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Practice Questions

The formation of a glycosidic bond during polysaccharide synthesis is a classic example of a

A. Hydrolysis reaction requiring a water molecule
B. Condensation reaction releasing a water molecule
C. Isomerization reaction rearranging functional groups
D. Redox reaction transferring electrons

The formation of a glycosidic bond between two monosaccharides is a dehydration synthesis (condensation). The hydroxyl group from one sugar's anomeric carbon combines with a hydrogen from a hydroxyl of another sugar, releasing one water molecule and forming a new covalent bond.

nmdcat.online BIO NMDCAT
Jun 27, 2026

In Benedict’s test, a positive result for a reducing sugar is indicated by the formation of a brick-red precipitate of

A. Copper(I) oxide (Cu₂O)
B. Copper(II) sulfate (CuSO₄)
C. Silver oxide (Ag₂O)
D. Lead acetate (Pb(C₂H₃O₂)₂)

The free aldehyde or ketone group of a reducing sugar reduces the blue Cu²⁺ ions (cupric) in Benedict's reagent to Cu⁺ ions (cuprous). These precipitate out of the alkaline solution as an insoluble brick-red solid, copper(I) oxide (Cu₂O).

nmdcat.online BIO NMDCAT
Jun 27, 2026

The function of cellulose as a structural polysaccharide in plant cell walls is directly related to its

A. Highly branched, amorphous structure
B. Linear chains of glucose linked by β-1,4 bonds, forming strong microfibrils via inter-chain hydrogen bonds
C. High solubility in water, allowing it to form a gel matrix
D. Ability to be easily hydrolyzed by a wide range of digestive enzymes

Cellulose is a linear, unbranched homopolymer of glucose. The β-1,4 linkage causes the chain to be straight. Adjacent chains align and form extensive inter-chain hydrogen bonds, creating rigid, high-tensile-strength microfibrils that provide structural integrity to the cell wall.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The carbohydrate storage form in animals, glycogen, is characterized by a structure that is

A. Less branched than amylopectin, allowing for denser packing
B. More highly branched than amylopectin, allowing for more rapid mobilization of glucose
C. Linear like amylose, but with a higher molecular weight
D. Composed of glucose units linked by β-1,4 glycosidic bonds

Glycogen is essentially the animal equivalent of amylopectin but is more extensively branched (branching every 8-12 residues). This extreme branching creates many non-reducing ends for glycogen phosphorylase to attack, enabling an extremely rapid release of glucose-1-phosphate to meet metabolic demands.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The structural difference between amylose and amylopectin, the two components of starch, is that amylopectin possesses

A. A linear, unbranched chain of glucose units only
B. A higher proportion of β-1,4 glycosidic linkages
C. Branch points formed by α-1,6 glycosidic bonds in addition to α-1,4 linkages
D. A triple helical structure that makes it water-insoluble

Amylose is a linear, helical polymer of glucose linked by α-1,4 bonds. Amylopectin is a highly branched polymer with an α-1,4 linked backbone and α-1,6 glycosidic bonds at branch points occurring approximately every 24-30 glucose units.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The primary reason that humans can digest starch but not cellulose is the specificity of human amylases for

A. The β-1,4 glycosidic bonds present in cellulose
B. The α-1,4 glycosidic bonds present in starch and glycogen
C. Peptide bonds that link starch monomers
D. Ester bonds in the starch polymer backbone

Human digestive enzymes (salivary and pancreatic amylase) can only hydrolyze the α-1,4 glycosidic bonds found in starch's amylose and amylopectin. Cellulose consists of glucose units linked by β-1,4 bonds, which require the enzyme cellulase, not produced in the human digestive tract.

nmdcat.online BIO NMDCAT
Jun 27, 2026

A characteristic feature distinguishing an oligosaccharide from a polysaccharide is that oligosaccharides typically contain

A. Only one type of monosaccharide unit
B. A large, highly branched structure with an average molecular weight over 100,000 Daltons
C. Chains of 2-10 monosaccharide units linked by glycosidic bonds
D. Exclusive β-linkages that make them indigestible to most animals

By definition, oligosaccharides (oligo = few) are short polymers of 2 to about 10 monosaccharides. Common examples are disaccharides (sucrose, lactose, maltose). Polysaccharides contain hundreds or thousands of monosaccharide units.

nmdcat.online BIO NMDCAT
Jun 27, 2026

Lactose, the primary sugar in milk, is a disaccharide composed of

A. Glucose and fructose linked by an α-1,2 bond
B. Galactose and glucose linked by a β-1,4 glycosidic bond
C. Two glucose units linked by an α-1,4 glycosidic bond
D. Glucose and galactose linked by an α-1,6 glycosidic bond

Lactose is a reducing disaccharide. It is formed from β-D-galactose linked to the C4 of D-glucose via a β-1,4 glycosidic linkage. The glucose unit has a free anomeric carbon, giving lactose its reducing properties.

nmdcat.online BIO NMDCAT
Jun 27, 2026

In living organisms, the disaccharide sucrose is classified as a non-reducing sugar because the glycosidic bond is formed between

A. The anomeric carbons of glucose and fructose, locking both carbonyl groups
B. The C-1 of one glucose and the C-4 of another glucose
C. The C-1 of galactose and the C-4 of glucose
D. The C-1 of glucose and the C-2 of fructose, where fructose is in a ketose open-chain form

