Revision guide
How to use NEET 2019 Biology PYQs
This set contains reviewed questions from the Code P1 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
- Microbes in Human Welfare
- Cell: The Unit of Life
- Environmental Issues
Best review method
Use this older paper to uncover stable NCERT facts. For every wrong answer, name the chapter and the line or diagram you need to revisit before moving on.
NCERT focus
Revise microbes and their products, cell-structure functions, environmental agreements, reproduction and endocrine terms. These are factual areas where one qualifier changes the answer.
Frequently asked questions
NEET 2019 Biology PYQ FAQs
Are these NEET 2019 Biology answers checked with the official key?
This page uses the NEET 2019 Code P1 Biology question paper and checks answers against the official final answer key before publication.
How should I use NEET 2019 Biology PYQs for revision?
Solve the question first, check the answer, then connect the explanation to the NCERT concept. Pay special attention to ecology, genetics, human physiology, and biotechnology questions.
Question 91 · Environmental Issues
The Earth Summit held in Rio de Janeiro in 1992 was called
Ato reduce CO₂ emissions and global warming
Bfor conservation of biodiversity and sustainable utilization of its benefits
Cto assess threat posed to native species by invasive weed species
Dfor immediate steps to discontinue use of CFCs that were damaging the ozone layer
Answer: B. for conservation of biodiversity and sustainable utilization of its benefits
The 1992 Earth Summit (United Nations Conference on Environment and Development – UNCED) held in Rio de Janeiro was a landmark global event focused on integrating environment and development. As per NCERT Class 12 Biology (Chapter 15: Biodiversity and Conservation), the summit led to the adoption of Agenda 21, the Rio Declaration, and the Convention on Biological Diversity (CBD). The CBD explicitly aims at conservation of biological diversity, sustainable use of its components, and fair and equitable sharing of benefits arising from genetic resources — directly aligning with option B. While climate change (option A) and ozone depletion (option D) were discussed, they were addressed under separate frameworks: the UNFCCC (established at Rio) laid groundwork for later climate agreements but did not mandate immediate CO₂ reductions; the Montreal Protocol (1987) had already targeted CFCs before Rio. Option C refers to invasive species — an important ecological concern, but not the primary objective of the Earth Summit. Thus, the summit’s core mandate was biodiversity conservation and sustainability, making option B the correct and NCERT-aligned answer.
Question 92 · Human Health and Disease / Lactation and Passive Immunity
Colostrum, the yellowish fluid secreted by the mother during the initial days of lactation, is very essential to impart immunity to the newborn infants because it contains:
ANatural killer cells
BMonocytes
CMacrophages
DImmunoglobulin A
Answer: D. Immunoglobulin A
Colostrum is the first milk secreted by mammary glands in the first 2–3 days after childbirth. It is rich in antibodies, especially Immunoglobulin A (IgA), which is secreted in dimeric form and remains stable in the acidic environment of the infant’s gut. IgA provides passive immunity by coating the mucosal surfaces of the gastrointestinal tract, preventing pathogen attachment and colonization. Unlike IgG (which crosses placenta prenatally), IgA is not transferred across the placenta; hence, colostrum serves as the critical postnatal source of mucosal immunity. While colostrum also contains lymphocytes, macrophages, and other immune cells, their role is secondary and less specific compared to the high concentration and functional significance of secretory IgA. NCERT Class 12 Biology (Chapter 8: Human Health and Disease) explicitly states that colostrum contains abundant IgA antibodies essential for protecting newborns against infections. Natural killer cells, monocytes, and macrophages—though present—are not the primary immunological agents responsible for this passive mucosal protection.
Question 93 · Anatomy of Flowering Plants
Grass leaves curl inwards during very dry weather. Select the most appropriate reason from the following:
AClosure of stomata
BFlaccidity of bulliform cells
CShrinkage of air spaces in spongy mesophyll
DTyloses in vessels
Answer: B. Flaccidity of bulliform cells
Grass leaves curl inward under water stress due to loss of turgor in specialized large, thin-walled, empty epidermal cells called bulliform (or motor) cells, located along the upper surface of the leaf lamina. When water is abundant, these cells become turgid and keep the leaf flat; during drought, they lose water and become flaccid, causing the leaf to fold or curl inward — a reversible mechanism that reduces surface area and minimizes transpiration. This adaptation is characteristic of monocots like grasses and is explicitly described in NCERT Class 11, Chapter 6 'Anatomy of Flowering Plants'. Closure of stomata (Option A) reduces water loss but does not cause leaf curling. Shrinkage of air spaces in spongy mesophyll (C) affects gas exchange but lacks mechanical role in rolling. Tyloses (D) are outgrowths in xylem vessels that block conduction during stress or aging — unrelated to leaf movement. Hence, flaccidity of bulliform cells is the direct and most appropriate cause.
Question 94 · Chromosomal structure and terminology
The shorter and longer arms of a submetacentric chromosome are referred to as
As-arm and l-arm respectively
Bp-arm and q-arm respectively
Cq-arm and p-arm respectively
Dm-arm and n-arm respectively
Answer: B. p-arm and q-arm respectively
In human cytogenetics, chromosomes are classified based on centromere position. A submetacentric chromosome has its centromere slightly off-centre, producing two unequal arms: a shorter arm and a longer arm. By international convention (ISCN), the shorter arm is designated as the 'p-arm' (from French 'petit', meaning small) and the longer arm as the 'q-arm' (simply the next letter after 'p'). This nomenclature is universally adopted in NCERT Class 11 Biology (Chapter 5: Cell – The Unit of Life, and Chapter 6: Cell Cycle and Cell Division) and is essential for describing chromosomal abnormalities, banding patterns, and gene mapping. Options using 's-arm/l-arm' or 'm-arm/n-arm' are non-standard and not recognized in cytogenetic literature. The 'q-arm' is never the shorter arm — that distinction exclusively belongs to the 'p-arm'. Hence, the correct designation is p-arm (shorter) and q-arm (longer), making option B the only scientifically accurate choice.
Question 95 · Respiration in Plants
Respiratory Quotient (RQ) value of tripalmitin is
Answer: B. 0.7
Tripalmitin is a triglyceride composed of glycerol and three palmitic acid residues. During aerobic respiration, its complete oxidation follows the reaction: C₅₁H₉₈O₆ + 72.5 O₂ → 51 CO₂ + 49 H₂O. Respiratory Quotient (RQ) is defined as the ratio of CO₂ produced to O₂ consumed (RQ = V_CO₂ / V_O₂). From the balanced equation, moles of CO₂ = 51 and moles of O₂ = 72.5; thus RQ = 51/72.5 ≈ 0.703, which rounds to 0.7. This value is characteristic of fats — lower than 1 because fats require more oxygen per molecule for oxidation compared to carbohydrates (RQ = 1) or organic acids (RQ > 1). As per NCERT Class 11 Biology (Chapter 14: Respiration in Plants, page 226), the RQ of tripalmitin is explicitly cited as 0.7, confirming it as the standard reference value for fats. Hence, option (2) — corresponding to choice B — is correct.
Question 96 · Microbes in Human Welfare
Which of the following is a commercial blood cholesterol-lowering agent?
ACyclosporin A
BStatin
CStreptokinase
DLipases
Answer: B. Statin
Statins are commercially produced fungal metabolites (e.g., lovastatin from Aspergillus terreus) that competitively inhibit HMG-CoA reductase—the rate-limiting enzyme in cholesterol biosynthesis in the liver. By blocking this enzyme, statins reduce intracellular cholesterol synthesis, prompting hepatocytes to upregulate LDL receptors and clear more LDL-cholesterol from blood—thus effectively lowering serum cholesterol levels. Cyclosporin A is an immunosuppressant used in organ transplantation; streptokinase is a thrombolytic enzyme that dissolves blood clots by activating plasminogen to plasmin; lipases are digestive enzymes that hydrolyse dietary fats but do not lower blood cholesterol. NCERT Class 12, Chapter 10 'Microbes in Human Welfare', explicitly lists statins under 'antibiotics and other bioactive molecules' as cholesterol-lowering agents. Their clinical use is well-established and aligns with NEET’s emphasis on application-based understanding of microbial products.
Question 97 · Digestion and Absorption / Chemical Coordination and Integration
Match the following structures with their respective location in organs:
(a) Crypts of Lieberkühn
(b) Glisson’s capsule
(c) Islets of Langerhans
(d) Brunner’s glands
(i) Pancreas
(ii) Duodenum
(iii) Small intestine
(iv) Liver
A(iii) (i) (ii) (iv)
B(ii) (iv) (i) (iii)
C(iii) (iv) (i) (ii)
D(iii) (ii) (i) (iv)
Answer: C. (iii) (iv) (i) (ii)
Crypts of Lieberkühn are simple tubular glands located in the mucosa of the small intestine (including duodenum, jejunum, ileum), so (a) matches (iii). Glisson’s capsule is a fibrous connective tissue sheath surrounding the liver, enclosing hepatic arteries, portal vein branches, and bile ducts — thus (b) → (iv). Islets of Langerhans are endocrine cell clusters embedded in the pancreas, responsible for insulin and glucagon secretion; hence (c) → (i). Brunner’s glands are submucosal serous glands found exclusively in the duodenum (first part of small intestine), secreting alkaline mucus to neutralize gastric acid — so (d) → (ii). Therefore, the correct matching is (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii), corresponding to option (3), i.e., C. This aligns precisely with NCERT Class 11 Biology Chapter 16 (Digestion and Absorption) and Chapter 22 (Chemical Coordination), where each structure is explicitly described with its anatomical site and function.
Question 98 · Biodiversity and its Conservation
Which of the following is the most important cause for animals and plants being driven to extinction?
AHabitat loss and fragmentation
BDrought and floods
CEconomic exploitation
DAlien species invasion
Answer: A. Habitat loss and fragmentation
According to NCERT Class 12 Biology (Chapter 15: Biodiversity and Conservation), habitat loss and fragmentation is the single largest driver of species extinction worldwide. Human activities such as deforestation, urbanisation, agricultural expansion, and infrastructure development directly destroy or divide natural habitats, leading to reduced population sizes, disrupted gene flow, increased edge effects, and diminished resource availability. While droughts, floods, economic exploitation (e.g., overharvesting), and alien species invasion are significant threats, they collectively rank lower in scale and impact compared to habitat degradation. The NCERT textbook explicitly states that 'the most important cause driving animals and plants to extinction is habitat loss and fragmentation', citing examples like tropical rainforest destruction and fragmentation of wildlife corridors. This aligns with global assessments by IUCN and CBD, which identify habitat alteration as the primary threat to >80% of threatened species. Hence, option A is scientifically accurate and NCERT-validated.
Question 99 · Neural Control and Coordination
Which part of the brain is responsible for thermoregulation?
ACerebrum
BHypothalamus
CCorpus callosum
DMedulla oblongata
Answer: B. Hypothalamus
The hypothalamus, a small but vital region located at the base of the diencephalon, acts as the body’s primary thermoregulatory centre. It contains temperature-sensitive neurons that monitor blood temperature and receive input from peripheral thermoreceptors in the skin. When body temperature deviates from the set point (~37°C), the hypothalamus initiates autonomic and behavioural responses: it triggers sweating and cutaneous vasodilation to dissipate heat, or shivering, vasoconstriction, and piloerection to conserve/generate heat. This integrative function aligns precisely with NCERT Class 11 Biology (Chapter 21: Neural Control and Coordination), which explicitly states, 'The hypothalamus controls body temperature, urge for eating and drinking, and several other vital functions.' In contrast, the cerebrum governs higher cognitive functions; the corpus callosum connects cerebral hemispheres; and the medulla oblongata regulates involuntary functions like respiration and heartbeat—but not thermoregulation. Hence, option B (Hypothalamus) is scientifically accurate and NCERT-aligned.
Question 100 · Animal Kingdom
Consider the following features: (a) Organ-system level of organisation, (b) Bilateral symmetry, (c) True coelomates with segmentation of body. Select the correct option of animal groups which possess all the above characteristics.
AAnnelida, Arthropoda and Chordata
BAnnelida, Arthropoda and Mollusca
CArthropoda, Mollusca and Chordata
DAnnelida, Mollusca and Chordata
Answer: A. Annelida, Arthropoda and Chordata
All three features—organ-system level of organisation, bilateral symmetry, and being true coelomates with body segmentation—are shared by Annelida, Arthropoda, and Chordata. Annelids (e.g., earthworm) exhibit metamerism (true segmentation), a well-developed coelom, bilateral symmetry, and complex organ systems (e.g., closed circulatory, nervous, excretory). Arthropods (e.g., cockroach) possess jointed appendages, a true coelom (though reduced in adults), clear segmentation (tagmosis), bilateral symmetry, and advanced organ systems (open circulatory, tracheal/respiratory, compound eyes). Chordates (e.g., human) have a dorsal hollow nerve cord, notochord, pharyngeal slits, bilateral symmetry, a true coelom (body cavity lined by mesoderm), and—in many groups like vertebrates—segmentation evident in vertebrae, somites, and nervous system organisation. Mollusca, though bilaterally symmetrical and coelomate, lack true segmentation (except rudimentary in some like cephalopods); their body plan is unsegmented with distinct regions (head-foot-visceral mass), disqualifying them from satisfying criterion (c). Hence, only option A fulfils all three criteria as per NCERT Class 11 Chapter 4.
