NEET (UG) 2026 re-examination, 21 June 2026, Test Booklet Code 50. Try each question first, then open the answer. The answer shown is NTA's final key; the explanation is ours, with the NCERT page that carries the fact.
The number of vertebrae in a human is ________.
- (1)7
- (2)12
- (3)26
- (4)206
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerThe human vertebral column is made of 26 serially arranged vertebrae (after fusion of the sacral and coccygeal vertebrae into single bones in the adult), running from the neck to the pelvis.
Why the other options are wrong- (1) 7 is the number of cervical vertebrae (neck region) only, not the whole column.
- (2) 12 is the number of thoracic vertebrae, the chest region alone; students confuse this single-segment count with the whole 26-vertebra column.
- (4) 206 is the total number of bones in the adult human skeleton, not the vertebral count -- a common mix-up.
NCERT: NCERT Class 11 Biology, Chapter 17 (Locomotion and Movement), page 225
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Revise the lesson: Locomotion Skeleton Joints Disorders
Symbiotic association between fungi and algae are called ________.
- (1)lichens
- (2)sponges
- (3)mycorrhiza
- (4)chrysophytes
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerA lichen is a symbiotic association between a fungal partner (mycobiont) and a photosynthetic alga or cyanobacterium (phycobiont); the alga makes food and the fungus provides shelter and absorbs water/minerals.
Why the other options are wrong- (2) Sponges (Porifera) are multicellular animals, not a fungus-alga association.
- (3) Mycorrhiza is a symbiotic fungus-root association with higher plants, not with algae.
- (4) Chrysophytes are a group of algae themselves (diatoms, golden algae), not a symbiosis.
NCERT: NCERT Class 11 Biology, Chapter 2 (Biological Classification), page 21
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Revise the lesson: Viruses Viroids Lichens
Cell theory was formulated by ______________.
- (1)Schleiden and Schwann
- (2)Robert Brown
- (3)Singer and Nicolson
- (4)Antonie Von Leeuwenhoek
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerSchleiden (a botanist) and Schwann (a zoologist) together proposed the cell theory, later extended by Virchow's principle that cells arise only from pre-existing cells.
Why the other options are wrong- (2) Robert Brown is known for discovering the cell nucleus, not for formulating cell theory; students confuse this famous cell-biology find with the theory's origin.
- (3) Singer and Nicolson proposed the fluid mosaic model of the membrane, unrelated to cell theory.
- (4) Leeuwenhoek first observed and described free-living cells (like bacteria), but did not formulate the theory.
NCERT: NCERT Class 11 Biology, Chapter 8 (Cell: The Unit of Life), page 103
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Revise the lesson: Cell Theory
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 Choose the correct answer from the options given below :
- (1)(b) and (c) only
- (2)(a) and (c) only
- (3)(a), (c) and (d) only
- (4)(a), (b) and (c) only
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerProkaryotic (70S) ribosomes are built of a 50S and a 30S subunit (a, true); many bacteria carry extra-chromosomal circular plasmid DNA (b, true); mesosomes are infoldings of the plasma membrane seen in prokaryotes (c, true). Peroxisomes are membrane-bound organelles found only in eukaryotic cells, so (d) is false. Correct set: (a), (b), (c).
Why the other options are wrong- (1) (a) is a true prokaryotic feature -- 70S ribosomes split into 50S and 30S subunits; this option wrongly drops it from the correct combination.
- (2) (b) plasmids do occur in many prokaryotes, a true statement; this option wrongly excludes it, keeping only two of the three correct statements.
- (3) Wrongly includes (d) peroxisomes, which prokaryotes lack (no membrane-bound organelles).
NCERT: NCERT Class 11 Biology, Chapter 8 (Cell: The Unit of Life), page 91
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Revise the lesson: Prokaryotic vs Eukaryotic
Which of the following is not a part of human central neural system?
- (1)Arachnoid
- (2)Dura mater
- (3)Pia mater
- (4)Pericardium
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerThe brain (part of the CNS) is wrapped in three meninges -- outer dura mater, middle arachnoid, and inner pia mater -- so options 1-3 are all CNS coverings. The pericardium is the membrane covering the heart, part of the circulatory system, not the CNS.
Why the other options are wrong- (1) Arachnoid is the middle cranial meninx, a genuine CNS covering.
- (2) Dura mater is the outer cranial meninx, a genuine CNS covering.
- (3) Pia mater is the inner meninx in direct contact with brain tissue.
NCERT: NCERT Class 11 Biology, Chapter 18 (Neural Control and Coordination), page 235
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Revise the lesson: Neural Human Nervous System
Mitochondrial inner membrane encloses ___________ .
- (1)matrix
- (2)cytosol
- (3)mucus
- (4)aqueous humor
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerThe mitochondrion has two membranes; the inner membrane folds inward as cristae and encloses the central fluid space called the matrix, which contains the enzymes of the Krebs cycle, mitochondrial DNA and ribosomes.
Why the other options are wrong- (2) Cytosol is the fluid of the whole cell outside organelles, not enclosed by the mitochondrial inner membrane.
- (3) Mucus is a secretion of epithelial cells, unrelated to mitochondria.
- (4) Aqueous humor is the fluid of the eye chamber, unrelated to mitochondria.
NCERT: NCERT Class 11 Biology, Chapter 8 (Cell: The Unit of Life), page 97
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Revise the lesson: Cell Organelles Mitochondria Plastid
Match List-I with List-II.
List-I List-II
A. Cristae I. Flat membrane sacs in stroma of chloroplast
B. Cisternae II. Infoldings in mitochondria
C. Thylakoids III. Cell membrane
D. Phospholipid IV. Disc shaped sacs in the Golgi apparatus
Choose the correct answer from the options given below:
- (1)A-III, B-IV, C-I, D-II
- (2)A-II, B-IV, C-I, D-III
- (3)A-II, B-IV, C-III, D-I
- (4)A-IV, B-III, C-I, D-II
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerCristae are infoldings of the mitochondrial inner membrane (A-II). Cisternae are the disc-shaped stacked sacs of the Golgi apparatus (B-IV). Thylakoids are flat membranous sacs inside the chloroplast stroma (C-I). Phospholipids are the basic structural molecules of the cell membrane (D-III).
Why the other options are wrong- (1) Swaps cristae with cell membrane and cisternae with mitochondria -- wrong pairing.
- (3) Puts thylakoids with cell membrane and phospholipid with stroma sacs -- wrong pairing.
- (4) Puts cristae with Golgi sacs and cisternae with cell membrane -- wrong pairing.
NCERT: NCERT Class 11 Biology, Chapter 8 (Cell: The Unit of Life), page 98
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Revise the lesson: Cell Organelles Mitochondria Plastid
The plastid that stores xanthophyll is known as __________.
- (1)chloroplast
- (2)chromoplast
- (3)aleuroplast
- (4)amyloplast
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerChromoplasts are the coloured plastids (other than green) that store fat-soluble pigments such as carotene and xanthophylls, giving flowers/fruits their yellow, orange or red colour.
Why the other options are wrong- (1) Chloroplasts store chlorophyll for photosynthesis, not primarily xanthophyll.
- (3) Aleuroplasts are colourless leucoplasts that store protein, not pigments; students mix up plastid names that all sound like 'stores something'.
- (4) Amyloplasts are colourless leucoplasts that store starch, easily confused with the pigment-storing chromoplast since both are specialised plastids.
NCERT: NCERT Class 11 Biology, Chapter 8 (Cell: The Unit of Life), page 97
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Revise the lesson: Cell Organelles Mitochondria Plastid
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 Choose the correct answer from the options given below :
- (1)(a) and (b) only
- (2)(a), (b) and (c) only
- (3)(c) and (d) only
- (4)(b) and (c) only
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerThe pituitary is anatomically divided into adenohypophysis and neurohypophysis (a, true). Its pars distalis secretes six trophic hormones including FSH (b, true) and prolactin, while pars intermedia secretes MSH (c, true). So (d), which claims no prolactin is secreted, is false -- prolactin IS one of the adenohypophyseal hormones. Correct set: (a), (b), (c).
Why the other options are wrong- (1) Leaves out (c), which is also correct (MSH is secreted by pars intermedia).
- (3) Wrongly includes (d), which is false since prolactin is secreted.
- (4) This option leaves out (a), the true fact that the pituitary is anatomically split into adenohypophysis and neurohypophysis, keeping only two of the three correct statements.
NCERT: NCERT Class 11 Biology, Chapter 19 (Chemical Coordination and Integration), page 241
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Revise the lesson: Endocrine Hypothalamus Pituitary Thyroid
Photorespiration reaction catalyzed by RuBisCo is shown below: RuBP + O2 → 3-Phosphoglycerate + X. Identify “X” from the given options:
- (1)Phosphoenolpyruvate
- (2)2-Phosphoglycolate
- (3)Oxaloacetate
- (4)Malate
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerRuBisCO's oxygenase activity adds O2 to RuBP, splitting it into one 3-carbon molecule of 3-phosphoglycerate and one 2-carbon molecule of phosphoglycolate; the latter is metabolised in the photorespiratory pathway.
Why the other options are wrong- (1) Phosphoenolpyruvate is the CO2 acceptor in C4 plants (PEP carboxylase reaction), unrelated to RuBisCO's oxygenase step.
- (3) Oxaloacetate is the first stable product of C4 photosynthesis, not of photorespiration.
- (4) Malate is a C4-pathway intermediate transported to bundle sheath cells, not a photorespiration product.
NCERT: NCERT Class 11 Biology, Chapter 11 (Photosynthesis in Higher Plants), page 147
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Revise the lesson: Photosynthesis C3 C4 CAM
Mad cow disease is caused by _______.
- (1)prions
- (2)viroids
- (3)Aspergillus sp.
- (4)Mycoplasma sp.
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerPrions are infectious proteinaceous particles with no nucleic acid; the most notable prion disease is bovine spongiform encephalopathy (mad cow disease) in cattle and its human analogue, Creutzfeldt-Jakob disease.
Why the other options are wrong- (2) Viroids are naked, low-molecular-weight infectious RNA (no protein coat) that cause plant diseases, not mad cow disease.
- (3) Aspergillus is a fungus causing infections like aspergillosis, not mad cow disease.
- (4) Mycoplasma are wall-less bacteria causing diseases like pneumonia, not mad cow disease.
NCERT: NCERT Class 11 Biology, Chapter 2 (Biological Classification), page 21
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Revise the lesson: Viruses Viroids Lichens
Which pigment has absorption peak at 700 nm in the photosynthetic reaction centre PS I (P700)?
- (1)Chlorophyll b
- (2)Chlorophyll a
- (3)Xanthophylls
- (4)Carotenoids
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerThe reaction-centre pigment of both photosystems is a special form of chlorophyll a. In PS I it absorbs maximally at 700 nm, hence named P700; in PS II the corresponding chlorophyll a absorbs at 680 nm (P680).
Why the other options are wrong- (1) Chlorophyll b is an accessory pigment that funnels light energy to chlorophyll a; it is not the reaction-centre pigment.