Sucrose consists of α-D-glucose and β-D-fructose linked via a glycosidic bond between their anomeric carbons (C1 of glucose and C2 of fructose). Since both anomeric carbons are involved, neither unit can open to expose a free carbonyl group, making it a non-reducing sugar.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The glycosidic bond in maltose, formed between two glucose units, is specifically an

A. α-1,2 glycosidic linkage
B. α-1,4 glycosidic linkage
C. β-1,4 glycosidic linkage
D. α-1,6 glycosidic linkage

Maltose is a reducing disaccharide formed from two D-glucose units linked by an α-1,4 glycosidic bond. The C1 of the first glucose (in α-configuration) is linked to the C4 of the second glucose. The second glucose retains a free anomeric carbon, making maltose a reducing sugar.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The transformation of α-D-glucose and β-D-glucose in an aqueous solution to an equilibrium mixture is a process termed

A. Epimerization
B. Mutarotation
C. Racemization
D. Tautomerization

When a pure anomer (α or β) is dissolved in water, the specific rotation of the solution changes over time until a constant value is reached. This is mutarotation, resulting from the ring opening and reclosing, establishing an equilibrium mixture of α (36%), β (64%), and the open-chain form (<0.1%).

nmdcat.online BIO NMDCAT
Jun 27, 2026

In the chair conformation of β-D-glucose, the most stable form, the hydroxyl groups and the hydroxymethyl group are predominantly oriented in the

A. Axial positions to minimize steric hindrance
B. Equatorial positions to minimize steric hindrance
C. Cis configuration relative to the ring oxygen
D. Random arrangement with no energy preference

In the chair conformation, bulky substituents preferentially occupy equatorial positions (pointing out from the ring) rather than axial positions (perpendicular to the ring). β-D-glucose has all its -OH and -CH₂OH groups in equatorial positions, making it the most stable and abundant hexose.

nmdcat.online BIO NMDCAT
Jun 27, 2026

Epimers are sugars that differ in configuration at only one chiral center. Glucose and galactose are identical in structure except for the orientation of the hydroxyl group on C-4, making them C-4 epimers. Glucose and mannose are C-2 epimers.

nmdcat.online BIO NMDCAT
Jun 27, 2026

Regarding the stereoisomerism of glucose, D-glucose and L-glucose are classified as enantiomers because they are

A. Mirror images of each other around all chiral centers
B. Identical in all physical and chemical properties
C. Structural isomers with different functional groups
D. Diastereomers that differ at only one chiral carbon

Enantiomers are a pair of molecules that are non-superimposable mirror images of each other. D-glucose and L-glucose are mirror images at all four chiral centers (C2, C3, C4, and C5), making them enantiomers. Diastereomers differ at one or more, but not all, chiral centers.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The functional group that defines a sugar as a reducing agent in reactions like Benedict’s test is a free

A. Phosphate group attached to C-6
B. Amino group on C-2
C. Aldehyde or ketone group capable of being oxidized
D. Methyl group on C-6 of the pyranose ring

A reducing sugar has a free anomeric carbon whose carbonyl group can be oxidized, thereby reducing another agent like Cu²⁺ in Benedict's reagent. The free aldehyde or α-hydroxyketone group is essential for this property. Non-reducing sugars lack this free group.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The nucleophilic addition of an alcohol (hydroxyl group) to a carbonyl group forms a hemiacetal (from an aldehyde) or a hemiketal (from a ketone). This intramolecular reaction converts the linear monosaccharide into its cyclic form, creating a new chiral center (the anomeric carbon).

nmdcat.online BIO NMDCAT
Jun 27, 2026

The general empirical formula for most carbohydrates is (CH₂O)n, and the functional groups that classify a monosaccharide as an aldose or a ketose are, respectively

A. A hydroxyl group and a carboxyl group
B. An aldehyde group and a ketone group
C. An amino group and a sulfhydryl group
D. A phosphate group and a methyl group

Monosaccharides are polyhydroxy carbonyl compounds. If the carbonyl group is at the end of the carbon chain (C1), it is an aldehyde and the sugar is an aldose (e.g., glucose). If the carbonyl group is on an internal carbon (C2), it is a ketone and the sugar is a ketose (e.g., fructose).

nmdcat.online BIO NMDCAT
Jun 27, 2026

The survival mechanism of certain freeze-tolerant fish in polar waters involves the synthesis of antifreeze glycoproteins. These proteins function by

A. Increasing the specific heat of the fish's blood
B. Binding to small ice crystals and inhibiting their growth by disrupting the orderly addition of water molecules
C. Decreasing the adhesive properties of water in the fish's tissues
D. Catalyzing the breakdown of water inside the fish's cells to generate heat

Antifreeze proteins adsorb to the surface of nascent ice nuclei. Their large, hydrophilic structure sterically hinders and geometrically mismatches the approach and orderly crystallization of further water molecules, effectively stopping ice crystal growth.

nmdcat.online BIO NMDCAT
Jun 27, 2026

During the process of cooking starchy food, the swelling and rupture of starch granules is primarily caused by the

A. Active transport of water into the starch granule
B. Hydrolysis of the starch into monosaccharides by water
C. Penetration of water into the granule, disrupting hydrogen bonds and hydrating the amylose and amylopectin
D. Formation of a hydrophobic core within the starch granule

Heat energy disrupts the internal hydrogen bonds of the starch granule. Water molecules then penetrate and form new H-bonds with the exposed -OH groups of the starch polymers (imbibition). This swelling, called gelatinization, eventually ruptures the granules, thickening the mixture.

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Jun 27, 2026
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