Q1. Match the amino acids in Column I with their nature in Column II:
Column I
Column II
A. Glutamic acid
(I) Neutral
B. Lysine
(II) Aromatic
C. Valine
(III) Acidic
D. Tyrosine
(IV) Basic
Correct Answer: (a)
Glutamic acid has an extra carboxyl group, making it acidic (III). Lysine has an extra amino group, making it basic (IV). Valine has a non‑polar aliphatic side chain, so it is neutral (I). Tyrosine contains a phenol ring, so it is aromatic (II). Thus, A‑III, B‑IV, C‑I, D‑II is the correct match.
Q2. Assertion: Adenine and Guanine are purines. Reason: They are substituted pyrimidines.
Correct Answer: (c)
Adenine and guanine are indeed purine bases, which have a two‑ring structure (pyrimidine ring fused to an imidazole ring). Pyrimidines (cytosine, thymine, uracil) have a single six‑membered ring. Thus, the reason that they are substituted pyrimidines is false; they are purines, not pyrimidines. Assertion true, reason false.
Q3. Match the base with its type:
Column I
Column II
A. Adenine
(I) Pyrimidine
B. Guanine
(II) Purine
C. Cytosine
(III) Purine
D. Uracil
(IV) Pyrimidine
Correct Answer: (a)
Adenine and guanine are purines (both have two rings). Cytosine and uracil are pyrimidines (one ring). The matching places adenine under II (Purine), guanine under III (Purine), cytosine under I (Pyrimidine), and uracil under IV (Pyrimidine). Thus, A‑II, B‑III, C‑I, D‑IV is correct.
Q4. Which are correct for fatty acids? (i) Palmitic acid has 16 carbons. (ii) Arachidonic acid has 20 carbons. (iii) Carboxyl carbon is excluded from the count.
Correct Answer: (a)
Palmitic acid has 16 carbons (including the carboxyl carbon) (i). Arachidonic acid has 20 carbons (ii). The carboxyl carbon is included in the total count, so (iii) is false. Thus, (i) and (ii) are correct.
Q5. Regarding the Fluid Mosaic Model: (i) Proposed by Singer and Nicolson in 1972. (ii) Quasi-fluid nature of lipid enables lateral movement of proteins. (iii) Fluidity is important for cell division and secretion.
Correct Answer: (a)
The fluid mosaic model was proposed by Singer and Nicolson in 1972 (i). The lipid bilayer's fluidity allows lateral diffusion of proteins (ii). Membrane fluidity is essential for cell division, secretion, and signaling (iii). All statements are correct.
Q6. Identify the correct enzyme class descriptions: (i) Isomerases catalyze inter-conversion of optical isomers. (ii) Lyases catalyze removal of groups leaving double bonds. (iii) Ligases catalyze hydrolysis of bonds.
Correct Answer: (a)
Isomerases catalyze isomerisation (including optical isomer interconversion) (i). Lyases remove groups to form double bonds (ii). Ligases catalyze joining of molecules, not hydrolysis (hydrolysis is by hydrolases) (iii false). Thus, (i) and (ii) are correct.
Q7. Which vitamin is present in co-enzymes NAD and NADP?
Correct Answer: (c)
NAD (nicotinamide adenine dinucleotide) and NADP (nicotinamide adenine dinucleotide phosphate) contain the vitamin niacin (vitamin B₃) as part of their structure. Niacin is incorporated into the nicotinamide moiety, which is the active part in redox reactions. Thiamine is part of TPP, riboflavin in FAD/FMN, and pyridoxine in PLP. Thus, niacin is the correct vitamin for NAD and NADP.
Q8. An active site of an enzyme is a ______ into which the substrate fits.
Correct Answer: (b)
The active site is typically a crevice or pocket on the enzyme surface, formed by the three‑dimensional arrangement of amino acid residues. This pocket provides a specific microenvironment where the substrate binds and the reaction takes place. The shape and chemical nature of the pocket determine substrate specificity. It is not a flat surface, nor a metal ion (though metals may be present), nor a prosthetic group.
Q9. Which of the following is a 'trihydroxy propane'?
Correct Answer: (b)
Glycerol is a three‑carbon alcohol with three hydroxyl groups, hence it is called trihydroxy propane (propane‑1,2,3‑triol). It forms the backbone of many lipids, including triglycerides and phospholipids. Lecithin is a phospholipid containing glycerol, but it is not the glycerol molecule itself. Cholesterol is a sterol with a different structure, and serine is an amino acid.
Q10. Regarding the zwitterionic form of an amino acid: (i) It has both positive and negative charges. (ii) It exists at a specific pH. (iii) It is formed due to the ionization of amino and carboxyl groups.
Correct Answer: (a)
A zwitterion carries both positive (NH₃⁺) and negative (COO⁻) charges (i). It predominates at the isoelectric point (specific pH) (ii). Formation depends on the ionizable nature of the amino and carboxyl groups (iii). All statements are correct.
Q11. Assertion: Polysaccharides are homopolymers in most cases. Reason: They are made of different types of monosaccharides.
Correct Answer: (c)
Most polysaccharides (e.g., starch, glycogen, cellulose, chitin) are homopolymers composed of a single type of monosaccharide (glucose or its derivatives). Heteropolysaccharides do exist (e.g., hyaluronic acid), but they are less common. The reason says they are made of different types, which is false for homopolymers. Thus, assertion true, reason false.
Q12. Amino acids are organic compounds containing an amino group and an acidic group on the:
Correct Answer: (b)
In standard amino acids, both the amino (−NH₂) and carboxyl (−COOH) groups are attached to the same carbon atom, known as the α‑carbon. This α‑carbon also bears a side chain (R group) and a hydrogen atom. This arrangement is characteristic of all 20 proteinogenic amino acids. The α‑carbon is chiral (except glycine), giving amino acids their optical activity.
