Chapter 4

Question 1

Epimers

Answer 1

Subtype of diastereomers are those that differ in configuration at one chiral center (C)
* Have slightly different chemical and physical properties

Question 2

Enantiomers

Answer 2

Same sugars in different optical families
* All chiral centers

Question 3

Stereoisomers

Answer 3

Compounds with the same chemical formula, but differ from one another in terms of spatial arrangements.

Question 4

Anomer

Answer 4

Two molecules differ at anomeric C

Question 5

a-Anomer

Answer 5

-OH group of C-1 trans to -CH2OH substituent (axial and down)
* Less stable

Question 6

B-anomer

Answer 6

-OH group of C-1 cis to the -CH2OH substituent (equatorial and up)
* More stable
* Undergoes less electron repulsion

Question 7

Esterification

Answer 7

Reaction where hydroxyl group reacts with either carboxylic acid or derivative to form an ester

Question 8

Glycoside Formation

Answer 8

Reaction between an alcohol and a hemiacetal group on a sugar to yield an alkoxy group

Question 9

Carbohydrate Oxidation

Answer 9

• Become oxidized when reducing other groups
• Aerobic metabolism requires reduced forms of electron carriers for oxidative phosphorylation
• Primary energy source so oxidized

Question 10

Amylose

Answer 10

Linear glucose polymer linked via a-1,4 glycosidic bonds
* How plants predominantly store starch

Question 11

Amylopectin

Answer 11

Starts off with same type of linkage of amylose and also has brances via a-1,6 glycosidic bonds
* More soluble in solution because of branched structure
* Highly branched structure decreases intermolecular bonding between polysaccharide polymers
* Increases interaction with surrounding solution

Question 12

Glycogen

Answer 12

Functions as main energy storage form for animals
* More branching than amylopectin so experiences higher rate of enzyme activity

Question 13

Aldonic Acids

Answer 13

• Oxidized aldoses
• Can be reduced to alditols (act as reducing agents)

Question 14

Mutarotation

Answer 14

Cyclic compounds shift from one anomeric form to another with straight chainform as an intermediate
a > B

Question 15

Tautomerization

Answer 15

Rearrangement of bonds to undergo keto-enol shifts
* Forms aldose that alllows them to act as reducing sugars

Question 16

B-amylase

Answer 16

Cleaves amylose at nonreducing end of polymer to yield only maltose

Question 17

a-Amylase

Answer 17

Cleaves amylose anywhere along chain to yield short polysaccharids, maltose, and glucose

Question 18

Debranching Enzymes

Answer 18

Removes oligosaccharides from branch in glycogen/starches

Question 19

Glycogen Phosphorylase

Answer 19

Yields glucose 1-phosphate

Question 20

Maltose

Answer 20

Digestible by humans because B-glycosidic linkages in cellobiose can’t be cleaved in human body