NEET Biology MCQ 2026booklet set 80 with Answers & Explanations

 - Dr.Sanjaykumar Pawar  

90 NEET Biology MCQs booklet code 80,2026 with answers and explanations — covering Human Physiology, Genetics, Ecology, Botany and Zoology.

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Cell: The Unit of Life NEET Mock Test – 45 MCQs — 

CBSE Class 11 Biology Notes: Introduction to Plant Physiology — anchor text: "revise Plant Physiology concepts"

Small Molecular Weight Organic Compounds in Living Tissues (Biomolecules Notes) — 

Biology MCQ Set — Q.91 to Q.180

Complete questions with correct answers and explanations

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Q91.

The number of vertebrae in a human is ____.

• (1) 12
• (2) 26 ✓
• (3) 206
• (4) 7

Answer: Option (2)

Explanation: The adult human vertebral column has 26 vertebrae (7 cervical + 12 thoracic + 5 lumbar + 1 sacrum (fused) + 1 coccyx (fused)). Note: 206 is the total number of bones in the human body, not vertebrae.

Q92.

Which one of the following statements is incorrect?

• (1) α-cells of pancreas secrete insulin ✓
• (2) Glucagon stimulates glycogenolysis
• (3) β-cells of pancreas secrete insulin
• (4) α-cells of pancreas secrete glucagon

Answer: Option (1)

Explanation: α-cells of the islets of Langerhans secrete glucagon, not insulin. Insulin is secreted by β-cells. Hence statement (1) is incorrect.

Q93.

Given below are two statements:
Statement I: The class name Reptilia refers to creeping or crawling mode of locomotion.
Statement II: All organisms belonging to Reptilia have three chambered heart.
In light of the above statements, choose the most appropriate answer.

• (1) Both Statement I and Statement II are incorrect
• (2) Statement I is correct but Statement II is incorrect ✓
• (3) Statement I is incorrect but Statement II is correct
• (4) Both Statement I and Statement II are correct

Answer: Option (2)

Explanation: Statement I is correct — 'Reptilia' comes from Latin 'repere' meaning to creep/crawl. Statement II is incorrect because most reptiles have a 3-chambered heart, but crocodiles have a 4-chambered heart.

Q94.

Which of the following statements related to pituitary gland are correct?
(a) It is divided anatomically into adenohypophysis and neurohypophysis
(b) It secretes follicle stimulating hormone
(c) It secretes melanocyte stimulating hormone
(d) It does not secrete prolactin

• (1) (a), (b) and (c) only ✓
• (2) (c) and (d) only
• (3) (b) and (c) only
• (4) (a) and (b) only

Answer: Option (1)

Explanation: The pituitary is divided into adenohypophysis (pars distalis + pars intermedia) and neurohypophysis (a, correct). It secretes FSH (b, correct) and MSH from pars intermedia (c, correct). It DOES secrete prolactin, so (d) is false.

Q95.

Match List I with List II.
List I: A. Starch, B. Antibody, C. Concanavalin A, D. Glut-4
List II: I. Fights infection, II. Energy storage, III. Glucose transport, IV. Lectin

• (1) A-II, B-I, C-IV, D-III ✓
• (2) A-II, B-I, C-III, D-IV
• (3) A-I, B-II, C-III, D-IV
• (4) A-I, B-II, C-IV, D-III

Answer: Option (1)

Explanation: Starch = energy storage polysaccharide (II); Antibody = fights infection (I); Concanavalin A = a lectin (IV); Glut-4 = glucose transporter protein (III).

Q96.

In water, frogs respire using ____.

• (1) buccal cavity
• (2) lungs
• (3) trachea
• (4) skin ✓

Answer: Option (4)

Explanation: Frogs are bimodal breathers. In water, gaseous exchange occurs mainly through the moist, vascularized skin (cutaneous respiration).

Q97.

Match List I with List II.
List I: A. Cristae, B. Cisternae, C. Thylakoids, D. Phospholipid
List II: I. Flat membrane sacs in stroma of chloroplast, II. Infoldings in mitochondria, III. Cell membrane, IV. Disc shaped sacs in the Golgi apparatus

• (1) A-II, B-IV, C-I, D-III ✓
• (2) A-II, B-IV, C-III, D-I
• (3) A-IV, B-III, C-I, D-II
• (4) A-III, B-IV, C-I, D-II

Answer: Option (1)

Explanation: Cristae = infoldings of the inner mitochondrial membrane (II); Cisternae = disc-shaped sacs of the Golgi (IV); Thylakoids = flat membranous sacs in the chloroplast stroma (I); Phospholipids are the main component of the cell membrane (III).

Q98.

Phyllotaxy is the pattern of arrangement of ____.

• (1) flowers
• (2) fruits
• (3) sepals
• (4) leaves ✓

Answer: Option (4)

Explanation: Phyllotaxy (phyllotaxis) refers to the pattern in which leaves are arranged on the stem/branch — alternate, opposite, or whorled.

Q99.

Match List I with List II.
List I: A. Spherical, B. Rod, C. Comma, D. Spirillum
List II: I. Vibrio, II. Cocci, III. Spirilla, IV. Bacilli

• (1) A-III, B-II, C-I, D-IV
• (2) A-II, B-I, C-IV, D-III
• (3) A-II, B-IV, C-I, D-III ✓
• (4) A-I, B-III, C-II, D-IV

Answer: Option (3)

Explanation: Spherical bacteria are Cocci (II); Rod-shaped are Bacilli (IV); Comma-shaped are Vibrio (I); Spirillum-shaped are Spirilla (III).

Q100.

Arrange the following elements in descending order of their contribution to the percentage weight of the human body.
(a) Oxygen (b) Carbon (c) Hydrogen (d) Nitrogen

• (1) (c), (a), (b), (d)
• (2) (b), (c), (d), (a)
• (3) (b), (a), (c), (d)
• (4) (a), (b), (c), (d) ✓

Answer: Option (4)

Explanation: By mass, the human body is composed of roughly Oxygen (~65%) > Carbon (~18%) > Hydrogen (~10%) > Nitrogen (~3%), so the descending order is O > C > H > N, i.e. (a) > (b) > (c) > (d).