- (3) Xanthophylls are accessory carotenoid pigments, not the reaction-centre pigment.
- (4) Carotenoids in general are accessory pigments, not the reaction-centre pigment.
NCERT: NCERT Class 11 Biology, Chapter 11 (Photosynthesis in Higher Plants), page 138
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Revise the lesson: Photosynthesis Light Reactions
In water, frogs respire using _____________.
- (1)skin
- (2)buccal cavity
- (3)lungs
- (4)trachea
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerFrogs are amphibians with dual respiration -- in water, the moist, vascularised skin acts as the respiratory surface for cutaneous respiration (dissolved O2 diffuses in through the skin); on land, they switch to buccal and pulmonary (lung) respiration.
Why the other options are wrong- (2) Buccal cavity respiration operates on land through air movement in the mouth, not the primary mode in water.
- (3) Lungs are used mainly for respiration on land, not the primary aquatic mode.
- (4) Frogs do not have a trachea-based tracheal respiratory system like insects.
NCERT: NCERT Class 11 Biology, Chapter 7 (Structural Organisation in Animals), page 82
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Revise the lesson: Frog Anatomy
Which of the following represents the correct sequence of arrangement of bones in the lower limb of humans?
- (1)Femur-tibia-patella-tarsal
- (2)Patella-femur-tibia-tarsal
- (3)Femur-patella-tibia-tarsal
- (4)Femur-tarsal-patella-tibia
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerProceeding proximal to distal down the hindlimb: the femur (thigh bone) articulates at the knee where the patella (kneecap) sits, then the tibia-fibula (leg bones), and finally the tarsals (ankle bones). This gives Femur-Patella-Tibia-Tarsal.
Why the other options are wrong- (1) Places patella after tibia, which is out of proximal-to-distal order (patella sits at the knee, before the tibia).
- (2) Starts with patella before femur, reversing the correct proximal-to-distal order.
- (4) Places tarsal (ankle) before patella and tibia, badly out of sequence.
NCERT: NCERT Class 11 Biology, Chapter 17 (Locomotion and Movement), page 226
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Revise the lesson: Locomotion Skeleton Joints Disorders
Phyllotaxy is the pattern of arrangement of ________.
- (1)leaves
- (2)flowers
- (3)fruits
- (4)sepals
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerPhyllotaxy (from Greek phyllon=leaf, taxis=arrangement) describes the pattern in which leaves are arranged on a stem or branch -- alternate, opposite, or whorled.
Why the other options are wrong- (2) Flower arrangement on the floral axis is called inflorescence, not phyllotaxy.
- (3) Fruit arrangement on an axis has no dedicated term like phyllotaxy; students wrongly stretch the leaf-arrangement word to other plant parts.
- (4) Sepal arrangement in the bud (aestivation) is a separate floral term, not phyllotaxy.
NCERT: NCERT Class 11 Biology, Chapter 5 (Morphology of Flowering Plants), page 61
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Revise the lesson: Morphology Flowering Plants
Match List-I with List-II.
List-I List-II
A. Starch I. Fights infection
B. Antibody II. Energy storage
C. Concanavalin A III. Glucose transport
D. Glut-4 IV. Lectin
Choose the correct answer from the options given below :
- (1)A-I, B-II, C-IV, D-III
- (2)A-II, B-I, C-IV, D-III
- (3)A-II, B-I, C-III, D-IV
- (4)A-I, B-II, C-III, D-IV
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerStarch is a storage polysaccharide used for energy storage (A-II). Antibodies are proteins that fight infection (B-I). Concanavalin A is a well-known lectin protein from jack bean (C-IV). GLUT-4 is a membrane transport protein that enables glucose transport into cells (D-III).
Why the other options are wrong- (1) Swaps starch/antibody roles and mismatches Concanavalin A with glucose transport.
- (3) This option swaps lectin Concanavalin A with the glucose-transporter Glut-4, mixing up a plant defence protein with a membrane transport protein.
- (4) Keeps starch/antibody wrong and mismatches lectin/transport pair.
NCERT: NCERT Class 11 Biology, Chapter 9 (Biomolecules), page 108
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Revise the lesson: Biomolecules Proteins Nucleic
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 the light of the above statements, choose the most appropriate answer from the options given below :
- (1)Both Statement I and Statement II are correct
- (2)Both Statement I and Statement II are incorrect
- (3)Statement I is correct but Statement II is incorrect
- (4)Statement I is incorrect but Statement II is correct
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerStatement I correctly states the textbook definition of radial symmetry. Statement II is false: in Echinodermata only the sessile/free-living adults are radially symmetrical, while their free-swimming larvae are bilaterally symmetrical -- a classic case of larval symmetry differing from the adult.
Why the other options are wrong- (1) Wrong because Statement II is false (larvae are bilateral, not radial).
- (2) Statement I is the standard definition of radial symmetry and is true, not false; only Statement II (about echinoderm larvae) is wrong, so calling both incorrect overcorrects.
- (4) Wrong because Statement I is actually correct, not incorrect.
NCERT: NCERT Class 11 Biology, Chapter 4 (Animal Kingdom), page 38
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Revise the lesson: Animal Kingdom Nonchordates
Endomembrane system includes _______.
- (1)endoplasmic reticulum, Golgi complex, lysosomes and vacuole
- (2)endoplasmic reticulum, chloroplast, peroxisomes and vacuole
- (3)mitochondria, chloroplast, peroxisomes and vacuole
- (4)Golgi complex, chloroplast, peroxisomes and vacuole
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerBecause their functions are coordinated through vesicle trafficking, the endoplasmic reticulum, Golgi complex, lysosomes and vacuole are grouped together as the endomembrane system. Mitochondria, chloroplasts and peroxisomes are excluded because their functions are not coordinated with this network.
Why the other options are wrong- (2) Wrongly includes chloroplast and peroxisomes, which are NOT part of the endomembrane system.
- (3) Lists only excluded organelles (mitochondria, chloroplast, peroxisomes), none of which belong to the endomembrane system.
- (4) Wrongly includes chloroplast and peroxisomes instead of ER and lysosomes.
NCERT: NCERT Class 11 Biology, Chapter 8 (Cell: The Unit of Life), page 95
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Revise the lesson: Cell Organelles Endomembrane
How many molecules of pyruvic acid are produced at the end of glycolysis from 206 molecules of glucose?
- (1)206
- (2)309
- (3)103
- (4)412
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerGlycolysis breaks down one molecule of glucose into two molecules of pyruvic acid. For 206 molecules of glucose, the yield is 206 x 2 = 412 molecules of pyruvic acid.
Why the other options are wrong- (1) 206 would be the yield only if 1 glucose gave 1 pyruvate, ignoring the 1:2 stoichiometry.
- (2) 309 does not correspond to any simple multiple of 206 tied to the correct 1:2 ratio.
- (3) 103 is half of 206, the reverse of the actual 1:2 relationship.
NCERT: NCERT Class 11 Biology, Chapter 12 (Respiration in Plants), page 155
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Revise the lesson: Respiration Glycolysis
Which of the following plant growth regulators is used as herbicide?
- (1)2, 4-D
- (2)Kinetin
- (3)Abscisic acid
- (4)Gibberellin
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answer2,4-Dichlorophenoxyacetic acid (2,4-D) is a synthetic auxin widely used as a herbicide -- it kills broad-leaved dicot weeds in monocot crop fields (e.g., lawns, cereal crops) without harming the monocots.
Why the other options are wrong- (2) Kinetin is a cytokinin used to promote cell division and delay senescence, not a herbicide.
- (3) Abscisic acid is the stress/dormancy hormone, used to induce dormancy, not as a herbicide.
- (4) Gibberellin promotes growth (bolting, seed germination); it is not used as a herbicide.
NCERT: NCERT Class 11 Biology, Chapter 13 (Plant Growth and Development), page 176
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Revise the lesson: Plant Hormones
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 the light of the above statements, choose the most appropriate answer from the options given below:
- (1)Both Statement I and Statement II are correct
- (2)Both Statement I and Statement II are incorrect
- (3)Statement I is correct but Statement II is incorrect
- (4)Statement I is incorrect but Statement II is correct
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerStatement I is correct: unlike bryophytes/pteridophytes, gymnosperm gametophytes have no independent free-living existence -- they remain retained within the sporangia on the sporophyte. Statement II is also correct: gymnosperm literally means 'naked seed' -- the ovules/seeds are not enclosed in an ovary/fruit wall.
Why the other options are wrong- (2) Wrong because both statements are individually verifiable as true.
- (3) Statement II is also true -- gymnosperm literally means 'naked seed', so the seeds are genuinely uncovered; this option wrongly rejects that correct fact.
- (4) Statement I is actually true -- gymnosperm gametophytes stay retained within the sporangia; this option mistakenly reverses which statement holds.
NCERT: NCERT Class 11 Biology, Chapter 3 (Plant Kingdom), page 33
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Revise the lesson: Plant Kingdom
Match List-I with List-II.
List-I List-II
A. Spherical I. Vibrio
B. Rod II. Cocci
C. Comma III. Spirilla
D. Spirillum IV. Bacilli
Choose the correct answer from the options given below :
- (1)A-I, B-III, C-II, D-IV
- (2)A-III, B-II, C-I, D-IV
- (3)A-II, B-I, C-IV, D-III
- (4)A-II, B-IV, C-I, D-III
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerThe four basic bacterial shapes are: spherical = coccus/cocci (A-II), rod-shaped = bacillus/bacilli (B-IV), comma-shaped = vibrio (C-I), and spiral = spirillum (D-III, matching itself with 'Spirilla' as the plural form/shape name).
Why the other options are wrong- (1) This option pairs spherical bacteria with vibrio (the comma shape) and rods with spirilla (the spiral shape), swapping two of the four shape names.
- (2) Mismatches spherical with spirilla and comma with vibrio pairing order.
- (3) This option pairs rod-shaped bacteria with vibrio and comma-shaped bacteria with bacilli, crossing the two shape-name pairs.
NCERT: NCERT Class 11 Biology, Chapter 8 (Cell: The Unit of Life), page 89
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Revise the lesson: Prokaryotic vs Eukaryotic
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 Choose the correct answer from the options given below :
- (1)(a), (b) and (c) only
- (2)(d) only
- (3)(a) and (b) only
- (4)(b), (c) and (d) only
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerSolanaceae flowers are bisexual, actinomorphic (radially symmetric) (a, true); calyx has five gamosepalous (united) sepals, persistent (b, true); androecium has five epipetalous stamens attached to the corolla tube (c, true). The gynoecium is bicarpellary, syncarpous with a SUPERIOR ovary, not inferior, so (d) is false.
Why the other options are wrong- (2) Wrongly isolates (d), which is actually false for Solanaceae (ovary is superior).
- (3) Drops the also-correct statement (c) about epipetalous stamens.
- (4) Wrongly includes false statement (d) and drops true statement (a).
NCERT: NCERT Class 11 Biology, Chapter 5 (Morphology of Flowering Plants), page 64
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Revise the lesson: Families Fabaceae Solanaceae
Select the correct sequence of experiments that led to a gradual understanding of photosynthesis in green plants.
