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

Question 1

To resolve a parental dispute, the Honourable Judge ordered for DNA-fingerprinting of the child. The probable DNA fingerprints of the child would be:

1. 100% similar to the DNA prints of both the parents (father and mother)
2. 100% similar to the DNA prints of the father.
3. 100% similar to the DNA print of the mother
4. About 50% of DNA prints would match either of the parents.

Correct Answer: 4

Explanation: A child inherits half of its DNA from each parent, so DNA fingerprinting shows about 50% similarity with the mother and 50% with the father. It is never 100% identical unless it is the same individual (like identical twins).

 

 

Question 2

The genetic codes are degenerative in nature. The degeneracy is observed primarily in the:

1. 1st base of the codon
2. 2nd base of the codon
3. Both 1st and the 2nd base of the codon
4. Both 3rd and the Wobble position base of the codon

Correct Answer: 4

Explanation: Genetic code is degenerate, meaning multiple codons code for the same amino acid. This degeneracy is mainly due to variation in the third base (wobble position), which often does not change the amino acid

 

 

Question 3

Consider a cross where some offspring in a generation have genotype ppqqrr. Which of the following genotypes CAN be a genotype of either of the parents?

1. PPQQRR
2. PPQqRr
3. PpQqRr
4. PpQQRr

Correct Answer: 3

Explanation: To produce offspring with genotype ppqqrr, both parents must carry recessive alleles for all genes. Only PpQqRr can produce all recessive combinations through segregation, making it the correct possibility.

 

 

Question 4

If Meselson and Stahl’s experiment on the mode of DNA replication is continued up to the sixth generation in bacteria, the ratio of Heavy strands 15N/15N : Hybrid 15N/14N : Light strand 14N/14N containing DNA in the sixth generation would be:

1. 1:1:1
2. 0:1:7
3. 0:1:15
4. 0:1:31

Correct Answer: 4

Explanation: In Meselson-Stahl experiment, after several generations, hybrid DNA decreases while light DNA increases. By the 6th generation, all DNA becomes light except a very small hybrid fraction, giving ratio 0 : 1 : 31.

 

 

Question 5

Assertion A: Indian Council of Agricultural Research (ICAR) is an autonomous body under the Ministry of Agriculture and Farmers’ Welfare, GOI.
Reason R: The Director General, ICAR is the President of the ICAR Society.
Options:

1. Both A and R are true.
2. Both A and R are false.
3. A is true but R is false.
4. A is false but R is true.

Correct Answer: 3

Explanation: ICAR is indeed an autonomous organization under the Ministry of Agriculture, but the President of ICAR Society is the Union Agriculture Minister, not the Director General. Hence, assertion is true but reason is false.

 

 

Question 6

Assertion A: University Education Commission (1948-49) recommended the establishment of Rural Universities in India.
Reason R: In 1960, the 1st Agricultural University was established in India in the pattern of the Land-Grant Universities of the USA.
Options:

1. Both A and R are true.
2. Both A and R are false.
3. A is true but R is false.
4. A is false but R is true.

Correct Answer: 1

Explanation: The University Education Commission (1948–49) recommended rural universities. Later, in 1960, the first agricultural university (Pantnagar) was established on the Land Grant model of the USA, so both statements are correct.

 

 

Question 7

As per the general model, the order of the activities in DNA replication in E. coli from the beginning to the end is:
A. Enzyme ‘primase’ synthesizes the ‘primer RNA’
B. Single-strand binding (SSB) proteins bind to the separated single strands and stabilize them.
C. DNA gyrase and DNA helicase unwind the complementary strands of DNA in the origin of replication (Ori C).
D. After digesting the RNA primers, the ‘Okazaki fragments’ are joined together by the enzyme ‘ligase’.
E. Synthesis of DNA starts and goes continuously in the ‘leading strand’ and discontinuously in the ‘lagging strand’
Options:

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

Correct Answer: 3

Explanation: DNA replication starts with unwinding (helicase, gyrase) → stabilization (SSB proteins) → primer synthesis (primase) → DNA synthesis (leading & lagging strands) → joining of fragments (ligase). Thus correct sequence is C, B, A, E, D.

