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1. Which disaccharide is formed by two glucose units joined through an $\alpha(1-4)$ glycosidic bond?
ⓐ. Sucrose (glucose + fructose; $\alpha(1-2)$)
ⓑ. Maltose (glucose + glucose; $\alpha(1-4)$)
ⓒ. Trehalose (glucose + glucose; $\alpha(1-1)$)
ⓓ. Lactose (galactose + glucose; $\beta(1-4)$)
Correct Answer: Maltose (glucose + glucose; $\alpha(1-4)$)
Explanation: Maltose is specifically composed of two glucose molecules linked by an $\alpha(1-4)$ glycosidic bond, where the anomeric carbon of one glucose connects to the fourth carbon of the second glucose. This linkage pattern commonly appears during the enzymatic breakdown of starch, so maltose is a frequent intermediate in carbohydrate digestion and hydrolysis reactions. The “$\alpha$” configuration is central because it matches the typical bond geometry seen in starch-derived fragments. Therefore, when the question fixes both monomers as glucose and the linkage as $\alpha(1-4)$, maltose is the unique match among common disaccharides.
2. A highly branched storage polysaccharide in animals, with an $\alpha(1-4)$ chain and frequent $\alpha(1-6)$ branches, is:
ⓐ. Amylose (starch fraction; mostly linear)
ⓑ. Cellulose (linear $\beta$-glucan; structural)
ⓒ. Pectin (heteropolysaccharide; cell wall matrix)
ⓓ. Glycogen (animal storage $\alpha$-glucan)
Correct Answer: Glycogen (animal storage $\alpha$-glucan)
Explanation: Glycogen is the principal storage carbohydrate in animals and is built from glucose units linked mainly by $\alpha(1-4)$ glycosidic bonds. Its key feature is frequent branching created by $\alpha(1-6)$ linkages, producing a compact molecule with many free ends. These multiple ends allow rapid addition and removal of glucose units, supporting quick energy release during activity or fasting. This branching architecture is a functional adaptation for fast mobilization rather than long-term rigid storage. Hence, an $\alpha(1-4)$ backbone with frequent $\alpha(1-6)$ branches identifies glycogen.
3. Cellulose is generally indigestible in humans primarily because it contains:
ⓐ. $\beta(1-4)$ glycosidic bonds not hydrolyzed by human enzymes
ⓑ. Glucose units held mainly by strong ionic interactions
ⓒ. Fructose units linked through $\alpha(1-2)$ bonds
ⓓ. Amino acids linked by repeating peptide bonds
Correct Answer: $\beta(1-4)$ glycosidic bonds not hydrolyzed by human enzymes
Explanation: Cellulose is a polymer of glucose, but the glucose units are connected by $\beta(1-4)$ glycosidic bonds that give the chain a straight, rigid structure. Humans lack effective enzymes to hydrolyze this specific $\beta(1-4)$ linkage, so cellulose is not broken down into absorbable glucose in the human digestive tract. The linear chains also align closely to form strong microfibrils, which further reduces access for digestive enzymes. As a result, cellulose largely functions as dietary fiber rather than an energy source. The core reason is the bond configuration and enzyme specificity.
4. At around neutral pH, most amino acids predominantly exist in which ionic form?
ⓐ. Only anionic form with $-COO^-$ and $-NH_2$
ⓑ. Fully unionized form with $-COOH$ and $-NH_2$
ⓒ. Zwitterionic form with $-COO^-$ and $-NH_3^+$
ⓓ. Only cationic form with $-COOH$ and $-NH_3^+$
Correct Answer: Zwitterionic form with $-COO^-$ and $-NH_3^+$
Explanation: Amino acids contain both an acidic carboxyl group and a basic amino group, so they behave as amphoteric molecules in water. Near neutral pH, the carboxyl group typically loses a proton to become $-COO^-$, while the amino group gains a proton to become $-NH_3^+$. This produces a dipolar ion called a zwitterion that has both charges but an overall neutral charge. The zwitterionic form explains why amino acids show buffering behavior and characteristic solubility patterns at physiological pH. Therefore, the dominant form around neutral pH is the zwitterion.
5. The covalent linkage formed between the $\alpha$-carboxyl group of one amino acid and the $\alpha$-amino group of another is:
ⓐ. Glycosidic bond (common in disaccharides)
ⓑ. Peptide bond ($-CO-NH-$ linkage in proteins)
ⓒ. Phosphodiester bond (common in nucleic acids)
ⓓ. Disulfide bond (common in certain proteins)
Correct Answer: Peptide bond ($-CO-NH-$ linkage in proteins)
Explanation: A peptide bond is formed when the carboxyl group of one amino acid reacts with the amino group of another, resulting in a condensation reaction with removal of water. The product is an amide linkage written as $-CO-NH-$, which creates the continuous backbone of peptides and proteins. This bond is fundamental to primary structure because it locks amino acids into a specific sequence. Its partial double-bond character gives it planarity and restricts rotation, influencing how polypeptides fold. Thus, the correct linkage joining amino acids into polypeptides is the peptide bond.
6. The protein structural level mainly stabilized by hydrogen bonding between backbone $C=O$ and $N-H$ groups to form $\alpha$-helices and $\beta$-sheets is:
ⓐ. Secondary structure (local backbone folding)
ⓑ. Primary structure (sequence arrangement only)
ⓒ. Tertiary structure (overall 3D folding)
ⓓ. Quaternary structure (subunit association)
Correct Answer: Secondary structure (local backbone folding)
Explanation: Secondary structure describes regular local folding patterns of the polypeptide backbone, most notably the $\alpha$-helix and $\beta$-pleated sheet. These structures are stabilized primarily by hydrogen bonds between the carbonyl oxygen ($C=O$) of one peptide bond and the amide hydrogen ($N-H$) of another peptide bond along the backbone. The repeating hydrogen-bond pattern creates stable coils or sheets without requiring the side chains to directly form the main stabilizing bonds. This level of structure forms a scaffold that supports higher-order folding. Therefore, $\alpha$-helices and $\beta$-sheets are features of secondary structure.
7. For an enzyme that follows Michaelis–Menten behavior, a lower $K_m$ value generally indicates:
ⓐ. Higher enzyme concentration in the reaction mixture
ⓑ. Irreversible binding of substrate to the active site
ⓒ. Lower $V_{max}$ because catalytic sites are blocked
ⓓ. Higher substrate affinity at low substrate concentration
Correct Answer: Higher substrate affinity at low substrate concentration
Explanation: The Michaelis constant, $K_m$, is the substrate concentration at which the reaction rate reaches half of $V_{max}$. When $K_m$ is lower, the enzyme reaches significant reaction rates at smaller substrate concentrations, which generally reflects higher effective affinity between enzyme and substrate under standard Michaelis–Menten assumptions. This is biologically important because many cellular substrates are present at low or fluctuating concentrations. A lower $K_m$ therefore suggests the enzyme can operate efficiently even when substrate availability is limited. Hence, a reduced $K_m$ is typically interpreted as higher substrate affinity in practical enzyme kinetics.
8. In classic competitive inhibition, the most expected kinetic change is:
ⓐ. Unchanged $K_m$ with decreased $V_{max}$
ⓑ. Decreased $K_m$ with unchanged $V_{max}$
ⓒ. Increased $K_m$ with unchanged $V_{max}$
ⓓ. Decreased $K_m$ with decreased $V_{max}$
Correct Answer: Increased $K_m$ with unchanged $V_{max}$
Explanation: Competitive inhibitors compete with the substrate for the same active site, reducing substrate binding at a given substrate concentration. Because more substrate is required to overcome this competition, the apparent $K_m$ increases, meaning half-maximal velocity is reached only at a higher substrate concentration. However, at sufficiently high substrate levels, the substrate can outcompete the inhibitor, allowing the enzyme to still achieve the same maximum velocity, so $V_{max}$ remains unchanged. This pattern is a hallmark of competitive inhibition and helps distinguish it from other inhibitor types. Therefore, increased $K_m$ with unchanged $V_{max}$ is the expected change.
9. In a nucleic acid strand, adjacent nucleotides are linked to form the sugar–phosphate backbone by:
ⓐ. Phosphodiester linkage between $3’$-OH and $5’$-phosphate
ⓑ. Glycosidic linkage between two sugars directly
ⓒ. Hydrogen bonding between nitrogen bases only
ⓓ. Peptide linkage between nitrogen and phosphate groups
Correct Answer: Phosphodiester linkage between $3’$-OH and $5’$-phosphate
Explanation: Nucleic acids are polymers of nucleotides, and their backbone is formed by strong covalent phosphodiester bonds. Specifically, the $3’$ hydroxyl group of the sugar in one nucleotide forms a bond with the phosphate group attached to the $5’$ carbon of the next nucleotide’s sugar. This repeating linkage produces a stable sugar–phosphate chain and gives the strand a defined direction, with distinct $5’$ and $3’$ ends. The backbone structure is essential for maintaining the integrity of genetic material and for processes like replication and transcription. Hence, the correct bond connecting nucleotides is the phosphodiester linkage.
10. Plant oils are usually liquid at room temperature mainly because they contain:
ⓐ. More long-chain saturated fatty acids with close packing
ⓑ. More cis-unsaturated fatty acids that reduce packing
ⓒ. More sterol-like components that increase rigidity
ⓓ. More wax-like esters that raise melting point
Correct Answer: More cis-unsaturated fatty acids that reduce packing
Explanation: Plant oils tend to be rich in cis-unsaturated fatty acids, which contain one or more cis double bonds in their hydrocarbon chains. These cis double bonds create bends that prevent the fatty acid chains from packing tightly together, reducing van der Waals interactions among the molecules. Because tight packing is a major factor that raises melting point, disrupted packing lowers the melting point and keeps the lipid mixture fluid at room temperature. This structural effect explains why many plant-derived lipids are oils rather than solid fats. Therefore, higher cis-unsaturation is the main reason plant oils remain liquid under common conditions.
11. In most actively living tissues, the approximate percentage of water in fresh mass is:
ⓐ. $50-60\%$
ⓑ. $70-90\%$
ⓒ. $60-70\%$
ⓓ. $90-98\%$
Correct Answer: $70-90\%$
Explanation: Most living tissues are water-rich because cellular reactions, transport, and diffusion occur in an aqueous environment. A standard exam-accepted range for fresh tissue water content is about $70-90\%$, reflecting the dominance of water in cytoplasm and body fluids. This high water level helps dissolve ions and polar metabolites so enzymes can access substrates efficiently. It also supports movement of molecules across membranes and within cells by diffusion and bulk flow. Water stabilizes internal conditions because of its high heat capacity and contributes to maintaining cell volume. The remaining fraction is “dry matter,” which contains the solid biochemical components.
12. Which biological material generally shows the lowest water percentage on a fresh-weight basis?
ⓐ. Fresh leaf mesophyll tissue (actively photosynthesizing)
ⓑ. Mammalian skeletal muscle (highly hydrated tissue)
ⓒ. Jelly-like aquatic animal body (very high water)
ⓓ. Mature dry seed (stored dormant state)
Correct Answer: Mature dry seed (stored dormant state)
Explanation: Mature dry seeds are designed for dormancy, so they contain very little water compared with active tissues. Low water content slows enzymatic activity and reduces overall metabolism, helping seeds remain viable for long periods. It also reduces the chances of unwanted biochemical reactions and limits microbial growth during storage. In contrast, leaf tissues and muscles need abundant water for continuous metabolic work and transport. Jelly-like aquatic animals typically contain extremely high water due to their body composition and habitat. Therefore, mature dry seeds have the lowest water percentage among the given options.