- (1)Absorption spectra of chlorophyll a and b → production of glucose → release of oxygen → role of air
- (2)Role of air → release of oxygen → production of glucose → absorption spectra of chlorophyll a and b
- (3)Release of oxygen → production of glucose → absorption spectra of chlorophyll a and b → role of air
- (4)Production of glucose → role of air → release of oxygen → absorption spectra of chlorophyll a and b
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerHistorically: Priestley (1770s) first showed plants restore something in air that is used up by burning/breathing (role of air); Ingenhousz then showed it is specifically the green parts, in sunlight, that release oxygen; Sachs demonstrated glucose (as starch) is produced during photosynthesis; Engelmann later used a prism and Cladophora with aerotactic bacteria to map the action/absorption spectrum of chlorophyll a and b to specific wavelengths. So the correct historical order is: role of air, release of oxygen, production of glucose, absorption spectra.
Why the other options are wrong- (1) Places the absorption-spectra work (a late, refined finding) first, before even the basic 'role of air' discovery -- reverses history.
- (3) Puts absorption spectra before the role-of-air experiment, which came first historically.
- (4) Puts glucose production before the role-of-air and oxygen-release experiments that actually preceded it.
NCERT: NCERT Class 11 Biology, Chapter 11 (Photosynthesis in Higher Plants), page 135
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Revise the lesson: Photosynthesis Chloroplast
The number of action potentials generated by sino-atrial node (SAN) in a healthy human is _______ per minute.
- (1)28 - 30
- (2)70 – 75
- (3)100 - 110
- (4)120 - 140
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerThe SAN is the natural pacemaker of the heart -- it generates the maximum number of action potentials among the nodal tissue, about 70-75 per minute, setting the normal resting heart rate (average ~72 beats/min).
Why the other options are wrong- (1) 28-30/min is far below normal SAN firing rate -- would represent severe bradycardia.
- (3) 100-110/min is above the normal resting rate, closer to mild tachycardia.
- (4) 120-140/min is well above normal resting heart rate, seen only during exertion/tachycardia.
NCERT: NCERT Class 11 Biology, Chapter 15 (Body Fluids and Circulation), page 199
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Revise the lesson: Circulation Heart Cardiac Cycle
How many turns of Calvin cycle are required for the formation of three molecules of glucose?
- (1)6
- (2)3
- (3)1
- (4)18
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerOne turn of the Calvin cycle fixes one CO2; producing a single molecule of glucose (6 carbons) requires the fixation of six CO2 molecules, i.e., six turns of the cycle. So three molecules of glucose need 3 x 6 = 18 turns.
Why the other options are wrong- (1) 6 turns of the Calvin cycle fix only enough CO2 for one glucose molecule; students stop at the per-glucose count instead of scaling it up to three.
- (2) 3 turns is far too few; not even one glucose can be made in 3 turns (needs 6).
- (3) 1 turn fixes only a single CO2, nowhere near enough carbon for even one glucose.
NCERT: NCERT Class 11 Biology, Chapter 11 (Photosynthesis in Higher Plants), page 144
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Revise the lesson: Photosynthesis C3 C4 CAM
Which of the following statements is incorrect?
- (1)Blood coagulates in response to an injury
- (2)Blood clot consists of fibrins
- (3)Fibrin is produced from fibrinogen
- (4)Fibrinogen is produced from fibrin
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerDuring coagulation, the enzyme thrombin converts soluble plasma fibrinogen into insoluble fibrin threads that form the clot network trapping blood cells -- the conversion is fibrinogen to fibrin, never the reverse. So statement 4, which reverses this direction, is the incorrect one.
Why the other options are wrong- (1) True -- coagulation is indeed the physiological response to vessel injury.
- (2) This statement is factually correct -- the clot's mesh is built of fibrin threads formed from fibrinogen; it does not fit the 'incorrect statement' the question asks for.
- (3) True -- this correctly states the actual direction of conversion (fibrinogen to fibrin).
NCERT: NCERT Class 11 Biology, Chapter 15 (Body Fluids and Circulation), page 196
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Revise the lesson: Circulation Blood Groups Coagulation
Match List-I with List-II.
List-I List-II
A. Family I. Sapindales
B. Genus II. Dicotyledonae
C. Class III. Anacardiaceae
D. Phylum IV. Angiospermae
E. Order V. Mangifera
Choose the correct answer from the options given below :
- (1)A-I, B-V, C-II, D-IV, E-III
- (2)A-II, B-I, C-III, D-IV, E-V
- (3)A-II, B-III, C-V, D-I, E-IV
- (4)A-III, B-V, C-II, D-IV, E-I
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerFor mango (Mangifera indica): Family = Anacardiaceae (A-III), Genus = Mangifera (B-V), Class = Dicotyledonae (C-II), Phylum/division = Angiospermae (D-IV), Order = Sapindales (E-I). Note that strictly, 'Angiospermae' is classified botanically as a division rather than a phylum, but the NTA-keyed option 4 is the one matching the standard NCERT taxonomic-category table for mango.
Why the other options are wrong- (1) Mismatches every category except D; wrong pairing throughout.
- (2) This option assigns Anacardiaceae to genus and Mangifera to family, reversing the true family-genus pair for mango's classification.
- (3) This option assigns genus to Anacardiaceae and family to Sapindales (which is really the order), mixing up several taxonomic ranks for mango.
NCERT: NCERT Class 11 Biology, Chapter 1 (The Living World), page 8
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Revise the lesson: Biodiversity Nomenclature
Arrange the following taxonomic categories in ascending order. (a) Genus (b) Class (c) Order (d) Phylum (e) Family (f) Kingdom (g) Species Choose the correct answer from the options given below :
- (1)(g), (a), (e), (c), (b), (d), (f)
- (2)(a), (c), (d), (g), (f), (b), (e)
- (3)(g), (c), (d), (b), (e), (a), (f)
- (4)(f), (c), (b), (g), (d), (e), (a)
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerThe standard taxonomic hierarchy, from the most specific (lowest rank, smallest group of organisms) to the most general (highest rank), is: Species -> Genus -> Family -> Order -> Class -> Phylum -> Kingdom. That is exactly (g),(a),(e),(c),(b),(d),(f).
Why the other options are wrong- (2) Starts with genus, skipping species, and scrambles the remaining order.
- (3) This sequence places order and phylum ahead of family and genus, breaking the true rank order species-genus-family-order-class-phylum-kingdom.
- (4) Runs the hierarchy essentially backwards, starting from kingdom.
NCERT: NCERT Class 11 Biology, Chapter 1 (The Living World), page 6
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Revise the lesson: Biodiversity Nomenclature
Match List-I with List-II.
List-I List-II
A. Marginal placentation I. Argemone
B. Axile placentation II. Tomato
C. Parietal placentation III. Primrose
D. Free central placentation IV. Pea
Choose the correct answer from the options given below :
- (1)A-II, B-IV, C-I, D-III
- (2)A-IV, B-II, C-III, D-I
- (3)A-IV, B-III, C-I, D-II
- (4)A-IV, B-II, C-I, D-III
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerMarginal placentation (ovules on the ridge along the ventral suture, single carpel) = pea (A-IV). Axile placentation (ovules on a central axis, multilocular ovary) = tomato (B-II). Parietal placentation (ovules on the inner ovary wall) = Argemone/mustard family (C-I). Free central placentation (ovules on a central axis without septa) = primrose/Dianthus (D-III).
Why the other options are wrong- (1) This option pairs marginal placentation with tomato (an axile example) and axile with pea (a marginal example), swapping the two.
- (2) This option pairs axile placentation with primrose (a free-central example) and free-central with pea, crossing those two types.
- (3) This option pairs free-central placentation with tomato (an axile example) and axile with primrose, reversing the two.
NCERT: NCERT Class 11 Biology, Chapter 5 (Morphology of Flowering Plants), page 65
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Revise the lesson: Morphology Flowering Plants
Sphenopsida class belongs to _____________.
- (1)bryophytes
- (2)angiosperms
- (3)gymnosperms
- (4)pteridophytes
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerPteridophytes are divided into four classes -- Psilopsida, Lycopsida, Sphenopsida (represented by Equisetum, the horsetail) and Pteropsida -- so Sphenopsida is a pteridophyte class.
Why the other options are wrong- (1) Bryophytes are the mosses/liverworts group, classified separately (not into Sphenopsida etc).
- (2) Angiosperms are flowering seed plants, not divided into Sphenopsida.
- (3) Gymnosperms are naked-seed plants, a separate group from pteridophyte classes.
NCERT: NCERT Class 11 Biology, Chapter 3 (Plant Kingdom), page 32
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Revise the lesson: Plant Kingdom
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 Choose the correct answer from the options given below:
- (1)(a) and (d) only
- (2)(c) and (d) only
- (3)(b) and (d) only
- (4)(a) and (b) only
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerPhotorespiration DOES occur in C3 plants (it is essentially a C3-plant phenomenon caused by RuBisCO's oxygenase activity), so (a) is false. Photorespiration takes up O₂ and gives off CO₂ (the reverse of statement b), so (b) is false. Phosphoglycolate is indeed the initial 2-carbon product (c, true). Unlike photosynthesis, photorespiration involves no synthesis of ATP or NADPH (d, true).
Why the other options are wrong- (1) Wrongly includes false statement (a); photorespiration does occur in C3 plants.
- (3) Wrongly includes false statement (b), which reverses the actual gas exchange.
- (4) Both (a) and (b) included here are false statements about photorespiration.
NCERT: NCERT Class 11 Biology, Chapter 11 (Photosynthesis in Higher Plants), page 147
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Revise the lesson: Photosynthesis C3 C4 CAM
Smooth endoplasmic reticulum ___________ .
- (1)has ribosomes attached to its surface
- (2)is the major site for the synthesis of lipids
- (3)is actively involved in protein synthesis
- (4)is a site for the synthesis of carbohydrates
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerSmooth ER (SER) lacks ribosomes on its surface (hence 'smooth') and is the major site for synthesis of lipids, including steroidal hormones; rough ER, by contrast, has ribosomes and is the main protein-synthesis site.
Why the other options are wrong- (1) Ribosome studding is the hallmark of ROUGH ER, not smooth ER.
- (3) Active protein synthesis is a rough-ER function, not smooth ER's.
- (4) Carbohydrate synthesis mainly occurs in chloroplasts (in plants), not smooth ER.
NCERT: NCERT Class 11 Biology, Chapter 8 (Cell: The Unit of Life), page 95
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Revise the lesson: Cell Organelles Endomembrane
Which one of the following statements is incorrect? (1) α-cells of pancreas secrete glucagon (2) α-cells of pancreas secrete insulin (3) Glucagon stimulates glycogenolysis (4) β-cells of pancreas secrete insulin
- (1)α-cells of pancreas secrete glucagon
- (2)α-cells of pancreas secrete insulin
- (3)Glucagon stimulates glycogenolysis
- (4)β-cells of pancreas secrete insulin
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerIn the Islets of Langerhans, the alpha-cells secrete glucagon and the beta-cells secrete insulin -- alpha-cells do NOT secrete insulin, so statement (2) is the incorrect one. Glucagon indeed stimulates glycogenolysis in the liver, raising blood glucose (hyperglycemic hormone).
