Mono-, Di - and Polysaccharides.

Question 1

What is the general formula of a polysaccharide?

Answer 1

(CH2O)n

Question 2

What are the smallest possible polysaccharides?

Answer 2

Trioses.

Question 3

All monosaccharides have a carbonyl group, but they can be divided into…

Answer 3

Aldoses and ketoses.

Question 4

What are the two possible trioses?

Answer 4

Glyceraldehyde (an aldotriose)

Dihydroxyacetone (a ketotriose).

Question 5

Monosaccharides have chiral centres. Which chiral centre is used to determine whether it is the D (dextrorotatory) or L (levorotatory) enantiomer?

Answer 5

The chiral centre furthest from the carbonyl group.

Question 6

Since it is impossible to tell which direction a monosaccharide will rotate plane polarised light from its structure alone, which standard molecule is used to assign D- or L-?

Answer 6

Glyceraldehyde.

Question 7

Which enantiomer has the -OH group on the furthest chiral carbon from the carbonyl group pointing to the right?

Answer 7

D-

Question 8

Which enantiomer has the -OH group on the furthest chiral carbon from the carbonyl group pointing to the left?

Answer 8

L-

Question 9

Which enantiomer of monosaccharides is nearly always found in nature, due to the stereospecificity of enzymes?

Answer 9

D-

Question 10

What determines the different biochemical properties of different monosaccharides, despite the fact that they have the same molecular formula?

Answer 10

The arrangement of atoms around chiral centres other than that furthest from the carbonyl group.

Question 11

What term is used to describe monosaccharides that differ in only one chiral centre?

Answer 11

Epimers.

Question 12

Give an example of an epimer of D-glucose. What carbon do they differ at?

Answer 12

D-galactose. They differ at C4.

Question 13

What reaction converts monosaccharides from linear to ring structures?

Answer 13

Intramolecular cyclisation reactions.

Question 14

Describe the intramolecular cyclisation reaction of D-glucose to form D-glucopyranose.

Answer 14

Rotation around the bond between C4 and C5 allows the hydroxyl group on C5 to act as a nucleophile, reacting with the aldehyde group on C1. This forms the six-membered pyranose ring. This creates a new chiral centre at C1, called the anomeric carbon.

Question 15

Why are two different anomers of D-glucopyranose formed?

Answer 15

The hydroxyl group can attack from above or below the plane of the carbonyl group, so an alpha- and beta- form are created.

Question 16

In alpha-D-glucopyranose, how are the H and OH groups arranged on C4 and C1?

Answer 16

They are oriented the same on both carbons, with the H atom at the top and the OH atom at the bottom.

Question 17

In beta-D-glucopyranose, how are the H and OH groups arranged on C4 and C1?

Answer 17

They are oriented on opposite sides on each carbon. At C4, H is at the top, and OH at the bottom. On C1, OH is at the top, and H at the bottom.

Question 18

In solution, the open chain and cyclic forms of glucose are in equilibrium with one another - what is the ratio between them at 30 degrees celsius?

Answer 18

2/3 is in the form of beta-D-glucopyranose, 1/3 is in the form of alpha-D-glucopyranose, and there are only trace amounts in the open chain form.

Question 19

What is the most stable conformation for cyclic monosaccharides, and why?

Answer 19

The chair form is most stable. There are fewer steric clashes between the hydroxyl groups, as they point outwards and are further apart.

Question 20

How are disaccharides formed?

Answer 20

Condensation reaction between two monosaccharides.

Question 21

D-alpha-glucose x 2 = ?

Answer 21

Maltose

Question 22

alpha-D-glucose + beta-D-glucose = ?

Answer 22

Beta- anomer of maltose.

Question 23

D-beta-glucose x 2 = ?

Answer 23

Cellobiose.

Question 24

What type of glycosidic bond is found in maltose?

Answer 24

alpha 1,4 glycosidic bond.

Question 25

What type of glycosidic bond is found in cellobiose?

Answer 25

beta 1,4 glycosidic bond.

Question 26

Why is the second monomer flipped in cellobiose?

Answer 26

It is not energetically stable for the glycosidic bond to join one monomer from above and one from below.

Question 27

What are the two forms of starch?

Answer 27

Amylopectin.

Amylose.

Question 28

Describe amylose.

Answer 28

Unbranched. All of the glycosidic bonds are alpha 1,4 linkages. It is compact and helical.

Question 29

Describe amylopectin.

Answer 29

Branched.
Primarily alpha 1,4 linkages, but with alpha 1,6 branch points approximately every 30 residues.
Slightly curved.

Question 30

What is the monosaccharide that makes up starch and glycogen?

Answer 30

Alpha-D-glucopyranose.

Question 31

What is the monosaccharide that makes up cellulose?

Answer 31

Beta-D-glucopyranose.

Question 32

Describe glycogen.

Answer 32

Highly branched.
Primarily alpha 1,4 linkages, but alpha 1,6 linkages approximately every 10 residues.
Due to branching, it does not pack together well, so is less compact.

Question 33

What is the benefit of glycogen’s high amount of branching?

Answer 33

It has lots of ends so can be hydrolysed more quickly, for faster energy release.

Question 34

Describe cellulose.

Answer 34

Unbranched.
Only beta-1,4 glycosidic bonds.
Chains pack together tightly and are cross-linked by hydrogen bonds, forming rigid structures.

Question 35

All monosaccharides are reducing sugars - why?

Answer 35

They have a free reactive carbonyl group that can react with oxidising agents.

Question 36

What is necessary for a disaccharide to be a reducing sugar?

Answer 36

They require a reducing end - a free anomeric carbon.

Question 37

Why is sucrose NOT a reducing sugar?

Answer 37

It does not have a free anomeric carbon.

Question 38

What property do the different ends of a polysaccharide give it?

Answer 38

Directionality.

Question 39

How can monosaccharides be modified?

Answer 39

Substitution of hydroxyl groups for other functional groups that alter the properties of the monosaccharide and its polysaccharides.

Question 40

Chitin is a beta-1,4 polymer of modified monosaccharides. What are the monomers of chitin?

Answer 40

N-acetylglucosamine.

Question 41

Where is chitin found?

Answer 41

The cell walls of fungi and the exoskeletons of arthropods.

Question 42

What is the group substituted for the hydroxyl group on carbon 2 in chitin?

Answer 42

Acetamide (ethanamide).

Question 43

What type of modified polysaccharide is found in synovial fluid between joints?

Answer 43

Glycosaminoglycans.

Question 44

What is hyaluronic acid?

Answer 44

A beta-1,4 linked polymer of a beta-1,3 linked dimer of D-glucuronic acid and N-acetyl-D-glucosamine.

Question 45

What makes hyaluronic acid good as part of synovial fluid?

Answer 45

The negatively charged carboxylate group on the D-glucuronic acid monomer (C6) means that chains of hyaluronic acid repel each other and are highly hydrated, making the fluid more viscous.

Question 46

What are oligosaccharides?

Answer 46

Polysaccharides with only a few monomers.

Question 47

What are the main roles of oligosaccharides?

Answer 47

Often attached to proteins to form glycoproteins, which are then used as receptors, protection against proteases and to help cells stick to one another.

Question 48

Why are oligosaccharides good for specificity?

Answer 48

They can be added to proteins in many different orders, so can provide many different receptors for different molecules.

Question 49

What is added to all proteins that enter the ER during protein synthesis?

Answer 49

A precursor oligosaccharide.