13. “Dry weight” of a tissue sample refers to the mass remaining after removal of mainly:
ⓐ. Water (moisture fraction)
ⓑ. Lipids (membrane and storage fats)
ⓒ. Proteins (enzymes and structural solids)
ⓓ. Mineral salts (ash fraction)
Correct Answer: Water (moisture fraction)
Explanation: Dry weight is measured after removing the water from a tissue sample, leaving behind all non-water solids. Because fresh tissues commonly contain a large majority of water, drying causes a major reduction in mass and reveals the true solid composition. The remaining dry matter includes organic macromolecules such as proteins, carbohydrates, lipids, and nucleic acids, along with many small metabolites. Inorganic ions and mineral elements also remain as part of the residue. This concept allows fair comparison between tissues that differ in hydration. Hence, removing water is the key step in obtaining dry weight.
14. In most cells, the largest fraction of dry mass is typically contributed by:
ⓐ. Carbohydrates (structural and storage pool)
ⓑ. Lipids (membrane and reserve pool)
ⓒ. Proteins (major cellular solids)
ⓓ. Nucleic acids (DNA and RNA pool)
Correct Answer: Proteins (major cellular solids)
Explanation: Proteins form the largest share of dry cellular mass because they drive most cellular structure and function. Enzymes that catalyze metabolism are proteins, and many essential cellular components like transporters, receptors, and cytoskeletal elements are also proteins. Cells require a large and diverse protein set to regulate pathways, maintain organization, and respond to signals. Even when carbohydrates or lipids are present, the total protein mass remains high due to constant functional demand. This is why composition summaries commonly place proteins as the dominant dry-weight component. Therefore, proteins contribute the largest fraction of dry mass in most cells.
15. In standard tissue composition summaries, the inorganic (ash) fraction of dry weight in most tissues is closest to:
ⓐ. $1\%$
ⓑ. $5\%$
ⓒ. $10\%$
ⓓ. $15\%$
Correct Answer: $5\%$
Explanation: The inorganic or ash fraction represents mineral ions and elemental constituents present in tissues after considering the organic biomolecules. In many exam-oriented summaries, this portion is approximated around $5\%$ of dry weight as a useful benchmark. These minerals include electrolytes important for osmotic balance, membrane potential, and proper functioning of enzymes. Several ions also serve as cofactors that enable catalytic activity and maintain protein stability. Although the fraction is small by mass, it is essential for normal physiology and cellular homeostasis. Thus, $5\%$ is the closest standard estimate for inorganic content in dry matter.
16. In a typical dry weight composition overview, carbohydrates are commonly approximated as about:
ⓐ. $5\%$
ⓑ. $10\%$
ⓒ. $25\%$
ⓓ. $15\%$
Correct Answer: $15\%$
Explanation: Carbohydrates contribute a significant but usually not dominant fraction of dry mass in many living tissues. They appear as storage molecules (like starch in plants and glycogen in animals), structural materials (notably in plant cell walls), and as components of surface and membrane-associated compounds. For broad composition summaries used in exams, carbohydrates are often placed around $15\%$ of dry weight as a reference value. This helps compare their contribution against proteins and lipids in a standardized way. While real tissues can vary, the approximate figure remains a common benchmark. Therefore, $15\%$ is the expected approximation.
17. In standard dry weight composition summaries, lipids are most often approximated as about:
ⓐ. $25\%$
ⓑ. $15\%$
ⓒ. $10\%$
ⓓ. $5\%$
Correct Answer: $10\%$
Explanation: Lipids are essential components of membranes and also act as energy reserves, so they contribute meaningfully to dry mass. Even in tissues not specialized for fat storage, membrane lipids ensure a baseline lipid fraction is always present. In many exam-style composition summaries, lipids are approximated around $10\%$ of dry weight to provide a simple comparative figure. This reflects a typical balance between structural membrane roles and variable storage roles across tissues. Some tissues can have higher values, but the standard overview uses a generalized estimate. Hence, $10\%$ is the commonly used approximation.
18. In standard dry weight composition summaries, nucleic acids are most often approximated as about:
ⓐ. $5\%$
ⓑ. $10\%$
ⓒ. $15\%$
ⓓ. $20\%$
Correct Answer: $5\%$
Explanation: Nucleic acids are indispensable for genetic information storage and expression, yet they typically form a smaller portion of dry mass compared with proteins. DNA quantity is limited by genome size, and although RNA levels can be substantial in actively synthesizing cells, the overall nucleic acid mass is still commonly lower than major protein pools. Therefore, broad composition summaries often approximate nucleic acids at about $5\%$ of dry weight. This value is used as a practical benchmark in exam contexts rather than a fixed constant for all tissues. The key idea is that nucleic acids are critical but not the largest mass component. Hence, $5\%$ is the expected approximation.
19. A key reason high water content supports cellular metabolism is that water mainly acts as a:
ⓐ. Long-term energy reserve molecule (storage fuel)
ⓑ. Permanent information template (genetic store)
ⓒ. Polar solvent for reactions (aqueous medium)
ⓓ. Structural polymer in cell walls (rigid support)
Correct Answer: Polar solvent for reactions (aqueous medium)
Explanation: Water’s polarity allows it to dissolve ions and many polar metabolites, creating an effective medium for biochemical reactions. Dissolved reactants can diffuse and collide, enabling enzymes to bind substrates efficiently and carry out catalysis. Water also supports correct macromolecular behavior, including hydrogen bonding and the hydrophobic effect that influence protein structure and function. Many cellular reactions involve water directly, especially hydrolysis steps in digestion and metabolism. In addition, water helps transport molecules within cells and across tissues through aqueous fluids. Therefore, water primarily supports metabolism by acting as a polar solvent and reaction medium.
20. Which option best represents a component mainly counted in the inorganic fraction of dry weight?
ⓐ. Ion such as $Na^+$ (electrolyte component)
ⓑ. Amino acid such as alanine (organic monomer)
ⓒ. Storage polymer such as glycogen (organic reserve)
ⓓ. Sterol such as cholesterol (organic lipid)
Correct Answer: Ion such as $Na^+$ (electrolyte component)
Explanation: The inorganic fraction of dry matter consists mainly of mineral ions and elemental components rather than carbon-based biomolecules. Ions like $Na^+$ function as electrolytes that regulate osmotic balance, help maintain membrane potential, and support transport mechanisms across membranes. They also influence enzyme activity by affecting ionic strength and charge interactions in cellular fluids. Such ions are therefore grouped under mineral or ash content in composition summaries. In contrast, amino acids, glycogen, and cholesterol are organic molecules built on carbon skeletons and belong to organic dry matter. Hence, $Na^+$ is the correct example of an inorganic dry-weight component.
21. A mineral element required in very small amounts (often in ppm range) but essential for normal metabolism is best termed:
ⓐ. Major element (needed in bulk)
ⓑ. Trace element (needed in minute)
ⓒ. Storage nutrient (kept as reserve)
ⓓ. Structural polymer (forms framework)
Correct Answer: Trace element (needed in minute)
Explanation: Trace elements are mineral elements required in extremely small quantities, often at ppm levels, yet they are indispensable for life. Their importance comes from specific biochemical roles, commonly as parts of enzymes, electron carriers, or regulatory proteins. Even though their total mass contribution to dry matter is tiny, missing them can disrupt key metabolic pathways. Many act as cofactors that enable enzyme active sites to function with correct shape and reactivity. Because the requirement is minute but essential, they are grouped as trace elements rather than bulk major elements. Hence, “Trace element” is the correct classification.
22. When the dry matter of a tissue is completely burned, the remaining “ash” mainly represents:
ⓐ. Mineral salts and ions
ⓑ. Carbon-rich organic residues
ⓒ. Denatured proteins and peptides
ⓓ. Stored polysaccharide granules
Correct Answer: Mineral salts and ions
Explanation: The ash left after complete burning of dry matter primarily consists of inorganic mineral components. Organic biomolecules such as proteins, carbohydrates, and lipids are largely combusted into gases, so they do not remain as solid residue. What persists is the inorganic fraction, including mineral salts and ionic forms of elements present in the tissue. This residue reflects the tissue’s mineral composition and is often used to estimate total inorganic content. Although small in proportion, these minerals are essential for physiology and enzyme function. Therefore, ash mainly represents mineral salts and ions.
23. A key reason high water content helps living tissues resist sudden temperature changes is water’s:
ⓐ. Low density property
ⓑ. High viscosity property
ⓒ. High electrical conductance
ⓓ. High specific heat capacity
Correct Answer: High specific heat capacity
Explanation: Water has a high specific heat capacity, meaning it can absorb or release a large amount of heat with only a small change in its own temperature. In living tissues, this property buffers rapid temperature fluctuations caused by external changes or internal metabolic heat. Such thermal stability protects enzyme activity and membrane structure, both of which are sensitive to temperature shifts. Because most cellular reactions occur in aqueous surroundings, stable temperature supports consistent reaction rates. This is one major advantage of maintaining a high water fraction in tissues. Hence, high specific heat capacity explains the buffering effect.
24. The trace element most directly associated with cytochrome oxidase function in cellular respiration is:
ⓐ. Zinc
ⓑ. Iodine
ⓒ. Copper
ⓓ. Sodium
Correct Answer: Copper
Explanation: Copper is a crucial trace element in several oxidative enzymes, including cytochrome oxidase in the electron transport chain. Its importance arises from its ability to participate in reversible electron transfer reactions, supporting efficient movement of electrons during respiration. This helps maintain the proton gradient used for ATP synthesis and overall energy production. Because the enzyme’s catalytic performance depends on the metal’s redox behavior, even small copper deficiency can reduce respiratory efficiency. Copper is required in minute amounts, but its functional impact is high. Therefore, copper is the correct trace element linked to cytochrome oxidase.
25. The trace element that forms the central component of the heme group in hemoglobin is:
ⓐ. Iron
ⓑ. Iodine
ⓒ. Zinc
ⓓ. Copper
Correct Answer: Iron
Explanation: Hemoglobin carries oxygen because its heme group contains iron at the center, which binds oxygen reversibly. The iron atom enables oxygen loading in respiratory surfaces and unloading in tissues according to physiological conditions. This role is highly specific: without iron in heme, the oxygen-binding property of hemoglobin is lost. Because iron is needed in relatively small amounts compared to major elements but is essential for oxygen transport, it is classified as a trace element in body composition contexts. Iron deficiency therefore commonly affects oxygen delivery capacity. Hence, iron is the correct answer.
26. The trace element most directly required for synthesis of thyroid hormones is:
ⓐ. Copper
ⓑ. Iodine
ⓒ. Sodium
ⓓ. Zinc
Correct Answer: Iodine
Explanation: Iodine is essential because thyroid hormones are iodinated compounds, and their synthesis depends on adequate iodine availability. These hormones regulate basal metabolic rate, growth, and development, so iodine deficiency can have wide systemic effects. The thyroid gland actively concentrates iodine to support hormone production, highlighting its specific physiological demand. Although required in minute amounts, iodine cannot be replaced by other common ions for this role. Its importance is therefore functional rather than based on large mass contribution. Hence, iodine is the trace element required for thyroid hormone synthesis.
