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

 

Question No. 1

Assertion (A): Sesamum phyllody is caused by Phytoplasma and develop Witches broom symptom.
Reason (R): All floral parts turn green, apical leaves also become very small and hence the name is phyllody. It is transmitted by both leaf hopper (Orosius albicinctus) and aphids.
Options:

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 completely correct explanation of (A).
3. (A) is correct but (R) is not correct.
4. (A) is not correct but (R) is correct.

Correct Answer:3
Explanation: (A) is correct: Sesamum phyllody is caused by Phytoplasma and witches’ broom is a symptom. (R) is incorrect because the name “phyllody” refers to floral parts turning green (leaf-like), not apical leaves becoming small. Also, the primary vector is leafhoppers, not aphids.

 

 

Question No. 2

During its growth through the style, the pollen tube’s cell wall is modified to facilitate its progression and interaction with the female tissues. Which of the following components is primarily involved in the strengthening and extension of the pollen tube cell wall?

1. Lignin
2. Cellulose and pectin
3. Suberin
4. Chitin

Correct Answer:2
Explanation: The pollen tube wall is primarily composed of cellulose (for strength) and pectin (for flexibility and extension). Lignin and suberin are for rigidity/waterproofing, chitin is in fungal walls.

 

 

Question No. 3

In Frederick Griffith’s transformation experiment (1928), the following activities were performed. Arrange the activities in proper sequential order from first to last.
(A) Injected mice with live R-strain: Mice lived.
(B) Injected mice with a mixture of heat-killed S-strain and live R-strain: Mice died.
(C) Injected mice with live S-strain: Mice died.
(D) Injected mice with heat-killed S-strain: Mice lived.

Options:

1. (A), (C), (B), (D)
2. (A), (C), (D), (B)
3. (B), (A), (D), (C)
4. (C), (A), (D), (B)

Correct Answer:4
Explanation: The correct order is: (C) live S → mice die, (A) live R → mice live, (D) heat-killed S → mice live, (B) heat-killed S + live R → mice die (transformation).

 

 

Question No. 4
Statement (I): Heterosis breeding involves crossing two genetically diverse inbred lines to produce F1 hybrids with superior traits.
Statement (II): The vigour observed in F1 hybrids remains unchanged for many subsequent generations if the hybrid is self-pollinated.

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:3
Explanation: Statement I is correct (definition of heterosis breeding). Statement II is false because heterosis declines in subsequent generations due to segregation and recombination.

 

 

Question No. 5:
Which of the following statements correctly describes the development of the female gametophyte in most angiosperms?

Options:

1. The functional megaspore undergoes a single meiotic division followed by cytokinesis to directly form the egg apparatus.
2. Following meiotic reduction, the surviving megaspore completes three successive mitotic divisions without immediate cellular partitioning, leading to an eight-nucleate, seven-celled embryo sac.
3. All four megaspores formed by meiosis contribute equally to the formation of a sixteen-nucleate mature female gametophyte.
4. The megaspore enters two rounds of meiosis followed by syngamy-independent nuclear fusion events forming a triploid embryo sac prior to fertilization.

Correct Answer: 2. 

Explanation:
This describes the Polygonum type of embryo sac development, which is the most common in angiosperms. The functional megaspore undergoes three mitotic divisions to form 8 nuclei, which organize into a 7-celled structure (egg apparatus, antipodals, and central cell).

 

 

Question No. 6
The random cup method of seed sampling techniques is particularly suitable for seeds that are not extremely chaffy, do not tend to bounce or roll and require a representative working sample of size (gram):

1. Up to 10 grams
2. Up to 50 grams
3. Up to 100 grams
4. Up to 150 grams

Correct Answer:1
Explanation: The random cup method is used for small working samples, typically up to 10 grams, for seeds that are free-flowing and not chaffy.

