CHEM 372 Week 2 Notes
CHEM 372 Week 2 Notes CHEM 372
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This 8 page Class Notes was uploaded by Joshua Torres on Tuesday April 12, 2016. The Class Notes belongs to CHEM 372 at California Polytechnic State University San Luis Obispo taught by Dr. Jones in Fall 2016. Since its upload, it has received 8 views. For similar materials see Metabolism in Chemistry at California Polytechnic State University San Luis Obispo.
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Date Created: 04/12/16
CHEM 372 Week 2 Notes Week 2: Day 1 ∆G < 0 exergonic, so ∆G > 0 Endergonic Catabolism – breaking down molecules and releasing energy Pathways o How to make sure you get one pathway and not the other pathway if you have both Cat./Ana. Regulation is the trick to get the reaction where you want it to go o Regulation – activation/inhibiting enzymes to introduce/remove a barrier to a pathway Glycolysis (downward) Makes one reaction faster than another Allosteric Inhibitors Separate pathways – share begin/end points o Activate first step of desired reaction o Inhibit first step of reverse reaction o Ex: Fatty acid metabolism o Ex: common pathway with only one step that differs, regulation occurs at this enzyme (Glycolysis vs. Gluconeogenesis) Catabolism of Carbon Compounds o Carbon ----- smaller carbon (energy output) ----- oxidized finally to C2 o Oxidize Carbon o Electron go onto electron carriers (NAD ) o Electron transferred to O ,2liberating energy o H 2, ADP, P -i-oxidized---O ,2ATP o Catabolism: electrons go to O ; el2ctron carrier is NAD /NADH, FAD/FADH o Anabolism: electrons go to carbon compounds; electron carrier is NADP /NADPH o NAD accepts a hydride anion (H: ) - Hydride comes from a fuel o Difference between the two electron carriers are differentiated by the enzyme: charge and size Specialization of cells (Metabolic) o Muscle, Heart, Brain are consumers of fuels (Acetly CoA, Glu.) - Catabolism o Liver and Adipose are producers of fuels – Anabolism at the same time that the others are undergoing Catabolism o ATP is strictly inside a cell; you cannot send ATP from one cell to another. Not outside of a cell, it does not travel outside a cell’s plasma membrane 4/7/2016 Glycolysis o First pathway that converts glucose to pyruvate (page 578) o Glucose + 2ADP + 2P + 2NiD ------- 2Pyruvate + + 2NADH + 2ATP + 2H o Know: Intermediates and structures, the enzymes, the regulated steps, know the balanced equation o Don’t bother: the mechanisms of reactions, 2 o Step 1: Glucokinase doesn’t bind to glucose as tightly as hexokinase o Step 3: Phosphofructokinase (PFK) is the most important enzyme Kinase transfers a phosphate from ATP to … This step is irreversible and the rate-limiting step (slow) First committed step – product only useful for glycolysis Very tightly/highly regulated regulation step Most rate-limiting steps are regulated, but not vis versa o Step 4: FBP spilt into two 3-carbon molecules, differ in position of ketone and hydroxyl o Step 5: is the end of the investment phase of Glycolysis o The next steps 6-10 are all doubled because of the two molecules of G3P o Step 6: Aldehyde oxidized to carboxylic acid and it also gets phosphorylated from a P i Enzyme: GAP Dehydrogenase – it’s a redox reaction; typically, electrons go onto the carrier o Step 7: breaking even point, ATP Phosphoglycerate Kinase (PGK) o Step 8: Mutates phosphate to 2 position Phosphoglycerate mutase (PGM) Type of isomerase o Step 9: eliminate water from 2PG via hydroxyl – PEP 3 Enolase enzyme PEP is high energy Phosphate group locks enol in place and prevents tautomerization o Steps 1, 3, and 10 are regulated in Glycolysis Irreversible under most conditions -∆G are the most regulated