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2. Stoichiometric Aspects of Metabolism

2. Stoichiometric Aspects of Metabolism. Hans V. Westerhoff Frank Bruggeman Thanks to Ferda Mavituna Vangelis Simeonidis. AIMS. Introduction to elemental balances for biological reactions Introduction to the concept and use of degree of reduction. Lecture contents.

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2. Stoichiometric Aspects of Metabolism

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  1. 2. Stoichiometric Aspects of Metabolism Hans V. Westerhoff Frank Bruggeman Thanks to Ferda Mavituna Vangelis Simeonidis

  2. AIMS Introduction to elemental balances for biological reactions Introduction to the concept and use of degree of reduction

  3. Lecture contents 2.1 Cell stoichiometry 2.2 Elemental balances 2.3 Degree of reduction, K

  4. 2.1 Cell stoichiometry Protein, RNA, DNA, Lipids, Lipopolysacchardes, Peptidoglycan, and Glycogen. • While cell composition may vary with cell‑type and physiological/environmental conditions, a typical cell can be assumed to contain:

  5. E. coli • 70% water • 15% protein • 7% nucleic acids • 3% polysaccharides • 3% lipids • 1% inorganic ions • 0.2% metabolites

  6. Macromolecular composition of a typical cell

  7. 2.1.1 Proteins • They are the most abundant organic molecules within the cell. • All proteins contain C (50%), H (7%), O (23%) and N (16%). They also contain sulphur (up to 3%) for formation of S‑S bonds. • Molecular weights range from 6000 to over 1 million.

  8. Proteins may serve a number of functions: • enzymes (biological catalysts) • regulatory proteins (e.g. insulin) • transport proteins (e.g. haemoglobin) • protective proteins in blood (e.g. antibodies) • toxins (e.g. proteins from Clostridium botulinum) • storage proteins (e.g. casein) • contractile proteins (e.g. flagella) • structural proteins (e.g. collagen)

  9. 2.1.2 RNA/DNA • Biological information is stored in DNA (MW: 2 x l09) and RNA (MW: 2.3 x 104 to 1.1 x 106). • The various RNAs which participate in normal cell function serve the purpose of reading and implementing the genetic instructions of DNA. Messenger RNAmolecules carry messages from DNA to other parts of the cell. These messages are read in the ribosome with the help of ribosomal RNA. Finally transfer RNA assists in the translation of the genetic code at the ribosome.

  10. 2.1.3 Other macromolecular components of the cell • The relative insolubility of lipids in water leads to their presence predominantly in the non‑aqueous biological phases such as the plasma and organelle membranes. • Fats serve as polymeric biological fuel storage. In addition, lipids constitute portions of more complex molecules, such as lipopolysaccharides.

  11. Lipopolysaccharides and peptidoglycans participate in the formation of the cell surface (membranes, envelopes) and are responsible for the cells' tendency to adhere to each other or to walls of reactors, pipes and separators. • They also dictate the cells' resistance to disruption by physical, enzymatic and chemical methods. • The number of building blocks necessary for cellular synthesis varies between 75 and 100 and these are synthesized from 12 precursor metabolites.

  12. 2.2 Elemental balances • Assuming that biomass consists of certain types of macromolecules (e.g. protein, RNA), it is possible to calculate an average elemental composition for biomass from the average content of the individual building blocks. • The following are typical values:

  13. Elemental composition of various cell components • Protein C H 1.58 O 0.31 N 0.27 S 0.004 • DNA C H 1.15 O 0.62 N 0.39 P 0.10 • RNA C H 1.23 O 0.75 N 0.38 P 0.11 • Carbohydrates C H 1.67 O 0.83 • Phospholipids C H 1.91 O 0.23N 0.02 P 0.02 • Neutral Fat C H 1.84 O 0.12 • BIOMASS C H 1.81 O 0.52 N 0.21

  14. Balance equations v4 Y P1 v2 v1 S X 2x v3 P2

  15. Balance equation • For every chemical compound • For every element • The balance must be closed at all times: • For energy • For redox (electrons)

  16. Example: SCP production • Write down a stoichiometric equation describing SCP production from methane. You may assume that the only metabolic products are carbon dioxide and water, and the nitrogen source is ammonia.

