ICAR JRF Plant Science Practice Series Memory Based 2024 (Module 1) (1 – 40 MCQ) 

Question 1

Which of the following is a Gram-positive, endospore-forming, rod-shaped bacterium commonly used as a biological control agent against plant pathogens?

1. Bacillus subtilis
2. Escherichia coli
3. Pseudomonas fluorescens
4. Rhizobium leguminosarum

Correct Answer: 1

 

Explanation:

• Bacillus subtilis– A Gram-positive, endospore-forming, rod-shaped It is widely used as a biological control agent against fungal plant pathogens (e.g., Fusarium, Rhizoctonia, Alternaria). It produces antibiotics (e.g., iturin, surfactin, bacillomycin) and forms resistant endospores, allowing long-term survival in formulations.
• Escherichia coli– Gram-negative, facultative anaerobe; not used as a biocontrol agent; primarily a model organism in molecular biology.
• Pseudomonas fluorescens– Gram-negative, does not form endospores; used as a biocontrol agent but does not fit the description “Gram-positive, endospore-forming”.
• Rhizobium leguminosarum– Gram-negative, does not form endospores; forms symbiotic nitrogen-fixing nodules on legumes, not a biocontrol agent.
• Therefore, the correct answer is Bacillus subtilis→ Option 1.

 

 

 

Question 2

Which of the following fungal pathogens is the causal agent of “Late blight of potato” and was responsible for the Irish Potato Famine (1845–1852)?

1. Alternaria solani
2. Phytophthora infestans
3. Fusarium oxysporum
4. Puccinia graminis

Correct Answer: 2

 

Explanation:

• Phytophthora infestans– An Oomycete (water mold) that causes late blight of potato and tomato. It was responsible for the Irish Potato Famine (1845–1852) , which led to mass starvation, death, and emigration from Ireland. The pathogen spreads rapidly via zoospores under cool, moist conditions.
• Alternaria solani– Causes early blight of potato and tomato, but not the Irish Famine.
• Fusarium oxysporum– Causes Fusarium wilt (vascular wilt) in many crops (e.g., tomato, banana, cotton), not late blight of potato.
• Puccinia graminis– Causes black stem rust of wheat (cereal rust), not potato disease.
• Key concept:Phytophthora infestans is a hemibiotrophic Oomycete with a high evolutionary potential, and it remains a major threat to potato production worldwide.
• Therefore, the correct answer is Phytophthora infestans→ Option 2.

 

 

 

Question 3

Match List I with List II:

List I	List II
A. Chitin	I. Archaeobacteria
B. Pseudomonas	II. Gram-positive bacteria
C. O-antigens	III. Fungal cell wall
D. Teichoic acid	IV. Gram-negative bacteria

Options:

1. A (III), B (I), C (II), D (IV)
2. A (I), B (III), C (II), D (IV)
3. A (I), B (III), C (IV), D (II)
4. A (III), B (I), C (IV), D (II)

Correct Answer: 4

 

Explanation:

• Chitin (A)– A polysaccharide composed of N-acetylglucosamine units. It is a major structural component of fungal cell walls (also found in arthropod exoskeletons). → A-III
• Pseudomonas(B) – A genus of Gram-negative bacteria (not Archaeobacteria). However, the matching shows B-I (Archaea). This appears to be an error in the question. In standard microbiology, Pseudomonas is Gram-negative, but the answer key assigns it to Archaea. → B-I (as per key)
• O-antigens (C)– The outermost portion of the lipopolysaccharide (LPS) layer in the cell wall of Gram-negative bacteria. Used for serotyping (e.g.,  coli O157). → C-IV
• Teichoic acid (D)– A polymer of glycerol or ribitol phosphate found in the cell wall of Gram-positive bacteria. It contributes to cell wall rigidity and antigenicity. → D-II
• Therefore, correct matching is A-III, B-I, C-IV, D-II→ Option 4.

 

 

 

Question 4

Fractional sterilization is also called:

1. Pasteurization
2. Tyndallization
3. Fractionation
4. Dry heat sterilization

Correct Answer: 2

 

Explanation:

• Tyndallization(also called fractional sterilization or intermittent sterilization) is a method of sterilizing heat-sensitive materials.
• Process:The material is heated at 100°C (boiling water or free-flowing steam) for 15–20 minutes on three consecutive days, with incubation periods in between to allow any surviving spores to germinate into heat-sensitive vegetative cells.
• Purpose:To kill bacterial endospores that survive the first heating. The vegetative cells from germinated spores are killed in subsequent heating cycles.
• Applications:Sterilization of media containing sugars, gelatin, or other heat-labile components that cannot withstand autoclaving.
• Other options:

• Pasteurization – Partial sterilization at 60–70°C to kill pathogenic bacteria in milk and other beverages (not complete sterilization).
• Fractionation – Separation of components, not a sterilization method.
• Dry heat sterilization – Oven at 160–180°C for 1–2 hours; kills by oxidation.
• Thus, fractional sterilization is called Tyndallization → Option 2.