Question 101 · Digestive system of cockroach
Select the correct sequence of organs in the alimentary canal of cockroach starting from mouth:
APharynx → Oesophagus → Crop → Gizzard → Ileum → Colon → Rectum
BPharynx → Oesophagus → Gizzard → Crop → Ileum → Colon → Rectum
CPharynx → Oesophagus → Gizzard → Ileum → Crop → Colon → Rectum
DPharynx → Oesophagus → Ileum → Crop → Gizzard → Colon → Rectum
Answer: A. Pharynx → Oesophagus → Crop → Gizzard → Ileum → Colon → Rectum
The alimentary canal of cockroach is a complete tube divided into three main regions: foregut, midgut, and hindgut. Starting from the mouth, food passes through the pharynx and then the oesophagus — both part of the foregut. Next comes the crop, a thin-walled sac for temporary food storage, followed by the gizzard (proventriculus), which has chitinous teeth for mechanical grinding. The midgut begins after the gizzard and includes the gastric caeca and the ileum — the anterior part of the midgut proper. The hindgut comprises the colon and rectum, where water reabsorption and waste compaction occur. Thus, the correct sequential order is: Pharynx → Oesophagus → Crop → Gizzard → Ileum → Colon → Rectum. This sequence aligns precisely with NCERT Class 11 Biology (Chapter 7: Structural Organisation in Animals), which describes the cockroach digestive tract as having a distinct foregut (pharynx, oesophagus, crop, gizzard), midgut (ileum, gastric caeca), and hindgut (colon, rectum). Option A matches this anatomical progression; all other options misplace either the crop or gizzard or disrupt the midgut–hindgut boundary.
Question 102 · Environmental Issues: Greenhouse Effect and Global Warming
Which of the following pairs of gases is mainly responsible for the greenhouse effect?
AOzone and Ammonia
BOxygen and Nitrogen
CNitrogen and Sulphur dioxide
DCarbon dioxide and Methane
Answer: D. Carbon dioxide and Methane
The greenhouse effect is a natural process where certain atmospheric gases trap infrared radiation emitted by Earth’s surface, preventing excessive heat loss and maintaining habitable temperatures. However, human activities have intensified this effect by increasing concentrations of key greenhouse gases. According to NCERT Class 12 Biology (Chapter 16: Environmental Issues), carbon dioxide (CO₂) and methane (CH₄) are among the most significant contributors — CO₂ accounts for the largest share due to fossil fuel combustion and deforestation, while CH₄, though less abundant, has ~28 times the global warming potential of CO₂ over 100 years and originates from paddy fields, livestock digestion, and landfills. Ozone (O₃) in the stratosphere protects against UV radiation but is not a primary greenhouse gas; ammonia (NH₃) is not a direct greenhouse gas and mainly contributes to aerosol formation. Oxygen (O₂) and nitrogen (N₂) constitute ~99% of the atmosphere but are non-greenhouse gases as they lack dipole moments and do not absorb infrared radiation. Sulphur dioxide (SO₂) causes acid rain and forms cooling aerosols, exerting a net cooling effect. Thus, only carbon dioxide and methane form the correct pair responsible for the enhanced greenhouse effect.
Question 103 · Disorders of Muscular and Skeletal System
Which of the following muscular disorders is inherited?
ATetany
BMuscular dystrophy
CMyasthenia gravis
DBotulism
Answer: B. Muscular dystrophy
Muscular dystrophy is a group of inherited genetic disorders characterized by progressive weakness and degeneration of skeletal muscles due to mutations in genes encoding structural muscle proteins—most commonly dystrophin in Duchenne muscular dystrophy. It follows X-linked recessive inheritance, predominantly affecting males, and is confirmed by family history, elevated serum creatine kinase, and genetic testing. In contrast, tetany results from hypocalcemia (e.g., due to parathyroid dysfunction or vitamin D deficiency), not inheritance. Myasthenia gravis is an autoimmune disorder where antibodies attack acetylcholine receptors at the neuromuscular junction, leading to fatigable muscle weakness—it is not genetically inherited though rare familial cases exist, it is not classified as an inherited muscular disorder per NCERT. Botulism is a toxin-mediated foodborne illness caused by Clostridium botulinum neurotoxin, resulting in flaccid paralysis—entirely acquired and non-genetic. NCERT Class 11 (Chapter 20: Locomotion and Movement) explicitly lists muscular dystrophy under 'inherited muscular disorders', while clearly distinguishing acquired conditions like tetany, myasthenia gravis, and botulism.
Question 104 · Epithelial tissue and its modifications
The ciliated epithelial cells are required to move particles or mucus in a specific direction. In humans, these cells are mainly present in
ABile duct and Bronchioles
BFallopian tubes and Pancreatic duct
CEustachian tube and Salivary duct
DBronchioles and Fallopian tubes
Answer: D. Bronchioles and Fallopian tubes
Ciliated epithelium consists of columnar or cuboidal cells bearing hair-like cilia on their free surface. These cilia beat rhythmically to propel mucus, dust, and trapped particles away from sensitive regions. According to NCERT Class 11 Biology (Chapter 7: Structural Organisation in Animals), ciliated epithelium is found in the respiratory tract — especially bronchioles — where it helps clear mucus and debris toward the pharynx. It is also present in the fallopian tubes (oviducts), where ciliary movement aids in transporting the ovum from the ovary toward the uterus. The bile duct lacks cilia and is lined by simple columnar epithelium without ciliation. Pancreatic ducts have simple columnar epithelium but are not predominantly ciliated. Salivary ducts are lined by stratified epithelium and lack cilia; the Eustachian (auditory) tube does possess ciliated epithelium, but salivary ducts do not — making option (3) incorrect. Option (1) incorrectly includes bile duct; option (2) wrongly includes pancreatic duct. Only option (4) — bronchioles and fallopian tubes — matches NCERT’s authoritative listing of major sites of ciliated epithelium in humans.
Question 105 · Cardiac Cycle and ECG
Match Column-I with Column-II:
Column-I
(a) P-wave
(b) QRS complex
(c) T-wave
(d) Reduction in the size of T-wave
Column-II
(i) Depolarisation of ventricles
(ii) Repolarisation of ventricles
(iii) Coronary ischemia
(iv) Depolarisation of atria
(v) Repolarisation of atria
Select the correct option.
A(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
B(a)-(iv), (b)-(i), (c)-(ii), (d)-(v)
C(a)-(ii), (b)-(i), (c)-(v), (d)-(iii)
D(a)-(ii), (b)-(iii), (c)-(v), (d)-(iv)
Answer: A. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
The electrocardiogram (ECG) records electrical activity of the heart. The P-wave represents depolarisation of the atria, initiating atrial contraction — matching (a) with (iv). The QRS complex reflects rapid depolarisation of the ventricles, triggering ventricular systole — so (b) pairs with (i). The T-wave corresponds to repolarisation of the ventricles, essential for ventricular relaxation — thus (c) matches (ii). A reduced T-wave amplitude is a classic ECG indicator of coronary ischemia, often due to inadequate blood supply to myocardial tissue — hence (d) correctly links with (iii). Option A aligns all these physiological correlations precisely. NCERT Class 11 (Chapter 18: Body Fluids and Circulation) explicitly states: 'P-wave — atrial depolarisation; QRS — ventricular depolarisation; T-wave — ventricular repolarisation', and notes that flattened or inverted T-waves suggest myocardial ischemia. Repolarisation of atria is not visible on standard ECG due to its overlap with the QRS complex, eliminating (v) as a valid match for any primary wave.
Question 106 · Biodiversity and Conservation
Which one of the following is not a method of in situ conservation of biodiversity?
ABiosphere Reserve
BWildlife Sanctuary
CBotanical Garden
DSacred Grove
Answer: C. Botanical Garden
In situ conservation refers to the protection of species within their natural habitats. Biosphere Reserves, Wildlife Sanctuaries, and Sacred Groves all conserve biodiversity in their native ecosystems — Biosphere Reserves integrate conservation with sustainable use across core, buffer, and transition zones; Wildlife Sanctuaries provide legal protection to wildlife in natural surroundings; Sacred Groves are community-protected forest patches preserving endemic flora and fauna. In contrast, Botanical Gardens represent ex situ conservation — they maintain living collections of plants outside their natural habitats, often for research, propagation, or education. NCERT Class 12 Biology (Chapter 15: Biodiversity and Conservation) explicitly distinguishes ex situ methods (e.g., botanical gardens, zoological parks, seed banks) from in situ ones (e.g., national parks, sanctuaries, biosphere reserves, sacred groves). Thus, Botanical Garden is the correct answer as it does not qualify as an in situ method.
Question 107 · Evolution and Natural Selection
In a species, the weight of newborns ranges from 2 to 5 kg. 97% of newborns with an average weight between 3 to 3.3 kg survive, whereas 99% of infants born with weight from 2 to 2.5 kg or 4.5 to 5 kg die. Which type of natural selection is operating?
ADirectional Selection
BStabilizing Selection
CDisruptive Selection
DCyclical Selection
Answer: B. Stabilizing Selection
This scenario exemplifies stabilizing selection — a mode of natural selection that favours intermediate phenotypes and selects against extreme variants. Here, newborns weighing 3–3.3 kg (near the population mean) show highest survival (97%), while those at either extreme — very low (2–2.5 kg) or very high (4.5–5 kg) birth weights — suffer near-total mortality (99% die). This reduces phenotypic variation over generations and maintains the optimal trait value, consistent with NCERT Class 12 Chapter 7 'Evolution', which states that stabilizing selection operates in stable environments where intermediate forms are best adapted. Directional selection would shift the mean toward one extreme; disruptive selection would favour both extremes simultaneously — neither fits the data. 'Cyclical selection' is not a standard term in NCERT or evolutionary biology and is therefore incorrect. Thus, option B (Stabilizing Selection) is scientifically accurate and fully aligned with NCERT’s treatment of selection types.
Question 108 · Cell Cycle and Cell Division
The correct sequence of phases of the cell cycle is:
AM → G₂ → G₁ → S
BG₁ → G₂ → S → M
CS → G₁ → G₂ → M
DG₁ → S → G₂ → M
Answer: D. G₁ → S → G₂ → M
According to NCERT Class 11 Biology (Chapter 10: Cell Cycle and Cell Division), the cell cycle consists of four sequential phases: G₁ (first gap phase, cell growth), S (synthesis phase, DNA replication), G₂ (second gap phase, preparation for division), and M (mitotic phase, nuclear and cytoplasmic division). The G₁ phase follows mitosis and precedes DNA synthesis; during S phase, each chromosome replicates to form two sister chromatids; G₂ allows final checks and protein synthesis before mitosis; and M phase includes karyokinesis and cytokinesis. Thus, the correct order is G₁ → S → G₂ → M. Option D matches this sequence exactly. Options A, B, and C misplace either G₁/G₂ order or insert S or M at incorrect positions — violating the fundamental, irreversible progression defined in the textbook. This sequence ensures genomic fidelity and is conserved across eukaryotes, as emphasized in NCERT’s description of interphase (G₁, S, G₂) followed by M phase.
Question 109 · Chemical Coordination and Integration
How does a steroid hormone influence cellular activities?
AChanging the permeability of the cell membrane
BBinding to DNA and forming a gene-hormone complex
CActivating cyclic AMP located on the cell membrane
DUsing aquaporin channels as second messenger
Answer: B. Binding to DNA and forming a gene-hormone complex
Steroid hormones are lipid-soluble and diffuse freely across the plasma membrane. Inside the target cell, they bind to specific intracellular receptors—typically in the cytoplasm or nucleus—forming a hormone-receptor complex. This complex undergoes conformational change, translocates to the nucleus (if not already there), and binds to specific regulatory sequences on DNA called hormone response elements (HREs). This binding modulates transcription of target genes—either enhancing or suppressing mRNA synthesis—thereby altering protein synthesis and long-term cellular responses. This genomic mechanism distinguishes steroid hormones from water-soluble hormones (e.g., epinephrine, glucagon) that act via membrane receptors and second messengers like cAMP. Options A, C, and D are incorrect: steroids do not alter general membrane permeability; cAMP is activated by G-protein-coupled receptors—not steroids—and aquaporins are water channels unrelated to steroid signaling. NCERT Class 12, Chapter 22 (Chemical Coordination and Integration), clearly states that steroid hormones 'interact with intracellular receptors and regulate gene expression'.
Question 110 · Cell organelles: Lysosomes and endomembrane system
Which of the following statements is not correct?
ALysosomes have numerous hydrolytic enzymes
BThe hydrolytic enzymes of lysosomes are active under acidic pH
CLysosomes are membrane-bound structures
DLysosomes are formed by the process of packaging in the endoplasmic reticulum
Answer: D. Lysosomes are formed by the process of packaging in the endoplasmic reticulum
Lysosomes are single-membrane-bound organelles containing over 60 types of acid hydrolases (e.g., proteases, nucleases, lipases) that function optimally at acidic pH (~4.5–5.0), maintained by proton pumps in their membrane. They originate from the Golgi apparatus—not the endoplasmic reticulum—through budding and packaging of enzymatically active proteins tagged with mannose-6-phosphate. While the rough ER synthesizes lysosomal enzymes as inactive precursors (proenzymes), their sorting, modification (e.g., phosphorylation of mannose residues), and final packaging into nascent lysosomes occur exclusively in the cis and trans regions of the Golgi complex. Hence, stating that lysosomes are formed by packaging in the endoplasmic reticulum is incorrect. NCERT Class 11 Biology (Chapter 8: Cell: The Unit of Life) explicitly states: 'Lysosomes are produced by the Golgi apparatus' and clarifies that ER is involved only in synthesis of proteins, not their final organelle assembly. This makes option D the only false statement.
