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Electron Flow in Biological proceses

Electron Flow in Biological proceses. This topic is covered largely by the computer learning activity - Oxidation states (Oxs). Slides in this section are mostly complimentory

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Electron Flow in Biological proceses

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  1. Electron Flow in Biological proceses This topic is covered largely by the computer learning activity - Oxidation states (Oxs). Slides in this section are mostly complimentory Successful study of this topic enables the student to understand oxidation and reduction half reactions and to stoichiometrically match electron donors to electron acceptors in redox reactions. Example: How much oxygen is needed for the complete oxidation of 1 mole of glucose to CO2?

  2. Electron carriers Why is focusing on Electron Flow Important? • Energy is conserved (ATP) from redox reactions • Redox reactions are characterised by a flow of electrons from e- doner to e- acceptor. • Thus energy metabolism = electron metabolism • The electron flow characterises the bioprocess: • endproducts formed • oxygen requirements • ATP producuced  biomass yield formed • Exception: hydrolysis reactions (e.g. cellulose to sugar)

  3. Overview of Energy Metabolismsimplifying FAD and ATP genration in TCA glucose TCA cycle glucolysis Cell glucose  12 NADH + 2 ATP ATP synthase 3H+  1ATP Keywords to look up: Electron carriers Proton gradient electron motive force Hydrogenation = Reduction Dehydrogenation = Oxidatioin ETC 38 ATP NADH  9 H+

  4. 1. Electron Flow: Which direction? Thermodynamics e- donor + e- acceptor  oxidized donor + reduced acceptor e.g. sugar + oxygen  CO2 H2O for other reactions it is less obvious what is e- donor and acceptor (e.g. sugar  CO2 + ethanol) 2. How powerful is the transfer? Thermodynamics Driving force of the reaction – Is it downhill (spontaneous important to estimate the heat generated and the potential biomass produced • 3. How rapidly is the electron flow? Kinetics • Substrate oxidation rate  Product formation rate  • Growth rate  Productivity (e.g. umax, kS) Interesting demonstration of electron flow Microbial Fuel Cell: Driving force (voltage) * Speed (amps)  Power (watts)

  5. 4. How many electrons? Stoichiometry depends on number of molecules reacting and number of electrons transferred per molecule allows to quantify the substrates and products of the reaction allows to establish the fermentation balance for : electrons carbon oxygen hydrogen charges Initially it requires to learn to assign oxidation states to atoms

  6. Example of the use of electron balance to examine bioprocesses: Case: Aerobic degradation of organic industrial wastewater Claim: 80% of organic waste degraded to CO2. Clean water and CO2 are the only endproducts Evidence: Only 20% of pollutants left in outflow Investigation showed: Aeration capacity (kLa) allowed only 25% degradation of waste Organic Waste CO2 Question raised: Could it be that bacteria used the oxygen in the water molecule to oxidise the organic pollutants? • Clean Water • • • • • • • • • O2

  7. Final result: Bacterial biomass (25%) plus flocculated waste (30%) accumulating in the reactor accounted for the missing organic carbon. Conclusion: Electron balance was essential in developing a clear understanding of the process

  8. Calculation of oxidation state and available electrons Methanol CH3OH 1. Non charged molecule  Sum of oxidation states is zero 2. Oxygen state of ligands: +IV-II = +II 3.  O.S. of C must be -II The number of electrons must be calculated by considering that a carbon of O.S. +IV (CO2) has zero electrons available. Every more negative O.S. carries a corresponding number of electrons (+III has 1 e-, -III has 7 e-). 4.  C in methanol (-II): 6 electrons available

  9. 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

  10. MSE 2011 • 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

  11. MSE 2011 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

  12. MSE 2011 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 = 12.5 mM/h  LUR = 4.17 mM/h

  13. 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

  14. 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

  15. 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.

  16. 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.

  17. 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.