Q101.

Length of the stem at time 0 is 20 cm. The arithmetic growth rate is 30 cm/day. What is the length of the stem at the end of the 7th day?

• (1) 170 cm
• (2) 230 cm ✓
• (3) 460 cm
• (4) 50 cm

Answer: Option (2)

Explanation: For arithmetic growth: L_t = L₀ + rt = 20 + (30 × 7) = 20 + 210 = 230 cm.

Q102.

Given below are two statements:
Statement I: In gymnosperms, the male and female gametophytes remain within the sporangia.
Statement II: In gymnosperms, seeds are not covered.
In light of the above statements, choose the most appropriate answer.

• (1) Both Statement I and Statement II are incorrect
• (2) Statement I is correct but Statement II is incorrect
• (3) Statement I is incorrect but Statement II is correct
• (4) Both Statement I and Statement II are correct ✓

Answer: Option (4)

Explanation: Both statements are true. In gymnosperms the gametophytes are retained within the sporangia and are dependent on the sporophyte; and the seeds are 'naked' (not enclosed within a fruit), which is exactly what the term 'gymnosperm' (naked seed) means.

Q103.

Smooth endoplasmic reticulum ____.

• (1) is the major site for the synthesis of lipids ✓
• (2) is actively involved in protein synthesis
• (3) is a site for the synthesis of carbohydrates
• (4) has ribosomes attached to its surface

Answer: Option (1)

Explanation: Smooth ER lacks ribosomes and is the principal site of lipid/steroid synthesis. Protein synthesis and ribosome attachment are features of rough ER.

Q104.

Which of the following is not a characteristic of chordates?

• (1) Central nervous system is dorsal
• (2) Absence of gills ✓
• (3) Presence of post anal part (tail)
• (4) Presence of notochord

Answer: Option (2)

Explanation: Chordates possess pharyngeal gill slits at some stage of life (present in fishes; present as embryonic pouches in others). Hence 'absence of gills' is NOT a chordate character.

Q105.

How many molecules of pyruvic acid are produced at the end of glycolysis from 206 molecules of glucose?

• (1) 309
• (2) 103
• (3) 412 ✓
• (4) 206

Answer: Option (3)

Explanation: Glycolysis converts each glucose molecule into 2 molecules of pyruvic acid. So 206 glucose × 2 = 412 pyruvic acid molecules.

Q106.

The correct sequence of adult cell cycle phases is ____.

• (1) G1-M-G2-S
• (2) G1-S-G2-M ✓
• (3) S-M-G2-G1
• (4) G1-S-G2-S-M

Answer: Option (2)

Explanation: The standard cell cycle sequence is G1 (growth) → S (DNA synthesis) → G2 (growth/preparation) → M (mitosis).

Q107.

Which of the following represents the correct sequence of arrangement of bones in the lower limb of humans?

• (1) Patella-femur-tibia-tarsal
• (2) Femur-patella-tibia-tarsal ✓
• (3) Femur-tarsal-patella-tibia
• (4) Femur-tibia-patella-tarsal

Answer: Option (2)

Explanation: From proximal to distal in the lower limb: Femur (thigh) → Patella (knee cap) → Tibia (shin bone) → Tarsals (ankle bones).

Q108.

Cell theory was formulated by ____.

• (1) Robert Brown
• (2) Singer and Nicolson
• (3) Antonie Von Leeuwenhoek
• (4) Schleiden and Schwann ✓

Answer: Option (4)

Explanation: The cell theory was proposed by Matthias Schleiden and Theodor Schwann (1838–39), later extended by Rudolf Virchow ('Omnis cellula-e cellula').

Q109.

Given below are two statements:
Statement I: Chromosomes are fully condensed at the end of prophase I.
Statement II: Meiosis I resembles mitosis.
In light of the above statements, choose the most appropriate answer.

• (1) Both Statement I and Statement II are false
• (2) Statement I is correct, but Statement II is false ✓
• (3) Statement I is incorrect, but Statement II is true
• (4) Both Statement I and Statement II are true

Answer: Option (2)

Explanation: By the end of prophase I, chromosomes are fully condensed and visible (Statement I true). Meiosis I is a reductional division (homologous chromosomes separate, chromosome number is halved) and is fundamentally different from mitosis; it is meiosis II that resembles mitosis. So Statement II is false.

Q110.

Which of the following statements regarding photorespiration are correct?
(a) Do not occur in C3 plants
(b) CO₂ is consumed and O₂ is generated
(c) Phosphoglycolate is formed
(d) No synthesis of ATP and NADPH

• (1) (c) and (d) only ✓
• (2) (b) and (d) only
• (3) (a) and (b) only
• (4) (a) and (d) only

Answer: Option (1)

Explanation: Photorespiration occurs in C3 plants when RuBisCO acts as an oxygenase (so (a) is false). It consumes O₂ and releases CO₂ — the opposite of (b), so (b) is false. RuBisCO oxygenase activity produces 2-phosphoglycolate (c, true) and the pathway yields no useful ATP/NADPH, unlike photosynthesis (d, true). Correct: (c) and (d).

Q111.

Which of the following are characteristics of prokaryotic cells?
(a) Ribosomes are made of 50S and 30S subunits
(b) They can have plasmids
(c) They contain mesosome
(d) They have peroxisomes

• (1) (a) and (c) only
• (2) (a), (c) and (d) only
• (3) (a), (b) and (c) only ✓
• (4) (b) and (c) only

Answer: Option (3)

Explanation: Prokaryotic ribosomes (70S) are made of 50S + 30S subunits (a, true). Plasmids are common in bacteria (b, true). Mesosomes are infoldings of the plasma membrane found in prokaryotes (c, true). Peroxisomes are membrane-bound organelles found only in eukaryotic cells, so (d) is false.