Question 112 · Biomolecules: Proteins and their functions
Concanavalin A is
Aan alkaloid
Ban essential oil
Ca lectin
Da pigment
Answer: C. a lectin
Concanavalin A (Con A) is a plant-derived glycoprotein isolated from the jack bean (Canavalia ensiformis). It belongs to the class of proteins known as lectins — carbohydrate-binding proteins that are highly specific for sugar moieties and do not modify the carbohydrates they bind. Lectins play roles in cell–cell recognition, immune responses, and mitogenic stimulation of lymphocytes. Con A is widely used in immunology and cell biology as a mitogen to activate T-lymphocytes. It is not an alkaloid (which are nitrogen-containing basic secondary metabolites like nicotine or morphine), nor an essential oil (volatile aromatic compounds such as eucalyptol), nor a pigment (e.g., chlorophyll or carotenoids). NCERT Class 11 Biology (Chapter 9: Biomolecules) explicitly lists lectins under 'proteins with specific functions' and mentions Con A as a classic example. Its structure includes metal ions (Ca²⁺ and Mn²⁺) essential for carbohydrate binding, reinforcing its classification as a functional protein — not a secondary metabolite.
Question 113 · Microbes in Industrial Production
Which one of the following equipments is essentially required for growing microbes on a large scale, for industrial production of enzymes?
ABOD incubator
BSludge digester
CIndustrial oven
DBioreactor
Answer: D. Bioreactor
For large-scale microbial cultivation aimed at industrial enzyme production, a bioreactor is indispensable. As per NCERT Class 12 Biology (Chapter 11: Biotechnology – Principles and Processes), a bioreactor is a sterile, controlled-environment vessel that provides optimal conditions—such as temperature, pH, oxygen supply, agitation, and nutrient feed—for microbial growth and metabolite synthesis. Unlike a BOD incubator (used only for measuring biochemical oxygen demand in water samples), a sludge digester (employed in wastewater treatment to decompose organic matter anaerobically), or an industrial oven (designed for drying or heating, not living cultures), a bioreactor enables precise regulation of fermentation parameters essential for high-yield enzyme production. Enzymes like amylase, protease, and lipase are commercially produced using bacteria or fungi grown in stirred-tank bioreactors under aerobic or anaerobic conditions. The scalability, sterility, and process control offered by bioreactors make them the only equipment among the options suitable for industrial microbial fermentation—hence, option D is correct.
Question 114 · Enzymes
Consider the following statements:
(A) A coenzyme or metal ion that is tightly bound to the enzyme protein is called a prosthetic group.
(B) A complete catalytically active enzyme with its bound prosthetic group is called an apoenzyme.
Select the correct option.
ABoth (A) and (B) are true.
B(A) is true but (B) is false.
CBoth (A) and (B) are false.
D(A) is false but (B) is true.
Answer: B. (A) is true but (B) is false.
According to NCERT Class 11 Biology (Chapter 9: Biomolecules), a prosthetic group is a non-protein, tightly bound cofactor—either an organic coenzyme (e.g., haem in peroxidase) or an inorganic metal ion (e.g., Zn²⁺ in carbonic anhydrase)—essential for enzyme activity. Statement (A) is therefore correct. However, statement (B) misnames the holoenzyme: the protein part alone, without its prosthetic group or cofactor, is called the apoenzyme; only when the apoenzyme binds its cofactor (prosthetic group, coenzyme, or metal ion) does it form the fully functional holoenzyme. Hence, (B) is false—it incorrectly labels the complete active enzyme as 'apoenzyme' instead of 'holoenzyme'. This distinction is explicitly clarified in NCERT on page 156 (2023–24 edition): 'The protein part is called apoenzyme and the non-protein part is called prosthetic group... The complete, catalytically active complex is called holoenzyme.' Thus, only statement (A) is true, making option (B) the correct choice.
Question 115 · Biomolecules: Nucleic Acids
Purines found both in DNA and RNA are
AAdenine and thymine
BAdenine and guanine
CGuanine and cytosine
DCytosine and thymine
Answer: B. Adenine and guanine
Purines are nitrogenous bases with a double-ring structure. The two purines in nucleic acids are adenine (A) and guanine (G). Both are present in DNA and RNA. In contrast, pyrimidines — cytosine (C), thymine (T), and uracil (U) — have a single-ring structure. Thymine is exclusive to DNA, while uracil replaces it in RNA; cytosine is common to both DNA and RNA but is a pyrimidine, not a purine. Therefore, among the given options, only adenine and guanine are purines shared by both DNA and RNA. Option A is incorrect because thymine is a pyrimidine and absent in RNA. Option C pairs guanine (purine) with cytosine (pyrimidine), violating the 'purines only' condition. Option D lists two pyrimidines. This distinction is clearly covered in NCERT Class 11, Chapter 9 'Biomolecules', which states: 'Adenine and guanine are purines whereas cytosine, uracil and thymine are pyrimidines' and emphasizes that 'DNA contains A, G, C, T while RNA contains A, G, C, U'. Hence, the correct pair of purines common to both is adenine and guanine.
Question 116 · Human Reproduction
Select the correct sequence for transport of sperm cells in the male reproductive system.
ATestis → Epididymis → Vasa efferentia → Rete testis → Inguinal canal → Urethra
BSeminiferous tubules → Rete testis → Vasa efferentia → Epididymis → Vas deferens → Ejaculatory duct → Urethra → Urethral meatus
CSeminiferous tubules → Vasa efferentia → Epididymis → Inguinal canal → Urethra
DTestis → Epididymis → Vasa efferentia → Vas deferens → Ejaculatory duct → Inguinal canal → Urethra → Urethral meatus
Answer: B. Seminiferous tubules → Rete testis → Vasa efferentia → Epididymis → Vas deferens → Ejaculatory duct → Urethra → Urethral meatus
Sperm production begins in the seminiferous tubules of the testis. From there, immature sperm move into the rete testis—a network of channels within the mediastinum testis. They then pass through the vasa efferentia (10–12 fine tubules) to reach the epididymis, where they mature and are stored. During ejaculation, sperm travel from the epididymis through the vas deferens (ductus deferens), which ascends via the inguinal canal into the pelvic cavity. The vas deferens joins the duct of the seminal vesicle to form the ejaculatory duct, which empties into the prostatic urethra. Sperm then traverse the entire urethra (prostatic → membranous → spongy) and exit via the external urethral orifice (urethral meatus). Option B correctly lists this anatomically and functionally accurate sequence. Options A, C, and D contain errors: A reverses rete testis and vasa efferentia; C skips rete testis and misplaces inguinal canal (a passage, not a storage/transport organ); D incorrectly places epididymis before vasa efferentia and treats inguinal canal as a conduit *after* ejaculatory duct—whereas vas deferens passes *through* it *before* joining the seminal vesicle.
Question 117 · Human Evolution
Match the hominids with their correct average brain size:
(a) Homo habilis
(b) Homo neanderthalensis
(c) Homo erectus
(d) Homo sapiens
(i) 900 cc
(ii) 1350 cc
(iii) 650–800 cc
(iv) 1400 cc
A(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
B(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
C(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
D(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
Answer: C. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
According to NCERT Class 12 Biology (Chapter 7: Evolution), brain size is a key criterion in hominid evolution. Homo habilis, the earliest tool-maker (~2.4–1.4 mya), had a cranial capacity of 650–800 cc — matching (iii). Homo erectus (~1.7 mya–100 kya), with improved bipedalism and Acheulean tools, averaged ~900 cc — matching (i). Homo neanderthalensis (~400–40 kya), adapted to cold climates, had a robust skull and brain size of ~1350–1400 cc; NCERT specifically cites 1350 cc as representative — matching (ii). Homo sapiens (~300 kya–present) shows an average cranial capacity of ~1350–1400 cc, but NCERT and standard references consistently list 1400 cc for modern humans — matching (iv). Thus, the correct pairing is: (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv) — which corresponds to option (3), i.e., choice C. Note: While Neanderthal and modern human brain sizes overlap, NCERT distinguishes them by citing 1350 cc for Neanderthals and 1400 cc for Homo sapiens — making option C unambiguously correct per syllabus.
Question 118 · Evolution: Mechanisms of Evolution
Variations caused by mutation, as proposed by Hugo de Vries, are
Arandom and directional
Brandom and directionless
Csmall and directional
Dsmall and directionless
Answer: B. random and directionless
Hugo de Vries, based on his experiments with evening primrose (Oenothera lamarckiana), proposed the Mutation Theory of evolution. He observed sudden, large, inheritable changes — which he termed 'mutations' — that could produce new species in a single generation. Unlike Darwin’s gradual variations, de Vries’ mutations were abrupt, random in occurrence (i.e., not induced by environmental needs), and directionless (i.e., not oriented toward adaptation or improvement). This contrasts with natural selection, which acts directionally on variation. NCERT Class 12 Biology (Chapter 7: Evolution) explicitly states: 'Mutations are random and directionless, while natural selection is directional.' Thus, option (2) — 'random and directionless' — correctly reflects de Vries’ view. Options suggesting 'directional' mutations misrepresent his theory, and 'small' variations refer to Darwinian continuous variation, not de Vries’ saltatory (jumping) mutations.
Question 119 · Cell: The Unit of Life
Which of the following pairs of organelles does not contain DNA?
AMitochondria and Lysosomes
BChloroplast and Vacuoles
CLysosomes and Vacuoles
DNuclear envelope and Mitochondria
Answer: C. Lysosomes and Vacuoles
DNA is present in semi-autonomous organelles that have their own genetic material and protein-synthesizing machinery. Mitochondria possess circular DNA and ribosomes, enabling limited self-replication and synthesis of some respiratory chain proteins. Chloroplasts also contain circular DNA and ribosomes, essential for photosynthesis-related protein synthesis. The nuclear envelope is a double-membrane structure continuous with the endoplasmic reticulum; though it surrounds the nucleus (which houses genomic DNA), the envelope itself contains no DNA — however, this option pairs it with mitochondria (which *does* contain DNA), making the pair invalid for 'does not contain DNA'. Lysosomes are membrane-bound vesicles containing hydrolytic enzymes; they lack DNA, RNA, and ribosomes. Vacuoles (especially in plant cells) are storage compartments filled with cell sap; they contain no genetic material. Thus, the pair 'Lysosomes and Vacuoles' contains *no DNA* in either organelle — making option (3) correct. This aligns with NCERT Class 11 Chapter 8 (Cell: The Unit of Life), which explicitly states that only mitochondria, chloroplasts, and nuclei contain DNA, while lysosomes, vacuoles, Golgi, ER, and ribosomes do not.
Question 120 · Respiratory Disorders
Due to increasing air-borne allergens and pollutants, many people in urban areas are suffering from a respiratory disorder causing wheezing due to
Abenign growth on mucous lining of nasal cavity
Binflammation of bronchi and bronchioles
Cproliferation of fibrous tissues and damage of the alveolar walls
Dreduction in the secretion of surfactants by pneumocytes
Answer: B. inflammation of bronchi and bronchioles
Wheezing is a high-pitched whistling sound during breathing, primarily caused by narrowed or obstructed airways. In urban populations exposed to airborne allergens (e.g., pollen, dust mites) and pollutants (e.g., PM2.5, NO₂), allergic inflammation triggers bronchoconstriction and oedema of the bronchial and bronchiolar mucosa — hallmark features of asthma. NCERT Class 11 (Chapter 17: Breathing and Exchange of Gases) explicitly states that asthma involves 'inflammation of bronchi and bronchioles', leading to spasm of smooth muscles, increased mucus secretion, and airway hyperresponsiveness — all culminating in wheezing. Option A describes nasal polyps (not typically wheeze-causing); Option C refers to emphysema (characterised by alveolar wall destruction and dyspnoea, not prominent wheezing); Option D relates to neonatal respiratory distress syndrome (due to surfactant deficiency), which presents with grunting and tachypnoea, not classic wheezing. Thus, only option B aligns with both pathophysiology and NCERT’s description.
Question 121 · Sex Determination
Select the incorrect statement.
AMale fruit fly is heterogametic.
BIn male grasshoppers, 50% of sperms have no sex chromosome.
CIn domesticated fowls, sex of progeny depends on the type of sperm rather than egg.
DHuman males have one of their sex chromosomes much shorter than the other.
Answer: C. In domesticated fowls, sex of progeny depends on the type of sperm rather than egg.
In domesticated fowls (chickens), sex determination follows the ZW system: females are heterogametic (ZW) and males are homogametic (ZZ). Therefore, the sex of the progeny depends on the type of *egg* (Z or W), not the sperm — all sperm carry Z. Hence, statement (3) is incorrect. In contrast, male fruit flies (Drosophila) are heterogametic (XY), making (1) correct. Grasshoppers use XO system: males have one X chromosome and no Y; meiosis produces 50% sperm with X and 50% with no sex chromosome — so (2) is correct. Human males are XY, and the Y chromosome is significantly smaller than the X — confirming (4) is correct. This aligns with NCERT Class 12 Chapter 5 'Principles of Inheritance and Variation', which explicitly states that in birds (including fowls), the female is the heterogametic sex and determines offspring sex via her gamete. Thus, option C is the only incorrect statement.
Question 122 · Biotechnology: Principles and Processes
DNA precipitation out of a mixture of biomolecules can be achieved by treatment with
AIsopropanol
BChilled ethanol
CMethanol at room temperature
DChilled chloroform
Answer: B. Chilled ethanol
DNA precipitation is a standard step in DNA isolation protocols to separate DNA from other cellular components like proteins, RNA, and carbohydrates. Ethanol—especially when chilled (−20°C)—reduces the dielectric constant of the solution, neutralizing the phosphate backbone’s negative charge and decreasing DNA solubility. This causes DNA to aggregate and precipitate as a visible white fibrous mass. Isopropanol can also precipitate DNA but requires lower volumes and often co-precipitates more salts; however, chilled ethanol is preferred in routine lab protocols due to higher purity and better recovery. Methanol is not used for DNA precipitation—it lacks sufficient dehydrating power and may denature DNA unpredictably. Chloroform is employed in phenol-chloroform extraction to remove proteins and lipids, but it does not precipitate DNA; instead, it partitions contaminants into the organic phase while DNA remains in the aqueous phase. NCERT Class 12 Biology (Chapter 11: Biotechnology – Principles and Processes) explicitly states that 'DNA is precipitated out by adding chilled ethanol'—making option B the scientifically accurate and syllabus-aligned choice.