enzymes o Usually the first step of every pathway is regulated Irreversible Uses ATP First reaction of glycolysis Reaction 1 o Hexokinase locks glucose in cell, no escape K 0.5= 0.1 mM [Glu] = 2.5 mM Velocity of enzyme = V max Moment a Glu enters a cell, Hexokinase is right there ready to bind right away 6 prime Hydroxyl deprotonated and attacks a phosphate in ATP ----> Glu-6-PO 32-+ ADP Regulation: competitive inhibitor Inhibited by its product (G6P) – product inhibition Negative feedback and a self-limiting reaction o Glucokinase present in pancreas and liver K0.5 = 10 mM 4 [Glu] = 5 mM Acts more like a Glu sensor, once Glu is high enough, then it converts sugar into other things Reaction 2 o G6P ---> F6P, ∆G’ = -3 kJ/mol o Make sugar more symmetric o Makes carbon one more easily phosphorylated o Histidine acting as a general acid and opens ring to an open-ring aldehyde o Lysine grabs a proton from carbon 1 and forms water. Water attacks open-ring aldehyde and deprotonates carbon 2 – Enediol Intermediate – directs formation of a ketone because the secondary carbocation is favored which leads to formation of the ring again Reaction 3 o F6P + ATP ---> F2,6BP o Irreversible, rate-limiting step, 1 committed step o Highly/tightly regulated o PFK is an allosteric enzyme – switched between R- state and T-state Inhibited by an allosteric inhibitor – ATP o Sensitive to the cell’s energy balance or energy charge (EC) o EC – ratio of ([ATP] + 0.5*[ADP])/([ATP]+[ADP]+ [AMP]) If all ATP, then EC = 1-- Higher EC = more ATP Less need for this reaction 5 o 2ADP <----> AMP + ATP, ∆G’ = 0, Adenylated Kinase o 2 active sites for ATP in F2,6BP, the active site and the regulatory site o Enzyme inhibited by its own substrate is uncommon Reaction 4 o FBP ---- G3P + DHAP Fully reversible, ∆G’ = 0 o Enzyme (FBPA) Lysine in active site is conserved. Nitrogen double bonded to the 2’-carbon on sugar. SHIFT BASE is the critical intermediate Ring structure broken – linear 6 carbon sugar Isolatable intermediate. Covalent bond between enzyme and sugar SHIFT BASE makes it possible to produce the two products Brakes C-C single bond to make an aldehyde Deprotonated Asp takes a proton from OH of 4’-Carbon to make a double bond with carbocation of 2’-Carbon GAP comes off first –the bond of the C-C bond broken (3 Carbons) Left with Enol Intermediate. Water attacks double bond through resonance – OH goes on 2’-C NaBH i4 an irreversible inhibitor because it turns alkene into a single bond – Inactive molecule – proves there must be a double bond in the intermediate after GAP leaves 6 Reaction 5 o DHAP --- G3P o Enzyme TPI switches hydroxyl “Time Barrel Structure” and active site is in the middle of the barrel Enzyme relies on protein dynamics to do its job Takes advantage that molecules are always fluxuating (bond length stretching, etc.) Glutamic acid conserved and is pushed right onto 1’-C Substrate is getting squeezed Enediol intermediate Base that gives up proton in first step now pulls of proton from carbon 1 and double bond takes proton from glutamate Now Glycolysis begins to produce ATP and everything is doubled Reaction 6 + o G3P + NAD + P --->iBPG + NADH o Oxidation reaction – least favorable, ∆G’ = 0 = ~6 kJ/mol o Oxidation of electron carriers is always favorable, + but NAD is being reduced, so ∆G’ is slightly above zero – break even here o 2 steps: A. Aldehyde is oxidized to carboxylic acid (unfavorable) + NAD reduced to NADH 7 B. Carboxylic acid gets phosphorylated by inorganic phosphate (favorable) o Not a protonation, but a reduction – not favorable o Intermediate is a Thiolester Hydrolysis of thiolester is very energetically favorable Phosphorylated thiolester 8
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