Call Us
9875526365
Call Us
9875526365
With one of the highest numbers of candidates registered for the NEET 2023 exam, the NEET exam previous year question paper 2023 serves as an essential reference for the medical entrance exams’ current competitive nature.
Answer: (1) Both A and R are correct and R is the correct explanation of A
Explanation:
Assertion A: True. The first stage of the moss gametophyte is the protonema stage.
Reason R: True. The protonema develops directly from the spores produced in the capsule of the sporophyte. The reason correctly explains the assertion.
Answer: (3) Synergids, Zygote and Primary endosperm nucleus
Explanation:
Synergids: These are haploid cells of the embryo sac (gametophyte).
Zygote: Formed by the fusion of a haploid male gamete and a haploid egg cell, making it diploid.
Primary Endosperm Nucleus (PEN): Formed by the fusion of a diploid secondary nucleus (fusion of two polar nuclei) with a haploid male gamete, making it triploid.
Answer: (4) Active Transport
Explanation:
Active transport is the movement of ions or molecules across a membrane against their concentration gradient (from lower to higher concentration). This process requires energy in the form of ATP.
Answer: (1) Insect pollinated plants
Explanation:
Large, colourful, and fragrant flowers with nectar are adaptations to attract biotic pollinators, especially insects. These characteristics are typical of entomophilous (insect-pollinated) plants.
Answer: (3) A single gene affecting multiple phenotypic expression
Explanation:
Pleiotropism is a phenomenon where a single gene influences two or more seemingly unrelated phenotypic traits. The gene responsible is called a pleiotropic gene.
Answer: (3) Ethylene
Explanation:
The plant hormone ethylene promotes rapid internode and petiole elongation in deep-water rice plants. This helps the plant keep its leaves above the water surface during flooding.
Answer: (1) Habitat loss and fragmentation
Explanation:
The “Evil Quartet” comprises the four major causes of biodiversity loss: (i) Habitat loss and fragmentation, (ii) Over-exploitation, (iii) Alien species invasions, and (iv) Co-extinctions. Habitat loss and fragmentation is considered the most important cause driving species to extinction.
Answer: (4) Bright orange colour
Explanation:
In gel electrophoresis, DNA fragments are stained with ethidium bromide. When exposed to ultraviolet (UV) radiation, the DNA bands appear as bright orange-coloured bands.
Answer: (1) Manganese
Explanation:
Manganese (Mn) is a micronutrient that plays a crucial role in the water-splitting reaction during the light reactions of photosynthesis (photolysis of water), which liberates oxygen.
Answer: (4) China rose, Petunia and Lemon
Explanation:
Axile placentation is found in plants like China rose (Hibiscus), Petunia, Tomato, and Lemon. In this type, the ovary is septate, and the ovules are borne on the central axis.
Answer: (2) Pachytene
Explanation:
The pachytene stage of prophase I in meiosis is characterized by the appearance of recombination nodules. These nodules are protein complexes where crossing over (genetic recombination) takes place between non-sister chromatids of homologous chromosomes.
Answer: (1) 680 nm
Explanation:
Photosystem II (PS II) has a reaction centre chlorophyll a molecule called P680, which has an absorption maximum at 680 nm.
Answer: (2) Alfred Hershey and Martha Chase
Explanation:
The unequivocal proof that DNA is the genetic material came from the Hershey and Chase experiment (1952), which used radioactive isotopes to show that DNA, not protein, is the genetic material in bacteriophages.
Answer: (2) S phase
Explanation:
In eukaryotes, DNA replication occurs during the S phase (Synthesis phase) of the cell cycle. The G1, G2, and M phases are not primarily for DNA replication.
Answer: (2) Dedifferentiation
Explanation:
In tissue culture, leaf mesophyll cells (which are differentiated and have lost the capacity to divide) regain their ability to divide and form a mass of cells called a callus. This phenomenon is known as dedifferentiation.
Answer: (3) It does not contain complex helices and hence cannot hold iodine molecules
Explanation:
Cellulose does not give a blue colour with iodine because its glucose units are linked by β-1,4 glycosidic bonds, forming a straight, extended chain without complex helices. Starch gives a blue colour because its helical structure can trap iodine molecules.
Answer: (2) Gibberellic Acid
Explanation:
Gibberellins (specifically Gibberellic Acid, GA₃) are known to hasten the maturity period of juvenile conifers, leading to early seed production.
Answer: (1) Both Statement I and Statement II are correct
Explanation:
Statement I: True. Transpiration can generate significant tension (negative pressure) capable of pulling a column of water over 130 meters high.
Statement II: True. Transpiration results in evaporative cooling, which can cool leaf surfaces by 10 to 15 degrees.
Answer: (1) Dialephous and Dithecous anthers
Explanation:
Fabaceae: Stamens are often diadelphous (fused into two bundles) and have dithecous anthers (two lobes).