 

 

 

Question 5

Match List I with List II:

List I	List II
A. Incineration	I. Culture media
B. Sodium hypochlorite	II. Sterilization of inoculation needle
C. Moist heat sterilization	III. Surface sterilization of seeds
D. Filter sterilization	IV. Thermo labile liquids

Options:

1. A (II), B (IV), C (III), D (I)
2. A (IV), B (II), C ( III), D (I)
3. A (II), B (III), C (I), D (IV)
4. A (III), B (I), C (IV), D (II)

Correct Answer: 3

 

Explanation:

• Incineration (A)– Direct burning of materials. Used for sterilization of inoculation needles (loops), wires, and other metal instruments in microbiology. → A-II
• Sodium hypochlorite (B)– A chemical sterilant (bleach). Used for surface sterilization of seeds and plant explants in tissue culture to kill surface microbes without damaging internal tissues. → B-III
• Moist heat sterilization (C)– Autoclaving (121°C, 15 psi, 15–20 min). Used for sterilizing culture media (heat-stable liquids and solids) that can withstand high temperatures. → C-I
• Filter sterilization (D)– Passing heat-sensitive liquids through membrane filters (0.22 µm pore size) to remove microorganisms. Used for thermo labile liquids (e.g., antibiotics, vitamins, serum). → D-IV
• Therefore, correct matching is A-II, B-III, C-I, D-IV→ Option 3.

 

 

 

Question 6

Transfer of a portion of DNA from one bacterium to another by a bacteriophage is called:

1. Transformation
2. Conjugation
3. Transduction
4. Translocation

Correct Answer: 3

 

Explanation:

• Transduction– The transfer of bacterial DNA from one bacterium to another via a bacteriophage (virus that infects bacteria).
• Mechanism:
• In the lytic cycle, the phage accidentally packages bacterial DNA instead of its own DNA (generalized transduction).
• In the lysogenic cycle, the prophage excises incorrectly, carrying adjacent bacterial genes (specialized transduction).
• The phage infects a recipient bacterium, transferring the bacterial DNA, which can recombine into the recipient’s genome.
• Other options:
  • Transformation– Uptake of free (naked) DNA from the environment by competent bacterial cells.
  • Conjugation– Direct cell-to-cell transfer of DNA through a pilus (requires cell contact). Mediated by F-plasmid.
  • Translocation– Movement of DNA within a cell or between organelles; not a method of bacterial gene transfer.
  • Thus, the correct answer is Transduction → Option 3.

 

 

 

Question 7

A type of mutualistic relationship in which individual of one species lives in close association with individual of another species is called:

1. Symbiosis
2. Parasitism
3. Syntrophism
4. Commensalism

Correct Answer: 1

 

Explanation:

• Symbiosis– A close, long-term association between two different species. It includes mutualism (both benefit), commensalism (one benefits, the other unaffected), and parasitism (one benefits, the other harmed). The question specifies “mutualistic relationship,” but the term symbiosis is often used broadly to mean mutualism.
• Parasitism– One organism (parasite) benefits at the expense of the host (harmed). This is a type of symbiotic relationship, but not mutualistic.
• Syntrophism– A type of mutualism where two organisms depend on each other for metabolism (e.g., one produces a nutrient that the other uses). Often used in microbial ecology.
• Commensalism– One organism benefits, the other is neither helped nor harmed.
• Since the question says “mutualistic relationship” and “lives in close association,” the most appropriate general term is Symbiosis → Option 1.

 

 

 

Question 8

Which of the following is a symbiotic nitrogen fixer?

1. Clostridium
2. Bradyrhizobium
3. Azospirillum
4. Azotobacter

Correct Answer: 2

 

Explanation:

• Bradyrhizobium– A symbiotic nitrogen-fixing bacterium that forms root nodules on leguminous plants (e.g., soybean). It fixes atmospheric N₂ into ammonia within the nodule.
• Clostridium– A free-living (non-symbiotic), anaerobic nitrogen-fixing bacterium found in soil. Does not form symbiosis.
• Azospirillum– A free-living, microaerophilic nitrogen-fixing bacterium associated with the rhizosphere of grasses (e.g., maize, wheat). It is associative, not truly symbiotic (no nodule formation).
• Azotobacter– A free-living, aerobic nitrogen-fixing bacterium found in soil. Does not form symbiosis.
• Therefore, the only symbiotic nitrogen fixer among the options is Bradyrhizobium → Option 2.

 

 

Question 9

Irradiation by UV light causes:

1. Dimerization of purines
2. Dimerization of pyrimidines
3. Dimerization of both purines and pyrimidines
4. Dimerization of either purines or pyrimidines

Correct Answer: 2

 

Explanation:

• UV light(especially UV-C, 254 nm) is absorbed maximally by DNA and causes formation of cyclobutane pyrimidine dimers.
• Mechanism:Adjacent pyrimidine bases (thymine or cytosine) on the same DNA strand form covalent bonds, creating dimers (most commonly thymine dimers).
• Effect:These dimers distort the DNA helix and block replication and transcription. If not repaired by photoreactivation or nucleotide excision repair, they can cause mutations.
• Purines(adenine and guanine) do not form dimers under UV irradiation in the same way; they are much less sensitive.
• Therefore, UV light causes dimerization of pyrimidines→ Option 2.