Why the other options are wrong- (1) This statement is factually correct -- alpha-cells do secrete glucagon; it is not the incorrect statement the question asks to identify.
- (3) This is a correct fact -- glucagon does trigger glycogenolysis in the liver; it does not fit the 'incorrect statement' being sought.
- (4) This statement is accurate -- insulin is secreted by beta-cells, not alpha-cells; picking it confuses a true fact for the required incorrect one.
NCERT: NCERT Class 11 Biology, Chapter 19 (Chemical Coordination and Integration), page 245
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Revise the lesson: Endocrine Adrenal Pancreas Gonads
Genus represents __________.
- (1)an individual plant or animal
- (2)a population of plants and animals
- (3)a group of closely related species
- (4)a group of closely related families
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerA genus comprises a group of closely related species that share more characters in common with each other than with species of other genera -- e.g., Panthera includes lion, tiger, leopard.
Why the other options are wrong- (1) An individual organism is described at the species/organism level, not a genus.
- (2) A population is a group of interbreeding individuals of one species, a different (lower) concept than genus.
- (4) A group of related families is the definition of an ORDER, not a genus.
NCERT: NCERT Class 11 Biology, Chapter 1 (The Living World), page 7
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Revise the lesson: Biodiversity Nomenclature
Which of the following is not a prokaryote?
- (1)Bacteria
- (2)Blue green algae
- (3)Mycoplasma
- (4)Fungi
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerProkaryotes include bacteria, blue-green algae (cyanobacteria), and Mycoplasma (wall-less bacteria) -- all lack a membrane-bound nucleus. Fungi, however, are eukaryotic organisms with a true nucleus and membrane-bound organelles.
Why the other options are wrong- (1) Bacteria are the textbook example of a prokaryote, lacking a membrane-bound nucleus; picking this confuses a prokaryote example with the 'not a prokaryote' answer being sought.
- (2) Blue-green algae (cyanobacteria) are prokaryotic despite the name 'algae'.
- (3) Mycoplasma is a wall-less prokaryote, the smallest known living cell.
NCERT: NCERT Class 11 Biology, Chapter 2 (Biological Classification), page 14
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Revise the lesson: Monera
Which of the following plant growth regulators promotes internode elongation prior to flowering in cabbage?
- (1)Abscisic acid
- (2)Gibberellin
- (3)Indole butyric acid
- (4)Ethephon
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerGibberellins cause 'bolting' -- rapid internode elongation just before flowering -- in rosette-habit plants such as cabbage and beet.
Why the other options are wrong- (1) Abscisic acid is the growth-inhibiting/stress hormone, promoting dormancy, not elongation.
- (3) Indole butyric acid is an auxin mainly used to promote rooting, not bolting.
- (4) Ethephon releases ethylene, associated with fruit ripening/senescence, not internode elongation.
NCERT: NCERT Class 11 Biology, Chapter 13 (Plant Growth and Development), page 176
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Revise the lesson: Plant Hormones
The correct sequence of adult cell cycle phases is ________.
- (1)G1-G2-S-M
- (2)G1-M-G2-S
- (3)G1-S-G2-M
- (4)S-M-G2-G1
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerInterphase is subdivided into G1 (Gap 1, cell growth), S (Synthesis, DNA replication) and G2 (Gap 2, further growth/preparation), after which the cell enters M phase (mitosis) for division. So the correct order is G1-S-G2-M.
Why the other options are wrong- (1) Places G2 before S, but DNA synthesis (S) must occur between G1 and G2.
- (2) Places M in the middle of interphase, but mitosis only happens after G2, at the end of the cycle.
- (4) This option lists the interphase-to-mitosis order in reverse, starting from S and ending at G1 instead of the true G1-S-G2-M sequence.
NCERT: NCERT Class 11 Biology, Chapter 10 (Cell Cycle and Cell Division), page 121
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Revise the lesson: Cell Cycle Phases
Match List-I with List-II.
List-I List-II
A. Fusion of protoplasms between gametes I. Meiosis
B. Fusion of two nuclei II. Plasmogamy
C. Generation of haploid spores III. Karyogamy
Choose the correct answer from the options given below :
- (1)A-II, B-III, C-I
- (2)A-II, B-I, C-III
- (3)A-III, B-II, C-I
- (4)A-I, B-III, C-II
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerFusion of protoplasms of two gametes is plasmogamy (A-II); subsequent fusion of the two nuclei is karyogamy (B-III); meiosis in the resulting zygote generates haploid spores (C-I).
Why the other options are wrong- (2) This option assigns karyogamy to spore formation and meiosis to nuclear fusion, swapping the true B and C pairings.
- (3) This option calls protoplasm fusion karyogamy and nuclear fusion plasmogamy, reversing the two fusion terms.
- (4) Mismatches all three -- reverses the logical sequence plasmogamy->karyogamy->meiosis.
NCERT: NCERT Class 11 Biology, Chapter 2 (Biological Classification), page 17
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Revise the lesson: Fungi
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 the light of the above statements, choose the most appropriate answer from the options given below:
- (1)Both Statement I and Statement II are correct
- (2)Both Statement I and Statement II are incorrect
- (3)Statement I is correct but Statement II is incorrect
- (4)Statement I is incorrect but Statement II is correct
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerStatement I is correct -- the class name Reptilia derives from Latin repere/reptum (to creep or crawl), reflecting their locomotion. Statement II is false: while most reptiles have a three-chambered heart, crocodiles are the exception with a four-chambered heart, so 'all' reptiles cannot be said to have three chambers.
Why the other options are wrong- (1) Wrong because Statement II is false (crocodiles are an exception).
- (2) Statement I is a verified true fact -- Reptilia's name does derive from the Latin for creeping or crawling; calling both statements false wrongly rejects it.
- (4) Statement I is actually correct, not false; this option wrongly reverses which of the two statements holds true.
NCERT: NCERT Class 11 Biology, Chapter 4 (Animal Kingdom), page 49
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Revise the lesson: Animal Kingdom Chordates
Given below are two statements : Statement I : Chromosomes are fully condensed at the end of prophase I. Statement II : Meiosis I resembles mitosis. In the light of the above statements, choose the most appropriate answer from the options given below :
- (1)Both Statement I and Statement II are true
- (2)Both Statement I and Statement II are false
- (3)Statement I is correct, but Statement II is false
- (4)Statement I is incorrect, but Statement II is true
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerStatement I is correct: diakinesis, the final sub-stage of prophase I, is marked by full chromosome condensation and disappearance of the nucleolus/nuclear envelope. Statement II is false: meiosis I is a REDUCTIONAL division involving pairing of homologues, crossing over and separation of homologous chromosomes -- fundamentally different from mitosis. It is meiosis II that resembles mitosis (equational division, sister chromatids separate).
Why the other options are wrong- (1) Statement II is false -- meiosis I is a reductional division unlike mitosis; it is meiosis II, not meiosis I, that resembles mitosis, so calling both statements true is wrong.
- (2) Wrong because Statement I is a correct, textbook fact about diakinesis.
- (4) Wrong because Statement I is actually correct, not incorrect.
NCERT: NCERT Class 11 Biology, Chapter 10 (Cell Cycle and Cell Division), page 126
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Revise the lesson: Meiosis Phases
Which of the following is not a characteristic of chordates?
- (1)Presence of notochord
- (2)Central nervous system is dorsal
- (3)Absence of gills
- (4)Presence of post anal part (tail)
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerChordates are fundamentally defined by a notochord, a dorsal, hollow nerve cord, paired pharyngeal gill slits, and a post-anal tail. So 'absence of gills' is actually the opposite of the true defining feature (pharyngeal gill slits ARE present at some stage), making it the characteristic that does NOT apply to chordates.
Why the other options are wrong- (1) A notochord is a genuine defining chordate feature, present at some life stage in every chordate; picking it confuses a true trait with the 'not a characteristic' answer sought.
- (2) A dorsal (not ventral) nerve cord is a genuine chordate feature.
- (4) A post-anal tail at some stage of life is a genuine chordate feature.
NCERT: NCERT Class 11 Biology, Chapter 4 (Animal Kingdom), page 45
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Revise the lesson: Animal Kingdom Chordates
Length of the stem at time 0 is 20 cm. The arithmetic growth rate is 30 cm per day. What is the length of the stem at the end of the 7th day?
- (1)50 cm
- (2)170 cm
- (3)230 cm
- (4)460 cm
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerArithmetic growth follows Lt = L0 + rt, where L0 is initial length, r is the constant growth rate, and t is time. Here Lt = 20 + 30 x 7 = 20 + 210 = 230 cm.
Why the other options are wrong- (1) 50 cm would result from only one day's growth added to L0 (20+30), ignoring the other 6 days.
- (2) 170 cm undercounts the total added growth (would come from a smaller/incorrect multiplier).
- (4) 460 cm overshoots -- it looks like doubling the correct 230 cm answer, not the linear formula's result.
NCERT: NCERT Class 11 Biology, Chapter 13 (Plant Growth and Development), page 170
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Revise the lesson: Plant Growth Phases
Arrange the following elements in descending order of their contribution to percentage weight of the human body. (a) Oxygen (b) Carbon (c) Hydrogen (d) Nitrogen Choose the correct answer from the options given below :
- (1)(a), (b), (c), (d)
- (2)(c), (a), (b), (d)
- (3)(b), (c), (d), (a)
- (4)(b), (a), (c), (d)
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerBy percentage weight of the human body, oxygen contributes about 65.0%, carbon about 18.5%, hydrogen about 9.5%, and nitrogen about 3.3%. So in descending order: Oxygen > Carbon > Hydrogen > Nitrogen, i.e., (a),(b),(c),(d).
Why the other options are wrong- (2) Places hydrogen (9.5%) ahead of oxygen (65%) and carbon (18.5%), badly out of order.
- (3) Starts with carbon, ranking it above oxygen, the actual highest-percentage element.
- (4) This ordering places carbon (about 18.5% by weight) ahead of oxygen (about 65%), reversing the two highest-contributing elements.
NCERT: NCERT Class 11 Biology, Chapter 9 (Biomolecules), page 105
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Revise the lesson: Biomolecules Carbohydrates Lipids
In frogs, the number of pairs of cranial nerves arising from the brain are ______.
- (1)6
- (2)9
- (3)10
- (4)12
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerIn frogs, ten pairs of cranial nerves arise directly from the brain (compared to twelve pairs in humans/mammals).
Why the other options are wrong- (1) 6 pairs undercounts the frog's cranial nerves; students may recall only a partial list of named nerves instead of the full ten pairs.
- (2) 9 pairs is close to but one short of the true count of ten cranial-nerve pairs in frogs, a common off-by-one slip.
- (4) 12 pairs is the number in humans/higher vertebrates, not frogs.
NCERT: NCERT Class 11 Biology, Chapter 7 (Structural Organisation in Animals), page 83
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Revise the lesson: Frog Anatomy
Which of the following is used as a clot buster?