Solanaceae: Stamens are polyandrous (free), epipetalous (attached to petals), and dithecous.
Liliaceae: Stamens are polyandrous, epiphyllous (attached to tepals), and dithecous.
Answer: (1) All genes that are expressed as RNA
Explanation:
Expressed Sequence Tags (ESTs) are short sequences of DNA that correspond to all the genes that are being expressed as RNA in a tissue or cell type at a given time.
Answer: (1) A, B, C only
Explanation:
A: Correct. Detritivores (detrivores) perform fragmentation of detritus.
B: Correct. Humus is further degraded by microbes during mineralization.
C: Correct. Water-soluble inorganic nutrients are precipitated into the soil by a process called leaching.
D: Incorrect. The detritus food chain begins with dead organic matter (detritus), not living organisms.
E: Incorrect. Earthworms break down detritus by a process called fragmentation, not catabolism. Catabolism is the breakdown of complex organic compounds by microbes into simpler inorganic substances.
Answer: (1) Dobson units
Explanation:
The thickness of the ozone layer in a column of air is measured in Dobson Units (DU).
Answer: (1) Both A and R are true and R is the correct explanation of A
Explanation:
Assertion A: True. Late wood (autumn wood) has fewer xylem elements and narrower vessels.
Reason R: True. The cambium is less active during the winter season.
The reason correctly explains the assertion, as the reduced activity of the cambium in winter leads to the formation of late wood with fewer and narrower vessels.
Answer: (3) Anaphase II
Explanation:
Centromere division (splitting) occurs during anaphase of mitosis and anaphase II of meiosis. In anaphase I, homologous chromosomes separate, but the centromere does not divide.
Answer: (2) 18 ATP and 12 NADPH₂
Explanation:
For the synthesis of one molecule of glucose (which requires 6 turns of the Calvin cycle):
Each turn fixes 1 CO₂ and requires 3 ATP and 2 NADPH₂.
For 6 turns (6 CO₂ → 1 glucose): 6 × 3 = 18 ATP and 6 × 2 = 12 NADPH₂.
Answer: (2) 1992
Explanation:
The historic Convention on Biological Diversity, also known as ‘The Earth Summit’, was held in Rio de Janeiro, Brazil, in the year 1992.
Answer: (3) Respiratory loss
Explanation:
In the equation GPP – R = NPP:
GPP = Gross Primary Productivity
NPP = Net Primary Productivity
R = Respiratory loss (the energy used by plants for respiration).
Answer: (2) DNA
Explanation:
During the isolation of genetic material (DNA) in recombinant DNA technology, the purified DNA is precipitated out by adding chilled ethanol. DNA is insoluble in ethanol and forms a visible precipitate.
Answer: (2) Transcription of tRNA, 5S rRNA and snRNA
Explanation:
In eukaryotes, RNA polymerase III is responsible for transcribing tRNA, 5S rRNA, and snRNA (small nuclear RNA).
Answer: (2) To trap pollen grains
Explanation:
The tassel of a corn cob represents the stigma and style of the female flower. They are long, silky, and feathery, and their function is to wave in the wind to trap pollen grains for pollination.
All Question Paper PDF NEET 2023 PYQ
| Section | NEET UG 2024 Questions Paper Solve |
|---|---|
| Physics | 2024-Physics Question Paper solve |
| Chemistry | 2024-Physics Question Paper solve |
Answer: (2) Selaginella and Salvinia
Explanation:
Heterosporous pteridophytes produce two different types of spores: microspores and megaspores. Selaginella and Salvinia are examples of heterosporous pteridophytes.
Answer: (3) Tungsten or gold
Explanation:
In the gene gun (biolistic) method, microparticles of tungsten or gold are used. These inert metals are coated with the foreign DNA and then bombarded into the host cells.
Answer: (4) Statement I is incorrect but Statement II is true
Explanation:
Statement I: Incorrect. Endarch and exarch are terms used for describing the position of primary xylem (protoxylem and metaxylem), not secondary xylem.
Statement II: True. Exarch condition (where protoxylem is outside and metaxylem is inside) is the most common feature of the root system.
Answer: (3) Alfred Sturtevant
Explanation:
Alfred Sturtevant, a student of Thomas Hunt Morgan, used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between them and created the first genetic map.
Answer: (1) Both A and R are true and R is the correct explanation of A
Explanation:
Assertion A: True. ATP is used at two steps in glycolysis.
Reason R: True. The two steps are the conversion of glucose to glucose-6-phosphate and fructose-6-phosphate to fructose-1,6-bisphosphate. The reason correctly explains the assertion.
Answer: (2) Algal blooms caused by excess of organic matter in water improve water quality and promote fisheries
Explanation:
This statement is incorrect. Algal blooms caused by excess nutrients (eutrophication) deteriorate water quality, deplete oxygen, and lead to fish mortality.
