NEET UGC Chemistry Question Paper for 2024

BOTANY SECTION-A

Question 1

Match List I with List II:

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

Question 2

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

Question 3

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

Question 4

Match List I with List II:

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

Question 5

A transcription unit of DNA is defined primarily by:

Answer: (4) Promoter, Structural gene, Terminator

Explanation:
A transcription unit has three regions in DNA:

1. Promoter – located towards 5′-end (upstream) of structural gene

2. Structural gene – the coding region

3. Terminator – located towards 3′-end (downstream) of coding strand

Question 6

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

Question 7

Major causes of biodiversity loss:

Answer: (4) A, B and D only

Explanation:
Major causes of biodiversity loss:

1. Habitat loss and fragmentation (largest cause)

2. Over-exploitation

3. Alien species invasions

4. Co-extinctions

Mutation and migration are natural processes, not major causes of biodiversity loss.

Question 8

Match List I with List II:

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)

Question 9

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

Question 10

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)

Question 11

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

Question 12

Match List I with List II:

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

Question 13

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

Question 14

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

Question 15

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

Question 16

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

Question 17

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

Question 18

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

Question 19

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)

Question 20

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

Question 21

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

Question 22

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

Question 23

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

Question 24

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

Question 25

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

Question 26

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)

Question 27

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”

Question 28

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

Question 29

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

Question 30

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)

Question 31

Not a criterion for classification of fungi:

Answer: (2) Mode of nutrition

Explanation:

• Basis for classification of fungi:

  1. Morphology of mycelium

  2. Mode of spore formation

  3. Fruiting body

• Mode of nutrition is NOT used as a criterion (all fungi are heterotrophic)

Question 32

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)

Question 33

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

Question 34

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

• Ratio: 2 Red : 2 Pink (50% each)

• No white flowers (rr) produced

Question 35

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

Question 36

Match List I with List II:

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

Question 37

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

Question 38

Match List I with List II:

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)

Question 39

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

Question 40

Match List I with List II:

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

Question 41

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

Question 42

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)

Question 43

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)

Question 44

Match List I with List II:

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)

Question 45

Match List I with List II:

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

Question 46

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

Question 47

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

Question 48

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

Question 49

Match List I with List II:

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

Question 50

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

ZOOLOGY SECTION-A

Question 51

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)

Question 52

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

Question 53

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

Question 54

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

Question 55

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

Question 56

Match List I with List II:

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)

Question 57

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

Question 58

Match sub-phases of Prophase I:

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)

Question 59

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

Question 60

Match List I with List II:

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)

Question 61

Match List I with List II:

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)

Question 62

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

Question 63

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)

Question 64

Match List I with List II:

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)

Question 65

Match List I with List II:

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)

Question 66

Match List I with List II:

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

Question 67

“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”

Question 68

Match List I with List II:

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)

Question 69

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)

Question 70

Correct sequence of stages of cell division:

Answer: (4) E-C-A-D-B

Explanation:

• E. Gap 1 phase (G₁) → Cell growth