Q13. Assertion: Starch forms helical secondary structures. Reason: Starch is a variant of cellulose and serves as a storehouse of energy in plant tissues.
Correct Answer: (b)
Starch does form helical structures (amylose helix) which can trap iodine, giving the blue colour. It is an energy store in plants, but it is not a variant of cellulose; they differ in glycosidic linkages (α vs β) and structure. Both statements are true, but the reason (being an energy store) does not explain why starch forms helices; the helices are due to the α‑(1→4) glycosidic bonds and the conformation. Thus, both true but reason not correct explanation.
Q14. Nucleic acids that behave like enzymes are called:
Correct Answer: (b)
Ribozymes are RNA molecules that possess catalytic activity, capable of accelerating specific biochemical reactions. They were discovered as self‑splicing introns and are now known to perform functions like RNA cleavage and ligation. This challenges the traditional view that all enzymes are proteins. Apoenzymes are inactive protein parts of enzymes, co‑enzymes are organic cofactors, and lyases are a class of enzymes.
Q15. Which of the following is NOT a polymer?
Correct Answer: (d)
Lipids are not polymers; they are small molecules (like fatty acids, glycerol, and sterols) that may aggregate but do not form long chain polymers. Proteins are polymers of amino acids, nucleic acids are polymers of nucleotides, and polysaccharides are polymers of monosaccharides. Although lipids can form complex structures like membranes, they are not built from repeating monomer units. This distinguishes them from true biological macromolecules.
Q16. How many carbons, including the carboxyl carbon, are present in Palmitic acid?
Correct Answer: (a)
Palmitic acid is a saturated fatty acid with 16 carbon atoms in total, counting the carboxyl carbon. Its chemical formula is CH₃(CH₂)₁₄COOH. The chain length is 16, with the terminal carboxyl carbon being part of that count. It is one of the most common saturated fatty acids in animals and plants.
Q17. The first amino acid in a polypeptide chain is called the:
Correct Answer: (c)
In a polypeptide, the amino acid at the beginning of the chain retains a free amino group (−NH₂) and is termed the N‑terminal amino acid. The last amino acid has a free carboxyl group (−COOH) and is the C‑terminal. This convention is used in protein sequencing and describes the direction of synthesis (from N‑terminus to C‑terminus). The R‑group or α‑carbon positions are not used to name the termini.
Q18. To analyze organic compounds in living tissue, it is grinded in which acid?
Correct Answer: (b)
Trichloroacetic acid (Cl₃CCOOH) is used because it precipitates proteins and other macromolecules, allowing separation of the acid‑soluble and acid‑insoluble fractions. It is a strong organic acid that denatures proteins and helps extract small molecules. Other acids like HCl or H₂SO₄ are not used for this specific biochemical extraction. The choice ensures that the soluble pool contains low‑molecular‑weight compounds, while macromolecules remain in the insoluble fraction.
Q19. Lecithin is an example of a:
Correct Answer: (c)
Lecithin is a phospholipid, composed of glycerol, two fatty acids, a phosphate group, and a nitrogenous base (choline). It is a major component of cell membranes, providing amphipathic properties. Simple lipids (like triglycerides) lack phosphate; lecithin is a complex lipid. Glycoproteins are proteins with carbohydrate chains, not lipids.
Q20. Adult human haemoglobin consists of how many subunits?
Correct Answer: (c)
Adult human haemoglobin (HbA) is a tetramer composed of four subunits: two α‑globin and two β‑globin chains. Each subunit contains a haem group with an iron ion that binds oxygen. The quaternary structure allows cooperative binding of O₂, essential for efficient oxygen transport. Thus, the correct number of subunits is four.
Q21. Correct statements about Polysaccharides: (i) They are long chains of sugars. (ii) They are like cotton threads. (iii) Building blocks are monosaccharides. (iv) They are found in the acid-soluble pool.
Correct Answer: (a)
Polysaccharides are long chains of monosaccharides (i) and have a thread‑like appearance (ii). Their monomers are monosaccharides (iii). They are macromolecules and thus found in the acid‑insoluble fraction, not the soluble pool (iv false). Thus, (i), (ii), and (iii) are correct.
Q22. Assertion: Lipids are not strictly macromolecules. Reason: Their molecular weight does not exceed 800 Da.
Correct Answer: (a)
Lipids have molecular weights below 800 Da (e.g., fatty acids ~256 Da, triglycerides ~800 Da), so by definition they are not macromolecules (which are >10,000 Da). However, they often appear in the acid‑insoluble fraction because they form aggregates or vesicles, but they are not true polymers. The reason correctly explains the assertion: low molecular weight excludes them from being considered macromolecules. Thus, both statements are true and the reason is the correct explanation.
Q23. The most abundant chemical in living organisms is:
Correct Answer: (b)
Water constitutes 70–90% of the total cellular mass, making it the most abundant chemical in living organisms. It serves as the universal solvent, participates in biochemical reactions, and helps maintain temperature and pH. Proteins, carbohydrates, and lipids are more abundant in dry weight, but overall water far exceeds them. This high water content is essential for life processes and cellular structure.
Q24. Which enzyme speeds up the formation of carbonic acid by 10 million times?
Correct Answer: (b)
Carbonic anhydrase is a zinc‑containing enzyme that catalyses the reversible hydration of CO₂ to carbonic acid (H₂CO₃) at a rate of about 10⁶ molecules per second, which is roughly 10 million times faster than the uncatalysed reaction. It plays a critical role in respiration, acid‑base balance, and CO₂ transport in blood. Trypsin digests proteins, pepsin is a stomach protease, and lipase breaks down fats. None of these have such a dramatic effect on carbonic acid formation.
