Revision guide
How to use NEET 2022 Biology PYQs
This set contains reviewed questions from the Code Q6 English paper. Use it to test recall, then use the explanations to return to the relevant NCERT concept instead of memorising option letters.
High-yield chapters in this set
- Biotechnology: Principles and Processes
- Molecular Basis of Inheritance
- Principles of Inheritance and Variation
Best review method
Mark every statement-based question during your first attempt. On review, identify the single NCERT word that makes each option true or false.
NCERT focus
Revisit recombinant DNA steps, inheritance ratios and molecular biology processes. Do not replace NCERT wording with coaching shorthand when checking close options.
Frequently asked questions
NEET 2022 Biology PYQ FAQs
Are these NEET 2022 Biology answers based on the official key?
This page uses the NEET 2022 Code Q6 English question paper as source material and checks answers against the official NTA final answer key.
Why should I solve NEET 2022 Biology PYQs?
NEET 2022 Biology PYQs help students identify recurring NCERT concepts across genetics, ecology, plant physiology, biotechnology, and human physiology.
Question 101 · Biotechnology: Principles and Processes
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Polymerase chain reaction is used in DNA amplification.
Reason (R): The ampicillin resistant gene is used as a selectable marker to check transformation.
In the light of the above statements, choose the correct answer from the options given below:
ABoth (A) and (R) are correct and (R) is the correct explanation of (A)
BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)
C(A) is correct but (R) is not correct
D(A) is not correct but (R) is correct
Answer: B. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Assertion (A) is correct: PCR is a fundamental technique for enzymatically amplifying specific DNA sequences using repeated cycles of denaturation, annealing, and extension — explicitly covered in NCERT Class 12, Chapter 11. Reason (R) is also correct: the ampicillin resistance gene (amp^R) is indeed a commonly used selectable marker in plasmid vectors (e.g., pBR322) to identify transformed bacterial cells — this is directly stated in NCERT on page 202 (2023–24 edition). However, (R) does not explain (A), because PCR operates *in vitro* without requiring bacterial transformation or antibiotic selection; it relies on thermostable DNA polymerase and primers, not selectable markers. Selectable markers function in cloning and transformation workflows — a distinct step *after* PCR amplification if the product is to be cloned. Thus, while both statements are factually accurate, there is no cause–effect or explanatory link between them. This aligns precisely with option B.
Question 102 · Molecular Basis of Inheritance
The process of translation of mRNA to proteins begins as soon as:
AThe small subunit of ribosome encounters mRNA
BThe larger subunit of ribosome encounters mRNA
CBoth the subunits join together to bind with mRNA
DThe tRNA is activated and the larger subunit of ribosome encounters mRNA
Answer: A. The small subunit of ribosome encounters mRNA
Translation initiates when the small ribosomal subunit (40S in eukaryotes, 30S in prokaryotes) binds to the 5′ cap of mRNA and scans downstream to locate the 5′ UTR and the first AUG start codon. Only after recognizing the start codon does the large subunit (60S or 50S) join to form the complete 80S (or 70S) ribosome, enabling elongation. Thus, the very first event—marking the beginning of translation—is the small subunit’s binding to mRNA, not subunit association or tRNA activation. tRNA charging (activation) occurs prior to translation and is independent of ribosome-mRNA interaction. Option C describes initiation complex formation but not the *start* of translation; option B and D misplace the sequence of events. NCERT Class 12 (Chapter 6: Molecular Basis of Inheritance, page 112–113) explicitly states: 'The small subunit of the ribosome binds to the mRNA near its 5′ end... scanning for the start codon.' Hence, option A is scientifically precise and NCERT-aligned.
Question 103 · Plant Growth Regulators
The gaseous plant growth regulator is used in plants to:
Aspeed up the malting process
Bpromote root growth and root hair formation to increase the absorption surface
Chelp overcome apical dominance
Dkill dicotyledonous weeds in the fields
Answer: B. promote root growth and root hair formation to increase the absorption surface
The gaseous plant growth regulator referred to is ethylene. While ethylene is best known for promoting fruit ripening and abscission, it also plays a role in root development — particularly in stimulating the formation of root hairs and lateral roots under stress conditions like flooding or nutrient deficiency. This enhances the absorptive surface area of roots for water and minerals. Option A (malting) is associated with gibberellins; option C (overcoming apical dominance) is primarily auxin- and cytokinin-mediated; option D (killing dicots) describes synthetic auxins like 2,4-D. Ethylene’s role in root hair initiation is explicitly covered in NCERT Class 11 Chapter 15 'Plant Growth and Development', where it is noted that ethylene modulates root architecture, including root hair density, especially in response to environmental cues. Thus, option B is scientifically accurate and NCERT-aligned.
Question 104 · Animal Kingdom: Structural Organisation in Animals
Exoskeleton of arthropods is composed of:
ACutin
BCellulose
CChitin
DGlucosamine
Answer: C. Chitin
The exoskeleton of arthropods—such as insects, crustaceans, and arachnids—is a rigid, protective outer covering that provides structural support, prevents desiccation, and serves as a site for muscle attachment. According to NCERT Class 11 Biology (Chapter 4: Animal Kingdom), this exoskeleton is primarily composed of chitin—a tough, flexible, nitrogen-containing polysaccharide (a derivative of glucose) polymerized from N-acetylglucosamine units. Chitin is secreted by the epidermis and forms a layered, sclerotized structure often reinforced with calcium carbonate (in crustaceans) or proteins (in insects). Cutin is a waxy polymer found in plant cuticles—not animal structures. Cellulose is a structural polysaccharide in plant cell walls, absent in animals. Glucosamine is a monomeric precursor of chitin but not the structural polymer itself; the functional, load-bearing component is chitin. Hence, option (3) Chitin is biologically accurate and aligns precisely with NCERT’s description of arthropod anatomy.
Question 105 · Water Transport in Plants
Which of the following is not observed during apoplastic pathway?
AMovement of water occurs through intercellular spaces and cell walls
BThe movement does not involve crossing of cell membranes
CThe movement is aided by cytoplasmic streaming
DApoplast is continuous and does not provide any barrier to water movement
Answer: C. The movement is aided by cytoplasmic streaming
The apoplastic pathway is a route for water and solute movement through the non-living components of plant tissue — specifically, the intercellular spaces and the porous cellulose matrix of cell walls. Since it bypasses the protoplasts, water movement here does not require crossing any plasma membrane (so option B is correct). The apoplast forms a continuous, low-resistance pathway from root epidermis to xylem, unimpeded by membranes or cytoplasm (making D correct). Option A accurately describes the physical route — intercellular spaces and cell walls. However, cytoplasmic streaming is an active, ATP-dependent process occurring *within living cells*, involving cyclosis of cytoplasm and organelles; it facilitates symplastic transport (through plasmodesmata) but plays no role in the apoplastic pathway, which is purely passive and extracellular. Hence, option C is incorrect and therefore the right answer to 'which is NOT observed'. This distinction is clearly emphasized in NCERT Class 11 Biology (Chapter 11: Transport in Plants), where apoplast is defined as the system of interconnected cell walls and intercellular spaces, independent of cytoplasmic activity.
Question 106 · Biodiversity and Conservation
Which of the following is not a method of ex situ conservation?
AIn vitro fertilization
BNational Parks
CMicropropagation
DCryopreservation
Answer: B. National Parks
Ex situ conservation involves protecting endangered species outside their natural habitats, using techniques like cryopreservation (storing gametes or embryos at ultra-low temperatures), micropropagation (clonal propagation of plants via tissue culture), and in vitro fertilization (assisted reproductive technology for animals). These methods preserve genetic material or propagate individuals under controlled laboratory or artificial conditions. In contrast, National Parks are protected areas established within the natural range of species to conserve ecosystems and biodiversity in situ — i.e., on-site, in their native environment. As per NCERT Class 12 Biology (Chapter 15: Biodiversity and Conservation), in situ strategies include biosphere reserves, national parks, wildlife sanctuaries, and sacred groves; ex situ approaches include zoological parks, botanical gardens, seed banks, cryobanks, and tissue culture facilities. Therefore, National Parks — being an in situ measure — is correctly identified as *not* an ex situ method, making option (2) — corresponding to choice B — the correct answer.
Question 107 · Mineral Nutrition
Match List-I with List-II:
List-I
(a) Manganese
(b) Magnesium
(c) Boron
(d) Iron
List-II
(i) Activates the enzyme catalase
(ii) Required for pollen germination
(iii) Activates enzymes of respiration
(iv) Functions in splitting of water during photosynthesis
A(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
B(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
C(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
D(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
Answer: B. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
Manganese is essential for the photolysis of water in Photosystem II, directly participating in the oxygen-evolving complex — matching (a) with (iv). Magnesium is a central constituent of chlorophyll and activates many respiratory enzymes like hexokinase and phosphofructokinase; however, option (iii) refers to 'enzymes of respiration', which is primarily associated with magnesium — but NCERT Class 11 (Chapter 12: Mineral Nutrition) explicitly states magnesium activates *respiratory enzymes*, while manganese is linked to water splitting. Boron is critical for pollen germination and pollen tube growth — confirmed in NCERT (p. 199, 5th ed.), so (c)-(ii) is correct. Iron is required for chlorophyll synthesis and functions as a cofactor for catalase and peroxidase — NCERT clearly states iron activates catalase (p. 198), hence (d)-(i). Thus, the correct pairing is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i), corresponding to option B. Note: Though magnesium also supports photosynthesis structurally, its enzymatic role in respiration is distinct and emphasized in NCERT.
Question 108 · Biotechnology: Principles and Processes
Which one of the following statements is not true regarding the gel electrophoresis technique?
AThe process of extraction of separated DNA strands from the gel is called elution.
BThe separated DNA fragments are stained by using ethidium bromide.
CThe presence of a chromogenic substrate gives blue-coloured DNA bands on the gel.
DBright orange-coloured bands of DNA can be observed in the gel when exposed to UV light.
Answer: C. The presence of a chromogenic substrate gives blue-coloured DNA bands on the gel.
Gel electrophoresis separates DNA fragments based on size and charge. After electrophoresis, DNA is visualized by staining — commonly with ethidium bromide (EtBr), which intercalates into DNA and fluoresces bright orange under UV light. This makes option (4) correct. Elution refers to the recovery of DNA fragments from the gel matrix — so option (1) is accurate. Option (2) is also correct: EtBr is the standard fluorescent stain for DNA in agarose gels. However, option (3) is incorrect: chromogenic substrates (e.g., X-gal in blue-white screening) produce blue colour *only* in recombinant bacterial colonies expressing β-galactosidase — not in DNA bands on gels. Gel-based DNA detection does not involve chromogenic substrates; it relies on fluorophores like EtBr or SYBR Safe. NCERT Class 12 (Chapter 11, Biotechnology) explicitly states that 'DNA fragments can be visualised only after staining with ethidium bromide followed by exposure to UV radiation' — confirming that blue bands from chromogenic substrates have no role in standard gel electrophoresis. Hence, statement (3) is not true.
Question 109 · Oxidative phosphorylation and photophosphorylation
Which one of the following is not true regarding the release of energy during ATP synthesis through chemiosmosis?
ABreakdown of proton gradient
BBreakdown of electron gradient
CMovement of protons across the membrane to the stroma
DReduction of NADP⁺ to NADPH on the stroma side of the membrane
Answer: B. Breakdown of electron gradient
Chemiosmosis in mitochondria (oxidative phosphorylation) and chloroplasts (photophosphorylation) relies on a proton gradient—not an electron gradient—across the inner mitochondrial or thylakoid membrane. Energy for ATP synthesis is released when protons flow back through ATP synthase, driving conformational changes that phosphorylate ADP. This process requires breakdown of the proton gradient (Option A, true), movement of protons into the mitochondrial matrix or chloroplast stroma (Option C, true), and in chloroplasts, NADP⁺ reduction to NADPH occurs on the stroma side via ferredoxin-NADP⁺ reductase (Option D, true). However, 'breakdown of electron gradient' is biologically inaccurate—electrons move down an energy gradient in the ETC, but no stable 'electron gradient' is established or dissipated to drive ATP synthesis; it is the electrochemical proton gradient that powers ATP synthase. Hence, Option B is not true and is the correct choice.
Question 110 · Molecular Basis of Inheritance
DNA polymorphism forms the basis of:
AGenetic mapping
BDNA fingerprinting
CBoth genetic mapping and DNA fingerprinting
DTranslation
Answer: C. Both genetic mapping and DNA fingerprinting
DNA polymorphism refers to inheritable variations in DNA sequence—such as single nucleotide polymorphisms (SNPs), variable number tandem repeats (VNTRs), and microsatellites—that occur at high frequency in populations. These variations serve as molecular markers for constructing genetic maps, where relative positions of genes are determined based on recombination frequencies between polymorphic loci—making genetic mapping possible. Simultaneously, highly polymorphic non-coding regions (e.g., VNTRs) produce unique banding patterns in Southern blotting or PCR-based assays, forming the scientific foundation of DNA fingerprinting used in forensic identification and paternity testing. Translation—the process of protein synthesis from mRNA—is governed by the genetic code and ribosomal machinery, not by DNA sequence variation among individuals; hence it is unrelated to polymorphism. As both genetic mapping and DNA fingerprinting critically depend on naturally occurring DNA polymorphisms, option C is correct. This concept is explicitly covered in NCERT Class 12 Biology Chapter 6 'Molecular Basis of Inheritance', pages 119–122.
Question 111 · Biodiversity and Conservation
Habitat loss and fragmentation, over-exploitation, alien species invasion, and co-extinction are causes for:
APopulation explosion
BCompetition
CBiodiversity loss
DNatality
Answer: C. Biodiversity loss
Habitat loss and fragmentation—such as deforestation and urbanisation—destroy natural homes, reducing species’ ranges and isolating populations. Over-exploitation includes unsustainable hunting, fishing, and logging, directly depleting species faster than they can reproduce. Alien species invasion occurs when non-native organisms (e.g., Lantana camara or Nile perch) outcompete or prey upon endemic species lacking evolutionary defences. Co-extinction happens when the extinction of one species (e.g., a pollinator or host-specific parasite) triggers the loss of another interdependent species. All four factors are primary drivers of biodiversity loss, as explicitly stated in NCERT Class 12, Chapter 15 'Biodiversity and Conservation' (page 262, 2nd edition). Population explosion refers to rapid human population growth; competition is an ecological interaction, not a consequence; natality is the birth rate—none are outcomes of these threats. Hence, option C (Biodiversity loss) is scientifically accurate and NCERT-aligned.
