Quantitative Metabolism 1. Quantitative Metabolism

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<ul><li> Slide 1 </li> <li> Quantitative Metabolism 1 </li> <li> Slide 2 </li> <li> Slide 3 </li> <li> Quantitative Metabolism </li> <li> Slide 4 </li> <li> Slide 5 </li> <li> Slide 6 </li> <li> Slide 7 </li> <li> ATP </li> <li> Slide 8 </li> <li> NAD(H) </li> <li> Slide 9 </li> <li> Quantitative Metabolism </li> <li> Slide 10 </li> <li> Slide 11 </li> <li> Slide 12 </li> <li> 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. </li> <li> Slide 13 </li> <li> Currency of the Cell Concentration of ATP in cell = 6 mol / g Cell Concentration of ADP in cell = 2.5 mol / g Cell Concentration of NAD(H) in cell = 1.5 mol / g Cell Concentration of NAD + in cell = 2.5 mol / g Cell Rate of ATP Consumption = /Y ATP = 0.4/10.5 mol / g Cell/h = 38,000 mol / g Cell / h Rate of NAD(H) Consumption = QO 2 * 2 = 4 mmol/ g Cell/h = 4,000 mol / g Cell / h If no ATP produced, the ATP within the cell would be used up in (6 mol / g Cell) /(38,000 mol / g Cell / h) = 0.56 sec If no NAD(H) produced, the NAD(H) within the cell would be used up in (1.5 mol / g Cell) /(4,000 mol / g Cell / h) = 1.35 sec Maximum Rate of ATP Production (Full Respiration)= Q S *4 = 59,200 mol / g Cell / h Maximum Rate of NAD(H) Production(Full Respiration) = Q S *12 = 177,600 mol / g Cell / h If no ATP used, the ADP within the cell would be used up in (2.5 mol / g Cell) /(59,200 mol / g Cell / h) = 0.15 sec If no NAD(H) used, the NAD + within the cell would be used up in (2.5 mol / g Cell) /(177,600 mol / g Cell / h) = 0.05 sec </li> <li> Slide 14 </li> <li> 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). </li> <li> Slide 15 </li> <li> 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 </li> <li> Slide 16 </li> <li> Slide 17 </li> <li> Quantitative Metabolism What determines whether a particular reaction is capable of generating ATP or NAD(H)??? </li> <li> Slide 18 </li> <li> 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 </li> <li> Slide 19 </li> <li> 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?? </li> <li> Slide 20 </li> <li> 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 </li> <li> Slide 21 </li> <li> Slide 22 </li> <li> 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 </li> <li> Slide 23 </li> <li> 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?? </li> <li> Slide 24 </li> <li> 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 </li> <li> Slide 25 </li> <li> Quantitative Metabolism </li> <li> Slide 26 </li> <li> Slide 27 </li> <li> Slide 28 </li> <li> Slide 29 </li> <li> Slide 30 </li> <li> Slide 31 </li> <li> Slide 32 </li> <li> Slide 33 </li> <li> Slide 34 </li> <li> Slide 35 </li> <li> Slide 36 </li> <li> Slide 37 </li> <li> Slide 38 </li> <li> End Product Formation </li> <li> Slide 39 </li> <li> Quantitative Metabolism </li> <li> Slide 40 </li> <li> Slide 41 </li> <li> Slide 42 </li> <li> Slide 43 </li> <li> Slide 44 </li> <li> Slide 45 </li> <li> 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 </li> <li> Slide 46 </li> <li> 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 </li> <li> Slide 47 </li> <li> 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 + </li> <li> Slide 48 </li> <li> 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. </li> <li> Slide 49 </li> <li> Slide 50 </li> <li> Quantitative Metabolism </li> </ul>

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