- (1)Streptokinase
- (2)Penicillin
- (3)Cyclosporin A
- (4)Statins
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerStreptokinase, produced by the bacterium Streptococcus and modified by genetic engineering, is used clinically as a 'clot buster' to dissolve blood clots in patients who have had a myocardial infarction (heart attack).
Why the other options are wrong- (2) Penicillin is an antibiotic that kills bacteria; it has no clot-dissolving action.
- (3) Cyclosporin A is an immunosuppressant used to prevent organ-transplant rejection, not a thrombolytic.
- (4) Statins lower blood cholesterol to slow plaque formation; they do not dissolve an existing clot.
NCERT: NCERT Class 12 Biology, Chapter 8 (Microbes in Human Welfare), page 153
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Revise the lesson: Microbes Household Industrial
The inactive form of Bt toxin is converted to the active form in the insect gut _______
- (1)due to alkaline pH
- (2)due to acidic pH
- (3)by proteases
- (4)by nucleases
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerBt toxin exists in the bacterium as an inactive protoxin crystal. Once an insect larva ingests it, the alkaline pH of the insect gut solubilises the crystal, converting the protoxin into its active, insecticidal form, which then binds to the gut epithelium and kills the insect.
Why the other options are wrong- (2) An acidic gut environment would not solubilise the crystal; it is the alkaline gut of the target insect that activates the toxin.
- (3) Proteases act on the toxin only after it has been solubilised; they are not what triggers the initial activation.
- (4) Nucleases act on nucleic acids and play no role in activating a protein toxin.
NCERT: NCERT Class 12 Biology, Chapter 10 (Biotechnology and its Applications), page 179
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Revise the lesson: Biotech Agriculture Bt GM
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 chromosomes.
In the light of the above statements, choose the correct answer from the options given below :
- (1)Both Statement I and Statement II are correct
- (2)Both Statement I and Statement II are incorrect
- (3)Statement I is correct but Statement II is incorrect
- (4)Statement I is incorrect but Statement II is correct
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerDown's syndrome is trisomy of chromosome 21 (an extra copy, 47 chromosomes total), not the loss of an X-chromosome. Turner's syndrome is caused by the loss of one X-chromosome (45, X0 / monosomy), not an additional chromosome copy. Both statements have swapped the two disorders' causes, so both are incorrect.
Why the other options are wrong- (1) Both statements have the disorders' causes swapped: Down's syndrome comes from an extra chromosome 21, and Turner's syndrome from a missing X, the reverse of what is claimed here.
- (3) Statement I is also incorrect, not just Statement II -- Down's syndrome is caused by an extra chromosome 21, not a missing X-chromosome.
- (4) Statement II is also incorrect, not just Statement I -- Turner's syndrome results from losing an X-chromosome, not gaining an extra chromosome.
NCERT: NCERT Class 12 Biology, Chapter 4 (Principles of Inheritance and Variation), page 75
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Revise the lesson: Mutations Genetic Disorders
Which of the following disease is not sexually transmitted?
- (1)Syphilis
- (2)Tuberculosis
- (3)Gonorrhoea
- (4)Genital warts
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerSyphilis, gonorrhoea and genital warts are classic sexually transmitted infections (STIs), caused respectively by Treponema pallidum, Neisseria gonorrhoeae and Human papilloma virus. Tuberculosis is caused by Mycobacterium tuberculosis and spreads through air (droplet infection), not sexual contact.
Why the other options are wrong- (1) Syphilis, caused by the bacterium Treponema pallidum, is a genuine sexually transmitted disease; picking it confuses an STI with the 'not sexually transmitted' answer sought.
- (3) Gonorrhoea, caused by Neisseria gonorrhoeae, is a well-known sexually transmitted disease; it does not fit the 'not sexually transmitted' answer the question asks for.
- (4) Genital warts, caused by human papilloma virus, spread through sexual contact; picking it confuses a genuine STI with the disease that is NOT sexually transmitted.
NCERT: NCERT Class 12 Biology, Chapter 3 (Reproductive Health), page 42
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Revise the lesson: Reproductive Health STDs
Sperm motility is due to ___________.
- (1)flagellar movement
- (2)ciliary movement
- (3)amoeboid movement
- (4)muscular movement
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerEach human sperm has a long tail (flagellum) whose whip-like lashing propels the sperm forward through the female reproductive tract, so sperm motility is due to flagellar movement.
Why the other options are wrong- (2) Ciliary movement (coordinated beating of many short cilia) moves fluid/ovum along the oviduct, not the sperm itself.
- (3) Amoeboid movement, using pseudopodia, is seen in cells like leucocytes, not in sperm.
- (4) Sperm have no muscle tissue; there is no muscular movement involved.
NCERT: NCERT Class 12 Biology, Chapter 2 (Human Reproduction), page 32
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Revise the lesson: Male Reproductive System
Natural selection can lead to _________. (a) stabilisation (b) genetic drift (c) directional change (d) disruption
Choose the correct answer from the options given below :
- (1)(a) only
- (2)(a), (c) and (d) only
- (3)(a), (b), (c) and (d)
- (4)(a) and (c) only
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerNCERT states natural selection can lead to stabilisation (more individuals acquire the mean character value), directional change (individuals shift towards one extreme) or disruption (individuals move towards both extremes). Genetic drift is a separate mechanism of evolution driven by chance sampling of alleles, not an outcome of natural selection itself, so (b) is excluded.
Why the other options are wrong- (1) Directional change and disruption are also valid outcomes of natural selection, so 'only (a)' is too narrow.
- (3) Genetic drift (b) is a distinct, chance-based evolutionary mechanism, not a consequence of natural selection.
- (4) Disruption (d) is also a recognised outcome of natural selection and should be included.
NCERT: NCERT Class 12 Biology, Chapter 6 (Evolution), page 121
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Revise the lesson: Natural Selection Drift
The method of directly injecting a sperm into ovum in assisted reproductive technology is called :
- (1)Gamete intra fallopian transfer (GIFT)
- (2)Zygote intra fallopian transfer (ZIFT)
- (3)Intra cytoplasmic sperm injection (ICSI)
- (4)Embryo transfer (ET)
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerIntra Cytoplasmic Sperm Injection (ICSI) is the assisted-reproduction procedure in which a sperm is directly injected into the cytoplasm of the ovum to form an embryo in the laboratory.
Why the other options are wrong- (1) GIFT places a donor's ovum in the fallopian tube of a woman who cannot produce ova but can support fertilisation and pregnancy; it does not involve direct sperm injection.
- (2) ZIFT is the transfer of a zygote or early embryo (up to 8 blastomeres) into the fallopian tube, done after normal fertilisation, not by injecting sperm.
- (4) Embryo transfer places an already-formed embryo into the uterus/fallopian tube; it does not describe the fertilisation step itself.
NCERT: NCERT Class 12 Biology, Chapter 3 (Reproductive Health), page 48
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Revise the lesson: Amniocentesis Art
Which of the following structure is not a part of the male reproductive system?
- (1)Rete testis
- (2)Epididymis
- (3)Vasa efferentia
- (4)Infundibulum
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerSperm leave the testis through a chain of ducts: rete testis, then vasa efferentia, then epididymis, then vas deferens. The infundibulum is the funnel-shaped, fimbriated opening of the fallopian tube that collects the ovum after ovulation, so it belongs to the female reproductive system, not the male.
Why the other options are wrong- (1) Rete testis is a network of tubules inside the testis that is part of the male duct system.
- (2) Epididymis is the coiled duct that stores and matures sperm, a male accessory duct.
- (3) Vasa efferentia are the ducts connecting the rete testis to the epididymis, part of the male system.
NCERT: NCERT Class 12 Biology, Chapter 2 (Human Reproduction), page 29
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Revise the lesson: Female Reproductive System
Arrange the following in descending order of number of species in the Amazonian rain forest.
(a) Plants (b) Birds (c) Fishes (d) Invertebrates (e) Mammals
Choose the correct answer from the options given below :
- (1)(c) > (b) > (d) > (e) > (a)
- (2)(d) > (a) > (c) > (b) > (e)
- (3)(e) > (b) > (a) > (c) > (d)
- (4)(b) > (a) > (d) > (c) > (e)
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerNCERT's Amazon rain forest figures are: more than 40,000 plant species, about 3,000 fish species, about 1,300 bird species, 427 mammal species, 378 reptile species and more than 1,25,000 invertebrate species. Ranking these five groups in descending order gives invertebrates > plants > fishes > birds > mammals, i.e. (d) > (a) > (c) > (b) > (e).
Why the other options are wrong- (1) Places fishes ahead of plants and invertebrates, which is incorrect; invertebrates (>1,25,000) and plants (>40,000) far outnumber fishes (~3,000).
- (3) Places mammals first, but mammals (427) are the least numerous of the five groups listed.
- (4) Places birds first and invertebrates too low; invertebrates vastly outnumber birds.
NCERT: NCERT Class 12 Biology, Chapter 13 (Biodiversity and Conservation), page 219
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Revise the lesson: Biodiversity Patterns
Given below are two statements:
Statement I : Ovulation is caused by LH surge leading to rupture of Graafian follicles.
Statement II: Graafian follicle remaining after ovulation transform into corpus luteum and secretes large amount of estrogen.
In the light of the above statements, choose the most appropriate answer from the options given below:
- (1)Both Statement I and Statement II are correct
- (2)Both Statement I and Statement II are incorrect
- (3)Statement I is correct but Statement II is incorrect
- (4)Statement I is incorrect but Statement II is correct
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerStatement I is correct: rapid secretion of LH (the LH surge) at mid-cycle induces rupture of the Graafian follicle, i.e. ovulation. Statement II is incorrect because the remnant of the Graafian follicle forms the corpus luteum, whose main hormone is progesterone, not estrogen (estrogen is secreted mainly by the growing follicle before ovulation).
Why the other options are wrong- (1) Statement II wrongly names estrogen instead of progesterone as the corpus luteum's secretion.
- (2) Statement I correctly describes the LH surge causing ovulation, so it cannot be called incorrect.
- (4) It is Statement I that holds true (the LH surge causes ovulation), not Statement II; this option wrongly swaps which statement is the correct one.
NCERT: NCERT Class 12 Biology, Chapter 2 (Human Reproduction), page 35
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Revise the lesson: Menstrual Cycle
Which of the following are primary consumers in a food chain?
- (1)Parasites
- (2)Predators
- (3)Herbivores
- (4)Carnivores
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerOrganisms that eat the producers (plants) directly occupy the second trophic level and are called primary consumers; these are the herbivores.
Why the other options are wrong- (1) Parasites live on/in a host and feed on it; they are not defined by trophic position as primary consumers.
- (2) Predators can occupy any consumer level depending on what they eat; the term itself does not mean primary consumer.
- (4) Carnivores eat other animals (secondary/tertiary consumers), not producers, so they are not primary consumers.
