1. What is Quantitative Metabolism?? Chemical Engineers are used to understanding: Rate of chemical reactions (HOW FAST IS THE DESIRED PRODUCT PRODUCED??) Yields of Chemical Reactions (HOW EFFICIENTLY IS THE DESIRED PRODUCT PRODUCED??) Simply put, Quantitative Metabolism is the extension of these concepts to a biological system

2. What are the main teaching and learning outcomes from these lectures?? 1.Understand the flow of carbon, energy and reducing power within the cell 2.Understand the origin and use of ATP and NAD(H) 3.Understand the stoichiometry of both catabolic and anabolic pathways 4.Understand how to undertake a complete mass and energy balance of a cell including the prediction of cell yield, product yield, heat evolution, oxygen uptake. 5.Be able to calculate specific metabolic rates including the specific rate of substrate uptake, specific rate of product formation, specific growth rate etc

3. Quantitative Metabolism 1

5. What happens to the carbon and Energy Source that are taken Up?

6. Quantitative Metabolism

8. ATP

9. NAD(H)

10. Quantitative Metabolism

13. Niether ADP/ATP nor NAD(H)/NAD + is allowed to accumulate in the cell and so the ADP/ATP and NAD(H )/NAD+ “pool” are very tightly controlled within the cell.

14. “Currency of the Cell” Concentration of ATP in cell = 6 µmol / g Cell Concentration of ADP in cell = 2.5 µmol / g Cell Concentration of NAD(H) in cell = 1.5 µmol / g Cell Concentration of NAD + in cell = 2.5 µmol / g Cell Rate of ATP Consumption = µ/Y ATP = 0.4/10.5 mol / g Cell/h = 38,000 µmol / g Cell / h Rate of NAD(H) Consumption = QO 2 * 2 = 4 mmol/ g Cell/h = 4,000 µmol / g Cell / h If no ATP produced, the ATP within the cell would be used up in (6 µmol / g Cell) /(38,000 µmol / g Cell / h) = 0.56 sec If no NAD(H) produced, the NAD(H) within the cell would be used up in (1.5 µmol / g Cell) /(4,000 µmol / g Cell / h) = 1.35 sec Maximum Rate of ATP Production (Full Respiration)= Q S *4 = 59,200 µmol / g Cell / h Maximum Rate of NAD(H) Production(Full Respiration) = Q S *12 = 177,600 µmol / g Cell / h If no ATP used, the ADP within the cell would be used up in (2.5 µmol / g Cell) /(59,200 µmol / g Cell / h) = 0.15 sec If no NAD(H) used, the NAD + within the cell would be used up in (2.5 µmol / g Cell) /(177,600 µmol / g Cell / h) = 0.05 sec

15. Conclusion The cell VERY TIGHTLY controls the ATP/ADP levels and the NAD(H)/NAD + levels to ensure that the pools of these intermediates keep within very fine tolerances. This is done by elaborate cellular controls which control the rate of formation and the rate of use of these intermediates – broadly by regulating energy substrate uptake (production) and cellular growth (use).

16. “Comparison” with the HKMA Foreign Currency Reserves = 122.3 billion US$ Exports = 628,137 million $HK = $US 80,530 million Imports = 576,328 million $HK =$US 73,888 million IF NO INCOME GENERATED (no exports), the cost of imports would drain the surplus in 122,300/78,888 = 1.65 years

18. Quantitative Metabolism What determines whether a particular reaction is “capable” of generating ATP or NAD(H)???

19. Quantitative Metabolism A  B  C  D  E  F  G  H  G 0 A  G 0 B  G 0 C  G 0 D  G 0 E  G 0 F  G 0 G  H F A  H F B  H F C  H F D  H F E  H F F  H F G  H R A  H R B  H R C  H C D  H R E  H R F  H R G  H C A  H C B  H C C  H C D  H C E  H C F  H C G Some measure of the amount of energy released is necessary

20. Quantitative Metabolism A requisite for ATp formation is that the enrgy released from a reaction is sufficient to “drive” the formation of ATP from ADP Questions: 1.Is this a sufficient requirement? 2.If there is sufficicent energy for 2 or nATP to be formed, will they be formed??