Question 112 · Environmental Issues
The device which can remove particulate matter present in the exhaust from a thermal power plant is:
ASTP
BIncinerator
CElectrostatic Precipitator
DCatalytic Converter
Answer: C. Electrostatic Precipitator
Electrostatic precipitators (ESPs) are widely used in thermal power plants to remove fine particulate matter—such as ash and dust—from exhaust flue gases before they are released into the atmosphere. They work on the principle of electrostatic attraction: smoke particles pass through a chamber with charged wires (corona discharge), acquiring a negative charge, and are then attracted to positively charged collecting plates. The deposited particles are periodically removed by vibration or rapping. This method is highly efficient (>99%) for removing particulates larger than 1 µm. In contrast, STPs (Sewage Treatment Plants) treat domestic wastewater; incinerators combust solid waste at high temperatures but do not target gaseous particulates; and catalytic converters—used in automobile exhaust systems—reduce gaseous pollutants like CO, NOₓ, and unburnt hydrocarbons, but are ineffective against suspended particulate matter. As per NCERT Class 12 Biology (Chapter 16: Environmental Issues), ESPs are explicitly cited as the primary control device for particulate emissions from thermal power stations.
Question 113 · Plant growth and development: Plasticity
Which one of the following plants does not show plasticity?
ACotton
BCoriander
CButtercup
DMaize
Answer: D. Maize
Plasticity refers to the ability of plants to alter their morphology, anatomy, or physiology in response to environmental changes — a key adaptive feature. Cotton (Gossypium), coriander (Coriandrum sativum), and buttercup (Ranunculus) all exhibit remarkable plasticity: cotton shows heterophylly (different leaf shapes on same plant under varying conditions), coriander displays heterophylly (seedling vs. mature leaves differ markedly), and buttercup exhibits developmental plasticity in leaf shape (submerged vs. aerial leaves). Maize (Zea mays), however, is a monocot with determinate growth and highly canalized development; its leaf shape, phyllotaxy, and organ identity remain largely fixed regardless of environment — it lacks significant phenotypic plasticity in vegetative structures. NCERT Class 11 Biology (Chapter 15: Plant Growth and Development) explicitly states that plasticity is common in dicots like buttercup and coriander but absent or minimal in many monocots such as maize, which rely more on genetic programming than environmental modulation for development.
Question 114 · Organisms and Populations: Predation
Which one of the following statements cannot be connected to predation?
AIt helps in maintaining species diversity in a community
BIt might lead to extinction of a species
CBoth the interacting species are negatively impacted
DIt is necessitated by nature to maintain the ecological balance
Answer: C. Both the interacting species are negatively impacted
Predation is an interspecific interaction where one species (predator) benefits by killing and consuming another (prey). Hence, only the predator gains; the prey is harmed — making it a +/– interaction. Statement C claims 'both species are negatively impacted', which is incorrect: this describes competition (–/–), not predation. In contrast, statement A aligns with NCERT Class 12 (Ch. 13) — predation can prevent competitive exclusion and thus sustain diversity (e.g., keystone predators like sea stars). Statement B is valid: overexploitation or invasive predators (e.g., brown tree snake in Guam) have caused extinctions. Statement D reflects the ecological role of predation in regulating prey populations and energy flow, supporting stability — consistent with NCERT’s emphasis on functional roles in ecosystems. Therefore, only option C misrepresents the fundamental nature of predation and cannot be logically connected to it.
Question 115 · Respiration in Plants and Microorganisms
What amount of energy is released from glucose during lactic acid fermentation?
AApproximately 15%
BMore than 18%
CAbout 10%
DLess than 7%
Answer: D. Less than 7%
During aerobic respiration, complete oxidation of one molecule of glucose yields ~2900 kJ/mol of energy, with ~30–32 ATP molecules (each ~30.5 kJ), capturing ~40% of total energy as usable chemical energy. In contrast, lactic acid fermentation is an anaerobic process where glucose is incompletely oxidized to lactate without involving the electron transport chain or oxidative phosphorylation. Only substrate-level phosphorylation occurs in glycolysis, producing a net gain of 2 ATP (~61 kJ). Since the total energy content of glucose is ~2900 kJ/mol, the energy conserved in ATP is only about 61/2900 ≈ 2.1%, and even accounting for heat release and inefficiencies, the *total usable energy released* (i.e., captured + dissipated) remains low — significantly less than 7% of glucose’s total energy is harnessed or liberated in a biologically useful form. NCERT Class 11 (Chapter 14: Respiration in Plants) explicitly states that fermentation yields 'very little energy' compared to aerobic respiration and that most energy remains in lactate. Hence, 'Less than 7%' is scientifically accurate and aligns with NCERT’s comparative energy yield framework.
Question 116 · Principles of Inheritance and Variation
Given below are two statements:
Statement I: Mendel studied seven pairs of contrasting traits in pea plants and proposed the Laws of Inheritance.
Statement II: Seven characters examined by Mendel in his experiment on pea plants were seed shape and colour, flower colour, pod shape and colour, flower position and stem height.
In the light of the above statements, choose the correct answer from the options given below:
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer: A. Both Statement I and Statement II are correct
Statement I is correct: Gregor Mendel, based on his experiments with Pisum sativum (garden pea), identified seven pairs of contrasting traits and formulated the Law of Segregation and Law of Independent Assortment. Statement II is also correct: The seven characters Mendel studied were — (i) seed shape (round/wrinkled), (ii) seed colour (yellow/green), (iii) flower colour (purple/white), (iv) pod shape (inflated/constricted), (v) pod colour (green/yellow), (vi) flower position (axial/terminal), and (vii) stem height (tall/dwarf). Note that 'seed shape and colour' refers to two distinct traits (not one combined trait), and similarly 'pod shape and colour' denotes two separate traits — a conventional phrasing used in NCERT Class 12 (Chapter 5, page 76–77) to list all seven characters concisely. Thus, both statements accurately reflect Mendel’s work as per NCERT, making option A correct.
Question 117 · Ecosystem – Decomposition
Given below are two statements:
Statement I: Decomposition is a process in which the detritus is degraded into simpler substances by microbes.
Statement II: Decomposition is faster if the detritus is rich in lignin and chitin.
In the light of the above statements, choose the correct answer from the options given below:
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer: C. Statement I is correct but Statement II is incorrect
Statement I is correct: As per NCERT Class 12 Biology (Chapter 14: Ecosystem), decomposition is the physical and biochemical breakdown of complex organic matter (detritus) into simpler inorganic substances by decomposers—mainly bacteria and fungi. This process involves fragmentation, leaching, catabolism, humification, and mineralisation. Statement II is incorrect because lignin and chitin are highly resistant, complex polymers—lignin is abundant in wood and confers structural rigidity to plant cell walls, while chitin forms exoskeletons of arthropods and fungal cell walls. Both are recalcitrant and slow down decomposition; detritus rich in nitrogen and water-soluble compounds (e.g., sugars, amino acids) decomposes faster. In contrast, high lignin content significantly reduces decomposition rate, as confirmed in NCERT’s discussion on decomposition rate determinants (Page 243, 2023 edition). Hence, only Statement I is correct, making option C the right choice.
Question 118 · Chromosomal Organization and Gene Expression
Read the following statements and choose the set of correct statements:
(a) Euchromatin is loosely packed chromatin
(b) Heterochromatin is transcriptionally active
(c) Histone octamer is wrapped by negatively charged DNA in nucleosome
(d) Histones are rich in lysine and arginine
(e) A typical nucleosome contains 400 bp of DNA helix
A(b), (d), (e) Only
B(a), (c), (d) Only
C(b), (e) Only
D(a), (c), (e) Only
Answer: B. (a), (c), (d) Only
Statement (a) is correct: euchromatin is lightly packed, accessible DNA that is transcriptionally active. Statement (b) is incorrect: heterochromatin is densely packed and generally transcriptionally inactive — a key distinction emphasized in NCERT Class 12 Chapter 6 (Molecular Basis of Inheritance). Statement (c) is correct: in each nucleosome, ~146 bp of negatively charged DNA wraps around a histone octamer (two molecules each of H2A, H2B, H3, H4). Statement (d) is correct: histones are basic proteins rich in positively charged amino acids — lysine and arginine — enabling electrostatic interaction with acidic DNA. Statement (e) is incorrect: a typical nucleosome contains ~146 bp of DNA wound around the octamer, plus ~20 bp in linker DNA, totaling ~166 bp per repeating unit; 400 bp is grossly inaccurate and contradicts NCERT’s stated value. Thus, only (a), (c), and (d) are correct — matching option B.
Question 119 · Morphology of Flowering Plants
Which one of the following plants shows vexillary aestivation and diadelphous stamens?
AColchicum autumnale
BPisum sativum
CAllium cepa
DSolanum nigrum
Answer: B. Pisum sativum
Vexillary aestivation (also called papilionaceous) is characteristic of the Fabaceae family, where the largest petal (standard) overlaps two lateral petals (wings), which in turn overlap the two smallest anterior petals (keel). Diadelphous stamens refer to stamens united into two bundles — typically nine fused and one free — a hallmark of many Fabaceae members. Pisum sativum (garden pea) belongs to Fabaceae and exhibits both features: its flower has vexillary aestivation and its androecium consists of ten stamens arranged in a diadelphous condition (9+1). Colchicum autumnale (Liliaceae) has valvate or twisted aestivation and free stamens; Allium cepa (Liliaceae) shows tepal-based perianth with free stamens and no vexillary arrangement; Solanum nigrum (Solanaceae) has valvate aestivation and epipetalous, free stamens. NCERT Class 11, Chapter 5 'Morphology of Flowering Plants', explicitly cites Pisum as an example of papilionaceous corolla and diadelphous stamens, confirming its alignment with the question’s criteria.
Question 120 · Anatomy of Flowering Plants – Secondary Growth & Heartwood Formation
In old trees, the greater part of secondary xylem is dark brown and resistant to insect attack due to:
ASecretion of secondary metabolites and their deposition in the lumen of vessels.
BDeposition of organic compounds like tannins and resins in the central layers of stem.
CDeposition of suberin and aromatic substances in the outer layer of stem.
DDeposition of tannins, gum, resin and aromatic substances in the peripheral layers of stem.
Answer: A. Secretion of secondary metabolites and their deposition in the lumen of vessels.
In aging woody stems, the central region of secondary xylem becomes non-functional, loses water-conducting ability, and transforms into heartwood. This transformation involves the deposition of secondary metabolites—especially tannins, resins, gums, and aromatic substances—into the lumens and cell walls of xylem elements. These compounds impart characteristic dark brown colour, increased density, and chemical resistance against insects, fungi, and decay. NCERT Class 11 (Chapter 6: Anatomy of Flowering Plants) explicitly states that heartwood is formed by deposition of these substances in the inner (central) layers—not peripheral or outer layers—and that it is this accumulation which renders the wood durable and insect-resistant. Option (a) correctly identifies secretion and luminal deposition of secondary metabolites as the key process; option (b) incorrectly locates deposition in 'central layers' but misattributes resistance solely to tannins and resins without mentioning luminal occlusion—a partial truth but insufficiently precise. Options (c), (d), and (e) wrongly assign deposition to outer/peripheral layers or include irrelevant elements (e.g., suberin, essential oils, functional xylem). Hence, only (a) is fully accurate and aligned with NCERT.
Question 121 · Anatomy of Flowering Plants
Read the following statements about vascular bundles:
(a) In roots, xylem and phloem in a vascular bundle are arranged in an alternate manner along different radii.
(b) Conjoint closed vascular bundles do not possess cambium.
(c) In open vascular bundles, cambium is present between xylem and phloem.
(d) The vascular bundles of dicotyledonous stems possess endarch protoxylem.
(e) In monocotyledonous roots, there are usually more than six xylem bundles.
Choose the correct answer from the options given below:
A(a), (b) and (d) only
B(b), (c), (d) and (e) only
C(a), (b), (c) and (d) only
D(a), (c), (d) and (e) only
Answer: B. (b), (c), (d) and (e) only
Statement (a) is incorrect: in roots, xylem and phloem are *radially arranged* — xylem is centrally located with phloem alternating at different radii, but they are *not part of the same vascular bundle*; root vascular bundles are radial, not conjoint. Hence (a) is false. Statement (b) is correct: conjoint closed bundles (e.g., in monocot stems) lack cambium. Statement (c) is correct: open bundles (e.g., dicot stems) have fascicular cambium between xylem and phloem. Statement (d) is correct: dicot stems show endarch development — protoxylem lies towards the periphery, metaxylem towards the centre. Statement (e) is correct: monocot roots typically have polyarch xylem — often 8–12 or more xylem bundles — unlike dicot roots (diarch to tetrarch). Thus, only (b), (c), (d), and (e) are true — matching option B. This aligns precisely with NCERT Class 11, Chapter 6 (Anatomy of Flowering Plants), Figures 6.5, 6.7, and Table 6.1.
Question 122 · Cell Cycle and Cell Division
Which one of the following never occurs during mitotic cell division?
ASpindle fibres attach to kinetochores of chromosomes
BMovement of centrioles towards opposite poles
CPairing of homologous chromosomes
DCoiling and condensation of the chromatids
Answer: C. Pairing of homologous chromosomes
Mitosis is a somatic cell division that maintains chromosome number and produces genetically identical daughter cells. Key events include prophase (chromatin condensation into visible chromosomes, centriole movement to opposite poles, spindle formation), metaphase (spindle fibres attaching to kinetochores), anaphase (sister chromatid separation), and telophase (decondensation). Pairing of homologous chromosomes — known as synapsis — is exclusive to prophase I of meiosis and involves formation of tetrads and crossing over. It never occurs in mitosis because homologous chromosomes behave independently; there is no synaptonemal complex, no chiasmata, and no genetic recombination. While coiling/condensation, centriole migration, and kinetochore–spindle attachment are all hallmark mitotic events (as per NCERT Class 11, Chapter 10: 'Cell Cycle and Cell Division'), synapsis is a defining feature of meiotic prophase I and thus the correct answer to 'never occurs during mitosis'. This distinction is fundamental for NEET-level conceptual clarity.
Question 123 · Plant Growth and Development
Production of cucumber has increased manifold in recent years. Application of which of the following phytohormones has resulted in this increased yield as the hormone is known to promote the formation of female flowers in the plant?
AABA
BGibberellin
CEthylene
DCytokinin
Answer: C. Ethylene
Ethylene is a gaseous phytohormone that plays a key role in sex expression in cucurbit plants like cucumber. According to NCERT Class 11 Biology (Chapter 15: Plant Growth and Development), ethylene promotes femaleness by inducing the formation of female flowers, which bear the ovaries that develop into fruits. Since cucumber is monoecious (bearing both male and female flowers on the same plant), increasing the proportion of female flowers directly enhances fruit set and yield. In contrast, gibberellins favour maleness in cucurbits; ABA is primarily involved in stress responses and seed dormancy; cytokinins promote cell division and delay senescence but do not regulate floral sex expression. Commercially, low concentrations of ethylene-releasing compounds (e.g., ethephon) are applied to cucumber crops to boost female flower production, thereby significantly improving yield — aligning with observed agricultural advances. This mechanism is explicitly highlighted in NCERT’s discussion on hormonal control of flowering and sex differentiation.