Q25. Compounds found in the acid-soluble pool have molecular weights ranging from:
Correct Answer: (a)
The acid‑soluble pool contains molecules with low molecular weights, typically between 18 and 800 daltons (Da). This includes amino acids, sugars, nucleotides, and other small metabolites. Molecules above 800 Da are generally considered macromolecules and are found in the acid‑insoluble fraction. The range 18–800 Da encompasses most primary and secondary metabolites.
Q26. The fraction obtained during chemical analysis that remains on the cheesecloth is called:
Correct Answer: (c)
After grinding tissue with trichloroacetic acid and filtering through cheesecloth, the material that does not pass through the cloth is the retentate. This retentate is also known as the acid‑insoluble fraction, containing proteins, nucleic acids, polysaccharides, and lipids. The filtrate, on the other hand, is the acid‑soluble pool containing small molecules like amino acids, sugars, and nucleotides. The term 'retentate' specifically refers to what is retained on the filter.
Q27. Identify the correct statements regarding Starch: (i) It is a storehouse of energy in plant tissues. (ii) It forms helical secondary structures. (iii) It can hold Iodine molecules in its helical portion. (iv) It gives a red colour with iodine.
Correct Answer: (a)
Starch is the primary energy storage molecule in plants (i). It forms helical structures (amylose helix) (ii) that can trap iodine molecules (iii). With iodine, starch gives a blue‑black colour, not red (iv false). Thus, statements (i), (ii), and (iii) are correct.
Q28. Assertion: Competitive inhibitors are used to control bacterial pathogens. Reason: They inhibit specific enzymes required by the pathogen by competing with its substrate.
Correct Answer: (a)
Many antibacterial drugs act as competitive inhibitors of essential bacterial enzymes (e.g., sulfonamides inhibit folate synthesis). These inhibitors resemble the natural substrate and bind reversibly to the active site, blocking the enzyme's action. This prevents the pathogen from carrying out vital metabolic pathways, thus controlling the infection. The reason correctly explains how competitive inhibitors work to control pathogens.
Q29. Which element's relative abundance is higher in living organisms compared to the earth's crust?
Correct Answer: (c)
Carbon and hydrogen are the primary building blocks of organic molecules, making them far more abundant in living tissues than in the earth's crust. In the crust, elements like silicon and oxygen dominate, while carbon is relatively scarce. Life is carbon-based, so its percentage in organisms is much higher than in the non‑living environment. This difference reflects the unique chemistry of life, where carbon forms stable covalent bonds with many elements.
Q30. Select the correct statements regarding Secondary Metabolites: (i) They are found in plant, fungal and microbial cells. (ii) Alkaloids, flavonoids, and rubber are examples. (iii) They always have identifiable functions in host organisms. (iv) Many are useful to human welfare.
Correct Answer: (a)
Secondary metabolites are present in plants, fungi, and microbes (i). Examples include alkaloids, flavonoids, and rubber (ii). Their functions in the host are not always clearly understood (iii false). Many are valuable to humans (e.g., drugs) (iv). Thus, (i), (ii), and (iv) are correct.
Q31. Match the lipid with its structure:
Lipid
Structure
A. Glycerol
(I) Phosphorylated organic compound
B. Phospholipid
(II) Trihydroxy propane
C. Palmitic acid
(III) 20 Carbons
D. Arachidonic acid
(IV) 16 Carbons
Correct Answer: (a)
Glycerol is trihydroxy propane (II). Phospholipids contain a phosphorylated organic compound (e.g., phosphate + choline) (I). Palmitic acid has 16 carbons (IV). Arachidonic acid has 20 carbons (III). Thus, A‑II, B‑I, C‑IV, D‑III is correct.
Q32. Which protein structure provides a 3-dimensional view and is necessary for biological activity?
Correct Answer: (c)
The tertiary structure is the overall three‑dimensional folding of a single polypeptide chain, driven by side‑chain interactions. It creates the specific active sites for enzymes and binding pockets for other molecules, making it essential for function. Primary structure is the linear sequence, secondary is local folding (helices/sheets), and quaternary is the assembly of multiple subunits. Without proper tertiary folding, most proteins lose their biological activity.
Q33. Malonate inhibits succinic dehydrogenase because it resembles:
Correct Answer: (b)
Malonate is a competitive inhibitor that closely resembles the substrate succinate (both have two carboxyl groups and a similar carbon skeleton). It binds to the active site of succinic dehydrogenase, preventing succinate from binding and thus inhibiting the enzyme. This is a classic example of competitive inhibition, where the inhibitor competes directly with the substrate. It does not resemble the enzyme, ATP, or zinc.
Q34. Assertion: Enzymes are damaged at high temperatures (above 40°C). Reason: Proteins are denatured by heat.
Correct Answer: (a)
Enzymes are proteins, and high temperatures disrupt the non‑covalent interactions that stabilise their folded structures, leading to denaturation. Denaturation causes loss of the active site conformation, hence loss of catalytic activity. Most enzymes from mesophilic organisms are denatured above ~40°C, though thermophilic enzymes are exceptions. The reason correctly explains why enzymes are damaged at high temperatures.
Q36. Which of the following is correct? (i) Adult human Hb has 4 subunits. (ii) Collagen is the most abundant animal protein. (iii) RuBisCO stands for Ribulose bisphosphate Carboxylase-Oxygenase.
Correct Answer: (a)
Adult haemoglobin is a tetramer (4 subunits) (i). Collagen is the most abundant protein in animals (ii). RuBisCO is indeed Ribulose‑1,5‑bisphosphate carboxylase‑oxygenase (iii). All statements are correct.
