ICAR JRF Plant Science Practice Series Memory Based 2024 (Module 1) (81 – 120 MCQ) 

 

Question No. 81 
Assertion (A): According to the ‘overdominance hypothesis’, the ‘heterosis’ is the result of the superiority of the heterozygote over its two homozygous parents.
Reason (R): The ‘overdominance hypothesis’ states that the superiority of the heterozygotes arises due to the masking of the deleterious effect of the recessive alleles by their corresponding dominant alleles.

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: 3
Explanation: (A) is correct (overdominance = heterozygote superiority). (R) describes the dominance hypothesis, not overdominance hypothesis. So (R) is incorrect for overdominance.

 

 

Question No. 82 
The family of the genus Xanthomonas was formerly known as:

1. Enterobacteriaceae
2. Rhizobacteriaceae
3. Pseudomonadaceae
4. Proteobacteriaceae

Correct Answer: 3
Explanation: Xanthomonas was formerly placed in Pseudomonadaceae before being reclassified into Xanthomonadaceae.

 

 

Question No. 83
Statement (I): Blue-green algae are a type of bacteria that are capable of photosynthesis.
Statement (II): Blue-green algae possess phycocyanin pigment that is responsible for the blue color.

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. Blue-green algae = cyanobacteria, photosynthetic, contain phycocyanin (blue pigment).

 

 

Question No. 84 
Which of the following statements about the scope of the ITPGRFA are correct?
(A) It covers crop species that are listed under Annex I.
(B) It includes all the world’s plant species across every biome.
(C) It emphasizes food security as a key motivation for its coverage.
(D) It aims to cover animals that support agriculture.

Options:

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

Correct Answer: 4
Explanation: ITPGRFA covers only Annex I crops (not all species), focuses on food security, and does not cover animals. So (A) and (C) are correct.

 

 

Question No. 85 
In endomycorrhizal associations, which of the following structures are formed intracellularly in the host plant:

1. Vesicles
2. Arbuscule
3. Spores
4. Sporangia

Correct Answer: 2
Explanation: Arbuscules are intracellular structures in endomycorrhizae (vesicles may be intercellular or intracellular depending on type). Spores and sporangia are external.

 

 

Question No. 86 

Match List-1 with List-2 (Crop and Planting material):
(A) Sugarcane → (II) Seed setts
(B) Banana → (IV) Suckers
(C) Sweet potato → (I) Vine cuttings
(D) Strawberry → (III) Runner plant

Options:

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

Correct Answer: 3

 

 

Question No. 87 
Which of the following rust is not caused by fungi:

1. Red rust of tea
2. Black rust of wheat
3. White rust of mustard
4. Yellow rust of wheat

Correct Answer: 1
Explanation: Red rust of tea is caused by an alga (Cephaleuros parasiticus), not a fungus. Black rust (Puccinia), white rust (Albugo – oomycete, fungus-like), yellow rust (Puccinia).

 

 

Question No. 88 

Statement (I): Mycoplasmas are not inhibited even by high doses of penicillin.
Statement (II): Mycoplasmas can be inhibited by antibiotics that affect protein synthesis.

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: Mycoplasmas lack cell wall, so penicillin (inhibits cell wall synthesis) is ineffective. They are inhibited by protein synthesis inhibitors (tetracycline, etc.).

 

 

Question No. 89 
Which of the following is NOT an assumption of the Hardy-Weinberg’s law of equilibrium?

1. Mutations
2. Random mating
3. Small population size
4. Selection

Correct Answer: 3
Explanation: Hardy-Weinberg assumes large population size (no genetic drift), no mutation, no selection, random mating. Small population size violates the assumption.

 

 

Question No. 90
Assertion (A): The Indian Council of Agricultural Research (ICAR) celebrates Agricultural Education Day every year on December 03.
Reason (R): It is celebrated to commemorate the birth anniversary of Dr. Rajendra Prasad, the first President of independent India and the Union Minister of Agriculture.

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: 3
Explanation: Agricultural Education Day is Dec 3, but it commemorates Dr. Rajendra Prasad’s birth anniversary? Actually Dec 3 is birth anniversary of Dr. Rajendra Prasad, but he was not Union Minister of Agriculture. So (R) is partially false.