Answer: (1) 80
Explanation:
A ribosome consists of structural RNAs (rRNAs) and about 80 different proteins.
Answer: (3) B and E only
Explanation:
B: True. Individuals with Klinefelter’s syndrome (47, XXY) have overall masculine development but some feminine development is also expressed.
E: True. Such individuals are sterile.
A: Incorrect. Langdon Down described Down’s syndrome, not Klinefelter’s.
C: Incorrect. Klinefelter individuals are generally tall, not short-statured.
D: Incorrect. Mental retardation is not a typical characteristic.
Answer: (The text for this question and its options is missing from the provided document.)
Answer: (The text for this question and its options is missing from the provided document.)
Answer: (3) A is true but R is false
Explanation:
Assertion A: True. In gymnosperms, the pollen grains are released from the microsporangium and are carried by air currents to the ovules.
Reason R: False. The pollen tube does not grow through the style and stigma (which are angiosperm structures). In gymnosperms, the pollen tube grows directly towards the archegonia within the ovule.
Answer: (1) A – II, B – IV, C – I, D – III
Explanation:
Cohesion (A): Mutual attraction among water molecules (II).
Adhesion (B): Attraction of water molecules to polar surfaces (IV).
Surface Tension (C): Water molecules are more attracted to each other in the liquid phase than to water in the gas phase (I).
Guttation (D): Loss of water in the liquid phase (III).
Answer: (1) Membrane, proton pump, proton gradient, ATP synthase
Explanation:
Chemiosmosis requires a membrane, a proton pump to create a proton gradient, the proton gradient itself (proton motive force), and ATP synthase to synthesize ATP using the energy from the gradient.
Answer: (1) Succinic dehydrogenase
Explanation:
Malonate is a competitive inhibitor of the enzyme succinic dehydrogenase. It resembles the substrate (succinate) in structure and binds to the enzyme’s active site, inhibiting its activity.
Answer: (2) A and D only
Explanation:
A: True. Lenticels are lens-shaped openings that permit gas exchange.
D: True. Bark is a non-technical term that refers to all tissues exterior to the vascular cambium, including the periderm and secondary phloem.
B: Incorrect. Bark formed early in the season is called soft bark.
C: Incorrect. “Bark” is a non-technical term.
E: Incorrect. Phellogen (cork cambium) is a single-layered meristem, but it produces multiple layers of phellem (cork).
Answer: (3) A – IV, B – I, C – II, D – III
Explanation:
M Phase (A): Equational division (IV).
G2 Phase (B): Proteins are synthesized (I) in preparation for mitosis.
Quiescent Stage (C): Inactive phase (II) where cells leave the cell cycle.
G1 Phase (D): Interval between mitosis and initiation of DNA replication (III).
Answer: (2) A – IV, B – I, C – II, D – III
Explanation:
Mutualism (A): (+, +) – Both species benefit (IV).
Commensalism (B): (+, 0) – One species benefits, the other is unaffected (I).
Amensalism (C): (-, 0) – One species is harmed, the other is unaffected (II).
Parasitism (D): (+, -) – One species benefits, the other is harmed (III).
Answer: (3) Statement I is correct but Statement II is false
Explanation:
Statement I: True. This is the correct statement of Gause’s Competitive Exclusion Principle.
Statement II: Incorrect. In general, herbivores (and plants) are more adversely affected by competition than carnivores.
Answer: (3) A – III, B – I, C – IV, D – II
Explanation:
Iron (A): Activator of catalase (III).
Zinc (B): Needed in the synthesis of auxin (I).
Boron (C): Required for cell elongation and differentiation (IV).
Molybdenum (D): Component of nitrate reductase (II).
Answer: (1) B, C, D, A
Explanation:
The correct sequence for the formation of recombinant DNA is:
B: Cutting of DNA at a specific location by restriction enzyme.
C: Isolation of the desired DNA fragment.
D: Amplification of the gene of interest using PCR.
A: Insertion of recombinant DNA into the host cell.
Answer: (2) A-III, B-IV, C-II, D-I
Explanation:
Vasectomy (A): Surgical method (III).
Coitus interruptus (B): Natural method (IV).
Cervical caps (C): Barrier method (II).
Saheli (D): Oral method (I) – a non-steroidal pill.
Answer: (1) Both Statement I and Statement II are true
Explanation:
Statement I: True. The vas deferens receives a duct from the seminal vesicle and opens into the urethra as the ejaculatory duct.
Statement II: True. The cavity of the cervix is called the cervical canal, which along with the vagina forms the birth canal.
Answer: (2) Biomagnification refers to increase in concentration of the toxicant at successive trophic levels
Explanation:
Statement 2 is correct. Biomagnification is the increase in the concentration of a toxicant at successive trophic levels.