Q37. Match the compounds in Column I with their molecular description in Column II:
Column I
Column II
A. Acid-soluble pool
(I) Proteins, Nucleic acids, Polysaccharides
B. Acid-insoluble fraction
(II) Thousands of organic compounds
C. Micromolecules
(III) Weights 18 to 800 Da
D. Biomacromolecules
(IV) Weights > 10,000 Da
Correct Answer: (a)
The acid‑soluble pool contains thousands of organic compounds (II). The acid‑insoluble fraction consists of macromolecules like proteins, nucleic acids, and polysaccharides (I). Micromolecules have molecular weights 18‑800 Da (III). Biomacromolecules have weights >10,000 Da (IV). Thus, A‑II, B‑I, C‑III, D‑IV is correct.
Q38. The most abundant protein in the whole of the biosphere is:
Correct Answer: (c)
RuBisCO (Ribulose‑1,5‑bisphosphate carboxylase‑oxygenase) is the most abundant protein on Earth because it is the key enzyme in photosynthesis, present in all plants and many microorganisms. Collagen is the most abundant protein in the animal kingdom, but globally RuBisCO surpasses it. Insulin and hemoglobin are important but not as abundant in total biomass. RuBisCO catalyses the fixation of CO₂ in the Calvin cycle.
Q39. Match the polysaccharide with its source:
Polysaccharide
Source
A. Starch
(I) Exoskeleton of Insects
B. Cellulose
(II) Liver/Muscle
C. Glycogen
(III) Plant tissues
D. Chitin
(IV) Cotton fibre
Correct Answer: (a)
Starch is a plant energy storage (III). Cellulose is the main component of cotton fibres and plant cell walls (IV). Glycogen is stored in animal liver and muscle (II). Chitin forms the exoskeleton of insects and arthropods (I). Thus, A‑III, B‑IV, C‑II, D‑I is correct.
Q40. The rate of a physical or chemical process doubles or decreases by half for every ______ change in temperature.
Correct Answer: (b)
As a general rule, the rate of many chemical reactions doubles or halves with every 10°C change in temperature (the Q₁₀ rule). This empirical observation applies to many biological and chemical processes, though enzyme‑catalysed reactions may deviate due to denaturation at extremes. The factor of 2 per 10°C is a convenient approximation for understanding temperature dependence. Other intervals like 5°C or 20°C do not match this common rule.
Q41. Match the Secondary Metabolites in Column I with their category in Column II:
Column I
Column II
A. Morphine
(I) Toxin
B. Monoterpenes
(II) Alkaloid
C. Abrin
(III) Lectin
D. Concanavalin A
(IV) Terpenoides
Correct Answer: (a)
Morphine is an alkaloid (II). Monoterpenes are terpenoids (IV). Abrin is a toxin (I). Concanavalin A is a lectin (III). Thus, A‑II, B‑IV, C‑I, D‑III is the correct match.
Q42. Assertion: Secondary metabolites like alkaloids and rubber are useful to human welfare. Reason: Their role in host organisms is clearly understood in all cases.
Correct Answer: (c)
Many secondary metabolites are indeed valuable to humans (e.g., alkaloids as drugs, rubber for industrial use). However, the biological roles of many secondary metabolites in the producing organisms are not fully understood; some may be defensive, others are pigments or attractants, but their functions are not always clear. Therefore, the reason is false. Assertion true, reason false.
Q43. Which of the following is a secondary metabolite used as a drug?
Correct Answer: (b)
Vinblastine is an alkaloid secondary metabolite obtained from the Madagascar periwinkle (Catharanthus roseus) and is used as an anticancer drug. Abrin and ricin are toxic proteins (lectins) and are not used as drugs. Anthocyanins are pigments with antioxidant properties but are not typically classified as drugs. Vinblastine interferes with microtubule assembly, inhibiting cell division in cancer cells.
Q44. Assertion: Tertiary structure is absolutely necessary for the biological activities of proteins. Reason: It brings distant amino acid side chains close together to form active sites.
Correct Answer: (a)
The tertiary structure is the overall 3D folding of a polypeptide, which creates specific pockets (active sites) where catalysis or binding occurs. During folding, residues far apart in the primary sequence come into close spatial proximity to form the functional site. Without this folding, the protein cannot carry out its biological function. Thus, the reason correctly explains why tertiary structure is necessary.
Q45. Assertion: The catalytic activity of an enzyme is lost when the co-factor is removed. Reason: Co-factors play a crucial role in the catalytic activity of the enzyme.
Correct Answer: (a)
Co‑factors (metal ions or organic molecules) are essential for the proper functioning of many enzymes. Removal of the co‑factor often inactivates the enzyme because the co‑factor participates directly in catalysis or stabilises the active conformation. Thus, the loss of activity upon co‑factor removal is due to their crucial role. Both assertion and reason are true, and the reason correctly explains the assertion.
Q46. Assertion: Water is the most abundant chemical in living organisms. Reason: It occupies 70-90% of total cellular mass.
Correct Answer: (a)
Water makes up 70‑90% of the total mass of most cells, making it the most abundant chemical. This high percentage directly explains why water is the most abundant. Thus, both assertion and reason are true and the reason correctly explains the assertion.
Q47. The non-protein constituent bound to an enzyme to make it active is called a:
Correct Answer: (b)
Co‑factors are non‑protein components (metal ions, organic molecules) required for the catalytic activity of some enzymes. When the co‑factor is removed, the protein part (apoenzyme) is inactive; together they form the active holoenzyme. Co‑factors can be tightly bound (prosthetic groups) or loosely associated (co‑enzymes). Substrates are reactants, and metabolites are products or intermediates.
Q48. Proteins are described as ______ of amino acids.
Correct Answer: (b)
Proteins are heteropolymers because they are made from 20 different types of amino acids linked in varying sequences. A homopolymer would consist of only one type of monomer, but proteins contain many different monomers. The diversity of amino acid sequences gives rise to the vast array of protein functions. Thus, proteins are not homopolymers or non‑polymeric; they are classic heteropolymers.