Q112.

The number of action potentials generated by the sino-atrial node (SAN) in a healthy human is ____ per minute.

• (1) 70 - 75 ✓
• (2) 100 - 110
• (3) 120 - 140
• (4) 28 - 30

Answer: Option (1)

Explanation: The SAN (the natural pacemaker of the heart) generates action potentials at the resting heart rate of about 70–75 beats per minute in a healthy adult.

Q113.

Endomembrane system includes ____.

• (1) endoplasmic reticulum, chloroplast, peroxisomes and vacuole
• (2) mitochondria, chloroplast, peroxisomes and vacuole
• (3) Golgi complex, chloroplast, peroxisomes and vacuole
• (4) endoplasmic reticulum, Golgi complex, lysosomes and vacuole ✓

Answer: Option (4)

Explanation: The endomembrane system consists of organelles that coordinate through direct contact or vesicle transport: ER, Golgi complex, lysosomes, and vacuoles. Mitochondria, chloroplasts, and peroxisomes are NOT part of the endomembrane system since their functions are not coordinated with it.

Q114.

Match List I with List II (taxonomic hierarchy of Mango).
List I: A. Family, B. Genus, C. Class, D. Phylum, E. Order
List II: I. Sapindales, II. Dicotyledonae, III. Anacardiaceae, IV. Angiospermae, V. Mangifera

• (1) A-II, B-I, C-III, D-IV, E-V
• (2) A-II, B-III, C-V, D-I, E-IV
• (3) A-III, B-V, C-II, D-IV, E-I ✓
• (4) A-I, B-V, C-II, D-IV, E-III

Answer: Option (3)

Explanation: Classification of Mango: Family – Anacardiaceae (III), Genus – Mangifera (V), Class – Dicotyledonae (II), Phylum/Division – Angiospermae (IV), Order – Sapindales (I).

Q115.

Which of the following plant growth regulators promotes internode elongation prior to flowering in cabbage?

• (1) Gibberellin ✓
• (2) Indole butyric acid
• (3) Ethephon
• (4) Abscisic acid

Answer: Option (1)

Explanation: Gibberellins are well known for causing bolting (rapid internode elongation) in plants like cabbage and beet just before flowering.

Q116.

The plastid that stores xanthophyll is known as ____.

• (1) chromoplast ✓
• (2) aleuroplast
• (3) amyloplast
• (4) chloroplast

Answer: Option (1)

Explanation: Chromoplasts store carotenoid pigments such as carotenes and xanthophylls, giving fruits and flowers their yellow, orange, or red colours.

Q117.

Photorespiration reaction catalysed by RuBisCo is shown below:
RuBP + O₂ → 3-Phosphoglycerate + X
Identify 'X' from the given options.

• (1) 2-Phosphoglycolate ✓
• (2) Oxaloacetate
• (3) Malate
• (4) Phosphoenolpyruvate

Answer: Option (1)

Explanation: In the oxygenase reaction of RuBisCO, RuBP + O₂ yields one molecule of 3-phosphoglycerate and one molecule of 2-phosphoglycolate.

Q118.

Match List I with List II.
List I: A. Marginal placentation, B. Axile placentation, C. Parietal placentation, D. Free central placentation
List II: I. Argemone, II. Tomato, III. Primrose, IV. Pea

• (1) A-IV, B-II, C-III, D-I
• (2) A-IV, B-III, C-I, D-II
• (3) A-IV, B-II, C-I, D-III ✓
• (4) A-II, B-IV, C-I, D-III

Answer: Option (3)

Explanation: Marginal placentation – Pea (IV); Axile placentation – Tomato (II); Parietal placentation – Argemone (I); Free central placentation – Primrose (III).

Q119.

Which of the following are characteristic features of Solanaceae family?
(a) Flowers are bisexual and actinomorphic
(b) Calyx have five sepals and are united
(c) Androecium have five stamens and are epipetalous
(d) Ovary is inferior

• (1) (d) only
• (2) (a) and (b) only
• (3) (b), (c) and (d) only
• (4) (a), (b) and (c) only ✓

Answer: Option (4)

Explanation: Solanaceae flowers are typically bisexual, actinomorphic, with 5 united sepals, 5 epipetalous stamens — (a), (b), (c) are correct. However, the ovary in Solanaceae is superior, not inferior, so (d) is incorrect.

Q120.

Which pigment has absorption peak at 700 nm in the photosynthetic reaction centre PS I (P700)?

• (1) Chlorophyll a ✓
• (2) Xanthophylls
• (3) Carotenoids
• (4) Chlorophyll b

Answer: Option (1)

Explanation: P700 refers to a special form of Chlorophyll a at the reaction centre of Photosystem I, which shows maximum light absorption near 700 nm.

Q121.

Mitochondrial inner membrane encloses ____.

• (1) cytosol
• (2) mucus
• (3) aqueous humor
• (4) matrix ✓

Answer: Option (4)

Explanation: The inner mitochondrial membrane, thrown into folds called cristae, encloses the mitochondrial matrix, which contains enzymes of the Krebs cycle, mitochondrial DNA, and ribosomes.

Q122.

Which of the following statements is incorrect?

• (1) Blood clot consists of fibrins
• (2) Fibrin is produced from fibrinogen
• (3) Fibrinogen is produced from fibrin ✓
• (4) Blood coagulates in response to an injury

Answer: Option (3)

Explanation: The correct sequence is fibrinogen (soluble plasma protein) → fibrin (insoluble threads), catalysed by thrombin. The reverse — 'fibrinogen is produced from fibrin' — is incorrect.

Q123.

Symbiotic association between fungi and algae are called ____.

• (1) sponges
• (2) mycorrhiza
• (3) chrysophytes
• (4) lichens ✓

Answer: Option (4)

Explanation: Lichens are a symbiotic/mutualistic association between a fungus (mycobiont) and a photosynthetic alga or cyanobacterium (phycobiont). (Mycorrhiza is a fungus-root association, not fungus-alga.)

