BIOL 302 1st Edition Lecture 25 Outline of Last Lecture I. CarbohydratesOutline of Current Lecture I. GLYCOLYSISII. IsomerizationIII. AldolaseIV. oxidation reactionCurrent LectureI. GLYCOLYSISA. Splitting of 6-carbon sugarB. Partial oxidation to yield ATP and reduced carriersII. Glycolysis is an ancient and conserved pathwayA. Glucose is an important fuel for many organisms. B. Practically all species use the process of glycolysis to generate ATP from the partial oxidation of glucose. C. Overall:1. glucose molecule (6-carbon poly-hydroxy aldehyde) is oxidized to 2. molecules of pyruvate (3-carbon -keto carboxylic acid) These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.D. Importantly, glycolysis does not require oxygen.III. How did glucose take such an important position in metabolism?A. It is one of several monosaccharides formed from formaldehyde under prebiotic conditionsB. It is quite stable and favors the ring form. C. D or α–glucose is the naturally occurring form.IV. How do we get glucose from our diet?A. We consume many mono-, di- and oligosaccharides in our dietB. -Amylases in our saliva and intestines break starch and glycogen downC. Specific di- and trisaccharides are further broken down by specific glycosidasesV. Glycolysis part 1A. Convert glucose to fructose 1,6 bisphosphateB. Hexose sugar is split into two interconvertible phosphorylated 3-C sugarsC. Notice that this phase costs ATPVI. Glycolysis part 2A. 2 molecules of glyceraldehyde-3-phosphate are produced from one molecule of glucose.B. G3P is then oxidized to pyruvate. In the process, 2 ATP are produced per molecule of G-3-P.C. Each glucose nets 2 ATPVII. Stage 1: Trapping & PreparationA. Phosphorylates glucose to trap it inside the cellB. Why can’t glucose-6-phosphate leave the cell?1. Isomerizes to a ketose sugar (fructose-6-P)2. Creates symmetry through phosphorylation to F-1,6-BPC. First energy-costing step in glycolysisD. Not a “committed” step towards glycolysisVIII. Induced fit in hexokinaseIX. Isomerization of glucose-6-phosphateA. phosphoglucose isomerase (PGI) is the enzyme that catalyzes all ofthese.B. How is fructose different from glucose?C. We will see in a moment how this is very importantX. Why isomerize to fructose? 2 reasonsA. makes the split chemically easier by making one of the splitting carbons alpha to a carbonylB. makes the terminal carbon an alcohol that can attack the terminal phosphate of ATP to be phosphorylated, creating symmetry beforesplitting the sugar1. First “committed” step down glycolytic pathway2. Therefore PFK is an important point of regulation in glycolysis and controls the pace of flow down this pathwayXI. Clicker Question: Based on what you know about regulatory mechanisms, do you think ATP activates or inhibits PFK?A. activatesB. inhibitsC. depends on the concentrations of substrateXII. Aldolase breaks the hexose into two 3-carbon sugarsA. Nicely illustrates why glucose-6-phosphate was isomerizedXIII. Next: Isomerization of 3-carbon piecesA. Triose phosphate isomerase isomerizes DHAP to G-3-PB. So after TPI, the yield is 2 G-3-PXIV.Isomerization= Ketone to AldehydeXV. Triose Phosphate Isomerase pt 1.A. Glutamate 165 acts as a base, abstracts proton from C1.B. His 95 acts as an acid, donating a proton to carbonyl O (C2).C. Enediol intermediate formedXVI. TPI pt 2A. Glutamic acid donates a proton (H) to C-2 while histidine (-) removes a proton from C-1 of enediol.XVII. TPI pt 3.G-3-P A. formed; glutamate and histidine return to their ionized and neutral formsXVIII. Clicker Question: Which of the following would affect catalysis by TPI?A. A CHANGE IN ALDOLASE ACTIVITYB. A SYNONYMOUS MUTATION IN THE GENE ENCODING GLU 165C. TREATMENT WITH UREAD. A CHANGE IN PH FROM PH 7 TO PH 9E. A CHANGE IN PH FROM PH 7 TO PH 5XIX. TPI SUPPRESSESA. AN UNWANTED ENEDIOL SIDE-REACTION BY TRAPPING IT1. Β-STRANDS (ORANGE) Α-HELICES (BLUE)2. ΑΒ BARREL: ALSO FOUND IN ALDOLASE, ENOLASE, AND PYRUVATE KINASE3. RED LOOP CLOSES OFF THE ACTIVE SITE WHILE ENEDIOL INTERMEDIATE IS INSIDE4. NOTE GLU165, HIS95 IN BARRELXX. Stage 3: Payoff PhaseA. Release energy from the oxidation of carbon B. Convert to high-phosphotransfer potential intermediatesC. Use these intermediates to generate ATPD. Also generates reduced electron carriersXXI. GAPDH (G3PDH)A. 2xXXII. The oxidation reaction GAPDHA. required to catalyze has a large negative ∆G, but the phosphorylation has a large positive ∆GB. So what does the cell do with these two reactions?1. GAPDH solves this problem by coupling the two processes using a thioester intermediate XXIII. Now we’ll have 2 high-energy phosphate bonds on each 3-carbon intermediateXXIV. GAPDH mechanism step-by-stepA. Cys 149 in GAPDH recruits the aldehyde substrate G-3-P. 1. Forms a hemithioacetalB. NAD+ is reduced and histidine is oxidized.1. Hydride ion transferred to NAD+2. Thioester intermediate is formed, serving as an activated carrierC. NADH is released and replaced by a second NAD+D. GAPDH’s1. thioester is polarized by NAD+, allowing attack by Pi2. His donates H, freeing Cys and forming 1,3-BPGXXV. Clicker Question: NAD+ plays which of the following roles in the GAPDH active siteA. ElectrophileB. electron acceptorC. electron donorD. A and BE. A, B, and CXXVI. Substrate-level phosphorylationA. 1,3-Bisphosphoglycerate has a high-energy phosphate bondB. Payback of the 2 ATP used in the trapping phaseC. Called “substrate level phosphorylation”D. Will contrast later with phosphorylation using ion gradientsXXVII. Payoff Phase ctdA. Steps toward making another high phosphate transfer molecule, PEP:1. Transfer of the phosphate group: Phosphoglycerate Mutase2. Then transfer of the phosphate group to ATP: Pyruvate KinaseXXVIII. Why is phosphoenolpyruvate such a good donor?XXIX. Alternate fates of pyruvateA. fermentation, acetyl-CoA1. Fermentation: Use of organic compounds as terminal electron acceptors2. The electron acceptors are usually the remnants of the very compounds from which the electrons were originally removedXXX. Ethanol FermentationA. Yeast and other microorganisms ferment pyruvate by first decarboxylating, then adding electrons back to acetaldehydeB. Is anaerobic, No energy yielded, loss of a reduced electron carrierXXXI. Lactic acid fermentationXXXII. NAD+ is regeneratedA. Now the cell can make ATP again!B. The early world