Q49. Enzymes isolated from ______ organisms are thermally stable up to 80-90°C.
Correct Answer: (c)
Thermophilic organisms thrive at high temperatures (often above 50°C), and their enzymes have evolved to maintain structure and function at 80‑90°C. These enzymes are used in industrial processes that require high‑temperature stability, such as PCR (Taq polymerase). Mesophilic enzymes are active at moderate temperatures (20‑40°C), psychrophilic at low temperatures, and halophilic in high salt. Thermostability is a key feature of thermophilic enzymes due to increased hydrophobic interactions and disulfide bridges.
Q50. Chemically, amino acids are considered as substituted:
Correct Answer: (b)
Amino acids are structurally derived from methane (CH₄) where four hydrogen atoms are replaced by four different groups: −NH₂, −COOH, −H, and the R‑group. Thus, they are substituted methanes. This tetrahedral arrangement around the α‑carbon is typical of amino acids. Ethane or benzene derivatives do not match this structural pattern.
Q51. Assertion: Cellulose does not give a blue colour with Iodine. Reason: Cellulose does not contain complex helices to hold Iodine molecules.
Correct Answer: (a)
Cellulose is a linear polymer of glucose linked by β‑(1→4) bonds, forming straight chains that do not adopt helical conformations. Iodine gives a blue colour with starch because starch (amylose) forms a helical structure that traps iodine molecules. Cellulose lacks such helices, so it cannot hold iodine molecules, hence no blue colour. Both assertion and reason are true, and the reason correctly explains why cellulose does not give the blue colour.
Q52. Which of the following is correct regarding the 'acid-insoluble fraction'? (i) It contains proteins and nucleic acids. (ii) It contains lipids. (iii) It contains polysaccharides. (iv) All compounds except lipids have molecular weights above 10,000 Da.
Correct Answer: (c)
The acid‑insoluble fraction includes proteins, nucleic acids, polysaccharides, and lipids (though lipids are not true macromolecules). All components except lipids have molecular weights exceeding 10,000 Da. Thus, all four statements are correct. Hence the answer is 'All are correct'.
Q53. Assertion: The primary structure of a protein gives its positional information. Reason: It specifies which amino acid is first, second, and so on.
Correct Answer: (a)
The primary structure is the linear sequence of amino acids, which indeed provides positional information (which residue is at each position). This sequence determines the order of amino acids from N‑terminus to C‑terminus. The reason correctly explains that the primary structure gives this information. Both true and correct explanation.
Q54. Which of the following is a nucleotide?
Correct Answer: (c)
Adenylic acid (AMP) is adenosine monophosphate, which includes adenine, ribose, and a phosphate group – hence a nucleotide. Adenosine and uridine are nucleosides (no phosphate). Cytidine is also a nucleoside. Nucleotides are the building blocks of nucleic acids and play key roles in energy transfer (ATP) and signaling (cAMP).
Q55. Match the metabolic products under different conditions in pathway:
Condition
Product
A. Skeletal muscle (Anaerobic)
(I) Pyruvic acid
B. Normal Aerobic
(II) Lactic acid
C. Yeast fermentation
(III) Ethanol
Correct Answer: (a)
In anaerobic skeletal muscle, pyruvate is reduced to lactic acid (II). Under normal aerobic conditions, pyruvate enters the TCA cycle (I). In yeast fermentation, pyruvate is converted to ethanol (III). Thus, A‑II, B‑I, C‑III is correct.
Q56. Assertion: The blood of humans contains 52% proteins and 40% lipids. Reason: This ratio is constant for all cell types.
Correct Answer: (c)
Human blood (plasma or whole blood dry weight) does have a high protein and lipid content, with approximately 52% protein and 40% lipid (values may vary). However, the composition of different cell types varies widely; for example, liver cells have different protein/lipid ratios than adipose tissue or nerve cells. Thus, the assertion is true (for blood), but the reason (constant ratio) is false. Correct option: Assertion true, reason false.
Q57. Regarding enzyme inhibition, which are correct? (i) Malonate is a competitive inhibitor. (ii) It inhibits succinic dehydrogenase. (iii) It resembles succinate in structure. (iv) Competitive inhibitors are used to control bacterial pathogens.
Correct Answer: (a)
Malonate is a classic competitive inhibitor of succinic dehydrogenase (i, ii). It resembles succinate and binds at the active site (iii). Competitive inhibitors are indeed used as drugs against bacterial enzymes (iv). All statements are correct.
Q58. Which polysaccharide gives a blue colour with iodine?
Correct Answer: (c)
Starch forms helical secondary structures that trap iodine molecules, resulting in a characteristic blue‑black colour. The iodine fits inside the amylose helix, causing a charge‑transfer complex that absorbs visible light in the blue region. Glycogen also gives a reddish‑brown colour with iodine, not blue. Cellulose and chitin do not have suitable helical structures to accommodate iodine.
Q59. Regarding protein functions: (i) Some transport nutrients across membranes. (ii) Some act as intercellular ground substance. (iii) Some fight infectious agents.
Correct Answer: (a)
Proteins are diverse in function: transport proteins move nutrients (i), collagen etc. form ground substance (ii), antibodies fight infections (iii). All three statements are correct.
Q60. Regarding Amino acid groups: (i) Lysine is basic. (ii) Valine is neutral. (iii) Glutamic acid is acidic. (iv) Alanine is aromatic.
Correct Answer: (a)
Lysine is basic (i). Valine is neutral (ii). Glutamic acid is acidic (iii). Alanine is neutral with a methyl group, not aromatic (iv false). Thus, (i), (ii), and (iii) are correct.