Q124.

Which of the following is not a part of the human central nervous system?

• (1) Dura mater
• (2) Pia mater
• (3) Pericardium ✓
• (4) Arachnoid

Answer: Option (3)

Explanation: Dura mater, pia mater, and arachnoid are the three meninges protecting the brain and spinal cord (CNS). The pericardium is the membrane covering the heart, unrelated to the nervous system.

Q125.

Which of the following plant growth regulators is used as a herbicide?

• (1) Kinetin
• (2) Abscisic acid
• (3) Gibberellin
• (4) 2,4-D ✓

Answer: Option (4)

Explanation: 2,4-Dichlorophenoxyacetic acid (2,4-D) is a synthetic auxin widely used as a selective herbicide to kill broad-leaved (dicot) weeds without harming monocot crops.

Q126.

How many turns of the Calvin cycle are required for the formation of three molecules of glucose?

• (1) 3
• (2) 1
• (3) 18 ✓
• (4) 6

Answer: Option (3)

Explanation: One turn of the Calvin cycle fixes one CO₂. Forming one glucose molecule (6 carbons) requires 6 turns. For 3 molecules of glucose: 6 × 3 = 18 turns.

Q127.

Arrange the following taxonomic categories in ascending order.
(a) Genus (b) Class (c) Order (d) Phylum (e) Family (f) Kingdom (g) Species

• (1) (a), (c), (d), (g), (f), (b), (e)
• (2) (g), (c), (d), (b), (e), (a), (f)
• (3) (f), (c), (b), (g), (d), (e), (a)
• (4) (g), (a), (e), (c), (b), (d), (f) ✓

Answer: Option (4)

Explanation: Ascending taxonomic hierarchy: Species → Genus → Family → Order → Class → Phylum → Kingdom, i.e. (g) → (a) → (e) → (c) → (b) → (d) → (f).

Q128.

In frogs, the number of pairs of cranial nerves arising from the brain are ____.

• (1) 9
• (2) 10 ✓
• (3) 12
• (4) 6

Answer: Option (2)

Explanation: Like most amphibians, frogs have 10 pairs of cranial nerves arising from the brain (unlike mammals, which have 12 pairs).

Q129.

Select the correct sequence of experiments that led to a gradual understanding of photosynthesis in green plants.

• (1) Role of air → release of oxygen → production of glucose → absorption spectra of chlorophyll a and b ✓
• (2) Release of oxygen → production of glucose → absorption spectra → role of air
• (3) Production of glucose → release of oxygen → absorption spectra → role of air
• (4) Absorption spectra of chlorophyll a and b → production of glucose → release of oxygen → role of air

Answer: Option (1)

Explanation: Historically: Priestley/Ingenhousz established the role of air and light; van Niel/others established O₂ release; further work established glucose production; and finally the action/absorption spectra of chlorophyll a and b were characterised.

Q130.

Sphenopsida class belongs to ____.

• (1) angiosperms
• (2) gymnosperms
• (3) pteridophytes ✓
• (4) bryophytes

Answer: Option (3)

Explanation: Pteridophytes are classified into four classes: Psilopsida, Lycopsida, Sphenopsida, and Pteropsida. Sphenopsida (e.g., Equisetum) is one of the pteridophyte classes.

Q131.

Match List I with List II.
List I: A. Fusion of protoplasms between gametes, B. Fusion of two nuclei, C. Generation of haploid spores
List II: I. Meiosis, II. Plasmogamy, III. Karyogamy

• (1) A-II, B-I, C-III
• (2) A-III, B-II, C-I
• (3) A-I, B-III, C-II
• (4) A-II, B-III, C-I ✓

Answer: Option (4)

Explanation: Fusion of protoplasms between two gametes = Plasmogamy (II); Fusion of two (haploid) nuclei = Karyogamy (III); Generation of haploid spores = Meiosis (I).

Q132.

Which of the following is not a prokaryote?

• (1) Blue green algae
• (2) Mycoplasma
• (3) Fungi ✓
• (4) Bacteria

Answer: Option (3)

Explanation: Blue-green algae (cyanobacteria), Mycoplasma, and Bacteria are all prokaryotes. Fungi are eukaryotes with a well-defined nucleus and membrane-bound organelles.

Q133.

Given below are two statements:
Statement I: When any plane passing through the central axis of the body divides the organism into two identical halves, it is called radial symmetry.
Statement II: In phylum Echinodermata, both adults and larvae are radially symmetrical.
In light of the above statements, choose the most appropriate answer.

• (1) Both Statement I and Statement II are incorrect
• (2) Statement I is correct but Statement II is incorrect ✓
• (3) Statement I is incorrect but Statement II is correct
• (4) Both Statement I and Statement II are correct

Answer: Option (2)

Explanation: Statement I correctly defines radial symmetry. Statement II is incorrect: adult echinoderms are radially symmetrical, but their larvae are bilaterally symmetrical.

Q134.

Genus represents ____.

• (1) a population of plants and animals
• (2) a group of closely related species ✓
• (3) a group of closely related families
• (4) an individual plant or animal

Answer: Option (2)

Explanation: A genus is a taxonomic rank comprising a group of closely related species that share similar characteristics and a common ancestor, e.g., Panthera includes lion, tiger, leopard.

Q135.

Mad cow disease is caused by ____.

• (1) viroids
• (2) Aspergillus sp.
• (3) Mycoplasma sp.
• (4) prions ✓

Answer: Option (4)

Explanation: Mad cow disease (Bovine Spongiform Encephalopathy) is caused by prions — misfolded infectious proteins that induce misfolding of normal proteins in the brain.

Q136.

The opening between the right atrium and the right ventricle is guarded by ____.

• (1) tricuspid valve ✓
• (2) semilunar valve
• (3) sino-atrial node
• (4) bicuspid valve

Answer: Option (1)

Explanation: The tricuspid valve (three cusps) guards the right atrioventricular opening. The bicuspid (mitral) valve guards the left atrioventricular opening.

