Chapter 18

Question 1

biochemisty

Answer 1

is the study of chemicals and their interactions within living organisms.
These chemicals are called biochemical substances and are divided into two groups:
1. bioinorganic
2. bioorganic

Question 2

monosaccharides

Answer 2

Contain single polyhydroxy aldehyde or ketone unit
Cannot be broken down into simpler substances by hydrolysis reactions
Contain 3–7 C atoms
5 and 6 carbon species are more common
Pure monosaccharides - Water soluble white, crystalline solids
Monosaccharides - Glucose and fructose

Question 3

disaccharides

Answer 3

Contain 2 monosaccharide units covalently bonded to each other
Crystalline and water soluble substances
Common disaccharides - Table sugar (sucrose) and milk sugar (lactose)
Upon hydrolysis, they produce 2 monosaccharide units

Question 4

Oligosaccharides

Answer 4

Contain three to ten monosaccharide units covalently bonded to each other
Free oligosaccharides are seldom encountered in biochemical systems
Usually found associated with proteins and lipids in complex molecules
Serve structural and regulatory functions

Question 5

mirror image

Answer 5

reflection of an object in a mirror

Question 6

achiral molecule

Answer 6

Superimposable mirror images: Images that coincide at all points when the images are laid upon each other
-do not have handedness

Question 7

chiral molecule

Answer 7

Nonsuperimposable mirror images: Images where not all points coincide when the images are laid upon each other.
-have handedness

Question 8

enantiomers

Answer 8

Stereoisomers whose molecules are nonsuperimposable mirror images of each other
Molecules with opposite configurations at all chiral centers

Question 9

diastereomers

Answer 9

Stereoisomers whose molecules are nonsuperimposable non-mirror images of each other
Molecules with opposite configurations at some chiral centers
Example: Cis-trans isomers
Molecules that contain more than one chiral center can also exist in diastereomeric as well as enantiomeric forms

Question 10

epimers

Answer 10

Diastereomers whose molecules differ only in the configuration at one chiral center

Question 11

dextrorotatory

Answer 11

compound: Chiral compound that rotates light towards right (clockwise; +)

Question 12

levorotatory

Answer 12

(L-) compound: Chiral compound that rotates light towards left (counterclockwise; -)

Question 13

d-glucose

Answer 13

Most abundant in nature
Most important source of human nutrition
Grape fruit and ripe fruits are good sources of glucose (20–30% by mass) 
Also named grape sugar
Other names 
Dextrose 
Blood sugar (70–100 mg/dL)
Six-membered cyclic form

Question 14

d- galactose

Answer 14

Milk sugar 
Synthesized in human beings 
Also called brain sugar
Part of brain and nerve tissue
Used to differentiate between blood types
Six-membered cyclic form

Question 15

D-Fructose

Answer 15

Ketohexose
Sweetest tasting of all sugars
Found in many fruits and in honey

Good dietary sugar due to
higher sweetness
Five-membered cyclic form

Question 16

D-Ribose

Answer 16

Part of a variety of complex molecules which include:
RNA
ATP
DNA
Five-membered cyclic form

Question 17

a form

Answer 17

the –OH of C1 and CH2OH of C5 are on opposite sides

Question 18

b form

Answer 18

where the –OH of C1 and CH2OH of C5 are on the same side

Question 19

anomers

Answer 19

Cyclic monosaccharide formation always produces two stereoisomers, an a and b form.
Cyclic monosaccharides that differ only in the position of the substituents on the anomeric carbon atom

Question 20

pyranose

Answer 20

Cyclic monosaccharide containing a six-atom ring

Question 21

furanose

Answer 21

Cyclic monosaccharide containing a five-atom ring

Question 22

haworth projection formula

Answer 22

Two-dimensional structural notation that specifies the three-dimensional structure of a cyclic form of a monosaccharide.
The hemiacetal ring is viewed “edge on” – the oxygen ring atom is always show as follows:
6-member ring  upper right corner
5-member ring  at the top