Q61. Match the following structural features of proteins:
Column I
Column II
A. Primary
(I) Hollow woolen ball
B. Secondary
(II) Positional info/Line
C. Tertiary
(III) Assembly of subunits
D. Quaternary
(IV) Helix/Revolving staircase
Correct Answer: (a)
Primary structure is the linear sequence (positional info) (II). Secondary structure includes α‑helices and β‑sheets (like a revolving staircase) (IV). Tertiary structure is the overall 3D folding, often compared to a hollow woolen ball (I). Quaternary structure is the assembly of multiple polypeptide subunits (III). Thus, A‑II, B‑IV, C‑I, D‑III is correct.
Q62. Assertion: Zinc is a co-factor for the enzyme carboxypeptidase. Reason: It is a prosthetic group tightly bound to the apoenzyme.
Correct Answer: (c)
Zinc is indeed a co‑factor for carboxypeptidase, where it plays a catalytic role. However, zinc is a metal ion co‑factor, not a prosthetic group. Prosthetic groups are organic molecules (e.g., haem) that are tightly and permanently bound. Zinc is a metal ion that may be loosely or tightly bound, but it is not classified as a prosthetic group. Thus, the assertion is true, but the reason is false.
Q63. In the absence of any enzyme, the formation of H₂CO₃ is about ______ molecules per hour.
Correct Answer: (a)
Without carbonic anhydrase, the hydration of CO₂ to form carbonic acid proceeds very slowly, at a rate of about 200 molecules per hour. This shows the huge catalytic power of enzymes, as carbonic anhydrase can accelerate the reaction to 600,000 molecules per second. The uncatalysed rate is extremely low, making the enzyme essential for rapid CO₂ transport and pH regulation. Thus, the correct value is 200 molecules per hour.
Q64. Identify the correct definitions for carbohydrate variants: (i) Glycogen has a reducing end (right) and a non-reducing end (left). (ii) Starch is the store house of energy in plant tissues. (iii) Glycogen is the energy store in animals. (iv) Cellulose is a heteropolymer.
Correct Answer: (a)
Glycogen is a branched polysaccharide with reducing and non‑reducing ends (i). Starch is plant energy storage (ii). Glycogen is animal energy storage (iii). Cellulose is a homopolymer of glucose, not heteropolymer (iv false). Thus, (i), (ii), and (iii) are correct.
Q65. Match the co-factor with its specific example:
Column I
Column II
A. Prosthetic group
(I) NAD / NADP
B. Co-enzyme
(II) Zinc
C. Metal ion
(III) Heme
Correct Answer: (a)
Heme is a tightly bound prosthetic group (III). NAD/NADP are co‑enzymes (I). Zinc is a metal ion co‑factor (II). Thus, A‑III, B‑I, C‑II is correct.
Q66. In the amino acid Glycine, the 'R' group is replaced by:
Correct Answer: (c)
Glycine is the simplest amino acid, where the R‑group is just a hydrogen atom (−H). This makes glycine the only amino acid that is not chiral, as the α‑carbon has two identical hydrogen substituents. Other amino acids have bulkier R‑groups; for example, alanine has a methyl group (–CH₃), and serine has a hydroxymethyl group (–CH₂OH). The absence of a side chain allows glycine to fit into tight protein structures.
Q67. Which of the following is true for protein structure? (i) Primary structure is a sequence of amino acids. (ii) Secondary structure only contains left-handed helices. (iii) Tertiary structure is like a hollow woolen ball. (iv) Quaternary structure involves assembly of multiple subunits.
Correct Answer: (a)
Primary structure is the linear sequence (i). Secondary structures include α‑helices (right‑handed) and β‑sheets; left‑handed helices are not typical in proteins (ii false). Tertiary structure is the overall 3D folding, often compared to a hollow woolen ball (iii). Quaternary structure is the assembly of multiple subunits (iv). Thus, (i), (iii), and (iv) are correct.
Q68. Which of the following is a basic amino acid?
Correct Answer: (b)
Lysine contains an additional amino group in its side chain, making it basic (positively charged at physiological pH). Basic amino acids include lysine, arginine, and histidine. Glutamic acid is acidic due to a second carboxyl group, while valine and alanine are neutral (non‑polar). The basic nature of lysine is important for protein‑DNA interactions and enzyme catalysis.
Q69. Assertion: A protein is a heteropolymer. Reason: It is made of only one type of amino acid monomer.
Correct Answer: (c)
A protein is indeed a heteropolymer because it consists of many different types of amino acids (20 common ones) in a specific sequence. The reason states that it is made of only one type of monomer, which is false because homopolymers are made of one type. Therefore, the assertion is true, but the reason is false. Hence, the correct option is 'Assertion is true but reason is false'.
Q70. When a nitrogenous base is attached to a sugar, it is called a:
Correct Answer: (b)
A nucleoside consists of a nitrogenous base (purine or pyrimidine) covalently bonded to a pentose sugar (ribose or deoxyribose) via a glycosidic bond. When a phosphate group is also attached, it becomes a nucleotide. Nucleic acids are polymers of nucleotides, and polynucleotides are long chains. Thus, the base‑sugar combination is specifically termed a nucleoside.
Q71. What is the material remaining after fully burning living tissue and removing all carbon compounds called?
Correct Answer: (c)
When living tissue is completely incinerated, all organic matter (carbon, hydrogen, oxygen, nitrogen) is oxidized to CO₂, H₂O, and other gases. The residual inorganic minerals (e.g., calcium, magnesium, phosphorus) remain as a grey/white powder called ash. This ash represents the mineral content of the tissue and is distinct from dry weight (which includes organic matter). It is used to estimate the total inorganic elements present in the organism.
Q72. Statements about co-factors: (i) Apoenzyme is the protein part of the enzyme. (ii) Prosthetic groups are tightly bound to apoenzyme. (iii) Co-enzymes association is only transient. (iv) Catalytic activity is retained even if co-factor is removed.