Q137.

The inactive form of Bt toxin is converted to the active form in the insect gut ____.

• (1) due to acidic pH
• (2) by proteases
• (3) by nucleases
• (4) due to alkaline pH ✓

Answer: Option (4)

Explanation: Bt toxin protein exists as an inactive protoxin in Bacillus thuringiensis; it is solubilised and activated in the alkaline pH of the insect gut, which then damages the gut epithelial cells.

Q138.

Match List I with List II (elements of a cloning vector).
List I: A. Transformation, B. Cloning site, C. Selection, D. Ori
List II: I. Restriction enzyme, II. Transfer DNA to host bacteria, III. Replication, IV. Antibiotic

• (1) A-I, B-II, C-IV, D-III
• (2) A-III, B-IV, C-I, D-II
• (3) A-IV, B-I, C-III, D-II
• (4) A-II, B-I, C-IV, D-III ✓

Answer: Option (4)

Explanation: Transformation = transfer of DNA into a host bacterium (II); Cloning site = cut by restriction enzyme (I); Selection = achieved using antibiotic resistance genes (IV); Ori = origin of replication, controls replication (III).

Q139.

How many theca are present in each lobe of a typical bilobed angiosperm anther?

• (1) 6
• (2) 8
• (3) 12
• (4) 2 ✓

Answer: Option (4)

Explanation: Each lobe of a typical bilobed anther has two theca (microsporangia), making a total of 4 theca (microsporangia) in the whole (dithecous, tetrasporangiate) anther.

Q140.

Which of the following hormones is not secreted by the human placenta?

• (1) Estrogen
• (2) Progesterone
• (3) LH ✓
• (4) hCG

Answer: Option (3)

Explanation: The placenta acts as an endocrine tissue and secretes hCG, hPL, estrogens, and progestogens. Luteinizing hormone (LH) is secreted by the anterior pituitary, not the placenta.

Q141.

Sperm motility is due to ____.

• (1) ciliary movement
• (2) amoeboid movement
• (3) muscular movement
• (4) flagellar movement ✓

Answer: Option (4)

Explanation: Each sperm has a whip-like tail (flagellum) whose lashing movements propel the sperm forward — flagellar movement.

Q142.

Given below are two statements: one labelled Assertion A and the other Reason R.
Assertion A: In an experiment, Mendel observed that the F1 progeny plants are all tall and none dwarf.
Reason R: Stem height is a contrasting trait, with tall being dominant and dwarf being recessive.
Choose the most appropriate answer.

• (1) Both A and R are correct but R is not the correct explanation of A
• (2) A is correct but R is not correct
• (3) A is not correct but R is correct
• (4) Both A and R are correct and R is the correct explanation of A ✓

Answer: Option (4)

Explanation: Mendel's monohybrid cross of tall x dwarf pea plants produced an F1 generation that was entirely tall, because tallness is the dominant trait and dwarfism is recessive — R correctly explains A.

Q143.

During PCR, primers bind to the DNA strands in the ____ step.

• (1) extension
• (2) annealing ✓
• (3) ligation
• (4) denaturation

Answer: Option (2)

Explanation: PCR has three steps: denaturation (strands separate at ~94°C), annealing (primers bind complementary single strands at ~50-65°C), and extension (Taq polymerase extends the primers).

Q144.

Which of the following plants produces non-albuminous seeds?

• (1) Maize
• (2) Barley
• (3) Pea ✓
• (4) Wheat

Answer: Option (3)

Explanation: Non-albuminous (exalbuminous) seeds lack persistent endosperm because it is completely consumed by the developing embryo, e.g., pea, gram, groundnut. Maize, barley, and wheat are albuminous (retain endosperm).

Q145.

For a person with blood group 'O', which of the following is not a possible combination of parents' blood group genotypes?

• (1) Father: I^Ai and Mother: I^Ai
• (2) Father: I^Bi and Mother: I^Bi
• (3) Father: I^AI^B and Mother: I^Ai ✓
• (4) Father: I^Ai and Mother: I^Bi

Answer: Option (3)

Explanation: A person with blood group O has genotype ii, requiring both parents to contribute an 'i' allele. A father with genotype I^AI^B (blood group AB) has no 'i' allele to pass on, so he can never father a group-O child — this combination is not possible.

Q146.

Which of the following is used as a clot buster?

• (1) Penicillin
• (2) Cyclosporin A
• (3) Statins
• (4) Streptokinase ✓

Answer: Option (4)

Explanation: Streptokinase, produced by the bacterium Streptococcus, is used clinically as a thrombolytic (clot-dissolving) agent in patients with myocardial infarction due to blood clots.

Q147.

Arrange the following in descending order of the number of species in the Amazonian rain forest.
(a) Plants (b) Birds (c) Fishes (d) Invertebrates (e) Mammals

• (1) (d) > (a) > (c) > (b) > (e) ✓
• (2) (e) > (b) > (a) > (c) > (d)
• (3) (b) > (a) > (d) > (c) > (e)
• (4) (c) > (b) > (d) > (e) > (a)

Answer: Option (1)

Explanation: Species richness data for the Amazon rain forest shows Invertebrates (insects etc., by far the most numerous) > Plants > Fishes > Birds > Mammals.

Q148.

Given below are two statements:
Statement I: Plasmids are autonomously replicating DNA.
Statement II: Plasmids are extrachromosomal DNA.
In light of the above statements, choose the most appropriate answer.

• (1) Both Statement I and Statement II are incorrect
• (2) Statement I is correct but Statement II is incorrect
• (3) Statement I is incorrect but Statement II is correct
• (4) Both Statement I and Statement II are correct ✓

Answer: Option (4)

Explanation: Plasmids are small, circular, extrachromosomal DNA molecules found in bacteria that replicate independently of the main chromosomal DNA — both statements are true.

Q149.