 

 

Question 8

Assertion A: In potato cultivation, the dehaulming is done to produce quality seed tubers.
Reason R: Dehaulming in potato seed crop is done 10-15 days before harvesting of the crop.
Options:

1. Both A and R are true.
2. Both A and R are false.
3. A is true but R is false.
4. A is false but R is true.

Correct Answer: 1

Explanation: Dehaulming (removal of foliage) in potato is done before harvest to improve skin hardening and seed quality and to reduce disease transmission. It is usually done 10–15 days before harvesting, so both statements are correct.

 

 

Question 9

The onion we consume is a:

1. Tuber
2. Corm
3. Bulb
4. Rhizome

Correct Answer: 3

Explanation: Onion is a bulb, which is a modified underground stem with fleshy scale leaves used for storage.

 

 

Question 10

For hybridization, in which of the following crops, the florets are clipped at the top to expose the anthers?

1. Barley
2. Jute
3. Sunnhemp
4. Sunflower

Correct Answer: 1

Explanation: In barley hybridization, florets are clipped at the top to expose anthers for emasculation and controlled pollination, ensuring successful crossing.

 

 

Question 11

Suppose, you wish to combine the mitochondria from species A with the chloroplast and nucleus of species B. Which of the following techniques is highly suited to achieve the objective?

1. Isolation and fusion of the protoplasts of both species A and B
2. Fusion of protoplast of species A with the cytoplast of species B
3. Fusion of protoplast of species B with the cytoplast of species A
4. Backcrossing program using species B as recurrent parent

Correct Answer: 3

Explanation: To combine mitochondria (cytoplasm) of species A with nucleus and chloroplast of species B, fusion of protoplast of B with cytoplast of A is required. Cytoplast provides cytoplasmic organelles without nucleus.

 

 

Question 12

An ‘artificial seed’ consists of gel enclosing one of the following components. Which one?

1. a zygotic embryo
2. an unfertilized ovule
3. a shoot bud or somatic embryo
4. a minute plantlet

Correct Answer: 3

Explanation: Artificial seeds are made by encapsulating somatic embryos or shoot buds in a gel matrix. These can develop into complete plants, mimicking true seeds.

 

 

Question 13

The test cross ratio for a trait governed by two genes with masking gene action will be:

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

Correct Answer: 2

Explanation: Masking gene action (epistasis) modifies Mendelian ratios. In a test cross, such interaction leads to 2:1:1 ratio, deviating from the typical 1:1:1:1 ratio.

 

 

Question 14

Which one of the following characteristics is NOT true for ‘genetic drift’?

1. It usually does not occur in a large random mating population
2. It is as like to genetic erosion
3. It leads to the loss of alleles
4. It is almost inevitable in a small population

Correct Answer: 2

Explanation: Genetic drift is a random change in allele frequency, more common in small populations, leading to loss of alleles. It is not exactly the same as genetic erosion, though both may result in loss of diversity.

 

 

Question 15

In seed technology, the term ‘dockage’ is commonly used to describe:

1. Genetic purity of the seeds
2. Genetic impurity of the seeds
3. Physical purity of the seeds
4. Physical impurity of the seeds

Correct Answer: 4

Explanation: Dockage refers to physical impurities in seed lots such as dirt, broken seeds, weed seeds, and other inert matter, which reduce seed quality but are not related to genetic purity.

 

 

Question 16

The process of mitosis in a cell consists of a series of events as given below. Arrange the events in order from first to the last as per the time of occurrence:
A. The highly condensed chromosomes move to the equatorial plate of the spindle
B. The centromere interacts with the spindle fiber apparatus
C. Division of the cytoplast
D. Sister chromatids move apart to the opposite poles
E. The nuclear membrane breaks down and the spindle-shaped structure of microtubules is organized
Options:

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

Correct Answer: 2

Explanation: Mitosis proceeds in a fixed sequence: first nuclear membrane breaks down and spindle forms (E), then chromosomes attach to spindle (B), followed by alignment at equatorial plate (A), then sister chromatids separate (D), and finally cytoplasm divides (C). This ensures equal distribution of genetic material.