27. The enzyme carbonic anhydrase typically requires which trace element for its catalytic activity?
ⓐ. Copper
ⓑ. Iron
ⓒ. Zinc
ⓓ. Iodine
Correct Answer: Zinc
Explanation: Zinc is a common metal cofactor for enzymes that catalyze reactions involving water and carbon dioxide, and carbonic anhydrase is a classic example. The enzyme accelerates the interconversion of $CO_2$ and bicarbonate, supporting gas transport and acid–base balance. Zinc contributes by stabilizing the active site environment and enabling efficient catalysis of the hydration reaction. Because this catalytic function depends on the presence of zinc, deficiency can impair processes linked to respiration and pH regulation. Zinc is needed in small amounts but has high biochemical significance. Therefore, zinc is the correct trace element for carbonic anhydrase.
28. In many biological systems, nitrate reductase activity is most closely associated with the trace element:
ⓐ. Sodium
ⓑ. Iron
ⓒ. Copper
ⓓ. Molybdenum
Correct Answer: Molybdenum
Explanation: Molybdenum serves as an essential trace element in enzymes involved in nitrogen metabolism, notably nitrate reductase. This enzyme catalyzes a key step in the utilization of nitrate, supporting nitrogen assimilation and downstream biomolecule synthesis. The metal’s presence in the enzyme complex is critical for proper electron transfer and catalytic turnover. Because the requirement is very small in quantity but essential for function, molybdenum is categorized as a trace element. Its deficiency can reduce nitrate utilization efficiency and impact overall growth and metabolism. Hence, molybdenum is the correct answer.
29. A fresh leafy tissue loses mass rapidly after harvest mainly due to loss of:
ⓐ. Proteins by degradation
ⓑ. Water by transpiration
ⓒ. Lipids by oxidation
ⓓ. Minerals by diffusion
Correct Answer: Water by transpiration
Explanation: Fresh leaves contain a high proportion of water (commonly around $70-90\%$), so even modest water loss can cause a noticeable drop in mass. After harvest, stomatal and surface evaporation can continue, leading to transpiration-driven water loss. This reduces turgor pressure, causing wilting and a decline in freshness even though the dry matter changes far less initially. Because water is the dominant component of fresh mass, it is the primary contributor to rapid weight reduction. The process is physical loss of water rather than immediate loss of major biomolecules. Therefore, water loss by transpiration best explains the rapid mass decrease.
30. An element contributing less than about $0.1\%$ of dry weight is generally grouped as a:
ⓐ. Micronutrient (trace element)
ⓑ. Macronutrient (bulk element)
ⓒ. Structural carbohydrate (major solid)
ⓓ. Storage lipid (major reserve)
Correct Answer: Micronutrient (trace element)
Explanation: Elements present below about $0.1\%$ of dry weight are typically classified as micronutrients, also called trace elements, because they are required in very small amounts. Their significance lies in specific biochemical roles, such as serving as enzyme cofactors, redox carriers, or components of key regulatory molecules. Despite low abundance, they can be essential for normal metabolism, growth, and physiological regulation. This classification distinguishes them from macronutrients, which are needed in larger quantities and form a greater share of dry matter. The “trace” label reflects quantity, not importance. Hence, such elements are grouped as micronutrients (trace elements).
31. Which option correctly identifies a ketohexose among common monosaccharides?
ⓐ. Glucose (aldohexose)
ⓑ. Ribose (aldopentose)
ⓒ. Fructose (ketohexose)
ⓓ. Deoxyribose (deoxypentose)
Correct Answer: Fructose (ketohexose)
Explanation: Fructose is classified as a ketohexose because it contains six carbon atoms and a ketone functional group in its open-chain form. This structural feature affects its chemistry, including how it cyclizes to form furanose rings in solution. Ketohexoses are important in metabolism, and fructose enters pathways after specific phosphorylation steps. In contrast, glucose is an aldohexose and has an aldehyde group, while ribose is a five-carbon aldose used in nucleic acids. Deoxyribose is also a five-carbon sugar but lacks one oxygen compared to ribose. Therefore, fructose is the correct ketohexose.
32. The key structural reason cellulose forms strong, rigid fibers in plant cell walls is that it is a glucose polymer with:
ⓐ. $\beta(1-4)$ glycosidic bonds forming straight chains
ⓑ. $\alpha(1-4)$ glycosidic bonds forming helical chains
ⓒ. $\alpha(1-6)$ branch bonds forming many chain ends
ⓓ. $\beta(1-6)$ branch bonds forming compact granules
Correct Answer: $\beta(1-4)$ glycosidic bonds forming straight chains
Explanation: Cellulose is made of glucose units linked by $\beta(1-4)$ glycosidic bonds, which keeps the polymer chains straight rather than coiled. Straight chains align closely and form extensive hydrogen bonding between neighboring chains, producing microfibrils with high tensile strength. This arrangement makes cellulose ideal for structural support in plant cell walls. The rigidity comes from both the bond type and the tight packing of parallel chains. Because the bonding pattern promotes linear geometry and strong intermolecular association, cellulose behaves as a structural polysaccharide. Hence, $\beta(1-4)$ linkage is the defining feature for strength and fiber formation.
33. In plant starch, which component is responsible for the branched architecture due to $\alpha(1-6)$ branch points?
ⓐ. Amylose (mostly unbranched starch part)
ⓑ. Cellulose (wall structural glucose polymer)
ⓒ. Chitin (nitrogen sugar polymer framework)
ⓓ. Amylopectin (branched starch part)
Correct Answer: Amylopectin (branched starch part)
Explanation: Amylopectin is the branched component of starch, built from glucose units mainly linked by $\alpha(1-4)$ bonds with branch points created by $\alpha(1-6)$ linkages. These branch points generate multiple chain ends, allowing faster enzymatic mobilization of glucose during storage use. This structure also makes starch granules more compact and functionally suitable for energy storage in plants. Amylopectin differs from amylose, which is largely unbranched and more helical. The presence of frequent $\alpha(1-6)$ branches is the signature feature of amylopectin. Therefore, amylopectin is the component responsible for starch branching.
34. A monosaccharide is called a “reducing sugar” mainly because it typically has:
ⓐ. Only ester linkages blocking reactive groups
ⓑ. A free anomeric carbon that can open to a reactive form
ⓒ. A peptide bond that donates electrons easily
ⓓ. A phosphate bridge that prevents ring opening
Correct Answer: A free anomeric carbon that can open to a reactive form
Explanation: Reducing sugars can act as reducing agents because they possess a free anomeric carbon in the cyclic form that can equilibrate with an open-chain form. In the open-chain form, an aldehyde (or an equivalent reactive group in certain ketoses via rearrangement) can participate in oxidation-reduction reactions. This chemical flexibility depends on the anomeric carbon not being locked in a glycosidic bond. When the anomeric position is free, the sugar can interconvert between cyclic and open forms in solution. That is why many monosaccharides show reducing behavior in standard biochemical tests. Hence, the presence of a free anomeric carbon explains reducing nature.
35. Which pair of monosaccharides are epimers differing specifically at the C-4 carbon position?
ⓐ. Ribose and deoxyribose (oxygen difference)
ⓑ. Glucose and fructose (aldose vs ketose)
ⓒ. Glucose and galactose (C-4 epimers)
ⓓ. Glucose and ribose (hexose vs pentose)
Correct Answer: Glucose and galactose (C-4 epimers)
Explanation: Epimers are stereoisomers that differ in configuration at only one specific chiral carbon atom. Glucose and galactose have the same molecular formula and the same overall carbon skeleton, but they differ in the orientation of the hydroxyl group at the C-4 position. This single-point stereochemical change is enough to alter how enzymes recognize the sugar and how it behaves in biological systems. The concept is important because many metabolic pathways are stereospecific and distinguish epimers. Since only C-4 configuration changes between these two, they are classic C-4 epimers. Therefore, glucose and galactose are the correct pair.
36. Sucrose is considered a non-reducing disaccharide mainly because:
ⓐ. It contains only $\beta(1-4)$ bonds in both units
ⓑ. It has no oxygen atom in the glycosidic linkage
ⓒ. Its rings cannot form cyclic structures in water
ⓓ. Both anomeric carbons are involved in the glycosidic bond
Correct Answer: Both anomeric carbons are involved in the glycosidic bond
Explanation: Sucrose is non-reducing because the glycosidic linkage connects the anomeric carbon of glucose to the anomeric carbon of fructose, leaving no free anomeric carbon. Without a free anomeric position, sucrose cannot readily open to an aldehyde-like reactive form that participates in typical reducing reactions. This structural “locking” prevents the key equilibrium between cyclic and open-chain forms that underlies reducing behavior. As a result, sucrose does not show the common reducing properties seen in sugars like glucose or maltose. The explanation depends on the involvement of both anomeric carbons in the linkage. Hence, sucrose is non-reducing for this specific structural reason.
37. The pentose sugar that forms part of RNA nucleotides is:
ⓐ. Glucose (aldohexose cellular fuel)
ⓑ. Ribose (pentose in RNA)
ⓒ. Fructose (ketohexose dietary sugar)
ⓓ. Galactose (aldohexose milk sugar part)
Correct Answer: Ribose (pentose in RNA)
Explanation: RNA nucleotides contain ribose as their pentose sugar, which provides the backbone for the RNA polymer. Ribose has a hydroxyl group at the $2’$ carbon, a key chemical feature that influences RNA stability and reactivity. This sugar connects to nitrogen bases and phosphate groups to form the repeating sugar-phosphate framework of RNA. The presence of the $2’$-OH is also why RNA is generally less chemically stable than DNA under alkaline conditions. Because RNA specifically uses ribose rather than deoxyribose, identifying ribose is a standard biomolecules concept. Therefore, ribose is the correct sugar in RNA.
38. Compared with plant starch, animal glycogen typically has:
ⓐ. More frequent branching, enabling faster mobilization
ⓑ. Only $\beta(1-4)$ bonds, making rigid fibers
ⓒ. No branches at all, making long straight chains
ⓓ. Only $\alpha(1-2)$ bonds, making non-reducing chains
Correct Answer: More frequent branching, enabling faster mobilization
Explanation: Glycogen is a storage polysaccharide in animals that is highly branched, with many $\alpha(1-6)$ branch points along $\alpha(1-4)$ chains. Greater branching creates more terminal ends from which enzymes can rapidly add or remove glucose units, allowing quick response to energy demands. This is especially important in muscle and liver where rapid glucose release supports activity and blood glucose regulation. Starch (particularly amylopectin) is also branched, but glycogen usually has branch points more frequently. The functional outcome is faster mobilization of stored carbohydrate. Hence, glycogen is characterized by more frequent branching for rapid use.