Given below are two statements: one labelled Assertion A and the other Reason R.
Assertion A: Forelimbs of human and bats are homologous.
Reason R: Forelimbs of humans and bats have similar anatomical structure.
Choose the most appropriate answer.

• (1) Both A and R are true, but R is not the correct explanation of A
• (2) A is true but R is false
• (3) A is false but R is true
• (4) Both A and R are correct and R is the correct explanation of A ✓

Answer: Option (4)

Explanation: Homologous organs have the same basic anatomical structure/origin but may perform different functions, arising from divergent evolution. Human and bat forelimbs have similar bone arrangement despite different functions (grasping vs flying), making them homologous — R correctly explains A.

Q150.

Which of the following statements about the reabsorption process in Henle's loop are correct?
(a) The descending limb of Henle's loop is permeable to water but almost impermeable to electrolytes
(b) Urine gets concentrated in Henle's loop
(c) Reabsorption of Na⁺ and water takes place in Henle's loop
(d) Active or passive transport of electrolytes occurs in the ascending limb of Henle's loop

• (1) (b), (c) and (d) only
• (2) (a), (b) and (c) only
• (3) (a), (b) and (d) only ✓
• (4) (a) and (b) only

Answer: Option (3)

Explanation: The descending limb is permeable to water and nearly impermeable to electrolytes (a, true), so filtrate becomes progressively concentrated as it descends (b, true). The ascending limb is impermeable to water but actively/passively transports Na⁺ and Cl⁻ out (d, true). Statement (c), attributing general Na⁺ and water reabsorption together within the loop, does not fit the loop's actual selective permeability pattern.

Q151.

Which of the following are secondary lymphoid organs?
(a) Bone marrow (b) Tonsils (c) Spleen (d) Thymus

• (1) (b) and (c) only ✓
• (2) (b) and (d) only
• (3) (a) and (d) only
• (4) (a) and (b) only

Answer: Option (1)

Explanation: Primary lymphoid organs (bone marrow, thymus) are where lymphocytes mature. Secondary lymphoid organs, where mature lymphocytes interact with antigens, include the spleen, tonsils, lymph nodes, and Peyer's patches. So (b) Tonsils and (c) Spleen are secondary lymphoid organs.

Q152.

Which of the following is the correct order of arrangement of the vertebral column from the head to toe?

• (1) Sacrum, lumbar vertebra, thoracic vertebra, cervical vertebra
• (2) Cervical vertebra, lumbar vertebra, thoracic vertebra, sacrum
• (3) Cervical vertebra, thoracic vertebra, lumbar vertebra, sacrum ✓
• (4) Cervical vertebra, thoracic vertebra, sacrum, lumbar vertebra

Answer: Option (3)

Explanation: From the head downward, the vertebral column is arranged as: Cervical (neck) → Thoracic (chest) → Lumbar (lower back) → Sacrum → Coccyx.

Q153.

If the diploid chromosome number of a typical angiosperm is 36, what would be the chromosome number in its endosperm?

• (1) 36
• (2) 54 ✓
• (3) 72
• (4) 18

Answer: Option (2)

Explanation: Diploid (2n) = 36, so haploid (n) = 18. The endosperm of angiosperms is triploid (3n), formed by fusion of one sperm with two polar nuclei: 3 × 18 = 54.

Q154.

Sponges exchange O₂ and CO₂ by ____.

• (1) moist cuticle
• (2) tracheal tubes
• (3) gills
• (4) simple diffusion over their entire body surface ✓

Answer: Option (4)

Explanation: Sponges lack specialised respiratory organs; gaseous exchange occurs by simple diffusion across the general body surface, aided by the constant flow of water through their pores/canal system.

Q155.

Which of the following diseases is not sexually transmitted?

• (1) Tuberculosis ✓
• (2) Gonorrhoea
• (3) Genital warts
• (4) Syphilis

Answer: Option (1)

Explanation: Gonorrhoea, genital warts, and syphilis are classic sexually transmitted infections (STIs). Tuberculosis is caused by Mycobacterium tuberculosis and spreads mainly through airborne droplets, not sexual contact.

Q156.

Which of the following in the female gametophyte of an angiosperm helps in guiding the pollen tube for fertilizing the eggs?

• (1) Synergids ✓
• (2) Central cells
• (3) Polar nucleus
• (4) Antipodal cells

Answer: Option (1)

Explanation: The two synergid cells flank the egg cell at the micropylar end of the embryo sac; one of them develops a filiform apparatus that helps guide/attract the entry of the pollen tube.

Q157.

Match List I with List II.
List I: A. Excess growth hormone, B. Luteinizing hormone, C. Vasopressin, D. Oxytocin
List II: I. Reabsorption of water and electrolytes in kidney, II. Contraction of uterus during child birth, III. Acromegaly, IV. Ovulation

• (1) A-III, B-IV, C-I, D-II ✓
• (2) A-II, B-IV, C-I, D-III
• (3) A-IV, B-III, C-I, D-II
• (4) A-III, B-IV, C-II, D-I

Answer: Option (1)

Explanation: Excess growth hormone causes Acromegaly (III); LH surge triggers Ovulation (IV); Vasopressin (ADH) promotes reabsorption of water/electrolytes in the kidney (I); Oxytocin causes uterine contraction during childbirth (II).

Q158.

The covering of the ovum at ovulation is ____.

• (1) zona radiata
• (2) zona pellucida ✓
• (3) chorion
• (4) endometrium

Answer: Option (2)

Explanation: At ovulation, the secondary oocyte is surrounded by a thick, transparent, non-cellular glycoprotein layer called the zona pellucida.

Q159.

Match List I with List II (ecological interactions).
List I: A. Both species are harmed, B. One species is harmed and the other is benefited, C. Both species are benefited, D. One is benefited while the other has no effect
List II: I. Predation, II. Mutualism, III. Competition, IV. Commensalism

• (1) A-I, B-II, C-III, D-IV
• (2) A-II, B-I, C-IV, D-III
• (3) A-III, B-I, C-II, D-IV ✓
• (4) A-IV, B-III, C-I, D-II

Answer: Option (3)

Explanation: Both harmed = Competition (III); one harmed, one benefited = Predation (I); both benefited = Mutualism (II); one benefited, other unaffected = Commensalism (IV).

