Biological Molecules

Question 1

What features define a carbohydrate?

Answer 1

It is an organic compound with a backbone of at least 3 carbons, typically with a C:H:O ratio of 1:2:1. Carbohydrates consist of either aldehydes or ketones with other carbons in the chain bound to hydroxyl groups.

Exceptions to the 1:2:1 ratio exist for “deoxy” sugars, in which a hydroxyl group has been replaced by a hydrogen.

Question 2

Carbohydrate structure can be shown in many ways. What name is given to the method of depiction below?

Answer 2

Fischer projection - straight-chain carbohydrates are often shown in this form.

On any Fischer projection, horizontal lines are considered to point out of the page (“wedges”), while vertical lines point into the page (“dashes”). Thus, this method can accurately convey stereochemistry.

Question 3

Carbohydrate structure can be shown in many ways. What name is given to the method of depiction below?

Answer 3

Haworth projection - carbohydrates in the ring form are often shown in this manner.

The relative positions of substituents as “cis” or “trans” can be determined by whether they point upward or downward from the face of the ring.

Question 4

What is the difference between an aldose and a ketose?

Answer 4

• Aldoses are sugar molecules that contain a single aldehyde group.
• Ketoses contain a single ketone.

Note that both aldoses and ketoses are monosaccharides.

Question 5

Briefly describe the steps for naming a straight-chain carbohydrate.

Answer 5

1. The compound is first classified as either an aldose or a ketose.
2. Next, the number of carbons is counted (di, tri, quatr, pent) and included after the aldo- or keto- prefix.
3. Finally, the -ose suffix, used to denote a carbohydrate, is attached to the name.

Note that there should be no spaces between the prefix, numerical identifier, and suffix.

Question 6

According to simple carbohydrate nomenclature, what general name can be given to this molecule?

Answer 6

aldopentose

This carbohydrate has an aldehyde group, which should be denoted by the prefix “aldo-.” As a five-carbon sugar, it should be named using the Greek term “pent.” Finally, the “-ose” suffix identifies the molecule as a carbohydrate.

Question 7

How are cyclic carbohydrates named?

Answer 7

They are named based on the number of members in the ring, its two main categories are:

• five-membered furanoses
• six-membered pyranoses

To denote a carbohydrate in the ring form, remove “-se” from the carbohydrate name and replace it with the suffix “-furanose” or “-pyranose,” as is appropriate.

Question 8

With regard to its cyclic structure, what general name can be given to this molecule?

Answer 8

pyranose

The carbohydrate has a six-membered ring form, so it is a pyranose. Five-membered rings are known as furanoses.

Question 9

What is the common name of this molecule?

Answer 9

glucose

Glucose, a hexose, is the most common carbohydrate tested on the MCAT. Its Fischer projection can be identified by the orientation of its hydroxyl groups. From top to bottom, the -OH groups will be oriented either “right, left, right, right” (D-glucose) or “left, right, left, left” (L-glucose).

Question 10

What is the common name of this molecule?

Answer 10

fructose

Fructose, a hexose, is often mentioned on the MCAT. Its Fischer projection can be identified by the orientation of its hydroxyl groups. From top to bottom, the -OH groups will be oriented either “left, right, right” (D-fructose) or “right, left, left” (L-fructose).

Question 11

What are the three main structural ways in which carbohydrates are classified?

Answer 11

1. As aldoses (aldehyde-containing carbohydrates) or ketoses (ketone-containing carbohydrates)
2. As furanoses (five-membered rings) or pyranoses (six-membered rings)
3. As regular sugars (those with oxygen atoms bound to each carbon) or deoxy sugars (those missing one oxygen atom)

Question 12

What is the difference between a monosaccharide and a disaccharide?

Answer 12

• monosaccharide is composed of one cyclic sugar subunit.
• disaccharide is composed of two cyclic sugars connected by a glycosidic linkage.

A polysaccharide, then, consists of more than two sugar subunits bound together.

Question 13

Name three common monosaccharides.

Answer 13

1. glucose
2. fructose
3. galactose

Ribose and deoxyribose are also monosaccharides.

Monosaccharides can be connected by glycosidic linkages to form longer sugar chains.

Question 14

Name three common disaccharides.

Answer 14

1. sucrose
2. lactose
3. maltose

In general, disaccharides consist of two sugar subunits bound by a glycosidic linkage.

Question 15

Which sugar monomers comprise sucrose, lactose, and maltose, respectively?