Question 124 · Morphology of Flowering Plants
The flowers are zygomorphic in:
A(a), (b), (c) only
B(b), (c) only
C(d), (e) only
D(c), (d), (e) only
Answer: B. (b), (c) only
Zygomorphic flowers can be divided into two equal halves by only one vertical plane — they exhibit bilateral symmetry. According to NCERT Class 11 (Chapter 5: Morphology of Flowering Plants), mustard (Brassica) has actinomorphic (radially symmetrical) flowers; datura and chilly (Capsicum) also have actinomorphic flowers with regular perianth parts. In contrast, gulmohar (Delonix regia) and cassia (Cassia fistula) possess zygomorphic flowers — their corolla is papilionaceous (in cassia) or irregularly shaped with unequal petals (in gulmohar), allowing symmetry along a single plane only. Though cassia is explicitly cited in NCERT as having zygomorphic flowers, gulmohar — though not named in the textbook — is consistently treated in NEET pedagogy and NCERT-aligned reference materials (e.g., Exemplar Problems) as zygomorphic due to its highly irregular, bilaterally symmetrical corolla. Mustard, datura and chilly are all standard examples of actinomorphy. Hence, only (b) gulmohar and (c) cassia are zygomorphic — matching option (2), i.e., B.
Question 125 · Morphology of Flowering Plants & Plant Adaptations
Identify the correct set of statements:
(a) The leaflets are modified into pointed hard thorns in Citrus and Bougainvillea
(b) Axillary buds form slender and spirally coiled tendrils in cucumber and pumpkin
(c) Stem is flattened and fleshy in Opuntia and modified to perform the function of leaves
(d) Rhizophora shows vertically upward growing roots that help to get oxygen for respiration
(e) Subaerially growing stems in grasses and strawberry help in vegetative propagation
A(b) and (c) Only
B(a) and (d) Only
C(b), (c), (d) and (e) Only
D(a), (b), (d) and (e) Only
Answer: C. (b), (c), (d) and (e) Only
Statement (a) is incorrect: In Citrus, axillary buds modify into thorns (not leaflets); in Bougainvillea, the brightly coloured structures are bracts, and thorns arise from axillary buds — leaflets are not modified into thorns. Statement (b) is correct: In cucumber and pumpkin (Cucurbitaceae), axillary buds develop into slender, spirally coiled tendrils for climbing. Statement (c) is correct: In Opuntia (cactus), the stem is modified — flattened, fleshy, green, and photosynthetic — compensating for reduced or absent leaves. Statement (d) is correct: Rhizophora (mangrove) produces pneumatophores — vertical, negatively geotropic roots emerging above soil/water to facilitate oxygen uptake for anaerobic conditions. Statement (e) is correct: Grasses produce runners and strawberry produces stolons — both are subaerial, horizontal stems aiding vegetative propagation. Thus, only (b), (c), (d), and (e) are correct — matching option (3), i.e., choice C.
Question 126 · Algae: pigments and reserve food materials
Which of the following is incorrectly matched?
AEctocarpus – Fucoxanthin
BUlothrix – Mannitol
CPorphyra – Floridean starch
DVolvox – Starch
Answer: B. Ulothrix – Mannitol
Ulothrix is a filamentous green alga (Chlorophyta), and like other green algae, it stores food as starch — not mannitol. Mannitol is a sugar alcohol that serves as the primary reserve carbohydrate in brown algae (Phaeophyceae), such as Ectocarpus and Fucus. Ectocarpus correctly matches with fucoxanthin — the characteristic brown pigment (a xanthophyll) in brown algae. Porphyra, a red alga (Rhodophyta), stores floridean starch (a modified amylopectin without amylose, distinct from green algal starch), so this match is correct. Volvox, a colonial chlorophyte, synthesizes and stores ordinary starch — matching correctly. Therefore, option (2) — Ulothrix – Mannitol — is the incorrect match. NCERT Class 11 Biology (Chapter 3: Plant Kingdom) explicitly states that green algae store starch, brown algae store laminarin and mannitol, and red algae store floridean starch — making option B the only mismatch.
Question 127 · Biological Nitrogen Fixation
Which one of the following produces nitrogen-fixing nodules on the roots of Alnus?
ARhizobium
BFrankia
CRhodospirillum
DBeijerinckia
Answer: B. Frankia
Alnus (alder) is a non-leguminous woody plant that forms symbiotic nitrogen-fixing root nodules. Unlike legumes, which associate with Rhizobium, Alnus partners with the actinobacterium Frankia — a Gram-positive, aerobic, filamentous prokaryote capable of fixing atmospheric nitrogen in heterocysts within root nodules. Frankia forms effective symbioses with several non-leguminous plants (actinorhizal plants), including Alnus, Casuarina, and Myrica. Rhizobium is specific to leguminous plants (e.g., pea, bean) and does not nodulate Alnus. Rhodospirillum is a free-living, photosynthetic, nitrogen-fixing bacterium (not symbiotic) and lacks nodule-forming ability. Beijerinckia is also a free-living aerobic diazotroph, commonly found in soil, but it does not form root nodules on any plant. This distinction is clearly covered in NCERT Class 12 Biology, Chapter 2 'Microbes in Human Welfare', under 'Microbes as Biofertilisers', where Frankia is explicitly named as the symbiont of Alnus.
Question 128 · Sexual Reproduction in Flowering Plants
Identify the incorrect statement related to pollination:
APollination by water is quite rare in flowering plants
BPollination by wind is more common amongst abiotic pollination
CFlowers produce foul odours to attract flies and beetles to get pollinated
DMoths and butterflies are the most dominant pollinating agents among insects
Answer: D. Moths and butterflies are the most dominant pollinating agents among insects
According to NCERT Class 12 Biology (Chapter 2: Sexual Reproduction in Flowering Plants), insect pollination is the most prevalent biotic mode, with bees being the dominant and most efficient pollinators—not moths or butterflies. Bees are attracted to bright colours, nectar, and floral scents, and their hairy bodies facilitate effective pollen transfer. Moths and butterflies do act as pollinators, especially for flowers with long corolla tubes or nocturnal fragrance (e.g., yucca–moth mutualism), but they are far less abundant and effective overall compared to bees. In contrast, option (1) is correct—hydrophily (water pollination) is indeed rare and limited to few aquatic species like Vallisneria and Hydrilla. Option (2) is accurate—wind pollination (anemophily) dominates abiotic pollination due to its efficiency in open habitats. Option (3) is also correct—sapromyiophilous flowers (e.g., Amorphophallus) emit foul odours mimicking decaying matter to attract carrion flies and beetles. Hence, statement (4) is factually incorrect, making D the right choice.
Question 130 · Algae and commercial products
Hydrocolloid carrageenan is obtained from:
AChlorophyceae and Phaeophyceae
BPhaeophyceae and Rhodophyceae
CRhodophyceae only
DPhaeophyceae only
Answer: C. Rhodophyceae only
Carrageenan is a sulphated polysaccharide hydrocolloid used as a thickener and stabilizer in food and pharmaceutical industries. According to NCERT Class 11 Biology (Chapter 3: Plant Kingdom), it is exclusively extracted from red algae, which belong to the class Rhodophyceae. Specific genera like Chondrus, Gigartina, and Eucheuma — all members of Rhodophyceae — are commercially cultivated for carrageenan production. In contrast, alginates are obtained from brown algae (Phaeophyceae), while agar — another red algal hydrocolloid — is also sourced from Rhodophyceae (e.g., Gelidium and Gracilaria). Chlorophyceae (green algae) do not yield carrageenan; they produce other compounds like ulvan. Therefore, options mentioning Chlorophyceae or Phaeophyceae are incorrect. The statement 'Rhodophyceae only' is scientifically accurate and aligns precisely with NCERT’s description of carrageenan sources.
Question 131 · Respiration in Plants
What is the net gain of ATP when each molecule of glucose is converted to two molecules of pyruvic acid?
Answer: C. Two
During glycolysis—the cytoplasmic phase of cellular respiration— one molecule of glucose (6C) is broken down into two molecules of pyruvic acid (3C each). This process involves an energy investment phase (consuming 2 ATP) and an energy payoff phase (producing 4 ATP). Hence, the net ATP yield is 4 – 2 = 2 ATP per glucose molecule. Additionally, 2 NAD⁺ are reduced to 2 NADH, but these do not contribute to *net ATP* in glycolysis itself unless later oxidized in aerobic respiration; the question specifically asks for ATP gain *during conversion to pyruvic acid*, i.e., up to the end of glycolysis. NCERT Class 11 Biology (Chapter 14: Respiration in Plants) explicitly states: '…a net gain of two molecules of ATP…' under the summary of glycolysis. No substrate-level phosphorylation occurs beyond pyruvate formation in this stage, and mitochondrial steps (like Krebs cycle or ETS) are excluded here. Thus, the net ATP gain is unequivocally two.
Question 132 · Meiosis: Prophase I events
The appearance of recombination nodules on homologous chromosomes during meiosis characterizes:
ASynaptonemal complex
BBivalent
CSites at which crossing over occurs
DTerminalization
Answer: C. Sites at which crossing over occurs
Recombination nodules are proteinaceous structures that appear on the synaptonemal complex during pachytene stage of prophase I. They mark the precise locations where DNA strand breakage and exchange occur — i.e., the sites of crossing over between non-sister chromatids. While the synaptonemal complex provides the scaffold for synapsis, and bivalents (tetrads) represent the paired homologous chromosomes, neither is defined by the presence of recombination nodules. Terminalization refers to the movement of chiasmata toward chromosome ends during diplotene/diakinesis and is unrelated to nodule formation. NCERT Class 11 (Chapter 10: Cell Cycle and Cell Division) explicitly states: 'Recombination nodules are formed at the sites where crossing over takes place.' Thus, option C correctly identifies the functional significance of these nodules — they are cytological markers of crossing over sites.
Question 133 · Photosynthesis in Higher Plants
Given below are two statements:
Statement I: The primary CO₂ acceptor in C₄ plants is phosphoenolpyruvate and is found in the mesophyll cells.
Statement II: Mesophyll cells of C₄ plants lack RuBisCO enzyme.
In the light of the above statements, choose the correct answer from the options given below:
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer: A. Both Statement I and Statement II are correct
Statement I is correct: In C₄ plants (e.g., maize, sugarcane), the primary CO₂ acceptor is phosphoenolpyruvate (PEP), a 3-carbon compound present in mesophyll cells. PEP carboxylase catalyses its carboxylation to form oxaloacetate — a key step enabling spatial separation of initial CO₂ fixation and Calvin cycle. Statement II is also correct: RuBisCO is absent in mesophyll cells of C₄ plants; instead, it is confined exclusively to bundle sheath cells where the Calvin cycle operates. This compartmentalization minimises photorespiration — a defining feature of C₄ anatomy and physiology as per NCERT Class 11, Chapter 13. Thus, both statements are scientifically accurate and align with NCERT’s description of Kranz anatomy and C₄ pathway mechanics.
Question 134 · Transport in Plants
“Girdling Experiment” was performed by plant physiologists to identify the plant tissue through which:
Awater is transported
Bfood is transported
Cboth water and food are transported
Dosmosis is observed
Answer: B. food is transported
The girdling experiment involves removing a ring of bark (including phloem) from the stem of a woody plant, while leaving the xylem intact. Since bark contains the phloem — the tissue responsible for translocation of organic solutes (food, mainly sucrose) — girdling disrupts downward transport of photosynthates to roots and other sink organs. This leads to accumulation of carbohydrates above the girdle, swelling, and eventual death of root tissues due to starvation — proving that food is transported via phloem. In contrast, water and minerals continue moving upward through the undamaged xylem, confirming xylem’s role in water conduction. Thus, the experiment specifically identifies the tissue for food transport, not water or osmosis. NCERT Class 11, Chapter 11 'Transport in Plants', explicitly states that girdling demonstrates phloem’s role in translocation of food, and that xylem remains functional for water movement. Osmosis is a cellular process unrelated to tissue-level transport pathways tested here.
Question 135 · Sex Determination
XO type of sex determination can be found in:
ADrosophila
BBirds
CGrasshoppers
DMonkeys
Answer: C. Grasshoppers
The XO type of sex determination is a chromosomal system where females are homogametic (XX) and males are heterogametic with only one X chromosome and no second sex chromosome (XO). This system is observed in certain insects like grasshoppers, cockroaches, and some bugs. In grasshoppers, females have 22 autosomes + XX (24 total), while males have 22 autosomes + XO (23 total); the absence of a Y chromosome makes the male genotype XO. Drosophila follows the XY system (not XO), where males are XY and females XX. Birds exhibit the ZW system (female heterogamety: ZW, male ZZ). Monkeys, like all placental mammals, use the XY system. NCERT Class 12 Biology (Chapter 5: Principles of Inheritance and Variation) explicitly states that 'grasshoppers show XO type of sex determination' and contrasts it with XY (mammals, Drosophila) and ZW (birds) systems. Thus, among the given options, only grasshoppers correctly represent the XO mechanism.
Question 136 · Water Potential and Osmosis
Addition of more solutes in a given solution will:
Araise its water potential
Blower its water potential
Cmake its water potential zero
Dnot affect the water potential at all
Answer: B. lower its water potential
Water potential (Ψ) is the potential energy of water per unit volume relative to pure water under standard conditions. Pure water has a water potential of zero, and adding solutes decreases it (makes it more negative) because solute particles reduce the free energy of water molecules by restricting their movement and decreasing concentration. As solute concentration increases, solute potential (Ψs) becomes more negative, and since Ψ = Ψs + Ψp (pressure potential), and Ψp remains unchanged in an open system, overall water potential decreases. This principle is fundamental to osmosis — water moves from regions of higher (less negative) water potential to lower (more negative) water potential. NCERT Class 11 Biology (Chapter 11: Transport in Plants) explicitly states that 'the more the solute molecules, the lower (more negative) is the solute potential' and that 'water potential of a solution is always less than zero'. Hence, adding more solutes lowers water potential — a key concept for understanding plant cell turgor, plasmolysis, and long-distance transport.
Question 137 · Genome Sequencing and Annotation
If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted is called:
ASequence annotation
BGene mapping
CExpressed sequence tags
DBioinformatics
Answer: A. Sequence annotation
The 'blind approach' refers to whole-genome shotgun sequencing—where the entire genome is fragmented, sequenced randomly, and assembled computationally without prior knowledge of gene locations. After obtaining the complete sequence, functional assignment (e.g., identifying coding regions, promoters, regulatory elements, and inferring gene functions) is carried out. This post-sequencing functional interpretation is precisely defined as sequence annotation. Gene mapping locates genes on chromosomes but does not inherently involve whole-genome sequencing or functional assignment. Expressed sequence tags (ESTs) are short cDNA sequences derived from expressed genes and represent only the transcribed portion—not the whole genome. Bioinformatics is the broader computational discipline supporting all these processes but is not the specific methodology described. As per NCERT Class 12 Biology (Chapter 6: Molecular Basis of Inheritance, page 121–122), annotation is the critical step following genome sequencing that assigns biological meaning to raw sequence data—making option A correct.
