Identifying Monosaccharides

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Biochemistry › Identifying Monosaccharides

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1

L-Glucose is found as 2 different hexose stereoisomers in solution: and glucose.

Suppose both and L-glucose were polymerized into long glycosidic linkages. Which of the two polymers could be catabolized (broken down into component parts) by a human?

Neither stereoisomer

CORRECT

-L-glucose

0

-L-glucose

0

Both stereoisomers

0

Explanation

In nature, L-sugars are indigestible by humans. In the case of L-glucose, even though our bodies can break down polymers of -D-glucose, it cannot digest L-glucose.

2

Identify the D-sugar and its anomeric designation (alpha or beta).

B galactose

Beta galactose

CORRECT

Alpha galactose

0

Alpha glucose

0

Beta glucose

0

Beta fructose

0

Explanation

The anomeric designation is beta because the hydroxyl group attached to carbon 1—the carbon adjacent to the heterocyclic oxygen with the highest priority substituent—is cis to the group attached to carbon 5.

The structure represents galactose because the hydroxyl groups attached to carbons 2, 3, and 4 are positioned trans, cis, and cis (respectively) with respect to the attached to carbon 5.

3

Identify the D-sugar and its anomeric designation (alpha or beta).

B fructose

Beta fructose

CORRECT

Alpha fructose

0

Beta glucose

0

Alpha galactose

0

Beta galactose

0

Explanation

The anomeric designation is beta because the hydroxyl group attached to carbon 2—the carbon adjacent to the heterocyclic oxygen with the highest priority substituent—is cis to the group attached to carbon 5.

The structure represents fructose because the hydroxyl groups attached to carbons 2, 3, and 4 of the pentose ring are positioned trans, cis, and trans (respectively) with respect to the attached to carbon 5 and there is a group attached to carbon 2 along with the hydroxyl group.

4

Identify the D-sugar and its anomeric designation (alpha or beta).

B glucose

Beta glucose

CORRECT

Alpha galactose

0

Alpha fructose

0

Beta fructose

0

Alpha glucose

0

Explanation

The anomeric designation is beta because the hydroxyl group attached to carbon 1—the carbon adjacent to the heterocyclic oxygen with the highest priority substituent—is cis to the group attached to carbon 5.

The structure represents glucose because the hydroxyl groups attached to carbons 2, 3, and 4 are positioned trans, cis, and trans (respectively) with respect to the attached to carbon 5.

5

Identify the D-sugar and its anomeric designation (alpha or beta).

A glucose

Alpha glucose

CORRECT

Alpha galactose

0

Beta glucose

0

Beta galactose

0

Alpha fructose

0

Explanation

The anomeric designation is alpha because the hydroxyl group attached to carbon 1—the carbon adjacent to the heterocyclic oxygen with the highest priority substituent—is trans to the group attached to carbon 5.

The structure represents glucose because the hydroxyl groups attached to carbons 2, 3, and 4 are positioned trans, cis, and trans (respectively) with respect to the attached to carbon 5.

6

Which of the following compounds is not a monosaccharide?

Lactose

CORRECT

Glucose

0

Ribose

0

Galactose

0

Fructose

0

Explanation

Out of all the sugars listed, lactose is the only one that is not a monosaccharide. Lactose is composed of a glucose and galactose bound together in a glycosidic linkage.

7

Which sugar is another name for "blood sugar?"

Glucose

CORRECT

Sucrose

0

Fructose

0

Lactose

0

Explanation

Glucose is also called "blood sugar" because it is the main sugar found in our blood. Sucrose is what we know as "table sugar" because it is commonly found in our kitchens and used in baking etc. Fructose is also known as "fruit sugar" because it occurs in naturally-occurring fruits, giving them their sweetness. Lactose is sometimes called "milk sugar" because it is found in milk.

8

Identify the D-sugar and its anomeric designation (alpha or beta).

A galactose

Alpha galactose

CORRECT

Beta galactose

0

Alpha glucose

0

Beta glucose

0

Beta fructose

0

Explanation

The anomeric designation is alpha because the hydroxyl group attached to carbon 1—the carbon adjacent to the heterocyclic oxygen with the highest priority substituent—is trans to the group attached to carbon 5.

The structure represents galactose because the hydroxyl groups attached to carbons 2, 3, and 4 are positioned trans, cis, and cis (respectively) with respect to the attached to carbon 5.

9

What is the formula for a polysaccharide made of 150 glucose monomers?

CORRECT

0

0

None of these

0

0

Explanation

A polysaccharide of 150 monosaccharides must have 149 glycosidic bonds. The formation of one glycosidic linkage results in the removal of one water molecule. We can find the answer by determining the number of each atom in 150 glucose molecules and subtracting the atoms found in 149 water molecules.

10

Identify the D-sugar and its anomeric designation (alpha or beta).

A fructose

Alpha fructose

CORRECT

Beta fructose

0

Alpha glucose

0

Beta galactose

0

Alpha galactose

0

Explanation

The anomeric designation is alpha because the hydroxyl group attached to carbon 2—the carbon adjacent to the heterocyclic oxygen with the highest priority substituent—is trans to the group attached to carbon 5.

The structure represents fructose because the hydroxyl groups attached to carbons 2, 3, and 4 of the pentose ring are positioned trans, cis, and trans (respectively) with respect to the attached to carbon 5 and there is a group attached to carbon 2 along with the hydroxyl group.