21. Quantitative Metabolism The concept of “Substrate Level Phosphorylation” is important here The formation of ATP at the molecular level within a certain reaction step requires a particular type of enzyme

23. Quantitative Metabolism ATP will ONLY be formed if the appropriate enzyme is present (an enzyme capable of substrate level phosphorylation) and the number of ATP formed is (almost?) always 1 Excess enrgy release is usually lost as HEAT

24. Quantitative Metabolism For NAD(H) production, the only requirement is for an oxidation reaction to occur,releasing one or more H + Questions: 1.Is this a sufficient requirement? 2.If there is sufficicent H + released for 2 or nNAD(H) to be formed, will they be formed??

25. Quantitative Metabolism Yes, this condition is both a requisite and sufficient condition. Since the H + exchange doe not occur via an enzyme similar to SLP, then more than one NAD(H) may be formed. The stoiciometry of this reaction is simply related to how many H + are relased in the coupled reaction. NAD + hydrogenases simply interact with the H + released and each H + released can inteact with a separate hydrogenase. This is unlike a SLP reaction, where both the reactant is bound to the enzyme in conjunction with the ADP form which the ATP is formed. Without an effective NAD + hydrogenase, the pH of the immediate environment of the reaction would fall very rapidly

26. Quantitative Metabolism

39. End Product Formation

40. Quantitative Metabolism

46. Many other electron acceptors may be used by microorganisms, including: Sulfate, Nitrate, Metal Ions etc.These all also use NADH 2- and NAD + as “linked” or “coupled” reactions. For example: SO 4 2- + 8H + + 8e - = S 2- + 4H 2 0 actually represents two reactions: 4NAD(H) + 4H + = 4NAD + + 8H + + 8e - SO 4 2- + 8H + + 8e - = S 2- + 4H 2 0 _______________________________________________ SO 4 2- + 4NAD(H) + 4H + = S 2- + 4NAD + +4H 2 0

47. Nitrification and Denitrification Nitrification is an aerobic process (requiring oxygen). The overall reactions are the following: Nitrification: NH 3 + 1.5O 2 = HNO 2 + H 2 0 HNO2 + 0.5O2 = HNO3 What actually happens in terms of H+ and e- is the following: Nitrification: NH 3 + 2H 2 O = HNO 2 + 6H + + 6e - HNO 2 + H 2 O = HNO 3 + 2H + + 2e - Denitrification: 2HNO 3 + 10H + + 10e - = N 2 + 6H 2 0

48. Hence Nitrification produces H + and e - and Dentrification requires H+ and e-. As usual, these H + and e - come from the reaction: NAD(H) + H + = NAD + + 2H + + 2e - The balanced reactions for nitrification and denitrification (in terms of NAD(H) and NAD + ) then become: Nitrification: NH 3 + 3NAD + + 2H 2 O = HNO 2 + 3NAD(H) + 3H + HNO 2 + NAD + + H 2 O = HNO 3 + NAD(H) + H + Denitrification: 2HNO 3 + 5NAD(H) + 5H + = N 2 + 6H 2 O + 5NAD +

49. In nitrification, oxygen is used to regenerate the NAD(H) formed: NAD(H) + H + = NAD+ + 2H + + 2e - 0.5O 2 + 2e - = O 2- O 2- + 2H + = H 2 O ----------------------------------------------------------- NAD(H) + 0.5 O 2 + H + = NAD + + H 2 O In nitrification, there is a nett use of NAD(H) from the energy generating pathways (using CO2 as a carbon source) and this is provided by the nitrification reaction. In denitrification, a carbon and energy source provides the NAD(H) required to drive the denitrification reaction.

51. Quantitative Metabolism