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With one of the highest numbers of candidates registered for the NEET 2024 exam, the NEET exam previous year question paper 2024 serves as an essential reference for the medical entrance exams’ current competitive nature.
Match List I with List II:
| List I | List II |
|---|---|
| A. Rhizopus | I. Mushroom |
| B. Ustilago | II. Smut fungus |
| C. Puccinia | III. Bread mould |
| D. Agaricus | IV. Rust fungus |
Answer: (1) A-III, B-II, C-IV, D-I
Explanation:
Rhizopus is a bread mould fungus (commonly seen on bread)
Ustilago causes smut disease in plants (smut fungi)
Puccinia causes rust disease in plants (rust fungi)
Agaricus is commonly called mushroom
Identify the part of the seed destined to form root when the seed germinates:
Answer: (3) C (Radicle)
Explanation:
The radicle is the embryonic root of the plant
During seed germination, the radicle emerges first and develops into the root system
In the diagram, ‘C’ represents the radicle
Lecithin, a small molecular weight organic compound found in living tissues, is an example of:
Answer: (2) Phospholipids
Explanation:
Lecithin is a phospholipid (contains phosphorus and a phosphorylated organic compound)
Found in cell membranes
Glycerides contain glycerol and fatty acids only
Amino acids and carbohydrates are separate biomolecule classes
Match List I with List II:
| List I | List II |
|---|---|
| A. Two or more alternative forms of a gene | I. Back cross |
| B. Cross of F1 progeny with homozygous recessive parent | II. Ploidy |
| C. Cross of F1 progeny with any of the parents | III. Alleles |
| D. Number of chromosome sets in plant | IV. Test cross |
Answer: (3) A-III, B-IV, C-I, D-II
Explanation:
Alleles = Two or more alternative forms of a gene
Test cross = Cross of F1 progeny with homozygous recessive parent
Back cross = Cross of F1 progeny with any of the parents
Ploidy = Number of chromosome sets in plant
A transcription unit of DNA is defined primarily by:
Answer: (4) Promoter, Structural gene, Terminator
Explanation:
A transcription unit has three regions in DNA:
Promoter – located towards 5′-end (upstream) of structural gene
Structural gene – the coding region
Terminator – located towards 3′-end (downstream) of coding strand
Auxin does not affect mature monocot plants because:
Answer: (3) does not affect mature monocotyledonous plants
Explanation:
Auxin does not affect mature monocot plants
In monocots (especially grasses), there is limited translocation of auxin
Rapid degradation of external auxin occurs in monocots
Major causes of biodiversity loss:
Answer: (4) A, B and D only
Explanation:
Major causes of biodiversity loss:
Habitat loss and fragmentation (largest cause)
Over-exploitation
Alien species invasions
Co-extinctions
Mutation and migration are natural processes, not major causes of biodiversity loss.
Match List I with List II:
| List I | List II |
|---|---|
| A. Clostridium butylicum | I. Ethanol |
| B. Saccharomyces cerevisiae | II. Streptokinase |
| C. Trichoderma polysporum | III. Butyric acid |
| D. Streptococcus sp. | IV. Cyclosporin-A |
Answer: (3) A-III, B-I, C-IV, D-II
Explanation:
Clostridium butylicum → Butyric acid
Saccharomyces cerevisiae → Ethanol (fermentation)
Trichoderma polysporum → Cyclosporin-A (immunosuppressant)
Streptococcus sp. → Streptokinase (clot buster)
Spindle fibers attach to kinetochores of chromosomes during:
Answer: (2) Metaphase
Explanation:
Spindle fibers attach to kinetochores of chromosomes during metaphase
This attachment ensures proper chromosome alignment at the metaphase plate
Prophase: Chromosomes condense
Anaphase: Sister chromatids separate
Telophase: Chromosomes decondense
Example of actinomorphic flower:
Answer: (1) Datura
Explanation:
Actinomorphic flower: Radially symmetrical (can be divided into equal halves in any plane)
Datura shows actinomorphic flowers
Cassia, Pisum, and Sesbania show zygomorphic flowers (bilateral symmetry)
Cofactor of the enzyme carboxypeptidase is:
Answer: (1) Zinc
Explanation:
Carboxypeptidase requires zinc (Zn²⁺) as its cofactor