Q160.

Which of the following structures is not a part of the male reproductive system?

• (1) Epididymis
• (2) Vasa efferentia
• (3) Infundibulum ✓
• (4) Rete testis

Answer: Option (3)

Explanation: The infundibulum is the funnel-shaped opening of the fallopian tube (oviduct) that receives the ovum, and is part of the female reproductive system. Epididymis, vasa efferentia, and rete testis are all male reproductive structures.

Q161.

Which of the following are primary consumers in a food chain?

• (1) Predators
• (2) Herbivores ✓
• (3) Carnivores
• (4) Parasites

Answer: Option (2)

Explanation: Primary consumers are organisms that feed directly on producers (plants) — i.e., herbivores. Carnivores/predators are secondary or tertiary consumers.

Q162.

Given below are two statements: one labelled Assertion A and the other Reason R.
Assertion A: Abingdon tortoise in Galapagos islands became extinct within a decade after goats were introduced.
Reason R: Goats were more efficient at browsing than Abingdon tortoise.
Choose the most appropriate answer.

• (1) Both A and R are correct but R is not the correct explanation of A
• (2) A is correct but R is not correct
• (3) A is not correct but R is correct
• (4) Both A and R are correct and R is the correct explanation of A ✓

Answer: Option (4)

Explanation: This is a classic example of competitive exclusion. Introduced goats outcompeted the native Abingdon tortoise for food (browsing more efficiently), driving the tortoise to extinction within about a decade — R correctly explains A.

Q163.

Which of the following statements about the lac-operon is correct?

• (1) Lactose activates repressor to bind to the operator
• (2) Genes i, z, y and a share a single common promoter
• (3) Galactose can act as an inducer of lac operon
• (4) Gene i is constitutively expressed ✓

Answer: Option (4)

Explanation: The regulatory gene 'i' (which codes for the repressor) is constitutively (continuously) expressed at a low, constant level, independent of the presence of lactose. (Allolactose, not galactose, is the actual inducer, and 'i' has its own separate promoter from z, y, a.)

Q164.

Which of the following statements is correct about Plasmodium?

• (1) Reproduces sexually in RBCs
• (2) Gametocytes develop in mosquito gut
• (3) Fertilization takes place in mosquito gut ✓
• (4) Reproduces sexually in liver cells

Answer: Option (3)

Explanation: Gametocytes formed in the human RBCs are taken up by the female Anopheles mosquito during a blood meal; fertilization (fusion of male and female gametes to form a zygote) actually occurs in the mosquito's gut. Plasmodium reproduces asexually (schizogony) in human RBCs and liver cells.

Q165.

A population of diploid organisms is at Hardy-Weinberg equilibrium. If the frequency of allele A is 0.1, the frequency of AA is ____.

• (1) 0.02
• (2) 0.10
• (3) 0.99
• (4) 0.01 ✓

Answer: Option (4)

Explanation: By the Hardy-Weinberg equation, frequency of AA = p², where p = frequency of allele A = 0.1. So p² = 0.1 × 0.1 = 0.01.

Q166.

Adaptive radiation in placental mammals and Australian marsupials leading to similarity between distant species is an example of ____.

• (1) convergent evolution ✓
• (2) founder effect
• (3) genetic drift
• (4) divergent evolution

Answer: Option (1)

Explanation: When unrelated species (placental mammals and marsupials) independently evolve similar body forms/adaptations in response to similar environments/niches, it is called convergent evolution.

Q167.

Colostrum, secreted by the mother during initial days of lactation, is abundant in ____.

• (1) IgM
• (2) IgA ✓
• (3) IgD
• (4) IgG

Answer: Option (2)

Explanation: Colostrum, the first milk secreted after childbirth, is rich in IgA antibodies, which provide passive immunity and protect the newborn from infections.

Q168.

Given below are two statements: one labelled Assertion A and the other Reason R.
Assertion A: In recombinant DNA technology, lysozyme is used for disrupting bacterial cells while cellulase is for plant cells.
Reason R: Isolation of genetic material needs disruption of cells.
Choose the most appropriate answer.

• (1) Both A and R are correct but R is not the correct explanation of A
• (2) A is correct but R is not correct
• (3) A is not correct but R is correct
• (4) Both A and R are correct and R is the correct explanation of A ✓

Answer: Option (4)

Explanation: To isolate DNA/genetic material, host cells must first be broken open. Lysozyme is used to digest bacterial cell walls, while cellulase digests plant cell walls — R correctly explains why these enzymes are needed (A).

Q169.

Given below are two statements:
Statement I: Ovulation is caused by an LH surge leading to rupture of Graafian follicles.
Statement II: The Graafian follicle remaining after ovulation transforms into corpus luteum and secretes a large amount of estrogen.
Choose the most appropriate answer.

• (1) Both Statement I and Statement II are incorrect
• (2) Statement I is correct but Statement II is incorrect ✓
• (3) Statement I is incorrect but Statement II is correct
• (4) Both Statement I and Statement II are correct

Answer: Option (2)

Explanation: A mid-cycle LH surge triggers rupture of the Graafian follicle and release of the ovum (Statement I true). After ovulation, the ruptured follicle transforms into the corpus luteum, which mainly secretes progesterone (not large amounts of estrogen) to maintain the endometrium — Statement II is false.

Q170.

Natural selection can lead to ____.
(a) stabilisation (b) genetic drift (c) directional change (d) disruption

• (1) (a), (c) and (d) only ✓
• (2) (a), (b), (c) and (d)
• (3) (a) and (c) only
• (4) (a) only

Answer: Option (1)

Explanation: Natural selection can act in three main ways: stabilising selection (favours average phenotypes), directional selection (shifts the mean toward one extreme), and disruptive selection (favours both extremes). Genetic drift is a distinct, random evolutionary mechanism, not a form of natural selection.

