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Steady State Concentration

1. A bioreactor with a k L a of 25 h-1 with active microbes is aerated resulting in a steady oxygen concentration of 1 mg/L. What is the microbial oxygen uptake rate (in mg/L/h) assuming the oxygen saturation concentration is 8 mg/L?.

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Steady State Concentration

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  1. 1. A bioreactor with a kLa of 25 h-1 with active microbes is aerated resulting in a steady oxygen concentration of 1 mg/L. What is the microbial oxygen uptake rate (in mg/L/h) assuming the oxygen saturation concentration is 8 mg/L?

  2. 2. The airflow to a Vinegar producing chemostat running at steady state was interrupted (at 90 sec. below) and oxygen data recorded. a. What is the kLa of the chemostat in h-1 ? b. What was the ethanol (CH3-CH2OH) to acetic acid (CH3-COOH) conversion rate of the process when it was at steady state?

  3. 4. List (in the box next to the molecule) the number of moles of oxygen needed for the complete oxidation to CO2 of the following compounds CH3-CH2-CH2OH   HOOC-COOH   CH3-CO-CH3 5. List (in the box next to the molecule) the number of moles of NAHD that can be generated from the complete oxidation to CO2 of the following compounds: Pentose (CH2O)5   CH3-COOH   H2CO3   6. List (in the box next to the molecule) the number of moles needed for an anaerobic microbe using these substances instead of oxygen as the electron acceptor for the complete oxidation to CO2 of ethanol (CH3-CH2OH): NO3- N2   SO42- H2S  Fe3+ Fe2+

  4. 7. Can microbes use the oxygen atom in the H2O molecule as an electron acceptor? Give reasons for your explanation and an example of the end product that would be formed (in the case you think it is feasible).

  5. 8. A 10L chemostat is operated with a flowrate of 0.6 L/h. An equilibrium is established with a constant oxygen, concentration, pH, biomass (3 g/L) and substrate concentration. What is the specific growth rate of the microbes in the chemostat and what is the biomass productivity R (g/L/h) of the chemostat? u = D = F/ V = 0.6L/h/ 10L = 0.06 h-1 Productivity R g/L/h can be calculated from X * D 3g/L *0.06 h-1 = 0.18 g/L/h

  6. 9. How would you determine the microbial Yield coefficient from a batch culture and a chemostat culture respectively?

  7. 10. Explain the effect of biomass feedback (recycle, retention) on the biomass concentration and productivity R of a chemostat. Use a plot of biomass (X) and productivity (R) versus the dilution rate to illustrate the point.

  8. Effect of biomass feedback (here 3 fold): • Dotted line no feedback: • Washout occuring early • 3-fold Feedback approximately: • 3*X 3*R  1/3* S • allows 1/3 reactor size to do same work • Feedback essential for pollutant removal. Can be used 100-fold  100-fold smaller treatment plant • Note: same assumed feed concentration (SR) SR X Steady State Concentration R S D Dcrit

  9. Effects of growth constants on steady state concentrations of biomass and substrate in a chemostat as a function of dilution rate (x-axis) Effect of ms Effect of decrease ks Effect of increased μmax Effect of increased Y

  10. 11. Sketch below an example graph of the specific growth rate of a microbe (Y-axis) dependent on the limiting substrate concentration (X-axis). Put a scale (numbers and units) on both axes. Point out in your graph (with an arrow) where 3 of the 4 growth constants can be read from and give their values and units as read from your example graph.

  11. µmax (h-1) µ (h-1) substrate limitation kS Substrate (g/L) Substrate limitation of microbial growth The two curves are described by two properties: The maximum specific growth rate obtained with no substrate limitation (umax (h-1)) and the half saturation constant (Michaelis Menten constat), giving the substrate concentratation at which half of the maximum u is reached (ks (g/L)). Growth- Michaelis Menten model

  12. Effect of Maintenance Coefficient (mS) on growth Rate The negative specific growth rate (µ) observed in the absence of substrate (when S = 0) (cells are starving, causing loss of biomass over time) is the decay rate mS*Ymax µ (h-1) 0 S(g/L) - mS*Ymax

