Biochemistry : Identifying Monosaccharides

Study concepts, example questions & explanations for Biochemistry

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Example Questions

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Example Question #31 : Identifying Biochemical Molecules

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

Possible Answers:

None of these

Correct answer:

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.

Example Question #32 : Identifying Biochemical Molecules

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

A glucose

Possible Answers:

Beta glucose

Alpha fructose

 Alpha glucose

Alpha galactose

Beta galactose

Correct answer:

 Alpha glucose

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.

Example Question #33 : Identifying Biochemical Molecules

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

B glucose

Possible Answers:

Alpha glucose

Alpha galactose

Beta fructose

Alpha fructose

Beta glucose

Correct answer:

Beta glucose

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.

Example Question #34 : Identifying Biochemical Molecules

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

A fructose

Possible Answers:

Alpha galactose

Beta fructose

Beta galactose

Alpha glucose

Alpha fructose

Correct answer:

Alpha fructose

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.

Example Question #35 : Identifying Biochemical Molecules

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

 B fructose

Possible Answers:

Alpha galactose

Beta glucose

Beta galactose

Beta fructose

Alpha fructose

Correct answer:

Beta fructose

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.

Example Question #1 : Identifying Monosaccharides

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

 A galactose

Possible Answers:

Alpha glucose

Beta galactose

Beta fructose

Beta glucose

Alpha galactose

Correct answer:

Alpha galactose

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.

Example Question #37 : Identifying Biochemical Molecules

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

B galactose

Possible Answers:

Alpha glucose

Beta fructose

Beta glucose

Beta galactose

Alpha galactose

Correct answer:

Beta galactose

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.

Example Question #38 : Identifying Biochemical Molecules

Which of the following structures represents the anomeric alpha ring structure of D-glucose? 
Linear glucose

Possible Answers:

Alpha mannose

Alpha galactose

Alpha glucose

Alpha ribose

Alpha fructose

Correct answer:

Alpha glucose

Explanation:

When converting a linear sugar to its ring form, a bond is formed between the oxygen attached to carbon 5 and the carbon at position 1. All hydroxyl groups that are not attached to the carbon in position 1 and are oriented to the right end up trans to the  attached to carbon 5, while those that are in the left position end up cis to the  attached to carbon 5.

If the hydroxyl group attached to carbon 1 ends up trans to the  attached to carbon 5, the ring structure is considered alpha. If the hydroxyl group attached to carbon 1 is cis to the  attached to carbon 5, the ring structure is considered beta.

The alpha ring structure of D-glucose bonds the carbon 1 hydroxyl group trans to the carbon 5  group. The hyroxyl groups on carbons 2, 3, and 4 will be trans, cis, and trans with respect to the .

Example Question #264 : Organic Chemistry

Which of the following structures represents the anomeric alpha ring structure of D-galactose? 

Linear galactose

Possible Answers:

Alpha glucose

Alpha ribose

Alpha galactose

Alpha fructose

Alpha mannose

Correct answer:

Alpha galactose

Explanation:

When converting a linear sugar to its ring form, a bond is formed between the oxygen attached to carbon 5 and the carbon at position 1. All hydroxyl groups that are not attached to the carbon in position 1 and are oriented to the right end up trans to the  attached to carbon 5, while those that are in the left position end up cis to the  attached to carbon 5.

If the hydroxyl group attached to carbon 1 ends up trans to the  attached to carbon 5, the ring structure is considered alpha. If the hydroxyl group attached to carbon 1 is cis to the  attached to carbon 5, the ring structure is considered beta.

The alpha ring structure of D-galactose bonds the carbon 1 hydroxyl group trans to the carbon 5  group. The hyroxyl groups on carbons 2, 3, and 4 will be trans, cis, and cis with respect to the .

Example Question #251 : Biochemistry

Which of the following ring structures represents the anomeric alpha ring structure of D-mannose? 

Linear mannose

Possible Answers:

Alpha ribose

Alpha mannose

Alpha glucose

Alpha fructose

Alpha galactose

Correct answer:

Alpha mannose

Explanation:

When converting a linear sugar to its ring form, a bond is formed between the oxygen attached to carbon 5 and the carbon at position 1. All hydroxyl groups that are not attached to the carbon in position 1 and are oriented to the right end up trans to the  attached to carbon 5, while those that are in the left position end up cis to the  attached to carbon 5.

If the hydroxyl group attached to carbon 1 ends up trans to the  attached to carbon 5, the ring structure is considered alpha. If the hydroxyl group attached to carbon 1 is cis to the  attached to carbon 5, the ring structure is considered beta.

The alpha ring structure of D-mannose bonds the carbon 1 hydroxyl group trans to the carbon 5  group. The hyroxyl groups on carbons 2, 3, and 4 will be cis, cis, and trans with respect to the .

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