ICAR JRF Plant Science Practice Series Memory Based 2024 (Module 1) (41 – 80 MCQ) 

 

 

Question 41

Presence of male and bisexual flowers in different crops:

1. Andromonoecious
2. Gynomonoecious
3. Monoecious
4. Dioecious

Correct Answer: 2

Explanation:

• Gynomonoecious– Presence of female (pistillate) and bisexual (hermaphrodite) flowers on the same plant.
• Andromonoecious– Male and bisexual flowers on the same plant.
• Monoecious– Male and female flowers separately on the same plant.
• Dioecious– Male and female flowers on different plants.
• The question asks for “male and bisexual” – that is andromonoecious. However, the answer key says 2 (Gynomonoecious). There may be a mismatch. Given the key, the answer is Gynomonoecious→ Option 2.

 

 

Question 42

Most common herbicide used in wheat:

1. Isoproturon
2. Atrazine
3. Simazine
4. Basalin

Correct Answer: 1

Explanation:

• Isoproturon– A selective, systemic herbicide widely used for post-emergence control of grassy and broadleaf weeds in wheat.
• Atrazine, Simazine– Used in maize, sugarcane.
• Basalin (Fluchloralin)– Used in cotton, pulses, oilseeds.
• Therefore, the correct answer is Isoproturon→ Option 1.

 

 

Question 43

First irrigation in wheat should be given at:

1. Tillering stage
2. Flowering stage
3. Panicle initiation stage
4. CRI stage

Correct Answer: 4

Explanation:

• CRI (Crown Root Initiation) stage– Occurs about 20-25 days after sowing. It is the most critical stage for the first irrigation in wheat.
• Tillering stage– Second irrigation.
• Flowering, panicle initiation– Later stages.
• Therefore, the correct answer is CRI stage→ Option 4.

 

 

Question 44

In groundnut intercultural operation should be avoided at:

1. Post flowering
2. Pre flowering
3. Pegging
4. Pod formation

Correct Answer: 3

Explanation:

• Pegging stage– After fertilization, the gynophore (peg) penetrates the soil for pod development. Intercultural operations (hoeing, weeding) at this stage can damage the pegs and reduce yield.
• Pre flowering, post flowering, pod formation– Less sensitive.
• Therefore, the correct answer is Pegging→ Option 3.

 

 

Question 45

Symbiotic N₂ fixing organism comes under the genus:

1. Rhizobium
2. Penicillium
3. Bacillus
4. Pseudomonas

Correct Answer: 1

Explanation:

• Rhizobium– A genus of Gram-negative bacteria that form symbiotic nitrogen-fixing nodules on legume roots.
• Penicillium– Fungus (not nitrogen-fixing).
• Bacillus– Some species fix nitrogen but are free-living (non-symbiotic).
• Pseudomonas– Some are nitrogen-fixing but not symbiotic.
• Therefore, the correct answer is Rhizobium→ Option 1.

 

 

Question 46

Vinegar production involves:

1. Aspergillus niger
2. Acetobacter aceti
3. gossypii
4. Propionibacterium

Correct Answer: 2

Explanation:

• Vinegar(acetic acid) is produced by the fermentation of ethanol by acetic acid bacteria (AAB), primarily Acetobacter aceti.
• Aspergillus niger– Used for citric acid production.
• Propionibacterium– Used for propionic acid and vitamin B₁₂ production.
• Therefore, the correct answer is Acetobacter aceti→ Option 2.

 

 

Question 47

In cotton red pigmentation is due to the glands secreting:

1. ABA
2. IAA
3. Gossypol
4. None of the above

Correct Answer: 3

Explanation:

• Gossypol– A polyphenolic compound produced in pigment glands of cotton (leaves, stems, seeds). It gives a reddish pigmentation and acts as a defense compound.
• ABA, IAA– Plant hormones, not responsible for red pigmentation.
• Therefore, the correct answer is Gossypol→ Option 3.

 

 

Question 48

Test cross ratio in supplementary epistasis:

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

Correct Answer: 2

Explanation:

• Supplementary epistasis(also called duplicate gene interaction with cumulative effect) – When dominant alleles at either of two loci produce the same trait, and the double recessive produces a different trait.
• F₂ ratio is 9:3:4(or 9:7 depending on type).
• Test cross ratio for supplementary epistasis is 1:1:2.
• Therefore, the correct answer is 1:1:2→ Option 2.

 

 

Question 49

Early blight in potato:

1. Phytophthora infestans
2. Alternaria solani
3. Rhizopus nigricans
4. Colletotrichum

Correct Answer: 2

Explanation:

• Early blight of potato– Caused by the fungus Alternaria solani (Deuteromycete). It causes dark, concentric ring-like lesions on leaves.
• Phytophthora infestans– Causes late blight.
• Rhizopus nigricans– Causes soft rot.
• Colletotrichum– Causes anthracnose.
• Therefore, the correct answer is Alternaria solani→ Option 2.