Niacin is associated with coenzymes NAD and NADP
Haem is the prosthetic group in peroxidase and catalase
Flavin is associated with FAD/FMN
Match List I with List II:
| List I | List II |
|---|---|
| A. Nucleolus | I. Site of formation of glycolipid |
| B. Centriole | II. Organization like the cartwheel |
| C. Leucoplasts | III. Site for active ribosomal RNA synthesis |
| D. Golgi apparatus | IV. For storing nutrients |
Answer: (1) A-III, B-II, C-IV, D-I
Explanation:
Nucleolus → Site of active ribosomal RNA synthesis
Centriole → Cartwheel-like organization (centrioles lie perpendicular to each other)
Leucoplasts → Colorless plastids for storing nutrients
Golgi apparatus → Site for formation of glycoproteins and glycolipids
Capacity to generate a whole plant from any cell:
Answer: (1) Totipotency
Explanation:
Totipotency is the capacity of a single cell to generate a whole plant
This property is used in plant tissue culture and micropropagation
Differentiation: Process of cells becoming specialized
Somatic hybridization: Fusion of protoplasts from different plants
Hind II recognition sequence consists of:
Answer: (2) 6 bp
Explanation:
Hind II was the first restriction endonuclease isolated
It recognizes and cuts at a specific sequence of 6 base pairs
This is a palindromic sequence
Other restriction enzymes may recognize 4, 6, or 8 bp sequences
Set of correct statements about water pollination:
Answer: (4) B, C, D and E only
Explanation:
A is incorrect: Flowers of Vallisneria are NOT colourful and do NOT produce nectar
B is correct: Water lily flowers are pollinated by insects/wind, not water
C is correct: Pollen grains of water-pollinated species are protected from wetting by mucilaginous covering
D is correct: Pollen grains of some hydrophytes are long and ribbon-like
E is correct: In some hydrophytes, pollen grains are carried passively by water currents
Statements about parenchyma, collenchyma, and gymnosperms:
Answer: (4) Statement I is false but Statement II is true
Explanation:
Statement I is false: Both parenchyma AND collenchyma are living tissues
Statement II is true: Gymnosperms lack xylem vessels (have tracheids only), while xylem vessels are characteristic of angiosperms
Formation of interfascicular cambium from parenchyma cells is:
Answer: (3) Dedifferentiation
Explanation:
Dedifferentiation: Process where differentiated cells regain the capacity to divide
Fully differentiated parenchyma cells dedifferentiate to form interfascicular cambium
Differentiation: Cells become specialized
Redifferentiation: Dedifferentiated cells mature into specific cell types
Maturation: Final stage of cell development
Malonate inhibits succinic dehydrogenase due to:
Answer: (3) Competitive inhibition
Explanation:
Malonate has close structural similarity to succinate (the substrate)
It competes with succinate for the active site of succinic dehydrogenase
This is a classic example of competitive inhibition
Feedback inhibition: End product inhibits its own synthesis
Cofactor inhibition: Not applicable here
Identify flower types based on ovary position:
Answer: (4) (a) Perigynous; (b) Perigynous
Explanation:
Perigynous flower: Gynoecium in the center, other parts on the rim of thalamus at the same level
Both diagrams (a) and (b) show perigynous condition
Hypogynous: Ovary superior (above other parts)
Epigynous: Ovary inferior (below other parts)
All Question Paper PDF NEET 2024 PYQ
| Section | NEET UG 2024 Questions Paper Solve |
|---|---|
| Physics | 2024-Physics Question Paper solve |
| Chemistry | 2024-Physics Question Paper solve |
To determine genotype of black seed plant (BB/Bb), cross with:
Answer: (2) bb
Explanation:
This is a test cross to determine if the black seed plant is homozygous (BB) or heterozygous (Bb)
Cross the black seed plant (unknown genotype) with white seed plant (bb – recessive)
If all offspring are black → Parent is BB
If 50% black and 50% white → Parent is Bb
Reasons for greatest species richness in tropical regions:
Answer: (1) A, C, D and E only
Explanation:
