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

 

Question 40

The wild progenitor of Oryza sativa:

1. rufipogon
2. nigistaminala(likely O. longistaminata)
3. barthii
4. nivara

Correct Answer: 4

Explanation:

• Oryza nivarais considered the wild progenitor of cultivated rice (Oryza sativa). It is a wild annual rice species from Asia.
• rufipogon– Also a wild relative, but O. nivara is often cited as the direct progenitor.
• barthii– Wild rice from Africa.
• Therefore, the correct answer is  nivara→ Option 4.

 

 

Question 41

The biological phenomenon that distinguishes most seed and fruit crops from their progenitors:

1. Pollination behaviour
2. Genome size
3. Domestication syndrome
4. Centre of origin

Correct Answer: 3

Explanation:

• Domestication syndrome– The set of phenotypic traits (e.g., larger seeds, loss of seed dormancy, non-shattering, reduced bitterness) that distinguish domesticated crops from their wild progenitors.
• Pollination behaviour, genome size, centre of origin– Not distinguishing features.
• Therefore, the correct answer is Domestication syndrome→ Option 3.

 

 

Question 42

Which of the following is not a ‘covered wheat’?

1. Triticale
2. Spelt
3. Einkorn
4. Emmer

Correct Answer: 1

Explanation:

• Covered wheats– Hulled wheats where the grain remains enclosed in glumes (hulls) after threshing (e.g., Spelt, Einkorn, Emmer).
• Triticale– A man-made hybrid of wheat and rye; it is a free-threshing (naked) grain, not a covered wheat.
• Therefore, the correct answer is Triticale→ Option 1.

 

 

Question 43

ISTA’s Rules for determining seed viability test suggest:

1. Four replication of 100 seeds
2. Three replication of 50 seeds
3. Four replication of 50 seeds
4. Three replication of 100 seeds

Correct Answer: 1

Explanation:

• According to ISTA (International Seed Testing Association)rules, the standard germination test uses four replications of 100 seeds each (400 seeds total).
• This ensures statistical reliability.
• Therefore, the correct answer is Four replication of 100 seeds→ Option 1.

 

 

Question 44

While developing core collection from a germplasm collection, the stratified sampling strategy which includes equal number of entries from all groups irrespective of number of accessions in each group is called:

1. Constant strategy
2. Proportional strategy
3. Logarithm strategy
4. Random strategy

Correct Answer: 1

Explanation:

• Constant strategy– A sampling strategy where an equal number of entries is taken from each group (stratum) regardless of the group size.
• Proportional strategy– Number sampled is proportional to group size.
• Logarithm strategy– Based on log transformation.
• Therefore, the correct answer is Constant strategy→ Option 1.

 

 

Question 45

While practising selection for yield improvement in crop plants, the genetic change that occurs in a population is:

1. Decreased adaptation
2. Loss of divergent alleles
3. Increase in genetic diversity
4. Increase in variability

Correct Answer: 2

Explanation:

• Selection for yield improvement typically leads to loss of divergent alleles(genetic erosion) because only the best-performing genotypes are selected, reducing genetic diversity.
• Increased variability/diversity– Not correct; selection reduces diversity.
• Therefore, the correct answer is Loss of divergent alleles→ Option 2.

 

 

Question 46

The central which possesses bulk feathery stigma which will be receptive before the anther dehiscence in the inflorescence is:

1. Holes
2. Finger millet
3. Sorghum
4. Pearl millet

Correct Answer: 4

Explanation:

• Pearl millet (bajra)– Has a bulk feathery stigma that is receptive before anther dehiscence (protogyny), promoting cross-pollination.
• Sorghum– Not feathery.
• Finger millet– Digitate spikes.
• Therefore, the correct answer is Pearl millet→ Option 4.

 

 

Question 47

Genetically localized genes of brown rust (Puccinia triticina) isolated from Triticum umbellulatum:

1. Lr9
2. Lr2a
3. Lr1
4. Pm21

Correct Answer: 1

Explanation:

• Lr9– A leaf rust (brown rust) resistance gene derived from the wild wheat species Triticum umbellulatum (syn. Aegilops umbellulata).
• Lr2a, Lr1– From other sources.
• Pm21– Powdery mildew resistance gene.
• Therefore, the correct answer is Lr9→ Option 1.