Correct Answer: (a)
Apoenzyme is the protein component (i). Prosthetic groups are tightly bound (ii). Co‑enzymes associate transiently (iii). Activity is lost upon removal of the co‑factor (iv false). Thus, (i), (ii), and (iii) are correct.
Q73. Match the elements with % weight in Earth's crust:
Element
% Weight
A. Silicon
(I) 46.6
B. Oxygen
(II) 27.7
C. Nitrogen
(III) very little
Correct Answer: (a)
Oxygen is the most abundant element in the Earth's crust (~46.6%) (I). Silicon is next (~27.7%) (II). Nitrogen is present in very small amounts (III). Thus, A‑II, B‑I, C‑III is correct.
Q74. The zwitterionic form of an amino acid exists because of:
Correct Answer: (b)
A zwitterion has both positive and negative charges on the same molecule, formed when the amino group gains a proton (−NH₃⁺) and the carboxyl group loses a proton (−COO⁻). This is possible because both groups are ionizable and their pKa values are such that at neutral pH the net charge is zero. The phenomenon is not due to R‑groups, peptide bonds, or molecular weight. The zwitterionic form is essential for the solubility and reactivity of amino acids in aqueous solutions.
Q75. Which are correct regarding Nitrogen bases? (i) Purines have two rings (Skeletal heterocyclic ring). (ii) Uracil is a purine. (iii) Thymine is found in DNA. (iv) Nitrogen bases are found in the acid-insoluble fraction only.
Correct Answer: (a)
Purines (adenine, guanine) have two fused rings (i). Uracil is a pyrimidine, not a purine (ii false). Thymine is present in DNA (iii). Free nitrogen bases are small molecules and thus found in the acid‑soluble pool, not only the insoluble fraction (iv false). Thus, (i) and (iii) are correct.
Q76. Identify correct statements for Lipids: (i) They are generally water insoluble. (ii) Glycerol is trihydroxy propane. (iii) Lecithin is a simple lipid. (iv) Neural tissues have complex lipids.
Correct Answer: (a)
Lipids are hydrophobic and water‑insoluble (i). Glycerol is indeed trihydroxy propane (ii). Lecithin is a phospholipid (complex lipid), not simple (iii false). Neural tissues are rich in complex lipids like sphingolipids (iv). Thus, (i), (ii), and (iv) are correct.
Q77. The transient state structure formed during an enzyme reaction is:
Correct Answer: (b)
The transition state is a high‑energy, short‑lived intermediate that occurs between substrate and product. It is highly unstable because bonds are partially broken and formed, with a high free energy. Enzymes stabilise this transition state, lowering the activation energy and speeding up the reaction. The final product is stable, and cofactors are separate components.
Q78. Which statements about enzymes are correct? (i) They are damaged at high temperatures above 40°C. (ii) They increase the rate of reactions by millions of times. (iii) They work by increasing the activation energy. (iv) They possess an active site.
Correct Answer: (a)
Enzymes are typically denatured above ~40°C (i). They accelerate reactions dramatically, often by millions of times (ii). They lower activation energy, not increase it (iii false). They have an active site for substrate binding (iv). Thus, (i), (ii), and (iv) are correct.
Q79. Assertion: Inorganic catalysts differ from enzymes. Reason: Inorganic catalysts work efficiently at high temperatures and pressures.
Correct Answer: (a)
Inorganic catalysts (e.g., metal oxides) are often more robust and can function under extreme conditions of temperature and pressure. Enzymes are biological catalysts that are highly specific and sensitive to temperature and pH, and they work under mild conditions. This difference in operating conditions is a key distinction between the two. Hence, both statements are true and the reason correctly explains the difference.
Q80. Regarding the structure of DNA and RNA, which statements are true? (i) DNA contains deoxyribose sugar. (ii) RNA contains ribose sugar. (iii) Both are made up of nucleotides only. (iv) Nitrogen bases are heterocyclic compounds.
Correct Answer: (a)
DNA has deoxyribose, RNA has ribose (i, ii). Both are polymers of nucleotides (iii). The nitrogenous bases (purines and pyrimidines) are heterocyclic rings (iv). All statements are correct.
Q81. Enzymes lower the ______ to make the transition of S to P easy.
Correct Answer: (b)
Enzymes function by lowering the activation energy (Eₐ) of the reaction – the energy barrier that must be overcome for the reaction to proceed. This is achieved by stabilising the transition state, thereby reducing the free energy of activation. They do not change the potential energy of the products or the temperature, nor do they alter the substrate concentration directly. A lower activation energy means more substrate molecules have sufficient energy to react, increasing the reaction rate.
Q82. Match the proteins in Column I with their functions in Column II:
Column I
Column II
A. Collagen
(I) Hormone
B. Insulin
(II) Intercellular ground substance
C. GLUT-4
(III) Fights infectious agents
D. Antibody
(IV) Enables glucose transport
Correct Answer: (a)
Collagen is the main component of intercellular ground substance (II). Insulin is a peptide hormone (I). GLUT‑4 is a glucose transporter (IV). Antibodies defend against pathogens (III). Thus, A‑II, B‑I, C‑IV, D‑III is correct.
Q83. Assertion: Oils remain as oil in winters. Reason: Oils have higher melting points than fats.
Correct Answer: (c)
Oils are unsaturated fats with one or more double bonds, giving them a lower melting point, so they remain liquid at room temperature and in winter. Fats are saturated and have higher melting points, making them solid at room temperature. The reason states the opposite (oils have higher melting points), which is false. Thus, the assertion is true, but the reason is false.
Q84. Assertion: The zwitterionic form of amino acids changes with pH. Reason: The amino and carboxyl groups in amino acids are ionizable.