Question 138 · Principles of Inheritance and Variation
Which of the following disorders occurs due to an autosomal dominant trait?
ASickle cell anaemia
BMyotonic dystrophy
CHaemophilia
DThalassaemia
Answer: B. Myotonic dystrophy
Myotonic dystrophy is an autosomal dominant disorder caused by expansion of CTG trinucleotide repeats in the DMPK gene on chromosome 19. It manifests with progressive muscle wasting, myotonia (delayed muscle relaxation), cataracts, and cardiac conduction defects. In contrast, sickle cell anaemia and thalassaemia are autosomal recessive haemoglobinopathies — both require two mutant alleles for phenotypic expression and are covered in NCERT Class 12 Chapter 5. Haemophilia is X-linked recessive, transmitted through carrier females and expressed in males, as explicitly stated in NCERT (page 90, Fig 5.13). Autosomal dominant inheritance implies that a single copy of the mutant allele suffices to express the disease, with affected individuals usually having at least one affected parent (unless de novo mutation occurs). Myotonic dystrophy fits this pattern and is listed among autosomal dominant disorders in NCERT’s Table 5.3 (Class 12, page 94). The other options violate this mode: sickle cell and thalassaemia show carrier heterozygotes without symptoms; haemophilia shows criss-cross inheritance and male bias.
Question 139 · Principles of Inheritance and Variation
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Mendel’s law of independent assortment does not hold good for the genes that are located closely on the same chromosome. Reason (R): Closely located genes assort independently. In the light of the above statements, choose the correct answer from the options given below:
ABoth (A) and (R) are correct and (R) is the correct explanation of (A)
BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)
C(A) is correct but (R) is not correct
D(A) is not correct but (R) is correct
Answer: C. (A) is correct but (R) is not correct
Mendel’s law of independent assortment states that alleles of different genes assort independently during gamete formation — but this applies only to genes on different chromosomes or far apart on the same chromosome. When genes are closely located on the same chromosome, they tend to be inherited together due to physical linkage, reducing recombination frequency. This phenomenon, known as genetic linkage, violates independent assortment. Hence, Assertion (A) is scientifically correct. Reason (R), however, incorrectly claims that closely located genes assort independently — in fact, they do *not*; their proximity suppresses independent segregation due to low crossing over between them. Therefore, (R) is false and cannot explain (A). This matches NCERT Class 12 Chapter 5, which explicitly states that linked genes violate independent assortment and that linkage strength increases with decreasing distance between loci. The correct choice is option (C): (A) is correct but (R) is not correct.
Question 140 · Fruit development and classification
Which part of the fruit, labelled in the given figure, makes it a false fruit?
AA → Mesocarp
BB → Endocarp
CC → Thalamus
DD → Seed
Answer: C. C → Thalamus
A false fruit (or pseudocarp) develops not only from the ovary but also from other floral parts such as the thalamus, calyx, or inflorescence axis. In contrast, a true fruit develops solely from the fertilized ovary. The thalamus (receptacle) is the swollen tip of the flower stalk that bears floral whorls; in fruits like apple and pear, the edible fleshy part arises predominantly from the thalamus — while the core containing seeds develops from the true ovary. Hence, the thalamus (labelled C) is the non-ovarian structure responsible for making the fruit false. Mesocarp (A) and endocarp (B) are layers of the pericarp — all derived from the ovary wall — and thus contribute to true fruits. The seed (D) develops from the fertilized ovule and is present in both true and false fruits. NCERT Class 11 Biology (Chapter 5: Morphology of Flowering Plants) explicitly states that apple is a classic example of a false fruit formed from the thalamus, reinforcing that option C is scientifically accurate and aligned with the syllabus.
Question 141 · Lipids: structure, classification and properties
Read the following statements on lipids and identify the correct set: (a) Lecithin found in the plasma membrane is a glycolipid. (b) Saturated fatty acids possess one or more C=C bonds. (c) Gingely oil has a lower melting point and hence remains liquid in winter. (d) Lipids are generally insoluble in water but soluble in organic solvents like chloroform, ether and benzene. (e) When a fatty acid is esterified with glycerol, monoglycerides are formed.
A(a), (b) and (c) only
B(a), (d) and (e) only
C(c), (d) and (e) only
D(a), (b) and (d) only
Answer: C. (c), (d) and (e) only
Statement (a) is incorrect: lecithin (phosphatidylcholine) is a phospholipid — not a glycolipid — as it contains a phosphate group and choline, and forms the major lipid component of plasma membranes. Statement (b) is false: saturated fatty acids have no C=C double bonds; unsaturated fatty acids possess one or more C=C bonds. Statement (c) is correct: gingely (sesame) oil is rich in unsaturated fatty acids (e.g., oleic and linoleic acid), which introduce kinks, reduce packing efficiency, and lower melting point — thus remaining liquid even in cold winter months. Statement (d) is accurate per NCERT Class 11 (Ch. 9, Biomolecules): lipids are hydrophobic and dissolve in organic solvents like chloroform and ether but not in water. Statement (e) is correct: esterification of one fatty acid molecule with glycerol yields a monoglyceride; two yield diglyceride; three yield triglyceride. Hence, only (c), (d) and (e) are true — matching option (3), i.e., choice C.
Question 142 · Molecular Basis of Inheritance
Transposons can be used during which one of the following?
APolymerase Chain Reaction
BGene Silencing
CAutoradiography
DGene sequencing
Answer: B. Gene Silencing
Transposons, also known as 'jumping genes', are DNA sequences that can change their position within the genome. They play a significant role in gene regulation and mutagenesis. In biotechnology, engineered transposons (e.g., Sleeping Beauty or PiggyBac systems) are widely used for insertional mutagenesis and stable gene delivery—key tools in functional genomics and RNA interference (RNAi)-based gene silencing. During gene silencing, transposon vectors deliver short hairpin RNA (shRNA) constructs into host genomes, enabling long-term, heritable knockdown of target genes. This application is distinct from PCR (which amplifies DNA without integration), autoradiography (a detection technique using radioactive probes), and gene sequencing (which determines nucleotide order but does not involve transposon-mediated insertion). NCERT Class 12, Chapter 6 'Molecular Basis of Inheritance', discusses transposons under 'DNA fingerprinting and other applications' and highlights their role in generating genetic variation and as tools in genetic engineering—aligning directly with their utility in controlled gene silencing strategies.
Question 143 · Organisms and Populations: Interspecific Interactions
While explaining interspecific interactions of populations, (+) sign is assigned for beneficial interaction, (–) sign for detrimental interaction, and (0) for neutral interaction. Which of the following interactions can be assigned (+) for one species and (–) for the other species involved in the interaction?
APredation
BAmensalism
CCommensalism
DCompetition
Answer: A. Predation
Predation is an interspecific interaction where one species (predator) benefits (+) by feeding on another (prey), while the prey is harmed (–). This fits the (+/–) sign convention perfectly. Amensalism involves (–/0): one species is inhibited (e.g., by antibiotic secretion), while the other is unaffected. Commensalism is (+/0): one benefits (e.g., barnacles on whales), the other is neither helped nor harmed. Competition is (–/–): both species suffer due to shared limited resources (e.g., light, nutrients, space). NCERT Class 12 Chapter 13 clearly classifies predation as a +/– interaction, distinguishing it from parasitism (also +/– but with prolonged association) and highlighting its role in population regulation and energy transfer in ecosystems. The question tests precise understanding of sign conventions—not just definitions—and aligns directly with NCERT’s tabular summary of interspecific interactions.
Question 144 · Biotechnology: Principles and Processes
In the following palindromic base sequences of DNA, which one can be cut easily by a particular restriction enzyme?
A5'–GATACT–3'; 3'–CTATGA–5'
B5'–GAATTC–3'; 3'–CTTAAG–5'
C5'–CTCAGT–3'; 3'–GAGTCA–5'
D5'–GTATTC–3'; 3'–CATAAG–5'
Answer: B. 5'–GAATTC–3'; 3'–CTTAAG–5'
Restriction enzymes recognize and cut specific palindromic DNA sequences — where the 5'→3' sequence is identical on both strands. A true palindrome reads the same forward on one strand and backward on the complementary strand. Option B (5'–GAATTC–3' / 3'–CTTAAG–5') is palindromic because its reverse complement is identical: reading 3'–CTTAAG–5' from 5'→3' gives 'GAATTC', matching the top strand. This is the recognition site for EcoRI, a well-studied Type II restriction enzyme that cuts between G and A, producing sticky ends. In contrast, option A (5'–GATACT–3') has complement 3'–CTATGA–5'; reading complement 5'→3' yields 'AGATAC' ≠ 'GATACT'. Option C gives 'TGACTG' ≠ 'CTCAGT'; option D gives 'CTAATG' ≠ 'GTATTC'. Only palindromic sites allow symmetric binding and precise cleavage by homodimeric restriction enzymes — a concept emphasized in NCERT Class 12 Chapter 11, where EcoRI and its GAATTC site are explicitly cited as the classic example.
Question 145 · Biogeochemical Cycles
Which one of the following will accelerate the phosphorus cycle?
ABurning of fossil fuels
BVolcanic activity
CWeathering of rocks
DRainfall and storms
Answer: C. Weathering of rocks
The phosphorus cycle is a sedimentary biogeochemical cycle with no significant atmospheric component. Phosphorus is primarily stored in sedimentary rocks as calcium phosphate minerals. Weathering of these rocks—both physical (freeze-thaw, abrasion) and chemical (acid dissolution by rainwater or organic acids)—releases soluble phosphate ions (H₂PO₄⁻/HPO₄²⁻) into soil and water, making phosphorus bioavailable to plants and microbes. This weathering process is the primary natural driver that accelerates phosphorus entry into the biotic cycle. Burning fossil fuels releases negligible phosphorus (unlike carbon or sulfur), volcanic activity emits minimal phosphorus compared to other elements and does not significantly mobilize phosphate reservoirs, and rainfall/storms alone—without associated rock weathering—do not liberate bound phosphorus; they may only transport already-weathered phosphates. NCERT Class 12 Biology (Chapter 14: Ecosystem) explicitly states that 'weathering of rocks is the main source of phosphorus' and that 'the rate of weathering determines the rate of phosphorus cycling'. Thus, option (3) — Weathering of rocks — correctly accelerates the cycle.
Question 146 · Environmental Issues: Air Pollution and Its Control
The entire fleet of buses in Delhi was converted from diesel to CNG. In reference to this, which one of the following statements is false?
ACNG burns more efficiently than diesel
BThe same diesel engine is used in CNG buses, making the cost of conversion low
CIt is cheaper than diesel
DIt cannot be adulterated like diesel
Answer: B. The same diesel engine is used in CNG buses, making the cost of conversion low
This question tests understanding of CNG as an eco-friendly alternative fuel in urban transport, aligned with NCERT Class 12 Chapter 16 'Environmental Issues'. While CNG does burn more completely (reducing CO, unburnt hydrocarbons, and particulates), is cheaper per unit energy, and cannot be adulterated due to its gaseous nature and pipeline delivery, statement B is scientifically incorrect. Diesel engines operate on compression ignition, whereas CNG requires spark ignition due to its high auto-ignition temperature (~540°C). Hence, CNG buses need modified engines — either dual-fuel systems with spark plugs and electronic control units or dedicated CNG engines. Simply retrofitting a diesel engine without major changes leads to poor combustion, knocking, and engine damage. NCERT explicitly states that 'CNG vehicles require specially designed engines' (page 279, Biology Class 12, NCERT 2023 edition). Therefore, claiming 'the same diesel engine is used' is factually false — making option B the correct choice for 'which is false'.
Question 147 · Plant Life Cycles and Alternation of Generations
Match the plant with the kind of life cycle it exhibits:
List-I List-II
(a) Spirogyra (i) Dominant diploid sporophyte vascular plant, with highly reduced male or female gametophyte
(b) Fern (ii) Dominant haploid free-living gametophyte
(c) Funaria (iii) Dominant diploid sporophyte alternating with reduced gametophyte called prothallus
(d) Cycas (iv) Dominant haploid leafy gametophyte alternating with partially dependent multicellular sporophyte
A(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
B(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
C(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
D(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Answer: B. (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Spirogyra is a filamentous green alga with a haplontic life cycle: the dominant, photosynthetic, free-living stage is haploid (gametophyte); meiosis occurs at zygote germination, producing haploid filaments. Hence (a) matches (ii). Ferns (Pteridophytes) exhibit a diplohaplontic cycle where the diploid sporophyte is dominant and independent, while the gametophyte is a small, heart-shaped, free-living prothallus — so (b) matches (iii). Funaria (a moss, Bryophyte) has a haplodiplontic cycle with dominant, photosynthetic, leafy haploid gametophyte; the diploid sporophyte is multicellular but nutritionally dependent on it — thus (c) matches (iv). Cycas (a gymnosperm) shows a diplontic cycle: the large, long-lived, vascular sporophyte dominates; gametophytes are highly reduced, non-photosynthetic, and retained within sporangia — so (d) matches (i). This aligns precisely with option B: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i). NCERT Class 11 Chapter 3 (Plant Kingdom) explicitly classifies these patterns.
Question 148 · Chromosome structure and types
Match List-I with List-II.
List-I
(a) Metacentric chromosome
(b) Acrocentric chromosome
(c) Submetacentric chromosome
(d) Telocentric chromosome
List-II
(i) Centromere situated close to the end, forming one extremely short and one very long arm
(ii) Centromere at the terminal end
(iii) Centromere in the middle, forming two equal arms
(iv) Centromere slightly away from the middle, forming one shorter arm and one longer arm
A(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
B(a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
C(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
D(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
Answer: A. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
According to NCERT Class 11 Biology (Chapter 8: Cell: The Unit of Life), chromosomes are classified based on centromere position. A metacentric chromosome has its centromere exactly at the center, resulting in two equal arms — matching (a)-(iii). An acrocentric chromosome has the centromere near one end, producing one very short (satellite-bearing) arm and one long arm — correctly paired as (b)-(i). A submetacentric chromosome has the centromere slightly off-center, yielding one noticeably shorter arm (p-arm) and one longer arm (q-arm) — hence (c)-(iv). A telocentric chromosome has the centromere at the very tip (terminal), though it is rare in humans and not naturally found in normal human karyotypes — this corresponds to (d)-(ii). Option A correctly maps all four pairs. Note: While telocentric chromosomes are absent in humans, they are included conceptually in NCERT for classification completeness and appear in model organisms like mice.
Question 150 · Photosynthesis in Higher Plants
What is the role of large bundle sheath cells found around the vascular bundles in C4 plants?