 

 

Question No. 91:
Arrange the following stages of mitosis in the correct chronological order:

(A) Metaphase
(B) Anaphase
(C) Telophase
(D) Prophase

Options:

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

Correct Answer: 1. 

 

 

Question No. 92 
The cell wall of Oomycetes is composed of:

1. Cellulose only
2. Chitin only
3. Chitosan only
4. Cellulose and glucans

Correct Answer: 4
Explanation: Oomycetes have cell walls composed of cellulose and β-glucans, not chitin (true fungi have chitin).

 

 

Question No. 93 
Assertion (A): Pigeonpea (Cajanus cajan) is considered as an ‘often cross-pollinated’ crop.
Reason (R): In Pigeonpea, although self-pollination can occur within the bud before the flower opens, cross-pollination (20-70%), often facilitated by insects like bees, is a common and significant factor in seed set.

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 why pigeonpea is often cross-pollinated.

 

 

Question No. 94

Blight disease in rice is caused by:

1. Xanthomonas oryzae pv. oryzae
2. Xylella fastidiosa
3. Pseudomonas syringae
4. Pseudomonas putida

Correct Answer: 1
Explanation: Bacterial blight of rice is caused by Xanthomonas oryzae pv. oryzae.

 

 

Question No. 95 
The detection and identification of a physiological race is done only by:

1. Selective culture medium
2. ELISA
3. Differential host range
4. Nested multiplex PCR

Correct Answer: 3
Explanation: Physiological races are distinguished by their differential interaction with host cultivars (differential host range), not by media or molecular tests alone.

 

 

Question No. 96 

Find out the events/factors that are exceptions to Mendelian principles:
(A) Complete dominance (B) Incomplete dominance (C) Co-dominance (D) Polygenic traits

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: 4
Explanation: Complete dominance follows Mendelian principles. Incomplete dominance, co-dominance, and polygenic traits are exceptions/modifications.

 

 

Question No. 97 
Statement (I): Actinobacteria are a large group of bacteria that produce asexual spores like fungi.
Statement (II): Some actinobacterial genera can be pathogenic to plants and animals.

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 true. Actinobacteria produce spores (e.g., Streptomyces), and some are pathogens (e.g., Mycobacterium, Streptomyces scabies).

 

 

Question No. 98

Arrange the general steps for seed processing in a typical sequence from start to end:
(A) Testing the seeds for quality parameters
(B) Harvesting, threshing and drying the seeds
(C) Packaging and labelling
(D) Thorough cleaning of the seeds

Options:

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

Correct Answer: 3
Explanation: Harvest/dry → clean → test → package/label.

 

Question No. 99 

Select the correct statements about heterosis:
(A) Heterosis typically manifests as superior performance of F1 hybrids.
(B) Dominance hypothesis explains heterosis by masking of deleterious recessive alleles.
(C) Overdominance hypothesis states heterozygote advantage.
(D) Narrow-sense heritability is consistently higher in F1 hybrids exhibiting strong heterosis.

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: 1
Explanation: (A), (B), (C) are correct. (D) is false: heterosis is not consistently associated with higher narrow-sense heritability.

 

 

Question No. 100 

Which of the following polymers are associated with plant cell wall:
(A) Chitin (B) Cellulose (C) Peptidoglycan (D) Hemicellulose (E) Pectin

Options:

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

Correct Answer: 3
Explanation: Plant cell walls contain cellulose, hemicellulose, and pectin. Chitin is in fungi, peptidoglycan in bacteria.

 

 

Question No. 101
The plant parasitic bacterium was first reported by:

1. Mikhail Woronin
2. T.J. Burrill
3. E.F. Smith
4. R.A. Fischer

Correct Answer: 2
Explanation: T.J. Burrill first reported a plant pathogenic bacterium (fire blight of pear caused by Erwinia amylovora) in 1878-1880.

 

 

Question No. 102 

The correct sequence of Koch’s postulates:
(A) Inoculation (B) Observation (C) Isolation (D) Pure culture (E) Re-isolation (F) Comparison

Options:

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

Correct Answer: 2
Explanation: Observe disease → Isolate pathogen → Obtain pure culture → Inoculate healthy host → Re-isolate → Compare.