 

 

Question 17

Which of the following constitutes ‘genetic emasculation’?
A. Genetic male sterility
B. Transgenic male sterility
C. Cytoplasmic-genetic male sterility
D. Chemical hybridizing agent
Options:

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

Correct Answer: 2

Explanation: Genetic emasculation means preventing self-pollination using genetic systems like male sterility (GMS, CGMS, transgenic). These eliminate the need for manual emasculation. Chemical agents are not genetic methods, so they are excluded

 

 

Question 18

In maize, the diploid chromosome number (2n) is 20. The number of chromosomes in endosperm cell (EC), pollen mother cell (PMC), pollen tube nucleus (PTN), and root tip cells (RTC) will be:
A. EC-30, PMC-20, PTN-20, RTC-40
B. EC-30, PMC-20, PTN-10, RTC-10
C. EC-20, PMC-20, PTN-20, RTC-10
D. EC-30, PMC-20, PTN-10, RTC-20
Options:

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

Correct Answer: 4

• Explanation:
• In maize (2n=20):
• Endosperm = 3n = 30
• PMC = 2n = 20
• Pollen tube nucleus = n = 10
• Root tip = 2n = 20
  This reflects fertilization (double fertilization) and haploid/diploid stages.

 

 

Question 19

The formation of a loop during the pachytene stage of meiosis indicates occurrence of:
A. Translocation
B. Deficiency
C. Inversion
D. Duplication
Options:

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

Correct Answer: 2

Explanation: Loop formation during pachytene occurs due to structural chromosomal changes like inversion, duplication, or deficiency. These cause improper pairing of homologous chromosomes.

 

 

Question 20

The use of synthetic varieties is more beneficial than single-cross or double-cross hybrids. The reasons are:
A. Yield of synthetic varieties is higher than hybrids
B. Farmers can save their own produce as seeds and no need to buy new seeds every year
C. Synthetic varieties have wider adaptability than hybrids
D. No need to make fresh crosses every year to produce seeds
Options:

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

Correct Answer: 2

Explanation: Synthetic varieties allow seed saving, wider adaptability, and no need for repeated hybrid seed production. However, their yield is generally lower than hybrids, so statement A is incorrect.

 

 

Question 21

The C4 plants usually have the following features:
A. The first stable product in the system is a 4-carbon compound
B. There are two CO2 acceptors, viz., PEP and RUBP
C. Oxygen does not have any inhibitory effect on the process.
D. RUBP carboxylase is present in the mesophyll.
Options:

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

Correct Answer: 2

Explanation: C4 plants fix CO₂ first into a 4-carbon compound (OAA) using PEP, then CO₂ enters Calvin cycle via RuBP. They avoid photorespiration, so oxygen has minimal effect. Rubisco is mainly in bundle sheath cells, not mesophyll.

 

 

Question 22

Match List I with List II (Micropropagation propagules and Crops):

List I (Propagules)	List II (Crops)
A. Protocorms	I. Oil palm
B. Bulblets	II. Orchids
C. Somatic embryos	III. Begonia
D. Adventitious shoot buds	IV. Lily

Options:

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

Correct Answer: 4

• Explanation:
• Protocorms → orchids
• Bulblets → lily
• Somatic embryos → oil palm
•  Adventitious buds → begonia
• These are used in micropropagation for rapid multiplication

 

 

Question 23

Match List I with List II (Disease and Causal Organism / Pathogen Type):

List I (Disease)	List II (Causal Organism / Type)
(A) Late blight of potato	(I) Xanthomonas oryzae pv. oryzae
(B) Bacterial blight of rice	(II) Phytophthora infestans
(C) White rust of crucifers	(III) Albugo candida
(D) Loose smut of wheat	(IV) Ustilago nuda tritici

Options:

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

Correct Answer: 2

 

 

Question 24

Match List I with List II (Compound and Target/Related insect-pests):

List I (Compound)	List II (Insect-pests)
A. Benzyl alcohol	I. Brown plant hopper
B. Silica content	II. Green bugs
C. Asparagine content	III. Cabbage aphid
D. Sinigrin content	IV. Rice stem borer

Options:

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

Correct Answer: 3

Explanation:

• Plant compounds affect insect resistance:
• Benzyl alcohol → affects green bugs
• Silica → strengthens plant, resists stem borers
• Asparagine → linked with insect feeding
• Sinigrin → attracts/affects cabbage aphid

 

 

Question 25

Match List I with List II (Terminology and Description):

List I (Term)	List II (Description)
A. Hybrid	I. Variety produced by mixing seeds of several phenotypically outstanding lines not tested for GCA
B. Pure line variety	II. Variety developed by intercrossing genotypes of known superior combining ability
C. Composite variety	III. First generation cross product grown commercially
D. Synthetic variety	IV. Progeny of a single, self-fertilized, homozygous plant

Options:

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

Correct Answer: 2

• Explanation:
•  Hybrid → F1 cross with high vigor
• Pure line → homozygous, uniform
• Composite → mixed seeds without GCA testing
•  Synthetic → intercrossed lines with known GCA Each type differs in genetic structure and seed production.