Question 123 · Microbes in Human Welfare
Select the correct group of biocontrol agents.
ABacillus thuringiensis, Tobacco mosaic virus, Aphids
BTrichoderma, Baculovirus, Bacillus thuringiensis
COscillatoria, Rhizobium, Trichoderma
DNostoc, Azospirillum, Nucleopolyhedrovirus
Answer: B. Trichoderma, Baculovirus, Bacillus thuringiensis
Biocontrol agents are living organisms used to manage pests and diseases in an eco-friendly manner. Bacillus thuringiensis (Bt) is a Gram-positive bacterium that produces insecticidal Cry proteins effective against lepidopteran larvae. Trichoderma species are free-living fungi that act as biofungicides by parasitizing plant-pathogenic fungi like Rhizoctonia and Fusarium. Baculoviruses (e.g., nucleopolyhedroviruses) are species-specific insect viruses used against caterpillars and other arthropod pests without harming beneficial insects or mammals. In contrast, Tobacco mosaic virus is a plant pathogen—not a biocontrol agent—while aphids are pests, not controllers. Oscillatoria and Nostoc are nitrogen-fixing cyanobacteria but not primarily used for pest control; Rhizobium and Azospirillum are symbiotic/associative nitrogen fixers, not biocontrol agents. Though nucleopolyhedrovirus is a type of baculovirus, option D incorrectly pairs it with non-biocontrol nitrogen fixers and omits key agents. Option B correctly groups three widely recognized, NCERT-mentioned biocontrol agents: Trichoderma (fungus), Baculovirus (virus), and Bacillus thuringiensis (bacterium).
Question 124 · Strategies for Enhancement of Food Production
Select the incorrect statement.
AInbreeding increases homozygosity.
BInbreeding is essential to evolve purelines in any animal.
CInbreeding selects harmful recessive genes that reduce fertility and productivity.
DInbreeding helps in accumulation of superior genes and elimination of undesirable genes.
Answer: C. Inbreeding selects harmful recessive genes that reduce fertility and productivity.
The question asks for the *incorrect* statement about inbreeding. Option (C) states: 'Inbreeding selects harmful recessive genes that reduce fertility and productivity.' This is misleading — inbreeding *exposes* harmful recessive alleles to selection by increasing homozygosity, but it does not *select for* them; rather, natural or artificial selection acts *against* them once expressed, leading to inbreeding depression. NCERT Class 12 (Chapter 9, 'Strategies for Enhancement of Food Production') clearly defines inbreeding depression as the reduction in fertility, vigor, and productivity due to increased homozygosity of deleterious recessive alleles — not because inbreeding actively 'selects' them. Option (A) is correct: inbreeding increases homozygosity. Option (B) is correct: inbreeding is essential for developing purelines (homozygous, true-breeding lines) in animals. Option (D) is conditionally correct: sustained inbreeding followed by selection can help fix superior alleles and eliminate undesirable ones — though it requires careful management to avoid depression. Thus, only (C) misrepresents the mechanism and is scientifically inaccurate.
Question 125 · Microbes in Human Welfare
Match the following organisms with the products they produce:
(a) Lactobacillus (i) Cheese
(b) Saccharomyces cerevisiae (ii) Curd
(c) Aspergillus niger (iii) Citric acid
(d) Acetobacter aceti (iv) Bread
(v) Acetic acid
Select the correct option.
A(a)-(ii), (b)-(iv), (d)-(v), (c)-(iii)
B(a)-(ii), (b)-(iv), (c)-(iii), (d)-(v)
C(a)-(iii), (b)-(iv), (c)-(v), (d)-(i)
D(a)-(ii), (b)-(i), (c)-(iii), (d)-(v)
Answer: B. (a)-(ii), (b)-(iv), (c)-(iii), (d)-(v)
Lactobacillus converts lactose in milk into lactic acid, causing coagulation and formation of curd — matching (a) with (ii). Saccharomyces cerevisiae (baker’s yeast) ferments sugars to produce CO₂ and ethanol; CO₂ makes dough rise, essential for bread-making — so (b) pairs with (iv). Aspergillus niger is a fungus widely used in industrial citric acid production via submerged fermentation — thus (c) matches (iii). Acetobacter aceti oxidizes ethanol to acetic acid under aerobic conditions, used in vinegar synthesis — hence (d) corresponds to (v). Option B correctly maps all four: (a)-(ii), (b)-(iv), (c)-(iii), (d)-(v). Note that cheese (i) involves complex microbial succession (e.g., Lactococcus, Penicillium), not primarily Lactobacillus; and while some Lactobacillus strains contribute to cheese ripening, NCERT class 12 explicitly attributes curd formation to Lactobacillus and bread to Saccharomyces cerevisiae. Citric acid and acetic acid are unambiguously linked to Aspergillus niger and Acetobacter aceti respectively per NCERT Table 10.1 (Ch. 10, Microbes in Human Welfare).
Question 127 · Asexual and Sexual Reproduction in Plants
In some plants, the female gamete develops into an embryo without fertilization. This phenomenon is known as
AAutogamy
BParthenocarpy
CSyngamy
DParthenogenesis
Answer: D. Parthenogenesis
Parthenogenesis is the development of an unfertilized egg (female gamete) into a new individual. In plants, this rare phenomenon occurs when the haploid egg cell undergoes mitotic divisions without syngamy to form a haploid embryo — observed in species like Nicotiana tabacum under experimental conditions. Autogamy refers to self-pollination within the same flower; parthenocarpy is fruit development without fertilization (yielding seedless fruits, e.g., banana), but no embryo forms; syngamy is the fusion of male and female gametes — synonymous with fertilization. NCERT Class 12 Biology (Chapter 1: Reproduction in Organisms) explicitly distinguishes parthenogenesis (embryo from unfertilized egg) from parthenocarpy (fruit without seeds). Though more common in animals (e.g., honeybee drones), its occurrence in certain angiosperms confirms it as the correct term for embryo formation sans fertilization. Hence, option D is scientifically precise and NCERT-aligned.
Question 128 · Seed structure and development
Persistent nucellus in the seed is known as
AChalaza
BPerisperm
CHilum
DTegmen
Answer: B. Perisperm
In angiosperm seeds, the nucellus is the diploid tissue surrounding the embryo sac in the ovule. Normally, it degenerates during seed development. However, in some plants like black pepper (Piper nigrum), beetroot (Beta vulgaris), and sugar beet, the nucellus persists and becomes a nutritive tissue in the mature seed — this persistent nucellus is called perisperm. It is distinct from endosperm (triploid, formed by double fertilization) and serves a similar reserve function but of maternal (sporophytic) origin. Chalaza is the region where integuments attach to the nucellus; hilum is the scar marking the point of seed detachment from the funiculus; tegmen is the inner integument that develops into the inner seed coat. NCERT Class 12, Chapter 2 'Sexual Reproduction in Flowering Plants', clearly states: 'In some seeds such as beet and black pepper, a portion of the nucellus persists and is called perisperm.' Thus, option B (Perisperm) is scientifically accurate and fully aligned with NCERT.
Question 129 · Principles of Inheritance and Variation
What map unit (centimorgan) is adopted in the construction of genetic maps?
AA unit of distance between two expressed genes representing 10% crossover.
BA unit of distance between two expressed genes representing 100% crossover.
CA unit of distance between genes on chromosomes, representing 1% crossover.
DA unit of distance between genes on chromosomes, representing 50% crossover.
Answer: C. A unit of distance between genes on chromosomes, representing 1% crossover.
A centimorgan (cM) is the standard unit used in genetic mapping to measure the distance between genes on a chromosome. According to NCERT Class 12 Chapter 5, one centimorgan corresponds to a 1% recombination frequency — meaning that two genes located 1 cM apart show crossing over in 1 out of every 100 meiotic events. This unit reflects the likelihood of recombination, not physical distance, and is derived from observed crossover frequencies in linkage studies. Option C correctly defines it as 'a unit of distance between genes on chromosomes, representing 1% crossover'. Options A and B incorrectly cite 10% and 100%, which misrepresent recombination frequency; 100% crossover would imply independent assortment (like unlinked genes), not a measurable map distance. Option D’s 50% crossover is equivalent to no linkage — genes behave as if unlinked, yielding a map distance of ∞ or undefined, not 50 cM. The term 'expressed genes' in options A and B is also biologically imprecise; map distance applies to any loci (including non-coding or silent alleles), not only expressed ones — reinforcing why C is both technically and pedagogically accurate.
Question 130 · Body Fluids and Circulation
What would be the heart rate of a person if the cardiac output is 5 L/min, blood volume in the ventricles at the end of diastole is 100 mL, and at the end of ventricular systole is 50 mL?
A50 beats per minute
B75 beats per minute
C100 beats per minute
D125 beats per minute
Answer: C. 100 beats per minute
Cardiac output (CO) is the volume of blood pumped by each ventricle per minute, calculated as CO = Heart Rate (HR) × Stroke Volume (SV). Stroke Volume is the difference between end-diastolic volume (EDV) and end-systolic volume (ESV): SV = EDV – ESV = 100 mL – 50 mL = 50 mL = 0.05 L. Given CO = 5 L/min, rearranging the formula: HR = CO / SV = 5 L/min ÷ 0.05 L/beat = 100 beats per minute. This aligns with NCERT Class 11 Biology (Chapter 18: Body Fluids and Circulation), which defines stroke volume as the amount of blood ejected by the left ventricle in one contraction and emphasizes that cardiac output depends on both heart rate and stroke volume. The question tests conceptual application—not memorization—and reflects standard NEET-level numerical reasoning grounded in NCERT fundamentals.
Question 131 · Microbial role in nitrogen cycle
Thiobacillus is a group of bacteria helpful in carrying out
ANitrogen fixation
BChemoautotrophic fixation
CNitrification
DDenitrification
Answer: D. Denitrification
Thiobacillus is a genus of aerobic, Gram-negative, chemoautotrophic bacteria that oxidize reduced sulfur compounds (e.g., elemental sulfur, thiosulfate) to obtain energy and fix CO₂. Crucially, certain species like Thiobacillus denitrificans perform heterotrophic denitrification under anaerobic conditions—reducing nitrate (NO₃⁻) to nitrogen gas (N₂) while using organic or inorganic electron donors. Though primarily known for sulfur oxidation, their documented role in denitrification is well-established in NCERT Class 12 Biology (Chapter 13: Organisms and Populations; also referenced in microbial ecology contexts of Chapter 9: Strategies for Enhancement in Food Production). Nitrogen fixation is carried out by Rhizobium, Azotobacter, and cyanobacteria—not Thiobacillus. Chemoautotrophic fixation is not a standard biological term; autotrophy refers to carbon fixation, not nitrogen. Nitrification involves conversion of NH₃ → NO₂⁻ → NO₃⁻ by Nitrosomonas and Nitrobacter—Thiobacillus does not perform this. Denitrification (NO₃⁻ → NO₂⁻ → NO → N₂O → N₂) is indeed mediated by facultative anaerobes including Thiobacillus denitrificans, making option (4) correct.
Question 132 · Excretory Products and their Elimination
Which of the following factors is responsible for the formation of concentrated urine?
ALow levels of antidiuretic hormone
BMaintaining hyperosmolarity towards the inner medullary interstitium in the kidneys
CSecretion of erythropoietin by juxtaglomerular apparatus
DHydrostatic pressure during glomerular filtration
Answer: B. Maintaining hyperosmolarity towards the inner medullary interstitium in the kidneys
Concentrated urine formation depends on the kidney’s ability to establish and maintain a hyperosmotic medullary interstitium — especially in the inner medulla — which serves as the osmotic gradient driving water reabsorption from the collecting duct. This gradient is built and sustained by the countercurrent multiplier system (loop of Henle) and the countercurrent exchanger (vasa recta), along with urea recycling. ADH enhances water permeability of the collecting duct, allowing water to move out into this hyperosmotic interstitium — thus concentrating urine. Low ADH (Option A) causes dilute urine; erythropoietin (Option C) regulates RBC production and is secreted by peritubular interstitial cells, not the juxtaglomerular apparatus (which secretes renin); hydrostatic pressure in glomerular filtration (Option D) determines filtration rate but does not concentrate urine. Hence, maintaining hyperosmolarity in the inner medullary interstitium is the fundamental prerequisite for urinary concentration — directly aligning with NCERT Class 11, Chapter 19.
Question 133 · Cell: The Unit of Life
Which of the following statements regarding mitochondria is incorrect?
AOuter membrane is permeable to monomers of carbohydrates, fats and proteins.
BEnzymes of electron transport are embedded in outer membrane.
CInner membrane is convoluted with infoldings.
DMitochondrial matrix contains single circular DNA molecule and ribosomes.
Answer: B. Enzymes of electron transport are embedded in outer membrane.
The outer mitochondrial membrane is porous due to porin proteins, allowing passive diffusion of small molecules (<5 kDa), including monomers like glucose, amino acids, and fatty acids — so statement (1) is correct. However, enzymes of the electron transport chain (ETC) — such as NADH dehydrogenase (Complex I), cytochrome bc₁ (Complex III), and cytochrome c oxidase (Complex IV) — are embedded exclusively in the inner mitochondrial membrane, not the outer membrane. Hence, statement (2) is incorrect. The inner membrane forms cristae — infoldings that increase surface area for oxidative phosphorylation — making (3) correct. The mitochondrial matrix houses a single, double-stranded, circular DNA molecule (similar to bacterial DNA), along with 70S ribosomes (55S in mammals), enabling semi-autonomous protein synthesis — confirming (4) is correct. This aligns precisely with NCERT Class 11, Chapter 8 'Cell: The Unit of Life', which states: 'The inner membrane bears the enzymes of the electron transport system' and describes matrix DNA and ribosomes.
