A of the bioreactor in m 3 h -1. T N,O2 = K L A (C O2 * - C O2 ) = K L A ( ) K L A = 1500 m 3 h -1

10 Question 4. (a-c; max. 21 points) The anaerobic fungus Amerigo multicoloria derives free energy for growth from the oxidation of glucose (C 6 H 12 O 6 ) to carbon dioxide (CO 2 ). A. multicoloria uses nitrate (NO 3 - ) as the electron acceptor and glucose as the sole carbon source. Growth of A. multicoloria in anaerobic, glucose-limited chemostat cultures has been extensively studied and can be described with the following parameters under the assumption of a growth-rate independent maintenance requirement and Monod-kinetics for the specific substrate consumption rate (q s ) of the growth limiting nutrient: K s = 0.50 g S /L q s max = 0.9 g S /g x /h q m = 0.06 g S /g x /h Y em = 0.30 g x /g S Researchers perform glucose-limited, anaerobic chemostat cultures of A. multicoloria with a subtrate concentration (C S,in or S in ) of 12 g s /L in the ingoing medium. All chemostats are assumed to be ideally mixed, with constant volume and equal in- and out flows. a. (max. 7 points) Calculate the maximum specific growth rate (μ max ) and the highest dilution rate (D highest ) at which a steady state can be obtained under the given conditions. At μ max we know that q s = q s max = q m + μ max / Y em. This gives: μ max = (q s max q m ) Y em μ max = ( ) * 0.30 = h -1 The ingoing substrate concentration (C s,in ) fo the described chemostat is 12 g/l. That is simultaneously the highest concentration possible in the bioreactor (C s ). According to Monod kinetics we can calculate the biomass specific substrate consumption rate at that concentration: q s = q s max * C s / (C s + K s ) = (0.9 * 12) / ( ) = g s /g x /h. The highest possible dilution rate in this experimental system, and thereby the highest possible growth rate, at which a steady state can be obtained is therefore: D = μ = (q s q m ) Y em = ( ) * 0.30 = h -1 Note: at this dilution rate the steady-state biomass concentration will be infinitely small. This specific steady state is experimentally virtually unachievable since small changes in controlled parameter would result in very large changes in all resulting parameters. b. (max. 7 points) Calculate the biomass concentration (C x or X) and the biomass yield (Y x/s ) in steady state chemostat cultures at a dilution rate of 0.05 h -1 under the above described conditions.

11 At a growth rate of 0.05 h -1 we know that: q s = q m + μ/ Y em = /0.30 = g s /g x /h With q s and the growth rate known we can calculate the yield: Y x/s = μ / q s = 0.05 / = g x /g S With q s and Monod we can also calculate the residual substrate concentration that is required to sustain that growth rate: q s = q s max * C s / (C s + K s ) C s = K s * q s / (q s max -q s ) = 0.50 * / ( ) = g/l From the steady-state mass balance on substrate with constant volume we can derive that: 0 = F in c s,in - F out c s - q s c x V L, which under the assumptions of constant volume (part of steady state definition), F in = F out, ideal mixing and via the definition of D and Y x/s results in: C x = (C s,in -C s )*Y x/s = ( )* = 2.61 g/l (Simply saying C s = 0 is wrong, a motivated decision to neglect the influence of C s on the answers is inaccurate (would give overestimation of C x at 2.65 g/l) but would be awarded some points). c. (max. 7 points) Give a graphical representation of the substrate concentration (C s or S), the biomass concentration (C x or X) and the biomass yield on glucose (Y x/s ) (all three on the y-axis) as a function of the dilution rate (D; x-axis) for steady state chemostat cultures of A. multicoloria as described above. Where possible, indicate relevant constants/parameters from the text above in the graph. In reality, each of these dilution rates would represent individually performed experiments.

12 In the figure you should indicate D highest (the highest μ which should has values in the figure) en Y em (The Y x/s on the secondary y-as that is approached asymptotically by the red line). On the primary axis, K S can be indicated (for C s ). Note: The (potential) overlap with question 2C was unintended (Sorry!).