 

 

Question 50

Germ theory was given by:

1. Robert Koch
2. Louis Pasteur
3. Edward Jenner
4. Joseph Lister

Correct Answer: 2

Explanation:

• Germ theory of disease– States that many diseases are caused by microorganisms. It was developed and popularized by Louis Pasteur (although others like Koch contributed).
• Robert Koch– Formulated Koch’s postulates.
• Edward Jenner– Developed smallpox vaccine.
• Joseph Lister– Introduced antiseptic surgery.
• Therefore, the correct answer is Louis Pasteur→ Option 2.

 

 

Question 51

Pigeon pea sterility mosaic virus was transmitted by:

1. Aphis gossypii
2. Aceria cajani
3. Thrips tabaci
4. All the above

Correct Answer: 2

Explanation:

• Pigeon pea sterility mosaic virus (PPSMV)is transmitted by the eriophyid mite, Aceria cajani (persistent transmission).
• Aphis gossypii– Cotton aphid.
• Thrips tabaci– Onion thrips.
• Therefore, the correct answer is Aceria cajani→ Option 2.

 

 

Question 52

Albinism in plant is controlled by:

1. Recessive lethal
2. Dominant lethal
3. Semi lethal
4. None of these

Correct Answer: 1

Explanation:

• Albinism(lack of chlorophyll) in plants is often caused by recessive lethal alleles (homozygous recessive condition leads to death due to inability to photosynthesize).
• Dominant lethal– Kills even in heterozygous condition (rare in plants).
• Semi lethal– Not all individuals die.
• Therefore, the correct answer is Recessive lethal→ Option 1.

 

 

Question 53

First division of meiosis:

1. Equational
2. Reductional
3. Recantation
4. None of these

Correct Answer: 2

Explanation:

• Meiosis I– The reductional division where homologous chromosomes separate, reducing the chromosome number from diploid (2n) to haploid (n).
• Meiosis II– The equational division where sister chromatids separate.
• Recantation– Not a term in meiosis.
• Therefore, the correct answer is Reductional→ Option 2.

 

 

Question 54

When two homozygous parents are crossed, no genetic variation is found in:

1. F₁
2. F₃
3. F₂
4. F₅

Correct Answer: 1

Explanation:

• When two homozygous parents(AA × aa) are crossed, the F₁ generation is uniform (all Aa) , showing no genetic variation (all progeny are identical).
• Genetic variation appears in F₂and subsequent generations due to segregation and recombination.
• Therefore, the correct answer is F₁→ Option 1.

 

 

Question 55

Initiation codon is:

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

Correct Answer: 4

Explanation:

• AUG– The start codon that initiates protein synthesis (codes for methionine in eukaryotes and formylmethionine in prokaryotes).
• UGA, UAG, UAA– Stop codons (opal, amber, ochre respectively).
• Therefore, the correct answer is AUG→ Option 4.

 

 

Question 56

Out of 24 species of rice, two species are cultivated, one is O. sativa and the other is:

1. nivara
2. perennis
3. glaberrima
4. longistaminata

Correct Answer: 3

Explanation:

• The two cultivated rice species are:
  • Oryza sativa(Asian rice) – grown worldwide.
  • Oryza glaberrima(African rice) – grown in West Africa.
• nivara, O. perennis, O. longistaminata– Wild species.
• Therefore, the correct answer is  glaberrima→ Option 3.

 

 

Question 57

Number of primary trisomics in wheat (T. aestivum):

1. 7
2. 14
3. 21
4. 42

Correct Answer: 3

Explanation:

• Primary trisomics– Aneuploids with an extra chromosome (2n+1). The number of possible primary trisomics equals the haploid chromosome number (n) .
• Wheat (Triticum aestivum)is hexaploid (2n = 42, n = 21) .
• Therefore, the number of primary trisomics is 21→ Option 3.

 

 

Question 58

Embryo is:

1. 2n
2. n
3. 3n
4. 4n

Correct Answer: 1

Explanation:

• The embryoin angiosperms develops from the zygote, which is formed by the fusion of male and female gametes (n + n = 2n).
• Endosperm– Triploid (3n).
• Therefore, the correct answer is 2n→ Option 1.

 

 

Question 59

Aleurone layer is:

1. n
2. 3n
3. 2n
4. 4n

Correct Answer: 2

Explanation:

• The aleurone layeris the outermost layer of the endosperm in cereal grains.
• It is triploid (3n)in angiosperms because it is part of the endosperm tissue.
• Therefore, the correct answer is 3n→ Option 2.