NCERT: NCERT Class 12 Biology, Chapter 12 (Ecosystem), page 209
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Revise the lesson: Ecosystem Structure Function
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)0.01
- (2)0.02
- (3)0.10
- (4)0.99
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerUnder Hardy-Weinberg equilibrium, genotype frequencies follow p^2 + 2pq + q^2 = 1, where p is the frequency of allele A. Given p = 0.1, the frequency of the AA genotype = p^2 = (0.1)^2 = 0.01.
Why the other options are wrong- (2) 0.02 would come from an arithmetic slip (e.g. doubling p rather than squaring it).
- (3) 0.10 is simply the allele frequency p itself, not the AA genotype frequency p^2.
- (4) 0.99 does not equal p^2; it is close to a miscalculation of q^2 or of 'not-A' proportions.
NCERT: NCERT Class 12 Biology, Chapter 6 (Evolution), page 121
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Revise the lesson: Hardy Weinberg Principle
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
Choose the correct answer from the options given below:
- (1)A-III, B-IV, C-II, D-I
- (2)A-III, B-IV, C-I, D-II
- (3)A-II, B-IV, C-I, D-III
- (4)A-IV, B-III, C-I, D-II
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerExcess growth hormone in adults causes acromegaly (A-III). Luteinizing hormone induces ovulation of the mature Graafian follicle (B-IV). Vasopressin (ADH) promotes reabsorption of water and electrolytes in the kidney (C-I). Oxytocin stimulates uterine contraction during child birth (D-II). This gives A-III, B-IV, C-I, D-II.
Why the other options are wrong- (1) Swaps C and D: vasopressin (not oxytocin) governs water/electrolyte reabsorption, and oxytocin (not vasopressin) causes uterine contraction.
- (3) Assigns acromegaly to oxytocin (D-III) and uterine contraction to growth hormone (A-II), both wrong.
- (4) Assigns ovulation to excess growth hormone (A-IV) and acromegaly to LH (B-III), both wrong.
NCERT: NCERT Class 11 Biology, Chapter 19 (Chemical Coordination and Integration), page 241
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Revise the lesson: Endocrine Hypothalamus Pituitary Thyroid
The opening between the right atrium and the right ventricle is guarded by _________.
- (1)bicuspid valve
- (2)tricuspid valve
- (3)semilunar valve
- (4)sino-atrial node
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerThe opening between the right atrium and right ventricle is guarded by a valve formed of three muscular flaps (cusps), called the tricuspid valve.
Why the other options are wrong- (1) The bicuspid (mitral) valve sits on the left side, between left atrium and left ventricle, so it cannot be the answer for the right side.
- (3) Semilunar valves guard the openings of the ventricles into the pulmonary artery and aorta, not the atrio-ventricular opening.
- (4) The sino-atrial node is the heart's pacemaker tissue in the right atrial wall, not a valve.
NCERT: NCERT Class 11 Biology, Chapter 15 (Body Fluids and Circulation), page 198
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Revise the lesson: Circulation Heart Cardiac Cycle
Sponges exchange O2 with CO2 by __________.
- (1)simple diffusion over their entire body surfaces
- (2)moist cuticle
- (3)tracheal tubes
- (4)gills
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerLower invertebrates such as sponges, coelenterates and flatworms exchange O2 and CO2 by simple diffusion over their entire body surface because they lack specialised respiratory organs.
Why the other options are wrong- (2) A moist cuticle is used by earthworms for gas exchange, not sponges.
- (3) Tracheal tubes are the gas-exchange system of insects; sponges lack these.
- (4) Gills are used by aquatic arthropods and molluscs for gas exchange, not by sponges.
NCERT: NCERT Class 11 Biology, Chapter 14 (Breathing and Exchange of Gases), page 183
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Revise the lesson: Breathing Respiratory Organs
How many theca are present in each lobe of a typical bilobed angiosperm anther ?
- (1)2
- (2)6
- (3)8
- (4)12
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerA typical angiosperm anther is bilobed, and each lobe has two theca (making it dithecous overall). A longitudinal groove runs lengthwise separating the two theca of a lobe.
Why the other options are wrong- (2) 6 theca per lobe does not correspond to any standard anther description in NCERT.
- (3) 8 theca per lobe is far in excess of the dithecous condition described for a typical anther.
- (4) 12 theca per lobe is not supported by the NCERT description of a bilobed, dithecous anther.
NCERT: NCERT Class 12 Biology, Chapter 1 (Sexual Reproduction in Flowering Plants), page 5
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Revise the lesson: Flower Structure Gametogenesis
Muscle contraction is initiated by a signal sent by the central nervous system by the release of _____ .
- (1)acetyl choline
- (2)acetyl coenzyme A
- (3)cyclic guanine monophosphate
- (4)cyclic adenine monophosphate
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerA signal from the CNS travels via a motor neuron to the neuromuscular junction (motor end plate), where the neurotransmitter acetylcholine is released. This generates an action potential in the sarcolemma, initiating muscle contraction.
Why the other options are wrong- (2) Acetyl coenzyme A is a metabolic intermediate (e.g. entering the Krebs cycle), not the neuromuscular transmitter.
- (3) Cyclic GMP is a second messenger in some signalling pathways (e.g. vision), not the neuromuscular junction transmitter.
- (4) Cyclic AMP is a second messenger for many hormones, not the neurotransmitter released at the neuromuscular junction.
NCERT: NCERT Class 11 Biology, Chapter 17 (Locomotion and Movement), page 222
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Revise the lesson: Locomotion Muscle Contraction
Which of the following statements about lac-operon is correct?
- (1)Gene i is constitutively expressed
- (2)Lactose activates repressor to bind to the operator
- (3)Genes i, z, y and a share single common promoter
- (4)Galactose can act as an inducer of lac operon
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerThe regulatory gene i of the lac operon is constitutively (always) expressed, continuously producing the repressor protein. When lactose (via allolactose) is present, it binds the repressor and inactivates it, so the repressor cannot bind the operator, allowing transcription of the structural genes.
Why the other options are wrong- (2) Lactose inactivates the repressor (prevents it from binding the operator); it does not activate the repressor to bind.
- (3) Gene i has its own separate promoter from the shared promoter of the structural genes z, y and a; they are not all under one common promoter.
- (4) Lactose (via allolactose), not galactose, acts as the inducer of the lac operon.
NCERT: NCERT Class 12 Biology, Chapter 5 (Molecular Basis of Inheritance), page 101
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Revise the lesson: Gene Regulation Lac Operon
Which of the following in female gametophyte of an angiosperm helps in guiding the pollen tube for fertilizing the eggs?
- (1)Antipodals
- (2)Synergids
- (3)Central cells
- (4)Polar nucleus
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerThe two synergid cells at the micropylar end of the embryo sac have a special cellular thickening called the filiform apparatus, which guides the entry of the pollen tube into one of the synergids so the male gametes can be discharged.
Why the other options are wrong- (1) Antipodals are three cells at the chalazal end of the embryo sac with no known role in guiding the pollen tube.
- (3) The central cell contains the two polar nuclei and is where triple fusion occurs, but it does not guide pollen tube entry.
- (4) The polar nucleus/nuclei fuse with a male gamete during triple fusion; they play no role in guiding the pollen tube.
NCERT: NCERT Class 12 Biology, Chapter 1 (Sexual Reproduction in Flowering Plants), page 11
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Revise the lesson: Pollination Fertilization
Which of the following plant produces non-albuminous seeds?
- (1)Wheat
- (2)Maize
- (3)Barley
- (4)Pea
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerNon-albuminous seeds have no residual endosperm because it is completely used up by the developing embryo; pea is the standard NCERT example. Wheat, maize and barley retain endosperm and are albuminous seeds.
Why the other options are wrong- (1) Wheat is a classic example of an albuminous (endosperm-retaining) seed.
- (2) Maize retains endosperm as the major food-storage tissue and is albuminous.
- (3) Barley retains endosperm as its food reserve at maturity, making it an albuminous seed like wheat and maize, not the non-albuminous example being asked for.
NCERT: NCERT Class 12 Biology, Chapter 1 (Sexual Reproduction in Flowering Plants), page 20
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Revise the lesson: Post Fertilization Seed Fruit
If the diploid chromosome number of typical angiosperm is 36, what would be the chromosome number in its endosperm?
- (1)18
- (2)36
- (3)54
- (4)72
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerThe angiosperm endosperm is triploid (3n), formed by triple fusion of one male gamete (n) with the two polar nuclei (n + n) of the central cell. Given 2n = 36, n = 18, so the endosperm's chromosome number = 3n = 3 x 18 = 54.
Why the other options are wrong- (1) 18 is just n (the haploid number), not the triploid endosperm number.
- (2) 36 is the diploid (2n) number of the parent sporophyte/zygote, not the endosperm, which is triploid.
- (4) 72 would be 4n, which does not correspond to the triploid nature of angiosperm endosperm.
NCERT: NCERT Class 12 Biology, Chapter 1 (Sexual Reproduction in Flowering Plants), page 18
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Revise the lesson: Post Fertilization Seed Fruit
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.
Choose the correct answer from the options given below:
- (1)(a) and (b) only
- (2)(b), (c) and (d) only
- (3)(a), (b) and (c) only
- (4)(a), (b) and (d) only
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerNCERT states the descending limb of Henle's loop is permeable to water but almost impermeable to electrolytes (a-true), and the counter-current arrangement of the loop concentrates the urine (b-true). NCERT also describes active/passive NaCl transport in the ascending limb (d-true). Statement (c), read as a blanket claim that Na+ AND water are reabsorbed together in the loop, does not match NCERT: water leaves only via the descending limb, while the ascending limb is impermeable to water and reabsorbs only Na+/electrolytes, so the two are not co-reabsorbed in the same segment. NCERT therefore supports (a), (b) and (d); we publish the NTA key, option 4, as given.
Why the other options are wrong- (1) Omits (d), which is a correct statement about ascending-limb electrolyte transport.
- (2) Includes (c), which wrongly implies simultaneous Na+ and water reabsorption in the same segment of the loop, and omits (a), a directly quoted NCERT fact.
- (3) Includes (c) for the same reason as above, and omits (d), a valid NCERT-supported statement.
NCERT: NCERT Class 11 Biology, Chapter 16 (Excretory Products and their Elimination), page 209
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Revise the lesson: Excretion Urine Formation
Which of the following is the correct order of arrangement of vertebrate column from the head to toe?
- (1)Cervical vertebra, thoracic vertebra, sacrum, lumbar vertebra
- (2)Sacrum, lumbar vertebra, thoracic vertebra, cervical vertebra
- (3)Cervical vertebra, lumbar vertebra, thoracic vertebra, sacrum
- (4)Cervical vertebra, thoracic vertebra, lumbar vertebra, sacrum
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerThe human vertebral column, from the skull downward, is differentiated into cervical (7), thoracic (12), lumbar (5), sacral (1, fused) and coccygeal (1, fused) regions, in that order, so the correct head-to-toe sequence is cervical, thoracic, lumbar, sacrum.
Why the other options are wrong- (1) Places sacrum before lumbar vertebrae, reversing their actual head-to-toe order.
- (2) Lists the regions in exactly the reverse (toe-to-head) order.