Correct Answer: (a)
Amino acids exist as zwitterions at their isoelectric point (pI), where the net charge is zero. As pH changes, the protonation state of the −NH₂ and −COOH groups changes, altering the net charge and the zwitterionic form. The ionizable nature of these groups is the basis for this pH‑dependent behaviour. Hence, both assertion and reason are true, and the reason correctly explains the change.
Q85. Which of the following are characteristics of amino acids? (i) They are α-amino acids. (ii) They are substituted methanes. (iii) Only 20 types occur in proteins. (iv) The R group in serine is a methyl group.
Correct Answer: (a)
Amino acids found in proteins are α‑amino acids (i). They are substituted methanes (ii). There are 20 standard amino acids in proteins (iii). In serine, the R group is hydroxymethyl (–CH₂OH), not a methyl group (iv false). Thus, (i), (ii), and (iii) are correct.
Q86. Match the carbohydrate with its monomer or type:
Column I
Column II
A. Cellulose
(I) Fructose polymer
B. Inulin
(II) Glucose homopolymer
C. Glycogen
(III) Nitrogen containing glucose derivative
D. Chitin
(IV) Animal variant of starch
Correct Answer: (a)
Cellulose is a glucose homopolymer (II). Inulin is a fructose polymer (I). Glycogen is the animal storage polysaccharide, analogous to starch (IV). Chitin is made of N‑acetylglucosamine, a nitrogen‑containing derivative (III). Thus, A‑II, B‑I, C‑IV, D‑III is correct.
Q87. Assertion: Chitin is a homopolymer. Reason: It is found in the exoskeletons of arthropods.
Correct Answer: (b)
Chitin is a homopolymer of N‑acetylglucosamine, so the assertion is true. It is indeed found in arthropod exoskeletons, so the reason is also true. However, the location (exoskeleton) does not explain why chitin is a homopolymer; the chemical structure (monomer composition) determines that. Thus, both true but reason is not the correct explanation.
Q88. High temperature destroys enzymatic activity because it causes ______ of proteins.
Correct Answer: (c)
Heat disrupts the weak bonds (hydrogen bonds, hydrophobic interactions, disulfide bridges) that maintain the tertiary and quaternary structures of proteins. This unfolding, called denaturation, leads to loss of the active site shape and therefore enzymatic activity. Denaturation is usually irreversible, though some proteins can refold. Ionisation changes may occur but are not the primary cause; saturation refers to substrate binding, and inhibition is caused by specific molecules.
Q89. Match the enzyme classes in Column I with their reaction type in Column II:
Column I
Column II
A. Dehydrogenases
(I) Linking 2 compounds
B. Transferases
(II) Oxidoreduction
C. Hydrolases
(III) Transfer of a group
D. Ligases
(IV) Hydrolysis of bonds
Correct Answer: (a)
Dehydrogenases catalyse oxidation‑reduction reactions (II). Transferases transfer functional groups (III). Hydrolases catalyse hydrolysis (IV). Ligases join two molecules (I). Thus, A‑II, B‑III, C‑IV, D‑I is correct.
Q90. Inulin is a polymer of:
Correct Answer: (b)
Inulin is a polymer of fructose units, linked by β‑(2→1) glycosidic bonds. It is a storage polysaccharide found in many plants, such as chicory and Jerusalem artichoke. Unlike starch (glucose polymer) or cellulose (also glucose), inulin is composed entirely of fructose. It is not digestible by humans and is used as a dietary fiber.
Q91. Enzymes that catalyse the linking together of two compounds are called:
Correct Answer: (b)
Ligases are enzymes that join two molecules together, often using the energy from ATP to form new bonds (e.g., DNA ligase). Lyases catalyse the removal of groups to form double bonds, isomerases rearrange molecules, and hydrolases break bonds by adding water. Linking reactions are essential in DNA replication, repair, and biosynthesis. Thus, the correct class is ligases.
Q92. A fatty acid is said to be unsaturated if it has:
Correct Answer: (b)
Unsaturated fatty acids contain one or more carbon‑carbon double bonds in their hydrocarbon chain. These double bonds introduce kinks in the chain, lowering the melting point and making the fatty acid liquid at room temperature (oils). Saturated fatty acids have only single bonds and are typically solid at room temperature. The presence of double bonds also affects the nutritional and stability properties of lipids.
Q93. Assertion: Ribozymes are enzymes. Reason: All enzymes are proteins.
Correct Answer: (c)
Ribozymes are RNA molecules with catalytic activity, so they are indeed enzymes (by function). However, the statement 'all enzymes are proteins' is false because ribozymes are nucleic acid enzymes. Thus, the assertion is true, but the reason is false. This demonstrates that not all enzymes are proteins.
Q94. Exoskeletons of arthropods have a complex polysaccharide called:
Correct Answer: (b)
Chitin is a long‑chain polymer of N‑acetyl‑β‑D‑glucosamine, a derivative of glucose. It provides structural support in the exoskeletons of arthropods (insects, crustaceans) and in fungal cell walls. Cellulose is found in plant cell walls, inulin is a fructose polymer, and glucosamine is a monomer. Chitin is the second most abundant polysaccharide in nature after cellulose.
Q95. Match the catalytic steps in order:
Step
Action
A. Step 1
(I) Breakage of chemical bonds
B. Step 2
(II) Release of product
C. Step 3
(III) Substrate binds to active site
D. Step 4
(IV) Enzyme alters shape
Correct Answer: (a)
The catalytic cycle: Step 1 – substrate binds to active site (III). Step 2 – enzyme changes shape (induced fit) (IV). Step 3 – chemical bonds are broken/formed (I). Step 4 – product is released (II). Thus, A‑III, B‑IV, C‑I, D‑II is correct.