A is correct: Tropical latitudes remain relatively undisturbed, allowing more time for diversification
B is incorrect: Tropical environments are less seasonal (not more)
C is correct: More solar energy is available in tropics
D is correct: Constant environments promote niche specialization
E is correct: Tropical environments are constant and predictable
In Verhulst-Pearl logistic growth equation, K indicates:
Answer: (3) Carrying capacity
Explanation:
Logistic growth equation: dN/dt = rN[(K-N)/K]
K = Carrying capacity (maximum population size environment can support)
r = Intrinsic rate of natural increase
N = Population density
Component with thin outer walls and highly thickened inner walls:
Answer: (1) C (Guard cells)
Explanation:
Guard cells of stomata have thin outer walls and thick inner walls
This differential thickening helps in the opening and closing of stomata
The thin outer wall stretches easily, while the thick inner wall is more rigid
Statements about leptotene and diplotene stages:
Answer: (1) Both Statement I and Statement II are true
Explanation:
Statement I correct: Chromosomes gradually become visible under light microscope during leptotene stage
Statement II correct: Beginning of diplotene is recognized by dissolution of synaptonemal complex
During diplotene, homologous chromosomes separate except at chiasmata sites
List of endangered species released by:
Answer: (4) IUCN
Explanation:
IUCN (International Union for Conservation of Nature) releases the Red List of endangered species
GEAC: Genetic Engineering Appraisal Committee
WWF: World Wide Fund for Nature
FOAM: Not a relevant organization
Requirements for dark reaction of photosynthesis:
Answer: (3) C, D and E only
Explanation:
Dark reaction (Calvin cycle) requires:
C. CO₂ (carbon source)
D. ATP (energy source)
E. NADPH (reducing power)
A. Light: Not required for dark reaction (though needed for light reaction)
B. Chlorophyll: Not directly required (pigments are for light absorption)
Type of conservation where threatened species are taken out of natural habitat:
Answer: (2) Biodiversity conservation (Ex-situ conservation)
Explanation:
Ex-situ conservation: Species are taken from natural habitat and placed in special settings (zoos, botanical gardens, seed banks, etc.)
In-situ conservation: Species protected in their natural habitat (national parks, sanctuaries)
The question’s answer options seem to have a typo – it should be “Ex-situ conservation”
Fate of DNA carrying gene of interest transferred into alien organism:
Answer: (3) B and C only
Explanation:
The DNA piece alone cannot multiply independently (A and E are incorrect)
B is correct: It may get integrated into the recipient genome
C is correct: It may multiply and be inherited along with host DNA
D is partially true but not one of the main fates
Lactose is transported into bacteria by:
Answer: (3) Permease
Explanation:
The y gene of lac operon codes for permease
Permease increases permeability of cell to β-galactosides (including lactose)
Beta-galactosidase: Breaks lactose into glucose and galactose
Acetylase: Involved in transacetylation
Polymerase: Involved in DNA replication/transcription
Mendel’s Law of Dominance – correct statements:
Answer: (2) A, C, D and E only
Explanation:
A is correct: One factor is dominant, the other is recessive
B is incorrect: This describes Law of Segregation, not Law of Dominance
C is correct: Factors occur in pairs in normal diploid plants
D is correct: Discrete unit controlling a character is called a factor (gene)
E is correct: Only one parental character is expressed in monohybrid cross (dominant trait)
Not a criterion for classification of fungi:
Answer: (2) Mode of nutrition
Explanation:
Basis for classification of fungi:
Morphology of mycelium
Mode of spore formation
Fruiting body
Mode of nutrition is NOT used as a criterion (all fungi are heterotrophic)
Statements about Bt toxins:
Answer: (3) Statement I is true but Statement II is false
Explanation:
Statement I is correct: Bt toxins are insect-group specific and coded by cry genes (cry IAc, cry IIAb, etc.)