 

 

 

Question 10

Match List I with List II:

List I	List II
A. Mycorrhizae	I. Autotrophic nitrification
B. Symbiosome	II. Microaerophilic conditions
C. Azospirillum	III. Symbiotic nitrogen fixation
D. Nitrosomonas	IV. Phosphorus nutrition

Options:

1. A (I), B (II), C (IV), D (III)
2. A (IV), B (III), C (II), D (I)
3. A (IV), B (II), C (III), D (I)
4. A (IV), B (I), C (III), D (II)

Correct Answer: 2

 

Explanation:

• Mycorrhizae (A)– Symbiotic associations between fungi and plant roots. They enhance phosphorus nutrition (and other minerals) for the plant in exchange for carbohydrates. → A-IV
• Symbiosome (B)– A membrane-bound organelle within root nodule cells of legumes that contains the nitrogen-fixing bacteria (rhizobia). It is the site of symbiotic nitrogen fixation. → B-III
• Azospirillum(C) – A free-living, microaerophilic nitrogen-fixing bacterium associated with grass roots. It requires low oxygen levels for nitrogenase activity. → C-II
• Nitrosomonas(D) – A chemoautotrophic bacterium responsible for the first step of nitrification: oxidation of ammonia (NH₃) to nitrite (NO₂⁻). This is autotrophic nitrification. → D-I
• Therefore, correct matching is A-IV, B-III, C-II, D-I → Option 2.

 

 

 

Question 11

Maximum assimilation of nitrate takes place in which part of the plant?

1. Stem
2. Root
3. Leaves
4. Fruit

Correct Answer: 3

 

Explanation:

• Nitrate assimilationis the process by which plants reduce nitrate (NO₃⁻) to ammonium (NH₄⁺) and incorporate it into amino acids.
• Steps:Nitrate (NO₃⁻) → Nitrite (NO₂⁻) → Ammonium (NH₄⁺) → Glutamine/Glutamate → Amino acids.
• Key enzymes:Nitrate reductase (NR) and Nitrite reductase (NiR).
• While nitrate can be assimilated in both roots and leaves, maximum assimilation occurs in leavesbecause:
  • Leaves contain high levels of nitrate reductase(light-inducible).
  • Nitrate reduction requires reducing power (NADH or NADPH) generated by photosynthesis.
  • Leaves are the primary site for carbon skeletons (from photosynthesis) needed for amino acid synthesis.
• Roots assimilate nitrate as well, but the majority of nitrate transported to shoots is assimilated in leaves.
• Therefore, maximum nitrate assimilation takes place in leaves→ Option 3.

 

 

 

Question 12

Statement I: Nitrogenase enzyme is sensitive to molecular oxygen.
Statement II: Bacteria of genus Azospirillum fix nitrogen under microaerophilic conditions.

Options:

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 II is correct but Statement I is false

Correct Answer: 1

 

Explanation:

• Statement I is true:Nitrogenase is the enzyme complex that reduces atmospheric N₂ to ammonia. It is extremely sensitive to molecular oxygen (O₂). Oxygen irreversibly damages the Fe-Mo cofactor and the Fe protein (dinitrogenase reductase). Nitrogen-fixing organisms have evolved various strategies to protect nitrogenase from O₂ (e.g., heterocysts in cyanobacteria, leghemoglobin in nodules, microaerophilic metabolism).
• Statement II is true:Azospirillum is a free-living, microaerophilic nitrogen-fixing bacterium. It grows best under low oxygen concentrations (2–5% O₂), which protect nitrogenase from inactivation. It is commonly found in the rhizosphere of grasses.
• Therefore, both statements are true → Option 1.

 

 

 

Question 13

Options:

1. UAG
2. UAA
3. AUG
4. UGA

Correct Answer: 3 (

 

Explanation:

• This question likely asks: Which of the following is the start (initiation) codon?
• AUG– The most common start codon. It codes for methionine in eukaryotes and formylmethionine (fMet) in prokaryotes. It signals the beginning of translation.
• UAG, UAA, UGA– These are stop (termination) codons (also called amber, ochre, and opal, respectively). They signal the end of translation and do not code for any amino acid.
• Therefore, the correct answer is AUG→ Option 3.

 

 

 

Question 14

Which of the following statements is NOT true for Koch’s postulates?
A. A specific organism can always be found in association with a given disease
B. The organism can be isolated from a diseased animal and grown in pure culture in the laboratory
C. The pure culture will produce the disease when inoculated into a resistant animal
D. The pure culture will not produce the disease when inoculated into a susceptible animal
E. It is possible to recover the organism in pure culture from the experimentally infected animal

Options:

1. C and D only
2. C, D and E only
3. C only
4. C and E only

Correct Answer: 1

 

Explanation:

• Koch’s postulates(Robert Koch, 1884) are the criteria for establishing a causal relationship between a microorganism and a disease.
• The true postulates are:
  • The microorganism must be present in every caseof the disease (A is true).
  • It must be isolatedfrom the diseased host and grown in pure culture (B is true).
  • The pure culture must reproduce the diseasewhen inoculated into a healthy, susceptible host (C and D are false as written: C says “resistant” should be “susceptible”; D says “will not produce” should be “will produce”).
  • The microorganism must be re-isolatedfrom the experimentally infected host and shown to be identical to the original (E is true).
• Statements that are NOT true:C (because it says “resistant animal” instead of “susceptible”) and D (because it says “will not produce” instead of “will produce”).
• Therefore, C and D onlyare NOT true → Option 1.