Statement 1 is incorrect because eutrophication is the increase in nutrients, not specifically domestic sewage.
Statement 3 is incorrect because an excess of nutrients promotes algal blooms.
Statement 4 is incorrect because algal blooms increase fish mortality.
Answer: (The images for the pedigree symbols are missing from the provided text.)
Answer: (2) Gonorrhoea
Explanation:
Gonorrhoea is a sexually transmitted disease (STD) caused by a bacterium (Neisseria gonorrhoeae). Bacterial STDs like gonorrhoea, syphilis, and chlamydia are completely curable if detected early and treated properly. The other options (Genital herpes, Hepatitis-B, HIV) are viral diseases.
Answer: (1) A-IV, B-III, C-II, D-I
Explanation:
CCK (Cholecystokinin) (A): Acts on the pancreas (IV) to stimulate enzyme secretion.
GIP (Gastric Inhibitory Peptide) (B): Acts on the gastric gland (III) to inhibit gastric secretion and motility.
ANF (Atrial Natriuretic Factor) (C): Released from the heart (II).
ADH (Anti-diuretic Hormone) (D): Acts on the kidney (I) to stimulate water reabsorption.
Answer: (2) B, C and D only
Explanation:
B: True. ADH facilitates water reabsorption to prevent diuresis.
C: True. ANF causes vasodilation.
D: True. ADH causes an increase in blood pressure.
A: Incorrect. Excessive loss of body fluid switches on osmoreceptors.
E: Incorrect. ADH increases GFR (or rather, it increases water reabsorption, which can indirectly affect GFR, but the statement as given is false).
Answer: (2) B and C only
Explanation:
B: True. The sarcoplasmic reticulum is a storehouse of calcium ions.
C: True. The striated appearance is due to the distribution pattern of actin and myosin proteins.
A: Incorrect. Muscle bundles (fascicles) are held together by a collagenous tissue layer called fascia, not fascicle. A fascicle is the bundle of muscle fibres itself.
D: Incorrect. The functional unit of contraction is the sarcomere, which is the region between two Z-lines, not the M-line.
Answer: (1) Helper T-lymphocytes
Explanation:
HIV (Human Immunodeficiency Virus) enters and replicates primarily in helper T-lymphocytes (also called CD4+ T cells). The virus destroys these cells, leading to a weakened immune system.
Answer: (2) Hemichordate
Explanation:
Radial symmetry is not found in adults of the phylum Hemichordata. Hemichordates are bilaterally symmetrical. Ctenophores, Coelenterates (Cnidarians), and adult Echinoderms are radially symmetrical.
Answer: (3) A is true but R is false
Explanation:
Assertion A: True. Nephrons are of two types based on their position: Cortical and Juxtamedullary.
Reason R: False. Juxtamedullary nephrons have a long loop of Henle that runs deep into the medulla. Cortical nephrons have a short loop of Henle.
Answer: (3) Motility
Explanation:
The cytoskeleton is involved in many functions, including mechanical support, motility (movement), and maintenance of cell shape.
Answer: (3) Ileo-caecal valve
Explanation:
The undigested food (faeces) passes from the ileum (the last part of the small intestine) into the caecum (the beginning of the large intestine) through the ileo-caecal valve, which prevents the backflow of the faecal matter.
Answer: (The question “Vital capacity of lung is” is incomplete in the provided text. The answer is not given.)
Answer: (4) A, C and D only
Explanation:
A: True. In non-primate mammals, the cyclical changes during reproduction are called the oestrus cycle.
C: True. Lack of menstruation (amenorrhoea) can be indicative of pregnancy.
D: True. Cyclic menstruation occurs between menarche and menopause.
B: Incorrect. The first menstrual cycle at puberty is called menarche, not menopause. Menopause is the cessation of the menstrual cycle.
Answer: (2) Numbat, Spotted cuscus, Flying phalanger
Explanation:
Numbat, spotted cuscus, and flying phalanger are all Australian marsupials that exhibit adaptive radiation. They have evolved from a common ancestor to occupy different ecological niches.
Answer: (2) Serum and Urine analysis
Explanation:
Conventional methods like serum and urine analysis are not helpful for early diagnosis of diseases. Techniques like Recombinant DNA Technology, PCR, and ELISA are used for early diagnosis.
Answer: (1) Both Statement I and Statement II are true
Explanation:
Statement I: True. RNA is unstable and mutates at a faster rate than DNA.
Statement II: True. Viruses with RNA genomes and shorter life spans mutate and evolve faster.
Answer: (1) A – II, B – III, C – IV, D – I
Explanation:
Ringworm (A): Caused by Trichophyton (II).
Filariasis (B): Caused by Wuchereria bancrofti (III).
Malaria (C): Caused by Plasmodium vivax (IV).
Pneumonia (D): Caused by Haemophilus influenzae (I).