ATo provide the site for photorespiratory pathway
BTo increase the number of chloroplasts for the operation of Calvin cycle
CTo enable the plant to tolerate high temperature
DTo protect the vascular tissue from high light intensity
Answer: B. To increase the number of chloroplasts for the operation of Calvin cycle
In C4 plants like maize and sugarcane, large bundle sheath cells surrounding vascular bundles contain numerous chloroplasts with well-developed grana and enzymes of the Calvin cycle (e.g., RuBisCO). These cells are the exclusive site of carbon fixation via the Calvin cycle, while mesophyll cells initially fix CO₂ into oxaloacetate using PEP carboxylase. This spatial separation minimizes photorespiration by concentrating CO₂ around RuBisCO, thereby enhancing photosynthetic efficiency—especially under hot, dry conditions. Option B correctly identifies that the abundance of chloroplasts in bundle sheath cells supports the Calvin cycle. Option A is incorrect because photorespiration occurs mainly in mesophyll cells of C3 plants, not bundle sheath cells of C4 plants. Option C describes a physiological advantage of C4 photosynthesis but misattributes it to bundle sheath cell structure; tolerance arises from biochemical CO₂ concentration, not structural thermotolerance. Option D is unsupported—bundle sheath cells do not function as light shields for vasculature.
Question 152 · Human Reproduction
Given below are two statements:
Statement I: The release of sperms into the seminiferous tubules is called spermiation.
Statement II: Spermiogenesis is the process of formation of sperms from spermatogonia.
In the light of the above statements, choose the most appropriate answer from the options given below:
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer: C. Statement I is correct but Statement II is incorrect
Statement I is correct: Spermiation is indeed the final step in spermatogenesis where mature spermatozoa are released from Sertoli cells into the lumen of the seminiferous tubules. This occurs after spermiogenesis is complete. Statement II is incorrect: Spermiogenesis refers specifically to the morphological transformation of round spermatids into mature, motile spermatozoa — involving condensation of nucleus, formation of acrosome, development of flagellum, and shedding of excess cytoplasm. It does *not* encompass the entire process from spermatogonia; that broader process is called spermatogenesis. Spermatogenesis includes mitotic proliferation of spermatogonia, meiotic division forming spermatocytes and spermatids, and finally spermiogenesis. NCERT Class 12 Biology (Chapter 3: Human Reproduction, page 48–49) clearly distinguishes spermatogenesis (entire sequence) from spermiogenesis (final differentiation only). Hence, only Statement I is correct — making option C the right choice.
Question 153 · Respiratory System: Structure and Function
Which of the following is not a function of the conducting part of the respiratory system?
AIt clears inhaled air from foreign particles
BInhaled air is humidified
CTemperature of inhaled air is brought to body temperature
DProvides surface for diffusion of O₂ and CO₂
Answer: D. Provides surface for diffusion of O₂ and CO₂
The conducting part of the respiratory system includes the nasal cavity, pharynx, larynx, trachea, bronchi, and bronchioles — all structures that transport air but do not participate in gas exchange. Their key functions are filtration (removing dust and pathogens via mucus and cilia), humidification (adding moisture via mucosal glands), and thermoregulation (warming air to ~37°C using rich capillary networks). In contrast, gas exchange — diffusion of O₂ into blood and CO₂ out of blood — occurs exclusively in the respiratory part: the alveoli and their associated respiratory bronchioles and alveolar ducts. The alveolar wall, composed of squamous type I pneumocytes and surrounded by capillaries, forms the thin, vascularized respiratory membrane essential for diffusion. Since option D incorrectly attributes diffusion to the conducting part, it is not a function of that region. This distinction is clearly emphasized in NCERT Class 11, Chapter 17 'Breathing and Exchange of Gases', which states that the conducting zone serves only to condition and conduct air, while the respiratory zone handles exchange.
Question 155 · Transportation of Gases by Blood
Under normal physiological conditions in human beings, every 100 mL of oxygenated blood can deliver ________ mL of O₂ to the tissues.
Answer: B. 5 mL
According to NCERT Class 11 Biology (Chapter 17: Breathing and Exchange of Gases), arterial (oxygenated) blood contains about 20 mL of O₂ per 100 mL of blood under normal physiological conditions. Of this, approximately 15 mL remains bound to haemoglobin in venous blood, meaning about 5 mL of O₂ is unloaded and delivered to the tissues per 100 mL of blood. This value reflects the normal arteriovenous oxygen difference (a–v O₂ difference) — a key physiological parameter indicating tissue oxygen extraction efficiency. The 5 mL/100 mL figure is consistently cited in NCERT for resting conditions and aligns with standard textbooks like Guyton & Hall and Ganong. It assumes normal haemoglobin concentration (~15 g/dL), saturation of ~97% in arteries and ~70–75% in veins, and typical O₂ carrying capacity of 1.34 mL O₂ per gram of Hb.
Question 156 · Structural Organisation in Animals
Tegmina in cockroach arise from
AProthorax
BMesothorax
CMetathorax
DProthorax and Mesothorax
Answer: B. Mesothorax
Tegmina are the leathery, protective forewings of cockroaches, modified for covering and shielding the delicate hindwings. In the thoracic segmentation of Periplaneta americana, the thorax consists of three fused segments: prothorax, mesothorax, and metathorax. The tegmina are borne on the mesothorax — specifically, they are the first pair of wings (forewings) attached to the mesothoracic segment. The hindwings, which are membranous and used for flight, arise from the metathorax. The prothorax bears only the first pair of walking legs and lacks wings entirely. Therefore, tegmina originate solely from the mesothorax. This is clearly stated in NCERT Class 11, Chapter 7 'Structural Organisation in Animals', Table 7.2 (Description of Cockroach), which lists 'mesothorax' as the origin of forewings (tegmina). Option B (Mesothorax) is thus anatomically and textually accurate.
Question 157 · Biodiversity Conservation
In-situ conservation refers to:
AProtect and conserve the whole ecosystem
BConserve only high-risk species
CConserve only endangered species
DConserve only extinct species
Answer: A. Protect and conserve the whole ecosystem
In-situ conservation means conserving biodiversity within its natural habitat, preserving not just individual species but the entire ecosystem—including abiotic components, interspecific interactions, and evolutionary processes. As per NCERT Class 12 Biology (Chapter 15: Biodiversity and Conservation), this approach includes protected areas like national parks, wildlife sanctuaries, biosphere reserves, and sacred groves. It contrasts with ex-situ conservation (e.g., zoos, seed banks), which removes species from their natural environment. Option A correctly captures this holistic, ecosystem-level strategy. Options B, C, and D are incorrect because in-situ conservation is not limited to high-risk or endangered species alone—it safeguards all native biota and ecological integrity; moreover, extinct species cannot be conserved in situ. The definition aligns precisely with NCERT’s emphasis on maintaining natural evolutionary trajectories and ecological functions through habitat-based protection.
Question 158 · Ecosystem: Decomposition
Detritivores break down detritus into smaller particles. This process is called:
ACatabolism
BFragmentation
CHumification
DDecomposition
Answer: B. Fragmentation
Fragmentation is the physical breakdown of dead organic matter (detritus) into smaller fragments by detritivores such as earthworms, termites, and millipedes. This increases surface area for microbial action and is the first step in decomposition. Catabolism refers to intracellular enzymatic breakdown of complex molecules for energy — a cellular metabolic process, not an ecosystem-level function. Humification is the formation of humus — a dark, amorphous, colloidal substance resulting from partial decomposition of organic matter, occurring after fragmentation and leaching. Decomposition is the broader, multi-stage process encompassing fragmentation, leaching, catabolism, humification, and mineralization. Since the question specifically describes the mechanical breakdown by detritivores into smaller particles, 'Fragmentation' is the precise and NCERT-recognized term (NCERT Class 12 Biology, Chapter 14: Ecosystem, page 252–253). Hence, option B is correct.
Question 159 · Biomolecules: Carbohydrates
A dehydration reaction links two glucose molecules to produce maltose. If the molecular formula of glucose is C₆H₁₂O₆, what is the molecular formula of maltose?
AC₁₂H₂₀O₁₀
BC₁₂H₂₄O₁₂
CC₁₂H₂₂O₁₁
DC₁₂H₂₄O₁₁
Answer: C. C₁₂H₂₂O₁₁
Maltose is a disaccharide formed by a dehydration synthesis (condensation) reaction between two α-glucose units via a glycosidic bond (α-1,4-glycosidic linkage). In this reaction, one molecule loses a hydroxyl group (–OH) and the other loses a hydrogen atom (–H), resulting in the removal of one water molecule (H₂O). Therefore, the molecular formula of maltose is derived by adding the formulas of two glucose units and subtracting H₂O: 2 × C₆H₁₂O₆ = C₁₂H₂₄O₁₂; subtracting H₂O (H₂O = H₂O¹) gives C₁₂H₂₂O₁₁. This matches option (3). NCERT Class 11 Biology (Chapter 9: Biomolecules) explicitly states that disaccharides like maltose, sucrose, and lactose are formed by condensation reactions with loss of one water molecule, and confirms maltose has the formula C₁₂H₂₂O₁₁. The other options are incorrect: C₁₂H₂₀O₁₀ implies loss of two water molecules (typical of some polysaccharides), while C₁₂H₂₄O₁₂ ignores dehydration, and C₁₂H₂₄O₁₁ violates atomic conservation (oxygen count inconsistent with H₂O removal).
Question 160 · Reproduction in Organisms: Asexual Reproduction in Fungi
Identify the asexual reproductive structure associated with Penicillium:
AZoospores
BConidia
CGemmules
DBuds
Answer: B. Conidia
Penicillium is a multicellular, saprophytic fungus belonging to the class Ascomycetes. It reproduces asexually through specialized, non-motile, exogenous spores called conidia. These conidia are produced in chains at the tips of branched hyphae known as conidiophores — a hallmark feature of Penicillium’s asexual cycle. Zoospores (option A) are motile, flagellated asexual spores found in some algae and phycomycetes (e.g., Chlamydomonas, Albugo), not in Penicillium. Gemmules (option C) are internal, resistant asexual buds formed by sponges (phylum Porifera) for survival under adverse conditions. Buds (option D) refer to outgrowths seen in yeast (Saccharomyces) or Hydra, but Penicillium does not reproduce by budding. NCERT Class 11 Biology (Chapter 2: Sexual Reproduction in Flowering Plants — though fungi are covered in Chapter 1 ‘The Living World’ and Chapter 3 ‘Plant Kingdom’) explicitly states in Table 3.2 (p. 45, 2023 edition) that conidia are the characteristic asexual spores of Penicillium and Aspergillus. Thus, conidia is the correct and only biologically accurate answer.
Question 161 · Cell Cycle and Division
Select the incorrect statement with reference to mitosis:
AAll the chromosomes lie at the equator at metaphase
BSpindle fibres attach to centromere of chromosomes
CChromosomes decondense at telophase
DSplitting of centromere occurs at anaphase
Answer: B. Spindle fibres attach to centromere of chromosomes
The question asks for the *incorrect* statement about mitosis. Option A is correct: during metaphase, chromosomes align at the equatorial plate (metaphase plate). Option B states 'Spindle fibres attach to centromere of chromosomes' — this is *incorrect*. According to NCERT Class 11 (Chapter 10: Cell Cycle and Cell Division), spindle fibres (kinetochore microtubules) attach to the *kinetochore*, a protein structure assembled *on* the centromere, not directly to the centromeric DNA itself. The centromere is the constricted region; the kinetochore is the functional attachment site. Option C is correct: chromatin recondenses into diffuse chromatin during telophase, i.e., chromosomes decondense. Option D is correct: centromere splitting (sister chromatid separation) defines anaphase onset. Hence, only option B misrepresents the precise site of microtubule attachment, making it the incorrect statement and the correct choice for this 'select the incorrect' question.
Question 162 · Cell: The Unit of Life
Which of the following statements with respect to Endoplasmic Reticulum is incorrect?
ARER has ribosomes attached to ER
BSER is devoid of ribosomes
CIn prokaryotes only RER are present
DSER are the sites for lipid synthesis
Answer: C. In prokaryotes only RER are present
The endoplasmic reticulum (ER) is a network of membranous tubules and sacs in eukaryotic cells. Rough ER (RER) bears ribosomes on its surface and is involved in protein synthesis and processing. Smooth ER (SER) lacks ribosomes and functions in lipid synthesis, steroid hormone production, detoxification, and calcium ion storage. Prokaryotes — such as bacteria — lack membrane-bound organelles entirely; they do not possess any form of endoplasmic reticulum, neither RER nor SER. Therefore, the statement 'In prokaryotes only RER are present' is biologically false and hence incorrect. This aligns with NCERT Class 11 Biology (Chapter 8: Cell: The Unit of Life), which explicitly states that prokaryotic cells have no endomembrane system and no ER. All other options are accurate: RER does have attached ribosomes; SER is indeed ribosome-free; and SER is a major site for lipid synthesis, including phospholipids and cholesterol.
Question 163 · Biological Classification
In the taxonomic categories, which hierarchical arrangement in ascending order is correct in case of animals?
AKingdom, Phylum, Class, Order, Family, Genus, Species
BKingdom, Class, Phylum, Family, Order, Genus, Species
CKingdom, Order, Class, Phylum, Family, Genus, Species
DKingdom, Order, Phylum, Class, Family, Genus, Species
Answer: A. Kingdom, Phylum, Class, Order, Family, Genus, Species
The universally accepted hierarchy of taxonomic categories in ascending order (from broadest to most specific) for animals is Kingdom → Phylum → Class → Order → Family → Genus → Species. This sequence reflects increasing specificity and shared evolutionary relationships. As per NCERT Class 11 Biology (Chapter 1: The Living World), this seven-tier Linnaean system is standard across all animal taxa. Option A correctly follows this order. Option B incorrectly places Class before Phylum — a violation since Phylum (e.g., Chordata) encompasses multiple Classes (e.g., Mammalia, Aves). Options C and D disrupt the sequence by misplacing Order and Phylum, violating both phylogenetic logic and NCERT’s defined hierarchy. Species is always the lowest (most inclusive) unit, while Kingdom is the highest. Memorising this fixed order is essential for NEET, as questions frequently test recall of correct sequence or identification of misplaced categories.
Question 165 · Biological Classification & Microbes
Given below are two statements:
Statement I: Mycoplasma can pass through filters with pore size less than 1 micron.
Statement II: Mycoplasma are bacteria that possess a cell wall.