• D-CH2OH is above ring
• L- CH2OH down

Question 23

Haworth - B configuration

Answer 23

both groups point same direction

Question 24

Haworth- A configuration

Answer 24

2 groups point in opposite directions

Question 25

Haworth; OH

Answer 25

to the right- points down

to the left- points up

Question 26

5 reactions of monosaccharides

Answer 26

1. Oxidation to acidic sugars
2. Reduction to sugar alcohols
3. Glycoside formation
4. Phosphate ester formation
5. Amino sugar formation

Question 27

Oxidation to produce acidic sugars: weak agent

Answer 27

oxidize the aldehyde end to give an aldonic acid
Glucose is oxidized to gluconic acid

In Tollens test glucose reduces the Ag+ ion to Ag. The aldose (glucose) acts as a reducing agent so it is called a reducing sugar.
A reducing sugar is a carbohydrate that gives a positive test with the Tollens solution.
Ketoses are also reducing sugars: in this situation, the ketose undergoes a structural rearrangement that produces an aldose – and then the aldose reacts.
Therefore, (almost) all monosaccharides, both aldoses and ketoses, are reducing sugars.

Question 28

Oxidation to produce acidic sugars: strong agent

Answer 28

can oxidize both ends of a monosaccharides at the same time. The carbonyl group and the terminal primary alcohol are both oxidized to a carboxylic acid group.
A polyhydroxy dicarboxylic acid is called an aldaric acid.
For glucose, this oxidation produces glucaric acid:

Question 29

Oxidation to produce acidic sugars: enzymes

Answer 29

will oxidize only the primary alcohol end of an aldose, such as glucose, without oxidizing the aldehyde group.
This type of acid is called an alduronic acid.
For glucose, this oxidation produces glucuronic acid:

Question 30

reduction to produce sugar alcohols

Answer 30

The carbonyl group in a monosaccharide (either an aldose or a ketose) is reduced to a hydroxyl group using hydrogen as the reducing agent
The product is the corresponding polyhydroxy alcohol called sugar alcohol or alditol

Question 31

Glycloside formation

Answer 31

Glycoside: An acetal formed from a cyclic monosaccharide by replacement of the hemiacetal carbon –OH group with an –OR group
Cyclic forms of monosaccharides are hemiacetals so they react with alcohols to form acetals.
Glycoside is an acetal formed from a cyclic monosaccharide by replacement of the hemiacetal carbon –OH group with an –OR group
Exist in both α and β forms

Question 32

Phosphate ester formation

Answer 32

• Hydroxyl groups of a monosaccharide can react with inorganic oxyacids to form inorganic esters
• Enzymes catalyze the esterification of the hemiacetal group (C1) and the primary alcohol group (C6) in glucose to produce the following compounds:
• Phosphate esters of various monosaccharides are stable in aqueous solution and play important roles in the metabolism of carbohydrates

Question 33

amino suga formation

Answer 33

Amino sugar - Formed when one of the hydroxyl groups of a monosaccharide is replaced with an amino group

In naturally occurring amino sugars (the three below), the C2 hydroxyl group is replaced by an amino group

Question 34

maltose

Answer 34

Structurally made of 2 D-glucose units, one of which must be α-D-glucose, linked via an α(14) glycosidic linkage
Digested easily by humans because of an enzyme that can break α(14) linkages
Baby foods are rich in maltose

Question 35

cellobiose

Answer 35

Produced as an intermediate in the hydrolysis of the polysaccharide cellulose
Contains two D-glucose monosaccharide units, one of which must have a β configuration, linked through a β(14) glycosidic linkage
Cannot be digested by humans

Question 36

lactose

Answer 36

Made up of β-D-galactose unit and a D-glucose unit (two different monosaccharides) joined by a β(14) glycosidic linkage
Milk is rich in the disaccharide lactose
Lactase hydrolyzes β(14) glycosidic linkages

Question 37

Sucrose (table suger and a disaccharide)

Answer 37

The most abundant of all disaccharides and found in plants
Produced from the juice of sugar cane and sugar beets
Sugar cane contains up to 20% by mass sucrose
Sugar beets contain up to 17% by mass sucrose
Glycosidic linkage is not a (14) linkage. Instead it is an a,b(12) glycosidic linkage.