 

 

Question 48

A distal segment of a chromosome that is separated from the rest of the chromosome by a chromatic filament:

1. Telomere
2. Centromere
3. Chromomere
4. Satellite

Correct Answer: 4

Explanation:

• Satellite– A distal segment of a chromosome separated from the rest by a secondary constriction (thin chromatic filament). Such chromosomes are called SAT-chromosomes.
• Telomere– End of chromosome.
• Centromere– Primary constriction.
• Chromomere– Bead-like structure on chromosomes.
• Therefore, the correct answer is Satellite→ Option 4.

 

 

Question 49

A region or land characterized by very little annual rainfall, usually from 250 to 500 mm:

1. Dry land
2. Semiarid
3. Garden land
4. Calcareous

Correct Answer: 2

Explanation:

• Semiarid region– Characterized by annual rainfall between 250–500 mm (or up to 750 mm in some definitions). It has a dry climate with limited agriculture.
• Dry land– Broader term, often includes arid (<250 mm).
• Therefore, the correct answer is Semiarid→ Option 2.

 

 

Question 50

Which of the following triplet code is not a stop codon?

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

Correct Answer: 2

Explanation:

• AUG– The start codon (codes for methionine), not a stop codon.
• UAA, UAG, UGA– Stop codons (ochre, amber, opal).
• Therefore, the correct answer is AUG→ Option 2.

 

 

Question 51

A cross involving one parent with three heterozygous gene pairs and another with three homozygous recessive gene pairs:

1. Top cross
2. Test cross
3. Back cross
4. Wide cross

Correct Answer: 2

Explanation:

• Test cross– Cross between an individual with dominant phenotype (unknown genotype) and a homozygous recessive Here, one parent is trihybrid (AaBbCc) and the other is triple recessive (aabbcc) – a classic test cross.
• Therefore, the correct answer is Test cross→ Option 2.

 

 

Question 52

The polymerization of ribonucleotides into a strand of RNA in a sequence complementary to that of a single strand of DNA which is mediated by a DNA dependent RNA polymerase:

1. Translation
2. Polymerization
3. Transfection
4. Transcription

Correct Answer: 4

Explanation:

• Transcription– The process of synthesizing RNA from a DNA template using DNA-dependent RNA polymerase.
• Translation– Protein synthesis from mRNA.
• Transfection– Introduction of nucleic acids into cells.
• Therefore, the correct answer is Transcription→ Option 4.

 

 

Question 53

A genome that is diploid but contains an extra chromosome, homologous with one of the existing pairs, so that one kind of chromosome is present in triplicate:

1. Trisomic
2. Triploidy
3. Triplex
4. Trioecious

Correct Answer: 1

Explanation:

• Trisomic (2n+1)– A diploid organism with an extra chromosome for one pair, so that chromosome is present in triplicate.
• Triploidy (3n)– Three complete sets.
• Triplex– Not standard.
• Trioecious– Having male, female, and bisexual flowers.
• Therefore, the correct answer is Trisomic→ Option 1.

 

 

Question 54

In the Central and Southern zones of cotton cultivation, the species of cotton that is grown under rainfed situation is:

1. Gossypium hirsutum
2. herbaceum
3. arboreum
4. barbadense

Correct Answer: 3

Explanation:

• Gossypium arboreum(Old World cotton) is grown under rainfed conditions in Central and Southern India (e.g., Madhya Pradesh, Maharashtra, Karnataka).
• hirsutum– Irrigated conditions.
• barbadense– Limited areas.
• Therefore, the correct answer is  arboreum→ Option 3.

 

 

Question 55

The method of breeding that gives the maximum opportunity for the breeder to use skill and judgement for the selection of plants is:

1. Mass selection
2. Pedigree method
3. Bulk method
4. Introduction and selection

Correct Answer: 2

Explanation:

• Pedigree method– Allows the breeder to use maximum skill and judgment by keeping detailed records of each plant’s ancestry and selecting individual plants across generations.
• Mass selection– Bulk selection without individual tracking.
• Bulk method– Natural selection, minimal breeder intervention.
• Therefore, the correct answer is Pedigree method→ Option 2.