Question 134 · Transport in Plants
Xylem translocates:
AWater only
BWater and mineral salts only
CWater, mineral salts and some organic nitrogen only
DWater, mineral salts, some organic nitrogen and hormones
Answer: D. Water, mineral salts, some organic nitrogen and hormones
Xylem is the vascular tissue responsible for upward conduction of water and dissolved minerals from roots to aerial parts. According to NCERT Class 11 (Chapter 11: Transport in Plants), xylem sap primarily contains water, inorganic ions (e.g., K⁺, Ca²⁺, NO₃⁻, PO₄³⁻), and small amounts of organic compounds — including amino acids, amides, and hormones like cytokinins and auxins synthesized in roots. Though phloem is the main pathway for organic solutes (e.g., sucrose, proteins), xylem does carry limited organic nitrogen (e.g., as ureides or amino acids) and root-derived hormones essential for shoot development. Option D correctly includes all these components. Options A and B are incomplete; option C omits hormones, which are experimentally verified to move via xylem (e.g., cytokinins from root tips regulate cell division in shoots). Thus, D is the most comprehensive and NCERT-aligned answer.
Question 135 · Cell Cycle and Cell Division
A cell in G₀ phase:
Aexits the cell cycle
Benters the cell cycle
Csuspends the cell cycle
Dterminates the cell cycle
Answer: C. suspends the cell cycle
The G₀ phase is a quiescent, non-dividing stage outside the active cell cycle (G₁ → S → G₂ → M). Cells enter G₀ from G₁ when they do not receive signals to proliferate—such as mature neurons, skeletal muscle cells, or some hepatocytes. In G₀, metabolic activity continues normally, but DNA replication and mitosis are halted; the cell remains viable and can re-enter the cycle upon appropriate stimuli (e.g., growth factors). This is not permanent exit (which implies terminal differentiation or senescence), nor entry (which occurs at G₁ start), nor termination (a biologically inaccurate term implying irreversible cessation or death). 'Suspends' accurately reflects the reversible, metabolically active pause—consistent with NCERT Class 11, Chapter 10: 'Some cells in adult animals do not divide, e.g., heart cells, and many others divide only occasionally to replace lost cells. These cells exit G₁ phase to enter an inactive stage called G₀.' Thus, option C is scientifically precise and NCERT-aligned.
Question 136 · Anatomy of Flowering Plants: Secondary Growth
Which of the following statements is not true about the formation of annual rings in trees?
AAnnual ring is a combination of spring wood and autumn wood produced in a year.
BDifferential activity of cambium causes light and dark bands of tissue — early wood and late wood, respectively.
CActivity of cambium depends upon variation in climate.
DAnnual rings are not prominent in trees of temperate region.
Answer: D. Annual rings are not prominent in trees of temperate region.
Annual rings (growth rings) form due to seasonal variation in secondary xylem production by the vascular cambium. In temperate regions, distinct climatic seasons cause cambium to be highly active in spring, producing large, thin-walled, light-coloured 'spring wood' (early wood), and less active in autumn, forming smaller, thick-walled, dark-coloured 'autumn wood' (late wood). Together, these constitute one annual ring. Thus, statement (1) is correct. The light–dark banding directly results from differential cambial activity — validating (2). Since cambial activity is regulated by environmental cues like temperature and rainfall, (3) is also true. Crucially, annual rings are *most prominent* in temperate trees due to sharp seasonal contrasts; they are *indistinct or absent* in tropical trees with uniform climates. Therefore, statement (4) — claiming annual rings are *not prominent* in temperate trees — is scientifically false and hence the correct choice for 'not true'. This aligns precisely with NCERT Class 11 Biology (Chapter 6: Anatomy of Flowering Plants), which states: 'In temperate regions, where growth is seasonal, the wood shows alternate concentric rings.'
Question 137 · Ecological Pyramids
Which of the following ecological pyramids is generally inverted?
APyramid of numbers in grassland
BPyramid of energy
CPyramid of biomass in a forest
DPyramid of biomass in a sea
Answer: D. Pyramid of biomass in a sea
Ecological pyramids depict trophic structure in ecosystems. The pyramid of energy is always upright because energy decreases progressively at each trophic level due to the 10% law (only ~10% transferred from one level to next). The pyramid of numbers is usually upright but can be inverted (e.g., in parasitic food chains), though not in grasslands — there it is upright due to abundant producers (grasses) supporting fewer herbivores and even fewer predators. The pyramid of biomass in forests is upright: large standing crop of trees (producers) supports smaller biomass of herbivores and still smaller carnivore biomass. However, in aquatic ecosystems like the sea, the pyramid of biomass is typically inverted — phytoplankton (producers) have high turnover rate but low standing biomass; they support a larger standing biomass of zooplankton (primary consumers), which in turn support fish. This inversion occurs because phytoplankton reproduce rapidly and are consumed continuously, so their instantaneous biomass is less than that of their slower-growing, longer-lived consumers. NCERT Class 12 Biology (Chapter 14: Ecosystem) explicitly states that the pyramid of biomass is inverted in marine ecosystems.
Question 139 · Environmental Issues
Which of the following protocols was aimed at reducing emissions of chlorofluorocarbons into the atmosphere?
AMontreal Protocol
BKyoto Protocol
CGothenburg Protocol
DGeneva Protocol
Answer: A. Montreal Protocol
The Montreal Protocol, adopted in 1987, is an international treaty specifically designed to protect the stratospheric ozone layer by phasing out the production and consumption of ozone-depleting substances (ODS), including chlorofluorocarbons (CFCs), halons, carbon tetrachloride, and methyl chloroform. CFCs were widely used in refrigerants, aerosol propellants, and foam-blowing agents; upon reaching the stratosphere, they release chlorine atoms that catalytically destroy ozone molecules. The Kyoto Protocol (1997) targets greenhouse gases like CO₂, CH₄, and N₂O to mitigate climate change—not ozone depletion. The Gothenburg Protocol (1999) addresses transboundary air pollution, focusing on sulphur dioxide, nitrogen oxides, VOCs, and ammonia—mainly for acid rain and eutrophication control. The Geneva Protocol (1925) prohibits chemical and biological weapons, unrelated to atmospheric pollutants. As per NCERT Class 12 Biology (Chapter 16: Environmental Issues), the Montreal Protocol is explicitly cited as the landmark agreement for CFC regulation and remains one of the most successful global environmental agreements.
Question 140 · Reproductive Health
Which of the following contraceptive methods involve a role of hormone?
ALactational amenorrhea, Pills, Emergency contraceptives
BBarrier method, Lactational amenorrhea, Pills
CCuT, Pills, Emergency contraceptives
DPills, Emergency contraceptives, Barrier methods
Answer: A. Lactational amenorrhea, Pills, Emergency contraceptives
Hormonal contraception works by altering hormonal balance to prevent ovulation, fertilization, or implantation. Oral contraceptive pills contain synthetic estrogen and/or progesterone that suppress gonadotropin secretion, thereby inhibiting ovulation. Emergency contraceptives (e.g., levonorgestrel or ulipristal acetate) act primarily by delaying or inhibiting ovulation through progesterone receptor modulation. Lactational amenorrhea is a natural postpartum contraceptive method where frequent and exclusive breastfeeding elevates prolactin levels, suppressing GnRH and thus preventing ovulation — hence it involves endogenous hormonal action. In contrast, barrier methods (e.g., condoms, diaphragms) physically prevent sperm entry and involve no hormones. CuT (copper-T) is a non-hormonal intrauterine device; copper ions create a toxic environment for sperm and impair implantation without hormonal involvement. Therefore, only option A — lactational amenorrhea, pills, and emergency contraceptives — correctly lists methods that involve hormonal mechanisms, aligning with NCERT Class 12 Biology Chapter 4 (Reproductive Health), which explicitly classifies lactational amenorrhea as a natural hormonal method.
Question 141 · Respiratory Volumes and Capacities
Tidal Volume and Expiratory Reserve Volume of an athlete are 500 mL and 1000 mL, respectively. What will be his Expiratory Capacity if the Residual Volume is 1200 mL?
A1500 mL
B1700 mL
C2200 mL
D2700 mL
Answer: A. 1500 mL
Expiratory Capacity (EC) is the total volume of air that can be exhaled after a normal inspiration. It is the sum of Tidal Volume (TV) and Expiratory Reserve Volume (ERV). According to NCERT Class 11 Biology (Chapter 17: Breathing and Exchange of Gases), EC = TV + ERV. Residual Volume (RV) — the air remaining in lungs after maximal expiration — is not part of EC; it contributes only to Functional Residual Capacity (FRC = ERV + RV) and Total Lung Capacity. Here, TV = 500 mL and ERV = 1000 mL, so EC = 500 + 1000 = 1500 mL. The given RV = 1200 mL is a distractor and irrelevant for calculating EC. This aligns precisely with NCERT’s definition and numerical examples, reinforcing conceptual clarity over rote memorization.
Question 142 · Double fertilization in angiosperms
What is the fate of the male gametes discharged in the synergid?
AOne fuses with the egg; the other(s) degenerate in the synergid.
BAll fuse with the egg.
COne fuses with the egg; the other(s) fuse(s) with synergid nucleus.
DOne fuses with the egg and the other fuses with the central cell nuclei.
Answer: D. One fuses with the egg and the other fuses with the central cell nuclei.
In angiosperms, double fertilization is a unique event where two male gametes from the pollen grain enter the embryo sac via the pollen tube. One male gamete fuses with the haploid egg cell to form the diploid zygote (2n), which develops into the embryo. The other male gamete fuses with the two polar nuclei (or the secondary nucleus, if fused) in the central cell to form the triploid primary endosperm nucleus (3n), which gives rise to the endosperm — a nutritive tissue for the developing embryo. This process ensures efficient resource allocation and evolutionary advantage. The synergids play a crucial role in guiding the pollen tube and facilitating discharge of male gametes, but they themselves do not participate in fertilization; their nuclei degenerate after pollen tube entry. Hence, options suggesting fusion with synergid nucleus or degeneration *only* in synergid (without specifying the second gamete’s role) are incorrect. NCERT Class 12 Biology (Chapter 2: Sexual Reproduction in Flowering Plants) explicitly states that one male gamete fertilizes the egg and the other fuses with the central cell.
Question 143 · Photoperiodism and Flowering
What is the site of perception of photoperiod necessary for induction of flowering in plants?
ALateral buds
BPulvinus
CShoot apex
DLeaves
Answer: D. Leaves
The site of photoperiod perception in plants is the leaves. Leaf mesophyll cells contain photoreceptors—primarily phytochrome—that detect day length (photoperiod) and initiate the floral transition. Upon sensing appropriate photoperiodic cues, leaves produce a mobile flowering stimulus called 'florigen', which is now known to involve FT (FLOWERING LOCUS T) protein transported via phloem to the shoot apical meristem. There, it triggers the conversion of vegetative meristem into floral meristem. Lateral buds are sites of branching, not photoperiod sensing; pulvinus mediates nastic movements like leaf folding in Mimosa; and while the shoot apex is the site of flower development, it does not perceive photoperiod—it responds to the signal generated elsewhere. This mechanism is explicitly covered in NCERT Class 12 Biology Chapter 15 'Plant Growth and Development', which states: 'It is now established that the site of perception of light for photoperiodic response is the leaves.'
Question 145 · Biotechnology: Principles and Processes
Following statements describe the characteristics of the enzyme restriction endonuclease. Identify the incorrect statement.
AThe enzyme cuts the DNA molecule at specific positions within the DNA.
BThe enzyme binds DNA at specific sites and cuts only one of the two strands.
CThe enzyme cuts the sugar-phosphate backbone at specific sites on each strand.
DThe enzyme recognizes a specific palindromic nucleotide sequence in the DNA.
Answer: B. The enzyme binds DNA at specific sites and cuts only one of the two strands.
Restriction endonucleases are bacterial enzymes that recognize specific palindromic DNA sequences (e.g., 5′-GAATTC-3′) and cleave both strands of the double helix at precise locations within or near those sites. They cut the sugar-phosphate backbone of *both* strands — not just one — generating either blunt or sticky ends. Statement (2) is therefore incorrect because it falsely claims the enzyme cuts only one strand; this is characteristic of some nucleases like RNase H or certain nicking enzymes, but *not* restriction endonucleases. In contrast, statements (1), (3), and (4) are accurate: they act at defined recognition sites (1), hydrolyze phosphodiester bonds in both strands’ backbones (3), and specifically bind palindromic sequences (4). As per NCERT Class 12 Biology (Chapter 11, 'Biotechnology: Principles and Processes'), restriction enzymes are indispensable tools in genetic engineering precisely because of their precise, double-stranded cleavage capability — a feature fundamental to recombinant DNA technology.
Question 146 · Plant Kingdom – Evolution of Embryophytes
From an evolutionary point of view, retention of the female gametophyte with the developing young embryo on the parent sporophyte for some time is first observed in:
ALiverworts
BMosses
CPteridophytes
DGymnosperms
Answer: C. Pteridophytes
Retention of the female gametophyte bearing the developing embryo on the parent sporophyte is a key evolutionary advancement leading to the seed habit. In bryophytes (liverworts and mosses), the embryo remains dependent on the gametophyte but is not retained *on the sporophyte* — the sporophyte itself is nutritionally dependent and short-lived. In pteridophytes (e.g., ferns), the embryo develops within the archegonium on the free-living gametophyte, not on the sporophyte; the sporophyte is independent but does not retain or nourish the embryo. Gymnosperms represent the first group where the female gametophyte (within the ovule) is retained *on the parent sporophyte*, and the embryo develops while still attached — a precursor to the seed. This feature — embryonic retention and nourishment by maternal sporophytic tissue — marks the origin of the seed habit and is absent in bryophytes and pteridophytes. NCERT Class 11 (Chapter 3: Plant Kingdom) explicitly states that gymnosperms show 'retention of the female gametophyte on the parent sporophyte' as a defining step toward seed evolution.