Answer: (2) Serum and Urine analysis
Explanation:
This is a duplicate of Question 67. Conventional methods like serum and urine analysis do not help in early diagnosis.
Answer: (1) A – III, B – I, C – IV, D – II
Explanation:
P-wave (A): Depolarisation of atria (III).
Q-wave (B): Beginning of systole (I) – Actually, it’s the beginning of ventricular depolarization.
QRS complex (C): Depolarisation of ventricles (IV).
T-wave (D): Repolarisation of ventricles (II).
Answer: (4) A is false but R is true
Explanation:
Assertion A: False. Amniocentesis is not a strategy of the Reproductive and Child Health Care (RCH) programme for sex determination. Its use for sex determination is banned in India.
Reason R: True. The ban on amniocentesis (for sex determination) helps check the increasing menace of female foeticide.
Answer: (2) A – II, B – III, C – IV, D – I
Explanation:
In the lac operon of E. coli:
Gene a: Codes for transacetylase (II).
Gene y: Codes for permease (III).
Gene i: Codes for repressor protein (IV).
Gene z: Codes for β-galactosidase (I).
Answer: (3) A – III, B – I, C – IV, D – II
Explanation:
Peptic cells (A): Secrete proenzyme pepsinogen (III).
Goblet cells (B): Secrete mucus (I).
Oxyntic cells (C): Secrete HCl and intrinsic factor for absorption of vitamin B₁₂ (IV).
Hepatic cells (D): Secrete bile juice (II).
Answer: (4) Probe
Explanation:
A probe is not a cloning vector. It is a single-stranded DNA or RNA sequence tagged with a radioactive or fluorescent molecule used to detect a specific gene or sequence. BAC (Bacterial Artificial Chromosome), YAC (Yeast Artificial Chromosome), and pBR322 are all cloning vectors.
Answer: (1) A – II, B – I, C – IV, D – III
Explanation:
Heroin (A): A depressant that slows down body functions (II).
Marijuana (B): Affects the cardiovascular system (I).
Cocaine (C): Interferes with the transport of dopamine (IV).
Morphine (D): Used as a painkiller (III).
Answer: (2) A, C & E only
Explanation:
The endomembrane system includes the Endoplasmic Reticulum (ER), Golgi complex, Lysosomes, and Vacuoles. Mitochondria (A), Chloroplasts (C), and Peroxisomes (E) are not part of this system.
Answer: (2) Both A and R are true but R is NOT the correct explanation of A
Explanation:
Assertion A: True. Endometrium is necessary for the implantation of the blastocyst.
Reason R: True. In the absence of fertilization, the corpus luteum degenerates, leading to the disintegration of the endometrium.
However, the reason does not explain the assertion. The reason explains menstruation, not the necessity of the endometrium for implantation.
Answer: (1) Both Statement I and Statement II are true
Explanation:
Statement I: True. Low temperature preserves enzymes in a temporarily inactive state, while high temperature denatures the protein.
Statement II: True. This is the correct definition of competitive inhibition.
Answer: (3) A – III, B – II, C – IV, D – I
Explanation:
Taenia (A – Tapeworm): Flame cells (protonephridia) for excretion (III).
Paramoecium (B): Contractile vacuole for osmoregulation and excretion (II).
Periplaneta (C – Cockroach): Ureose glands (IV) – actually, malpighian tubules are the main excretory organs in cockroaches.
Pheretima (D – Earthworm): Nephridia for excretion (I).
Answer: (1) A-I, B-II, C-III, D-IV
Explanation:
Leopard and Lion (A): Competition (I).
Cuckoo laying eggs in Crow’s nest (B): Brood parasitism (II).
Fungi and root of a higher plant in Mycorrhizae (C): Mutualism (III).
Cattle egret and Cattle (D): Commensalism (IV).
Answer: (4) Statement I is incorrect but Statement II is true
Explanation:
Statement I: Incorrect. In prokaryotes, the negatively charged DNA is held with some positively charged proteins in a region called the nucleoid.
Statement II: True. In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form a nucleosome.
Answer: (1) Down syndrome
Explanation:
Broad palm with a single palmar crease (simian crease) is a characteristic symptom of Down syndrome (Trisomy 21).
Answer: (2) Hemichordate
Explanation:
This is a duplicate of Question 60. Hemichordates are bilaterally symmetrical animals. Radial symmetry is not found in them.
Answer: (1) A-II, B-I, C-III, D-IV
Explanation:
Logistic growth (A): Limited resource availability condition (II).
Exponential growth (B): Unlimited resource availability condition (I).
Expanding age pyramid (C): Pre-reproductive age group is largest, followed by reproductive and post-reproductive (III).
Stable age pyramid (D): Pre-reproductive and reproductive age groups are the same (IV).
Answer: (2) B and C only
Explanation:
Characteristics of chordates include:
B: True. Presence of a closed circulatory system.