 

 

Question No. 7
The fourth rule of Koch’s postulates was contributed by:

1. Robert Koch
2. Antoine van Leeuwenhoeck
3. F. Smith
4. Julius Oscar Brefeld

Correct Answer:3
Explanation: F. Smith, a plant pathologist, added the fourth postulate (the pathogen must be re-isolated from the experimentally infected host and shown to be identical to the original) for plant diseases.

 

Question No. 8:
In angiosperms, during megasporogenesis, what is typically formed inside the ovule?

Options:

1. Four microspores, of which usually one is functional
2. Four megaspores, of which usually one is functional
3. A diploid embryo sac
4. A triploid endosperm

Correct Answer: 2

Explanation:
During megasporogenesis in angiosperms, a megaspore mother cell undergoes meiosis to produce four haploid megaspores. Out of these, typically only one survives and develops into the female gametophyte (embryo sac), while the remaining three degenerate.

 

 

Question No. 9
In plant species exhibiting alternation of generations (sporophyte, gametophyte), the sporophytic generation is characterized by:

1. Haploid chromosome number and gamete production via mitosis.
2. Diploid chromosome number and spore production via meiosis.
3. Haploid chromosome number and spore production via meiosis.
4. Diploid chromosome number and gamete production via mitosis.

Correct Answer:2
Explanation: Sporophyte is diploid (2n) and produces haploid spores via meiosis. Gametophyte is haploid (n) and produces gametes via mitosis.

 

Question No. 10
Which of the following statements is CORRECT about Nucleus Seed?

1. Nucleus seed is produced solely by private seed companies under minimal supervision.
2. Nucleus seed is a mix of different varieties bred for broad adaptation.
3. Nucleus seed is the genetically pure seed maintained by the original breeder or institution.
4. Nucleus seed is always distributed directly to farmers for commercial planting.

Correct Answer:3
Explanation: Nucleus seed is the most genetically pure seed stock, produced and maintained by the original plant breeder or institution, and is the source for breeder seed.

 

 

Question No. 11

Integrated plant disease management is necessary for sustainability and to avoid adverse effects on the environment. Hence, step-wise one should follow the plant disease management principles like:
(A) Eradication (B) Avoidance (C) Exclusion (D) Protection (E) Resistance (F) Therapy
Options:

1. (B), (A), (C), (E), (D), (F)
2. (B), (C), (A), (E), (D), (F)
3. (A), (B), (C), (E), (D), (F)
4. (A), (D), (F), (B), (C), (E)

Correct Answer:2
Explanation: Logical sequence: Avoidance → Exclusion → Eradication → Resistance → Protection → Therapy.

 

 

Question No. 12
Assertion (A): Trichoderma harzianum is used as a biocontrol agent against Rhizoctonia solani.
Reason (R): Trichoderma harzianum parasitize Rhizoctonia by coiling around and penetrating its hyphae thus disrupting the cytoplasm and causing suppression.
Options:

1. Both (A) and (R) are true 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.

Correct Answer:1
Explanation: Both are true. Trichoderma harzianum is a well-known mycoparasite of Rhizoctonia solani, and the mechanism involves coiling, penetration, and cytoplasmic disruption.

 

 

Question No. 13
Arrange the following crops in ascending order of their area under cultivation (mha) during 2023-24 (based on Government of India data):
(A) Groundnut (B) Rapeseed and Mustard (C) Soybean (D) Sunflower
Options:

1. (A), (B), (C), (D)
2. (B), (C), (D), (A)
3. (C), (B), (A), (D)
4. (D), (A), (B), (C)

Correct Answer:4 (approx. order: Sunflower < Groundnut < Rapeseed & Mustard < Soybean)
Explanation: As per recent data (2023-24): Sunflower (~0.3 mha), Groundnut (~4-5 mha), Rapeseed & Mustard (~8-9 mha), Soybean (~11-12 mha). Ascending order: D, A, B, C.

