What is Quantitative Metabolism??

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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

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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

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Quantitative Metabolism 1
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What happens to the carbon and Energy Source that
are taken Up?
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ATP
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NAD(H)
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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.
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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/YATP 0.4/10.5 mol /
  g Cell/h
• 38,000 µmol / g Cell / h
• Rate of NAD(H) Consumption QO2 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) QS4 59,200 µmol / g Cell / h
• Maximum Rate of NAD(H) Production(Full
  Respiration) QS12 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

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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).

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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

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Quantitative Metabolism

• What determines whether a particular reaction is
  capable of generating ATP or NAD(H)???

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• A ? B ? C ? D ? E ? F ? G ? H
• ?G0A ?G0B ?G0C ?G0D ?G0E ?G0F ?G0G
• ?HFA ?HFB ?HFC ?HFD ?HFE ?HFF ?HFG
• ?HRA ?HRB ?HRC ?HCD ?HRE ?HRF ?HRG
• ?HCA ?HCB ?HCC ?HCD ?HCE ?HCF ?HCG
• Some measure of the amount of energy released is
  necessary

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

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• 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

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• 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

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

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• 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

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End Product Formation
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Many other electron acceptors may be used by
microorganisms, including Sulfate, Nitrate,
Metal Ions etc .These all also use NADH2- and
NAD as linked or coupled reactions. For
exampleSO42- 8H 8e- S2-
4H20actually represents two reactions4NAD(H)
4H 4NAD 8H 8e-SO42- 8H 8e-
S2- 4H20____________________________________
___________SO42- 4NAD(H) 4H S2-
4NAD 4H20
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Nitrification and DenitrificationNitrification
is an aerobic process (requiring oxygen). The
overall reactions are the followingNitrificatio
nNH3 1.5O2 HNO2 H20HNO2 0.5O2
HNO3What actually happens in terms of H and 
e- is the followingNitrificationNH3 2H2O
HNO2 6H  6e-HNO2 H2O HNO3  2H 
2e-Denitrification2HNO3 10H 10e-  N2
6H20
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Hence Nitrification produces H and e- and
Dentrification requires H and e-.As usual,
these H and e - come from the reactionNAD(H) 
  H    NAD  2H  2e-The balanced
reactions for nitrification and denitrification
(in terms of NAD(H) and NAD) then
becomeNitrificationNH3 3NAD  2H2O 
HNO2  3NAD(H) 3HHNO2 NAD H2O HNO3
NAD(H) HDenitrification2HNO3  5NAD(H)
5H  N2   6H2O  5NAD
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In nitrification, oxygen is used to regenerate
the NAD(H) formedNAD(H) H NAD 2H
2e-0.5O2 2e-  O2-O2-   2H 
H2O----------------------------------------------
-------------NAD(H) 0.5 O2  H  NAD 
H2OIn 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.
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