Statement II is false: Bt toxin exists as inactive protoxin, but after ingestion, it gets converted to active form due to alkaline pH of insect gut (not acidic)
Bulliform cells in grasses cause:
Answer: (1) Inward curling of leaves in monocots
Explanation:
Bulliform cells are large, empty, colorless cells in grass leaves
When turgid: Leaves are exposed (water available)
When flaccid (water stress): Leaves curl inward to minimize water loss
This is an adaptation for water conservation
Cross between pink and red flowered Snapdragon:
Answer: (2) Red flowered as well as pink flowered plants
Explanation:
Snapdragon shows incomplete dominance
Pink flower = Rr (heterozygous)
Red flower = RR (homozygous dominant)
Cross: RR × Rr
| R | R | |
|---|---|---|
| R | RR (Red) | RR (Red) |
| r | Rr (Pink) | Rr (Pink) |
Ratio: 2 Red : 2 Pink (50% each)
No white flowers (rr) produced
ATP and NADPH required for CO₂ fixation in Calvin cycle:
Answer: (4) 3 molecules of ATP and 2 molecules of NADPH
Explanation:
For every 1 molecule of CO₂ fixed in Calvin cycle:
3 ATP molecules required
2 NADPH molecules required
These are consumed in the reduction phase of the Calvin cycle
Match List I with List II:
| List I | List II |
|---|---|
| A. Robert May | I. Rivet popper hypothesis |
| B. Alexander von Humboldt | II. Species-area relationship |
| C. Paul Ehrlich | III. Global species diversity (~7 million) |
| D. David Tilman | IV. Long term ecosystem experiments |
Answer: (2) A-III, B-II, C-I, D-IV
Explanation:
Robert May: Placed global species diversity at about 7 million
Alexander von Humboldt: Gave species-area relationship
Paul Ehrlich: Rivet popper hypothesis (role of species in ecosystem)
David Tilman: Long-term ecosystem experiments using outdoor plots
Diagram shows a wind-pollinated plant with:
Answer: (1) Wind pollinated plant inflorescence showing flowers with well exposed stamens
Explanation:
Wind-pollinated flowers have well-exposed stamens for easy pollen dispersal
Compact inflorescence is shown
Stamens are exposed, so complete autogamy does NOT occur
Match List I with List II:
| List I | List II |
|---|---|
| A. Frederick Griffith | I. Genetic code |
| B. Francois Jacob & Jacques Monod | II. Semi-conservative mode of DNA replication |
| C. Har Gobind Khorana | III. Transformation |
| D. Meselson & Stahl | IV. Lac operon |
Answer: (2) A-III, B-IV, C-I, D-II
Explanation:
Frederick Griffith: Transformation experiment (1918)
Francois Jacob & Jacques Monod: Lac operon model
Har Gobind Khorana: Developed chemical method for genetic code determination
Meselson & Stahl: Semi-conservative DNA replication (1958)
Spraying sugarcane with which PGR increases stem length:
Answer: (2) Gibberellin
Explanation:
Gibberellins increase stem elongation
Sugarcane stores sugar in stems; spraying gibberellins increases stem length → increases yield
Auxin: Promotes root growth, apical dominance
Cytokinin: Promotes cell division
Abscisic acid: Stress hormone, promotes dormancy
Match List I with List II:
| List I | List II |
|---|---|
| A. Rose | I. Twisted aestivation |
| B. Pea | II. Perigynous flower |
| C. Cotton | III. Drupe |
| D. Mango | IV. Marginal placentation |
Answer: (1) A-II, B-IV, C-I, D-III
Explanation:
Rose: Perigynous flower (half-inferior ovary)
Pea: Marginal placentation (ovules borne on ridge along ventral suture)
Cotton: Twisted aestivation
Mango: Drupe fruit
Section | Number of Questions | Marks for Correct Answer | Negative Marking | Total Marks |
Physics | 45 | +4 | −1 | 180 |
Chemistry | 45 | +4 | −1 | 180 |
Biology (Botany + Zoology) | 90 | +4 | −1 | 360 |
Total | 180 | 720 |
Correct statements about Phaeophyceae:
Answer: (3) A, C, D and E only
Explanation:
A is correct: Asexual reproduction by biflagellate zoospores