39. Chitin, a structural polysaccharide in many organisms, is primarily a polymer of:
ⓐ. Glucose units linked by $\alpha(1-4)$ bonds
ⓑ. Ribose units linked by phosphodiester bonds
ⓒ. $N$-acetylglucosamine units linked by $\beta(1-4)$ bonds
ⓓ. Fructose units linked by $\alpha(1-2)$ bonds
Correct Answer: $N$-acetylglucosamine units linked by $\beta(1-4)$ bonds
Explanation: Chitin is a strong structural polysaccharide built from repeating $N$-acetylglucosamine units. These monomers are linked by $\beta(1-4)$ glycosidic bonds, producing straight chains that can align and form extensive hydrogen bonding, similar to cellulose. The acetylated amino group adds additional chemical features that contribute to toughness and structural stability. This design makes chitin suitable for protective and supportive roles in organisms where it occurs. The key identity points are the monomer (acetylated amino sugar) and the $\beta(1-4)$ linkage. Therefore, chitin is correctly described as a $\beta(1-4)$ polymer of $N$-acetylglucosamine.
40. In plant starch, the mostly unbranched component that tends to form helical chains is:
ⓐ. Amylose (mostly unbranched starch part)
ⓑ. Amylopectin (branched starch part)
ⓒ. Cellulose (wall structural glucose polymer)
ⓓ. Pectin (matrix heteropolysaccharide)
Correct Answer: Amylose (mostly unbranched starch part)
Explanation: Amylose is the relatively unbranched fraction of starch, composed mainly of glucose units connected by $\alpha(1-4)$ glycosidic bonds. Because the chain is largely linear, it can coil into a helical structure, a feature often linked to how amylose interacts with iodine in common identification tests. This helical, less-branched architecture contrasts with amylopectin, which has many $\alpha(1-6)$ branch points and a more tree-like structure. Amylose therefore represents the simpler, largely unbranched storage polysaccharide component in starch. The structural property of helix formation follows from the bonding pattern and limited branching. Hence, amylose is the correct answer.
41. Which disaccharide is the most common transport form of carbohydrate in many plants?
ⓐ. Maltose (starch digestion intermediate)
ⓑ. Lactose (milk-associated sugar)
ⓒ. Sucrose (major transport sugar)
ⓓ. Trehalose (stress-protective sugar)
Correct Answer: Sucrose (major transport sugar)
Explanation: Sucrose is widely used in plants as the principal transport sugar because it is stable, soluble, and efficiently moved through phloem from source tissues (like leaves) to sink tissues (like roots, fruits, and seeds). Its stability is linked to being a non-reducing disaccharide, which reduces unwanted side reactions during long-distance transport. Plants can load sucrose into phloem and later hydrolyze it where energy or carbon skeletons are needed. This makes sucrose ideal for distributing photosynthetic products throughout the plant body. Therefore, sucrose is the characteristic transport disaccharide in many plants.
42. A disaccharide that typically gives a positive reducing-sugar test without prior hydrolysis is:
ⓐ. Maltose (reducing disaccharide)
ⓑ. Sucrose (non-reducing disaccharide)
ⓒ. Trehalose (non-reducing disaccharide)
ⓓ. Cellobiose-like unit (structural linkage context)
Correct Answer: Maltose (reducing disaccharide)
Explanation: Maltose is a reducing disaccharide because one of its glucose units retains a free anomeric carbon that can open to a reactive form in solution. This allows maltose to participate in redox-based tests used to detect reducing sugars. Structurally, maltose has glucose units linked so that only one anomeric carbon is involved in the glycosidic bond, leaving the other available for ring opening. This chemical feature is why maltose can show reducing behavior directly. As a result, maltose typically tests positive in standard reducing-sugar assays without needing hydrolysis first.
43. Which disaccharide is classically non-reducing because both anomeric carbons are involved in the linkage?
ⓐ. Maltose (one free anomeric carbon)
ⓑ. Lactose (one free anomeric carbon)
ⓒ. Cellobiose (one free anomeric carbon)
ⓓ. Sucrose (both anomeric involved)
Correct Answer: Sucrose (both anomeric involved)
Explanation: Sucrose is non-reducing because the glycosidic bond connects the anomeric carbon of glucose to the anomeric carbon of fructose. With both anomeric positions locked in the linkage, sucrose cannot readily form an open-chain reactive form needed for typical reducing behavior. This structural “blocking” makes sucrose chemically more stable during transport and storage. The key concept is not the presence of oxygen or ring formation, but the absence of any free anomeric carbon. Therefore, sucrose is correctly identified as the non-reducing disaccharide for this reason.
44. Lactose is best described as a disaccharide composed of:
ⓐ. Glucose + fructose (common transport pair)
ⓑ. Glucose + galactose (milk sugar pair)
ⓒ. Fructose + galactose (dietary uncommon pair)
ⓓ. Glucose + glucose (starch breakdown pair)
Correct Answer: Glucose + galactose (milk sugar pair)
Explanation: Lactose is the characteristic sugar of milk and is formed by combining one glucose unit with one galactose unit. This composition is important because digestion requires the enzyme lactase, which splits lactose into these two monosaccharides for absorption. Galactose derived from lactose can be converted into glucose derivatives and enter normal metabolic pathways. The glucose component provides an immediate energy source after absorption, while the galactose component reflects specific carbohydrate handling in the body. Hence, the correct identification of lactose is glucose plus galactose.
45. Maltose consists of two glucose units linked most commonly by:
ⓐ. $\alpha(1-4)$ glycosidic bond
ⓑ. $\beta(1-4)$ glycosidic bond
ⓒ. $\alpha(1-6)$ glycosidic bond
ⓓ. $\alpha(1-2)$ glycosidic bond
Correct Answer: $\alpha(1-4)$ glycosidic bond
Explanation: Maltose is formed when two glucose molecules join through an $\alpha(1-4)$ glycosidic linkage, where the anomeric carbon of one glucose connects to the fourth carbon of the second glucose. This linkage type is common in starch segments, so maltose frequently appears during starch hydrolysis and digestion. The $\alpha$ configuration also influences the overall shape and enzymatic recognition of the molecule. Because the bond is specifically $\alpha(1-4)$ in maltose, it fits the classic disaccharide structure taught in biomolecules. Therefore, the correct linkage in maltose is $\alpha(1-4)$.
46. A person develops bloating and watery stools after drinking milk due to low intestinal lactase activity. The immediate cause is:
ⓐ. Reduced protein digestion in the stomach
ⓑ. Excess fat absorption in the intestine
ⓒ. Lactose remains unhydrolyzed in the gut lumen
ⓓ. Complete breakdown of lactose into amino acids
Correct Answer: Lactose remains unhydrolyzed in the gut lumen
Explanation: Lactase is required to hydrolyze lactose into absorbable monosaccharides, so low lactase activity leaves lactose undigested in the intestinal lumen. Undigested lactose increases osmotic load, drawing water into the intestine and contributing to watery stools. It also becomes available for microbial fermentation, producing gases that cause bloating and discomfort. The problem is therefore not a defect in protein or fat digestion, but the failure to split lactose into glucose and galactose. This directly explains the symptoms occurring after milk intake. Hence, unhydrolyzed lactose in the gut lumen is the immediate cause.
47. During starch digestion, the disaccharide commonly produced as a key intermediate is:
ⓐ. Sucrose (plant transport form)
ⓑ. Maltose (starch breakdown intermediate)
ⓒ. Lactose (milk-associated carbohydrate)
ⓓ. Trehalose (insect hemolymph sugar)
Correct Answer: Maltose (starch breakdown intermediate)
Explanation: Starch is primarily a glucose polymer with many $\alpha(1-4)$ linkages, and enzymatic cleavage of these bonds commonly releases maltose units. Because maltose itself is made of two glucose molecules connected by an $\alpha(1-4)$ bond, it naturally appears when starch chains are cut into smaller pieces. This intermediate can then be further hydrolyzed by specific enzymes to yield glucose for absorption and metabolism. The concept links polymer structure to digestion products, which is a common exam focus. Therefore, maltose is the typical disaccharide intermediate formed during starch digestion.
48. Trehalose is a non-reducing disaccharide because its two glucose units are linked by:
ⓐ. $\alpha(1-4)$ bond with one free anomeric carbon
ⓑ. $\beta(1-4)$ bond forming straight structural chains
ⓒ. $\alpha(1-2)$ bond joining glucose and fructose
ⓓ. $\alpha(1-1)$ bond involving both anomeric carbons
Correct Answer: $\alpha(1-1)$ bond involving both anomeric carbons
Explanation: Trehalose is composed of two glucose units joined through an $\alpha(1-1)$ glycosidic bond, meaning both anomeric carbons participate in the linkage. Because no free anomeric carbon remains, trehalose cannot easily form an open-chain reactive form, so it behaves as a non-reducing sugar. This structural stability helps trehalose function in certain organisms as a protective sugar under stress conditions. The key exam point is that non-reducing behavior comes from blocking the anomeric positions. Therefore, the $\alpha(1-1)$ linkage explains trehalose being non-reducing.
49. Sucrose does not show typical reducing behavior mainly because:
ⓐ. It has no free anomeric carbon available
ⓑ. It lacks oxygen in its glycosidic linkage
ⓒ. It cannot dissolve in water under normal conditions
ⓓ. It is made only of nitrogen-containing sugars
Correct Answer: It has no free anomeric carbon available
Explanation: Reducing behavior in sugars depends on having a free anomeric carbon that can open to a reactive form in solution. In sucrose, the glycosidic bond connects the anomeric carbon of glucose with the anomeric carbon of fructose, so neither anomeric position remains free. This prevents the structural equilibrium needed to generate the reactive open-chain form that participates in reducing reactions. As a result, sucrose is chemically more stable and does not behave as a reducing sugar in standard tests. The explanation is entirely structural and centers on anomeric carbon involvement. Hence, sucrose is non-reducing because it lacks a free anomeric carbon.
50. On complete hydrolysis, sucrose yields:
ⓐ. Glucose + galactose (milk sugar products)
ⓑ. Two glucose molecules (maltose products)
ⓒ. Glucose + fructose (transport sugar products)
ⓓ. Two fructose molecules (rare digestion products)
Correct Answer: Glucose + fructose (transport sugar products)
Explanation: Sucrose is a disaccharide built from one glucose and one fructose unit, so hydrolysis splits it into these two monosaccharides. The linkage in sucrose involves the anomeric carbons of both units, but hydrolysis breaks the glycosidic bond and releases free glucose and fructose. These monosaccharides can then enter cellular metabolism through their respective phosphorylation and pathway entry steps. This hydrolysis outcome is a standard biomolecules fact tested in board and competitive exams. Therefore, sucrose yields glucose and fructose upon complete hydrolysis.
51. Which polysaccharide is the principal storage carbohydrate in many plants?
ⓐ. Glycogen (animal storage $\alpha$-glucan)
ⓑ. Cellulose (plant wall structural $\beta$-glucan)
ⓒ. Starch (plant storage polysaccharide)
ⓓ. Chitin (structural polymer in many fungi)
Correct Answer: Starch (plant storage polysaccharide)
Explanation: Starch is the primary storage polysaccharide in many plants, accumulated in organs such as seeds, tubers, and storage tissues. It serves as a reserve of glucose units that can be mobilized when the plant needs energy for growth, germination, or respiration. Starch is built from glucose and stored in the form of granules, making it compact and suitable for long-term storage. Its chemical design allows enzymes to hydrolyze it efficiently into smaller sugars when required. This storage role is the defining feature that distinguishes it from structural polysaccharides. Therefore, starch is the correct plant storage carbohydrate.