  13. Relationship between oxidation state and electron equivalents of carbon atoms • The electron equivalents (EE) on a carbon atom is 4 minus the oxidation state (OS) : • EE = 4-OS • Note: Electron equivalent= Reducing equivalent= (degree of reduction) OS EE Example +4 0 CO2 +3 1 -COOH +2 2 HCOOH, CO, -CO- +1 3 -CHO 0 4 -CHOH- -1 5 -CH2OH -2 6 -CH2-, CH3OH -3 7 -CH3 -4 8 CH4

  14. MSE 2011 1) A bioreactor with a kLa of 20 h-1 with active microbes is aerated resulting in a steady oxygen concentration of 2 mg/L. What is the microbial oxygen uptake rate (in mg/L/h) assuming the oxygen saturation concentration is 8 mg/L? OUR= 20h-1 *(8-2 mg/L) = 120 mg/L/h 2) The airflow to a chemostat running at steady state DO of 5 mg/L (cS was 8 mg/L) was temporarily interrupted. The oxygen concentration decreased steadily by 0.05 mg/L every second. What is the kLa of the chemostat in h-1 ? kLA = 180 mg/L/g / (8-5 mg/L) = 60 h-1 3) What is the maximum possible rate (in mM/h) of lactate (CH3-CHOH-COOH) oxidation to CO2 by an aerobic reactor that is limited by an oxygen supply due to a kLa of 50 h-1 assuming an oxygen saturation concentration of 8 mg/L? Lac = 12 e-  1 Lac reacts with 3 O2 OUR = 50 h-1* 8mg/L = 400 mg/L/h = 25 mM/h  LUR = 4.17 mM/h

  15. MSE 2011 1) A bioreactor with a kLa of 20 h-1 with active microbes is aerated resulting in a steady oxygen concentration of 2 mg/L. What is the microbial oxygen uptake rate (in mg/L/h) assuming the oxygen saturation concentration is 8 mg/L? OUR= 20h-1 *(8-2 mg/L) = 120 mg/L/h 2) The airflow to a chemostat running at steady state DO of 5 mg/L (cS was 8 mg/L) was temporarily interrupted. The oxygen concentration decreased steadily by 0.05 mg/L every second. What is the kLa of the chemostat in h-1 ? kLA = 180 mg/L/g / (8-5 mg/L) = 60 h-1 3) What is the maximum possible rate (in mM/h) of lactate (CH3-CHOH-COOH) oxidation to CO2 by an aerobic reactor that is limited by an oxygen supply due to a kLa of 50 h-1 assuming an oxygen saturation concentration of 8 mg/L? Lac = 12 e-  1 Lac reacts with 3 O2 OUR = 50 h-1* 8mg/L = 400 mg/L/h = 25 mM/h  LUR = 4.17 mM/h

  16. List (in the box next to the molecule) the number of moles of oxygen needed for the complete oxidation to CO2 of the following compounds: CH3-CH2-CH2OH   4.5 HOOC-COOH  0.5 CH3-CO-CH3   4 List the four growth constants with their units. State in one short sentence what this growth constant means by referring to its units. Ymax gX/gS umax gX/L/h / /gX/L = h-1 ms gS/gX/h = h-1 kS = gS/L

  17. How much NADH can be produced from the complete oxidation to CO2 of the following compounds: CH3-CHOH-CH2-CH2OH  11 CHOOH  1 benzoate (aromatic ring with a COOH group attached to one of the carbons  15 Can microbes use the oxygen atom in the H2O molecule as an electron acceptor? Give reasons for your explanation and an example of the end product that would be formed (in the case you think it is feasible). A chemostat is used to produce microbial biomass for the purpose of recombinant protein production. Lactate (CH3-CHOH-COOH) from dairy wastewater is used as the substrate. The yield coefficient of the recombinant strain is 0.3 g of cells per g of lactate degraded. When interrupting the air flow the oxygen concentration decreased as follows (time is time in sec after interruption): 0 sec: 3 mg/L, 2 sec: 2.5 mg/L, 4 sec: 2 mg/L, 8 sec: 1 mg/L, 12 sec 0.2 mg/L. What is the a) lactate oxidation rate, b) the biomass productivity (mg biomass formed/L/h)? OUR = 0.25 mg/L/s = 900 mg/L/h = 28.1 mmol/L/h (MW = 32 mg/mmol)/  LUR 9.38 mmol/L/h 0.3 g X/ g Lac degraded  Needed LUR in mg/L/h  LUR (3*12 + 3* 16 +6= 90mg/mmol)= 844.5 mg/L/h  Productivity = 844.5 * 0.3 = 253.3 mg/L/h

