5. Glycolysis is a metabolic pathway that converts glucose into pyruvate while synthesizing high-energy molecules. During one step of Glycolysis, glucose-6 phosphate (an aldose) is converted to fructose-6-phosphate (a ketose) via an ene-diol intermediate. Draw arrow pushing mechanism to describe this isomerization. Use B: and HA as surrogates for basic/acidic amino acid residues. ОН ОН О 왜왜요 OP ОН ОН ОН ОН ОН OP OH ОН О

5. Glycolysis is a metabolic pathway that converts glucose into pyruvate while synthesizing high-energy molecules. During one step of Glycolysis, glucose-6 phosphate (an aldose) is converted to fructose-6-phosphate (a ketose) via an ene-diol intermediate. Draw arrow pushing mechanism to describe this isomerization. Use B: and HA as surrogates for basic/acidic amino acid residues. ОН ОН О 왜왜요 OP ОН ОН ОН ОН ОН OP OH ОН О

Introduction to General, Organic and Biochemistry

11th Edition

ISBN:9781285869759

Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres

Chapter20: Carbohydrates

Section: Chapter Questions

Problem 20.81P

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Bicarbonate (shown below) functions as a conjugate base within the plasma's bicarbonate buffer system. Bicarbonate also was important for the: OI (10 OH O process of lipogenesis (triacylglycerol synthesis) rate-limiting step in fatty acid metabolism O process of lipolysis (triacylglycerol metabolism) committed step in fatty acid synthesis J Bicarbonate (shown below) functions as a conjugate base within the plasma's bicarbonate buffer system. Bicarbonate also was important for the: C10 OH O process of lipogenesis (triacylglycerol synthesis) Ⓒrate-limiting step in fatty acid metabolism O process of lipolysis (triacylglycerol metabolism) committed step in fatty acid synthesis
rivatives: Nucleophilic Acyl Substitution Reactions - EOC O C-O [References] One step in the gluconeogenesis pathway for the biosynthesis of glucose is the partial reduction of 3-phosphoglycerate to give glyceraldehyde 3-phosphate. The process occurs by phosphorylation with ATP to give 1,3-bisphosphoglycerate, reaction with a thiol group on the enzyme to give an enzyme-bound thioester, and reduction with NADH. H-C-OH ATP ADP CH₂OPO32- 3-phosphoglycerate 0= OPO32- C H-C-OH CH₂OPO32 1,3-bisphosphoglycerate Enz-SH PO43- O= C S-Enz H-C-OH CH₂OPO3² (Enzyme-bound thioester) NADH/H+ O=C NAD*, Enz-SH H H-C-OH CH₂OPO3²- glyceraldehyde 3-phosphate Propose a structure for the first intermediate in the reaction of the enzyme-bound thioester with NADH to form glyceraldehyde 3-phosphate. To simplify the drawing process, substitute the structure below for the enzyme-bound thioester. S-CH3 substitute for the CH2CH3 enzyme-bound thioester You do not have to consider stereochemistry. • You do not have…
ACTIVITY-4. Refer to the structures below. Draw the Haworth formula (pyranose) -OH но H- FOH FOH- но но H- H FOH H FOH- H- CH2OH CH2OH Alpha-D- glucopyranose Beta-D-glucopyranose (Haworth) Beta-D-glucopyranose Alpha-D- glucopyranose (Haworth) ACTIVITY-5. Draw the Haworth formula for D-mannose and D-galactose. Label the alpha and beta anomer. CYCLIC FORMS OF FRUCTOSE D-fructose, a ketohexose sugar, can be presented using furanose and pyranose. Cyclic Fisher (furanose) 1 2OH Open chain Fisher Haworth formula (furanose) 1 CH2OH HOH,C- 1 2 F0 НО 3_H. ÇH2OH ÇH2OH но з—н 4 H- H но 2 H-4 HO- H- 5. OH 13 H-5 FOH 6 CH,OH 14 OH H. 6 CH2OH D-fructose Alpha form Alpha form Cyclic Fisher (pyranose) Haworth formula (pyranose) 1 HOH2C-2 3. 1 HO- CH2OH Но H. 5 он 2 H-4 HO- Но 4 OH 3 H-5 FHO- OH H-6 H Alpha form of D-fructose Alpha form of D-fructose ACTIVITY-6. Draw the Fisher and Haworth formula for beta-D-fructose (furanose and pyranose form). Label the beta furanose and beta pyranose.