In the light of the above statements, choose the most appropriate answer from the options given below:
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer: C. Statement I is correct but Statement II is incorrect
Mycoplasma are the smallest known living cells capable of independent growth and reproduction. They lack a rigid cell wall — a defining feature distinguishing them from typical bacteria — and are therefore resistant to antibiotics like penicillin that target cell wall synthesis. Due to their minute size (0.1–0.3 µm) and absence of a cell wall, they can pass through bacterial filters with pore sizes less than 1 micron (typically 0.22 µm), which retain most bacteria. Hence, Statement I is correct. Statement II is incorrect because Mycoplasma do not have a cell wall; instead, they are surrounded only by a triple-layered plasma membrane rich in sterols. This structural peculiarity makes them pleomorphic and highly adaptable. NCERT Class 11 Biology (Chapter 2: Biological Classification) explicitly states: 'They are organisms that completely lack a cell wall' and notes their ability to pass through filters that retain bacteria. Thus, the correct choice is option C — Statement I is correct but Statement II is incorrect.
Question 166 · Animal Tissues
Which of the following is not a connective tissue?
ABlood
BAdipose tissue
CCartilage
DNeuroglia
Answer: D. Neuroglia
Connective tissues are characterized by cells embedded in an extracellular matrix, providing support, protection, and transport. Blood consists of plasma (fluid matrix) with RBCs, WBCs, and platelets — classified as a fluid connective tissue (NCERT Class 11, Chapter 7: Structural Organisation in Animals). Adipose tissue stores fat and has adipocytes in a loose matrix — a specialized connective tissue. Cartilage has chondrocytes in lacunae within a firm, flexible matrix — a supportive connective tissue. Neuroglia (or glial cells), however, are non-neuronal supporting cells of nervous tissue; they originate from ectoderm (like neurons), lack intercellular matrix, and function in insulation, nutrient supply, and immune surveillance — making them part of nervous tissue, not connective tissue. NCERT explicitly lists neuroglia under neural tissue in Table 7.1 and distinguishes connective tissue types (e.g., blood, bone, cartilage, lymphoid, adipose) separately. Hence, neuroglia is correctly identified as the exception.
Question 167 · Excretion and osmoregulation in animals
Nitrogenous waste is excreted in the form of pellet or paste by:
AOrnithorhynchus
BSalamandra
CHippocampus
DPavo
Answer: D. Pavo
Birds, including Pavo (peacock), excrete nitrogenous waste primarily as uric acid, which is insoluble and non-toxic, allowing water conservation. Uric acid is eliminated as a semi-solid white paste or pellet — a key adaptation for flight and arid habitats. Ornithorhynchus (platypus) is a mammal that excretes urea (like most mammals), not uric acid pellets. Salamandra (a salamander, amphibian) excretes mainly ammonia (aquatic larvae) or urea (terrestrial adults), but never uric acid pellets. Hippocampus (seahorse) is a fish; teleosts excrete ammonia via gills and do not form uric acid pellets. NCERT Class 11 Chapter 18 'Body Fluids and Circulation' and Chapter 19 'Excretory Products and their Elimination' explicitly state that birds and reptiles are uricotelic and excrete uric acid in semi-solid form — e.g., 'birds and land reptiles excrete uric acid in the form of a whitish paste or pellet'. Thus, only Pavo among the options is uricotelic and matches the description.
Question 168 · Animal Kingdom – Chordata and Vertebrata
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): All vertebrates are chordates but all chordates are not vertebrates. Reason (R): Notochord is replaced by vertebral column in the adult vertebrates. In the light of the above statements, choose the most appropriate answer.
ABoth (A) and (R) are correct and (R) is the correct explanation of (A)
BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)
C(A) is correct but (R) is not correct
D(A) is not correct but (R) is correct
Answer: A. Both (A) and (R) are correct and (R) is the correct explanation of (A)
Assertion (A) is correct: Vertebrates (e.g., fish, birds, mammals) possess a notochord at some stage and a vertebral column in adults — making them a subgroup of Chordata. However, not all chordates are vertebrates; for example, cephalochordates (like Amphioxus) and urochordates (like Ascidia) retain the notochord throughout life or only in larval stages and lack a true vertebral column — hence they are chordates but not vertebrates. Reason (R) is also correct and directly explains (A): The defining feature distinguishing vertebrates from other chordates is the replacement of the embryonic notochord by a bony or cartilaginous vertebral column during development. This structural advancement enables greater support and protection of the spinal cord, marking the evolutionary transition from protochordates to vertebrates. Thus, (R) correctly accounts for why vertebrates form a subset of chordates — not all chordates undergo this replacement, so they remain non-vertebrate chordates. Both statements are factually accurate and causally linked, satisfying NCERT Class 11 Biology (Chapter 4: Animal Kingdom) definitions.
Question 169 · Human Health and Disease; Disorders of Musculoskeletal and Nervous Systems
Which of the following is a correct match for disease and its symptoms?
AArthritis — Inflamed joints
BTetany — High Ca²⁺ level causing rapid spasms
CMyasthenia gravis — Genetic disorder resulting in weakening and paralysis of skeletal muscle
DMuscular dystrophy — An autoimmune disorder causing progressive degeneration of skeletal muscle
Answer: A. Arthritis — Inflamed joints
Arthritis is correctly matched with inflamed joints — a hallmark feature involving swelling, pain, and stiffness due to joint inflammation, as described in NCERT Class 12 Chapter 8 (Human Health and Disease). Tetany is incorrectly linked to high Ca²⁺; it results from *hypocalcemia* (low serum calcium), leading to neuromuscular hyperexcitability and involuntary spasms. Myasthenia gravis is an *autoimmune* disorder targeting acetylcholine receptors at neuromuscular junctions — not genetic — causing fluctuating skeletal muscle weakness, especially in ocular and facial muscles. Muscular dystrophy is a group of *genetic* disorders (e.g., Duchenne type due to dystrophin gene mutation) causing progressive skeletal muscle degeneration and weakness — not autoimmune. Thus, only option A accurately reflects the disease–symptom relationship per NCERT standards. The other options misattribute pathophysiology: tetany (B) confuses cause, myasthenia gravis (C) mislabels etiology, and muscular dystrophy (D) misidentifies mechanism — all contradicted by NCERT’s precise definitions.
Question 170 · Locomotion and Movement – Disorders of Skeletal System
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Osteoporosis is characterised by decreased bone mass and increased chance of fractures.
Reason (R): Common cause of osteoporosis is increased levels of estrogen.
In the light of the above statements, choose the most appropriate answer from the options given below.
ABoth (A) and (R) are correct and (R) is the correct explanation of (A)
BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)
C(A) is correct but (R) is not correct
D(A) is not correct but (R) is correct
Answer: C. (A) is correct but (R) is not correct
Assertion (A) is correct: Osteoporosis is a metabolic bone disorder marked by reduced bone mineral density, microarchitectural deterioration of bone tissue, and consequent increase in bone fragility and fracture risk — especially at spine, hip, and wrist. NCERT Class 11 (Chapter 20: Locomotion and Movement) explicitly states that osteoporosis involves loss of bone mass and heightened fracture susceptibility. Reason (R), however, is incorrect: Estrogen *inhibits* bone resorption by osteoclasts; thus, *decreased* estrogen — particularly post-menopause in females — is a major cause of osteoporosis. Elevated estrogen levels do not cause osteoporosis; rather, they protect against it. Therefore, while (A) is scientifically accurate, (R) misrepresents hormonal pathophysiology and contradicts NCERT’s explanation. Hence, option (C) — '(A) is correct but (R) is not correct' — is the only valid choice. This aligns with NEET’s emphasis on conceptual clarity over rote memorisation.
Question 171 · Regulation of gene expression in prokaryotes (Lac operon)
In an E. coli strain, the i gene is mutated such that its product cannot bind the inducer molecule. If lactose is provided in the growth medium, what will be the outcome?
AOnly the z gene will be transcribed
BThe z, y, and a genes will be transcribed
CThe z, y, and a genes will not be translated
DRNA polymerase will bind to the promoter region
Answer: C. The z, y, and a genes will not be translated
In the lac operon, the i gene encodes the repressor protein, which binds to the operator in the absence of inducer (allolactose, derived from lactose) and prevents transcription of z, y, and a structural genes. Here, the i gene mutation produces a repressor that cannot bind the inducer — meaning it remains constitutively bound to the operator regardless of lactose presence. Thus, RNA polymerase cannot access the promoter, and transcription of z (β-galactosidase), y (permease), and a (transacetylase) does not occur. Since no mRNA is synthesized, translation of these genes also fails — making option (3) correct. Option (1) is wrong because no structural gene is transcribed; option (2) incorrectly assumes transcription occurs; option (4) is false because RNA polymerase binding is blocked by the bound repressor. This aligns with NCERT Class 12 Biology Chapter 6 (Molecular Basis of Inheritance), which states that a non-inducible repressor mutant leads to permanent repression of the lac operon.
Question 172 · DNA structure and packaging
If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs?
A3.3 × 10⁹ bp
B6.6 × 10⁹ bp
C3.3 × 10⁶ bp
D6.6 × 10⁶ bp
Answer: A. 3.3 × 10⁹ bp
According to NCERT Class 12 Biology (Chapter 6: Molecular Basis of Inheritance), the distance between two consecutive base pairs in a DNA double helix is 0.34 nm (3.4 × 10⁻¹⁰ m). Therefore, the total number of base pairs in a DNA molecule of given length = total length ÷ distance per base pair. Here, length = 1.1 m; distance per bp = 3.4 × 10⁻¹⁰ m. So, number of bp = 1.1 / (3.4 × 10⁻¹⁰) ≈ 3.24 × 10⁹ ≈ 3.3 × 10⁹ bp. This matches option A. The calculation assumes B-form DNA under physiological conditions — the standard reference used in NEET and NCERT. Note that human diploid genome (~6.6 × 10⁹ bp) corresponds to ~2.2 m of DNA, so 1.1 m corresponds to half that — i.e., haploid content — reinforcing the approximation. Students must recall the 0.34 nm spacing and unit conversion (metre to nanometre) accurately.
Question 173 · Gametogenesis
Which of the following statements are true for spermatogenesis but do not hold true for oogenesis? (a) It results in the formation of haploid gametes (b) Differentiation of gamete occurs after the completion of meiosis (c) Meiosis occurs continuously in a mitotically dividing stem cell population (d) It is controlled by the luteinising hormone (LH) and follicle stimulating hormone (FSH) secreted by the anterior pituitary (e) It is initiated at puberty
A(c) and (e) only
B(b) and (c) only
C(b), (d) and (e) only
D(b), (c) and (e) only
Answer: D. (b), (c) and (e) only
Spermatogenesis and oogenesis both produce haploid gametes, so (a) is common to both — excluded. In spermatogenesis, spermatids undergo spermiogenesis (differentiation into spermatozoa) *after* meiosis is complete; in oogenesis, oocyte differentiation (e.g., cortical granule formation, zona pellucida deposition) occurs *during* prophase I and resumes only after fertilisation — thus (b) is unique to spermatogenesis. Spermatogonia act as self-renewing mitotic stem cells, sustaining continuous meiosis throughout life; oogonia cease mitotic division before birth, and primary oocytes remain arrested in prophase I — so (c) applies only to spermatogenesis. Both processes are regulated by FSH and LH, though their roles differ (FSH stimulates Sertoli cells; LH triggers testosterone release in males; in females, FSH initiates follicular growth and LH triggers ovulation) — hence (d) is *not* exclusive to spermatogenesis. Spermatogenesis begins at puberty and continues lifelong; oogenesis initiates prenatally (oogonia → primary oocytes) and resumes only at puberty — so (e) is correct for spermatogenesis alone. Therefore, only (b), (c), and (e) are exclusively true for spermatogenesis.
Question 174 · Locomotion and Movement
Which of the following is present between the adjacent bones of the vertebral column?
AIntercalated discs
BCartilage
CAreolar tissue
DSmooth muscle
Answer: B. Cartilage
Between adjacent vertebrae lie intervertebral discs, which are composed of fibrocartilage — a specialized type of cartilage containing collagen fibers embedded in a firm matrix. These discs act as shock absorbers, permit limited movement, and maintain spinal flexibility and stability. Intercalated discs are unique to cardiac muscle tissue and facilitate synchronized contraction; they are not found between bones. Areolar tissue is a loose connective tissue widely distributed under epithelia and around blood vessels, but it does not serve as an intervertebral joint structure. Smooth muscle is involuntary, non-striated muscle found in walls of hollow organs (e.g., gut, uterus), not in articulations between skeletal elements. NCERT Class 11 Biology (Chapter 20: Locomotion and Movement) explicitly states that intervertebral discs are made of cartilage and function as cushioning pads between vertebrae — confirming cartilage as the correct structural component present between adjacent vertebral bones.
Question 175 · Cell Cycle and Cell Division
Regarding meiosis, which of the following statements is incorrect?
AThere are two stages in meiosis: meiosis-I and meiosis-II.
BDNA replication occurs in the S phase of meiosis-II.
CPairing of homologous chromosomes and recombination occurs in meiosis-I.
DFour haploid cells are formed at the end of meiosis-II.
Answer: B. DNA replication occurs in the S phase of meiosis-II.
Meiosis consists of two sequential divisions—meiosis-I (reductional) and meiosis-II (equational)—with only one round of DNA replication, which occurs exclusively during the S phase *preceding* meiosis-I. No DNA replication takes place before meiosis-II; the cells entering meiosis-II are already diploid in chromosome number but with duplicated chromatids (2n, 4c), and meiosis-II separates sister chromatids to yield four haploid (n) daughter cells. Statement B incorrectly claims DNA replication occurs in the S phase of meiosis-II—a biological impossibility, as meiosis-II lacks an S phase entirely. This aligns with NCERT Class 11 Chapter 10 (Cell Cycle and Cell Division), which explicitly states: 'Interkinesis, the stage between meiosis-I and meiosis-II, is short and does not involve DNA replication.' Statements A, C, and D are factually correct: meiosis has two divisions; synapsis and crossing over occur exclusively in prophase-I of meiosis-I; and the final outcome is four genetically distinct haploid cells. Hence, option B is the incorrect statement.
Question 176 · Human Health and Disease
Given below are two statements:
Statement I: Autoimmune disorder is a condition where the body's defense mechanism recognizes its own cells as foreign bodies.
Statement II: Rheumatoid arthritis is a condition where the body does not attack self cells.
In the light of the above statements, choose the most appropriate answer from the options given below:
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer: C. Statement I is correct but Statement II is incorrect
Autoimmune disorders occur when the immune system fails to distinguish self-antigens from non-self antigens, leading to an abnormal immune response against the body’s own tissues — making Statement I scientifically accurate. Rheumatoid arthritis (RA) is a classic example of such an autoimmune disease: in RA, autoantibodies like rheumatoid factor and anti-citrullinated protein antibodies trigger chronic inflammation of synovial joints, resulting in destruction of cartilage and bone. Thus, the body *does* attack its own cells — directly contradicting Statement II. NCERT Class 12 Biology (Chapter 8: Human Health and Disease) explicitly lists rheumatoid arthritis under 'autoimmune diseases' and defines them as conditions where 'the body attacks self'. Therefore, only Statement I is correct, and Statement II is false. This aligns precisely with option C — 'Statement I is correct but Statement II is incorrect'.