  18. A 20L chemostat is operated with a flowrate of 0.6 L/h. An equilibrium is established with a constant oxygen, concentration, pH, biomass (2g/L) and substrate concentration. What is the specific growth rate of the microbes in the chemostat and what is the biomass productivity R (g/L/h) of the chemostat? •   D= 0.03 h-1  u = 0.03 h-1 • X= 2 g/L  R = 0.06 gX/L/h • In the absence of oxygen, many bacteria can use nitrate (NO3-) as electron acceptor and produce N2 as the endproduct (nitrate respiration or denitrification). What rate of nitrate reduction to N2 would you expect of a reactor that was switched from aerobic (aerated) conditions to nitrate reducing conditions, if the aerobic reactor had an oxygen uptake rate of 80 mg/L/h? •  NO3-  N2 requires 5 e- while O2  H2O requires 4 e- • NUR= 4/5 OUR (molar) • OUR= 80mg/L/h / 32 mg/mmol = 2.5 mmol/L/h  NUR = 2 mmol/L/h • Contrast batch culture against chemostat culture by pointing out advantages and limitations. •  Chem +: higher productivity, easier automation, ideal for study • Chem-: not for secondary metabolites, prone to cont from outside and backmutations • How can you calculate the productivity of a chemostat? Give 3 examples of how the productivity of a chemostat can be approximately doubled by the operator and one statement for each example how this works.

  19. How can you calculate the productivity of a chemostat? Give 3 examples of how the productivity of a chemostat can be approximately doubled by the operator and one statement for each example how this works. R (gX/L/h) = D (h-1) * X (g/L) Can be increased by operator by increasing either D or X D: Double flowrate  X: Double SR X: Retain bacteria by recycle or filter to twice the concentration

  20. Growth- Simplified Scheme of Energy preservation as ATP Biological growth requires ATP as the energy source (energy rich phosphate-phosphate bond). ATP is generated mostly during Respiration (Dissimilation) ATP then drives the biomass synthesis (Assimilation) How is it generated ? How much is generated ?

  21. Energy preservation as ATP • Four steps for aerobic ATP generation from glucose: • Glycolysis : sugar  acetate (C2)) • TCA cycle: acetate  CO2 + 4 NADH • NADH + O2  NAD + proton gradient • Proton gradient runs a nano-scale “turbine” called ATP synthase

  22. Growth- Overview of Energy Metabolism simplifying FAD and ATP genration in TCA CO2 glucose glucolysis 2 acetate TCA cycle Cell 1 ATP  3 H+ 2 NADH 8 NAD+ ATP synthase 8 NADH ETC Overall: 36 ATP (+2) allowing growth O2 2 NADH 1NADH  9 H+

  23. Growth- Simplified Scheme of Energy preservation as ATP Important Quantities: ATP-synthase: 3H+  1 ATP ETC: 1 NADH  3*3 = 9 H+ 2 NADH reduce 1 O2 glycolysis: 1 glucose  12 NADH 1 glucose  12*9 = 108 H+ = 36 ATP + 2 ATP generated from glycolysis via substrate level phosphorylation = 38 ATP 1NADH  3 ATP

  24. Energy Source for Growth Electron flow: • is critical for the understanding of microbial product formation • allows to understand fermentations • the rate of electron flow determines the metabolic activity • Which direction?  Thermodynamics • How powerful ? Thermodynamics • How rapid ?  Kinetics • How many ?  Stoichiometry, mass balance, fermentation balance

  25. Growth- Simplified Scheme of Energy preservation as ATP How does ATP synthase work? A mechanical turbine that generates a energy rich phopspate bond driven by a proton gradient across the cell membrane See animated clip.