 

 

Question 60

How many mitotic divisions should occur in a cell to give rise to 256 cells?

1. 128
2. 61
3. 8
4. 32

Correct Answer: 3

Explanation:

• Starting with 1 cell, after nmitotic divisions, the number of cells = 2ⁿ.
• To get 256 cells: 2ⁿ = 256 → n = 8 (since 2⁸ = 256).
• Therefore, the correct answer is 8→ Option 3.

 

 

Question 61

Linkage group of rice is:

1. 12
2. 24
3. 11
4. 42

Correct Answer: 1

Explanation:

• Linkage group– The number of linkage groups equals the haploid chromosome number (n) .
• Rice (Oryza sativa)has 2n = 24, n = 12 .
• Therefore, the number of linkage groups in rice is 12→ Option 1.

 

 

Question 62

1 ha-cm of water is equal to:

1. 10,000 L
2. 1,00,000 L
3. 1,000 L
4. None of the above

Correct Answer: 2

Explanation:

• 1 hectare-cm (ha-cm)= 1 hectare area × 1 cm depth of water.
• 1 hectare = 10,000 m²
• 1 cm = 0.01 m
• Volume = 10,000 m² × 0.01 m = 100 m³
• 1 m³ = 1000 litres
• Therefore, 100 m³ = 1,00,000 litres→ Option 2.

 

 

Question 63

Formation of embryo from synergid/egg cells is called:

1. Apogamy
2. Apospory
3. Adventive embryo
4. Apomixis

Correct Answer: 1

Explanation:

• Apogamy– Development of an embryo from a cell of the embryo sac other than the egg (e.g., synergid, antipodal) without fertilization.
• Apospory– Development of a gametophyte from sporophytic cells without meiosis.
• Adventive embryony (nucellar embryony)– Embryos develop from nucellus or integuments.
• Apomixis– General term for asexual reproduction through seeds.
• Therefore, the correct answer is Apogamy→ Option 1.

 

 

Question 64

Cell sap is found in:

1. Mitochondria
2. Chloroplast
3. Cytosol
4. All the above

Correct Answer: 3

Explanation:

• Cell sap– The fluid found inside the central vacuole of plant cells (not in cytosol). However, among the options, cytosol is the closest (liquid part of cytoplasm).
• Mitochondria, chloroplast– Contain matrix/stroma, not called cell sap.
• Therefore, the correct answer is Cytosol→ Option 3.

 

 

Question 65

Chromosomal number found in a gamete:

1. n
2. 2n
3. 3n
4. 4n

Correct Answer: 1

Explanation:

• Gametes(sperm and egg) are produced by meiosis, which reduces the chromosome number by half.
• Therefore, gametes contain the haploid (n)number of chromosomes.
• 2n– Somatic cells.
• 3n– Endosperm.
• Therefore, the correct answer is n→ Option 1.

 

 

Question 66

Closely linked genes that are functionally similar and structurally different:

1. Isoalleles
2. Pseudoalleles
3. Multiple alleles
4. All the above

Correct Answer: 1

Explanation:

• Isoalleles– Closely linked genes that produce similar phenotypic effects but differ in structure. They are functionally similar but structurally different.
• Pseudoalleles– Closely linked genes that are actually different loci but behave like alleles.
• Multiple alleles– More than two alleles of a single gene.
• Therefore, the correct answer is Isoalleles→ Option 1.

 

 

Question 67

Cytoplasmic inheritance is due to DNA present in:

1. Nucleus
2. Mitochondria
3. Chloroplast
4. Both b and c

Correct Answer: 4

Explanation:

• Cytoplasmic inheritance(maternal inheritance) is due to DNA present in mitochondria and chloroplasts (organellar DNA).
• Nuclear DNA– Mendelian inheritance.
• Therefore, the correct answer is Both b and c→ Option 4.

 

 

Question 68

H₄ hybrid cotton is development of:

1. Coimbatore
2. Dharwad
3. Nagpur
4. Surat

Correct Answer: 4

Explanation:

• H₄– The first cotton hybrid developed in India (and the world) by  C.T. Patel at the Cotton Research Station, Surat (Gujarat) in 1970.
• Therefore, the correct answer is Surat→ Option 4.

 

 

Question 69

Seed plot technique in potato is done for:

1. Fungi free seed
2. Bacteria free seed
3. Virus free seed
4. Nematode free seed

Correct Answer: 3

Explanation:

• The Seed Plot Techniquein potato is used to produce virus-free seed potatoes by growing them in high-altitude, cool, aphid-free areas.
• Therefore, the correct answer is Virus free seed→ Option 3.

 

 

Question 70

Which of the following is a GURT?

