BMSC 200 Module 7 Exam Questions
and Answers
D-sugars have the same configuration as? - ANSWER-D-glyceraldehyde at the
chiral carbon most distant from the carbonyl carbon
molecules with n chiral carbons will have ______ stereoisomers? - ANSWER-2n
4 chiral centers = - ANSWER-16 stereoisomers
2 chiral centers = - ANSWER-4 stereoisomers
what are epimers? - ANSWER-sugars that differ at only one of several chiral centers
in solution carbohydrates of 5 carbons and longer tend to be? - ANSWER-cyclized,
will only see them in the linear form when you are metabolizing them
formation of these cyclized structures is the result of a reaction of an alcohol with: -
ANSWER-- an aldehyde (aldoses) to form a hemiacetal
- a ketone (ketoses) to form a hemiketal
what are the two ring forms cyclized carbohydrates can be found in? - ANSWER--
six membered sugar ring (pyranose)
- five membered sugar ring (furanose)
cyclization of glucose involves? - ANSWER-reaction of the free hydroxyl of C-5 with
the aldehyde group of C-1
cyclization renders? - ANSWER-C-1 asymmetric, producing two stereoisomers, a
and B
the carbon that becomes chiral as a consequence of cyclization is the? - ANSWER-
anomeric carbon
the a and B forms are? - ANSWER-anomers of each other
cyclization produces a new ________ carbon at the _______ carbon with ____ new
stereoisomers - ANSWER-asymmetric, anomeric, two
what stereoisomer will it be if hydroxyl of anomeric carbon is below the plane of the
sugar? - ANSWER-alpha
what stereoisomer will it be if hydroxyl of anomeric carbon is above the plane of the
sugar? - ANSWER-beta
and Answers
D-sugars have the same configuration as? - ANSWER-D-glyceraldehyde at the
chiral carbon most distant from the carbonyl carbon
molecules with n chiral carbons will have ______ stereoisomers? - ANSWER-2n
4 chiral centers = - ANSWER-16 stereoisomers
2 chiral centers = - ANSWER-4 stereoisomers
what are epimers? - ANSWER-sugars that differ at only one of several chiral centers
in solution carbohydrates of 5 carbons and longer tend to be? - ANSWER-cyclized,
will only see them in the linear form when you are metabolizing them
formation of these cyclized structures is the result of a reaction of an alcohol with: -
ANSWER-- an aldehyde (aldoses) to form a hemiacetal
- a ketone (ketoses) to form a hemiketal
what are the two ring forms cyclized carbohydrates can be found in? - ANSWER--
six membered sugar ring (pyranose)
- five membered sugar ring (furanose)
cyclization of glucose involves? - ANSWER-reaction of the free hydroxyl of C-5 with
the aldehyde group of C-1
cyclization renders? - ANSWER-C-1 asymmetric, producing two stereoisomers, a
and B
the carbon that becomes chiral as a consequence of cyclization is the? - ANSWER-
anomeric carbon
the a and B forms are? - ANSWER-anomers of each other
cyclization produces a new ________ carbon at the _______ carbon with ____ new
stereoisomers - ANSWER-asymmetric, anomeric, two
what stereoisomer will it be if hydroxyl of anomeric carbon is below the plane of the
sugar? - ANSWER-alpha
what stereoisomer will it be if hydroxyl of anomeric carbon is above the plane of the
sugar? - ANSWER-beta
