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Example Questions
Example Question #31 : Identifying Biochemical Molecules
What is the formula for a polysaccharide made of 150 glucose monomers?
None of these
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 #1 : Identifying Monosaccharides
Identify the D-sugar and its anomeric designation (alpha or beta).
Alpha fructose
Beta galactose
Beta glucose
Alpha galactose
Alpha glucose
Alpha glucose
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 #32 : Identifying Biochemical Molecules
Identify the D-sugar and its anomeric designation (alpha or beta).
Alpha glucose
Alpha fructose
Beta fructose
Beta glucose
Alpha galactose
Beta glucose
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 #3 : Identifying Monosaccharides
Identify the D-sugar and its anomeric designation (alpha or beta).
Alpha galactose
Alpha glucose
Beta fructose
Beta galactose
Alpha fructose
Alpha fructose
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 #2 : Identifying Monosaccharides
Identify the D-sugar and its anomeric designation (alpha or beta).
Beta galactose
Beta fructose
Beta glucose
Alpha fructose
Alpha galactose
Beta fructose
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 #33 : Identifying Biochemical Molecules
Identify the D-sugar and its anomeric designation (alpha or beta).
Alpha glucose
Beta glucose
Beta galactose
Alpha galactose
Beta fructose
Alpha galactose
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 #34 : Identifying Biochemical Molecules
Identify the D-sugar and its anomeric designation (alpha or beta).
Alpha galactose
Alpha glucose
Beta glucose
Beta galactose
Beta fructose
Beta galactose
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 #2 : Help With Organic Carbohydrates
Which of the following structures represents the anomeric alpha ring structure of D-glucose?
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 #3 : Help With Organic Carbohydrates
Which of the following structures represents the anomeric alpha ring structure of D-galactose?
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 #2 : Help With Organic Carbohydrates
Which of the following ring structures represents the anomeric alpha ring structure of D-mannose?
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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