52. The major structural polysaccharide that provides tensile strength to plant cell walls is:
ⓐ. Cellulose (linear $\beta(1-4)$ glucose polymer)
ⓑ. Starch (granular plant storage glucan)
ⓒ. Glycogen (highly branched animal storage)
ⓓ. Chitin (amino-sugar polymer in fungi)
Correct Answer: Cellulose (linear $\beta(1-4)$ glucose polymer)
Explanation: Cellulose is the main structural polysaccharide of plant cell walls and is responsible for high tensile strength and rigidity. Its glucose units are linked by $\beta(1-4)$ bonds, producing straight chains that align closely. Parallel chains form extensive hydrogen bonding, creating microfibrils that act like reinforcing fibers in the wall. This organized packing makes the wall resistant to stretching and helps cells maintain shape under turgor pressure. The key point is that the structure is optimized for support rather than easy mobilization for energy. Hence, cellulose is the correct structural polysaccharide in plant cell walls.
53. In glycogen, the glycosidic bond that typically creates branch points is:
ⓐ. $\beta(1-4)$
ⓑ. $\alpha(1-2)$
ⓒ. $\beta(1-6)$
ⓓ. $\alpha(1-6)$
Correct Answer: $\alpha(1-6)$
Explanation: Glycogen is a highly branched glucose polymer used for energy storage in animals, and its branch points are formed by $\alpha(1-6)$ glycosidic bonds. The main chains are connected by $\alpha(1-4)$ linkages, but branching occurs when a glucose unit links from its anomeric carbon to the C-6 position of another glucose. These branches create many terminal ends, allowing enzymes to add or remove glucose rapidly during energy demand. This architecture supports quick glucose mobilization, especially in liver and muscle. Therefore, the branching linkage in glycogen is $\alpha(1-6)$.
54. Which component of starch is mostly unbranched and commonly associated with a deep blue iodine reaction?
ⓐ. Glycogen (highly branched animal storage)
ⓑ. Amylose (mostly unbranched $\alpha(1-4)$ chains)
ⓒ. Amylopectin (branched starch with $\alpha(1-6)$ points)
ⓓ. Cellulose (linear $\beta(1-4)$ wall polymer)
Correct Answer: Amylose (mostly unbranched $\alpha(1-4)$ chains)
Explanation: Amylose is the largely unbranched fraction of starch, composed mainly of glucose units linked by $\alpha(1-4)$ bonds. Its long, relatively linear chains tend to coil into helical structures, which can trap iodine molecules and produce a characteristic deep blue color in standard tests. This reaction is a practical indicator of amylose-rich starch content. The structural feature is the dominance of $\alpha(1-4)$ linkages with minimal branching, enabling stable helix formation. Because the question focuses on “mostly unbranched” starch and iodine blue color, amylose best fits. Hence, amylose is the correct answer.
55. Chitin is best described as a polymer made of:
ⓐ. $N$-acetylglucosamine units linked by $\beta(1-4)$ bonds
ⓑ. Glucose units linked by $\alpha(1-4)$ bonds in long chains
ⓒ. Glucose units linked by $\alpha(1-6)$ bonds at many points
ⓓ. Fructose units linked by $\alpha(1-2)$ bonds repeatedly
Correct Answer: $N$-acetylglucosamine units linked by $\beta(1-4)$ bonds
Explanation: Chitin is a structural polysaccharide found in many fungi and in the exoskeletons of several organisms. It is composed of repeating $N$-acetylglucosamine units joined by $\beta(1-4)$ glycosidic bonds, producing straight chains. These chains align and form extensive hydrogen bonding, creating strong, tough structural material similar in organization to cellulose. The presence of an acetylated amino group adds chemical features that support rigidity and durability. Because both the monomer identity and the $\beta(1-4)$ linkage define chitin, option A matches precisely. Therefore, chitin is a $\beta(1-4)$ polymer of $N$-acetylglucosamine.
56. Compared with amylopectin, glycogen generally shows:
ⓐ. Fewer branches, giving fewer terminal ends
ⓑ. No branching, forming fully straight chains
ⓒ. Only $\beta(1-4)$ linkages, making rigid fibers
ⓓ. More frequent branching, enabling faster mobilization
Correct Answer: More frequent branching, enabling faster mobilization
Explanation: Glycogen and amylopectin are both branched storage polysaccharides made of glucose, but glycogen is typically more highly branched. Branch points created by $\alpha(1-6)$ linkages occur more frequently, increasing the number of chain ends available for rapid enzymatic action. More terminal ends allow faster release or addition of glucose units, which is critical for animals needing quick energy supply during activity or fasting. This functional need explains why glycogen’s structure favors greater branching density than plant starch. The difference is therefore not about being unbranched or having $\beta$ linkages, but about branch frequency. Hence, glycogen generally shows more frequent branching for faster mobilization.
57. Humans digest starch efficiently but not cellulose mainly because humans:
ⓐ. Lack enzymes to split $\alpha(1-4)$ bonds in starch
ⓑ. Lack enzymes to hydrolyze $\beta(1-4)$ bonds in cellulose
ⓒ. Cannot absorb glucose released from polysaccharides
ⓓ. Cannot break any glycosidic bond in carbohydrates
Correct Answer: Lack enzymes to hydrolyze $\beta(1-4)$ bonds in cellulose
Explanation: The key difference between starch and cellulose is the glycosidic bond configuration: starch contains mainly $\alpha(1-4)$ linkages, while cellulose contains $\beta(1-4)$ linkages. Human digestive enzymes are well-suited for hydrolyzing common $\alpha$ linkages in dietary starch, releasing glucose for absorption. However, humans generally do not produce effective cellulases to cleave the $\beta(1-4)$ bonds of cellulose, so it remains largely undigested. This bond-specific enzyme limitation explains why cellulose acts as dietary fiber rather than a glucose source. The concept depends on enzyme specificity to bond geometry. Therefore, lack of enzymes for $\beta(1-4)$ hydrolysis is the correct reason.
58. Which option is a heteropolysaccharide rather than a homopolysaccharide?
ⓐ. Cellulose (glucose-only structural polymer)
ⓑ. Glycogen (glucose-only storage polymer)
ⓒ. Hyaluronic acid (repeating disaccharide polymer type)
ⓓ. Amylose (glucose-only starch fraction)
Correct Answer: Hyaluronic acid (repeating disaccharide polymer type)
Explanation: A heteropolysaccharide is made from more than one type of monosaccharide arranged in a repeating pattern. Hyaluronic acid is built from repeating disaccharide units, meaning two different sugar derivatives alternate along the chain, which is the hallmark of heteropolymeric design. Such polymers often play roles in extracellular matrices and biological lubrication because their mixed monomer composition affects hydration and viscosity. In contrast, cellulose, glycogen, and amylose are homopolysaccharides because they are composed only of glucose units, even though their linkages and branching differ. The defining criterion is monomer diversity, not storage versus structure. Hence, hyaluronic acid is the heteropolysaccharide among the options.
59. Storage polysaccharides are often favored over free glucose for storage mainly because polymers:
ⓐ. Reduce osmotic effect compared with many free monomers
ⓑ. Cannot be hydrolyzed back to glucose when needed
ⓒ. Are always structural and never used for energy
ⓓ. Are insoluble in water in every tissue type
Correct Answer: Reduce osmotic effect compared with many free monomers
Explanation: Storing carbohydrates as large polymers like starch or glycogen helps minimize osmotic pressure inside cells. If the same amount of carbohydrate were stored as free glucose monomers, the number of particles in solution would be extremely high, drawing water in and disrupting cell volume and ionic balance. Polymerization packs many glucose units into fewer molecules, greatly reducing the osmotic effect while still keeping an accessible energy reserve. Cells can then hydrolyze the polymer when glucose is needed for metabolism. This strategy supports stable cellular conditions and efficient storage. Therefore, storage polysaccharides are favored because they reduce osmotic effects.
60. Cellulose microfibrils gain major strength mainly due to:
ⓐ. Dense $\alpha(1-6)$ branching that compacts the polymer
ⓑ. Covalent cross-links between every glucose unit
ⓒ. Complete insolubility caused only by lipid coating
ⓓ. Hydrogen bonding between parallel straight chains
Correct Answer: Hydrogen bonding between parallel straight chains
Explanation: Cellulose chains are linear due to $\beta(1-4)$ linkages, allowing them to align side-by-side in parallel arrays. This alignment permits extensive hydrogen bonding between hydroxyl groups on adjacent chains, bundling them into strong microfibrils. The large number of hydrogen bonds collectively provides high tensile strength and resistance to stretching, which is essential for plant cell wall support. This strength arises from orderly packing and intermolecular bonding rather than heavy branching or universal covalent cross-linking. The result is a fiber-like architecture optimized for structure. Hence, hydrogen bonding between parallel straight chains is the main basis of cellulose microfibril strength.
61. A glycosidic bond in carbohydrates is best defined as a covalent bond formed between:
ⓐ. Two fatty acids through esterification
ⓑ. Two amino acids through dehydration
ⓒ. Anomeric carbon of a sugar and a hydroxyl group of another molecule
ⓓ. A phosphate group and a nitrogen base in a nucleotide
Correct Answer: Anomeric carbon of a sugar and a hydroxyl group of another molecule
Explanation: A glycosidic bond is the covalent linkage that connects sugar units (or a sugar to a non-sugar group) by involving the anomeric carbon of a monosaccharide. When the anomeric carbon reacts with a hydroxyl group of another molecule, it forms a stable acetal-type linkage that builds disaccharides and polysaccharides. This bond is central to carbohydrate polymer formation and determines whether the linkage is $\alpha$ or $\beta$ based on stereochemistry at the anomeric carbon. The position numbers like $1-4$ or $1-6$ specify which carbon atoms participate. Therefore, the correct definition centers on the anomeric carbon forming a covalent link to an -OH group.
62. In starch, the predominant glycosidic linkage along the straight chain of amylose is:
ⓐ. $\alpha(1-4)$
ⓑ. $\beta(1-4)$
ⓒ. $\alpha(1-6)$
ⓓ. $\alpha(1-2)$
Correct Answer: $\alpha(1-4)$
Explanation: Amylose is the mostly unbranched component of starch and is built from glucose units connected mainly by $\alpha(1-4)$ glycosidic bonds. This specific linkage causes the chain to adopt a coiled or helical tendency, which is one reason amylose shows characteristic interactions in common identification tests. The $\alpha(1-4)$ bond also matches the linkage pattern that digestive enzymes commonly target when breaking down starch into smaller sugars. Because the question asks specifically for the straight-chain linkage within amylose, $\alpha(1-4)$ is the defining structural feature. Hence, $\alpha(1-4)$ is the correct answer.
63. The glycosidic bond responsible for branch points in amylopectin is:
ⓐ. $\beta(1-4)$
ⓑ. $\alpha(1-2)$
ⓒ. $\beta(1-6)$
ⓓ. $\alpha(1-6)$
Correct Answer: $\alpha(1-6)$
Explanation: Amylopectin is the branched fraction of starch, where most glucose units are linked in chains by $\alpha(1-4)$ bonds but branching occurs at specific junctions. These branch points are created when the anomeric carbon of a glucose forms a bond to the C-6 hydroxyl group of another glucose, producing an $\alpha(1-6)$ linkage. Branching increases the number of chain ends, which supports more efficient enzymatic access when the stored carbohydrate is mobilized. This branching pattern is a key way amylopectin differs from amylose. Therefore, $\alpha(1-6)$ is the correct branch-forming bond.