C: True. Presence of paired pharyngeal gill slits.
A: Incorrect. The nerve cord is dorsal, hollow, and single, not solid and double.
D: Incorrect. Chordates generally have a ventral heart (except in some), not a dorsal heart.
E: Incorrect. Chordates are triploblastic coelomates, not pseudocoelomates.
Section | Number of Questions | Marks for Correct Answer | Negative Marking | Total Marks |
Physics | 45 | +4 | −1 | 180 |
Chemistry | 45 | +4 | −1 | 180 |
Biology (Botany + Zoology) | 90 | +4 | −1 | 360 |
Total | 180 | 720 |
Correct statements about Phaeophyceae:
Answer: (3) A, C, D and E only
Explanation:
A is correct: Asexual reproduction by biflagellate zoospores
B is incorrect: Sexual reproduction can be oogamous, isogamous, or anisogamous
C is correct: Stored food is mannitol or laminarin
D is correct: Major pigments: chlorophyll a, c, carotenoids, xanthophylls
E is correct: Cellulosic wall covered by gelatinous coating of algin
Step in TCA cycle that does NOT involve oxidation:
Answer: (3) Succinyl-CoA → Succinic acid
Explanation:
Oxidation involves loss of electrons (increase in oxidation state)
Succinyl-CoA → Succinic acid is a hydrolysis reaction (no oxidation)
All other options involve oxidation:
Malic acid → Oxaloacetic acid (oxidation)
Succinic acid → Malic acid (oxidation)
Isocitrate → α-ketoglutaric acid (oxidation)
Statements about C3 and C4 plants:
Answer: (3) Statement I is true but Statement II is false
Explanation:
Statement I is correct: In C3 plants, O₂ competes with CO₂ at RuBisCO (photorespiration), decreasing CO₂ fixation
Statement II is false: Photorespiration does NOT occur in C4 plants (they lack RuBisCO in mesophyll cells)
Match List I with List II:
| List I | List II |
|---|---|
| A. GLUT-4 | I. Hormone |
| B. Insulin | II. Enzyme |
| C. Trypsin | III. Intercellular ground substance |
| D. Collagen | IV. Enables glucose transport into cells |
Answer: (1) A-IV, B-I, C-II, D-III
Explanation:
GLUT-4: Enables glucose transport into cells (glucose transporter)
Insulin: Hormone (regulates blood glucose)
Trypsin: Enzyme (digests proteins)
Collagen: Intercellular ground substance (structural protein)
Match List I with List II:
| List I | List II |
|---|---|
| A. Monoadelphous | I. Citrus |
| B. Diadelphous | II. Pea |
| C. Polyadelphous | III. Lily |
| D. Epiphyllous | IV. China-rose |
Answer: (1) A-IV, B-II, C-I, D-III
Explanation:
Monoadelphous (stamens united in one bundle) → China-rose
Diadelphous (stamens united in two bundles) → Pea
Polyadelphous (stamens united in many bundles) → Citrus
Epiphyllous (stamens attached to petals) → Lily
DNA present in chloroplast is:
Answer: (2) Circular, double stranded
Explanation:
Chloroplast DNA is:
Circular (not linear)
Double-stranded
Similar to prokaryotic DNA and mitochondrial DNA
This supports the endosymbiotic theory
GPP of third trophic level if NPP of first trophic level is 100x:
Answer: (3) 10x (kcal m⁻²) yr⁻¹
Explanation:
Only about 10% energy is transferred from one trophic level to next
NPP at first trophic level = GPP for second trophic level = 100x
NPP at second trophic level = 10% of 100x = 10x = GPP for third trophic level
Therefore, GPP of third trophic level = 10x
Fused in somatic hybridization:
Answer: (3) Protoplasts
Explanation:
Somatic hybridization: Fusion of protoplasts (cells without cell wall) from two different plant varieties/species
This produces somatic hybrids
Callus: Mass of undifferentiated cells
Somatic embryos: Embryos from somatic cells
Pollens: Male gametophytes
Match List I with List II:
| List I | List II |
|---|---|
| A. Citric acid cycle | I. Cytoplasm |
| B. Glycolysis | II. Mitochondrial matrix |
| C. Electron transport system | III. Intermembrane space of mitochondria |
| D. Proton gradient | IV. Inner mitochondrial membrane |
Answer: (2) A-II, B-I, C-IV, D-III
Explanation:
Citric acid cycle → Mitochondrial matrix
Glycolysis → Cytoplasm/Cytosol
Electron transport system → Inner mitochondrial membrane
Proton gradient → Intermembrane space of mitochondria
Correct statement about replication in E. coli:
Answer: (4) DNA dependent DNA polymerase catalyses polymerization in 5’→3′ direction
Explanation:
DNA polymerase catalyses polymerization in 5’→3′ direction only
It cannot polymerize in 3’→5′ direction
The enzyme requires a primer to initiate synthesis
DNA-dependent RNA polymerase is involved in transcription, not replication
Which is not a steroid hormone?