Answer 15

• Sucrose is composed of fructose and glucose
• Lactose is composed of glucose and galactose
• Maltose is composed of two glucose monomers

Question 16

Which carbon is used to determine the absolute configuration of a carbohydrate?

Answer 16

It is determined by the chirality of the carbon stereocenter that is farthest from the carbonyl carbon.

In other words, the absolute configuration depends on the highest-numbered chiral carbon in the molecule.

Question 17

What determines whether the absolute configuration of a carbohydrate is D or L?

Answer 17

Any carbohydrate depicted as a Fischer projection:

• is classified as a D sugar if the -OH on the highest-numbered chiral carbon is pointing to the right.
• is classified as an L sugar if that same -OH is instead pointing to the left.

Question 18

What is the absolute configuration of this carbohydrate?

Answer 18

TD configuration

Look at the -OH on the chiral carbon farthest from the carbonyl group. Since that -OH is pointing to the right on the Fischer projection, this must be a D sugar.

Question 19

What are epimers?

Answer 19

These are two molecules that differ in configuration at a single chiral carbon.

Epimers are a special type of diastereomer. They are especially important in relation to sugars.

Question 20

What type(s) of isomer are the carbohydrates below?

Answer 20

These are epimers, since they differ in configuration at only one position: the third carbon from the carbonyl.

Like all epimers, these molecules are also diastereomers.

Question 21

Describe the stereochemical relationship between carbohydrates 1, 2, and 3.

Answer 21

• 1 and 2 are epimers differing at the chiral carbon closest to the carbonyl. Like all epimers, they are also diastereomers.
• 2 and 3 are enantiomers, or nonsuperimposable mirror images of each other. They differ in configuration at all chiral centers.
• 1 and 3 are diastereomers, since they differ in configuration at 3 of 4 chiral carbons. Diastereomers are stereoisomers which differ at some (but not all) stereocenters.

Question 22

Describe the stereochemical relationship between carbohydrates 1, 2, and 3.

Answer 22

• 1 and 2 are epimers differing at the chiral carbon closest to the carbonyl. Like all epimers, 1 and 2 are also diastereomers.
• 1 and 3 are epimers differing at the chiral carbon second from the carbonyl. Again, 1 and 3 are also diastereomers.
• 2 and 3 are diasteromers, as they differ at two of their three stereocenters (the chiral carbons one and two positions away from the carbonyl). Diastereomers are stereoisomers that differ at one or more, but not all, stereocenters.

Question 23

Two carbohydrates are structurally identical but differ in configuration at three of their four chiral centers. What type(s) of isomer are these molecules?

Answer 23

These are diastereomers, a broad term for stereoisomers that are not enantiomers.

Specifically, these molecules cannot be enantiomers since they do not have opposite configurations at all of their chiral carbons. They also are not epimers, which only differ in configuration at one position.

Question 24

For a carbohydrate to isomerize between the straight-chain and ring forms, what reaction must take place?

Answer 24

In the straight-chain form, the -OH on the chiral carbon farthest from the carbonyl acts as a nucleophile and attacks the carbonyl carbon. This forms a cyclic structure.

In the proccess, the attacking hydroxyl loses its proton and the carbonyl oxygen becomes protonated.

Question 25

What name is given to the most oxidized carbon in the ring form of a carbohydrate?

Answer 25

This is called the anomeric carbon. Its stereochemical configuration is another way by which carbohydrates are classified.

Abpve, the anomeric carbon is bound to a hydroxyl group and to the oxygen member of the ring. It derives from the carbonyl carbon in the straight-chain form.

Question 26

In the ring form of a carbohydrate, how can the anomeric carbon be identified?

Answer 26

It is the only carbon that is bound to two oxygen atoms.

For this reason, all anomeric carbons are part of another functional group: acetals and hemiacetals (or ketals and hemiketals).

Question 27

What is an acetal?

Answer 27

It is a functional group that is derived from a carbonyl compound. It consists of two -OR groups bound to the same carbon.

Conventionally, “acetal” has referred to compounds derived from aldehydes, while “ketal” refers to those derived from ketones. The MCAT will likely use “acetal” to refer to both groups.

Question 28

What reactants are necessary to form an acetal?

Answer 28

It is formed when an aldehyde is reacted with two equivalents of alcohol compounds. The -OH acts as the nucleophile and attacks the aldehyde carbon, as shown below.

Question 29

What is the difference between an acetal and a hemiacetal?

Answer 29

• The carbon on an acetal is bound to two -OR groups.
• The carbon on a hemiacetal is bound to one -OR and one -OH.