 

 

Question 26

Match List I with List II (Phytochemical and Plant species):

List I (Phytochemical)	List II (Plant species)
A. BOAA	I. Brassica species
B. Erucic acid	II. Flax or Linseed
C. Gossypol	III. Lathyrus
D. Linolenic acid	IV. Cotton

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

Correct Answer: 4

• Explanation:
• BOAA → Lathyrus (toxin causing lathyrism)
• Erucic acid → Brassica (oil quality issue)
• Gossypol → Cotton (toxic pigment)
• Linolenic acid → Linseed (oil component) These are important phytochemicals in crops.

 

 

Question 27

Match List I with List II (Terminology and Genetical formula):

List I (Term)	List II (Genetical formula)
A. Double monosomy	1. 2N+2+2
B. Trisomy	2. 2N-2
C. Nullisomy	3. 2N-1-1
D. Double tetrasomy	4. 2N+1

Options:

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

Correct Answer: 1

• Explanation:
• Chromosomal variations:
• Double monosomy → 2n-1-1
• Trisomy → 2n+1
• Nullisomy → 2n-2
• Double tetrasomy → 2n+2+2
  These represent aneuploid conditions affecting gene balance.

 

 

Question 28

Match List I with List II (Parameter and Activity):

List I (Parameter)	List II (Activity)
A. Germplasm conservation	I. Virus infection, mutation
B. Clonal degeneration	II. Cryoprotectant
C. DMSO	III. Cryotherapy
D. Virus elimination	IV. In-situ, ex-situ, cryopreservation

Options:

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

Correct Answer: 3

• Explanation:
• Germplasm conservation → in situ, ex situ, cryopreservation
• Clonal degeneration → due to virus/mutation
• DMSO → cryoprotectant
• Virus elimination → cryotherapy These are key tools in plant biotechnology.

 

 

Question 29

The mating schemes that increase homozygosity in the plants are:
A. Genetic assortative mating
B. Genetic disassortative mating
C. Phenotypic assortative mating
D. Phenotypic disassortative mating
Options:

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

Correct Answer: 2

Explanation: Assortative mating (genetic or phenotypic) pairs similar individuals, increasing homozygosity. Disassortative mating increases heterozygosity, so only assortative types are correct.

 

 

Question 30

In maize, the opaque 2 mutants are found to have some of the following quality parameters. Identify:

1. High methionine and high lysine
2. High methionine and high tryptophan
3. High lysine and high tryptophan
4. High lysine and high proline

Correct Answer: 3

Explanation: The opaque-2 (o₂) mutant in maize improves protein quality by increasing lysine and tryptophan content, which are normally deficient in maize. This mutation reduces zein protein synthesis, thereby improving nutritional quality (Quality Protein Maize – QPM).

 

 

Question 31

Sclerotia of Claviceps purpurea responsible for the infamous St. Anthony’s Fire is believed to contain:

1. Ethyl Alcohol
2. Methyl Alcohol
3. Lysergic acid diethylamide
4. Triethylamine

Correct Answer: 3

Explanation: Sclerotia of Claviceps purpurea contain alkaloids derived from lysergic acid, including LSD-like compounds. These cause ergotism (St. Anthony’s Fire), leading to hallucinations, convulsions, and gangrene due to vasoconstriction.

 

 

Question 32

In which of the following disease you will find both sign and symptoms together?

1. Southern blight infected by Athelia rolfsii
2. Wilt infected by Fusarium
3. Gall infected by Agrobacterium
4. Blight infected by Xanthomonas

Correct Answer: 1

Explanation:

• “Signs” = visible pathogen structures; “Symptoms” = host response. In southern blight (Athelia rolfsii), both are visible:
• Signs → fungal mycelium and sclerotia
• Symptoms → plant wilting/rotting
  Thus both appear together.

