University:
Amherst College
Course:
CHEM 330 | Biochemical Principles of Life at the Molecular Level
Academic year:

2011
Views:

456

Pages:

18
Author:

Amani Galvan

BioChem 330 - Course Outline October 25-Nov 3, 2011 • Bio-molecular Structure/Function (I cont’d) – NUCLEIC ACID • DNA sequence and structure • Protein/nucleic acid interactions – CARBOHYDRATES • Sugars - mono and disaccharides • Polysaccharides • Glycerides and glycerol – FATS AND LIPIDS • Chemistry and nomenclature for fatty acids • Saturated and unsaturated fatty acids • Fluid mosaic model of membrane structure BioChem 330 - Course Outline • Metabolism and Bioenergetics (II) – ENZYME CATALYSIS: • kinetic constants kcat, Km • Catalytic strategies, the serine proteases – CATABOLISM (breakdown) • Carbohydrates – – – – Glycolysis Tricarboxylic Acid Cycle Electron Transport Chemiosmosis and ATPase • Fatty acids and amino acids Carbohydrate Metabolism October 25-Nov 3, 2011 • Intro to Metabolism –ATP, the energy currency of the cell –sugar structure •Glycolysis Phase I – gly 1-5 •Glycolysis Phase II – gly 6-10 •Control in Glycolysis Metabolism • Catabolism – Degradation reactions – More complex to simpler molecules • Anabolism – Biosynthesis – Less complex to more complex molecules • Entropy: – Catabolism is favorable, and should have DGo’ < 0 – Anabolism is unfavorable, and should have DGo’ > 0 Over 1000 different reactions are being catalyzed in an e.coli cell, we’ll focus on about two dozen of them which form the blue central line of carbohydrate metabolism. Why must a living organism metabolize nutrients such as fats, proteins, carbohydrates? ANS: To generate ATP, the fuel that drives all cellular events  ATP-4 + H2O ADP-3 + PO4-3 + 2H+ DGo’ = -30.5 kJ/mole ATP hydrolysis liberates energy making it available for biological processes that require energy Phosphate Hydrolysis Potential Energy • Molecules higher in table can transfer a phosphate group to other molecules below them in table (so net ΔGo’ < 0) • A reaction far away from its equilibrium ratio of prod/react and from the standard state can have additional free energy. • In the case of ATP hydrolysis, cells can be many orders of magnitude away from equilibrium and are certainly NOT at the standard state. How does this affect the free energy available from hydrolysis? Which phosphate bonds in ATP and ADP are “high-energy” bonds? • Breakdown of O-P phosphoanhydride bond at Pγ of ATP produces ADP and Pi and energy. • Breakdown of O-P phosphoanhydride bond at Pβ of ADP produces AMP and Pi and energy. • Breakdown of O-P phosphoanhydride bond at Pβ from ATP produced PPi and AMP and energy and then some. Why is the phosphoanhydride bond a “high-energy” bond? • LOTS of energy is released due to – destabilization of reactants (electrostatic repulsion due to multiple negative charges) – stabilization of products (two resonance forms rather than 1 plus products are more easily hydrated b/c charge is more spread out) Destabilized by electrostatic repulsion of negative charges at O- :OH2 attacks at Pg or Pb better resonance stabilization Energetics of making ATP • ADP + PO4-3 + H+ = ATP + H2O DGo’ = +30.5 kJ/mole It takes 30.5 kJ to make one mole of ATP under biological standard state conditions (1 M reactants and products and pH 7.0 295 K) . At equilibrium………. K’ = exp (- D Go’/RT) = 5 x 10-6 M-1 = [Prod] / [Reactants] = [ATP] / [ADP][Pi] = 1/200,000 But how far are cells away from equilibrium with regard to the synthesis of ATP? Energetics of making ATP under cellular conditions • ADP + PO4-3 + H+ = ATP + H2O DG = ? •Cells are far away from equilibrium and far away from standard state conditions. We have much more ATP than would be dictated by equilibrium; the ratio of ATP to ADP+Pi in some cells is as high as 200/1 rather than 1/200,000. •This means that a cell can be far from equilibrium w.r. to this ratio, and now, through metabolism, we are going to make EVEN MORE ATP. •Under these conditions, thermodynamics wants the system instead to hydrolyze the ATP and reach equilibrium, but instead, metabolic reactions will need to fight equilibrium; let’s calculate exactly how much more energy it will take to make this ATP: Energetics of making ATP under cellular conditions • ADP + PO4-3 + H+ = ATP + H2O DG = ? To calculate this energy, remember from thermodynamics: DG = DGo’ + RT ln Q* (*At equilibrium, this equation simplifies to the more familiar eqn. DGo’ = - RT ln K because DG = 0) Where ΔG is the free energy available to the system not at equilibrium, and Q is the mass action ratio: Q = [Products]/[Reactants] when a cell is away from equilibrium…. so what is this value in an average cell? Energetics of making ATP under cellular conditions • ADP + PO4-3 + H+ ⇌ ATP + H2O DG = ? In human erythrocytes for example: Q= [ATP] / [ADP][Pi] = 2.25x10-3/(0.25x10-3 x 1.65x10-3) = 5.45 x 103 M-1 DG = DGo’ + RT ln Q = 30.5 kJ/mole + 21 kJ/mole = +51.5 kJ/mole Overview Of Glycolysis • Stage I – Uses 2 ATP’s per glucose • Stage II – Generates 4 ATP’s PHASE I PHASE II Glycolysis: a biochemical analogue of the movie “Roots” Glucose+2Pi+2ADP+2NAD+ = 2pyruvate+2ATP+2NADH+2H++2H2O DGo’ = -85 kJ/mole (Note above NET reaction of glycolysis book keeps important biological molecules but is not charge balanced) Phase I, Priming Phase II, Cashing In Phase I Glucose + 2 ATP = 2 GAP + 2 ADP Priming the Pump in five steps * This phase of glycolysis requires the investment of two molecules of ATP. * Gly3, PFK, is the committed step for glycolysis. * No oxidation and no involvement of molecular oxygen.

BioChem 330 - Course Outline 2011 Part 1

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