 

 

 

Question 15

Starting from interior to exterior, arrange the following in correct order:
A. Periplasmic space
B. Nucleus
C. Plasma membrane
D. Cell wall
E. Endoplasmic reticulum

Options:

1. B, E, A, C, D
2. B, E, C, D, A
3. B, E, C, A, D
4. B, A, E, C, D

Correct Answer: 3

Explanation:

• The correct order from interior (center)to exterior (outside) in a prokaryotic cell (since periplasmic space and cell wall are mentioned, indicating bacteria):
  • Nucleus– In prokaryotes, the nucleoid region (not a true nucleus) is the central area containing DNA.
  • Endoplasmic reticulum– Prokaryotes do not have ER; this suggests the question may be mixing concepts. However, based on the answer key, the order is B → E → C → A → D.
  • Plasma membrane– The innermost membrane surrounding the cytoplasm.
  • Periplasmic space– The space between the plasma membrane and the cell wall in Gram-negative bacteria.
  • Cell wall– The outermost layer providing structural support.
• Therefore, the correct sequence is B (Nucleus) → E (ER) → C (Plasma membrane) → A (Periplasmic space) → D (Cell wall)→ Option 3 (B, E, C, A, D).

 

 

 

Question 16

The degree of compost maturity can be assessed by:

1. Seed germination test only
2. Infrared technique only
3. Moisture content only
4. Both, seed germination test and infrared technique

Correct Answer: 4

 

Explanation:

• Compost maturityindicates that the compost is stable, has completed active decomposition, and is safe for plant use (no phytotoxicity or nitrogen immobilization).
• Seed germination test– A bioassay where seeds (e.g., cress, radish) are sown in compost extract. High germination percentage and root growth indicate low phytotoxicity and maturity. Immature compost contains phytotoxic compounds (e.g., organic acids, phenols).
• Infrared technique (FTIR, NIR)– Non-destructive method to assess compost maturity by analyzing chemical composition (e.g., C:N ratio, humification indices, lignin content). NIR spectroscopy can predict germination index and maturity parameters.
• Moisture content only– Not sufficient to determine maturity; immature compost can have similar moisture content.
• Therefore, both seed germination test and infrared techniqueare used to assess compost maturity → Option 4.

 

 

 

Question 17

Conjugation between two bacteria is mediated through:

1. F-plasmid
2. R-plasmid
3. C-plasmid
4. D-plasmid

Correct Answer: 1

 

Explanation:

• Conjugationis the direct transfer of DNA from one bacterium to another through cell-to-cell contact via a pilus.
• F-plasmid (Fertility plasmid)– Contains genes required for conjugation, including those for pilus synthesis (tra genes). The F⁺ (donor) cell transfers a copy of the F-plasmid to the F⁻ (recipient) cell.
• R-plasmid (Resistance plasmid)– Carries antibiotic resistance genes and can be transferred by conjugation (if it has tra genes), but the primary mediator of conjugation is the F-plasmid or other conjugative plasmids.
• C-plasmid and D-plasmid– Not standard terms for conjugation mediators.
• The question asks for the classicmediator of conjugation, which is the F-plasmid → Option 1.

 

 

 

Question 18

Biogas production from cattle manure involves which of the following types of bacteria?
A. Hydrolytic
B. Methanogenic
C. Transitional

Options:

1. A only
2. B only
3. C only
4. A, B and C

Correct Answer: 4

 

Explanation:

• Biogas production(anaerobic digestion) involves a consortium of microorganisms in four stages:
  1. Hydrolysis– Hydrolytic bacteria (e.g., Clostridium, Bacillus) break down complex polymers (cellulose, hemicellulose, proteins, fats) into simpler monomers (sugars, amino acids, fatty acids).
  2. Acidogenesis– Acidogenic bacteria convert monomers into volatile fatty acids (VFAs), alcohols, H₂, and CO₂.
  3. Acetogenesis– Acetogenic bacteria convert VFAs and alcohols into acetic acid, H₂, and CO₂.
  4. Methanogenesis– Methanogenic archaea (e.g., Methanosarcina, Methanobacterium) convert acetic acid, H₂, and CO₂ into methane (CH₄) and CO₂.
• Transitional bacteria– Referred to as syntrophic or acetogenic bacteria that bridge the acidogenic and methanogenic stages (e.g., Syntrophomonas).
• Therefore, all three types(hydrolytic, methanogenic, and transitional) are involved → Option 4 (A, B and C).

 

 

 

Question 19

Statement I: In bacterial transformation, a segment of naked DNA from the surroundings passes through the cell wall and plasma membrane of a recipient cell.
Statement II: Only a few competent recipient bacterial cells can take up the DNA from the surroundings.

Options:

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

Correct Answer: 1

 

Explanation:

• Statement I is true:Transformation is the process by which a bacterial cell takes up naked (free) DNA from its surroundings. The DNA passes through the cell wall and plasma membrane (via competence proteins) and integrates into the recipient’s genome by homologous recombination.
• Statement II is true:Only competent bacterial cells can take up DNA. Competence can be natural (e.g., Bacillus subtilis, Streptococcus pneumoniae, Neisseria gonorrhoeae) or artificially induced (e.g., by chemical treatment or electroporation). In a population, only a fraction of cells are naturally competent at any given time.
• Therefore, both statements are true → Option 1.

 

 

 

Question 20

Scientific name of Oil palm is:

1. Olea europea
2. Elaeis guineensis
3. Ricinus communis
4. Sesamum indicum

Correct Answer: 2

 

Explanation:

• Elaeis guineensis– The African oil palm, native to West Africa. It is the primary source of palm oil (from the mesocarp) and palm kernel oil. It is a major commercial oil crop in tropical regions (especially Indonesia and Malaysia).
• Olea europea– Olive, grown for olive oil and table olives.
• Ricinus communis– Castor bean, source of castor oil (used in industry and medicine).
• Sesamum indicum– Sesame (til), source of sesame oil.
• Therefore, the scientific name of Oil palm is Elaeis guineensis→ Option 2.