  17. Example: SCP production Single Cell Protein or SCP refers to proteinaceous materials which are dried cells of micro‑organisms. Example species which have been cultivated for use in animal or human foods include algae, actinomycetes, bacteria, yeasts, molds and higher fungi. While human consumption of microbial protein is ancient in origin, more recent food products involve microbial growth in aerated bioreactors using substrates such as natural gas and paraffins.

  18. Example: SCP production b. How many independent experimental measurements would you need in order to fully describe the system? c. Assume that the oxygen consumption is 1.35 mol oxygen per mol methane. Calculate the maximum SCP yield in g SCP /g methane.

  19. Solution: SCP production • The following reaction scheme can be assumed: CH4 +a O2 + b NH3 c C H 1.81 O 0.52N 0.21 + d CO 2+ e H2O

  20. CH4 +a O2 + b NH3c C H 1.81 O 0.52N 0.21 + d CO 2+ e H2O • The following balances can be written: • Carbon: 1 = c + d • Hydrogen: 4 + 3b = 1.81 c + 2e • Nitrogen: b = 0.21c • Oxygen: 2a = 0.52c + 2d + e

  21. Degrees of freedom • Number of unknowns: 5 • Number of equations: 4 • Degrees of freedom: 1 • One equation is still needed to fully describe the system. In this particular case, the missing equation is: experimental data:a = 1.35 (mol oxygen used per mol methane used)

  22. Solution of elemental balances • Solving the above equations gives: a = 1.35 (Given) b = 0.13 c = 0.63 d = 0.37 e = 1.63

  23. 2.3 Degree of reduction, K • Any solution to the set of N elemental balances is also a solution to a linear combination of these N elemental balance equations. • This fact is used to derive a single equation which may be more convenient to use in the calculation of stoichiometric coefficients (in elemental balances).

  24. Degree of reduction, K • This equation is thedegree of reduction balancewhich simply states thatthe sum of the degrees of reduction of reactants of a reaction is equal to the sum of the degrees of reduction of the products.

  25. 2.3.1 Generalised Degree of Reduction • The generalised degree of reduction for a compound is the number of electrons available for transfer to oxygen on combustion of the compound.

  26. Generalised Degree of Reduction • k is the number of electrons available for transfer to oxygen on combustion of compound i. • k does not necessarily have a physical meaning.

  27. Generalised Degree of Reduction of the Carbon • Multiplication factors account for the valencies of individual elements as follows. They correspond to the number of electrons on the atoms: C = +4 (+2 O2- 4e -+ CO2) H = +1 ( H+ +1e -) O = - 2 (+2e- O2-) Now attribute these electrons to the carbon atoms

  28. Generalised Degree of Reduction; carbohydrate • k normalised to one carbon equivalent. • e.g. C6H12O6 = CH2O and = 4 + 2 – 2 = 4

  29. Calculating degrees of reduction: κ = number of electrons on carbon + 4 • Methane: CH4; 4H+ =4+; C=4+: κ=8 • Methanol: CH3OH 4H+ =4+ ; O2-=2-; C=4+: κ=6 • Methanal (Formaldehyde): CH2O 2H+ =2+; O2-=2-; C=4+: κ=4 • Methylic acid (formic acid): CO2H2 2H+ =2+, 2O2-=4-; C=4+: κ=2 • Carbon dioxyde: CO2, 2O2-=4-; C=4+: κ=0 • Ethane: C2H6; 6H+ =6+: 6/2=3+; C=4+: κ=7 • Ethene: C2H4; 4H+ =4+: 4/2=2+; C=4+: κ=6 • Ethyne: C2H2; 2H+ =2+: 2/2=1+; C=4+: κ=5 • Ethanol: C2H5OH; 6H+ =6+, O2-=2-: 4+/2=2+ C=4+; κ=6 • Acetaldehyde: C2H4O; 4H+ =4+, O2-=2-: 2+/2=1+; C=4+: κ=5 • Acetate: C2H3OOH; 4H+ =4+, 2O2-=4-: 0/2=0; C=4+: κ=4 • Glycerol: C2H6O2 ; 6H+ =6+, 2O2-=4-: 2+/2=1+; C=4+: κ=5

  30. Consider aerobic growth of a microorganism Biomass: CHaxObxNcxSdxPex Substrates: • carbon source, CHasObs • nitrogen source, CfNHaNObNNcN • sulphate, H2SO4 • and phosphate, H3PO4

  31. Other products of aerobic metabolism • In addition to biomass, carbon dioxide, and water; • a metabolite product CHapObpNcp is produced.