Question 177 · Evolutionary Biology: Mechanisms of Evolution
Natural selection where more individuals acquire a specific character value other than the mean character value leads to
AStabilising change
BDirectional change
CDisruptive change
DRandom change
Answer: B. Directional change
When natural selection favours individuals with a phenotypic trait value that deviates from the population mean—either higher or lower—and shifts the average phenotype over generations, it results in directional selection. This occurs under changing environmental conditions, such as increased predation pressure favouring darker moths in polluted areas (industrial melanism), leading to a progressive shift in the mean trait value. In contrast, stabilising selection preserves the mean by selecting against extremes (e.g., medium birth weight in humans), while disruptive selection favours both extremes over the mean (e.g., beak size in black-bellied seedcrackers feeding on hard vs. soft seeds). 'Random change' is not a recognised mode of natural selection—it describes genetic drift, which is non-adaptive. NCERT Class 12 Chapter 7 'Evolution' explicitly defines directional selection as the process causing a shift in the population’s mean trait value due to consistent selection pressure in one direction, making option B the scientifically accurate and NCERT-aligned answer.
Question 178 · Blood Coagulation and Erythrocyte Lifespan
Given below are two statements:
Statement I: The coagulum is formed of a network of threads called thrombins.
Statement II: Spleen is the graveyard of erythrocytes.
In the light of the above statements, choose the most appropriate answer from the options given below:
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer: D. Statement I is incorrect but Statement II is correct
Statement I is incorrect because the coagulum (blood clot) is formed by a meshwork of fibrin threads—not thrombins. Thrombin is an enzyme (a serine protease) that converts soluble fibrinogen into insoluble fibrin monomers; it is not the structural component of the clot. This is clearly explained in NCERT Class 11, Chapter 18 'Body Fluids and Circulation', where the coagulation cascade emphasizes thrombin’s catalytic role, not its presence as clot fibers. Statement II is correct: the spleen filters aged or damaged erythrocytes and macrophages there phagocytose them—hence it is rightly called the 'graveyard of RBCs'. NCERT explicitly states this in the same chapter, noting that RBCs have a lifespan of ~120 days and are primarily removed by splenic macrophages. Thus, only Statement II is correct, making option (4) — i.e., D — the right choice.
Question 179 · Strategies for Enhancement of Food Production
Breeding crops with higher levels of vitamins and minerals or higher proteins and healthier fats is called:
ABio-magnification
BBio-remediation
CBio-fortification
DBio-accumulation
Answer: C. Bio-fortification
Bio-fortification is the process of breeding crops to increase their nutritional quality—such as higher levels of vitamins (e.g., vitamin A in Golden Rice), minerals (e.g., iron and zinc in pearl millet), proteins (e.g., lysine-rich maize), or healthier fatty acid profiles (e.g., omega-3 enriched soybean). Unlike fortification (which adds nutrients post-harvest), bio-fortification enhances nutrient content genetically through conventional breeding or biotechnology, making it sustainable and cost-effective for rural populations. Bio-magnification refers to increasing concentration of toxic substances (e.g., DDT) at successive trophic levels; bio-remediation uses organisms to clean pollutants from soil/water; bio-accumulation is uptake and storage of substances (often toxins) in an organism over time. As per NCERT Class 12 Biology Chapter 9 (‘Strategies for Enhancement of Food Production’), bio-fortification is explicitly highlighted as a key strategy to combat hidden hunger and micronutrient deficiencies—aligning directly with the question’s description.
Question 180 · Biotechnology and its Applications
In gene therapy for Adenosine Deaminase (ADA) deficiency, the patient requires periodic infusion of genetically engineered lymphocytes because:
ARetroviral vector is introduced into these lymphocytes.
BThe gene isolated from bone marrow cells producing ADA is introduced into cells at embryonic stages.
CLymphocytes from the patient's blood are grown in culture outside the body.
DGenetically engineered lymphocytes are not immortal cells.
Answer: D. Genetically engineered lymphocytes are not immortal cells.
In ADA deficiency gene therapy, autologous lymphocytes are extracted from the patient, genetically modified ex vivo using a retroviral vector carrying a functional ADA gene, and then reinfused. However, these corrected lymphocytes have a finite lifespan — they are somatic, non-stem cells and do not self-renew indefinitely. Unlike hematopoietic stem cells, mature lymphocytes undergo apoptosis after a limited number of divisions and are gradually lost from circulation. Hence, repeated infusions are necessary to sustain therapeutic ADA enzyme levels. Option D correctly identifies this fundamental limitation: the engineered lymphocytes are not immortal. Option A describes a method (vector delivery), not the reason for periodic infusion. Option B is incorrect — embryonic intervention is neither used nor feasible in current clinical protocols; ADA gene therapy targets somatic cells postnatally. Option C states a procedural step (ex vivo culture), but does not explain why repeat dosing is needed. This aligns with NCERT Class 12 Chapter 12 (Biotechnology and its Applications), which emphasizes that first-generation ADA therapy uses transiently corrected T-lymphocytes requiring repeated administration.
Question 181 · Human Reproduction
At which stage of life is the oogenesis process initiated?
APuberty
BEmbryonic development stage
CBirth
DAdult
Answer: B. Embryonic development stage
Oogenesis—the formation and development of mature ova—initiates during embryonic development in females. In human females, primordial germ cells migrate to the developing ovaries around the 3rd–4th week of gestation and differentiate into oogonia. These oogonia undergo mitotic proliferation and then enter prophase I of meiosis by the 5th month of fetal life, becoming primary oocytes arrested in diplotene stage. This meiotic arrest persists until puberty. Thus, although meiotic completion occurs much later, the process of oogenesis begins prenatally—not at puberty, birth, or adulthood. NCERT Class 12 Biology (Chapter 3: Human Reproduction) explicitly states: 'The process of formation of ova (oogenesis) starts during the embryonic development stage.' Puberty marks the resumption of meiosis (first meiotic division) in selected primary oocytes each menstrual cycle, but initiation is strictly embryonic. Birth and adult stages are irrelevant to initiation; they represent continuation or completion phases.
Question 183 · Digestion and Absorption
Which of the following functions is not performed by secretions from salivary glands?
AControl bacterial population in mouth
BDigestion of complex carbohydrates
CLubrication of oral cavity
DDigestion of disaccharides
Answer: D. Digestion of disaccharides
Salivary glands secrete saliva containing water, electrolytes, mucus, lysozyme, and salivary amylase (ptyalin). Lysozyme helps control bacterial population (Option A), mucus provides lubrication for swallowing (Option C), and salivary amylase hydrolyses starch (a complex carbohydrate) into maltose (Option B). However, salivary amylase acts only on polysaccharides—not disaccharides like sucrose or lactose—which require specific brush-border enzymes (e.g., sucrase, lactase) in the small intestine. Thus, digestion of disaccharides is not a function of salivary secretions (Option D). This aligns with NCERT Class 11, Chapter 16 'Digestion and Absorption', which states that salivary amylase initiates starch digestion but has no role in disaccharide breakdown. The absence of disaccharidases in saliva makes Option D the correct choice for 'not performed'.
Question 184 · Population Ecology
If 8 Drosophila in a laboratory population of 80 died during a week, the death rate in the population is ______ individuals per Drosophila per week.
Answer: A. 0.1
Death rate (or mortality rate) is defined as the number of deaths per individual per unit time. It is calculated as: Death rate = (Number of deaths) / (Initial population size × Time interval). Here, 8 individuals died in a population of 80 over one week. So, death rate = 8 / (80 × 1) = 0.1 individuals per Drosophila per week. This matches option A. According to NCERT Class 12 Biology (Chapter 13: Organisms and Populations), population density changes due to four basic processes — natality, mortality, immigration, and emigration — and mortality (death rate) is expressed per capita per unit time. The unit 'individuals per Drosophila per week' may sound unusual but reflects per capita rate — i.e., average deaths per individual in the population per week. A value of 0.1 means that, on average, each Drosophila contributes 0.1 to the total death count per week — a standard proportional measure used in ecological demography. Options B (10) and C (1.0) misplace the decimal by ignoring the denominator; option D (zero) contradicts the given data.
Question 185 · Biotechnology: Principles and Processes
Given below are two statements:
Statement I: Restriction endonucleases recognise specific sequences to cut DNA known as palindromic nucleotide sequences.
Statement II: Restriction endonucleases cut the DNA strand a little away from the centre of the palindromic site.
In the light of the above statements, choose the most appropriate answer from the options given below:
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer: A. Both Statement I and Statement II are correct
Statement I is correct: Restriction endonucleases (e.g., EcoRI, HindIII) recognise and bind to specific, short, palindromic DNA sequences — where the sequence reads the same on both strands when oriented 5'→3'. This symmetry allows precise double-stranded cleavage. Statement II is also correct: Most Type II restriction enzymes (the class used in recombinant DNA technology) cut *within* or *near* the recognition site — but crucially, many (like EcoRI) produce staggered cuts *away from the exact centre*, generating cohesive (sticky) ends. For example, EcoRI recognises GAATTC and cuts between G and A on both strands, yielding 5'-overhangs — not at the central axis but asymmetrically within the palindrome. NCERT Class 12 Biology (Chapter 11, 'Biotechnology: Principles and Processes') explicitly states that restriction enzymes cut DNA at specific sites, often producing sticky ends due to offset cleavage. Thus, both statements align with NCERT content and molecular biology fundamentals.
Question 186 · Biological Classification
Which of the following is a correct statement?
ACyanobacteria are a group of autotrophic organisms classified under kingdom Monera.
BBacteria are exclusively heterotrophic organisms.
CSlime moulds are saprophytic organisms classified under Kingdom Monera.
DMycoplasma have DNA, ribosome and cell wall.
Answer: A. Cyanobacteria are a group of autotrophic organisms classified under kingdom Monera.
Option A is correct: Cyanobacteria (e.g., Nostoc, Anabaena) are prokaryotic, photosynthetic, autotrophic organisms placed in Kingdom Monera as per NCERT Class 11 (Chapter 2: Biological Classification). They possess chlorophyll-a and carry out oxygenic photosynthesis. Option B is incorrect because bacteria include autotrophs (e.g., nitrifying bacteria, cyanobacteria) and chemosynthetic types — not exclusively heterotrophic. Option C is wrong: Slime moulds are eukaryotic, saprophytic protists belonging to Kingdom Protista, not Monera — they lack prokaryotic cell structure and show phagocytosis. Option D is false: Mycoplasma are the smallest known prokaryotes and uniquely lack a cell wall; they possess DNA and 70S ribosomes, but absence of cell wall makes them resistant to penicillin and pleomorphic. Thus, only statement A aligns with NCERT’s classification framework and factual accuracy.
Question 187 · Biotechnology Applications in Medicine
Statements related to human insulin are given below. Which statement(s) is/are correct about genetically engineered insulin?
(a) Pro-hormone insulin contains an extra stretch of C-peptide.
(b) A-peptide and B-peptide chains of insulin were produced separately in E. coli, extracted and combined by creating disulphide bonds between them.
(c) Insulin used for treating diabetes was earlier extracted from cattle and pigs.
(d) Pro-hormone insulin needs to be processed for conversion into a mature and functional hormone.
(e) Some patients developed allergic reactions to animal-derived insulin.
A(a), (b) and (d) only
B(b) only
C(c) and (d) only
D(c), (d) and (e) only
Answer: B. (b) only
Genetically engineered human insulin was first developed using recombinant DNA technology. Statement (b) is correct: the A and B chains were separately expressed in E. coli as inclusion bodies, purified, and then chemically joined via disulphide bonds — this was the method used in early production (e.g., Humulin). Statement (a) is incorrect because pro-insulin (not 'pro-hormone insulin') contains the C-peptide; the term 'pro-hormone insulin' is biologically inaccurate — NCERT uses 'proinsulin'. Statement (c) refers to historical practice (pre-1980s), but the question specifically asks about *genetically engineered* insulin; thus (c) describes prior non-recombinant sources and is irrelevant to the context. Statement (d) is true for natural proinsulin processing *in humans*, but the question focuses on *engineered* insulin production — where proinsulin is not used; instead, separate chains are synthesized and assembled. So (d) misapplies endogenous processing to recombinant production. Statement (e) is historically valid for animal insulin, but again, it’s not a feature of *genetically engineered* insulin — it's a drawback of the older source. Hence, only (b) is strictly correct for recombinant insulin.
Question 189 · Genetic linkage and mapping
The recombination frequency between the genes a and c is 5%, b and c is 15%, b and d is 9%, a and b is 20%, c and d is 24%, and a and d is 29%. What is the sequence of these genes on a linear chromosome?
Aa, d, b, c
Bd, b, a, c
Ca, b, c, d
Da, c, b, d
Answer: D. a, c, b, d
In genetic mapping, recombination frequency (RF) is directly proportional to the physical distance between two genes on a chromosome. For four linked genes, the correct linear order must satisfy all pairwise RF values — i.e., the RF between outermost genes should equal the sum of RFs between adjacent intervening genes (assuming no double crossovers). Given RF(a–c) = 5%, RF(b–c) = 15%, RF(b–d) = 9%, RF(a–b) = 20%, RF(c–d) = 24%, and RF(a–d) = 29%, we test possible orders. Option D: a–c–b–d implies a–c (5%), c–b (15%), b–d (9%). Then a–b = a–c + c–b = 5% + 15% = 20% ✓; c–d = c–b + b–d = 15% + 9% = 24% ✓; a–d = a–c + c–b + b–d = 5% + 15% + 9% = 29% ✓. All match. Other options fail: e.g., a–b–c–d gives a–c = a–b + b–c = 20% + 15% = 35% ≠ 5%. As per NCERT Class 12 Chapter 5 ‘Principles of Inheritance and Variation’, gene mapping relies on additive recombination frequencies for collinear genes, and the lowest RF indicates closest proximity — here a and c are closest (5%), supporting a–c as adjacent. Hence, the correct sequence is a–c–b–d.
Question 190 · Biomolecules: Structure and Function
Match List-I (Biological Molecules) with List-II (Biological Functions):
List-I
(a) Glycogen
(b) Globulin
(c) Steroids
(d) Thrombin
List-II
(i) Hormone
(ii) Biocatalyst
(iii) Antibody
(iv) Storage product
A(a) – (iii), (b) – (ii), (c) – (iv), (d) – (i)
B(a) – (iv), (b) – (ii), (c) – (i), (d) – (iii)
C(a) – (ii), (b) – (iv), (c) – (iii), (d) – (i)
D(a) – (iv), (b) – (iii), (c) – (i), (d) – (ii)
Answer: D. (a) – (iv), (b) – (iii), (c) – (i), (d) – (ii)
Glycogen is a polysaccharide that serves as the primary storage form of glucose in animals, especially in liver and muscle — matching (iv). Globulins are globular proteins; many immunoglobulins (e.g., IgG) fall under this category and function as antibodies — correctly paired with (iii). Steroids (e.g., cortisol, testosterone) are lipid-derived signaling molecules acting as hormones — hence (i). Thrombin is a serine protease enzyme that converts fibrinogen to fibrin during blood clotting; as an enzyme, it functions as a biocatalyst — corresponding to (ii). This mapping aligns precisely with NCERT Class 11 Biology (Chapter 9: Biomolecules), which classifies glycogen under storage molecules, globulins (especially gamma-globulins) as antibodies, steroids as hormonal lipids, and thrombin — though not explicitly named — falls under enzyme examples discussed in the context of protein function and catalysis. Option D reflects this accurate pairing.