B is incorrect: Sexual reproduction can be oogamous, isogamous, or anisogamous
C is correct: Stored food is mannitol or laminarin
D is correct: Major pigments: chlorophyll a, c, carotenoids, xanthophylls
E is correct: Cellulosic wall covered by gelatinous coating of algin
Step in TCA cycle that does NOT involve oxidation:
Answer: (3) Succinyl-CoA → Succinic acid
Explanation:
Oxidation involves loss of electrons (increase in oxidation state)
Succinyl-CoA → Succinic acid is a hydrolysis reaction (no oxidation)
All other options involve oxidation:
Malic acid → Oxaloacetic acid (oxidation)
Succinic acid → Malic acid (oxidation)
Isocitrate → α-ketoglutaric acid (oxidation)
Statements about C3 and C4 plants:
Answer: (3) Statement I is true but Statement II is false
Explanation:
Statement I is correct: In C3 plants, O₂ competes with CO₂ at RuBisCO (photorespiration), decreasing CO₂ fixation
Statement II is false: Photorespiration does NOT occur in C4 plants (they lack RuBisCO in mesophyll cells)
Match List I with List II:
| List I | List II |
|---|---|
| A. GLUT-4 | I. Hormone |
| B. Insulin | II. Enzyme |
| C. Trypsin | III. Intercellular ground substance |
| D. Collagen | IV. Enables glucose transport into cells |
Answer: (1) A-IV, B-I, C-II, D-III
Explanation:
GLUT-4: Enables glucose transport into cells (glucose transporter)
Insulin: Hormone (regulates blood glucose)
Trypsin: Enzyme (digests proteins)
Collagen: Intercellular ground substance (structural protein)
Match List I with List II:
| List I | List II |
|---|---|
| A. Monoadelphous | I. Citrus |
| B. Diadelphous | II. Pea |
| C. Polyadelphous | III. Lily |
| D. Epiphyllous | IV. China-rose |
Answer: (1) A-IV, B-II, C-I, D-III
Explanation:
Monoadelphous (stamens united in one bundle) → China-rose
Diadelphous (stamens united in two bundles) → Pea
Polyadelphous (stamens united in many bundles) → Citrus
Epiphyllous (stamens attached to petals) → Lily
DNA present in chloroplast is:
Answer: (2) Circular, double stranded
Explanation:
Chloroplast DNA is:
Circular (not linear)
Double-stranded
Similar to prokaryotic DNA and mitochondrial DNA
This supports the endosymbiotic theory
GPP of third trophic level if NPP of first trophic level is 100x:
Answer: (3) 10x (kcal m⁻²) yr⁻¹
Explanation:
Only about 10% energy is transferred from one trophic level to next
NPP at first trophic level = GPP for second trophic level = 100x
NPP at second trophic level = 10% of 100x = 10x = GPP for third trophic level
Therefore, GPP of third trophic level = 10x
Fused in somatic hybridization:
Answer: (3) Protoplasts
Explanation:
Somatic hybridization: Fusion of protoplasts (cells without cell wall) from two different plant varieties/species
This produces somatic hybrids
Callus: Mass of undifferentiated cells
Somatic embryos: Embryos from somatic cells
Pollens: Male gametophytes
Match List I with List II:
| List I | List II |
|---|---|
| A. Citric acid cycle | I. Cytoplasm |
| B. Glycolysis | II. Mitochondrial matrix |
| C. Electron transport system | III. Intermembrane space of mitochondria |
| D. Proton gradient | IV. Inner mitochondrial membrane |
Answer: (2) A-II, B-I, C-IV, D-III
Explanation:
Citric acid cycle → Mitochondrial matrix
Glycolysis → Cytoplasm/Cytosol
Electron transport system → Inner mitochondrial membrane
Proton gradient → Intermembrane space of mitochondria
Correct statement about replication in E. coli:
Answer: (4) DNA dependent DNA polymerase catalyses polymerization in 5’→3′ direction
Explanation:
DNA polymerase catalyses polymerization in 5’→3′ direction only
It cannot polymerize in 3’→5′ direction
The enzyme requires a primer to initiate synthesis
DNA-dependent RNA polymerase is involved in transcription, not replication
Which is not a steroid hormone?