  18. C H O O C O C H H H H C C

  19. C H O O H H H H C H H H O H C H

  20. C H O O

  21. Stoichiometric equations Correct Stoichiometry of biochemical Redox Reactions: How much ethanol acid can be formed from the fermentation of 1 mole of glucose? (CH2O)6  x CH3-CH2OH 1. 24 e-  12 e- (2 ethanol will be produced) (CH2O)6  2CH3-CH2OH (can’t be 3 ethanol !) 2. 6C  4 C (2 bicarbonate are produced) (CH2O)6  2 CH3-CH2OH + 2 HCO3- 3. 6O  8O (2 water are consumed) (CH2O)6 + 2 H2O  2 CH3-CH2OH + 2 HCO3- 4. 16H  14H (2 protons are produced) (CH2O)6 + 2 H2O  2 CH3-CH2OH + 2 HCO3- + 2H+ 5. By balancing hydrogen, the charges should be also balanced  The reaction will decrease the pH (however some acidity will escape as CO2 (as HCO3- + H+  CO2 + H2O)

  22. Background carbonate equlibrium The Carbonate equilibrium: Essential Inorganic Chemistry Background CO2 in water: acidic condtions CO2 + H2O  H2CO3 at neutral pH (taking protons away): H2CO3  HCO3- + H+ at basic pH (taking more protons away): HCO3-  CO3-- + H+ Overall: the production of CO2 has the capacity to supply protons. For neutral pH use HCO3- rather than CO2.

  23. Alternatively the number of electrons available can be determined from the number of reduced bonds (C-C, C-H). Each reduced bond corresponds to 2 electrons available. Because of its high electro negativity oxygen atoms “posses the electron couple bonding to the carbon atom. OH H C H H Use the oxidation state of compounds when the molecule is very large or the structure formula is not known.

  24. 1. Glucose formula: (CH2O)6 2. Oxidation state of H and O atoms: 0 3. O.S. of each C atom: 0 4. Electrons available per C atom: 4 5. Total number of electrons available 24 6. Electrons required to reduce O2 (to H2O): 4 7. The oxidation of one glucose to 6 CO2 requires (24/4) 6 O2 Since the electrons available per molecule is critical to know for fermentation balances organic molecules can be characterised by the number of electrons and carbons. For glucose this is 24 6. The oxidation of glucose to gluconic acid removes 2 electrons. Thus gluconic acid C6H12O7 can be characterised as 22 6.

  25. Gluconic acid, C6H12O7 1. Temporarily assign 4 electrons to each carbon by neglecting all ligands (this would be correct if all bonds were C-C (e.g. or when total O.S. of all ligands were zero) 2. Calculate O.S. of all ligands: -14+12= -2 3. Add number of O.S. of ligands to number of electrons assigned under 1): 24+ (-2)= 22 Succinate, COOH-CH2-CH2-CHOOH, C4H6 O4 1. 4*4 = 16 electr., 2. O.S. ligands: -8+6= -2, 3. Assigned electr. (16)+ O.S. of ligands: (-2) = 14 Fumarate, COOH-CH=CH-COOH, C4H4O4 1. 4*4 = 16 electr., 2. O.S. ligands: -8+4= -4, 3. 16+(-4) = 12

  26. Malate, COOH-CH2OH-CH2OH-COOH, C4H8O6 1. 4*4 = 16 electr., 2. O.S. ligands: -12+8= -4, 3. 16+(-4) = 12 Ethanol, CH3-CH2OH, C2H6O 1. 2*4 = 8 electr., 2. O.S. ligands: -2+6= +4, 3. 8 + 4 = 12

  27. Glucose Gluconate HCO HOCH HOCH HOCH HOCH HOCH H OOCH HOCH HOCH HOCH HOCH HOCH H

  28. Rote learning approach to linking electron numbers (reducing equivalents) with oxidation states of organics H -2 • The electrons in NADH as the most importan electron carrier can also be visualised 0 -1 +1 NADH/NAD+ as electron carrier

  29. How do electrons get from donor to acceptor? By electron carriers Examples: NAHD, CoQ, FADH, Ferredoxin Electron carriers are present in the cell in its reduced and oxidised form. The ratio of reduced to oxidised e- carrier reflects the energy situation (“starving” to “overfed”) Reduced e- carriers are also called reducing equivalents. Definition of reducing equivalents: 1 reducing equivalent = 1 electron or one electron equivalent in form of a hydrogen atom

  30. Significance of reducing equivalents for the microbial cell. Advantage or disadvantage? Reducing equivalents must be produced and consumed during microbial metabolism. Consumption is by using other compounds as electron acceptors. Electron acceptors can be either: providing the conservation of ATP (electron transport phosphorylation, respiration) or just a dumping ground (fermentations using internal electron acceptors