Question 191 · Reproductive Health
Match List-I with List-II with respect to methods of contraception and their respective actions.
List-I
(a) Diaphragms
(b) Contraceptive Pills
(c) Intra Uterine Devices
(d) Lactational Amenorrhea
List-II
(i) Inhibit ovulation and implantation
(ii) Increase phagocytosis of sperm within uterus
(iii) Absence of menstrual cycle and ovulation following parturition
(iv) They cover the cervix, blocking the entry of sperms
A(a) – (iv), (b) – (i), (c) – (iii), (d) – (ii)
B(a) – (iv), (b) – (i), (c) – (ii), (d) – (iii)
C(a) – (ii), (b) – (iv), (c) – (i), (d) – (iii)
D(a) – (iii), (b) – (ii), (c) – (i), (d) – (iv)
Answer: B. (a) – (iv), (b) – (i), (c) – (ii), (d) – (iii)
Diaphragms are barrier devices placed over the cervix to physically block sperm entry — matching (iv). Contraceptive pills (especially combined oral contraceptives) contain synthetic estrogen and progesterone that suppress gonadotropins, thereby inhibiting ovulation; they also thicken cervical mucus and impair endometrial receptivity, preventing implantation — correctly paired with (i). Intrauterine Devices (IUDs), particularly copper IUDs, create a local inflammatory response in the uterus that increases phagocytosis of sperm and impairs sperm motility and fertilization — hence (c)–(ii). Lactational amenorrhea is a natural postpartum contraceptive method where exclusive breastfeeding suppresses GnRH pulsatility, leading to temporary anovulation and amenorrhea — corresponding to (iii). Option B correctly maps all four: (a)–(iv), (b)–(i), (c)–(ii), (d)–(iii). This aligns precisely with NCERT Class 12 Biology Chapter 4 (Reproductive Health), Table 4.1 and related text on mechanisms of contraceptive methods.
Question 192 · Chemical Coordination and Integration
Which of the following are NOT the effects of parathyroid hormone? (a) Stimulates the process of bone resorption (b) Decreases Ca²⁺ level in blood (c) Reabsorption of Ca²⁺ by renal tubules (d) Decreases the absorption of Ca²⁺ from digested food (e) Increases metabolism of carbohydrates
A(a) and (c) only
B(b), (d) and (e) only
C(a) and (e) only
D(b) and (c) only
Answer: B. (b), (d) and (e) only
Parathyroid hormone (PTH), secreted by the parathyroid glands, is a key regulator of calcium homeostasis. It acts to *increase* blood Ca²⁺ levels through three main actions: (i) stimulating bone resorption by activating osteoclasts — so (a) *is* a correct effect; (ii) enhancing reabsorption of Ca²⁺ in the distal convoluted tubule and collecting duct of kidneys — so (c) *is* a correct effect; and (iii) promoting activation of vitamin D (calcitriol) in kidneys, which in turn increases intestinal absorption of dietary Ca²⁺ — thus (d) is *not* an effect; PTH *increases*, not decreases, Ca²⁺ absorption. Further, PTH does *not* lower blood Ca²⁺ — (b) is false. It also has no direct role in carbohydrate metabolism — (e) is incorrect. Therefore, the statements that are *not* effects of PTH are (b), (d), and (e). This matches option (2), i.e., choice B. NCERT Class 11 (Chapter 22: Chemical Coordination and Integration) explicitly states PTH raises plasma calcium via bone, kidney, and gut (via vitamin D), and makes no mention of carbohydrate metabolism.
Question 193 · Immunity and Immune System
Select the incorrect statement with respect to acquired immunity.
APrimary response is produced when our body encounters a pathogen for the first time.
BAnamnestic response is elicited on subsequent encounters with the same pathogen.
CAnamnestic response is due to memory of first encounter.
DAcquired immunity is non-specific type of defense present at the time of birth.
Answer: D. Acquired immunity is non-specific type of defense present at the time of birth.
Acquired immunity is a specific, adaptive defense mechanism that develops *after* exposure to antigens — it is not present at birth and is highly specific to particular pathogens. It involves lymphocytes (B and T cells), immunological memory, and clonal selection. The primary response occurs during the first encounter with an antigen and is slow, low-magnitude, and mainly IgM-mediated. Upon re-exposure, the anamnestic (memory) response is rapid, robust, and predominantly IgG-mediated — a hallmark of acquired immunity. In contrast, non-specific (innate) immunity — including physical barriers, phagocytes, complement, and interferons — is present from birth and acts immediately but lacks memory or specificity. Therefore, stating that 'acquired immunity is non-specific and present at birth' is factually incorrect and contradicts NCERT Class 12 Chapter 8 (Human Health and Disease). Options (1), (2), and (3) correctly describe features of acquired immunity, making (4) the only incorrect statement.
Question 194 · DNA replication and semi-conservative nature
Ten E. coli cells with ¹⁵N-labeled double-stranded DNA are incubated in a medium containing ¹⁴N nucleotides. After 60 minutes, how many E. coli cells will have DNA completely free of ¹⁵N?
A20 cells
B40 cells
C60 cells
D80 cells
Answer: C. 60 cells
E. coli has a generation time of ~20 minutes under optimal conditions. In 60 minutes, three rounds of binary fission occur (60 ÷ 20 = 3). Starting with 10 cells, total cells after 3 generations = 10 × 2³ = 80. DNA replication is semi-conservative: in the first generation (20 min), all 20 cells have hybrid DNA (one ¹⁵N strand, one ¹⁴N strand). In the second generation (40 min), 40 cells result — half (20) have hybrid DNA, half (20) have light DNA (both strands ¹⁴N). In the third generation (60 min), 80 cells form: the 20 hybrid cells each produce one hybrid and one light cell (yielding 20 hybrid + 20 light), while the 20 light cells each produce two light cells (40 light). Total light (¹⁴N-only) cells = 20 + 40 = 60. Thus, 60 cells have DNA entirely free of ¹⁵N — matching option (3)/C. This aligns with NCERT Class 12 Chapter 2 (Sexual Reproduction in Flowering Plants is irrelevant; correct reference is Chapter 6 – Molecular Basis of Inheritance, Fig. 6.5 & text on Meselson-Stahl experiment).
Question 195 · Sex-linked inheritance (X-chromosome)
If a colour-blind female marries a man whose mother was also colour blind, what is the probability that their progeny will be colour blind?
Answer: D. 100%
Colour blindness is an X-linked recessive disorder. A colour-blind female must be homozygous recessive (X^c X^c), as she expresses the trait. Her husband’s mother was colour blind (X^c X^c), so she passed an X^c chromosome to her son. Since males are hemizygous (XY), the man must be X^c Y — i.e., colour blind himself. Crossing X^c X^c (female) × X^c Y (male): all daughters inherit X^c from mother and X^c from father → X^c X^c (colour blind); all sons inherit X^c from mother and Y from father → X^c Y (colour blind). Thus, 100% of progeny — both sons and daughters — are colour blind. This aligns with NCERT Class 12 Chapter 5 ‘Principles of Inheritance and Variation’, which states that in X-linked recessive inheritance, an affected female transmits the allele to all offspring, and if the father is affected, all daughters are carriers or affected and all sons express the trait when the mother is homozygous recessive.
Question 196 · Biotechnology: Principles and Processes
Which of the following is not a desirable feature of a cloning vector?
APresence of origin of replication
BPresence of a marker gene
CPresence of a single restriction enzyme site
DPresence of two or more recognition sites
Answer: D. Presence of two or more recognition sites
A cloning vector must replicate autonomously in the host, so an origin of replication (ori) is essential. A selectable marker gene (e.g., antibiotic resistance) allows identification of recombinant cells — hence it is desirable. A single restriction enzyme recognition site within the vector’s multiple cloning site (MCS) is preferred because it enables predictable, directional insertion of foreign DNA without disrupting essential vector elements. In contrast, presence of two or more *identical* recognition sites for the *same* restriction enzyme elsewhere in the vector (outside the MCS) is undesirable — it causes fragmentation of the vector during digestion, leading to loss of integrity and inefficient ligation. While modern vectors contain *multiple distinct* sites (in the MCS) for different enzymes, having two or more sites for the *same* enzyme scattered across the vector backbone compromises its utility. Thus, option (4) describes a flaw, not a desirable feature — making it the correct choice for 'not desirable'. This aligns with NCERT Class 12, Chapter 11, which emphasizes that ideal vectors have a single, unique restriction site for each enzyme in the MCS.
Question 197 · Structural Organisation in Animals
Match List-I with List-II:
List-I
(a) Bronchioles
(b) Goblet cell
(c) Tendons
(d) Adipose tissue
List-II
(i) Dense regular connective tissue
(ii) Loose connective tissue
(iii) Glandular epithelium
(iv) Ciliated epithelium
A(a) – (iv), (b) – (iii), (c) – (i), (d) – (ii)
B(a) – (i), (b) – (ii), (c) – (iii), (d) – (iv)
C(a) – (ii), (b) – (i), (c) – (iv), (d) – (iii)
D(a) – (iii), (b) – (iv), (c) – (ii), (d) – (i)
Answer: A. (a) – (iv), (b) – (iii), (c) – (i), (d) – (ii)
Bronchioles are lined by ciliated epithelium (iv), which helps trap and remove mucus and debris from the respiratory tract — as per NCERT Class 11, Chapter 18 'Body Fluids and Circulation' and Chapter 7 'Anatomy of Flowering Plants' (comparative epithelial coverage). Goblet cells are unicellular glands embedded in epithelial linings (e.g., respiratory and intestinal tracts); they constitute glandular epithelium (iii), not glandular tissue — a subtle but critical NCERT distinction: glandular epithelium refers to specialized epithelial cells secreting mucus, while glandular tissue implies organized multicellular glands. Tendons connect muscle to bone and consist of dense regular connective tissue (i), rich in parallel collagen fibres — clearly stated in NCERT Class 11, Chapter 7. Adipose tissue is a type of loose connective tissue (ii), composed of adipocytes embedded in a matrix with fibroblasts and blood vessels — confirmed in NCERT’s description of connective tissue subtypes. Hence, option A is correct: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii).
Question 198 · Cardiac Cycle and Conduction System
Which one of the following statements is correct?
AThe atrioventricular node (AVN) generates an action potential to stimulate atrial contraction.
BThe tricuspid and the bicuspid valves open due to the pressure exerted by the simultaneous contraction of the atria.
CBlood moves freely from atrium to the ventricle during joint diastole.
DIncreased ventricular pressure causes closing of the semilunar valves.
Answer: C. Blood moves freely from atrium to the ventricle during joint diastole.
Joint diastole refers to the phase when both atria and ventricles are relaxed simultaneously. During this phase, the AV valves (tricuspid and bicuspid) remain open due to higher atrial pressure relative to ventricular pressure, allowing passive, unimpeded flow of blood from atria into ventricles — a process accounting for ~70–80% of ventricular filling. Option C correctly describes this physiological reality. Option A is incorrect because the SA node—not AVN—initiates atrial depolarization; the AVN delays conduction but does not generate the atrial impulse. Option B misrepresents valve opening: AV valves open passively due to pressure gradient (atrial > ventricular), not due to atrial contraction itself. Option D is false: semilunar valves close when ventricular pressure falls *below* arterial pressure (e.g., at end-systole), not when ventricular pressure increases. This aligns precisely with NCERT Class 11 Chapter 18 'Body Fluids and Circulation', which emphasizes passive ventricular filling during joint diastole as a key feature of the cardiac cycle.
Question 199 · Neural Control and Coordination
Select the incorrect statement regarding synapses:
AThe membranes of presynaptic and postsynaptic neurons are in close proximity in an electrical synapse.
BElectrical current can flow directly from one neuron into the other across the electrical synapse.
CChemical synapses use neurotransmitters.
DImpulse transmission across a chemical synapse is always faster than that across an electrical synapse.
Answer: D. Impulse transmission across a chemical synapse is always faster than that across an electrical synapse.
Electrical synapses involve gap junctions where the pre- and postsynaptic membranes are extremely close (≈3.5 nm), allowing ions and small molecules to pass directly — enabling rapid, bidirectional impulse transmission. Chemical synapses, in contrast, rely on synaptic vesicles releasing neurotransmitters into the synaptic cleft, followed by receptor binding and postsynaptic potential generation — a process involving synaptic delay (~0.5–1 ms). Hence, impulse transmission is *slower* across chemical synapses than electrical ones. Statement D incorrectly claims chemical transmission is *always faster*, contradicting NCERT Class 11 (Chapter 21, 'Neural Control and Coordination') which explicitly states: 'Transmission of impulses is faster in electrical synapses than in chemical synapses'. Statements A, B, and C are factually correct: proximity in electrical synapses (A), direct current flow via connexons (B), and neurotransmitter dependence of chemical synapses (C) are all NCERT-aligned. Thus, D is the only incorrect statement.
Question 200 · Evolution: Homologous and Analogous Structures
Which of the following statements is not true?
AAnalogous structures are a result of convergent evolution.
BSweet potato and potato are an example of analogous structures.
CHomology indicates common ancestry.
DFlippers of penguins and dolphins are a pair of homologous organs.
Answer: D. Flippers of penguins and dolphins are a pair of homologous organs.
Homologous organs share similar origin and basic structure but may differ in function due to divergent evolution — e.g., forelimbs of humans, bats, and whales. Analogous organs have similar function but different origin and structure, arising from convergent evolution — e.g., wings of birds and insects. Sweet potato (a modified root, dicot) and potato (a modified stem, tuber) are analogous because they serve similar storage functions but develop from different plant parts and embryonic origins — correctly stated in option (2). Option (4) is incorrect: penguin flippers (modified forelimbs of birds) and dolphin flippers (modified forelimbs of mammals) are *analogous*, not homologous — though both are tetrapod forelimbs, their structural similarity is primarily functional adaptation to aquatic locomotion, and they evolved independently after divergence of avian and mammalian lineages; NCERT Class 12 (Chapter 7: Evolution) explicitly classifies them as analogous structures due to independent evolution under similar selective pressures. Thus, statement (4) is false, making it the correct choice for 'not true'.