 

 

Question 56

Genetic organisation by which the harmful effects of unfavourable recessive alleles are masked by their dominant alleles as a result of which they are retained in the population is:

1. Genetic load
2. Heterozygous balance
3. Genetic drift
4. Lethality

Correct Answer: 2

Explanation:

• Heterozygous balance(Balanced polymorphism) – Harmful recessive alleles are masked by dominant alleles in heterozygotes, allowing them to be retained in the population.
• Genetic load– The burden of harmful alleles.
• Genetic drift– Random changes in allele frequency.
• Therefore, the correct answer is Heterozygous balance→ Option 2.

 

 

Question 57

Important factor which determines the structure and cohesiveness of populations:

1. Gene frequency
2. Gene flow
3. Homozygosity
4. Heterozygosity

Correct Answer: 2

Explanation:

• Gene flow(migration) – The movement of alleles between populations, which determines the structure and cohesiveness of populations by reducing genetic differentiation.
• Gene frequency– Changes over time.
• Homozygosity/heterozygosity– Not primary determinants of population structure.
• Therefore, the correct answer is Gene flow→ Option 2.

 

 

Question 58

A highly mathematical field that examines the statistical relationships between genes and the traits they encode:

1. Molecular genetics
2. Quantitative genetics
3. Population genetics
4. Classical genetics

Correct Answer: 2

Explanation:

• Quantitative genetics– A highly mathematical field that examines the statistical relationships between genes (genotypes) and quantitative traits (phenotypes), including heritability, genetic variance, and response to selection.
• Population genetics– Focuses on allele frequencies.
• Molecular genetics– Focuses on DNA structure and function.
• Therefore, the correct answer is Quantitative genetics→ Option 2.

 

 

Question 59

Recombination occurs during the following phase of the cell division:

1. Telophase I
2. Metaphase I
3. Prophase I
4. Anaphase I

Correct Answer: 3

Explanation:

• Prophase I of meiosis– Crossing over (recombination) occurs during the pachytene stage of prophase I.
• Metaphase I– Chromosomes align.
• Anaphase I– Homologous chromosomes separate.
• Therefore, the correct answer is Prophase I→ Option 3.

 

 

Question 60

In plant breeding, selection of genotypes in favour of two extremes is called:

1. Cyclic selection
2. Disruptive selection
3. Directional selection
4. Stabilizing selection

Correct Answer: 2

Explanation:

• Disruptive selection– Selection in favor of two extreme phenotypes (both extremes are selected) against the intermediate.
• Directional selection– Selection towards one extreme.
• Stabilizing selection– Selection against extremes, favoring the mean.
• Therefore, the correct answer is Disruptive selection→ Option 2.

 

 

Question 61

If two or more characters are inherited together and continue to be inherited together in next two or more generations, then those characters are said to be:

1. Co-segregating
2. Dominant traits
3. Partially linked
4. Completely linked

Correct Answer: 4

Explanation:

• Completely linked– Genes located very close together on the same chromosome that are always inherited together because crossing over does not occur between them.
• Partially linked– Some crossing over occurs, producing some recombinants.
• Co-segregating– Traits that are inherited together (general term).
• Therefore, the correct answer is Completely linked→ Option 4.

 

 

Question 62

The most dreaded disease of sugarcane that has caused the elimination of several important sugarcane varieties from cultivation is:

1. Smut
2. Red Rot
3. Grassy Shoot Disease
4. Ratoon Stunting Disease

Correct Answer: 2

Explanation:

• Red Rot of sugarcane(caused by Colletotrichum falcatum) – One of the most destructive diseases, responsible for eliminating many important sugarcane varieties.
• Smut– Important but less devastating.
• Grassy shoot, ratoon stunting– Significant but not as historically destructive.
• Therefore, the correct answer is Red Rot→ Option 2.

 

 

Question 63

The genetic measure calculated by dividing the observed number of double recombinant chromosomes by the expected number is:

1. Coefficient of coincidence
2. Chromosome interference
3. Interference
4. Linkage

Correct Answer: 1

Explanation:

• Coefficient of coincidence (CoC)– Observed double recombinants / Expected double recombinants.
• Interference= 1 – CoC (measures the degree to which one crossover prevents another).
• Therefore, the correct answer is Coefficient of coincidence→ Option 1.

