MCQs

11262 questions found

Practice Questions

In the context of carbohydrate classification, a triose sugar serves as a critical intermediate in metabolic pathways. The simplest aldose and ketose trioses are, respectively

A. Erythrose and Erythrulose
B. Glyceraldehyde and Dihydroxyacetone ✓
C. Ribose and Ribulose
D. Glucose and Fructose

Trioses are 3-carbon monosaccharides. The simplest aldose (aldehyde-containing) triose is glyceraldehyde. The simplest ketose (ketone-containing) triose is dihydroxyacetone. Both are key intermediates in glycolysis and photosynthesis.

nmdcat.online BIO NMDCAT
Jun 27, 2026

Sucrose is commonly known as “invert sugar” after its hydrolysis, because the resulting mixture of glucose and fructose

A. Precipitates out of solution as a solid
B. Changes the direction of plane-polarized light from dextrorotatory to levorotatory ✓
C. Has a higher boiling point than the original sucrose solution
D. Absorbs visible light and becomes colorless

Sucrose is dextrorotatory (+66.5°). Upon hydrolysis, the resulting fructose is strongly levorotatory (-92°), which outweighs the dextrorotation of glucose (+52.7°). The net optical rotation of the mixture (invert sugar) is negative (-19.8°), thus the rotation is inverted.

nmdcat.online BIO NMDCAT
Jun 27, 2026

A common intermediate in the metabolic pathways of both starch digestion and cellulose synthesis is

A. Glucose-1-phosphate ✓
B. Fructose-2,6-bisphosphate
C. Ribose-5-phosphate
D. Erythrose-4-phosphate

Starch digestion hydrolyzes starch to glucose, which is then phosphorylated to glucose-6-phosphate and isomerized to glucose-1-phosphate for entry into glycolysis. Cellulose is synthesized in plants from the activated monomer UDP-glucose, which is derived from glucose-1-phosphate.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The main reason that the structure of glycogen is more suitable for rapid energy mobilization in animal tissues than starch in plants is its

A. Lower molecular weight, allowing for faster diffusion
B. Higher degree of branching, which provides more non-reducing ends for enzymatic attack ✓
C. Exclusive presence of α-1,4 linkages, which are easier to hydrolyze
D. Association with lipid droplets in the cytoplasm

Glycogen's extreme branching creates a compact, highly soluble granule with thousands of terminal non-reducing ends. Enzymes like glycogen phosphorylase and debranching enzyme can work simultaneously at multiple ends, releasing glucose-1-phosphate far faster than from the less branched amylopectin of starch.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The iodine test is a specific qualitative test used to detect the presence of starch. The characteristic blue-black color is a result of the

A. Oxidation of iodine by the aldehyde groups of starch
B. Formation of a covalent bond between iodine and glucose monomers
C. Trapping of polyiodide ions (I₃⁻, I₅⁻) within the helical structure of amylose ✓
D. Reduction of iodine to iodide by the reducing end of amylopectin

Amylose forms a left-handed helix. Iodine (as I₃⁻ or I₅⁻ ions) fits into the central hydrophobic channel of this helix. The resulting charge-transfer complex absorbs light strongly, giving a characteristic deep blue-black color.

nmdcat.online BIO NMDCAT
Jun 27, 2026

Regarding the structure of chitin, the polysaccharide forming the exoskeleton of arthropods, its monomer is a modified glucose where the C-2 hydroxyl is replaced by

A. A hydrogen atom, forming deoxyglucose
B. An acetylamino group (-NHCOCH₃) ✓
C. A carboxyl group, forming glucuronic acid
D. A sulfate group, forming glucosamine sulfate

Chitin is a linear homopolymer of N-acetyl-D-glucosamine, which is glucose with an N-acetylamino group at the C-2 position. It is linked by β-1,4 glycosidic bonds, analogous to the structure of cellulose.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The observation that a solution of pure α-D-glucose has a specific rotation of +112°, which slowly changes to +52.7° upon standing, is an example of

A. Epimerization at C-2
B. Mutarotation involving the interconversion of anomers ✓
C. Tautomerization between keto and enol forms
D. Hydrolysis of the ring structure

This change in optical rotation is due to mutarotation. α-D-glucose (+112°) undergoes ring opening and reclosing to form an equilibrium mixture of β-D-glucose (+18.7°) and α-D-glucose, with the overall specific rotation settling at +52.7°.

nmdcat.online BIO NMDCAT
Jun 27, 2026

A six-membered ring (pyranose) forms when the carbonyl carbon (C1 in aldoses) reacts with the -OH group on C5. A five-membered ring (furanose) would involve a reaction with the -OH on C4. Glucose predominantly forms a pyranose ring.

nmdcat.online BIO NMDCAT
Jun 27, 2026

The digestion of lactose in the human small intestine requires the activity of the enzyme lactase. A deficiency in this enzyme leads to lactose intolerance, where undigested lactose

A. Is absorbed directly into the bloodstream and acts as a toxin
B. Is fermented by gut bacteria, producing gases and organic acids that cause gastrointestinal distress ✓
C. Crystallizes in the kidneys, leading to kidney stones
D. Inhibits the absorption of all other disaccharides

In lactose intolerance, lactase (β-galactosidase) is deficient. Undigested lactose passes to the colon, where gut microbiota ferment it. This produces gases like H₂, CO₂, and methane (causing bloating) and short-chain fatty acids, which draw water into the bowel (osmotic diarrhea).

nmdcat.online BIO NMDCAT
Jun 27, 2026

In the context of stereochemistry, the D or L configuration of a monosaccharide is determined by the orientation of the hydroxyl group on the

A. Anomeric carbon (C-1) in its ring form
B. Penultimate (farthest chiral) carbon from the carbonyl group ✓
C. Carbonyl carbon (C-1 or C-2)
D. Carbon bearing the free amino group

The D/L classification is based on the configuration of the chiral carbon farthest from the most oxidized end (the carbonyl group). If the -OH on this carbon is on the right in a Fischer projection, it is D; if on the left, it is L. For glucose, this is C-5.

nmdcat.online BIO NMDCAT
Jun 27, 2026

Cellulose is a homopolymer of β-D-glucose and constitutes the majority of plant biomass. Its annual production is estimated to be over 10¹¹ tons, making it the most abundant organic compound on Earth. Starch and glycogen are storage, not structural, polymers.

nmdcat.online BIO NMDCAT
Jun 27, 2026

Among the following statements, the one that correctly describes the products of acid-catalyzed or enzymatic hydrolysis of sucrose is

A. Only glucose, as fructose is destroyed
B. An equimolar mixture of glucose and fructose ✓
C. Glucose and galactose in a 2:1 ratio
D. Only fructose, as glucose is further broken down

Sucrose is a disaccharide of glucose and fructose. Hydrolysis of the glycosidic bond (e.g., by sucrase/invertase) yields one molecule of D-glucose and one molecule of D-fructose. This 1:1 mixture is known as invert sugar.

nmdcat.online BIO NMDCAT
Jun 27, 2026

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
Page 166 of 593
Jump to:

🏆 Top Contributors

  • N

    nmdcat.online

    11260 MCQs

  • N

    NMDCAT.ONLINE

    1 MCQ

  • G

    GULABsb

    1 MCQ

Categories

View all →