BIOL 302 1st Edition Lecture 28Outline of Last Lecture I. Electron transport and OxidativePhosphorylationOutline of Current Lecture I. Nicotinamide Electron CarriersII. biosynthetic reactionsIII. Pentose Phosphate PathwayVII. Oxidative decarboxylationCurrent I. Plants can use photosynthesis to generate NADPH,but non-photosynthetic organisms also need NADPHA. An alternative oxidative pathway shunts hexoses from glycolysis to generate NADPH and 5-carbon sugars before (sometimes) funneling intermediates back into glycolysisB. This pentose phosphate pathway is AKA the phosphogluconate pathway or hexose monophosphate shunt1. glucose-6-phosphate + 2 NADP + + H2O ribulose-5-phosphate + 2 NADPH + 2 H+ + CO2II. Nicotinamide Electron CarriersA. NADH is produced in _glycolysis and _the CA cycle__B. - Its primary purpose is _oxidation by the ETC for ATP_C. - Therefore it is mainly used in __catabolic__ processesD. NADPH is produced in the pentose phosphate pathwayE. - Its primary purpose is _reducing power__F. - Therefore it is mainly used in _biosynthetic__ processesThese 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.III. Two main functions for NADPHA. Reduction potential in biosynthetic reactions1. Calvin Cycle 2. Fatty acid biosynthesis3. Cholesterol biosynthesis4. Nucleotide biosynthesisB. Protection of the cell from oxidative damage1. Reduction of glutathioneIV. Two Phases of the Pentose Phosphate PathwayA. Oxidative Phase1. Responsible for the generation of NADPH2. Payoff upfrontB. Non-oxidative Phase1. Shuffles carbons around into different sugars depending on cellular needs2. Can feed precursors into the synthesis of RNA, DNA, ATP, NAD(P)H, FADH, FMN, Coenzyme AV. PPP: Oxidative PhaseA. The pentose phosphate pathway is an alternative shunt out of glycolysisB. Glucose-6-phosphate and NADP+ are required. Rate of PPP α [NADP+].C. Glucose-6-phosphate is oxidized and decarboxylated by glucose 6-phosphate dehydrogenase, yielding NADPH and CO2D. 6-carbon sugar to 5-carbon sugar(hexose to pentose)1. Glucose-6-Phosphate + 2 NADP+ + H2O  Ribulose-5-Phosphate + 2 NADPH + CO2VI. PPP: Oxidative PhaseA. Glucose-6-phosphate is oxidized to 6-Phosphoglucono--lactone, producing NADPHB. Lactonase promotes hydrolysis: water hydrolyzes the lactone, opening the ring and yielding 6-phosphogluconateC. 6-Phosphogluconate then undergoes oxidative decarboxylation to yield Ribulose-5-phosphate (6 C to 5 C) AND NADPHVII. Oxidative decarboxylation of 6-phosphogluconateA. We’ve seen oxidative decarboxylation a number of timesB. A hydroxyl group is oxidized by NADP+ to form the ketoneC. Decarboxylation leads to the formation CO2 and another intermediate (2)D. Tautomerization reforms the ketone to make ribulose-5phosphate (3)E. The end result is reduction ofNADP+ and release of CO2F. Which structure is NADP+ and how does itaccommodate the hydride ion? pyridine ring. Onee- goes to the +vely charged N; other e- to C4opposite this N)VIII. Non-oxidative PhaseA. First Ribulose-5-phosphate is isomerized into Ribose-5-phosphate and alternatively epimerized into Xylulose-5-phosphateB. This produces an aldose or ketose sugarC. Multi-carbon units are then shuffled around to produce biosynthetic precursors or glycolytic intermediatesD. PPP’s non-oxidative phase is controlled primarily by substrate availability1. TransketolaseC5 + C5  C3 + C72. TransaldolaseC3 + C7  C6 + C4 3. TransketolaseC4 + C5  C6 + C3E. What is different about the transaldolase reaction - transfers 3 carbons.transketolase transfers 2 carbonsIX.Clicker Question: Which of the following best describes the net activate carrier producedformed during the oxidation of one molecule of glucose 6-phosphate to ribose 5-phosphate and CO2 via the pentose phosphate pathway?A. 1 NADPHB. 2 NAD+C. 2 NADP+D. 2 NADHE. 2 NADPHX. Clicker Question: You are doing an experiment in the lab measuring the rate of ribulose-5-phosphate synthesis via the pentose phosphate pathway and notice the reaction has slowed dramatically. Which of the following is the most likely reason?A. Reaction has very little O2 lefB. Reaction has very little glucose-6-phosphate lefC. Reaction has very little NADP+ lef NADP+ controls the rate of this pathway)D. Reaction should have been set up in an aqueousenvironmentE. None of the above is a plausible reason for yourreaction to have slowedXI. C3-C7 sugars can be made via the PPPA. transketolase: 2-C transferB. transaldolase: 3-C transfer C. A second transketolasecatalyzedstep occurs tomake fructose-6-phosphate andglyceraldehyde-3-phosphate from thesetwo sugars. (C4and C5)XII. Transketolase #1A. Transketolase transfers a 2-carbon unit from a 5-carbon ketose to a 5-carbon aldose yielding a 3-carbon aldose and a 7-carbon ketoseXIII. Transketolase MechanismA. What other important mechanism have we seen that uses TPP to break C-C bondsXIX. TransketolaseA. TPP ionizes to form a cationB. carbanion of TPP attacks the ketose, which was xylulose-5-phosphate in the last slideC. Cleavage of a C-C glycoaldehyde intermediate; frees aldose product. D. activated glycoaldehyde intermediate attacks aldose substrate to make new C-C bondE. Ketose product (sedoheptulose 7-PO4) is released, freeing TPP for the next cycleXX. Pyruvate DehydrogenaseA. Very similar reaction mechanism – electrons from broken C-C bond arepushed to electron sink before eventually being used to form a new bondB. Note the leaving group is a carboxylic acid rather than an alcohol like intransketolaseXXI. TransaldolaseA. Transaldolase takes the 3-carbon and 7-carbon products of Transketolase #1and performs a 3-carbon transfer to make 6-carbon and 4-carbon productsB. Notice the conversion from odd-numbered to even-numbered sugarsXXII. Transaldolase MechanismA. The same principles of chemistry are used between transaldolase and transketolase:the electron sinkXXIII. Transaldolase MechanismA. Lysine residue in transaldolase acts as a Schiff base, binding the ketose substrate.B. Protonation of the Schiff base leads to release of the aldose product andC. A 3-C fragment is attached to the lysine residueD. This fragment is added to the aldose substrate to form a new C-C bondE. Deprotonation and 6. hydrolysis of the Schiff base releases the ketose product from the lysine side chainXXIV. transaldolase: 3-C transferXXV. Electron SinksXXVI. Transketolase #2A.