 

 

 

Question 21

The National Genebank Facility to conserve plant genetic resources in India is located at:

1. Chennai (Tamil Nadu)
2. Ahmedabad (Gujarat)
3. New Delhi (Delhi)
4. Hyderabad (Telangana)

Correct Answer: 3

 

Explanation:

• The National Genebankof India is located at the National Bureau of Plant Genetic Resources (NBPGR) in New Delhi.
• It was established in 1986and is the primary facility for ex-situ conservation of plant genetic resources (seeds, tissue culture, cryopreserved material).
• The genebank conserves over 5 lakh accessionsof crop plants, including landraces, wild relatives, and elite breeding lines.
• Other locations:
  • Chennai– Has a regional station of NBPGR, but not the main genebank.
  • Ahmedabad– No major NBPGR genebank facility.
  • Hyderabad– Hosts ICRISAT (international genebank), not the national genebank.

Thus, the National Genebank is located in New Delhi → Option 3.

 

 

 

Question 22

Which of the following crops are considered to have originated in South American Centre?

1. Potato, Tomato, Tobacco, Rice
2. Potato, Tomato, Tobacco, Maize
3. Potato, Tomato, Sugarcane, Wheat
4. Potato, Tomato, Soybean, Brinjal

Correct Answer: 2

 

Explanation:

• According to I. Vavilov’s Centers of Origin:
  • South American Centre(Peru-Ecuador-Bolivia-Chile) includes: Potato, Tomato, Tobacco, Maize, Cassava, Peanut, Cotton (Gossypium barbadense), Lima bean, etc.
• Rice(Option 1) – Originated in Southeast Asia (Indo-Burma or China centre).
• Sugarcane(Option 3) – Originated in Southeast Asia/New Guinea.
• Wheat(Option 3) – Originated in Near East (Southwest Asia).
• Soybean(Option 4) – Originated in China (East Asian centre).
• Brinjal (Eggplant)(Option 4) – Originated in Southeast Asia (Indo-Burma centre).
• Maize(Option 2) – Originated in South America/Mexico (part of the South American centre in Vavilov’s classification).
• Therefore, the correct set is Potato, Tomato, Tobacco, Maize → Option 2.

 

 

 

Question 23

Which one of the following species has an inflorescence of terminal digitate spikes of spikelets?

1. Setaria italica
2. Eleusine coracana
3. Panicum sumatrense
4. Panicum miliaceum

Correct Answer: 2

Explanation:

• Eleusine coracana(Finger millet / Ragi) has an inflorescence consisting of terminal digitate spikes (finger-like spikes arranged like fingers on a hand at the top of the stem). The spikes are arranged in a whorl or digitately.
• Setaria italica(Foxtail millet) – Inflorescence is a dense, bristly panicle (not digitate spikes).
• Panicum sumatrense(Little millet) – Inflorescence is an open panicle with spreading branches.
• Panicum miliaceum(Proso millet) – Inflorescence is an open or compact panicle.
• Key feature:Eleusine coracana is characterized by its digitate inflorescence (2–7 spikes clustered at the stem apex).
• Thus, the correct answer is Eleusine coracana → Option 2.

 

 

 

Question 24

Who among the following scientist is considered as ‘Father of Green Revolution in India’?

1. Dr B P Pal
2. Dr R S Paroda
3. Dr M S Swaminathan
4. Dr Gurudev S Khush

Correct Answer: 3

 

Explanation:

• Dr M S Swaminathan(Monkombu Sambasivan Swaminathan) is widely known as the “Father of Green Revolution in India”.
• He collaborated with Norman Borlaug(Father of Green Revolution globally) to introduce high-yielding semi-dwarf wheat varieties (e.g., Lerma Rojo, Sonora 64) into India in the 1960s.
• He played a key role in the development and adoption of high-yielding varieties of wheat and rice, helping India achieve food self-sufficiency.
• Dr B P Pal– Former Director General of ICAR; contributed to wheat breeding but not primarily known as Father of Green Revolution.
• Dr R S Paroda– Former DG of ICAR; contributed to plant genetic resources and agricultural research policy.
• Dr Gurudev S Khush– Worked on rice breeding at IRRI (Philippines); developed IR36, not directly associated with Indian Green Revolution.
• Thus, the correct answer is Dr M S Swaminathan → Option 3.

 

 

 

Question 25

Options:

1. Turkey
2. Marquis
3. Federation
4. Latisail

Correct Answer: 1 (Turkey wheat)

 

Explanation:

• Turkey wheatis a landrace variety of wheat introduced into the USA from Russia (Crimean region) in the late 19th century.
• It became a major parent of many hard red winter wheat varieties in the Great Plains.
• The question likely asks for a specific wheat variety or category; based on the answer key, Turkeyis correct.
• Thus, the correct answer is Turkey → Option 1.

 

 

 

Question 26

Options:

1. Turkey
2. Marquis
3. Federation
4. Latisail

Correct Answer: 4 (Latisail)

Explanation:

• Latisailis a variety of rice (not wheat). It was one of the popular varieties in India before the Green Revolution.
• The question likely asks for a rice variety or specific crop; based on the answer key, Latisailis correct.
• Thus, the correct answer is Latisail → Option 4.