64. The main enzyme that initiates starch digestion by cleaving internal $\alpha(1-4)$ glycosidic bonds is:
ⓐ. Lipase
ⓑ. Amylase
ⓒ. Lactase
ⓓ. Pepsin
Correct Answer: Amylase
Explanation: Amylase is the enzyme specialized to hydrolyze $\alpha(1-4)$ glycosidic bonds within starch molecules, producing smaller carbohydrates such as maltose and oligosaccharides. This “internal cleavage” is important because it rapidly reduces large starch polymers into units that can be further processed to glucose. The specificity for $\alpha(1-4)$ linkages fits the dominant bond type found in amylose and much of amylopectin’s linear segments. Because starch digestion depends on targeting these common bonds, amylase is the correct initiating enzyme. The question focuses on starch and $\alpha(1-4)$ cleavage, which directly matches amylase function.
65. Maltose is formed when two glucose units are linked by which glycosidic bond?
ⓐ. $\alpha(1-2)$
ⓑ. $\beta(1-4)$
ⓒ. $\alpha(1-4)$
ⓓ. $\alpha(1-1)$
Correct Answer: $\alpha(1-4)$
Explanation: Maltose is a disaccharide composed of two glucose molecules connected through an $\alpha(1-4)$ glycosidic bond. This linkage is the same type commonly found along the main chains of starch, which is why maltose frequently appears as an intermediate during starch hydrolysis. The bond involves the anomeric carbon of one glucose and the C-4 hydroxyl of the second glucose, creating a stable disaccharide that can still be further hydrolyzed to glucose. Because the question asks for the bond specifically defining maltose, $\alpha(1-4)$ is the key structural identifier. Hence, $\alpha(1-4)$ is correct.
66. Cellulose differs from starch mainly because cellulose contains glucose units linked by:
ⓐ. $\alpha(1-4)$ bonds forming helical chains
ⓑ. $\alpha(1-6)$ bonds forming frequent branches
ⓒ. $\alpha(1-2)$ bonds joining two anomeric carbons
ⓓ. $\beta(1-4)$ bonds forming straight chains
Correct Answer: $\beta(1-4)$ bonds forming straight chains
Explanation: Cellulose is a structural polysaccharide in plant cell walls built from glucose units connected by $\beta(1-4)$ glycosidic bonds. This bond geometry makes the chains straight, allowing them to align closely and form strong microfibrils through extensive hydrogen bonding between neighboring chains. In contrast, starch uses mainly $\alpha$-type linkages that produce a different chain shape and support storage rather than tensile strength. The $\beta(1-4)$ linkage is therefore the defining reason cellulose behaves as a rigid structural fiber. Because the question asks for the key linkage difference, $\beta(1-4)$ is the correct answer.
67. A polysaccharide made of glucose with $\alpha(1-4)$ chains and $\alpha(1-6)$ branches is best identified as:
ⓐ. Glycogen
ⓑ. Cellulose
ⓒ. Chitin
ⓓ. Pectin
Correct Answer: Glycogen
Explanation: Glycogen is the major storage polysaccharide in animals and is composed solely of glucose units. Its main chains use $\alpha(1-4)$ glycosidic bonds, while branching points are created by $\alpha(1-6)$ linkages, producing a highly branched structure. This design generates many terminal ends, enabling rapid glucose release when energy is needed, especially in liver and muscle. The specific combination of $\alpha(1-4)$ backbone plus $\alpha(1-6)$ branching is a classic structural signature for glycogen. Therefore, the polymer described is glycogen.
68. A disaccharide is non-reducing most directly when:
ⓐ. It contains only glucose units
ⓑ. Its glycosidic bond involves both anomeric carbons
ⓒ. It has at least one $\alpha(1-4)$ linkage
ⓓ. It dissolves completely in water
Correct Answer: Its glycosidic bond involves both anomeric carbons
Explanation: Reducing behavior depends on the presence of a free anomeric carbon that can open to a reactive form in solution. When a disaccharide’s glycosidic linkage uses both anomeric carbons, neither unit retains a free anomeric carbon, so the open-chain reactive form is not readily available. This structural locking makes the sugar non-reducing in typical biochemical tests and often increases stability during transport or storage roles. The key condition is therefore the involvement of both anomeric positions in the bond, not simply the monomers present or solubility. Hence, a bond involving both anomeric carbons is the direct reason.
69. In starch, the term “glycosidic bond” most specifically refers to the covalent linkage connecting:
ⓐ. Amino acids within enzymes that digest starch
ⓑ. Fatty acids within membrane lipids of plastids
ⓒ. Phosphate groups within ATP used in starch synthesis
ⓓ. Glucose units within amylose and amylopectin
Correct Answer: Glucose units within amylose and amylopectin
Explanation: Starch is a polysaccharide built by linking many glucose monomers into long chains and branched structures. The covalent link that joins these glucose units is the glycosidic bond, mainly $\alpha(1-4)$ along chains and $\alpha(1-6)$ at branch points in amylopectin. This bond type determines starch’s polymer architecture and how enzymes recognize and hydrolyze it during digestion or mobilization. Because starch is fundamentally a glucose polymer, the glycosidic bond in this context is the linkage between glucose units. Therefore, it refers to the covalent connections joining glucose monomers in amylose and amylopectin.
70. Which statement correctly links glycosidic bond type to the function of starch as a storage polysaccharide?
ⓐ. $\beta(1-4)$ bonds create rigid fibers for mechanical support
ⓑ. $\alpha(1-2)$ bonds prevent any enzymatic hydrolysis in plants
ⓒ. $\alpha(1-4)$ bonds allow polymer formation that can be hydrolyzed to glucose when needed
ⓓ. $\alpha(1-1)$ bonds form only structural chains in plant walls
Correct Answer: $\alpha(1-4)$ bonds allow polymer formation that can be hydrolyzed to glucose when needed
Explanation: Starch functions as a storage polymer because it forms large glucose-containing molecules that can be efficiently broken down when energy or carbon is required. The predominant $\alpha(1-4)$ glycosidic bonds create long chains that enzymes can hydrolyze to release smaller sugars and ultimately glucose. This makes starch a reversible reserve: stable enough to store but accessible for mobilization. The storage role depends on the bond type supporting polymerization and controlled enzymatic breakdown rather than producing rigid, purely structural fibers. Therefore, linking $\alpha(1-4)$ bonds with hydrolysis to glucose captures the key functional logic of starch storage.
71. In cellulose, the glycosidic bond that links successive glucose units is:
ⓐ. $\alpha(1-4)$
ⓑ. $\alpha(1-6)$
ⓒ. $\alpha(1-2)$
ⓓ. $\beta(1-4)$
Correct Answer: $\beta(1-4)$
Explanation: Cellulose is a linear polymer of glucose in which each glucose unit is joined to the next by a $\beta(1-4)$ glycosidic bond. This $\beta$ linkage places successive glucose residues in an alternating orientation, producing straight, extended chains rather than coiled ones. Straight chains can align in parallel and pack closely, which is essential for forming strong cellulose fibers in plant cell walls. The specific bond type also determines which enzymes can hydrolyze the polymer, as many common digestive enzymes target $\alpha$ linkages instead. Therefore, identifying cellulose requires recognizing the $\beta(1-4)$ glycosidic linkage as its defining structural bond.
72. The exceptional tensile strength of cellulose microfibrils is most directly due to:
ⓐ. Extensive hydrogen bonding between aligned chains
ⓑ. Many $\alpha(1-6)$ branch points creating compact granules
ⓒ. Covalent cross-links between every glucose in adjacent chains
ⓓ. A lipid coating that permanently waterproofs the polymer
Correct Answer: Extensive hydrogen bonding between aligned chains
Explanation: Cellulose chains are straight and can align side-by-side, allowing many hydrogen bonds to form between hydroxyl groups of neighboring chains. These numerous intermolecular hydrogen bonds collectively stabilize the bundle, creating microfibrils with high tensile strength. The strength arises from orderly packing and repeated weak interactions acting together over long chain lengths. This is why cellulose behaves like a fiber-reinforcement material in plant cell walls. The key requirement for this bonding network is the parallel alignment made possible by cellulose’s linear chain geometry. Hence, extensive hydrogen bonding between aligned chains is the primary basis of microfibril strength.
73. The disaccharide produced on complete hydrolysis of cellulose (repeating unit released first) is:
ⓐ. Maltose
ⓑ. Cellobiose
ⓒ. Sucrose
ⓓ. Lactose
Correct Answer: Cellobiose
Explanation: Cellulose is built from glucose units linked by $\beta(1-4)$ bonds, and the immediate repeating disaccharide unit associated with this linkage is cellobiose. When cellulose is hydrolyzed stepwise, cleavage of the polymer commonly yields cellobiose fragments before they are further broken into glucose. This is a structural consequence of how glucose residues are connected in the cellulose chain. The identity of the disaccharide is important because enzymes that digest cellulose often act on cellobiose as an intermediate. Therefore, the characteristic disaccharide linked to cellulose breakdown is cellobiose.
74. Humans generally cannot digest cellulose efficiently because they lack enzymes that hydrolyze:
ⓐ. $\alpha(1-2)$ glycosidic bonds
ⓑ. $\alpha(1-6)$ glycosidic bonds
ⓒ. $\alpha(1-4)$ glycosidic bonds
ⓓ. $\beta(1-4)$ glycosidic bonds
Correct Answer: $\beta(1-4)$ glycosidic bonds
Explanation: The defining linkage in cellulose is the $\beta(1-4)$ glycosidic bond, and efficient digestion requires enzymes that specifically cleave this bond type. Humans typically do not produce cellulase enzymes that hydrolyze $\beta(1-4)$ linkages, so cellulose passes largely undigested through the gut. This bond specificity matters because common carbohydrate-digesting enzymes are adapted mainly to $\alpha$ linkages found in starch-related carbohydrates. As a result, cellulose serves largely as dietary fiber in humans rather than a glucose source. The key reason is enzyme specificity for the glycosidic bond configuration. Hence, the undigested bond type is $\beta(1-4)$.
75. The straight-chain nature of cellulose arises mainly because $\beta(1-4)$ linkages cause:
ⓐ. Permanent branching at regular intervals
ⓑ. A coiled helix similar to storage starch
ⓒ. Alternate flipping of glucose units along the chain
ⓓ. A lock that prevents any hydrogen bonding between chains
Correct Answer: Alternate flipping of glucose units along the chain
Explanation: In cellulose, each $\beta(1-4)$ bond connects glucose units in a way that orients successive residues in an alternating pattern, often described as a flipped arrangement. This repeating alternation produces an extended, straight chain rather than a compact helix. The straight geometry is crucial for close alignment of many chains, which then permits strong hydrogen bonding and microfibril formation. This structural outcome directly links bond stereochemistry to polymer shape and function in cell walls. The key point is that the $\beta$ configuration enforces the alternating orientation. Therefore, alternate flipping of glucose units explains the straight-chain character.