 

 

Question 33

Which of the following pathogen produces sporodochium?

1. Graphium
2. Epicoccum
3. Phoma
4. Fusarium

Correct Answer: 4 (Note: Key shows 6370/6372, indicating both 2 and 4? But Fusarium is correct for sporodochium)

Explanation: Sporodochium is a cushion-like asexual fruiting body. It is typically produced by Fusarium, where conidia are formed in clusters on these structures.

 

 

Question 34

Sequence of pathogenesis events:
A. Invasion
B. Dissemination
C. Reproduction
D. Infection
E. Inoculation
F. Penetration
Options:

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

Correct Answer: 1

• Explanation:
• Correct sequence of pathogenesis:
• Inoculation → Penetration → Infection → Invasion → Reproduction → Dissemination
• This reflects the complete disease cycle from pathogen arrival to spread.

 

 

Question 35

1. Porins allow bacteria to move nutrients and waste across the cell membrane
   B. Type III Secretion System (TTSS) allows bacteria to move nucleoprotein complex across the cell membrane
   C. EPS is a virulence factor in wilt-causing bacteria
   D. LPS allows bacteria to bind to host lectins
   E. Hydrolytic enzymes are not virulence factors for bacteria
   Options:
2. B and E only
3. B only
4. A, C and D only
5. D only

Correct Answer: 3

• Explanation:
• Porins help transport molecules across bacterial membrane
•  EPS (extracellular polysaccharide) aids in blocking xylem in wilt diseases
• LPS helps in host recognition and interaction
• Hydrolytic enzymes ARE virulence factors (so statement E is false), Thus A, C, D are correct.

 

 

Question 36

• Assertion (A): Systemic Acquired Resistance (SAR) is a broad-spectrum, nonspecific defense response in plants that is effective against a wide range of pathogens.
• Reason (R): SAR is induced only by pathogenic infection and is mediated by salicylic acid, but it is not effective against biotrophic pathogens.

In light of the above statements, choose the correct answer from the options given below:

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: 3

• Explanation:
• Assertion (A) is true: SAR is indeed a broad-spectrum, nonspecific defense response effective against fungi, bacteria, and viruses.
• Reason (R) is false: SAR is actually effective against biotrophic pathogens, and it is mediated by salicylic acid. It is not ineffective against biotrophs. Also, SAR can be induced by both pathogenic infection and certain chemical treatments (e.g., salicylic acid or its synthetic analogs).

Therefore, (A) is true but (R) is false.

 

 

Question 37

1. Necrosis manifest in the form of blotch
   B. Abnormal growth manifest in the form of warts
   C. Necrosis manifest in the form of leaf curl
   D. Abnormal growth manifest in the form of anthracnose
   E. Witches broom is caused only by phytoplasma
   Options:
2. A and B only
3. C and D only
4. D and E only
5. C and E only

Correct Answer: 1

• Necrosis → appears as blotches (dead tissue)
• Abnormal growth → appears as warts (tumor-like growth) Leaf curl is not necrosis, and anthracnose is necrotic, not abnormal growth.
  Thus only A and B are correct.

 

 

Question 38

Match List I with List II (Character and Pathogen):

List I (Character)	List II (Pathogen)
A. Spores/Cells with two anterior, unequal, whiplash flagella	I. Magnaporthe
B. Spores/Cells with one posterior whiplash flagellum	II. Phytophthora
C. Spores/Cells with two flagella; one anterior and one posterior	III. Synchytrium
D. Spores/Cells without any flagella	IV. Plasmodiophora

Options:

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

Correct Answer: 4

• Explanation:
  Flagella types help identify pathogens:
• Posterior whiplash → Synchytrium
• Two flagella (anterior + posterior) → Plasmodiophora
• No flagella → Magnaporthe
• Anterior flagella → oomycetes like Phytophthora
  These are key diagnostic features.

 

 

Question 39

Match List I with List II (Powdery mildew Tribe and Genera):
Options:

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

Correct Answer: 1

Explanation: Powdery mildew fungi are classified into tribes and genera based on morphology and host specificity, which helps in identification and disease management.

 

 

Question 40

Match List I with List II (Society and Journal):
Options:

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

Correct Answer: 1

Explanation: Scientific societies are linked with specific journals for publishing research. Matching ensures understanding of academic organizations and their publications in plant pathology.