Biochemistry Chapter 6: Carbohydrate Structure

Question 1

Anomeric carbon glucose

Answer 1

C1 (where the carbonyl group is)

Question 2

Anomeric carbon fructose

Answer 2

C2 (where the carbonyl group is)

Question 3

Enantiomers

Answer 3

are mirror images, different in orientation at every chiral carbon

Question 4

diastereomers

Answer 4

They are molecules that have the same molecular formula and sequence of bonded atoms (constitution) but differ in the spatial arrangement of their atoms in a way that they are not mirror images of each other.

Question 5

Epimers

Answer 5

Diastereomers thats differ at 1 chiral center

Question 6

Carbons numbered starting from the

Answer 6

Terminal C=O or the end closest to it

Question 7

Answer 7

Hemiacetal

Question 8

Answer 8

Hemiketal

Question 9

Pentoses and hexoses can form cyclic forms featuring what kind of group?

Answer 9

Hemiacetal or hemiketal

Question 10

Which carbons connect to form a pyranose?

Answer 10

C1 and C5

Question 11

Which carbons join to form a furanose?

Answer 11

C1 and C4

Question 12

OH groups points _____ in alpha-anomers

Answer 12

DOWN

Question 13

OH groups points _____ in beta anomers

Answer 13

UP

Question 14

An anomer

Answer 14

is a type of stereoisomer that arises from the formation of cyclic structures in sugars. Specifically, anomers differ in the configuration at the anomeric carbon, which is the carbon derived from the carbonyl group (aldehyde or ketone) during the cyclization of a sugar.

Question 15

Glycosidic bond creates what?

Answer 15

Ketal/Acetal

Question 16

Sucrose is made up of what monosaccarides?

Answer 16

glucose + fructose, α1→β2 glycosidic bond.

Question 17

Lactose is made up of what monosaccarides?

Answer 17

galactose + glucose, β(1→4) glycosidic bond.

Question 18

Maltose is made up of what monosaccarides?

Answer 18

glucose + glucose, α(1→4) glycosidic bond.

Question 19

Amylose

Answer 19

linear glucose chain with α(1→4) bonds.

Question 20

Polysaccarides

Answer 20

long linear chains of glucose

Question 21

Amylopectin

Answer 21

like amylose, but with α(1→6) branches every 24–30 units.

Question 22

Starch (in foods) is made of what

Answer 22

20%–30% amylose + 70%–80% amylopectin.

Question 23

Glycogen

Answer 23

Polysaccaride. (energy storage in humans): α(1→4) bonds with α(1→6) branches every 8–12 units.

Question 24

Cellulose

Answer 24

indigestible fiber in plants; connected by β(1→4) bonds.

Question 25

Most common carb stereochemistry system

Answer 25

D/L system, based on orientation of D-glyceraldehyde — in a linear Fischer projection, a D-carbohydrate is one with the bottom-most -OH group pointing to the right.

Question 26

Virtually all biological carbohydrates are (D or L)?

Answer 26

D

Question 27

How to find number of stereoisomers

Answer 27

2^X = number of stereoisomers where X = number of chiral centers

Question 28

Each hexose will have how many enatiomers, epimers and diastereomers?

Answer 28

one enantiomer (differing by D vs. L). four epimers (one epimer for each stereocenter). 14 diastereomers.

Question 29

Reducing sugars

Answer 29

terminal C=O in aldoses can easily be oxidized to carboxylic acids. Aldoses are reducing agents/sugars. Reducing sugars = reducing agents

Question 30

disaccharides where an element can be converted into a linear aldose are

Answer 30

reducing sugars

Question 31

Are Ketose monosaccharides reducing sugars?

Answer 31

Yes because they can tautomerize to aldoses

Question 32

Is sucrose a reducing sugar?

Answer 32

No

Question 33

Are starch and polymers reducing sugars?

Answer 33

No because only one end of the sugar gets reduced

Question 34

Why sucrose alpha1 - beta2 glycosidic bond?

Answer 34

The glucose (C1) is in the α-configuration (OH group on the anomeric carbon is trans to the CH₂OH group). The fructose (C2) is in the β-configuration (OH group on the anomeric carbon is cis to the CH₂OH group).

Question 35

Anomeric Carbon

Answer 35

An anomeric carbon is a carbon atom in a sugar molecule that forms a stereocenter when the sugar cyclizes In the open-chain form of a sugar, the anomeric carbon is the carbonyl carbon of the aldehyde or ketone functional group. In the cyclic form of a sugar, the anomeric carbon is the carbon atom bonded to the ring oxygen and a hydroxyl group.

Question 36

The stereochemical designators α and β distinguish between:

Answer 36

epimers at anomeric carbon

Question 37

Difference between AMP and cAMP

Answer 37

AMP = standard nucleotide with 1 phosphate group cAMP = A modified form of AMP where the phosphate group forms a cyclic bond with both the 3' and 5' carbons of the ribose, creating a ring-like structure. AMP = Plays a role in energy metabolism and nucleotide synthesis. cAMP = second messenger t is synthesized from ATP by the enzyme adenylyl cyclase in response to signals like hormones (e.g., epinephrine, glucagon) and activates protein kinase A (PKA), leading to various cellular responses.

Question 38

α (Alpha) anomer → The OH on the anomeric carbon (C1) is

Answer 38

opposite (trans) to the CH₂OH group (C6).

Question 39

β (Beta) anomer → The OH on the anomeric carbon (C1) is on the

Answer 39

same side (cis) as the CH₂OH group (C6).

Question 40

Fischer projection: L Glucose

Answer 40

The D- or L- configuration is determined by the position of the OH group on the highest-numbered chiral carbon (which is C5 in glucose). L-glucose → OH on C5 is on the left.

Question 41

Fischer projection: D Glucose

Answer 41

The D- or L- configuration is determined by the position of the OH group on the highest-numbered chiral carbon (which is C5 in glucose). D-glucose → OH on C5 is on the Right.

Question 42

Hawthorn projection: L Glucose

Answer 42

In the Haworth projection, cyclization changes the orientation of the CH₂OH (C6) group L-glucose → CH₂OH (C6) is below the ring.

Question 43

Hawthorn projection: D Glucose

Answer 43

In the Haworth projection, cyclization changes the orientation of the CH₂OH (C6) group D-glucose → CH₂OH (C6) is above the ring.

Question 44

Difference between L, D and alpha beta for naming monosaccarides

Answer 44

The D- or L- configuration is determined by the position of the OH group on the highest-numbered chiral carbon (which is C5 in glucose). Alpha (α) and Beta (β) anomers are determined by the orientation of the OH group on the anomeric carbon (C1) when the sugar cyclizes into a ring form.

Question 45

bond that links monosaccharides together in an oligosaccharide is a special type of acetal linkage known as a

Answer 45

glycoside bond