  32. This aerobic growth is represented by the following stoichiometric equation: CHasObs + aN CfNHaNObNNcN + ao O2 + asH2SO4 + apH3PO4 g CHaxObxNcxSdxPex + bp CHapObpNcp + bc CO2 + bw H2O

  33. Convention for stoichiometric coefficients

  34. Convention for elemental subscripts

  35. The same reaction can also be written as g CHaxObxNcxSdxPex + bp CHapObpNcp + bc CO2 + bw H2O – CHasObs – aN CfNHaNObNNcN – ao O2 –as H2SO4 – ap H3PO4=0

  36. We forget about N, S and P for the moment g CHaxObx + bp CHapObp+ bc CO2 + bw H2O – CHasObs – ao O2 =0

  37. Aerobic growth equation • There are five stoichiometric coefficients and three elemental balances (C, H, O; or C, O, and electrons). • Two coefficients are to be determined experimentally • These have to do with two degree of freedom: • How much product per biomass • How much of the substrates is combusted to deliver the free energy for growth and production

  38. The generalised degree of reduction for biomass is given by (g CHaxObx)

  39. Similarly, the degree of reduction for the carbon source ( CHasObs):

  40. and the degree of reduction for the metabolic product (bp CHapObp):

  41. Degree of reduction (number of electrons) must be conserved CHasObs + ao O2 g CHaxObx + bp CHapObp+ bc CO2 + bw H2O

  42. Advantage: no need to keep track of carbon dioxide, water, oxygen as O2- • for molecular oxygen: • for water: • for nitrogen source: • for carbon dioxide:

  43. Red blood cells and anaerobic yeast • Glucose as substrate • Anaerobic • No growth • Hence product must have same degree of reduction as substrate: lactate: CH3CHOHCOOH=C3O3H6 (κ=4), • Or (yeast): 2/3alcohol C2OH6 (κ=4+2=6), plus 1/3CO2 (κ=0): 4 CH2O + 0 O2 0 CHaxObx + bp CHapObp+ bc CO2 + bw H2O

  44. Advantage: no need to keep track of Oxygen, water CHasObs + ao O2 g CHaxObx + bp CHapObp+ bc CO2 + bw H2O 5 stoichiometric coefficients; 2 additional element balance equations Hence 2 degrees of freedom

  45. How about with nitrogen? Has various redox states

  46. Generalised Degree of Reduction of the Carbon; the effect of nitrogen • Multiplication factors account for the valencies of individual elements as follows: C=4 (CO2); H = 1 (H+) O = - 2 (-›O2-) N = ‑3 (NH3) or N = 0 (N2) or N = 5 (HNO3)

  47. Convention for stoichiometric coefficients • The stoichiometric coefficient for the main carbon source is taken to be one (1). • The nitrogen source is written for fN atoms of carbon. If NH3 is the nitrogen source, fN=0.

  48. Full stoichiometric equation CHasObs + aN CfNHaNObNNcN + ao O2 + asH2SO4 + apH3PO4 g CHaxObxNcxSdxPex + bp CHapObpNcp + bc CO2 + bw H2O

  49. The same reaction can also be written as g CHaxObxNcxSdxPex + bp CHapObpNcp + bc CO2 + bw H2O – CHasObs – aN CfNHaNObNNcN – ao O2 –as H2SO4 – ap H3PO4=0

  50. Neglect S and P g CHaxObxNcx + bp CHapObpNcp + bc CO2 + bw H2O – CHasObs – aN CfNHaNObNNcN – ao O2 =0

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