 

 

Question No. 103 
In the context of ex situ conservation of plant genetic resources, what is the primary benefit of integrating molecular marker technologies during germplasm characterization and evaluation?

1. They entirely eliminate the need for field-based phenotypic assessments.
2. They enable accurate and efficient evaluation of genetic diversity and help identify genetically redundant accessions.
3. They ensure that conserved germplasm is completely free from all harmful genetic mutations.
4. They significantly reduce the cost and labour required for maintaining live accessions in field gene banks.

Correct Answer: 2
Explanation: Molecular markers assess genetic diversity and identify duplicates (redundancy), but do not eliminate phenotypic assessment or ensure mutation-free status.

 

 

Question No. 104 

Statement (I): AFLP combines restriction enzyme digestion with selective PCR amplification of genomic fragments.
Statement (II): AFLP typically functions as a co-dominant marker system, providing easy heterozygote detection.

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) true. (II) false: AFLP is a dominant marker (presence/absence), not co-dominant.

 

 

Question No. 105 

Arrange the steps of a tetrazolium test in appropriate order from first to last:
(A) Immersing seeds in Tetrazolium salt solution and incubating
(B) Rinsing seeds in fresh water
(C) Soaking seeds to hydrate tissues
(D) Examining seeds for intensity and location of staining

Options:

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

Correct Answer: 4
Explanation: Soak (hydrate) → Immerse in TTC solution → Rinse → Examine.

 

 

Question No. 106 

Choose the combination of correct statements about crossing over and recombination:
(A) Crossing over takes place in pachytene stage of meiotic prophase I, where unequal exchange occurs.
(B) Recombination frequency depends on the distance between a pair of genes.
(C) Recombination frequency in general does not go beyond 50%.
(D) Genes on different chromosomes exhibit independent assortment.

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: 4
Explanation: (A) is false because crossing over in pachytene is equal exchange, not unequal. (B), (C), (D) are correct.

 

 

Question No. 107

Arrange the schemes in ascending order of their launching/introduction:
(A) PM Fasal Bima Yojana
(B) National mission on edible oils – Oil palm
(C) Soil Health Card
(D) PM Kisan Sanman Nidhi

Options:

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

Correct Answer: 1? Approx order: Soil Health Card (2015), PM Fasal Bima (2016), PM Kisan (2019), Oil palm mission (2021). So (C), (A), (D), (B).

 

 

Question No. 108 
Assertion (A): Gene deployment strategy in wheat is the most effective strategy in restricting the spread of rust disease.
Reason (R): Monoculture of one popular variety but susceptible to a disease may lead to quick epidemic situation. Therefore, different resistant varieties are grown in a mosaic pattern covering large area.

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) explains why gene deployment (mosaic of resistant varieties) works.

 

 

Question No. 109 
Two genes, G and H, are located on the same chromosome and have a recombination frequency of 20%. An individual is GgHh in cis-configuration (GH/gh). What fraction of that individual’s gametes is expected to be parental versus recombinant?

1. 50% parental, 50% recombinant
2. 60% parental, 40% recombinant
3. 80% parental, 20% recombinant
4. 90% parental, 10% recombinant

Correct Answer: 3
Explanation: Recombination frequency = 20% recombinant gametes, so 80% parental.

 

 

Question No. 110 

Choose the correct statements about the ‘Dominance hypothesis’ of ‘Heterosis’:
(A) Proposed by Davenport in 1908.
(B) Heterosis results from heterozygosity.
(C) Heterosis is directly proportional to the number of dominant genes each parent contributes.
(D) In heterozygous state, deleterious effects of recessive alleles are masked by dominant alleles.

Options:

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

Correct Answer: 4? Actually dominance hypothesis (Davenport 1908) states: heterosis = masking of deleterious recessives (D), heterozygosity (B), more dominant genes (C). (A) is correct but often (B),(C),(D) are key. Check standard: (A) is true but not always included. Answer likely 4: (B),(C),(D).

 

 

Question No. 111

In self-pollinated crops, what is the most important step to make sure the seeds produced stay genetically pure and true to type?

1. Letting different varieties of the same crop grow side-by-side.
2. Keeping the seed field isolated and using seeds from only one uniform plant type (genotype).
3. Mixing seeds from different varieties to increase diversity.
4. Using machines to harvest the crop and cleaning the seeds after harvesting.