 

 

 

Question 27

Options:

1. Turkey
2. Marquis
3. Federation
4. Latisail

Correct Answer: 3 (Federation)

Explanation:

• Federationis a famous wheat variety developed in Australia. It was released in 1900 and became widely grown in Australia and other regions.
• It was derived from a cross between Purple Strawand Yandilla.
• The question likely asks for a wheat variety; based on the answer key, Federationis correct.
• Thus, the correct answer is Federation → Option 3.

 

 

 

Question 28

Which of the following statements are correct with respect to fibres or fibre yielding plants?
A. Vegetable fibres are mainly composed of cellulose
B. Vegetable fibres are mainly composed of proteins
C. Fibre durability is conditioned by the diameter of the fibre or fibrous aggregate
D. The flexible wall of cotton fibre contains no lignin

Options:

1. A, B, C only
2. A, C, D only
3. B, C, D only
4. A, B, D only

Correct Answer: 2

 

Explanation:

• A is correct:Vegetable fibres (e.g., cotton, jute, flax, hemp) are mainly composed of cellulose, a polysaccharide that provides strength and flexibility.
• B is incorrect:Vegetable fibres are not mainly composed of proteins. Protein fibres are from animal sources (e.g., wool, silk).
• C is correct:Fibre durability is influenced by diameter (thicker fibres are generally more durable) and the fibrous aggregate structure.
• D is correct:Cotton fibre is almost pure cellulose (about 90–95%) and contains very little or no lignin (unlike woody fibres). This gives cotton its flexible, soft, and non-rigid nature.
• Therefore, the correct statements are A, C, and D→ Option 2 (A, C, D only).

 

 

 

Question 29

What is the role of antibiotics in disease management?
A. Prevent the growth of pathogens
B. Prevent the development of pathogens
C. Accelerate the growth of pathogens
D. Accelerate the development of pathogens

Options:

1. A, D only
2. B, C only
3. A, B only
4. C, D only

Correct Answer: 3

 

Explanation:

• Antibioticsare antimicrobial agents that inhibit the growth (bacteriostatic) or kill (bactericidal) of bacteria and other pathogens.
• A is correct:Antibiotics prevent the growth of pathogens by interfering with essential processes (e.g., cell wall synthesis, protein synthesis, nucleic acid synthesis).
• B is correct:Antibiotics prevent the development of pathogens by halting their multiplication and spread.
• C and D are incorrect:Antibiotics do not accelerate growth or development; they suppress or destroy pathogens.
• Therefore, the correct statements are A and B→ Option 3 (A, B only).

 

 

 

Question 30

Which of the following listed items do NOT come under the purview of the National Plant Protection Organization (NPPO) under the International Plant Protection Convention (IPPC)?
A. Surveillance and inspection
B. Controlling pests (example administering treatments, preventing spread, disinfection or disinfection)
C. Conduction of pest risk analysis
D. Conservation and regeneration of plant genetic resources
E. Multiplication of plant genetic resources

Options:

1. A, B only
2. B, C only
3. A, E only
4. D, E only

Correct Answer: 4

 

Explanation:

• NPPO(National Plant Protection Organization) is the official service responsible for plant quarantine and phytosanitary measures under the IPPC (International Plant Protection Convention).
• Functions of NPPO include:
  • Surveillance and inspection(A) – Monitoring for pests and diseases.
  • Controlling pests(B) – Administering treatments, disinfection, preventing spread.
  • Conduction of pest risk analysis (PRA)(C) – Assessing risk of pest introduction.
• D and E (Conservation and regeneration of plant genetic resources, Multiplication of plant genetic resources)– These are NOT functions of NPPO. They fall under the purview of PGR (Plant Genetic Resources) organizations like NBPGR, seed certification agencies, and genebanks.
• Therefore, the items NOTunder NPPO are D and E → Option 4 (D, E only).

 

 

 

Question 31

Which of the statements below do NOT stand true about the ‘Convention on Biological Diversity (CBD)’?
A. CBD deals with all biological resources
B. One of the objectives of CBD is the conservation and sustainable use of the component of biodiversity
C. The CBD affirms the sovereign rights of the nation over its genetic resources
D. CBD is an instrument for harmonizing phytosanitary measures in international trade
E. CBD provides a model for securing protection for plant breeder rights for plant varieties

Options:

1. A, B only
2. B, C only
3. C, D only
4. D, E only

Correct Answer: 4

 

Explanation:

• CBD(Convention on Biological Diversity, 1992) has three main objectives:
  1. Conservation of biological diversity.
  2. Sustainable use of its components.
  3. Fair and equitable sharing of benefits arising from genetic resources.
• A is true:CBD deals with all biological resources (ecosystems, species, genetic resources).
• B is true:Conservation and sustainable use are core objectives.
• C is true:CBD affirms sovereign rights of nations over their genetic resources.
• D is false:Harmonizing phytosanitary measures is the role of IPPC (International Plant Protection Convention) and WTO-SPS, not CBD.
• E is false:Protection of plant breeder rights is covered by UPOV (International Union for the Protection of New Varieties of Plants) and PPV&FR Act (India), not CBD.
• Therefore, the statements NOT trueabout CBD are D and E → Option 4 (D, E only).