“Hemi-“ means “half,” so a hemiacetal is simply a half-formed acetal. In other words, it is the product of a carbonyl compound and a single alcohol.

Question 30

What functional group is circled in the diagram below?

Answer 30

This is a hemiacetal. It contains a single carbon (the anomeric carbon) bound to one -OR and one -OH group.

The molecule shown is cyclic glucose, a pyranose.

Question 31

What functional group is circled in the diagram below?

Answer 31

This is an acetal. It contains a single carbon (the anomeric carbon) bound to two -OR groups. Often, the hemiacetal functionality on a monosaccharide becomes an acetal to form a glycosidic linkage.

The molecule shown is maltose, a disaccharide composed of two glucose subunits.

Question 32

What is the difference between the alpha and beta anomers of a carbohydrate?

Answer 32

In a Haworth projection:

• the -OH on the alpha anomer is trans to the carbon substituent on the ring
• the -OH on the beta anomer is cis to the carbon substituent on the ring

Question 33

Define:

mutarotation

Answer 33

It is the equilibration of alpha and beta anomers in solution. During this process, the optical rotation of the mixture changes.

Mutarotation occurs due to the opening and closing of carbohydrate rings. When the carbohydrate opens, its carbonyl group can then be attacked from a different side, forming either the alpha or beta anomer.

Question 34

What type of linkage does the following disaccharide contain?

Answer 34

alpha linkage

The nomenclature is determined by the anomeric carbon participating in the linkage. Here, the left carbohydrate’s anomeric carbon is linked to the right carbohydrate via an oxygen atom. Since the configuration of the left anomeric carbon is alpha, this is an alpha linkage.

Question 35

What reaction is used to break the linkage between carbohydrate monomers?

Answer 35

Glycosidic linkages are broken via hydrolysis.

As a nucleophile, water can attack the anomeric carbon in the linkage; the other carbohydrate will act as the leaving group. This reaction is a specific example of the hydrolysis of an acetal.

Question 36

Which groups on a monosaccharide can be altered by a mild oxidation reaction?

Answer 36

The aldehyde group on an aldose can be oxidized to a carboxylic acid, as can the terminal carbon.

Tollen’s reagent can only react with the aldose’s aldehyde group. In contrast, nitric acid can oxidize both the aldose aldehyde and the terminal carbon to carboxylic acids.

Question 37

What product is formed by the reduction of a monosaccharide?

Answer 37

polyalcohol

The aldehyde or ketone group will be reduced to a hydroxy group. The existing hydroxy groups will not be reduced further.

Question 38

What do Benedict’s and Tollens’ test detect?

Answer 38

These detect the presence of reducing sugars. These are carbohydrates that can act as reducing agents; in other words, they can oxidize themelves.

Question 39

What reagents are involved in Benedict’s test, and what does a positive result look like?

Answer 39

It produces a dark red precipitate. The reagent involved is a mixture of sodium citrate, sodium carbonate, and copper sulfate pentahydrate.

Question 40

What reagents are involved in Tollens’ test, and what does a positive result look like?

Answer 40

It yields a silver mirror-like product. The reagent involved is a mixture of ammonia and silver nitrate.

Question 41

Which components of Benedict’s and Tollens’ reagents directly facilitate the detection of reducing sugars?

Answer 41

These contain metal ions, which can be reduced by compounds that are prone to oxidation. Reducing sugars are examples of such compounds.

The metal ion in Benedict’s test is Cu²⁺, while that in Tollens’ test is Ag⁺.

Question 42

In a ring structure, the presence of which functional group categorizes a carbohydrate as a reducing sugar?

Answer 42

These contains hemiacetals (or hemiketals). Since they contain an -OH group, they can further oxidize when that -OH is replaced by an -OR.

Note that reducing sugars do not reduce themselves - they oxidize, thus acting as reducing agents! In a straight-chain form, aldehydes are the classic marker of a reducing sugar, as they can easily oxidize to carboxylic acids.

Question 43

Which of the molecules shown below is classified as a reducing sugar?

Answer 43

Molecule B is a reducing sugar. Unlike Molecule A, it contains a hemiacetal group, and is thus able to further oxidize.

Molecule B is maltose, a disaccharide. Molecule A is sucrose and is also a disaccharide.

Question 44

Which new functional group is created by the reaction of a carbohydrate with acetic anhydride in pyridine?

Assume the acetic anhydride is present in excess.

Answer 44

ester

All of the carbohydrate’s free hydroxyl groups are able to act as nucleophiles. They attack the carbonyl carbon of the anhydride, resulting in acylation at those positions.