- (3) Places lumbar vertebrae before thoracic vertebrae, which is incorrect; thoracic precedes lumbar.
NCERT: NCERT Class 11 Biology, Chapter 17 (Locomotion and Movement), page 225
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Revise the lesson: Locomotion Skeleton Joints Disorders
Which of the following is not evidence for evolution?
- (1)Convergent evolution of traits like wings of birds and butterflies
- (2)Paleontological evidence from fossil records
- (3)Embryological support for evolution as proposed by Ernst Heckel
- (4)Divergent evolution of anatomical structures such as forelimbs
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerHeckel's embryological 'biogenetic law' (that embryonic stages recapitulate ancestral adult forms) was proposed as support for evolution but was later disapproved after careful study by Karl Ernst von Baer, so it is not valid evidence for evolution. Fossil records, and divergent evolution producing homologous structures, are genuine, currently accepted lines of evidence.
Why the other options are wrong- (1) Convergent evolution of analogous structures (wings of birds/insects) does illustrate adaptive evolution by natural selection, unlike Heckel's disproved recapitulation idea.
- (2) Fossil records are a standard, currently accepted line of paleontological evidence for evolution.
- (4) Divergent evolution producing homologous structures (e.g. forelimbs) is a core, currently accepted evidence for common ancestry.
NCERT: NCERT Class 12 Biology, Chapter 6 (Evolution), page 113
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Revise the lesson: Evidences Evolution
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.
In the light of the above statements, choose the most appropriate answer from the options given below :
- (1)Both Statement I and Statement II are correct
- (2)Both Statement I and Statement II are incorrect
- (3)Statement I is correct but Statement II is incorrect
- (4)Statement I is incorrect but Statement II is correct
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerNCERT states Homo sapiens arose in Africa, not Australia, and moved across continents, so Statement I is incorrect. Modern Homo sapiens arose during the ice age between 75,000 and 10,000 years ago, so Statement II is correct.
Why the other options are wrong- (1) Statement I wrongly places the origin of Homo sapiens in Australia instead of Africa.
- (2) Statement II is factually correct per NCERT's stated time range, so it cannot be called incorrect.
- (3) It is Statement I that is wrong -- Homo sapiens arose in Africa, not Australia; this option mistakenly flags the correct time-range statement (Statement II) instead.
NCERT: NCERT Class 12 Biology, Chapter 6 (Evolution), page 125
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Revise the lesson: Hominid Evolution
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)1 million
- (2)5 million
- (3)10 million
- (4)100 million
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerPopulation growth rate r = (b - d) = 0.002 - 0.002 = 0. Since dN/dt = rN = 0, the population size does not change over time regardless of the number of generations, so after 10 generations the population remains 10 million cells.
Why the other options are wrong- (1) 1 million would imply the population declined to a tenth, which contradicts r = 0 (no net change).
- (2) 5 million would imply the population halved, again inconsistent with a zero net growth rate.
- (4) 100 million would imply tenfold growth, but with birth rate equal to death rate there is no net growth.
NCERT: NCERT Class 12 Biology, Chapter 11 (Organisms and Populations), page 194
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Revise the lesson: Population Attributes Growth
During PCR, primers bind to the DNA strands in the ______step.
- (1)denaturation
- (2)extension
- (3)annealing
- (4)ligation
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerPCR has three steps -- denaturation (strand separation by heat), annealing (primers bind/anneal to the single-stranded template) and extension (Taq polymerase extends the primers). Primer binding to the DNA strands occurs specifically in the annealing step.
Why the other options are wrong- (1) Denaturation is the heating step that separates the double-stranded DNA into single strands; primers are not bound yet at this stage.
- (2) Extension is when the polymerase synthesises new DNA from the already-bound primers, not when primers first bind.
- (4) Ligation is a separate step (used e.g. when joining DNA fragments/vectors), not part of the three-step PCR cycle.
NCERT: NCERT Class 12 Biology, Chapter 9 (Biotechnology: Principles and Processes), page 172
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Revise the lesson: rDNA Process PCR Blotting
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The logistic growth model of populations is considered more realistic than the exponential growth model.
Reason R : Resources are finite.
In the light of the above statements, choose the most appropriate answer from the options given below :
- (1)Both A and R are correct and R is the correct explanation of A
- (2)Both A and R are correct but R is not the correct explanation of A
- (3)A is correct but R is not correct
- (4)A is not correct but R is correct
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerAssertion A is correct -- the logistic model, which incorporates a carrying capacity, is considered more realistic than the unrestricted exponential model. Reason R is also correct and is exactly why: since resources for growth are finite and become limiting, growth cannot continue exponentially forever, which is precisely the assumption built into the logistic model.
Why the other options are wrong- (2) R directly explains why the logistic model is more realistic (finite resources cap growth), so it is the correct explanation, not an unrelated true fact.
- (3) Reason R (finite resources) is a true, NCERT-stated premise, not an incorrect one.
- (4) Assertion A is also correct, not false; the logistic model genuinely is considered more realistic.
NCERT: NCERT Class 12 Biology, Chapter 11 (Organisms and Populations), page 196
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Revise the lesson: Population Attributes Growth
Adaptive radiation in placental mammals and Australian Marsupials leading to similarity between distant species is an example of ________.
- (1)divergent evolution
- (2)convergent evolution
- (3)founder effect
- (4)genetic drift
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerAustralian marsupials underwent adaptive radiation in isolation to fill ecological niches similar to those filled by placental mammals elsewhere. Because this independent, parallel radiation in geographically isolated areas produced superficially similar (but not closely related) forms, it is termed convergent evolution.
Why the other options are wrong- (1) Divergent evolution describes one ancestral form giving rise to different structures/species adapting to different needs (e.g. Darwin's finches), not two separate lineages converging on similar forms.
- (3) Founder effect is genetic drift arising when a small group starts a new, isolated population; it does not describe the parallel adaptive radiation of marsupials and placental mammals.
- (4) Genetic drift is a chance-driven change in allele frequency in small populations, unrelated to this large-scale radiation comparison.
NCERT: NCERT Class 12 Biology, Chapter 6 (Evolution), page 117
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Revise the lesson: Evidences Evolution
Which of the following are secondary lymphoid organs? (a) Bone marrow (b) Tonsils (c) Spleen (d) Thymus
Choose the correct answer from the options given below:
- (1)(a) and (b) only
- (2)(b) and (c) only
- (3)(b) and (d) only
- (4)(a) and (d) only
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerLymphocytes are made and mature in bone marrow and the thymus, which is why these two count as primary lymphoid organs. Tonsils and spleen are secondary lymphoid organs, where mature lymphocytes interact with antigen and proliferate into effector cells, so the correct pair is (b) and (c).
Why the other options are wrong- (1) Bone marrow (a) is a primary, not secondary, lymphoid organ.
- (3) Thymus is where T-lymphocytes mature, making it a primary lymphoid organ; students confuse it with secondary organs since both terms relate to lymphocyte development.
- (4) Bone marrow (a) and thymus (d) are where lymphocytes form and mature, i.e. primary, not secondary, lymphoid organs; neither belongs in the secondary category.
NCERT: NCERT Class 12 Biology, Chapter 7 (Human Health and Disease), page 138
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Revise the lesson: Immunity Innate Acquired
Which of the following hormone is not secreted by human placenta?
- (1)hCG
- (2)Estrogen
- (3)Progesterone
- (4)LH
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerDuring pregnancy the placenta also works as a hormone-producing organ, releasing hCG, human placental lactogen (hPL), estrogens and progestogens. Luteinizing hormone (LH), by contrast, is secreted by the anterior pituitary, not the placenta.
Why the other options are wrong- (1) hCG (human chorionic gonadotropin) is a classic placental hormone, used as the basis of pregnancy tests.
- (2) Estrogen is secreted by the placenta during pregnancy, among other sources.
- (3) Progesterone is secreted by the placenta (as progestogens) to maintain pregnancy.
NCERT: NCERT Class 12 Biology, Chapter 2 (Human Reproduction), page 37
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Revise the lesson: Fertilization Pregnancy
Which of the following enzymes synthesizes precursor mRNA?
- (1)RNA polymerase I
- (2)RNA polymerase II
- (3)RNA polymerase III
- (4)DNA polymerase
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerIn eukaryotes, RNA polymerase II transcribes the precursor of mRNA, the heterogeneous nuclear RNA (hnRNA), which is subsequently processed into mature mRNA.
Why the other options are wrong- (1) RNA polymerase I transcribes rRNA (28S, 18S and 5.8S), not precursor mRNA.
- (3) RNA polymerase III transcribes tRNA, 5S rRNA and other small nuclear RNAs, not precursor mRNA.
- (4) DNA polymerase synthesises DNA during replication; it does not transcribe RNA at all.
NCERT: NCERT Class 12 Biology, Chapter 5 (Molecular Basis of Inheritance), page 95
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Revise the lesson: Transcription Translation
Given below are two statements :
Statement I : Plasmids are autonomously replicating DNA.
Statement II : Plasmids are extrachromosomal DNA.
In the light of the above statements, choose the most appropriate answer from the options given below :
- (1)Both Statement I and Statement II are correct
- (2)Both Statement I and Statement II are incorrect
- (3)Statement I is correct but Statement II is incorrect
- (4)Statement I is incorrect but Statement II is correct
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerPlasmids are circular, extrachromosomal DNA molecules found in bacterial cytoplasm that can replicate independently of the bacterial chromosome (autonomously replicating). Both statements correctly describe plasmids, so both are correct.
Why the other options are wrong- (2) Both statements are accurate descriptions of plasmids, so they cannot both be incorrect.
- (3) Statement II is also correct (plasmids are indeed extrachromosomal), not incorrect.
- (4) Statement I is also correct (plasmids do replicate autonomously), not incorrect.
NCERT: NCERT Class 12 Biology, Chapter 9 (Biotechnology: Principles and Processes), page 164
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Revise the lesson: rDNA Tools Restriction Ligase
For a person with blood group 'O', which of the following is not a possible combination of parents' blood group genotypes?
- (1)Father : IAi and Mother : IBi
- (2)Father : IAi and Mother : IAi
- (3)Father : IBi and Mother : IBi
- (4)Father : IAIB and Mother : IAi
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerAn 'O' blood group child (genotype ii) needs an 'i' allele from each parent. A father with genotype IAIB has no 'i' allele at all to pass on, so this parental combination can never produce an 'O' (ii) child, unlike the other three combinations, each of which has an 'i' allele available from both parents.
Why the other options are wrong- (1) IAi (father) x IBi (mother) can each contribute an i allele, giving an ii ('O') child -- this combination is possible.
- (2) IAi x IAi can both contribute an i allele, giving an ii child -- possible.
- (3) IBi x IBi can both contribute an i allele, giving an ii child -- possible.
NCERT: NCERT Class 12 Biology, Chapter 4 (Principles of Inheritance and Variation), page 62
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Revise the lesson: Incomplete Dominance Codominance
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Forelimbs of human and bats are homologous.
Reason R : Forelimbs of humans and bats have similar anatomical structure.
In the light of the above statements, choose the most appropriate answer from the options given below.