 

 

Question No. 14
Which of the following pigment contributes to the yellow colour of the maize kernel:

1. Cryptocyanin
2. Anthocyanin
3. Zeaxanthin
4. Pyocyanin

Correct Answer:3
Explanation: Zeaxanthin is a carotenoid pigment responsible for yellow color in maize kernels. Anthocyanin gives purple/red, pyocyanin is bacterial, cryptocyanin is not relevant here.

 

Question No. 15
Seed germination is a tightly regulated physiological process that requires the weakening of the seed coat to allow the emergence of the radicle. Which of the following hormonal interactions is primarily responsible for initiating this seed coat weakening during the early stages of germination?

1. Synergistic action of GA and ethylene stimulating the production of cell wall-degrading enzymes.
2. ABA and cytokinin working together to weaken the seed coat.
3. Auxin and salicylic acid inhibiting cell wall reinforcement.
4. Ethylene and ABA collaborating to induce seed coat lignification.

Correct Answer:1
Explanation: Gibberellins (GA) and ethylene synergistically promote cell wall-degrading enzymes (e.g., endo-β-mannanase) that weaken the seed coat. ABA inhibits germination.

 

 

Question No. 19
The inheritance of mitochondrial DNA (mtDNA) in most sexually reproducing organisms exhibits a unique pattern primarily due to:

1. High rates of homologous recombination during oogenesis.
2. Predominant paternal inheritance of mitochondrial genomes.
3. Maternal inheritance of mitochondrial genomes.
4. Strict biparental inheritance with equal contribution from both parents.

Correct Answer:3
Explanation: mtDNA is typically inherited maternally because the egg contributes cytoplasm (including mitochondria) to the zygote, while sperm contributes little to no mitochondria.

 

 

Question No. 20
Which is NOT a multi-parent mapping population?

1. MAGIC population
2. NAM population
3. Association mapping panel
4. F2,3 plants

Correct Answer:4
Explanation: F2:3 plants are derived from a biparental cross (two parents). MAGIC, NAM, and association mapping panels involve multiple parents.

 

 

Question No. 21
Assertion (A): Bacillus thuringiensis, a soil bacterium possessing Cry-1Ac gene, kills the larvae of boll worms.
Reason (R): Cry-1Ac gene codes for crystal protein which is responsible for the insecticidal effect.
Options:

1. Both (A) and (R) are true 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.

Correct Answer:1
Explanation: Both true, and (R) correctly explains (A): Cry1Ac produces crystalline toxin that kills lepidopteran larvae like bollworms.

 

 

Question No. 22
In hybrid sunflower seed production, what is the typical ratio of the A-line (female parent) to the R-line (male parent)?

1. 1:1
2. 2:1
3. 3:1
4. 4:1

Correct Answer:4
Explanation: In hybrid sunflower production, the typical ratio of A-line (female, male-sterile) to R-line (male, restorer) is 4:1 to ensure adequate pollen supply.

 

 

Question No. 23
Consider a hypothetical repressible operon under negative control. A mutation in the gene encoding the repressor protein results in a non-functional repressor that cannot bind to the operator. What would be the expected expression pattern of the structural genes in this operon?

1. Expression would be induced only in the presence of the corepressor.
2. Expression would be repressed under all conditions.
3. Expression would occur constitutively, regardless of the presence or absence of the corepressor.
4. Expression would be completely abolished.

   Correct Answer:3
   Explanation: In a repressible operon (e.g., trp operon), the repressor is inactive alone and requires corepressor to bind operator. Non-functional repressor → no repression → constitutive expression.

 

 

Question No. 24:
Which of the following represents an aneuploid condition in plants?

Options:

1. A plant with 12 chromosomes
2. A plant with 24 chromosomes
3. A plant with 48 chromosomes
4. A plant with 25 chromosomes

Correct Answer: 4. 