Correct Answer: 2
Explanation: In self-pollinated crops, genetic purity is maintained by preventing cross-pollination (isolation) and starting with uniform, true-to-type seed (genotype). Side-by-side planting increases crossing; mixing reduces purity; mechanical cleaning does not ensure genetic purity.

 

 

Question No. 112
Assertion (A): Nitrogen deficiency significantly reduces a plant’s capacity for photosynthesis.
Reason (R): Nitrogen is an essential constituent of chlorophyll, and hence its deficiency affects photosynthesis.

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. Nitrogen is a key component of chlorophyll molecules; deficiency reduces chlorophyll synthesis, directly reducing photosynthetic capacity.

 

 

Question No. 113 
In a Chi-square (χ²) test with 2 phenotypic classes for plant height, what would be the degree of freedom (df)?

1. 0
2. 1
3. 2
4. 4

Correct Answer: 2? Wait: Degrees of freedom = number of classes − 1. For 2 classes, df = 2 − 1 = 1. So correct answer is 1.
Explanation: The chosen option (2) is incorrect. Standard formula: df = (number of phenotypic classes) − 1. For 2 classes, df = 1.

 

 

Question No. 114 

In loose smut of wheat, spore balls are formed due to conversion of: (A) Flag leaf (B) Floral parts (C) All leaves (D) Apical stems

Options:

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

Correct Answer: 4
Explanation: Loose smut of wheat (Ustilago nuda tritici) converts floral parts (spikelets, grains) into black spore masses; leaves and stems are not converted into spore balls.

 

 

Question No. 115 

Match List-1 with List-2 (Specialty and Crop Name):

(A) Golden fibre → (I) Jute
(B) White gold → (III) Cotton
(C) King of cereals → (IV) Wheat
(D) Queen of cereals → (II) Maize
(E) Camel crop → (V) Sorghum

Options:

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

Correct Answer: 3

 

 

Question No. 116 
Citrus tristeza virus is detected by using the indicator plant:

1. Sweet orange
2. Rangpur Lime
3. Trifoliate orange
4. Kagzi Lime

Correct Answer: 2 (Rangpur Lime) or 4? Standard: Mexican lime (Kagzi lime) is the most sensitive indicator for CTV. So Kagzi Lime is correct.
Explanation: Kagzi lime (Citrus aurantifolia) shows characteristic vein-clearing and stem pitting symptoms upon CTV infection and is widely used as an indicator plant.

 

 

Question No. 117 
Options:

1. Puducherry (Pondicherry) under the TNAU, Chennai
2. Puducherry under the TNAU, Coimbatore
3. Kudumiyanmalai under the TNAU, Coimbatore
4. Puducherry under the TNAU, Madurai

Correct Answer: 2 (Likely refers to a research station location; Puducherry falls under TNAU, Coimbatore jurisdiction)

 

Question No. 118
As per the seed certification standard for sugarcane, the age of the seed cane crop at harvest for seed purposes in the tropics should be:

1. 4-6 months
2. 6-8 months
3. 8-10 months
4. 10-12 months

Correct Answer: 3 (8-10 months)
Explanation: For seed cane in tropical conditions, the ideal age is 8-10 months to ensure optimum germination and vigour. 10-12 months is too old, leading to reduced viability.

 

 

Question No. 119 

Match List-1 with List-2 (Seed tests and Scientist concerned):
(A) Tetrazolium Test → (IV) George Lakon (1942)
(B) Indoxyl Acetate Test → (I) Mills and Gherna (1987)
(C) Electrical Conductivity Test for seed vigor in Pea → (II) Fick and Hibbard (1925)
(D) Excised Embryo Test → (III) Florence Flemion (1934)

Options:

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

Correct Answer: 3

 

Question No. 120 
A plant with both staminate (male) and pistillate (female) flowers on separate branches of the same plant can best be described as:

1. Cleistogamous
2. Dioecious
3. Monoecious
4. Apomictic

Correct Answer: 3
Explanation: Monoecious = male and female flowers on the same plant (e.g., maize, cucumber). Dioecious = male and female on separate plants. Cleistogamous = self-pollination in unopened flowers. Apomictic = asexual seed production.