Answer: (4) Glucagon
Explanation:
Glucagon is a proteinaceous/peptide hormone secreted by pancreas (alpha cells)
Cortisol: Steroid hormone (adrenal cortex)
Testosterone: Steroid hormone (testes)
Progesterone: Steroid hormone (ovaries)
Statements about hymen:
Answer: (3) Statement I is true but Statement II is false
Explanation:
Statement I is correct: Presence or absence of hymen is NOT a reliable indicator of virginity
Statement II is false: Hymen can be broken by:
Sudden jolt
Insertion of vaginal tampon
Active participation in sports
In some women, hymen persists even after coitus
Incorrect statement about bioreactors:
Answer: (3) Bio-reactors are used to produce small scale bacterial cultures
Explanation:
Bioreactors are used for large volumes (100-1000 litres) of culture
Small volumes cannot yield appreciable quantities of products
Bioreactors provide optimal growth conditions
Most commonly used bioreactors are of stirring type
Have agitator system, oxygen delivery system, and foam control system
Product of DNA dependent RNA polymerase for given template:
Answer: (1) 5’AUGUACCGUUUUAUGGUUAAGU3′
Explanation:
Template: 3’TACATGGCAATATCCATTCAS’
mRNA is complementary and antiparallel to template strand
RNA polymerase synthesizes mRNA in 5’→3′ direction
mRNA sequence: 5’AUGUACCGUUUUAUGGUUAAGU3′
Note: In RNA, T is replaced by U
Assertion and Reason about FSH:
Answer: (4) A is false but R is true
Explanation:
Assertion A is false: FSH acts on ovarian follicles in females, but in males, LH (not FSH) acts on Leydig cells
Reason R is true: Growing ovarian follicles secrete estrogen in females; interstitial cells (Leydig) secrete androgens in males
LH stimulates Leydig cells to produce testosterone
Match List I with List II:
| List I | List II |
|---|---|
| A. Typhoid | I. Fungus |
| B. Leishmaniasis | II. Nematode |
| C. Ringworm | III. Protozoa |
| D. Filariasis | IV. Bacteria |
Answer: (2) A-IV, B-III, C-I, D-II
Explanation:
Typhoid → Bacteria (Salmonella typhimurium)
Leishmaniasis → Protozoa (Leishmania donovani)
Ringworm → Fungus (Microsporum, Trichophyton, Epidermophyton)
Filariasis → Nematode (Wuchereria bancrofti/malayi)
Correct pathway of conduction of action potential through heart:
Answer: (1) E-C-A-D-B
Explanation:
SA node (E) → AV node (C) → AV bundle (A) → Bundle branches (D) → Purkinje fibres (B)
SA node: Pacemaker of heart
AV node: Delays impulse slightly
AV bundle (Bundle of His): Conducts impulse to ventricles
Purkinje fibres: Distribute impulse throughout ventricles
Match sub-phases of Prophase I:
| List I | List II |
|---|---|
| A. Diakinesis | I. Synaptonemal complex formation |
| B. Pachytene | II. Completion of terminalisation of chiasmata |
| C. Zygotene | III. Chromosomes look like thin threads |
| D. Leptotene | IV. Appearance of recombination nodules |
Answer: (3) A-II, B-IV, C-I, D-III
Explanation:
Diakinesis: Completion of terminalisation of chiasmata
Pachytene: Appearance of recombination nodules (crossing over occurs)
Zygotene: Synaptonemal complex formation (pairing of homologous chromosomes)
Leptotene: Chromosomes look like thin threads (start condensing)
Identify correct matching of muscles with location:
Answer: (2) (a) Skeletal – Triceps, (b) Smooth – Stomach, (c) Cardiac – Heart
Explanation:
Figure (a): Skeletal muscle (striated, multinucleated) → Triceps/Biceps
Figure (b): Smooth muscle (non-striated, spindle-shaped) → Stomach, intestine, blood vessels
Figure (c): Cardiac muscle (striated, branched, intercalated discs) → Heart
Match List I with List II:
| List I | List II |
|---|---|
| A. Non-medicated IUD | I. Multiload 375 |
| B. Copper releasing IUD | II. Progestogens |
| C. Hormone releasing IUD | III. Lippes loop |
| D. Implants | IV. LNG-20 |
Answer: (4) A-III, B-I, C-IV, D-II
Explanation:
Non-medicated IUD: Lippes loop
Copper releasing IUD: Multiload 375, Cu-T
Hormone releasing IUD: LNG-20 (releases levonorgestrel)
Implants: Progestogens (e.g., Norplant)
Match List I with List II:
| List I | List II |