76. Which statement best describes the difference in glycosidic bond type between starch and cellulose?
ⓐ. Starch mainly uses $\alpha(1-4)$, cellulose uses $\beta(1-4)$
ⓑ. Starch mainly uses $\beta(1-4)$, cellulose uses $\alpha(1-4)$
ⓒ. Both mainly use $\beta(1-6)$ for branching and strength
ⓓ. Both mainly use $\alpha(1-2)$ to avoid reducing ends
Correct Answer: Starch mainly uses $\alpha(1-4)$, cellulose uses $\beta(1-4)$
Explanation: Starch and cellulose are both glucose polymers, but their primary linkages differ in stereochemistry at the anomeric carbon. Starch contains mainly $\alpha(1-4)$ bonds (and branching via $\alpha(1-6)$ in amylopectin), which supports a storage role and easy enzymatic mobilization. Cellulose contains $\beta(1-4)$ bonds, which produce straight chains that align to form strong microfibrils for structural support. This single difference in bond configuration leads to major differences in shape, packing, and digestibility. The comparison is a standard conceptual trap because both are glucose-based polymers. Hence, starch uses $\alpha(1-4)$ while cellulose uses $\beta(1-4)$.
77. In a cellulose chain, the glycosidic bond is formed between:
ⓐ. Two phosphate groups on adjacent nucleotides
ⓑ. Two amino groups on adjacent amino acids
ⓒ. An anomeric carbon and a hydroxyl group on another glucose
ⓓ. Two fatty acids through ester bond formation
Correct Answer: An anomeric carbon and a hydroxyl group on another glucose
Explanation: Glycosidic bonds are carbohydrate linkages formed when the anomeric carbon of one monosaccharide reacts with a hydroxyl group of another molecule. In cellulose, this occurs between glucose residues to create the $\beta(1-4)$ linkage that builds the polymer chain. The “anomeric carbon” is the carbon derived from the carbonyl carbon during ring formation and is the key reactive center for forming glycosidic bonds. This bonding creates a stable covalent connection that defines polymer structure and properties. In cellulose, repeated formation of this bond yields long, linear chains. Therefore, the bond forms between an anomeric carbon and a hydroxyl group on another glucose.
78. Which enzyme specifically hydrolyzes the glycosidic bonds of cellulose in organisms that can digest it?
ⓐ. Amylase
ⓑ. Cellulase
ⓒ. Lactase
ⓓ. Sucrase
Correct Answer: Cellulase
Explanation: Cellulose contains $\beta(1-4)$ glycosidic bonds, and enzymes that hydrolyze cellulose must recognize and cleave this specific linkage. Cellulase is the enzyme group specialized for breaking $\beta(1-4)$ bonds in cellulose, enabling conversion of the polymer into smaller sugars and ultimately glucose. This is why certain microbes and symbiotic systems can digest plant cell walls effectively. The enzyme’s specificity matches the structural chemistry of cellulose rather than starch-type $\alpha$ linkages. Therefore, cellulose digestion requires cellulase activity. Hence, cellulase is the correct enzyme.
79. Which observation best supports cellulose being a structural polysaccharide rather than a storage polysaccharide?
ⓐ. It is composed of fructose units in alternating forms
ⓑ. It forms compact granules designed for rapid hydrolysis
ⓒ. It forms aligned fibers with high tensile strength in cell walls
ⓓ. It dissolves readily to maintain high osmotic pressure
Correct Answer: It forms aligned fibers with high tensile strength in cell walls
Explanation: Structural polysaccharides are characterized by architecture that provides mechanical support and resistance to deformation. Cellulose chains align and bundle into microfibrils, creating fibrous material with high tensile strength that reinforces plant cell walls. This fibrous arrangement is ideal for maintaining cell shape and resisting internal turgor pressure. The structural role is evident from how cellulose organizes into strong, oriented frameworks rather than compact, readily mobilized reserves. The key link is between polymer packing and mechanical function. Therefore, forming aligned fibers with high tensile strength supports cellulose’s structural classification.
80. A “glycosidic bond” in cellulose is best described as linking glucose units mainly by:
ⓐ. $\beta(1-4)$ covalent linkages along the chain
ⓑ. $\alpha(1-6)$ covalent linkages at frequent branch points
ⓒ. $\alpha(1-2)$ covalent linkages joining two anomeric carbons
ⓓ. $\beta(1-6)$ covalent linkages producing compact granules
Correct Answer: $\beta(1-4)$ covalent linkages along the chain
Explanation: The cellulose polymer is defined by glucose residues connected through $\beta(1-4)$ glycosidic bonds repeated along the length of the chain. These covalent linkages produce a linear, unbranched structure, which allows many chains to align closely. The aligned arrangement supports extensive hydrogen bonding between chains, forming strong microfibrils in plant cell walls. Because the key structural identity of cellulose is its linear backbone, the bond type along the chain is the most important description. This directly ties the glycosidic bond type to cellulose’s rigidity and structural role. Hence, cellulose mainly uses $\beta(1-4)$ covalent linkages along the chain.
81. Glycogen is best described as a glucose polymer with a backbone mainly of:
ⓐ. $\beta(1-4)$ bonds forming rigid straight chains
ⓑ. $\beta(1-6)$ bonds forming compact fibers
ⓒ. $\alpha(1-2)$ bonds joining two anomeric carbons
ⓓ. $\alpha(1-4)$ bonds forming storage chains
Correct Answer: $\alpha(1-4)$ bonds forming storage chains
Explanation: Glycogen is a storage polysaccharide in animals made entirely of glucose units. The main chain consists of glucose residues linked by $\alpha(1-4)$ glycosidic bonds, which is the common linkage that supports easy enzymatic hydrolysis to release glucose when energy is needed. This backbone design is compatible with rapid metabolic mobilization, especially in liver and muscle. The chain structure is not rigid like cellulose because it does not use $\beta(1-4)$ linkages. The storage function depends on having bonds that metabolic enzymes can efficiently cleave. Therefore, the glycogen backbone is mainly $\alpha(1-4)$.
82. The glycosidic bond that creates branch points in glycogen is:
ⓐ. $\alpha(1-6)$
ⓑ. $\alpha(1-2)$
ⓒ. $\beta(1-4)$
ⓓ. $\beta(1-6)$
Correct Answer: $\alpha(1-6)$
Explanation: Glycogen is highly branched, and its branching occurs when a glucose unit forms a glycosidic bond from its anomeric carbon to the C-6 hydroxyl group of another glucose residue. This creates an $\alpha(1-6)$ linkage at the branch point. Branching increases the number of terminal ends, providing many sites for enzymes to rapidly add or remove glucose units. This structural feature supports quick glucose mobilization in response to energy demand. The key identification point is the C-6 involvement with an $\alpha$ configuration. Hence, the branch-forming bond in glycogen is $\alpha(1-6)$.
83. Compared with amylopectin, glycogen usually has:
ⓐ. Less branching and fewer terminal ends
ⓑ. More frequent branching and more terminal ends
ⓒ. Only $\beta(1-4)$ linkages for strength
ⓓ. No branching, forming long straight chains
Correct Answer: More frequent branching and more terminal ends
Explanation: Both glycogen and amylopectin are branched glucose polymers, but glycogen generally shows a higher density of branches. More frequent branching creates more non-reducing ends, which are the sites where enzymes rapidly remove or add glucose residues. This increases the speed of glycogen mobilization, which is important for animals that need rapid energy release during activity or fasting. The structural difference is functional, favoring quick response rather than long-term static storage. Glycogen’s high branching also makes it more compact and readily accessible to enzymes. Therefore, glycogen typically has more frequent branching and more terminal ends.
84. The key storage advantage of glycogen’s branching pattern is that it:
ⓐ. Prevents any enzymatic hydrolysis of glucose units
ⓑ. Increases the number of end points for rapid glucose release
ⓒ. Converts glucose into cellulose-like rigid fibers
ⓓ. Eliminates the need for glucose in blood entirely
Correct Answer: Increases the number of end points for rapid glucose release
Explanation: Glycogen branching creates many terminal ends, and these ends are the primary sites where enzymes can remove glucose units quickly. Because multiple enzymes can work simultaneously on different ends, the cell can mobilize glucose rapidly when energy demand rises. This is especially important in muscle for immediate ATP generation and in liver for maintaining blood glucose. The branching therefore increases both speed and flexibility of storage usage without changing the basic glucose composition. This design is a major reason glycogen is well suited as a fast-access energy reserve. Hence, branching increases the number of endpoints for rapid glucose release.
85. In glycogen, the glucose units at most terminal ends are connected by:
ⓐ. $\alpha(1-4)$ bonds along the chain
ⓑ. $\beta(1-4)$ bonds for straight fiber formation
ⓒ. $\alpha(1-2)$ bonds joining two anomeric positions
ⓓ. $\beta(1-6)$ bonds forming branch points
Correct Answer: $\alpha(1-4)$ bonds along the chain
Explanation: In glycogen, the main chain segments are formed by $\alpha(1-4)$ glycosidic bonds between glucose units. Even though branching points are created by $\alpha(1-6)$ linkages, the majority of glucose residues in glycogen lie within the $\alpha(1-4)$ linked chains. Terminal ends are typically part of these $\alpha(1-4)$ chain segments, which is why enzymes that cleave $\alpha(1-4)$ bonds can rapidly remove glucose from many ends. The polymer’s overall architecture is therefore dominated by $\alpha(1-4)$ linkages with occasional $\alpha(1-6)$ junctions. Hence, most terminal-end connections are $\alpha(1-4)$.
86. A single statement that correctly links glycogen to its biological role is:
ⓐ. Glycogen uses $\beta(1-4)$ bonds to strengthen plant cell walls
ⓑ. Glycogen is a branched $\alpha$-glucan suited for rapid energy supply
ⓒ. Glycogen is a fructose polymer used mainly for plant transport
ⓓ. Glycogen is a non-reducing sugar because it has no terminal ends
Correct Answer: Glycogen is a branched $\alpha$-glucan suited for rapid energy supply
Explanation: Glycogen is composed of glucose units linked mainly by $\alpha(1-4)$ bonds with frequent $\alpha(1-6)$ branching, making it a branched $\alpha$-glucan. This branching produces many chain ends, allowing rapid enzymatic mobilization of glucose when energy demand is high. The structure is therefore directly adapted to fast energy release rather than mechanical support. It is stored prominently in liver and muscle, matching its role in maintaining blood glucose and supplying quick fuel for contraction. The key concept is that structure (branching) supports function (rapid access). Hence, glycogen is a branched $\alpha$-glucan suited for rapid energy supply.
87. The term “glycosidic bond” in glycogen refers to the covalent linkage between:
ⓐ. Nitrogen bases in DNA strands
ⓑ. Amino acids in polypeptides
ⓒ. Glucose residues within the glycogen polymer
ⓓ. Fatty acids and glycerol in triglycerides
Correct Answer: Glucose residues within the glycogen polymer
Explanation: Glycogen is a carbohydrate polymer built by joining glucose molecules into long chains and branches. The covalent link that connects one glucose residue to another is the glycosidic bond, which in glycogen is mainly $\alpha(1-4)$ along the chain and $\alpha(1-6)$ at branch points. These bonds determine glycogen’s architecture and its metabolic behavior during synthesis and breakdown. Because the question asks what the bond links within glycogen, the answer is the glucose residues themselves. This is a fundamental point: glycosidic bonds are carbohydrate-to-carbohydrate linkages. Therefore, in glycogen, glycosidic bonds connect glucose residues within the polymer.