  26. Energy Source for Growth • Microbes catalyse redox reactions (electron transfer reactions) • A redox reaction oxidises one compound while reducing another compound • The electron flow represents the energy source for growth • An energy source must have an electron donor and electron acceptor oxidation Electron donor (Reductand) Electron Carrier reduction Electron acceptor (Oxidant) Electron flow (arrows) electron donor to electron acceptor

  27. Energy Source for Growth Electron flow: • Which direction?  Thermodynamics • How powerful ? Thermodynamics • How rapid ?  Kinetics • How many ?  Stoichiometry, mass balance, fermentation balance oxidation Electron donor (Reductand) Electron Carrier reduction Electron acceptor (Oxidant) Electron flow (arrows) electron donor to electron acceptor

  28. Energy Source for Growth • What are electron carriers? • A redox couple that mediates between donor and acceptor • A redox couple consists of the oxidised and the reduced form (e.g. NADH and NAD+) • acts also as reducing equivalents buffer • What are suitable electron donors and acceptors? oxidation Electron donor (Reductand) Electron Carrier reduction Electron acceptor (Oxidant) Electron flow (arrows) electron donor to electron acceptor

  29. Growth- Simplified Scheme of Energy preservation as ATP What do electron carriers look like?

  30. Working principle of electron carriers • What are electron carriers? • A redox couple that mediates between donor and acceptor • A redox couple consists of the oxidised and the reduced form (e.g. NADH and NAD+) • electron buffer • What are suitable electron donors and acceptors? OH O OH O Electron carriers exist as a couple

  31. Working principle of electron carriers • What are electron carriers? • A redox couple that mediates between donor and acceptor • A redox couple consists of the oxidised and the reduced form (e.g. NADH and NAD+) • electron buffer • What are suitable electron donors and acceptors? OH O OH O Electron carriers exist as a couple

  32. Working principle of electron carriers (EC) OH • What is the most important difference between the two forms? • Different number of double bonds • OH instead of =O O OH O Quinone and hydroquinone as central pieces of Ubiquinone

  33. Working principle of electron carriers (EC) OH • Which form carries electrons? • The reduced form! • Which is the reduced form? • The oxidation states will tell! • Which carbon atoms changed their oxidation state? O OH O Quinone and hydroquinone as central pieces of Ubiquinone

  34. Working principle of electron carriers (EC) OH • Which carbon atoms changed their oxidation state? • All carbons that have just one H bonded maintain OS of -1 • The top and bottom C have changed their OS. H H O H H H H OH H H O Quinone and hydroquinone as central pieces of Ubiquinone

  35. Working principle of electron carriers (EC) OH • Which carbon atoms changed their oxidation state? • All carbons that have just one H bonded maintain OS of -1 • The top and bottom C have changed their OS. • The reduced form carries two more electrons than the oxidised form • Where are they? +1 H H O H H +2 +1 H H OH H H +2 O Quinone and hydroquinone as central pieces of Ubiquinone

  36. Working principle of electron carriers (EC) OH • Which carbon atoms changed their oxidation state? • All carbons that have just one H bonded maintain OS of -1 • The top and bottom C have changed their OS. • The reduced form carries two more electrons than the oxidised form • Where are they? H H O H H +1 H H OH H H +2 O Quinone and hydroquinone as central pieces of Ubiquinone

  37. Working principle of electron carriers (EC) OH • How many electrons are carried ? • 2 • What else is carried? • a proton • Together the electron and the proton make one H • The reduced electron carrier can also be called a hydrogen carrier? • Hydrogenation = adding hydrogen or electrons to another compound = reducing the compound H H O H H +1 H H OH H H +2 O Quinone and hydroquinone as central pieces of Ubiquinone

  38. Working principle of electron carriers (EC) OH • What can a reduced EC do? • Does a cell also need oxidised EC? H H O H H +1 H H OH H H +2 O Quinone and hydroquinone as central pieces of Ubiquinone

  39. Working principle of electron carriers (EC) H -1 • The electrons in NADH as the most importanT electron carrier can also be visualised • as N is more electronegative than C it is allocated the electrons of C-N bonds (similar to oxygen) R H +1 H H N H H -2 R H R 0 H H N R NADH/NAD+ as electron carrier

  40. Main advantage of reducing power (NADH) aerobic conditions, NADH =  ATP generation: NADH + H+ 0.5O2 +3 ADP + 3Pi NAD+ +3 ATP +4 H2O Respiration balance: combination of ETC and ATP synthase reaction How useful is NADH without O2 ?