|---|---|
| A. Lipase | I. Peptide bond |
| B. Nuclease | II. Ester bond |
| C. Protease | III. Glycosidic bond |
| D. Amylase | IV. Phosphodiester bond |
Answer: (3) A-II, B-IV, C-I, D-III
Explanation:
Lipase: Breaks ester bonds (in lipids)
Nuclease: Breaks phosphodiester bonds (in nucleic acids)
Protease: Breaks peptide bonds (in proteins)
Amylase: Breaks glycosidic bonds (in carbohydrates/starch)
Assertion and Reason about breast-feeding:
Answer: (1) Both A and R are correct and R is the correct explanation of A
Explanation:
Assertion A: Breast-feeding during initial period is recommended for healthy baby
Reason R: Colostrum contains antibodies essential for developing resistance
Colostrum is rich in IgA antibodies (passive immunity)
This protects the newborn from infections
R correctly explains A
Correct sequence of human evolution (Past to Recent):
Answer: (4) A-D-C-B
Explanation:
A. Homo habilis (2.4-1.4 mya) → First tool maker
D. Homo erectus (1.9-0.1 mya) → First to use fire, migrate
C. Homo neanderthalensis (400,000-40,000 years ago)
B. Homo sapiens (300,000 years ago to present)
Match List I with List II:
| List I | List II |
|---|---|
| A. Axoneme | I. Centriole |
| B. Cartwheel pattern | II. Cilia and flagella |
| C. Crista | III. Chromosome |
| D. Satellite | IV. Mitochondria |
Answer: (4) A-II, B-I, C-IV, D-III
Explanation:
Axoneme: Found in cilia and flagella (9+2 arrangement of microtubules)
Cartwheel pattern: Seen in centrioles (9 triplets arrangement)
Crista: Found in mitochondria (inner membrane folds)
Satellite: Present in chromosomes (secondary constriction)
Match List I with List II:
| List I | List II |
|---|---|
| A. Pterophyllum | I. Hag fish |
| B. Myxine | II. Saw fish |
| C. Pristis | III. Angel fish |
| D. Exocoetus | IV. Flying fish |
Answer: (2) A-III, B-I, C-II, D-IV
Explanation:
Pterophyllum → Angel fish (freshwater fish)
Myxine → Hag fish (jawless fish)
Pristis → Saw fish (cartilaginous fish)
Exocoetus → Flying fish (bony fish)
Match List I with List II:
| List I | List II |
|---|---|
| A. α-1 antitrypsin | I. Cotton bollworm |
| B. Cry IAb | II. ADA deficiency |
| C. Cry IAc | III. Emphysema |
| D. Enzyme replacement therapy | IV. Corn borer |
Answer: (3) A-III, B-IV, C-I, D-II
Explanation:
α-1 antitrypsin → Treatment of emphysema
Cry IAb gene → Controls corn borer
Cry IAc gene → Controls cotton bollworms
Enzyme replacement therapy → Treatment option for ADA deficiency
“Ti plasmid” of Agrobacterium tumefaciens stands for:
Answer: (3) Tumor inducing plasmid
Explanation:
Ti plasmid = Tumor inducing plasmid
Contains T-DNA (transfer DNA) which causes tumors in dicot plants
Used as a vector in plant genetic engineering
Not “tumour inhibiting” or “temperature independent”
Match List I with List II:
| List I | List II |
|---|---|
| A. Pleurobrachia | I. Mollusca |
| B. Radula | II. Ctenophora |
| C. Stomochord | III. Osteichthyes |
| D. Air bladder | IV. Hemichordata |
Answer: (2) A-II, B-I, C-IV, D-III
Explanation:
Pleurobrachia → Phylum Ctenophora (comb jellies)
Radula → Phylum Mollusca (rasping feeding organ)
Stomochord → Phylum Hemichordata (rudimentary notochord-like structure)
Air bladder → Class Osteichthyes (provides buoyancy)
Which is not a natural/traditional contraceptive method?
Answer: (4) Vaults
Explanation:
Vaults are barrier methods (rubber cap covering cervix)
Natural methods include:
Coitus interruptus (withdrawal)
Periodic abstinence (avoiding coitus during fertile period)
Lactational amenorrhea (no ovulation during intense breastfeeding)
Correct sequence of stages of cell division:
Answer: (4) E-C-A-D-B
Explanation:
E. Gap 1 phase (G₁) → Cell growth
Biology Solve Questions Paper 2026 >
Physics Solve Questions Paper 2026 >
Chemistry Solve Questions Paper 2026 >
Biology Solve Questions Paper 2023 >
Physics Solve Questions Paper 2023 >
Chemistry Solve Questions Paper 2023 >
Biology Solve Questions Paper 2022 >
Physics Solve Questions Paper 2022 >
Chemistry Solve Questions Paper 2022 >