Which of the following correctly describes the linkages found in adenosine triphosphate N-C glycosidic bond, phosphoester bond and phosphodiester bonds N-C glycosidic bond, phosphoester bond and phosphoanhydrides bonds phosphoester bond and phosphoanhydrides bonds N-C glycosidic bond, and phosphoanhydrides bonds

Which of the following correctly describes the linkages found in adenosine triphosphate N-C glycosidic bond, phosphoester bond and phosphodiester bonds N-C glycosidic bond, phosphoester bond and phosphoanhydrides bonds phosphoester bond and phosphoanhydrides bonds N-C glycosidic bond, and phosphoanhydrides bonds

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter3: The Chemistry Of Life: Organic Compounds

Section: Chapter Questions

Problem 4TYU: The synthetic process by which monomers are covalently linked is (a) hydrolysis (b) isomerization...

See similar textbooks

Similar questions

Which of the following is NOT correctly paired? cellulose: beta-1,4-glycosidic linkage amylose: alpha-1,4-glycosidic linkage chitosan: alpha-1,4-glycosidic linkage cellubiose: beta-1,4-glycosidic linkage Which is NOT correctly paired?Immersive Reader fibrous protein: spherical alpha helix protein: right-handed coil beta pleated sheet: fully extended peptide chain globular protein: soluble in water
guanosine and a water bind in the active site, does not matter in which order. The scissile N-1 glycosidic bond between the anomeric carbon and N9 of the base is broken resulting in an oxocarbenium transition state. The leaving group electrons are not protonated, but instead swing up to form a double bond between N9 and C8. The double bond that is already present on the purine swings out on N7 and is protonated by an active site histidine acting as a general acid. An active site aspartate acts as a general base, and deprotonates the water substrate molecule. The deprotonated water is then free to attack the transition state in a manner that will generate an αribose. Draw the active site and show how the mechanism works. You will need to draw 3 or 4 different active site “snap-shots” to show the entire process. Be sure to clearly show the transition state and how it might be stabilized
Which of the following statements about sugar polymers and glycosaminoglycans is/are true? Glucosamine, an amino sugar, would be positively charged physiologically unless it is part of an amide bond. Chitin is a homopolymer of GlcNac in beta 1--> 4 linkages. Amylose is a homopolymer of glucose containing alpha 1--> 6 linkages. Amylopectin in a non-reducing sugar polymer consisting solely of glucose monomers. Agarose is a highly charged sugar polymer used in electrophoresis. A core of hyaluronan protein is linked to an aggrecan polysaccharide in the extracellular matrix around joints and tendons. Mucins are secreted proteins that contain branched oligosaccharides linked to serine residues. None of the options shown is true. The glycosaminoglycan "hyaluronic acid" would carry more negative charge overall when compared to a polymer of "chondroitin" of the same length. Please answer very soon will give rating surely Complete Answer needed
Which of the following is the correct glycosidic bond linkage of the disaccharide shown? β-(1→6) glycosidic bond α-(1→6) glycosidic bond α-(1→1) glycosidic bond β-(1→1) glycosidic bond
The amount of branching (number of (α1→6) glycosidic bonds) in amylopectin can be determined by the following procedure. A sample of amylopectin is exhaustively methylated—treated with a methylating agent (methyl iodide) that replaces the hydrogen of every sugar hydroxyl witha methyl group, converting —OH to —OCH3 . All the glycosidic bonds in the treated sample are then hydrolyzed in aqueous acid, and the amount of 2,3-di-O-methylglucose so formed is determined.(a) Explain the basis of this procedure for determining the number of (α1→6) branch points in amylopectin. What happens to the unbranched glucose residues in amylopectin during the methylation and hydrolysis procedure?(b) A 258 mg sample of amylopectin treated as described above yielded 12.4 mg of 2,3-di-O-methylglucose. Determine what percentage of the glucose residues in the amylopectin contained an (α1→6) branch. (Assume that the average molecular weight of a glucose residue in amylopectin is 162 g/mol.)
In the structure in the picture, the monosaccharide units (alpha-D-galactopyranosyl and beta-D-allopyranose) are linked via alpha-1->3 glycosidic bond. Then, the formed disaccharide units are linked via beta-1->4 glycosidic bond. Then, an Oligosaccharide is formed which has 10 monosaccharide units, meaning that it contains 5 disaccharide units. Question: Is this oligosaccharide a good substrate for glycolysis? Provide two reasons for your answer.
Show anabolism of two molecules of serine. Be sure to clearly identify each molecule and bond/linkage involved. Hand-draw the diagram
Give an example of each of the following:a. epimerb. glycosidic linkagec. reducing sugard. monosaccharidee. anomerf. diastereomer
The synthetic process by which monomers are covalently linked is (a) hydrolysis (b) isomerization (c) condensation (d) glycosidic linkage (e) ester linkage
what is the difference between glucosidic linkage and glycosidic lingake
Penicillin reacts with the cysteine residue of penicillin binding protein. Show the reaction here.
Arrange the following: Formation of nucleoside which is the joining of nitrogen base and a 5-carbon sugar. Condensation reaction between the phosphoric acid and the alcohol group on carbon number 5. The nucleotides are covalently linked through 5’ phosphate-group to the 3’- hydroxyl group of the next nucleotide creating a phosphodiester linkage. a. None of the choices b. II, I, III c. I, II, III d. II, III, I