88. Which bond type would be most directly targeted during rapid glycogen breakdown at the chain ends?
ⓐ. $\beta(1-4)$
ⓑ. $\beta(1-6)$
ⓒ. $\alpha(1-2)$
ⓓ. $\alpha(1-4)$
Correct Answer: $\alpha(1-4)$
Explanation: During glycogen breakdown, enzymes act primarily at the non-reducing ends of the polymer where glucose units are most accessible. The main chain linkages at these ends are $\alpha(1-4)$ glycosidic bonds, so cleaving these bonds rapidly releases glucose units for energy use. Branch points exist, but the bulk of glucose removal proceeds by repeated cleavage of $\alpha(1-4)$ linkages until a branch is approached. This arrangement allows fast, stepwise mobilization from multiple ends at once. The structural dominance of $\alpha(1-4)$ linkages explains why they are the primary targets during rapid breakdown. Hence, $\alpha(1-4)$ is the bond type targeted at chain ends.
89. Which feature most clearly distinguishes glycogen from cellulose at the bond level?
ⓐ. Glycogen uses $\alpha$ linkages; cellulose uses $\beta(1-4)$ linkages
ⓑ. Glycogen has only $\beta(1-4)$ linkages; cellulose has only $\alpha(1-4)$
ⓒ. Glycogen has no glycosidic bonds; cellulose has many glycosidic bonds
ⓓ. Glycogen is made of fructose; cellulose is made of glucose
Correct Answer: Glycogen uses $\alpha$ linkages; cellulose uses $\beta(1-4)$ linkages
Explanation: Glycogen is an animal storage polymer of glucose built with $\alpha(1-4)$ bonds in the main chain and $\alpha(1-6)$ bonds at branch points. Cellulose is a plant structural polymer of glucose built with $\beta(1-4)$ bonds that produce straight chains capable of strong alignment. The stereochemistry of the glycosidic bond is therefore the most direct structural difference that drives their contrasting functions and digestibility. This bond-level distinction explains why glycogen is readily mobilized for energy while cellulose provides mechanical strength and is generally indigestible to humans. Hence, glycogen uses $\alpha$ linkages while cellulose uses $\beta(1-4)$ linkages.
90. Glycogen is stored predominantly in which organs for rapid physiological use?
ⓐ. Skin and bone marrow
ⓑ. Kidney and spleen
ⓒ. Lungs and pancreas
ⓓ. Liver and skeletal muscle
Correct Answer: Liver and skeletal muscle
Explanation: Glycogen is stored mainly in liver and skeletal muscle because these tissues have high and immediate energy demands and key roles in glucose regulation. Liver glycogen can be mobilized to help maintain blood glucose levels, especially between meals, supporting organs that depend on steady glucose supply. Muscle glycogen serves as a local energy reserve that can be rapidly broken down to fuel muscle contraction during activity. The polymer’s branching architecture supports this quick mobilization, aligning with the metabolic roles of these tissues. This distribution is a standard fact tied to glycogen’s function as a rapid-access reserve. Therefore, liver and skeletal muscle are the main storage sites.
91. In an $\alpha$-amino acid, the amino group and carboxyl group are attached to:
ⓐ. The same $\alpha$-carbon
ⓑ. Two different carbon atoms in chain
ⓒ. Two different rings in chain
ⓓ. Two different phosphate groups
Correct Answer: The same $\alpha$-carbon
Explanation: An $\alpha$-amino acid is defined by having both the amino group and the carboxyl group bonded to the same central carbon, called the $\alpha$-carbon. This carbon also carries a hydrogen atom and a variable side chain (R group), giving the general form $H_2N-CH(R)-COOH$. Because the two key functional groups share the same carbon, amino acids show predictable acid–base behavior and can link in a consistent head-to-tail manner during peptide formation. This shared $\alpha$-carbon framework is the structural basis for protein diversity, since only the R group changes while the backbone remains constant. Therefore, “same $\alpha$-carbon” is the defining structural feature.
92. At about neutral pH, most amino acids exist mainly as:
ⓐ. Only $-NH_2$ and only $-COOH$
ⓑ. Only $-NH_3^+$ and only $-COOH$
ⓒ. Zwitterion $-NH_3^+$ and $-COO^-$
ⓓ. Only $-NH_2$ and only $-COO^-$
Correct Answer: Zwitterion $-NH_3^+$ and $-COO^-$
Explanation: In water, amino acids behave as amphoteric molecules because they contain both an acidic carboxyl group and a basic amino group. Around neutral pH, the carboxyl group tends to lose a proton to become $-COO^-$, while the amino group tends to accept a proton to become $-NH_3^+$. This produces a dipolar ion called a zwitterion that has both charges but an overall net charge that depends on pH and the side chain. The zwitterionic form explains typical solubility behavior and buffering capacity of amino acids in biological fluids. Since this state is most stable in aqueous conditions near neutral pH, it is the predominant form.
93. A peptide bond is formed by joining:
ⓐ. Two R groups directly
ⓑ. Two amino groups directly
ⓒ. Two carboxyl groups directly
ⓓ. One $-COOH$ with one $-NH_2$
Correct Answer: One $-COOH$ with one $-NH_2$
Explanation: A peptide bond forms when the carboxyl group of one amino acid reacts with the amino group of another amino acid. This creates an amide linkage in the backbone, written as $-CO-NH-$, which is the repeating covalent connection in polypeptides. The reaction is specific: the carbonyl carbon from $-COOH$ and the nitrogen from $-NH_2$ become directly connected in the new bond. This head-to-tail linkage produces a directional chain with distinct ends and a consistent backbone structure across all proteins. Because this bond is the fundamental link that builds peptides and proteins, the correct joining partners are one carboxyl group and one amino group.
94. Formation of a peptide bond between two amino acids is a:
ⓐ. Oxidation reaction without water
ⓑ. Condensation reaction releasing water
ⓒ. Reduction reaction consuming water
ⓓ. Neutralization reaction producing salt
Correct Answer: Condensation reaction releasing water
Explanation: Peptide bond formation occurs through a condensation process in which a molecule of water is eliminated as the bond forms. Specifically, the $-OH$ from the carboxyl group and an $-H$ from the amino group are removed to produce $H_2O$, while the remaining carbonyl carbon and nitrogen become linked as $-CO-NH-$. This is why peptide bond formation is also called a dehydration reaction. The chemistry is central to building polypeptide chains from amino acid monomers and explains why the reaction is reversible by hydrolysis under suitable conditions. Therefore, peptide bond formation is correctly classified as condensation releasing water.
95. A key structural property of the peptide bond is that it:
ⓐ. Is planar due to partial double bond
ⓑ. Rotates freely like single bond
ⓒ. Breaks spontaneously in water
ⓓ. Forms only at high $pH$ always
Correct Answer: Is planar due to partial double bond
Explanation: The peptide bond shows partial double-bond character because of resonance between the carbonyl group and the nitrogen atom. This resonance restricts rotation around the $-CO-NH-$ bond, making the peptide bond region relatively rigid and planar. As a result, the atoms involved in the peptide bond tend to lie in the same plane, which strongly influences how polypeptide chains fold into stable shapes. This planarity is a foundational reason proteins adopt regular structural patterns and why backbone angles are constrained. The property is not about easy breaking in water, but about bonding geometry and stability. Hence, the peptide bond is planar due to partial double-bond character.
96. The $N$-terminus of a polypeptide chain refers to the end that has:
ⓐ. A free carboxyl group only
ⓑ. A free phosphate group only
ⓒ. A free amino group ($-NH_2$ or $-NH_3^+$)
ⓓ. A free sugar group only
Correct Answer: A free amino group ($-NH_2$ or $-NH_3^+$)
Explanation: Polypeptides are directional because peptide bonds join amino acids in a consistent head-to-tail manner. One end retains an unreacted amino group, and that end is called the $N$-terminus. Depending on pH, this terminal group may be present as $-NH_2$ or in protonated form as $-NH_3^+$. The opposite end retains an unreacted carboxyl group and is called the $C$-terminus. This directionality matters for naming sequences, predicting processing, and understanding how proteins are synthesized. Therefore, the $N$-terminus is the end with the free amino group.
97. A dipeptide formed from two amino acids contains how many peptide bonds?
ⓐ. 0
ⓑ. 1
ⓒ. 2
ⓓ. 3
Correct Answer: 1
Explanation: A dipeptide is a molecule made by linking exactly two amino acid residues into a single chain. Joining two amino acids requires forming one linkage between the carboxyl group of one and the amino group of the other, creating a single $-CO-NH-$ connection. That single connection is the peptide bond, and it defines the dipeptide backbone. Additional peptide bonds are only present when three or more amino acids are linked, producing tripeptides and longer polypeptides. Since only two residues are present in a dipeptide, only one bond is needed to connect them. Hence, a dipeptide contains 1 peptide bond.
98. At the isoelectric point ($pI$) of an amino acid, the molecule has:
ⓐ. Maximum positive charge
ⓑ. Maximum negative charge
ⓒ. No ionic groups at all
ⓓ. Net charge equal to 0
Correct Answer: Net charge equal to 0
Explanation: The isoelectric point, $pI$, is the pH at which an amino acid exists predominantly in a form where its overall (net) charge is zero. At this pH, positive and negative charges present on the molecule balance each other, typically as a zwitterion with $-NH_3^+$ and $-COO^-$ contributions summing to zero net charge. This concept is important because solubility and mobility in an electric field depend strongly on net charge, and many amino acids show minimum solubility near $pI$. The molecule is still ionic at $pI$, but the charges cancel overall. Therefore, at $pI$, the net charge is 0.
99. The part of an amino acid that mainly determines its specific identity and properties is the:
ⓐ. R group (side chain)
ⓑ. $\alpha$-carbon (backbone carbon)
ⓒ. Amino group (common group)
ⓓ. Carboxyl group (common group)
Correct Answer: R group (side chain)
Explanation: All standard amino acids share the same basic backbone: an $\alpha$-carbon attached to an amino group and a carboxyl group. What differs among amino acids is the side chain, called the R group, and this variation controls size, polarity, charge, and chemical reactivity. Because protein structure and function depend on how side chains interact (for example, hydrophobic packing, ionic attractions, and specific binding), the R group is the main determinant of an amino acid’s behavior. The amino and carboxyl groups are largely common to all amino acids and therefore do not define uniqueness. Thus, the R group is the component that primarily determines identity and properties.
100. Hydrolysis of a peptide bond results in:
ⓐ. Formation of a longer polypeptide chain
ⓑ. Removal of all R groups from amino acids
ⓒ. Splitting into smaller peptides or amino acids
ⓓ. Conversion of peptide into a lipid molecule
Correct Answer: Splitting into smaller peptides or amino acids
Explanation: Hydrolysis is the chemical process that breaks a bond by adding the components of water across it. For a peptide bond, hydrolysis adds $-OH$ to the carbonyl carbon side and $-H$ to the nitrogen side, reversing the condensation that originally formed the $-CO-NH-$ linkage. This cleavage converts a peptide into shorter peptide fragments, and with sufficient hydrolysis, into free amino acids. The reaction is fundamental in digestion and in many cellular protein turnover processes. It does not create longer chains; instead, it dismantles existing peptide connections. Therefore, hydrolysis of a peptide bond splits the chain into smaller peptides or amino acids.