Q171.

The method of directly injecting a sperm into the ovum in assisted reproductive technology is called ____.

• (1) Zygote intra fallopian transfer (ZIFT)
• (2) Intra cytoplasmic sperm injection (ICSI) ✓
• (3) Embryo transfer (ET)
• (4) Gamete intra fallopian transfer (GIFT)

Answer: Option (2)

Explanation: ICSI involves directly injecting a single sperm into the cytoplasm of an ovum in vitro, commonly used to treat severe male infertility.

Q172.

Which of the following is used as an effective sedative and painkiller for treating post-surgery patients?

• (1) Antibiotics
• (2) Morphine ✓
• (3) Anti-retroviral drugs
• (4) Interferon

Answer: Option (2)

Explanation: Morphine is a powerful opioid analgesic/sedative widely used clinically to relieve severe pain, including in post-surgical patients (though also prone to abuse in unregulated use).

Q173.

Given below are two statements:
Statement I: Down's syndrome is caused by the absence of one of the X-chromosomes.
Statement II: Turner's syndrome is caused by the presence of an additional copy of the X-chromosome.
Choose the correct answer.

• (1) Both Statement I and Statement II are incorrect ✓
• (2) Statement I is correct but Statement II is incorrect
• (3) Statement I is incorrect but Statement II is correct
• (4) Both Statement I and Statement II are correct

Answer: Option (1)

Explanation: Down's syndrome is actually caused by trisomy of chromosome 21 (an extra copy), not the absence of an X-chromosome. Turner's syndrome is actually caused by the absence (monosomy) of an X-chromosome (45, X0), not an additional copy. Both statements, as worded, describe the wrong condition, so both are incorrect.

Q174.

Which of the following is not evidence for evolution?

• (1) Paleontological evidence from fossil records
• (2) Embryological support for evolution as proposed by Ernst Heckel ✓
• (3) Divergent evolution of anatomical structures such as forelimbs
• (4) Convergent evolution of traits like wings of birds and butterflies

Answer: Option (2)

Explanation: Ernst Haeckel's 'biogenetic law' / recapitulation theory (embryological support as he proposed it) has largely been discredited by modern embryology and is not accepted as valid evidence for evolution, unlike fossil records, homologous (divergent) structures, and analogous (convergent) structures.

Q175.

Given below are two statements: one labelled Assertion A and the other Reason R.
Assertion A: The logistic growth model of populations is considered more realistic than the exponential growth model.
Reason R: Resources are finite.
Choose the most appropriate answer.

• (1) Both A and R are correct but R is not the correct explanation of A
• (2) A is correct but R is not correct
• (3) A is not correct but R is correct
• (4) Both A and R are correct and R is the correct explanation of A ✓

Answer: Option (4)

Explanation: In nature, resources (food, space) are always limited, which is why populations cannot grow exponentially forever and instead follow a logistic (S-shaped) growth pattern with a carrying capacity — R correctly explains why the logistic model is more realistic (A).

Q176.

Given below are two statements:
Statement I: Modern Homo sapiens arose in Australia and moved across continents.
Statement II: Homo sapiens arose around 75000 to 10000 years ago.
Choose the most appropriate answer.

• (1) Both Statement I and Statement II are incorrect
• (2) Statement I is correct but Statement II is incorrect
• (3) Statement I is incorrect but Statement II is correct ✓
• (4) Both Statement I and Statement II are correct

Answer: Option (3)

Explanation: Fossil and genetic evidence indicates that modern Homo sapiens arose in Africa, not Australia, and then migrated across continents — Statement I is incorrect. The approximate timeframe given for the emergence and spread of modern humans in Statement II is accepted as correct.

Q177.

Consider a population of 10 million cells. Given the per-capita birth rate of 0.002 (per unit time) and the per-capita death rate of 0.002 (per unit time), the expected number of cells after 10 generations is ____.

• (1) 5 million
• (2) 10 million ✓
• (3) 100 million
• (4) 1 million

Answer: Option (2)

Explanation: The intrinsic rate of increase, r = birth rate − death rate = 0.002 − 0.002 = 0. With r = 0, the population neither grows nor declines, so it remains at 10 million after any number of generations.

Q178.

Which of the following statements are correct?
(a) Energy flow from producers to consumers is unidirectional
(b) Energy pyramid can never be inverted
(c) Transfer of energy follows the 1% law

• (1) (a) and (b) only ✓
• (2) (a) and (c) only
• (3) (b) and (c) only
• (4) (a), (b) and (c)

Answer: Option (1)

Explanation: Energy flow in an ecosystem is always unidirectional (a, true), and unlike pyramids of number or biomass, the pyramid of energy is always upright and can never be inverted (b, true). However, energy transfer between trophic levels follows the 10% law, not a 1% law, so (c) is false.

Q179.

Muscle contraction is initiated by a signal sent by the central nervous system through the release of ____.

• (1) acetyl coenzyme A
• (2) cyclic guanine monophosphate
• (3) cyclic adenine monophosphate
• (4) acetyl choline ✓

Answer: Option (4)

Explanation: A motor neuron releases the neurotransmitter acetylcholine at the neuromuscular junction, which generates an action potential in the muscle fibre and initiates contraction.

Q180.

Which of the following enzymes synthesizes precursor mRNA (hnRNA)?

• (1) RNA polymerase II ✓
• (2) RNA polymerase III
• (3) DNA polymerase
• (4) RNA polymerase I

Answer: Option (1)

Explanation: In eukaryotes, RNA polymerase II transcribes structural genes to produce heterogeneous nuclear RNA (hnRNA), the precursor of mRNA, which is later processed (capping, tailing, splicing) into mature mRNA.

Correct option is highlighted in green with a checkmark. Answers compiled based on standard NCERT Biology concepts.