Answer: (4) Glucagon
Explanation:
Glucagon is a proteinaceous/peptide hormone secreted by pancreas (alpha cells)
Cortisol: Steroid hormone (adrenal cortex)
Testosterone: Steroid hormone (testes)
Progesterone: Steroid hormone (ovaries)
Statements about hymen:
Answer: (3) Statement I is true but Statement II is false
Explanation:
Statement I is correct: Presence or absence of hymen is NOT a reliable indicator of virginity
Statement II is false: Hymen can be broken by:
Sudden jolt
Insertion of vaginal tampon
Active participation in sports
In some women, hymen persists even after coitus
Incorrect statement about bioreactors:
Answer: (3) Bio-reactors are used to produce small scale bacterial cultures
Explanation:
Bioreactors are used for large volumes (100-1000 litres) of culture
Small volumes cannot yield appreciable quantities of products
Bioreactors provide optimal growth conditions
Most commonly used bioreactors are of stirring type
Have agitator system, oxygen delivery system, and foam control system
Product of DNA dependent RNA polymerase for given template:
Answer: (1) 5’AUGUACCGUUUUAUGGUUAAGU3′
Explanation:
Template: 3’TACATGGCAATATCCATTCAS’
mRNA is complementary and antiparallel to template strand
RNA polymerase synthesizes mRNA in 5’→3′ direction
mRNA sequence: 5’AUGUACCGUUUUAUGGUUAAGU3′
Note: In RNA, T is replaced by U
Assertion and Reason about FSH:
Answer: (4) A is false but R is true
Explanation:
Assertion A is false: FSH acts on ovarian follicles in females, but in males, LH (not FSH) acts on Leydig cells
Reason R is true: Growing ovarian follicles secrete estrogen in females; interstitial cells (Leydig) secrete androgens in males
LH stimulates Leydig cells to produce testosterone
Match List I with List II:
| List I | List II |
|---|---|
| A. Typhoid | I. Fungus |
| B. Leishmaniasis | II. Nematode |
| C. Ringworm | III. Protozoa |
| D. Filariasis | IV. Bacteria |
Answer: (2) A-IV, B-III, C-I, D-II
Explanation:
Typhoid → Bacteria (Salmonella typhimurium)
Leishmaniasis → Protozoa (Leishmania donovani)
Ringworm → Fungus (Microsporum, Trichophyton, Epidermophyton)
Filariasis → Nematode (Wuchereria bancrofti/malayi)
Correct pathway of conduction of action potential through heart:
Answer: (1) E-C-A-D-B
Explanation:
SA node (E) → AV node (C) → AV bundle (A) → Bundle branches (D) → Purkinje fibres (B)
SA node: Pacemaker of heart
AV node: Delays impulse slightly
AV bundle (Bundle of His): Conducts impulse to ventricles
Purkinje fibres: Distribute impulse throughout ventricles
Match sub-phases of Prophase I:
| List I | List II |
|---|---|
| A. Diakinesis | I. Synaptonemal complex formation |
| B. Pachytene | II. Completion of terminalisation of chiasmata |
| C. Zygotene | III. Chromosomes look like thin threads |
| D. Leptotene | IV. Appearance of recombination nodules |
Answer: (3) A-II, B-IV, C-I, D-III
Explanation:
Diakinesis: Completion of terminalisation of chiasmata
Pachytene: Appearance of recombination nodules (crossing over occurs)
Zygotene: Synaptonemal complex formation (pairing of homologous chromosomes)
Leptotene: Chromosomes look like thin threads (start condensing)
Identify correct matching of muscles with location:
Answer: (2) (a) Skeletal – Triceps, (b) Smooth – Stomach, (c) Cardiac – Heart
Explanation:
Figure (a): Skeletal muscle (striated, multinucleated) → Triceps/Biceps
Figure (b): Smooth muscle (non-striated, spindle-shaped) → Stomach, intestine, blood vessels
Figure (c): Cardiac muscle (striated, branched, intercalated discs) → Heart
Match List I with List II:
| List I | List II |
|---|---|
| A. Non-medicated IUD | I. Multiload 375 |
| B. Copper releasing IUD | II. Progestogens |
| C. Hormone releasing IUD | III. Lippes loop |
| D. Implants | IV. LNG-20 |
Answer: (4) A-III, B-I, C-IV, D-II
Explanation:
Non-medicated IUD: Lippes loop