 

 

Question 64

Which of the following enzymes is used to degrade DNA molecules?

1. DNA polymerase
2. Nuclease
3. Ligase
4. Kinase

Correct Answer: 2

Explanation:

• Nuclease– An enzyme that cleaves phosphodiester bonds in nucleic acids, degrading DNA molecules (e.g., DNase, restriction enzymes).
• DNA polymerase– Synthesizes DNA.
• Ligase– Joins DNA fragments.
• Kinase– Adds phosphate groups.
• Therefore, the correct answer is Nuclease→ Option 2.

 

 

Question 65

The mechanism which promotes self pollination in crop plants is:

1. Monoecy
2. Dichogamy
3. Chasmogamy
4. Heterostyly

Correct Answer: 3

Explanation:

• Chasmogamy– Flowers open, allowing pollination. In self-pollinated crops, chasmogamy still occurs but pollination happens within the same flower.
• Monoecy, Dichogamy, Heterostyly– Promote cross-pollination.
• Therefore, the correct answer is Chasmogamy→ Option 3.

 

 

Question 66

What are all the assumptions of Hardy-Weinberg equilibrium?

1. Small population size, random mating, no selection, no migration, no mutation
2. Large population size, random mating, no selection, no migration, no mutation
3. Large population, like individuals mate, no selection, no migration, no mutation
4. Large population size, random mating, heterozygotes survive the best, no migration, no mutation

Correct Answer: 2

Explanation:

• The Hardy-Weinberg equilibriumassumptions are:
  • Large population size(prevents genetic drift)
  • Random mating
  • No selection(all genotypes equally fit)
  • No migration(no gene flow)
  • No mutation(alleles stable)
• Therefore, the correct answer is Large population size, random mating, no selection, no migration, no mutation→ Option 2.

 

 

Question 67

The value of an individual, judged by the mean value of its progeny is called:

1. Average effect
2. Breeding value
3. Recombination value
4. Genetic drift

Correct Answer: 2

Explanation:

• Breeding value– The value of an individual as a parent, judged by the mean value of its progeny (the deviation of progeny mean from population mean, doubled).
• Average effect– The effect of a single allele substitution.
• Recombination value– Frequency of recombination.
• Therefore, the correct answer is Breeding value→ Option 2.

 

 

Question 68

During photosynthesis, O₂ is released from:

1. CO₂
2. H₂O
3. C₆H₁₂O₆
4. N₂

Correct Answer: 2

Explanation:

• During the light reactionsof photosynthesis, water (H₂O) is split (photolysis) by photosystem II, releasing oxygen (O₂) , protons, and electrons.
• CO₂– Fixed into glucose.
• Therefore, the correct answer is H₂O→ Option 2.

 

 

Question 69

In aerobic respiration, the gross number of ATP produced is:

1. 32
2. 34
3. 36
4. 38

Correct Answer: 4

Explanation:

• The gross ATP yieldfrom complete oxidation of one glucose molecule in aerobic respiration is 38 ATP (in prokaryotes) and 36 ATP (in eukaryotes due to transport costs). The question likely expects 38 as the gross number.
• Net ATP(after subtracting transport costs) – 36 in eukaryotes.
• Therefore, the correct answer is 38→ Option 4.

 

 

Question 70

Cell wall is absent in:

1. Algae
2. Mycoplasma
3. Bacteria
4. Fungi

Correct Answer: 2

Explanation:

• Mycoplasma– Prokaryotic organisms that lack a cell wall; they are enclosed only by a plasma membrane (pleomorphic).
• Algae, bacteria, fungi– Have cell walls (cellulose, peptidoglycan, chitin respectively).
• Therefore, the correct answer is Mycoplasma→ Option 2.

 

 

Question 71

In which group of plants, the endosperm is 3N?

1. Pteridophytes
2. Angiosperms
3. Gymnosperms
4. Fungi

Correct Answer: 2

Explanation:

• In Angiosperms, the endosperm is triploid (3n)formed by double fertilization (fusion of one sperm with two polar nuclei).
• Gymnosperms– Endosperm is haploid (n) (female gametophyte).
• Pteridophytes, fungi– No endosperm.
• Therefore, the correct answer is Angiosperms→ Option 2.