 

 

 

Question 32

With regard to the cultivation of groundnut, which of the following statements are correct?
A. Groundnut is a warm-season crop requiring abundant sunshine
B. Well-drained sandy loams are most favourable for growing groundnut
C. Groundnut is propagated through whole pods, decorticated seeds or from seedlings
D. Groundnut is widely grown in coastal and deltaic regions
E. Groundnut takes six to nine months to mature

Options:

1. A, B, C only
2. B, C, D only
3. C, D, E only
4. A, D, E only

Correct Answer: 1

 

Explanation:

• A is correct:Groundnut (peanut) is a warm-season crop requiring high temperatures (25–30°C) and abundant sunshine for proper growth, flowering, and pod development.
• B is correct:Well-drained sandy loam soils are most favourable because they allow easy peg penetration, pod development, and harvesting. Heavy clay soils cause pod deformation and rotting.
• C is correct:Groundnut is propagated through whole pods (farmer-saved seed), decorticated seeds (shelled kernels), or seedlings (rare). Whole pods maintain seed viability longer.
• D is incorrect:Groundnut is not widely grown in coastal and deltaic regions (high humidity and waterlogging cause diseases like tikka leaf spot and pod rot). It is grown in rainfed, well-drained uplands.
• E is incorrect:Groundnut takes about 3–5 months (90–150 days) to mature, depending on the variety (bunch types mature earlier, spreading types later). 6–9 months is too long.
• Therefore, the correct statements are A, B, and C→ Option 1 (A, B, C only).

 

 

 

Question 33

Which of the following statements are true with respect to tomato?
A. Tomato plants are not sensitive to frost
B. Typically tomato seeds are first sown in a nursery and transplanted later into the field
C. Tomato is a native crop of Asia
D. Under normal conditions, tomato assumes spreading habit
E. Tomato is an annual crop

Options:

1. A, C, E only
2. B, D, E only
3. D, A, C only
4. A, B, D only

Correct Answer: 2

 

Explanation:

• A is false:Tomato is highly sensitive to frost. It is a warm-season crop and cannot tolerate freezing temperatures (frost kills the plants).
• B is true:Typically, tomato seeds are sown in a nursery (protected environment) and transplanted into the field after 4–6 weeks. This ensures uniform stand and early establishment.
• C is false:Tomato is native to South America (Peru-Ecuador region), not Asia. It was introduced to Asia after European colonization.
• D is true:Under normal (unpruned) conditions, tomato assumes a spreading habit (indeterminate types vine and spread; determinate types are more compact but still spreading).
• E is true:Tomato is grown as an annual crop (completes life cycle in one season), though it is botanically a perennial but grown as an annual for fruit production.
• Therefore, the true statements are B, D, and E→ Option 2 (B, D, E only).

 

 

 

Question 34

Arrange these Conventions and Treaties in ascending order of year they entered into force:
A. World Trade Organization-Trade Related Aspect of Intellectual Property Rights (WTO-TRIPS)
B. Convention on Biological Diversity (CBD)
C. Nagoya Protocol on Access and Benefit Sharing (NP)
D. Geographical Indication of Goods (Registration and Protection), GI Act
E. International Treaty on Plant Genetic Resources for Food and Agriculture (ITPGRFA)

Options:

1. B, D, A, E, C
2. A, B, D, E, C
3. E, C, B, A, D
4. D, B, A, E, C

Correct Answer: 4

Explanation:

• The correct ascending order(earliest to latest) of the year they entered into force:

Instrument	Year entered into force
D. GI Act (India)	1999 (Geographical Indications of Goods (Registration and Protection) Act, 1999 – enacted in 1999, came into force in 2003? Actually, the Act was passed in 1999 but rules notified in 2002. For this question, D is placed first.)
B. Convention on Biological Diversity (CBD)	1993 (Adopted 1992, entered into force 29 December 1993)
A. WTO-TRIPS	1995 (WTO established 1 January 1995; TRIPS agreement part of it)
E. ITPGRFA	2004 (Adopted 2001, entered into force 29 June 2004)
C. Nagoya Protocol (NP)	2014 (Adopted 2010, entered into force 12 October 2014)

Chronological order: D (1999/2002) → B (1993)Wait, this needs correction. Actually:

• CBD (1993)is earlier than GI Act (1999). But the answer key shows D, B, A, E, C.
• Possibly the GI Act (India) is considered as 1999(enactment year) and CBD as 1993, so CBD should come before D. However, the answer key has D first. This may be based on a different interpretation (e.g., Indian GI Act implementation year 2003 vs CBD 1993).
• Given the answer key, the correct order is D, B, A, E, C→ Option 4.

 

 

 

Question 35

Match List I with List II:

List I (Common name)	List II (Botanical name)
A. Rosewood	I. Shorea robusta
B. Sal	II. Diospyros ebenum
C. Ebony	III. Dalbergia latifolia
D. Teak	IV. Tectona grandis

Options:

1. A-II, B-III, C-IV, D-I
2. A-IV, B-II, C-III, D-I
3. A-I, B-II, C-III, D-IV
4. A-III, B-I, C-II, D-IV

Correct Answer: 4

 

Explanation:

• Rosewood (A)– Dalbergia latifolia (Indian rosewood), a tropical hardwood used for furniture and musical instruments. → A-III
• Sal (B)– Shorea robusta, a hardwood tree native to the Indian subcontinent, used for timber. → B-I
• Ebony (C)– Diospyros ebenum, a tropical hardwood known for its dense black heartwood, used for ornamental items. → C-II
• Teak (D)– Tectona grandis, a tropical hardwood prized for its durability, water resistance, and use in shipbuilding and furniture. → D-IV
• Therefore, correct matching is A-III, B-I, C-II, D-IV→ Option 4.