Question 148 · Glycolysis
Conversion of glucose to glucose-6-phosphate, the first irreversible reaction of glycolysis, is catalyzed by
AAldolase
BHexokinase
CEnolase
DPhosphofructokinase
Answer: B. Hexokinase
The phosphorylation of glucose to glucose-6-phosphate is the committed first step of glycolysis and is irreversible under cellular conditions. This reaction requires ATP and is catalyzed by hexokinase (in most tissues) or glucokinase (in liver and pancreatic beta cells). Hexokinase has a high affinity for glucose and is inhibited by its product, glucose-6-phosphate — a key regulatory feature. Aldolase cleaves fructose-1,6-bisphosphate into glyceraldehyde-3-phosphate and dihydroxyacetone phosphate; enolase converts 2-phosphoglycerate to phosphoenolpyruvate; and phosphofructokinase-1 (PFK-1) catalyzes the phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate — the major rate-limiting step of glycolysis. As per NCERT Class 11 Biology (Chapter 14: Respiration in Plants), hexokinase is explicitly identified as the enzyme responsible for the initial phosphorylation of glucose, making it the correct answer. This step traps glucose inside the cell and primes it for subsequent metabolic processing.
Question 149 · Drugs and their effects
The drug 'heroin' is synthesized by
Amethylation of morphine
Bacetylation of morphine
Cglycosylation of morphine
Dnitration of morphine
Answer: B. acetylation of morphine
Heroin, chemically known as diacetylmorphine, is semi-synthetically derived from morphine — a natural opioid alkaloid obtained from the opium poppy (Papaver somniferum). The synthesis involves acetylating both the phenolic (–OH) group at position 3 and the alcoholic (–OH) group at position 6 of morphine using acetic anhydride. This double acetylation increases lipid solubility, enabling faster crossing of the blood-brain barrier and enhancing potency compared to morphine. Methylation yields codeine (a less potent analgesic), glycosylation is not involved in opioid modification and is typical for plant secondary metabolites like flavonoids, while nitration is irrelevant in this context and would likely degrade the molecule. As per NCERT Class 12 Biology (Chapter 8: Human Health and Disease), heroin is explicitly described as 'diacetyl morphine' formed by acetylation — confirming option B as correct. Its high addiction potential, respiratory depression, and CNS depressant effects stem from this structural modification, making it a Schedule H drug under Indian law and a substance of abuse.
Question 150 · Reproductive Health
Select the hormone-releasing Intra-Uterine Devices.
AVaults, LNG-20
BMultiload 375, Progestasert
CProgestasert, LNG-20
DLippes Loop, Multiload 375
Answer: C. Progestasert, LNG-20
Hormone-releasing Intra-Uterine Devices (IUDs) release synthetic hormones—primarily progesterone or its derivatives—to prevent pregnancy by thickening cervical mucus, inhibiting sperm motility, and suppressing endometrial proliferation. Progestasert releases progesterone and is effective for one year; LNG-20 (Levonorgestrel-releasing IUD) releases levonorgestrel, a potent progestin, and remains effective for up to 5 years. In contrast, non-hormonal IUDs like Lippes Loop (made of plastic and stainless steel) and Multiload 375 (copper-containing) work by inducing a sterile inflammatory response and releasing copper ions that are spermicidal. Vaults are diaphragm-like barrier devices—not IUDs—and do not release hormones. Thus, only Progestasert and LNG-20 are hormone-releasing IUDs, matching option (3), which corresponds to choice C. This aligns with NCERT Class 12 Biology Chapter 4 'Reproductive Health', which explicitly lists Progestasert and LNG-20 as examples of hormonal IUDs, while distinguishing them from copper-based and barrier methods.
Question 151 · Hardy-Weinberg Principle
A gene locus has two alleles, A and a. If the frequency of the dominant allele A is 0.4, what are the frequencies of homozygous dominant (AA), heterozygous (Aa), and homozygous recessive (aa) individuals in the population?
A0.36(AA); 0.48(Aa); 0.16(aa)
B0.16(AA); 0.24(Aa); 0.36(aa)
C0.16(AA); 0.48(Aa); 0.36(aa)
D0.16(AA); 0.36(Aa); 0.48(aa)
Answer: C. 0.16(AA); 0.48(Aa); 0.36(aa)
According to the Hardy-Weinberg principle, for a gene with two alleles A and a, where p = frequency of A and q = frequency of a, p + q = 1. Given p = 0.4, then q = 1 − 0.4 = 0.6. Genotype frequencies are calculated as: AA = p² = (0.4)² = 0.16; Aa = 2pq = 2 × 0.4 × 0.6 = 0.48; aa = q² = (0.6)² = 0.36. This matches option (3). The principle assumes no evolutionary forces—no mutation, migration, natural selection, genetic drift, or non-random mating—and applies only to large, randomly mating populations. NCERT Class 12 Chapter 7 (Evolution) explicitly states these conditions and derives genotype frequencies using p² + 2pq + q² = 1. Students must recognize that allele frequency ≠ genotype frequency and that heterozygotes constitute the largest proportion when p and q are intermediate—here, 0.48 > both 0.16 and 0.36—confirming biological plausibility.
Question 152 · Biotechnology and its Applications
Which of the following is true for Golden rice?
AIt is Vitamin A enriched, with a gene from daffodil
BIt is pest resistant, with a gene from Bacillus thuringiensis
CIt is drought tolerant, developed using Agrobacterium vector
DIt has yellow grains, because of a gene introduced from a primitive variety of rice
Answer: A. It is Vitamin A enriched, with a gene from daffodil
Golden rice is a genetically modified variety developed to combat vitamin A deficiency. It contains beta-carotene — a precursor of vitamin A — which imparts a golden-yellow hue to the grains. This trait was achieved by introducing two genes: phytoene synthase (psy) from daffodil (Narcissus pseudonarcissus) and carotene desaturase (crtI) from the soil bacterium Erwinia uredovora. The psy gene catalyses the first committed step in carotenoid biosynthesis, while crtI enables efficient conversion to beta-carotene. Contrary to option B, pest resistance via Bt toxin is characteristic of Bt cotton or Bt corn, not Golden rice. Option C misattributes drought tolerance and Agrobacterium-mediated transformation — though Agrobacterium was used in early research, drought tolerance is not its defining feature. Option D is incorrect because the yellow colour arises from beta-carotene accumulation due to transgenes, not introgression from primitive rice varieties. NCERT Class 12 Biology (Chapter 12: Biotechnology and its Applications) explicitly states Golden rice as an example of biofortification for vitamin A enrichment.
Question 153 · Plant Kingdom – Gymnosperms and Mycorrhiza
Pinus seed cannot germinate and establish without fungal association. This is because:
Aits embryo is immature.
Bit has obligate association with mycorrhizae.
Cit has a very hard seed coat.
Dits seeds contain inhibitors that prevent germination.
Answer: B. it has obligate association with mycorrhizae.
Pinus, a gymnosperm, forms an obligate symbiotic relationship with fungi known as ectomycorrhizae. The fungal hyphae envelop the root tips and greatly enhance absorption of water and minerals—especially phosphorus and nitrogen—from nutrient-poor, acidic soils typical of coniferous habitats. In return, the plant supplies carbohydrates to the fungus. Crucially, Pinus seeds lack sufficient nutrient reserves in their cotyledons and depend entirely on this fungal partnership for early seedling establishment; without mycorrhizae, seedlings fail to absorb adequate nutrients and die. While Pinus embryos are mature at dispersal and the seed coat, though tough, is permeable and not the primary barrier, the dependency is physiological—not mechanical or chemical. NCERT Class 11 (Chapter 4: Plant Kingdom) explicitly states that 'the roots of Pinus form mycorrhizal associations' and highlights their essential role in nutrient uptake and seedling survival. Thus, the correct explanation is the obligate mycorrhizal association.
Question 154 · Molecular Basis of Inheritance
Which of the following features of the genetic code allows bacteria to produce human insulin using recombinant DNA technology?
AGenetic code is not ambiguous
BGenetic code is redundant
CGenetic code is nearly universal
DGenetic code is specific
Answer: C. Genetic code is nearly universal
The key feature enabling bacteria to synthesize functional human insulin is the near-universality of the genetic code. As per NCERT Class 12 (Chapter 6), almost all living organisms—from bacteria to humans—use the same codons to specify the same amino acids. This universality means that when the human insulin gene is inserted into bacterial plasmids via recombinant DNA technology, the bacterial transcription and translation machinery correctly reads the human mRNA sequence and assembles the exact human insulin polypeptide chain. While redundancy (multiple codons for same amino acid) and non-ambiguity (each codon specifies only one amino acid) are essential properties, they alone do not permit cross-species gene expression. Specificity refers to the precise codon–amino acid correspondence, but it is the universality — the shared 'language' across domains — that makes heterologous protein production feasible. Without this feature, bacterial ribosomes would misread human codons or incorporate wrong amino acids, yielding nonfunctional protein. Hence, option (3) — 'Genetic code is nearly universal' — is scientifically accurate and directly explains the success of recombinant insulin production.
Question 155 · Human Health and Disease
Which of the following sexually transmitted diseases is not completely curable?
AGonorrhoea
BGenital warts
CGenital herpes
DChlamydiasis
Answer: C. Genital herpes
Among the listed STDs, genital herpes—caused by Herpes Simplex Virus (HSV), primarily HSV-2—is not completely curable. While antiviral drugs like acyclovir can suppress viral replication, reduce symptom severity, and decrease transmission risk, they cannot eliminate latent virus from sensory nerve ganglia. Once established, HSV remains dormant and may reactivate periodically, causing recurrent outbreaks. In contrast, gonorrhoea (Neisseria gonorrhoeae) and chlamydiasis (Chlamydia trachomatis) are bacterial infections effectively cured with appropriate antibiotics (e.g., ceftriaxone + azithromycin). Genital warts, caused by certain strains of human papillomavirus (HPV), are clinically manageable—warts can be removed via cryotherapy, laser, or topical agents—but HPV infection itself may persist; however, the question asks for 'not completely curable', and NCERT Class 12 (Chapter 8: Human Health and Disease) explicitly states that genital herpes is 'incurable' due to viral latency, distinguishing it from treatable bacterial STDs and even HPV-related lesions which are often self-limiting or removable. Hence, option C (Genital herpes) is correct.
Question 156 · Biological Classification & Infectious Agents
Which of the following statements is incorrect?
AViroids lack a protein coat.
BViruses are obligate parasites.
CInfective constituent in viruses is the protein coat.
DPrions consist of abnormally folded proteins.
Answer: C. Infective constituent in viruses is the protein coat.
The incorrect statement is that the infective constituent in viruses is the protein coat. According to NCERT Class 11 (Chapter 2: Biological Classification), the genetic material — either DNA or RNA — is the infective component of a virus; the protein coat (capsid) only protects the genome and aids in host cell attachment and entry, but it is not itself infectious. Viroids (option A) are indeed naked RNA molecules without any protein coat, making them simpler than viruses. Viruses (option B) are obligate intracellular parasites because they lack cellular machinery for replication and depend entirely on host cells. Prions (option D) are misfolded, pathogenic isoforms of normal cellular proteins (e.g., PrP^C → PrP^Sc) and cause neurodegenerative diseases like scrapie and Creutzfeldt-Jakob disease — a fact confirmed in NCERT’s discussion on non-living infectious agents. Thus, only option C contradicts established NCERT content and is therefore the correct choice for 'incorrect statement'.
Question 157 · Animal Kingdom: Phylum-specific diagnostic features
Match the following organisms with their respective characteristics:
(a) Pila (i) Flame cells
(b) Bombyx (ii) Comb plates
(c) Pleurobrachia (iii) Radula
(d) Taenia (iv) Malpighian tubules
A(i), (ii), (iii), (iv)
B(iii), (iv), (ii), (i)
C(ii), (iv), (iii), (i)
D(iii), (ii), (iv), (i)
Answer: B. (iii), (iv), (ii), (i)
Pila (apple snail) belongs to Mollusca and possesses a radula — a rasping, tongue-like organ used for feeding; hence (a)-(iii). Bombyx (silkworm moth) is an insect (Arthropoda) and excretes nitrogenous waste via Malpighian tubules; thus (b)-(iv). Pleurobrachia (a ctenophore) moves using comb plates (ciliated rows); so (c)-(ii). Taenia (tapeworm) is a platyhelminth with flame cells as excretory structures; therefore (d)-(i). This matching aligns precisely with NCERT Class 11 Chapter 4 'Animal Kingdom', which explicitly states: 'Molluscs have radula', 'Insects possess Malpighian tubules', 'Ctenophores bear eight rows of comb plates', and 'Flame cells are characteristic of Platyhelminthes'. Option B — (iii), (iv), (ii), (i) — correctly pairs all four. Other options misassign at least one feature: e.g., option A wrongly assigns flame cells to Pila, and option D incorrectly gives radula to Taenia.
Question 158 · Biotechnology Principles and Processes
Expressed Sequence Tags (ESTs) refer to:
AGenes expressed as RNA
BPolypeptide expression
CDNA polymorphism
DNovel DNA sequences
Answer: A. Genes expressed as RNA
Expressed Sequence Tags (ESTs) are short, single-read sequences derived from cDNA libraries — that is, complementary DNA synthesized from mRNA. Since mRNA represents only the transcriptionally active (expressed) portion of the genome, ESTs serve as identifiers for genes that are actively transcribed into RNA in a particular tissue or developmental stage. They are not full-length genes but unique tags representing expressed regions. Option B is incorrect because ESTs correspond to RNA transcripts, not translated polypeptides. Option C is wrong as DNA polymorphism refers to variations like SNPs or VNTRs, unrelated to expression profiling. Option D is misleading: while ESTs may sometimes reveal novel genes, their defining feature is expression-derived origin, not novelty per se. As per NCERT Class 12, Chapter 11 'Biotechnology: Principles and Processes', ESTs are explicitly described as 'sequences of cDNA clones' used to identify expressed genes — aligning precisely with option A.
