. Some anaerobic bacteria use an alternative pathway for glucosecatabolism, shown below, that converts glucose to acetate ratherthan to pyruvate. The first part of this pathway (glucose to fructose-1,6-bisphosphate) is identical to the glycolytic pathway. In the secondpart of the alternative pathway, Enzymes 1-6 all havemechanisms/activities analogous to enzymes in glycolysis. Note that there are two C-C bond cleavage reactions in this new pathway:A-B + C(Enzyme 1) and C → B + D. All the steps where ATP isconsumed or generated have been shown; however, the addition orloss of NAD* /NADH, P, H,0, or H* has not been shown explic-itly. Draw the structures for the intermediates B, F, G, H, and I, andinclude other reaction participants as needed.HKPGIPFKGlucoseGlucose-6-PFructose-6-PFructose-1,-6-bisPАТРADPАТРADP"Isomerase"CH,-0-PO,-CH,-0-PO,2-ATPOHHC-OHHC-OHC=0C=0"Kinase"CH,OHHC-OHHC-OHCH,-0-PO,2-ADPB"Dehydrogenase"ADPATPADP6H"Dehydrogenase""Kinase""Kinase"AcetateATP

The intermediates have to be given,

Answered: . Some anaerobic bacteria use an…

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter29: The Organic Chemistry Of Metabolic Pathways

Section29.SE: Something Extra

Problem 24MP: One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of xylulose...

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The enzyme d-amino acid oxidase has avery high turnover number because the d-amino acids are potentially toxic. The KM for the enzyme is in the range of 1 to2 mM for the aromatic amino acids and in the range of 15 to20 mM for such amino acids as serine, alanine, and the acidicamino acids. Which of these amino acids are the preferred substrates for the enzyme?
1 In relation to the redox below from electron transport chain (SuccinateDehydrogenase) Show calculations that proof how oxidation of FADH2 by ubuiqinone(Q) theroteically liberates enough enrgy to drive synthesis of ATP. △G°‘ or ATP synthesis from ADP and π is +30.5kj/mol. Explain calculation in terms of how the oxidation of FAD Hz by ubiquinone supplies sufficient energy for ATP synthesis from ADP and π.
Which of the reactions are spontaneous (favorable)? 1)DHAP↽−−⇀glyceraldehyde-3-phosphateΔ?=3.8 kJ/mol 2)L-malate+NAD+⟶oxaloacetate+NADH+H+Δ?=29.7 kJ/mol- 3) glutamate+NAD++H2O⟶NH+4+α-ketoglutarate+NADH+H+Δ?=3.7 kcal/ 4) C6H13O9P+ATP⟶C6H14O12P2+ADPΔ?=−14.2KJ/mol 5) −Rh(I)C2H6Δ?=−150.97 kJ/molC2H4+H2→Rh(I)C2H6ΔG=−150.97 kJ/mol 6) C4H4O5⟶C4H2O4+H2OΔ?=3.1 kJ/mol
You are called in as a consult for a patient who suffers from acidosis. Urine levels of both lactic acid and isocitric acid are elevated. Mitochondria isolated from the patient’s fibroblasts oxidize pyruvate very slowly whereas α-ketoglutarate and malate are oxidized normally. a. At what enzyme would you predict the patient has a defect? b. Why? What evidence do you have to choose that enzyme? c. How do you account for the increase in lactic acid?
When some sugars dissolve in water they spontaneously undergo changes in optical rotation called mutarrotation. The Mutarrotation of D-glucopyranose is catalyzed by acid and bases. 2-Hydroxypyridine is a more effective catalyst than phenol and pyridine for this reaction because: a.Both oxygen and N in 2-hydroxypyridine act as bases increasing the rapid interconversion of sugar b. The OH of 2-hydroxypyridine serves as the base while the current N as the acid. c. 2-hydroxypyridine acts both as a base to remove the proton from the hydroxyl group in the hemiacetal and as an acid to provide a proton to the oxygen in the hemiacetal. d.Phenol and pyridine are very expensive.
What would be the concentration of pyrophosphate that would remain in equilibrium without hydrolyzed when, by the action of inorganic pyrophosphatase, the concentration of orthophosphate was 20 mM. (∆Gº’ pyrophosphate hydrolysis = - 30.5 kJ / mol)
TPP is a coenzyme for transketolase, the enzyme that catalyzes the conversion of a ketopentose (xylulose5-phosphate) and an aldopentose (ribose-5-phosphate) to an aldotriose (glyceraldehyde-3-phosphate) and a ketoheptose (sedoheptulose-7-phosphate). Notice that the total number of carbons in the reactants and products is the same (5 + 5 = 3 + 7). Propose a mechanism for this reaction.
Review the structure of the standard amino acids and listthose that are capable of acting as acids or bases in enzymecatalysis. Are there any that can function as either acids orbases?
Imidazoleglycerol‑phosphate dehydratase is an enzyme in the histine biosynthesis pathway. It catalyzes the E1 dehydration of D‑erthyro‑imidazole‑glycerol phosphate to imidazole acetol‑phosphate. This is a rare example of a biological E1 reaction, as most biological elimination reactions occur through E1cB instead. In this reaction, D‑erthyro‑imidazole‑glycerol phosphate is first protonated to form a good leaving group. Then, the leaving group is ejected to form the resonance‑stabilized carbocation shown. Draw curved arrows forming the most stable resonance structure to explain why this reaction goes through an E1 mechanism. Draw curved arrows to form the most stable resonance structure.
If the hydrolysis of 1 M glucose 6-phosphate catalyzed by glucose 6-phosphatase has a delta G’^o of -11.386 kJ/mol at 25 °C, what percentage of substrate remains once the reaction reaches equilibrium assuming no product was initially present? (Round answer to the nearest whole number) (R = 8.315 J/mol · K)
One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of xylulose 5-phosphate with ribose 5-phosphate in the presence of a transketolase to give glyceraldehyde 3-phosphate and sedoheptulose 7-phosphate. (a) The first part of the reaction is nucleophilic addition of thiamin diphosphate (TPP) ylide to xylulose 5-phosphate, followed by a retro-aldol cleavage to give glyceraldehyde 3-phosphate and a TPPcontaining enamine. Show the structure of the enamine and the mechanism by which it is formed. (b) The second part of the reaction is addition of the enamine to ribose 5-phosphate followed by loss of TPP ylide to give sedoheptulose 7-phosphate. Show the mechanism.
When trying to calculate the Kav for glutamate dehydrogenase I keep getting a negative value using the given values along with those in the graph but I fear that that's not correct

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