Explanation:
Aneuploidy refers to a condition where the chromosome number is not an exact multiple of the haploid set (n). Numbers like 12, 24, and 48 represent euploid conditions (exact multiples), whereas 25 is not a multiple of the base number, indicating aneuploidy.

 

 

Question No. 25
Key experiments that led to the understanding that DNA is the genetic material:
(A) Hershey and Chase (1952)
(B) Griffith (1928)
(C) Fraenkel-Conrat (1957)
(D) Avery, MacLeod, McCarty (1944)
Options:

1. (A), (B) and (C) only
2. (A), (B) and (D) only
3. (A), (C) and (D) only
4. (B), (C) and (D) only

Correct Answer:2
Explanation: Griffith (transformation principle), Avery et al. (DNA as transforming principle), Hershey-Chase (DNA as genetic material in phage). Fraenkel-Conrat showed RNA as genetic material in TMV.

 

 

Question No. 26
Statement (I): Population of Actinobacteria is higher in alkaline soils.
Statement (II): Actinobacteria plays an important role in the decomposition of recalcitrant compounds such as chitin, lignin and cellulose.
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: Both statements are true. Actinobacteria prefer neutral to alkaline pH and are key decomposers of complex organic matter.

 

 

Question No. 27
Statement (I): Transformation is a natural process in which a bacterium can take up naked DNA from the environment.
Statement (II): Transformation does not cause genetic recombination.
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:3
Explanation: (I) is correct. (II) is false because transformation can cause genetic recombination when the taken-up DNA integrates into the bacterial chromosome.

 

 

Question No. 28
Match List-1 with List-2:
(A) Mitochondria → (III) Generates most of the cell’s supply of ATP
(B) Endoplasmic reticulum (rough) → (I) Site of protein synthesis and modification for secretion
(C) Golgi apparatus → (IV) Modifies, sorts, and packages proteins and lipids
(D) Lysosomes → (II) Contains digestive enzymes to break down waste materials
Options:

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

Correct Answer:3

 

 

Question No. 29
Which one of the following is a rapid test to detect the mechanical damage to legume seeds?

1. Topographical Tetrazolium Test
2. X-ray Photography Test
3. Lactophenol Test
4. Ferric Chloride Test

Correct Answer:4
Explanation: Ferric chloride test detects mechanical damage in legume seeds by reacting with phenolic compounds released from damaged tissues, giving a dark color.

 

Question No. 30
Match List-1 with List-2:
(A) Fluorescent pseudomonads → (III) King’s B medium
(B) Ralstonia solanacearum → (I) Tetrazolium chloride agar
(C) Xanthomonas species → (IV) Yeast extract nutrient agar
(D) Fungi and bacteria → (II) Potato dextrose agar

Options:

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

Correct Answer:2

 

Question No. 31

Match List-1 with List-2:
(A) CPCRI → (IV) Kasaragod, Kerala
(B) CAZRI → (III) Jodhpur, Rajasthan
(C) CTCRI → (II) Thiruvananthapuram, Kerala
(D) CSWRI → (I) Avikanagar, Rajasthan

Options:

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

Correct Answer:4

 

Question No. 32
The likelihood of an individual in a population carrying two specific alleles of a human DNA marker, each of which has a frequency of 0.3, will be:

1. 03
2. 06
3. 18
4. 09

Correct Answer:4
Explanation: Assuming Hardy-Weinberg equilibrium and independent assortment, probability = 0.3 × 0.3 = 0.09.

 

Question No. 33

Match List-1 with List-2:
(A) Coining the term “Gene” → (III) Wilhelm Johannsen
(B) Construction of first accurate physical model of DNA → (I) James Watson and Francis Crick
(C) Confirmation of semi-conservative replication of DNA → (IV) Matthew Meselson and Franklin Stahl
(D) Discovery of Chargaff’s Rules → (II) Erwin Chargaff

Options:

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

Correct Answer:4

 

Question No. 34
Assertion (A): Synchronization of flowering between parental lines ensures effective hybrid seed set.
Reason (R): For effective synchronization of flowering, the seed parental line is sown earlier than the pollen parent.