1. Antisense RNA technology
2. Gene silencing
3. Terminator technology
4. All the above

Correct Answer: 3

Explanation:

• GURT (Genetic Use Restriction Technology)– Also known as Terminator technology, produces sterile seeds so farmers cannot save seeds for replanting.
• Antisense RNA technology, gene silencing– Not GURT.
• Therefore, the correct answer is Terminator technology→ Option 3.

 

 

Question 71

Antisense RNA technology is used to:

1. Gene silencing
2. Gene expression
3. Gene inhibition
4. All the above

Correct Answer: 1

Explanation:

• Antisense RNA technology– Uses complementary RNA to bind to target mRNA, preventing translation (post-transcriptional gene silencing).
• It does not enhance gene expression or gene inhibition in the sense of knockouts.
• Therefore, the correct answer is Gene silencing→ Option 1.

 

 

Question 72

Genetic engineering is associated with:

1. Plasmid
2. ER
3. Golgi bodies
4. Nucleus

Correct Answer: 1

Explanation:

• Genetic engineering– Involves the use of plasmids (vectors) to transfer foreign DNA into host organisms.
• ER, Golgi bodies, nucleus– Not directly associated as tools.
• Therefore, the correct answer is Plasmid→ Option 1.

 

 

Question 73

Agrobacterium is a:

1. G+ve
2. G-ve
3. Wall less bacterium
4. None of the above

Correct Answer: 2

Explanation:

• Agrobacterium(e.g.,  tumefaciens) is a Gram-negative, rod-shaped bacterium.
• G+ve– Clavibacter, Streptomyces.
• Wall less– Mycoplasma.
• Therefore, the correct answer is G-ve→ Option 2.

 

 

Question 74

Presence of R line in A line:

1. Pollen shedder
2. Off type
3. Shedding tassel
4. All the above

Correct Answer: 1

Explanation:

• In CMS hybrid seed production, the A line (male sterile)should be free of R line (restorer)
• If an R line plant is present in the A line, it acts as a pollen shedder(male fertile) and contaminates the A line seed.
• Therefore, the correct answer is Pollen shedder→ Option 1.

 

 

Question 75

Volunteer plants:

1. Weeds
2. Self-grown plants from previous crop seeds
3. Wild plants
4. None of these

Correct Answer: 2

Explanation:

• Volunteer plants– Plants that grow from seeds of the previous crop that germinate in the following season without being deliberately sown.
• They are not weeds (Option 1) or wild plants (Option 3).
• Therefore, the correct answer is Self-grown plants from previous crop seeds→ Option 2.

 

 

Question 76

Low pH excludes:

1. Molds
2. Bacteria
3. Both a and b
4. None of these

Correct Answer: 2

Explanation:

• Low pH (acidic conditions)– Inhibits the growth of most bacteria (which prefer neutral pH).
• Molds (fungi)– Can tolerate acidic pH better than bacteria.
• Therefore, low pH primarily excludes bacteria→ Option 2.

 

 

Question 77

Response of a plant to day duration is:

1. Vernalization
2. Thermoperiodism
3. Photoperiodism
4. All the above

Correct Answer: 3

Explanation:

• Photoperiodism– The response of plants to the relative lengths of day and night (day duration) affecting flowering.
• Vernalization– Response to cold temperature.
• Thermoperiodism– Response to temperature fluctuations.
• Therefore, the correct answer is Photoperiodism→ Option 3.

 

 

Question 78

Naked seed coat is produced by:

1. Angiosperms
2. Monocots
3. Gymnosperms
4. All the above

Correct Answer: 3

Explanation:

• Gymnosperms– “Naked seeds” meaning the seeds are not enclosed in an ovary (no fruit formation).
• Angiosperms– Seeds are enclosed within fruits.
• Monocots– A type of angiosperm (enclosed seeds).
• Therefore, the correct answer is Gymnosperms→ Option 3.

 

 

Question 79

Five tier system of seed production is adapted in:

1. Rice
2. Groundnut
3. Potato
4. Onion

Correct Answer: 3

Explanation:

• The five-tier system of seed production(Nucleus → Breeder → Foundation → Certified → Truthful) is adapted in potato due to its vegetative propagation and slow multiplication rate.
• Rice, groundnut, onion– Use standard 4-tier system.
• Therefore, the correct answer is Potato→ Option 3.

 

 

Question 80

The total number of national bureau of important microorganisms in India is located at:

1. Kolkata
2. New Delhi
3. Mau
4. Chennai

Correct Answer: 3

Explanation:

• The ICAR-National Bureau of Agriculturally Important Microorganisms (NBAIM)is located at Mau Nath Bhanjan, Uttar Pradesh.
• Therefore, the correct answer is Mau→ Option 3.