Copper releasing IUD: Multiload 375, Cu-T
Hormone releasing IUD: LNG-20 (releases levonorgestrel)
Implants: Progestogens (e.g., Norplant)
Match List I with List II:
| List I | List II |
|---|---|
| A. Lipase | I. Peptide bond |
| B. Nuclease | II. Ester bond |
| C. Protease | III. Glycosidic bond |
| D. Amylase | IV. Phosphodiester bond |
Answer: (3) A-II, B-IV, C-I, D-III
Explanation:
Lipase: Breaks ester bonds (in lipids)
Nuclease: Breaks phosphodiester bonds (in nucleic acids)
Protease: Breaks peptide bonds (in proteins)
Amylase: Breaks glycosidic bonds (in carbohydrates/starch)
Assertion and Reason about breast-feeding:
Answer: (1) Both A and R are correct and R is the correct explanation of A
Explanation:
Assertion A: Breast-feeding during initial period is recommended for healthy baby
Reason R: Colostrum contains antibodies essential for developing resistance
Colostrum is rich in IgA antibodies (passive immunity)
This protects the newborn from infections
R correctly explains A
Correct sequence of human evolution (Past to Recent):
Answer: (4) A-D-C-B
Explanation:
A. Homo habilis (2.4-1.4 mya) → First tool maker
D. Homo erectus (1.9-0.1 mya) → First to use fire, migrate
C. Homo neanderthalensis (400,000-40,000 years ago)
B. Homo sapiens (300,000 years ago to present)
Match List I with List II:
| List I | List II |
|---|---|
| A. Axoneme | I. Centriole |
| B. Cartwheel pattern | II. Cilia and flagella |
| C. Crista | III. Chromosome |
| D. Satellite | IV. Mitochondria |
Answer: (4) A-II, B-I, C-IV, D-III
Explanation:
Axoneme: Found in cilia and flagella (9+2 arrangement of microtubules)
Cartwheel pattern: Seen in centrioles (9 triplets arrangement)
Crista: Found in mitochondria (inner membrane folds)
Satellite: Present in chromosomes (secondary constriction)
Match List I with List II:
| List I | List II |
|---|---|
| A. Pterophyllum | I. Hag fish |
| B. Myxine | II. Saw fish |
| C. Pristis | III. Angel fish |
| D. Exocoetus | IV. Flying fish |
Answer: (2) A-III, B-I, C-II, D-IV
Explanation:
Pterophyllum → Angel fish (freshwater fish)
Myxine → Hag fish (jawless fish)
Pristis → Saw fish (cartilaginous fish)
Exocoetus → Flying fish (bony fish)
Match List I with List II:
| List I | List II |
|---|---|
| A. α-1 antitrypsin | I. Cotton bollworm |
| B. Cry IAb | II. ADA deficiency |
| C. Cry IAc | III. Emphysema |
| D. Enzyme replacement therapy | IV. Corn borer |
Answer: (3) A-III, B-IV, C-I, D-II
Explanation:
α-1 antitrypsin → Treatment of emphysema
Cry IAb gene → Controls corn borer
Cry IAc gene → Controls cotton bollworms
Enzyme replacement therapy → Treatment option for ADA deficiency
“Ti plasmid” of Agrobacterium tumefaciens stands for:
Answer: (3) Tumor inducing plasmid
Explanation:
Ti plasmid = Tumor inducing plasmid
Contains T-DNA (transfer DNA) which causes tumors in dicot plants
Used as a vector in plant genetic engineering
Not “tumour inhibiting” or “temperature independent”
Match List I with List II:
| List I | List II |
|---|---|
| A. Pleurobrachia | I. Mollusca |
| B. Radula | II. Ctenophora |
| C. Stomochord | III. Osteichthyes |
| D. Air bladder | IV. Hemichordata |
Answer: (2) A-II, B-I, C-IV, D-III
Explanation:
Pleurobrachia → Phylum Ctenophora (comb jellies)
Radula → Phylum Mollusca (rasping feeding organ)
Stomochord → Phylum Hemichordata (rudimentary notochord-like structure)
Air bladder → Class Osteichthyes (provides buoyancy)
Which is not a natural/traditional contraceptive method?
Answer: (4) Vaults
Explanation:
Vaults are barrier methods (rubber cap covering cervix)
Natural methods include:
Coitus interruptus (withdrawal)
Periodic abstinence (avoiding coitus during fertile period)
Lactational amenorrhea (no ovulation during intense breastfeeding)
Correct sequence of stages of cell division:
Answer: (4) E-C-A-D-B
Explanation:
E. Gap 1 phase (G₁) → Cell growth