Question 159 · Biological Classification – Kingdom Fungi
Which of the following statements is incorrect?
AMorels and truffles are edible delicacies.
BClaviceps is a source of many alkaloids and LSD.
CConidia are produced exogenously and ascospores endogenously.
DYeasts have filamentous bodies with long thread-like hyphae.
Answer: D. Yeasts have filamentous bodies with long thread-like hyphae.
Yeasts (e.g., Saccharomyces cerevisiae) are unicellular, oval-shaped fungi that reproduce asexually by budding or fission; they do not form true mycelia with long, thread-like hyphae. In contrast, filamentous fungi like Aspergillus or Rhizopus possess hyphal networks. Statement D is therefore incorrect. Statement A is correct: Morels (Morchella) and truffles (Tuber) are ascomycetes prized as gourmet edibles. Statement B is accurate: Claviceps purpurea, an ascomycete parasite on rye, produces ergot alkaloids—including lysergic acid diethylamide (LSD)—in its sclerotia. Statement C correctly distinguishes fungal spore formation: conidia arise externally on conidiophores (exogenous), while ascospores develop inside sac-like asci (endogenous), a hallmark of Ascomycota. This aligns precisely with NCERT Class 11 Chapter 2 'Biological Classification', which emphasizes structural and reproductive differences among fungal groups.
Question 160 · Biological Classification & Microbes in Human Welfare
Match Column I with Column II:
Column I
(a) Saprophyte
(b) Parasite
(c) Lichens
(d) Mycorrhiza
Column II
(i) Symbiotic association of fungi with plant roots
(ii) Decomposition of dead organic materials
(iii) Living on living plants or animals
(iv) Symbiotic association of algae and fungi
A(1) (i) (ii) (iii) (iv)
B(2) (iii) (ii) (i) (iv)
C(3) (ii) (i) (iii) (iv)
D(4) (ii) (iii) (iv) (i)
Answer: D. (4) (ii) (iii) (iv) (i)
Saprophytes obtain nutrients by decomposing dead organic matter — correctly matched with (ii). Parasites derive nutrition from living hosts without conferring benefit — matched with (iii). Lichens are classic symbiotic associations between algae (photosynthetic partner) and fungi (absorptive, protective partner) — matched with (iv). Mycorrhiza is a mutualistic symbiosis where fungal hyphae associate with plant roots, enhancing mineral and water absorption — matched with (i). This aligns precisely with NCERT Class 11 Chapter 2 'Biological Classification' and Chapter 10 'Microbes in Human Welfare', which define saprophytism, parasitism, lichens as dual organisms, and mycorrhizae as root-fungal symbioses. Option (4) — (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) — is therefore fully consistent with NCERT terminology and conceptual understanding.
Question 161 · Transport and Cell Signalling
Which of the following glucose transporters is insulin-dependent?
AGLUT I
BGLUT II
CGLUT III
DGLUT IV
Answer: D. GLUT IV
GLUT IV is the primary insulin-dependent glucose transporter in humans. It is predominantly expressed in insulin-sensitive tissues — skeletal muscle, cardiac muscle, and adipose tissue. Under basal (fasting) conditions, GLUT IV is sequestered intracellularly in vesicles. Upon insulin binding to its receptor, a cascade involving IRS-1, PI3K, and Akt/PKB triggers translocation of GLUT IV vesicles to the plasma membrane, enabling facilitated diffusion of glucose into cells. This mechanism is critical for postprandial glucose clearance and maintaining blood glucose homeostasis. In contrast, GLUT I (widely distributed, e.g., blood-brain barrier), GLUT II (liver, pancreatic beta cells, intestine; low-affinity, high-capacity), and GLUT III (neurons, placenta; high affinity) are all insulin-independent. Their expression and activity are constitutive and not regulated by insulin signalling. NCERT Class 12 Biology (Chapter 18: Body Fluids and Circulation; also referenced in Chapter 11: Transport in Plants — comparative context on transport proteins) emphasizes GLUT IV’s role in insulin-mediated glucose uptake as a key physiological adaptation in mammals.
Question 162 · Immunity and Transplantation
Which of the following immune responses is responsible for rejection of a kidney graft?
AAuto-immune response
BHumoral immune response
CInflammatory immune response
DCell-mediated immune response
Answer: D. Cell-mediated immune response
Kidney graft rejection is primarily mediated by T-lymphocytes, making it a classic example of cell-mediated immunity. Unlike humoral immunity (which involves antibodies produced by B-cells), graft rejection occurs when recipient cytotoxic T-cells recognize foreign MHC class I antigens on donor kidney cells and directly destroy them. Helper T-cells also amplify this response via cytokine release. While antibodies (humoral response) and inflammation may contribute secondarily—especially in chronic or hyperacute rejection—the primary and most critical mechanism in acute allograft rejection is T-cell-driven cell-mediated immunity. NCERT Class 12 Biology (Chapter 8: Human Health and Disease) explicitly states that 'cell-mediated immune response is responsible for graft rejection', distinguishing it from autoimmune disorders (which target self-antigens) and general inflammatory responses (non-specific, innate). Autoimmunity is not involved here, as the graft is foreign—not self—and inflammation alone lacks antigen specificity. Thus, option (4), i.e., cell-mediated immune response, is scientifically accurate and NCERT-aligned.
Question 163 · Excretory Products and Their Elimination
Use of an artificial kidney during hemodialysis may result in:
(a) Nitrogenous waste build-up in the body
(b) Non-elimination of excess potassium ions
(c) Reduced absorption of calcium ions from gastro-intestinal tract
(d) Reduced RBC production
Which of the following options is the most appropriate?
A(a) and (b) are correct
B(b) and (c) are correct
C(c) and (d) are correct
D(a) and (d) are correct
Answer: C. (c) and (d) are correct
Hemodialysis uses an artificial kidney (dialyzer) to remove nitrogenous wastes, excess fluids, and electrolytes like potassium from blood when kidneys fail. Thus, (a) — nitrogenous waste build-up — is incorrect, as dialysis prevents it. (b) — non-elimination of excess K⁺ — is also incorrect; hyperkalemia is actively corrected during dialysis. However, chronic kidney disease (CKD) and repeated dialysis impair vitamin D activation in renal tubular cells, reducing intestinal calcium absorption — validating (c). Additionally, failing kidneys produce less erythropoietin (EPO), a hormone essential for RBC production in bone marrow; dialysis does not restore EPO synthesis, leading to anemia — confirming (d). Hence, only (c) and (d) are correct consequences directly linked to renal failure and dialysis dependence. NCERT Class 11 (Chapter 19) explicitly states that CKD causes hypocalcemia due to impaired calcitriol synthesis and anemia due to EPO deficiency — both persist despite hemodialysis.
Question 165 · Chromosomal Theory of Inheritance and Genetic Mapping
The frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes was explained by:
AT.H. Morgan
BGregor J. Mendel
CAlfred Sturtevant
DSutton and Boveri
Answer: C. Alfred Sturtevant
Alfred Sturtevant, a student of T.H. Morgan, pioneered genetic mapping in 1913 by analyzing recombination frequencies from Drosophila crosses. He reasoned that the farther apart two genes are on a chromosome, the higher the chance of crossing over occurring between them — thus, recombination frequency directly correlates with physical distance. Using this principle, he constructed the first linear genetic map of the X chromosome, assigning relative positions to genes in map units (centiMorgans). While Morgan discovered linkage and coined the term 'recombination', and Sutton–Boveri proposed the chromosomal basis of inheritance, it was Sturtevant who quantitatively linked recombination frequency to gene distance. Mendel, working before chromosomes were understood, established laws of segregation and independent assortment but had no concept of linkage or chromosomal mapping. Hence, Sturtevant is rightly credited with developing the method of gene mapping using recombination data — a foundational concept in modern genetics covered in NCERT Class 12 Chapter 5 'Principles of Inheritance and Variation'.
Question 166 · Regulation of gene expression in prokaryotes
Match the following genes of the Lac operon with their respective products:
(a) i gene (i) β-galactosidase
(b) z gene (ii) Permease
(c) a gene (iii) Repressor
(d) y gene (iv) Transacetylase
Select the correct option.
A(i) (iii) (ii) (iv)
B(iii) (i) (ii) (iv)
C(iii) (i) (iv) (ii)
D(iii) (iv) (i) (ii)
Answer: C. (iii) (i) (iv) (ii)
In the Lac operon of E. coli, the i gene encodes the repressor protein, which binds to the operator and prevents transcription in the absence of lactose. The z gene codes for β-galactosidase, an enzyme that hydrolyses lactose into glucose and galactose. The y gene encodes permease, a membrane protein facilitating lactose uptake into the cell. The a gene produces transacetylase, which detoxifies certain thiogalactosides by acetylating them — though its exact physiological role is secondary. Thus, the correct matching is: (a)–(iii), (b)–(i), (c)–(iv), (d)–(ii). This aligns precisely with option C. NCERT Class 12 Biology (Chapter 6: Molecular Basis of Inheritance) explicitly states these gene–product relationships and emphasizes that the i gene product is the repressor, z encodes β-galactosidase, y encodes permease, and a encodes transacetylase — confirming the sequence (iii)(i)(iv)(ii). Understanding this mapping is essential for grasping operon regulation, a high-yield NEET concept rooted in Jacob and Monod’s model.
Question 167 · Plant Growth Regulators and Photoperiodism
It takes a very long time for pineapple plants to produce flowers. Which combination of hormones can be applied to artificially induce flowering in pineapple plants throughout the year to increase yield?
AAuxin and Ethylene
BGibberellin and Cytokinin
CGibberellin and Abscisic acid
DCytokinin and Abscisic acid
Answer: A. Auxin and Ethylene
Pineapple (Ananas comosus) is a short-day plant but exhibits ethylene-sensitive floral induction. Ethylene, particularly its precursor ethephon (which releases ethylene), is widely used commercially to synchronize flowering in pineapple. Auxins like naphthaleneacetic acid (NAA) enhance ethylene biosynthesis by stimulating ACC synthase activity, thereby synergistically promoting flowering. This auxin–ethylene interaction is well documented in NCERT Class 12 Biology (Chapter 15: Plant Growth and Development) as a key example of hormonal crosstalk in flowering control. Gibberellins promote stem elongation and bolting in rosette plants but do not induce flowering in pineapple; cytokinins delay senescence and promote cell division but lack floral initiation role here; abscisic acid is a growth inhibitor involved in dormancy and stress response — it antagonizes flowering. Hence, only auxin + ethylene reliably triggers uniform, off-season flowering in pineapple, boosting yield and harvest scheduling.
Question 168 · Digestion and Absorption
Identify the cells whose secretion protects the lining of the gastrointestinal tract from various enzymes.
AChief Cells
BGoblet Cells
COxyntic Cells
DDuodenal Cells
Answer: B. Goblet Cells
Goblet cells are unicellular, mucus-secreting glands found in the epithelium of the stomach, small intestine, and large intestine. Their primary function is to secrete mucin, which upon hydration forms a viscous, alkaline mucus layer that coats the luminal surface of the GI tract. This mucus acts as a physical and chemical barrier — it prevents autodigestion by shielding the epithelial lining from proteolytic enzymes (e.g., pepsin, trypsin) and acidic gastric juice. Chief cells secrete pepsinogen (inactive precursor of pepsin), oxyntic (parietal) cells secrete HCl and intrinsic factor, and 'Duodenal cells' is not a standard histological term — Brunner’s glands (in duodenum) secrete alkaline mucus, but they are multicellular glands, not 'duodenal cells'. NCERT Class 11 Biology (Chapter 16: Digestion and Absorption) explicitly states: 'Goblet cells secrete mucus which helps in lubrication and protection of the intestinal mucosa.' Thus, only goblet cells directly provide enzyme-mediated protection via mucus secretion.
Question 169 · Microbes in Human Welfare
Which of the following can be used as a biocontrol agent in the treatment of plant disease?
ATrichoderma
BChlorella
CAnabaena
DLactobacillus
Answer: A. Trichoderma
Trichoderma is a free-living, cellulolytic fungus commonly found in soil and root ecosystems. It acts as an effective biocontrol agent against several plant-pathogenic fungi—including Rhizoctonia, Fusarium, and Pythium—by mechanisms such as mycoparasitism (direct parasitism on fungal hyphae), competition for nutrients and space, and production of antifungal metabolites like gliotoxin and viridin. NCERT Class 12, Chapter 10 'Microbes in Human Welfare', explicitly lists Trichoderma as a key example of a biological control agent used in organic farming to manage soil-borne plant diseases. In contrast, Chlorella is a unicellular green alga used as a nutritional supplement; Anabaena is a nitrogen-fixing cyanobacterium beneficial in paddy fields but not for disease control; Lactobacillus is a lactic acid bacterium used in food fermentation and gut health, with no established role in plant disease management. Thus, only Trichoderma qualifies as a biocontrol agent for plant diseases.