4) 4- Customized synthesis of carbohydrate compounds is a popular area of research. Complete the following steps, as required, from a recent report, which begins with a (D)-glucopyranose. но HO Ph Ph HO OH A HO C. ÓCH3 HO HO HO HO 4. OH HOYH hosTH OCH3 OCH3 4. 4. NABH;CN HCI 1 equivalent of TsCl /py 1. CI -C. CI DMSO Ph OH 2) (CH3CH2);N Ph OCH3 (Swern reaction) 4. 4. NaN3 NaBH4 HO HƆ stereoselective formationof an isomer of compound F 4.
(d) Use the diagram below to complete the cyclic alpha form of structure V (e) Circle the hemiacetal in cyclic alpha form of structure V. (f) Redraw the cyclic alpha form of structure V but replace the OH group on the anomericcarbon with a methoxy group. Is this modified monosaccharide a reducing sugar or anonreducing sugar?
H NH3 H₂N CO₂ Lysine CO₂ a-ketoglutarate NADPH/H* H H NH3* NADPH CO₂ Saccharopine The first step in the biological degradation of lysine is reductive amination with a-ketoglutarate to give saccharopine. Nicotinamide adenine dinucleotide phosphate (NADPH), a relative of NADH, is the reducing agent. Draw curved arrows to show the movement of electrons in this step of the reaction mechanism. Arrow-pushing Instructions H₂N H NH3+ CO₂ H-A CO₂ OH *H3N H :A CO₂ جائے
Need help. The oxidation of glyceraldehyde 3-phosphate to 1,3-bisphospho-glycerate, catalyzed by glyceraldehyde 3-phosphate dehydroge-nase, proceeds with an unfavorable equilibrium constant (K9eq 5 0.08; DG98 5 6.3 kJ/mol), yet the flow through this point in the glycolytic pathway proceeds smoothly. How does the cell over-come the unfavorable equilibrium?
3a. 3b. 3c 3d. 3e. CO₂ clavulanic acid CH₂-OH H Answer the following questions about clavulanic acid. Does clavulanic acid inhibit D-alanyl-D-alanine transpeptidase? Does clavulanic acid contain a ß-lactam? Does clavulanic acid contain a thiazolium ring? What is the result of the treatment of penicillinase with clavulanic acid? Does clavulanic acid form a covalent acyl-enzyme intermediate with penicillinase?
II. Analysis. Given below is a schematic diagram for a simple analysis of a novel tetrasaccharide isolated from a new species of red algae. Analyze each step and the accompanied results of tests, and answer the succeeding questions: A H B H c H 0 a1,4-endoglycosidase treatment B Benedict's test: (+) Seliwanoff's test: (-) Fehling's test: (+) Seliwanoff's test: (-) Bial's-Orcinol test: (+) Bial's-Orcinol test: (-) Osazone test: needle- like crystals, 2N HCI, 100°C, 60 mins. formed 5 mins. D | Benedict's test (-) conc. HNO3: rhombic crystals Osazone test: broad, flat crystals, formed 15 mins. Fehling's test: (+) Seliwanoff's test: (+) Bial's-Orcinol test: (-) Osazone test: needle- like crystals, formed 2 mins. TLC Profile Fehling's test: (+) ++ rib gal glc fru man xyl Give the identity of the monosaccharide components of the tetrasaccharide: A: B: С: D:
Draw the organic products formed in the attached reaction
6. Identifv each of the following reactions as an organic oxidation, reduction or neither. Pt + H, b. COO ÇO0 но -с — н C=0 Malate dehydrogenase CH; + NAD CH; + NADH + H COO COO Malate Oxaloacetate с. OH ACP + NADPH + H* ACP + NADP* d. OH + но
Draw the organic products formed in the following reaction.