Identifying Monosaccharides - Biochemistry
Card 1 of 68
Which of the following is not a monosaccharide?
Which of the following is not a monosaccharide?
Tap to reveal answer
While glucose, galactose, and fructose are all monosaccharides, lactose is a disaccharide comprised of two monosaccharides, glucose and galactose, joined by a 
-1,4-glycosidic bond.
While glucose, galactose, and fructose are all monosaccharides, lactose is a disaccharide comprised of two monosaccharides, glucose and galactose, joined by a
← Didn't Know|Knew It →
What is the formula for a polysaccharide made of 150 glucose monomers?
What is the formula for a polysaccharide made of 150 glucose monomers?
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
Identify the D-sugar and its anomeric designation (alpha or beta).
Identify the D-sugar and its anomeric designation (alpha or beta).
Tap to reveal answer
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.
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
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
← Didn't Know|Knew It →
Identify the D-sugar and its anomeric designation (alpha or beta).
Identify the D-sugar and its anomeric designation (alpha or beta).
Tap to reveal answer
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.
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
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
← Didn't Know|Knew It →
Identify the D-sugar and its anomeric designation (alpha or beta).
Identify the D-sugar and its anomeric designation (alpha or beta).
Tap to reveal answer
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.
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
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
← Didn't Know|Knew It →
Identify the D-sugar and its anomeric designation (alpha or beta).
Identify the D-sugar and its anomeric designation (alpha or beta).
Tap to reveal answer
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.
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
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
← Didn't Know|Knew It →
Identify the D-sugar and its anomeric designation (alpha or beta).
Identify the D-sugar and its anomeric designation (alpha or beta).
Tap to reveal answer
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.
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
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
← Didn't Know|Knew It →
Identify the D-sugar and its anomeric designation (alpha or beta).
Identify the D-sugar and its anomeric designation (alpha or beta).
Tap to reveal answer
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.
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
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
← Didn't Know|Knew It →
Which of the following structures represents the anomeric alpha ring structure of D-glucose?
Which of the following structures represents the anomeric alpha ring structure of D-glucose?
Tap to reveal answer
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 
.
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
If the hydroxyl group attached to carbon 1 ends up trans to the
The alpha ring structure of D-glucose bonds the carbon 1 hydroxyl group trans to the carbon 5
← Didn't Know|Knew It →
Which of the following structures represents the anomeric alpha ring structure of D-galactose?
Which of the following structures represents the anomeric alpha ring structure of D-galactose?
Tap to reveal answer
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 .
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
If the hydroxyl group attached to carbon 1 ends up trans to the
The alpha ring structure of D-galactose bonds the carbon 1 hydroxyl group trans to the carbon 5
← Didn't Know|Knew It →
Which of the following ring structures represents the anomeric alpha ring structure of D-mannose?
Which of the following ring structures represents the anomeric alpha ring structure of D-mannose?
Tap to reveal answer
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 .
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
If the hydroxyl group attached to carbon 1 ends up trans to the
The alpha ring structure of D-mannose bonds the carbon 1 hydroxyl group trans to the carbon 5
← Didn't Know|Knew It →
Identify the aldose pictured, including its alpha or beta designation.
Identify the aldose pictured, including its alpha or beta designation.
Tap to reveal answer
The structure pictured is mannose because the hydroxyl groups at carbons 2, 3, and 4 are situated cis, cis, and trans (respectively) to the 
attached to carbon 5.
The mannose pictured is in alpha form because the hydroxyl group at carbon 1 is trans to the 
attached to carbon 5.
The structure pictured is mannose because the hydroxyl groups at carbons 2, 3, and 4 are situated cis, cis, and trans (respectively) to the
The mannose pictured is in alpha form because the hydroxyl group at carbon 1 is trans to the
← Didn't Know|Knew It →
The Fischer projection pictured is a form of glucose. The carbon labeled "x" is the chiral carbon farthest away from carbon 1 and the hydroxyl group connected to carbon "x" is on the right. This fact designates that the glucose as what configuration?
The Fischer projection pictured is a form of glucose. The carbon labeled "x" is the chiral carbon farthest away from carbon 1 and the hydroxyl group connected to carbon "x" is on the right. This fact designates that the glucose as what configuration?
Tap to reveal answer
The chiral carbon farthest away from carbon 1 is designated as "D" if its hydroxyl group is on the right side in the Fischer projection. In other words, this is D-glucose because the hyroxyl group on carbon "x" is oriented to the right.
The chiral carbon farthest away from carbon 1 is designated as "D" if its hydroxyl group is on the right side in the Fischer projection. In other words, this is D-glucose because the hyroxyl group on carbon "x" is oriented to the right.
← Didn't Know|Knew It →
What is the name of the aldose pictured in this Fischer projection?
What is the name of the aldose pictured in this Fischer projection?
Tap to reveal answer
The structure is D-ribose because it is a five-carbon aldose with the hydroxyl groups on carbons 2, 3, and 4 all on the right in the Fischer projection.
The structure is D-ribose because it is a five-carbon aldose with the hydroxyl groups on carbons 2, 3, and 4 all on the right in the Fischer projection.
← Didn't Know|Knew It →
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?
L-Glucose is found as 2 different hexose stereoisomers in solution:
Suppose both
Tap to reveal answer
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.
In nature, L-sugars are indigestible by humans. In the case of L-glucose, even though our bodies can break down polymers of
← Didn't Know|Knew It →
Which of the following compounds is not a monosaccharide?
Which of the following compounds is not a monosaccharide?
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
Which sugar is another name for "blood sugar?"
Which sugar is another name for "blood sugar?"
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
What is the formula for a polysaccharide made of 150 glucose monomers?
What is the formula for a polysaccharide made of 150 glucose monomers?
Tap to reveal answer
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.
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.
← Didn't Know|Knew It →
Identify the D-sugar and its anomeric designation (alpha or beta).
Identify the D-sugar and its anomeric designation (alpha or beta).
Tap to reveal answer
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.
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
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
← Didn't Know|Knew It →
Identify the D-sugar and its anomeric designation (alpha or beta).
Identify the D-sugar and its anomeric designation (alpha or beta).
Tap to reveal answer
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.
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
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
← Didn't Know|Knew It →