Options:

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.

Correct Answer:2
Explanation: Both are correct, but (R) is not always true; sometimes pollen parent is sown earlier or staggered sowings are done depending on relative maturity.

 

 

Question No. 35

Match List-1 with List-2:
(A) RFLP → (II) Polymorphisms detected by restriction enzymes and Southern blot
(B) RAPD → (I) Uses short arbitrary primers under low-stringency PCR
(C) AFLP → (IV) Combines restriction digestion with selective PCR amplification
(D) SSR (Microsatellite) → (III) Tandem repeats amplified by specific flanking primers

Options:

• (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
• (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
• (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
• (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Correct Answer:1

 

Question No. 36

Arrange the following legal instruments in chronological order of their adoption/enactment:
(A) Convention on Biological Diversity
(B) Protection of Plant Varieties and Farmers’ Rights Act, India
(C) Biological Diversity Act, India
(D) International Treaty on Plant Genetic Resources for Food and Agriculture

Options:

1. A → B → C → D
2. A → D → B → C
3. B → A → D → C
4. C → A → B → D

Correct Answer:2
Explanation: Correct order: CBD (1992) → ITPGRFA (2001) → PPV&FR Act (2001) → Biological Diversity Act (2002).

 

 

Question No. 37
Certain plant species produce recalcitrant seeds that are highly sensitive to desiccation and low temperatures. Which of the following ex situ conservation approaches is most appropriate for long-term preservation of such species, and why?

1. Establishing field gene banks, because natural conditions are best replicated in situ.
2. Conventional seed storage in dry, cold conditions, as it maximizes longevity.
3. In vitro tissue culture combined with cryopreservation, as it circumvents the drying intolerance.
4. Botanical gardens, since they provide a semi-controlled outdoor environment.

Correct Answer:3
Explanation: Recalcitrant seeds cannot be dried or stored at low temperatures. Cryopreservation of in vitro tissues (e.g., embryos) is the only long-term option.

 

 

Question No. 38
Which of the following best describes the primary mechanism by which histone modifications contribute to heritable changes in gene activity without altering the DNA sequence?

1. They directly alter the base pairing within the DNA double helix.
2. They recruit DNA repair enzymes that introduce specific mutations.
3. They influence chromatin compaction and accessibility to transcription factors.
4. They lead to the degradation of specific mRNA transcripts.

Correct Answer:3
Explanation: Histone modifications (acetylation, methylation, etc.) alter chromatin structure, affecting transcription factor access and gene expression (epigenetic regulation).

 

 

Question No. 39
The classic ‘Operon’ structure, which consists of a cluster of genes under the control of a single promoter, and is transcribed together into a single polycistronic mRNA molecule, is primarily found in:

1. Viruses only
2. Eukaryotes only
3. Prokaryotes (Bacteria and Archaea)
4. Both prokaryotes and eukaryotes.

Correct Answer:3
Explanation: Operons are characteristic of prokaryotes (bacteria and archaea). Eukaryotes generally have monocistronic transcription units.

 

Question No. 40
Seed certification processes typically involve several key steps. Arrange the steps in the correct sequence order from the start to the end:
(A) Supervision at post-harvest stages, including processing and packaging.
(B) Receipt and scrutiny of application, and verification of seed source.
(C) Seed sampling and testing for genetic purity and seed health, and grant of the certificate.
(D) Field inspection to verify compliance with the prescribed field standard.
Options:

1. (B), (D), (C), (A)
2. (A), (B), (C), (D)
3. (B), (D), (A), (C)
4. (C), (B), (D), (A)

Correct Answer:3
Explanation: Correct sequence: Application scrutiny (B) → Field inspection (D) → Post-harvest supervision (A) → Sampling, testing, certification (C).