 

 

 

Question 36

Match List I with List II:

List I (Crop)	List II (Edible part)
A. Cauliflower	I. Root
B. Chilli	II. Leaves
C. Carrot	III. Inflorescence
D. Spinach	IV. Fruits

Options:

1. A-I, B-II, C-III, D-IV
2. A-III, B-IV, C-I, D-II
3. A-II, B-I, C-III, D-IV
4. A-I, B-III, C-IV, D-II

Correct Answer: 2

 

Explanation:

• Cauliflower (A)– The edible part is the inflorescence (immature flower buds and stem). → A-III
• Chilli (B)– The edible part is the fruit (botanically a berry, used as a spice or vegetable). → B-IV
• Carrot (C)– The edible part is the root (taproot, rich in β-carotene). → C-I
• Spinach (D)– The edible part is the leaves (leafy vegetable). → D-II
• Therefore, correct matching is A-III, B-IV, C-I, D-II→ Option 2.

 

 

 

Question 37

Match List I with List II:

List I (Plant)	List II (Product)
A. Hevea brasiliensis	I. Henna
B. Achras zapota	II. Tannin
C. Lawsonia inermis	III. Rubber
D. Schinopsis lorentzii	IV. Chewing gum

Options:

1. A-III, B-IV, C-I, D-II
2. A-I, B-III, C-II, D-IV
3. A-II, B-I, C-III, D-IV
4. A-IV, B-II, C-III, D-I

Correct Answer: 1

 

Explanation:

• Hevea brasiliensis(A) – The Pará rubber tree, the primary source of natural rubber (latex). → A-III
• Achras zapota(B) – Also known as Manilkara zapota (sapodilla). The latex (chicle) is used to make chewing gum. → B-IV
• Lawsonia inermis(C) – The henna The leaves are dried and ground to produce henna dye for body art and hair coloring. → C-I
• Schinopsis lorentzii(D) – A hardwood tree native to South America, known as quebracho. It is a rich source of tannin (used for leather tanning). → D-II
• Therefore, correct matching is A-III, B-IV, C-I, D-II→ Option 1.

 

 

 

Question 38

Statement I: Mera Gaon Mera Gaurav Scheme under extension education was launched by Prime Minister Sh. Narender Modi with the theme ‘Lab to Land’ by asking scientists to adopt villages and remain in touch with them providing technical information to farmers and other aspects.
Statement II: The scheme Mera Gaon Mera Gaurav functions with the help of Krishi Vigyan Kendra (KVKs) and Agriculture Technology Management Agencies (ATMA).

Options:

1. Both Statement I and Statement II are true
2. Both Statement I and Statement II are false
3. Statement I is true but Statement II is false
4. Statement I is false but Statement II is True

Correct Answer: 1

 

Explanation:

• Statement I is true:Mera Gaon Mera Gaurav (MGMG) scheme was launched by Prime Minister Narendra Modi on 25 July 2015. The theme was “Lab to Land”, where agricultural scientists adopted villages to provide technical knowledge and support to farmers.
• Statement II is true:The scheme functions with the help of Krishi Vigyan Kendras (KVKs) and Agriculture Technology Management Agencies (ATMA). KVKs provide frontline extension, and ATMA coordinates extension activities at the district level.
• Therefore, both statements are true → Option 1.

 

 

 

Question 39

Statement I: Active collection in the seed genebank is conserved for a medium period of 2-10 years. They are regularly used for research and distribution to bonafide users.
Statement II: Base collections in the seed genebank are conserved for a long duration, 15 years or more for safety and prosperity.

Options:

1. Both Statement I and Statement II are true
2. Both Statement I and Statement II are false
3. Statement I is true but Statement II is false
4. Statement I is false but Statement II is True

Correct Answer: 1

 

Explanation:

• Statement I is true:Active collections (also called working collections) are maintained for medium-term conservation (2–10 years) at reduced temperature and humidity. They are regularly used for research, characterization, evaluation, and distribution to bona fide users (breeders, researchers).
• Statement II is true:Base collections are maintained for long-term conservation (15 years or more, often up to 50–100 years) under optimal conditions (e.g., -20°C, low humidity). They serve as safety backup to prevent genetic erosion and ensure long-term preservation of genetic resources.
• Therefore, both statements are true → Option 1.

 

 

Question 40

Statement I: The process of CTC in tea stands for Crushing, Tearing and Curling.
Statement II: The process of CTC in tea stands for Curling, Tearing and Crushing.

Options:

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

Correct Answer: 3

 

Explanation:

• CTCstands for Crushing, Tearing, and Curling – a mechanical process used in tea manufacturing to produce black tea (especially for tea bags).
• Process:Tea leaves are passed through cylindrical rollers with sharp teeth that crush, tear, and curl the leaves into small, even pellets. This increases surface area and accelerates oxidation (fermentation).
• Statement I is correct(Crushing, Tearing, Curling).
• Statement II is false(Curling, Tearing, Crushing is incorrect order).
• Therefore, Statement I is correct but Statement II is false → Option 3.