Biochemistry: The Molecular Basis of Life
6th Edition
ISBN: 9780190209896
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 7, Problem 3Q
Summary Introduction
To review:
The structure of a
Introduction:
Glucose is a monosaccharide thatacts as a primary source of fuel in livingbeings. Threonine is an essential amino acid that plays a vital role in living beings. It is obtained by the living being through diet, as it cannot be synthesized in the living being’s body. A new linkage is formed when the two forms of monosaccharid, eemiacetal or hemiketal form, react with alcohol. The linkage thus formed, is called a glycosidic linkage.
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Chapter 7 Solutions
Biochemistry: The Molecular Basis of Life
Ch. 7 - Prob. 1QCh. 7 - Prob. 2QCh. 7 - Prob. 3QCh. 7 - Prob. 4QCh. 7 - Prob. 5QCh. 7 - Prob. 6QCh. 7 - Prob. 7QCh. 7 - Prob. 1RQCh. 7 - Prob. 2RQCh. 7 - Prob. 3RQ
Ch. 7 - Prob. 4RQCh. 7 - Prob. 5RQCh. 7 - Prob. 6RQCh. 7 - Prob. 7RQCh. 7 - Prob. 8RQCh. 7 - Prob. 9RQCh. 7 - Prob. 10RQCh. 7 - Prob. 11RQCh. 7 - Prob. 12RQCh. 7 - Prob. 13RQCh. 7 - Prob. 14RQCh. 7 - Prob. 15RQCh. 7 - Prob. 16RQCh. 7 - Prob. 17RQCh. 7 - Prob. 18RQCh. 7 - Prob. 19RQCh. 7 - Prob. 20RQCh. 7 - Prob. 21RQCh. 7 - Prob. 22RQCh. 7 - Prob. 23RQCh. 7 - Prob. 24RQCh. 7 - Prob. 25RQCh. 7 - Prob. 26RQCh. 7 - Prob. 27RQCh. 7 - Prob. 28FBCh. 7 - Prob. 29FBCh. 7 - Prob. 30FBCh. 7 - Prob. 31FBCh. 7 - Prob. 32FBCh. 7 - Prob. 33FBCh. 7 - Prob. 34FBCh. 7 - Prob. 35FBCh. 7 - Prob. 36FBCh. 7 - Prob. 37FBCh. 7 - Prob. 38SACh. 7 - Prob. 39SACh. 7 - Prob. 40SACh. 7 - Prob. 41SACh. 7 - Prob. 42SACh. 7 - Prob. 43TQCh. 7 - Prob. 44TQCh. 7 - Prob. 45TQCh. 7 - Prob. 46TQCh. 7 - Prob. 47TQCh. 7 - Prob. 48TQCh. 7 - Prob. 49TQCh. 7 - Prob. 50TQCh. 7 - Prob. 51TQCh. 7 - Prob. 52TQCh. 7 - Prob. 53TQCh. 7 - Prob. 54TQCh. 7 - Prob. 55TQCh. 7 - Prob. 56TQCh. 7 - Prob. 57TQCh. 7 - Prob. 58TQCh. 7 - Prob. 59TQCh. 7 - Prob. 60TQCh. 7 - Prob. 61TQCh. 7 - Prob. 62TQCh. 7 - Prob. 63TQCh. 7 - Prob. 64TQCh. 7 - Prob. 65TQCh. 7 - Prob. 66TQCh. 7 - Prob. 67TQCh. 7 - Prob. 68TQCh. 7 - Prob. 69TQCh. 7 - Prob. 70TQ
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- Provide an explanation for the fact that a-D-mannose is more stable than B-D-mannose, whereas the opposite is truc for glucose.arrow_forwardDraw the open-chain structure of a ketoheptose.arrow_forwardDraw the structures of the activated carbon groups bound to thiamine pyrophosphate in three enzymes that contain this coenzyme. Hint: Keto–enol tautomerism may enter into the picture.arrow_forward
- Why is it reasonable to expect that glucose6-phosphate will be oxidized to a lactone rather than to an open-chain compound?arrow_forwardWhich of an alpha-D-2,3-di-O-methylglucopyranose or alpha-D-2,3, 6-tri-O-methylglucopyranose represents a glucose unit in glycogen which was originally carrying an (alpha 1-->6) glycosidic bond?arrow_forwardGlucose-1-phosphate has a higher phosphoryl group transferpotential than does glucose-6-phosphate. Review the structures of these molecules and suggest a reason for thisphenomenon.arrow_forward
- Talose is the C4 epimer of mannose.draw the chair conformation of D-talopyranosearrow_forwardAn oligosaccharide is a repeating unit of a-D-galactopyranosyl-(a-1 >3)-allopyranoside. Each disaccharide unit is linked via B-1 --->4 glycosidic bond. The oligosaccharide has 10 monosaccharide residues. Required: Is this oligosaccharide a good substrate for glycolysis? Why or why not? Provide two reasons and discuss corn prehensively.arrow_forwardDiscuss the difference between an ⍺-D-glucose and a β-D-glucosearrow_forward
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