  31. Electron carriers 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

  32. Electron carriers 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 oxidation Electron donor (Reductand) Electron Carrier reduction Electron acceptor (Oxidant) Electron flow (arrows) electron donor to electron acceptor

  33. Electron carriers 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

  34. Electron carriers What is a redox reaction? The most significant biological reactions relevant to bioprocesses are “redox reactions”. Redox reactions are characterised by an electron transfer from an electron donor to an electron acceptor. The electron donor is oxidised by loosing electrons while the electron acceptor is reduced when it receives the electrons. As often together with the electrons also a proton is transferred, biochemists refer to an oxidation as dehydrogenation (loss of e- and H+ meaning loss of H) and to reduction as hydrogenation. Consequently enzyme names can be confusingly called dehydrogenases and hydrogenases instead of oxidases and reductases, respectively. Strictly speaking an oxidation by itself will not occur as it is only an electrochemical half reaction. Half reactions are characterised by either showing electrons as a reactant or a product (e.g. Fe 2+  Fe3+ + e-). The reaction can only exist in the real world if it is coupled with a suitable opposing half reaction: The oxidising half reaction needs to be coupled with a reducing half reaction to become a full redox reaction. How can we decide whether a compound is an electron donor or acceptor? Depending on the half reaction it is coupled with, many compounds can be an electron donor in one moment and an electron acceptor at a different moment. For example pyruvate can accept electrons to be reduced to lactate (lactate dehyrogenase in lactic fermentation) or it can release electrons (and CO2) to be oxidised to acetate. To know whether a compound is likely to accept or give electrons one would need to know the other player in the overall redox reaction. If it is a strong oxidising agent (tending pull electrons away from other compounds) it will turn the first compound into an electron donor and if it is a strong reducing agent it may force electrons onto the first compound turning it into an electron acceptor. It can be exactly prodicted under which conditions a compound will be an acceptor or donor of electrons. This area of chemistry is thermodynamics. The redox potential and the Gibbs Free Energy change (Delta G) are the useful measures to predict what will happen in the reaction. Enzyme names. Enzyme names can be confusing. It is not easy to give enzymes a proper name that describes its metabolic activity. This is because enzymes typically catalyse both the forward and the backward reaction. For example the enzyme that catalyses the reduction of pyruvate to lactate could either be called pyruvate reductase or lactate oxidase. As a proton is transferred together with the electron, it is actually called lactate dehydrogenase instead of pyruvate reductase. By realising the reversibility of enzyme reactions and that hydrogenases are reductases it should not be too difficult to derive the role of certain enzymes from their names. Organic acids dissociation Lactic acid  Lactate- + H+

  35. Electron carriers 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+) • electron 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

  36. Electron carriers 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

  37. 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

  38. 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

  39. Electron carriers 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

  40. Electron carriers 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

  41. 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

  42. Electron carriers 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

  43. Electron carriers 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

  44. Electron carriers Working principle of electron carriers (EC) H -1 • The electrons in NADH as the most importan electron carrier can also be visualised H R +1 H H N H H R -2 H R H H 0 N R NADH/NAD+ as electron carrier

  45. O C O C C C O C C C C O Ubiquinone as electron carrier Oxidised +II 2 electrons Reduced 2 protons OH +II +I C O C C C O C C C C +I OH

  46. H -I C H C C R H C H C +I +N R NADH as electron carrier Oxidised 2 electrons Reduced 1 protons H H -II C H C C R H C H C ±0 N R

  47. Stoichiometric equations Correct Stoichiometry of biochemical Redox Reactions: How much lactic acid can be formed from the fermentation of 1 mole of glucose and what is the effect on the pH? (CH2O)6  CH3-CHOH-COO- 1. 24 e-  12 e- (two lactate will be produced) (CH2O)6  2CH3-CHOH-COO- 2. 6C  6C (no CO2 is produced) (CH2O)6  2 CH3-CHOH-COO- 3. 6O  6O (water is neither consumed or produced) (CH2O)6  2 CH3-CHOH-COO- 4. 12H  10H (2 protons are produced) (CH2O)6  2 CH3-CHOH-COO- + 2H+ 2 protons are produced, hence the reaction will lower the pH.

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