 

 

Question 72

Transpiration takes place through:

1. Cuticle
2. Lenticels
3. Stomata
4. All of these

Correct Answer: 4

Explanation:

• Transpiration occurs through three main pathways:
  • Stomata– Stomatal transpiration (major, 80-90%).
  • Cuticle– Cuticular transpiration.
  • Lenticels– Lenticular transpiration (minor).
• Therefore, the correct answer is All of these→ Option 4.

 

 

Question 73

Nitrosomonas bacteria convert ammonia to:

1. Nitrate
2. Nitrogen
3. Nitrite
4. Ammonia

Correct Answer: 3

Explanation:

• Nitrosomonas– Chemoautotrophic bacteria that oxidize ammonia (NH₃) to nitrite (NO₂⁻) (first step of nitrification).
• Nitrobacter– Converts nitrite to nitrate.
• Therefore, the correct answer is Nitrite→ Option 3.

 

 

Question 74

Symbiotic nitrogen fixing bacteria occur in the genus:

1. Xanthomonas
2. Pseudomonas
3. Rhizobium
4. Azotobacter

Correct Answer: 3

Explanation:

• Rhizobium– Symbiotic nitrogen-fixing bacteria that form nodules on legume roots.
• Azotobacter– Free-living nitrogen fixer.
• Pseudomonas, Xanthomonas– Not nitrogen fixers.
• Therefore, the correct answer is Rhizobium→ Option 3.

 

 

Question 75

Sauerkraut is a fermented food product obtained from:

1. Onion
2. Olives
3. Cabbage
4. Cauliflower

Correct Answer: 3

Explanation:

• Sauerkraut– A fermented food made from cabbage (finely shredded) through lactic acid fermentation by Leuconostoc and Lactobacillus
• Therefore, the correct answer is Cabbage→ Option 3.

 

 

Question 76

Which among the following is a free living nitrogen fixing bacteria:

1. Azotobacter
2. Azospirillum
3. Rhizobium
4. Agrobacterium

Correct Answer: 1

Explanation:

• Azotobacter– Free-living, aerobic nitrogen-fixing bacterium found in soil.
• Azospirillum– Associative nitrogen fixer (in rhizosphere).
• Rhizobium– Symbiotic (nodules).
• Agrobacterium– Not a nitrogen fixer.
• Therefore, the correct answer is Azotobacter→ Option 1.

 

 

Question 77

Golden rice is a rich source of:

1. Vitamin B
2. Ascorbic acid
3. Vitamin A
4. Vitamin C

Correct Answer: 3

Explanation:

• Golden rice– Genetically modified rice containing beta-carotene, which is converted to Vitamin A in the human body. It is designed to address vitamin A deficiency.
• Therefore, the correct answer is Vitamin A→ Option 3.

 

 

Question 78

The fruit of rapeseed and mustard plant is known as:

1. Caryopsis
2. Grain
3. Capsule
4. Siliqua

Correct Answer: 4

Explanation:

• Siliqua (silique)– The characteristic fruit (capsule) of the family Brassicaceae (mustard family), including rapeseed and mustard.
• Caryopsis– Grain of grasses.
• Therefore, the correct answer is Siliqua→ Option 4.

 

 

Question 79

Which of the following schemes is exclusively meant for self-employment of rural youths?

1. NREP
2. TRYSEM
3. IRDP
4. DPAP

Correct Answer: 2

Explanation:

• TRYSEM (Training of Rural Youth for Self Employment)– A scheme exclusively for training rural youth in skills for self-employment.
• NREP, IRDP, DPAP– Other rural development programs.
• Therefore, the correct answer is TRYSEM→ Option 2.

 

 

Question 80

Sowing depth in dwarf wheat is related to:

1. Moisture
2. Length of radicle
3. Size of seed
4. Length of coleoptile

Correct Answer: 4

Explanation:

• Coleoptile length– Determines the maximum sowing depth in dwarf wheat. The coleoptile must reach the soil surface for emergence. Dwarf wheat varieties have shorter coleoptiles, so they must be sown shallower.
• Therefore, the correct answer is Length of coleoptile→ Option 4.