Question 170 · Anatomy of Flowering and Non-flowering Plants
Phloem in gymnosperms lacks:
AAlbuminous cells and sieve cells
BSieve tubes only
CCompanion cells only
DBoth sieve tubes and companion cells
Answer: D. Both sieve tubes and companion cells
Gymnosperms possess phloem composed of sieve cells (not sieve tubes) and albuminous cells (functionally analogous to companion cells but not derived from the same mother cell). Unlike angiosperms, gymnosperms lack true sieve tubes (which are formed by longitudinal fusion of sieve elements) and do not have companion cells—these are exclusive innovations of angiosperms. Sieve cells in gymnosperms are associated with albuminous cells, which assist in loading/unloading but differ structurally and developmentally from companion cells. NCERT Class 11 (Chapter 6: Anatomy of Flowering Plants) explicitly states: 'In gymnosperms, the phloem is without companion cells and sieve tubes; it has sieve cells and albuminous cells.' Therefore, option (4)—'Both sieve tubes and companion cells'—is correct. Options (1) and (2) are incorrect because sieve cells *are* present, and albuminous cells *are* present. Option (3) is incomplete—while companion cells are indeed absent, sieve tubes are also absent, making D the most accurate and complete answer.
Question 171 · Human Reproduction
Extrusion of the second polar body from the egg nucleus occurs:
Aafter entry of sperm but before fertilization
Bafter fertilization
Cbefore entry of sperm into ovum
Dsimultaneously with first cleavage
Answer: A. after entry of sperm but before fertilization
The extrusion of the second polar body occurs only after sperm entry triggers completion of meiosis II in the secondary oocyte. In human females, oogenesis arrests at metaphase II until fertilization. Upon sperm penetration, the secondary oocyte completes meiosis II, producing one mature haploid ovum and a second polar body. This event happens *after* sperm entry but *before* karyogamy (fusion of male and female pronuclei), i.e., before fertilization is complete. Thus, option A is correct. Option B is incorrect because fertilization is defined as the fusion of pronuclei — by which time the second polar body has already been extruded. Option C is wrong since meiosis II remains arrested until sperm contact; no second polar body forms pre-sperm entry. Option D is incorrect: first cleavage is a mitotic division occurring ~30 hours post-fertilization, long after second polar body extrusion (~2–3 hours post-sperm entry). NCERT Class 12, Chapter 2 'Human Reproduction', clearly states that 'the secondary oocyte undergoes meiosis II only when a sperm enters it, resulting in formation of a second polar body and a mature ovum'.
Question 172 · Molecular basis of inheritance: Genetic code and frameshift mutations
Under which of the following conditions will there be no change in the reading frame of the following mRNA? 5′-AACAGCGGUGCUAUU-3′
AInsertion of G at 5′ position
BDeletion of G from 5′ position
CInsertion of A and G at 4th and 5th positions respectively
DDeletion of GGU from 7th, 8th and 9th positions
Answer: D. Deletion of GGU from 7th, 8th and 9th positions
The genetic code is read in non-overlapping triplets (codons) starting from a fixed 5′ initiation point. Any insertion or deletion not divisible by three disrupts the reading frame — a frameshift mutation — altering all downstream amino acids. Here, the mRNA sequence is 5′-AAC AGC GGU GCU AUU-3′ (grouped in codons). Option D deletes GGU — a complete codon (positions 7–9) — removing exactly three nucleotides. Since the reading frame resets after every triplet, deleting one full codon preserves the frame for subsequent codons; only that single amino acid (glycine) is lost. In contrast, option A (inserting G at 5′ end) shifts all downstream triplets; option B (deleting first G — but note: 5′-AACAGC... means position 1=A, 2=A, 3=C, 4=A, 5=G — so deleting G at position 5 removes first base of second codon, causing frameshift); option C inserts two nucleotides (A and G) — not a multiple of three — also causing frameshift. NCERT Class 12 Chapter 6 explicitly states that frameshifts occur with insertions/deletions of 1 or 2 bases, but not 3 (or multiples), as they preserve codon boundaries.
Question 173 · Cell: The Unit of Life
The concept of "Omnis cellula e cellula" regarding cell division was first proposed by
ARudolf Virchow
BTheodor Schwann
CMatthias Schleiden
DAristotle
Answer: A. Rudolf Virchow
The phrase "Omnis cellula e cellula" (meaning 'every cell arises from a pre-existing cell') is a foundational principle of modern cell theory. While Matthias Schleiden (1838) and Theodor Schwann (1839) proposed that plants and animals are composed of cells, they initially believed cells could form de novo from non-living material. Rudolf Virchow, in 1855, critically revised this view based on extensive microscopic observations and physiological reasoning, asserting that cells only originate from pre-existing cells — thereby completing the classical cell theory. Though Virchow popularized the Latin phrase, he acknowledged earlier insights by Robert Remak; however, historical consensus and NCERT Class 11 Biology (Chapter 8, 'Cell: The Unit of Life') explicitly credit Virchow for formally establishing this principle. Aristotle, predating microscopy by centuries, contributed to early natural philosophy but had no concept of cells or cellular reproduction. Thus, option A is scientifically and syllabus-accurate.
Question 174 · Biotechnology and its Applications
What triggers activation of protoxin to active Bt toxin of Bacillus thuringiensis in boll worm?
ABody temperature
BMoist surface of midgut
CAlkaline pH of gut
DAcidic pH of stomach
Answer: C. Alkaline pH of gut
The Bt toxin produced by Bacillus thuringiensis is initially synthesized as an inactive protoxin. When ingested by susceptible insects like the boll worm, the protoxin reaches the midgut, where the alkaline pH (typically pH 9.5–10.5) causes solubilization and conformational change. This alkaline environment facilitates proteolytic cleavage by gut proteases, converting the protoxin into its active toxic form. The activated toxin then binds to specific receptors on the brush border membrane of midgut epithelial cells, forming pores that cause cell lysis and insect death. In contrast, mammals and birds have acidic stomachs (pH ~1.5–3.5) and lack the appropriate receptors and alkaline gut conditions, making Bt safe for non-target organisms. NCERT Class 12 Biology (Chapter 12: Biotechnology and its Applications) explicitly states that 'the activated toxin binds with the surface of midgut epithelial cells and creates pores... this happens in the alkaline medium of the gut of the insect'. Thus, alkaline pH—not body temperature, moisture, or acidic pH—is the critical trigger.
Question 175 · Human Health and Disease
Identify the correct pair representing the causative agent of typhoid fever and the confirmatory test for typhoid.
APlasmodium vivax — UTI test
BStreptococcus pneumoniae — Widal test
CSalmonella typhi — Anthrone test
DSalmonella typhi — Widal test
Answer: D. Salmonella typhi — Widal test
Typhoid fever is a systemic bacterial infection caused exclusively by Salmonella enterica serovar Typhi (commonly referred to as Salmonella typhi in NCERT Class 12, Chapter 8). It is transmitted via contaminated food or water and manifests as sustained high fever, abdominal pain, and intestinal involvement. The Widal test is a serological agglutination test that detects antibodies against O (somatic) and H (flagellar) antigens of S. typhi in the patient’s serum — it remains the standard confirmatory diagnostic tool taught in NCERT despite its limitations in specificity and timing. Plasmodium vivax causes malaria, not typhoid; Streptococcus pneumoniae causes pneumococcal pneumonia and meningitis; the Anthrone test is used for carbohydrate estimation (e.g., glucose), not typhoid diagnosis; and UTI test refers to urinalysis for urinary tract infections — unrelated to typhoid. Hence, only option D correctly pairs the pathogen and its confirmatory test, aligning precisely with NCERT’s treatment of infectious diseases.
Question 176 · Human Genetics and Chromosomal Disorders
What is the genetic disorder in which an individual has overall masculine development, gynaecomastia, and is sterile?
ATurner's syndrome
BKlinefelter's syndrome
CEdward syndrome
DDown's syndrome
Answer: B. Klinefelter's syndrome
Klinefelter's syndrome is caused by the presence of an extra X chromosome (47,XXY karyotype) in males. Affected individuals have male external genitalia and overall masculine development due to functional testes producing testosterone during early life, but spermatogenesis fails due to seminiferous tubule hyalinization — leading to azoospermia and sterility. Elevated gonadotropins (FSH/LH) and relative estrogen excess cause gynaecomastia. In contrast, Turner’s syndrome (45,X) presents with female phenotype, short stature, and ovarian dysgenesis; Edward syndrome (trisomy 18) and Down’s syndrome (trisomy 21) involve severe multisystem congenital anomalies and intellectual disability but not the specific triad of masculine development, gynaecomastia, and sterility. NCERT Class 12 Biology (Chapter 5: Principles of Inheritance and Variation) explicitly describes Klinefelter’s syndrome as characterized by tall stature, small testes, gynaecomastia, and infertility — aligning precisely with the question.
Question 177 · Environmental Issues: Solid Waste Management
Polyblend, a fine powder of recycled modified plastic, has proved to be a good material for
AMaking plastic sacks
BUse as a fertilizer
CConstruction of roads
DMaking tubes and pipes
Answer: C. Construction of roads
Polyblend is a patented eco-friendly product developed by Ahmed Khan of Bangalore, made by mixing recycled modified plastic with bitumen. When used in road construction, it enhances the water-repellent property and tensile strength of bitumen, thereby increasing road durability and reducing maintenance costs. NCERT Class 12 Biology (Chapter 16: Environmental Issues) explicitly states that Polyblend is blended with bitumen to lay roads — a successful example of recycling plastic waste and mitigating environmental pollution. It is not used for making new plastic sacks (which would perpetuate single-use plastic), nor does it function as a fertilizer (lacking nutrient composition or microbial activity). While recycled plastic can be extruded into pipes or tubes, Polyblend specifically refers to the bitumen-blended formulation optimized for road surfacing — not general plastic fabrication. Its adoption aligns with India’s Swachh Bharat Mission and sustainable development goals by converting non-biodegradable waste into infrastructure assets.
Question 178 · Environmental Issues – Radioactive Waste Management
Which of the following methods is the most suitable for disposal of nuclear waste?
AShoot the waste into space
BBury the waste under Antarctic ice-cover
CDump the waste within rocks under deep ocean
DBury the waste within rocks deep below the Earth's surface
Answer: D. Bury the waste within rocks deep below the Earth's surface
According to NCERT Class 12 Biology (Chapter 16: Environmental Issues), the most scientifically accepted and safest method for disposing of high-level nuclear waste is deep geological disposal — i.e., burying it within stable, impermeable rock formations several hundred metres below the Earth’s surface. This isolates radioactive material from the biosphere for thousands of years, minimising risk of groundwater contamination or human exposure. Shooting waste into space is prohibitively expensive, technologically unreliable, and poses catastrophic launch failure risks. Burying under Antarctic ice violates the Antarctic Treaty System and risks meltwater transport due to climate change. Ocean dumping is banned under the London Convention (1972) and risks bioaccumulation in marine food chains. Deep underground repositories — like those planned in countries such as Finland (Onkalo) and India (proposed sites in granite shields of Karnataka and Andhra Pradesh) — rely on multiple engineered and natural barriers (e.g., bentonite clay, rock matrix) ensuring long-term containment. Hence, option D aligns precisely with NCERT’s recommended strategy.
Question 179 · Chemical Coordination and Integration
Match the following hormones with the respective disease:
(a) Insulin
(b) Thyroxin
(c) Corticoids
(d) Growth Hormone
(i) Addison's disease
(ii) Diabetes insipidus
(iii) Acromegaly
(iv) Goitre
(v) Diabetes mellitus
A(a)-(v), (b)-(i), (c)-(ii), (d)-(iii)
B(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
C(a)-(v), (b)-(iv), (c)-(i), (d)-(iii)
D(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Answer: C. (a)-(v), (b)-(iv), (c)-(i), (d)-(iii)
Insulin deficiency causes diabetes mellitus (v), characterized by hyperglycemia and glycosuria. Thyroxin (thyroid hormone) deficiency leads to goitre (iv) — an enlarged thyroid gland due to TSH overstimulation in iodine deficiency or hypothyroidism. Corticoids (glucocorticoids like cortisol) deficiency results in Addison’s disease (i), marked by fatigue, hypotension, and hyperpigmentation. Excess growth hormone (GH) in adults causes acromegaly (iii), featuring enlarged hands, feet, and facial bones. Note: Diabetes insipidus (ii) is due to ADH deficiency or renal resistance — unrelated to insulin, thyroxin, corticoids, or GH. Acromegaly is correctly matched only with GH excess; goitre with thyroxin deficiency; Addison’s with corticoid deficiency; and diabetes mellitus with insulin deficiency — confirming option C as correct per NCERT Class 11 Chapter 22.
Question 180 · Axial Skeleton – Ribs and Thoracic Cage
Select the correct option.
A8th, 9th and 10th pairs of ribs articulate directly with the sternum.
B11th and 12th pairs of ribs are connected to the sternum with the help of hyaline cartilage.
CEach rib is a flat thin bone and all the ribs are connected dorsally to the thoracic vertebrae and ventrally to the sternum.
DThere are seven pairs of vertebrosternal, three pairs of vertebrochondral and two pairs of vertebral ribs.
Answer: D. There are seven pairs of vertebrosternal, three pairs of vertebrochondral and two pairs of vertebral ribs.
According to NCERT Class 11 Biology (Chapter 20: Locomotion and Movement), human ribs are classified based on their attachment. The first 7 pairs (ribs 1–7) are 'true ribs' or vertebrosternal ribs — they attach dorsally to thoracic vertebrae and ventrally to the sternum via their own hyaline costal cartilages. Ribs 8–10 are 'false ribs' or vertebrochondral ribs — they attach dorsally to vertebrae but ventrally join the costal cartilage of rib 7, not the sternum directly. Ribs 11–12 are 'floating ribs' or vertebral ribs — they attach only dorsally to vertebrae and have no anterior connection. Option D correctly states there are seven vertebrosternal, three vertebrochondral, and two vertebral ribs — matching NCERT’s classification. Option A is wrong because ribs 8–10 do not articulate directly with sternum. Option B is incorrect — floating ribs lack sternal connection entirely. Option C is false — only true ribs attach ventrally to sternum; floating ribs lack ventral attachment.