- (1)Both A and R are correct and R is the correct explanation of A
- (2)Both A and R are true, but R is not the correct explanation of A
- (3)A is true but R is false
- (4)A is false but R is true
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerHuman and bat forelimbs share the same basic bone arrangement (humerus, radius, ulna, carpals, metacarpals, phalanges) despite performing different functions, which is exactly what makes them homologous -- structures with a common origin/basic plan that diverged for different needs. So A is true, R is true, and R is the correct explanation of A.
Why the other options are wrong- (2) R is not merely a coincidental true fact -- shared anatomical structure is precisely the criterion that defines homology, so it does explain A.
- (3) Reason R is also true, not false; the similar bone arrangement is well documented.
- (4) Assertion A is also true; forelimb homology between humans and bats is a standard NCERT example of divergent evolution.
NCERT: NCERT Class 12 Biology, Chapter 6 (Evolution), page 114
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Revise the lesson: Evidences Evolution
Colostrum, secreted by mother during initial days of lactation, is abundant in ________.
- (1)IgG
- (2)IgM
- (3)IgA
- (4)IgD
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerColostrum, the milk secreted during the initial days of lactation, is rich in IgA antibodies, which provide passive immunity to the newborn.
Why the other options are wrong- (1) IgG is the antibody that crosses the placenta to give the fetus passive immunity before birth, not the antibody characteristic of colostrum.
- (2) IgM is typically the first antibody produced in a primary immune response, not the one highlighted as abundant in colostrum.
- (4) IgD functions mainly as a B-cell surface receptor and is not associated with colostrum.
NCERT: NCERT Class 12 Biology, Chapter 2 (Human Reproduction), page 38
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Revise the lesson: Fertilization Pregnancy
Given below are two statements : one is labelled as Assertion A and the other is labelled as 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.
In the light of the above statements, choose the most appropriate answer from the options given below:
- (1)Both A and R are correct and R is the correct explanation of A
- (2)Both A and R are correct but R is not the correct explanation of A
- (3)A is correct but R is not correct
- (4)A is not correct but R is correct
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerNCERT gives this exact case as an example of interspecific competition: the Abingdon tortoise in the Galapagos Islands went extinct within a decade of goats being introduced, and this was apparently because the goats were more efficient browsers/competitors for the same food resource. So A is true, R is true, and R is the correct explanation of A.
Why the other options are wrong- (2) R is not a separate, unrelated true fact -- NCERT explicitly attributes the tortoise's extinction to the goats' greater browsing efficiency, so R does explain A.
- (3) Reason R is also correct, as directly stated in NCERT, not incorrect.
- (4) Assertion A is also correct -- the extinction of the Abingdon tortoise following goat introduction is the NCERT example itself.
NCERT: NCERT Class 12 Biology, Chapter 11 (Organisms and Populations), page 199
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Revise the lesson: Population Interactions
The covering of ovum at ovulation is _________.
- (1)endometrium
- (2)zona radiata
- (3)zona pellucida
- (4)chorion
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerAfter the Graafian follicle ruptures at ovulation, the released secondary oocyte (ovum) is surrounded by a membrane called the zona pellucida.
Why the other options are wrong- (1) Endometrium is the innermost lining layer of the uterine wall, not a covering of the ovum.
- (2) Zona radiata is not the term used by NCERT for the ovum's covering at ovulation; the text specifically names the zona pellucida.
- (4) Chorion is the outermost extraembryonic membrane surrounding the developing embryo, formed later, not the ovum's covering at ovulation.
NCERT: NCERT Class 12 Biology, Chapter 2 (Human Reproduction), page 33
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Revise the lesson: Fertilization Pregnancy
Match List-I with List-II.
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
Choose the correct answer from the options given below:
- (1)A-III, B-IV, C-II, D-I
- (2)A-I, B-II, C-III, D-IV
- (3)A-II, B-I, C-IV, D-III
- (4)A-III, B-I, C-II, D-IV
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerNCERT defines: both species harmed = competition (A-III); one harmed, other benefited = predation (B-I); both species benefited = mutualism (C-II); one benefited, other unaffected = commensalism (D-IV). This gives A-III, B-I, C-II, D-IV.
Why the other options are wrong- (1) Pairs B with commensalism and D with predation, both mismatched relative to the NCERT definitions.
- (2) Pairs A with predation and B with mutualism, which reverses the harm/benefit pattern each interaction actually involves.
- (3) Pairs A with mutualism (both benefited) when A actually describes both species being harmed (competition).
NCERT: NCERT Class 12 Biology, Chapter 11 (Organisms and Populations), page 197
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Revise the lesson: Population Interactions
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : In an experiment, Mendel observed that the F1 progeny plants are all tall and none are dwarf.
Reason R : Stem height is a contrasting trait, with tall being dominant and dwarf being recessive.
In the light of the above statements, choose the most appropriate answer from the options given below :
- (1)Both A and R are correct and R is the correct explanation of A
- (2)Both A and R are correct but R is not the correct explanation of A
- (3)A is correct but R is not correct
- (4)A is not correct but R is correct
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerMendel crossed pure-breeding tall and dwarf pea plants and found all F1 progeny were tall, none dwarf (A is correct). This happens precisely because tall is the dominant allele and dwarf the recessive one for this contrasting trait, so the dominant trait alone is expressed in F1 (R correctly explains A).
Why the other options are wrong- (2) R is exactly the mechanism (dominance) that produces the F1 result described in A, so it is the correct explanation, not an unrelated fact.
- (3) Reason R correctly states the dominant/recessive relationship for stem height as established by Mendel's experiments.
- (4) Assertion A correctly reports Mendel's observed F1 result (all tall).
NCERT: NCERT Class 12 Biology, Chapter 4 (Principles of Inheritance and Variation), page 58
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Revise the lesson: Mendel Inheritance
Given below are two statements : one is labelled as Assertion A and the other is labelled as 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.
In the light of the above statements, choose the most appropriate answer from the options given below :
- (1)Both A and R are correct and R is the correct explanation of A
- (2)Both A and R are correct but R is not the correct explanation of A
- (3)A is correct but R is not correct
- (4)A is not correct but R is correct
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerSince DNA is enclosed within cell membranes and associated with other macromolecules, cells must be broken open to release and isolate the DNA (R). This is done using enzymes such as lysozyme for bacterial cells and cellulase for plant cells (A), so R correctly explains why A's enzymatic treatments are needed.
Why the other options are wrong- (2) R is not a coincidental fact -- it is precisely why enzymes like lysozyme and cellulase are used, so it does explain A.
- (3) Reason R is a true, NCERT-stated premise about needing to disrupt cells to isolate genetic material.
- (4) Assertion A correctly matches lysozyme to bacteria and cellulase to plant cells as per NCERT.
NCERT: NCERT Class 12 Biology, Chapter 9 (Biotechnology: Principles and Processes), page 171
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Revise the lesson: rDNA Principles
Which of the following is used as an effective sedative and painkiller for treating post-surgery patients?
- (1)Interferon
- (2)Antibiotics
- (3)Morphine
- (4)Anti-retroviral drugs
Show answer and explanation
Answer (NTA final key): Option (3)
Why this is the answerMorphine, derived from the opium poppy, is a very effective sedative and painkiller and is widely used for patients who have undergone surgery.
Why the other options are wrong- (1) Interferons are cytokines that act as part of the innate immune barrier against viral infection, not surgical painkillers.
- (2) Antibiotics restrict bacterial growth/infection; they have no sedative or analgesic action.
- (4) Anti-retroviral drugs are used against retroviral infections such as HIV, not as sedatives or painkillers.
NCERT: NCERT Class 12 Biology, Chapter 7 (Human Health and Disease), page 143
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Revise the lesson: Drugs Addiction
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
Choose the correct answer from the options given below :
- (1)(a), (b) and (c)
- (2)(a) and (b) only
- (3)(a) and (c) only
- (4)(b) and (c) only
Show answer and explanation
Answer (NTA final key): Option (2)
Why this is the answerEnergy flows only one way, from producers to consumers/decomposers (a-true). The pyramid of energy is always upright and can never be inverted, since energy is lost as heat at every transfer (b-true). However, energy transfer between trophic levels follows the 10 per cent law, not a 1% law, so (c) is false.
Why the other options are wrong- (1) Includes (c), which misstates the 10 per cent law as a '1% law'.
- (3) Includes (c), the incorrect '1% law' claim, though (a) is correctly included.
- (4) Omits (a), which is a correct NCERT statement, and includes the incorrect (c).
NCERT: NCERT Class 12 Biology, Chapter 12 (Ecosystem), page 211
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Revise the lesson: Ecosystem Productivity Decomposition
Which of the following statements is correct about Plasmodium?
- (1)Reproduces sexually in liver cells
- (2)Reproduces sexually in RBCs
- (3)Gametocytes develop in mosquito gut
- (4)Fertilization takes place in mosquito gut
Show answer and explanation
Answer (NTA final key): Option (4)
Why this is the answerPlasmodium reproduces asexually in liver cells and then in RBCs; the sexual stages (gametocytes) form in RBCs and are taken up by the female Anopheles mosquito during a blood meal, and fertilisation (fusion of gametes) then takes place in the mosquito's gut.
Why the other options are wrong- (1) Plasmodium reproduces asexually, not sexually, in liver cells.
- (2) Plasmodium reproduces asexually in RBCs, causing their rupture; sexual fusion happens later in the mosquito.
- (3) Gametocytes develop in human RBCs, not in the mosquito gut; it is fertilization that occurs in the mosquito gut after the gametocytes are taken up.
NCERT: NCERT Class 12 Biology, Chapter 7 (Human Health and Disease), page 131
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Revise the lesson: Health Disease Pathogens
Match List-I with List-II.
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
Choose the correct answer from the options given below:
- (1)A-II, B-I, C-IV, D-III
- (2)A-I, B-II, C-IV, D-III
- (3)A-III, B-IV, C-II, D-I
- (4)A-IV, B-I, C-III, D-II
Show answer and explanation
Answer (NTA final key): Option (1)
Why this is the answerTransformation is the process of transferring DNA (a vector/plasmid) into a host bacterium (A-II). The cloning site is where restriction enzymes cut the vector to insert foreign DNA (B-I). Selection of transformed cells (recombinants) is done using antibiotic resistance markers (C-IV). Ori (origin of replication) is the sequence where the host's machinery begins replicating the plasmid (D-III). This gives A-II, B-I, C-IV, D-III.
Why the other options are wrong- (2) Swaps A and B: transformation is about DNA transfer into the host, not about restriction-enzyme recognition, which belongs to the cloning site.
- (3) Assigns replication to transformation (A-III) and antibiotic selection to cloning site (B-IV), both mismatched.
- (4) Assigns antibiotic function to transformation (A-IV) and restriction enzyme to Ori (B-I is actually correct here but C and D are swapped incorrectly relative to the rest of the mapping), which does not match the NCERT roles overall.
NCERT: NCERT Class 12 Biology, Chapter 9 (Biotechnology: Principles and Processes), page 170
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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Revise the lesson: rDNA Tools Restriction Ligase