  41. Consequences of O2 depletion on cells Consequences of O2 depletion: • No ATP generation • NAHD accumulates and NAD+ is depleted • TCA cycle (requiring NAD+) can’t run • glucose uptake stops NADH (or NADPH) can also be used for anabolism (assimilation) but in addition to reducing power also ATP is needed for assimilation Without O2 NADH is a problem rather than advantage Anaerobic organisms have developed special metobolic pathways to re-oxidise NADH (fermentations and anerobic respirations)

  42. Energy Metabolism Scheme simplifying FAD and ATP genration in TCA CO2 glucose glucolysis 2 acetate TCA cycle Cell 1 ATP  3 H+ 2 NADH 8 NAD+ ATP synthase 8 NADH ETC Overall: 36 ATP (+2) allowing growth O2 2 NADH 1NADH  9 H+

  43. Electron flow in fermentations. Anaerobic fermentations (strict sense) make use of internal organic electron acceptors . The electron flow in anaerobic fermentations can be easily demonstrated by documenting the changes in carbon numbers and electron numbers. For example glucose (CH2O)6 contains 6 carbons with an oxidation state of zero (4 electrons/carbon). Glucose can be presented as 6 C, 24 e-

  44. Lactic acid fermentation . Anaerobic fermentations (strict sense) make use of internal organic electron acceptors . The electron flow in anaerobic fermentations can be easily demonstrated by documenting the changes in carbon numbers and electron numbers. For example glucose (CH2O)6 contains 6 carbons with an oxidation state of zero (4 electrons/carbon). Glucose can be presented as 6 C, 24 e-

  45. Lactic Fermentation - Electron and carbon flow - 2 2 2 12 12 12 10 24 10 24 10 0 0 0 6 3 6 3 3 3 3 3 ATP ATP LDH LDH lactate = glucose (CH2O)6 = 2 red. equiv. = pyruvate (CH3-CO-COOH) = hydroxy propanoate =lactate (CH3-CHOH-COOH) LDH = Lactate dehydrogenase enzyme

  46. Notes on origin of enzyme namesWith 2 electrons also 2 protons are transferred electron transfer= hydrogen transfer:Remove e-/H2: Dehydrogenation = oxidationAdd e-/H2: Hydrogenation = reduction Pyruvate + 2e-  LactatePossible names for the enzyme catalysing the equilibrium (forward and backward reaction):Lactate dehydrogenaseLactate oxidasePyruvate hydrogenasePyruvate reductase

  47. Quizz: Glucose(6 carbons) is fermented to 2 lactate(CH3-CHOH-COOH) 123If instead ethanol (CH3-CH2OH) 122 is the end product, how many can be formed?Carbon balance would suggest 3 (2 carbons)!Electron balance suggests 2 (12 electrons) Electrons are relevant, not carbon.If electrons are balanced any extra carbon must be in the form of CO2.

  48. Ethanolic Fermentation - Electron and carbon flow - 2 2 0 0 2 10 10 12 12 12 24 10 10 10 24 10 1 0 1 0 0 2 2 2 2 2 3 6 6 2 3 3 glucose ATP ATP PDC PDC EDH EDH = glucose ethanol = 2 red. equiv. = pyruvate Key enzymes: PDC = pyruvate decarboxylase EDH = Ethanol dehydrogenase = acetaldehyde = ethanol

  49. The Entner Doudoroff (KDPG) pathway of ethanolic fermentation 2 2 0 0 0 10 24 10 10 12 12 12 12 12 24 24 10 22 10 10 0 0 1 1 1 2 6 3 3 3 6 2 6 3 2 2 2 3 2 2 Orgainism: Zymonas mobilis = glucose = gluconate = GAP = pyruvate ATP = CO2. = acetaldehyde = ethanol

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