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

 

Question 1 

In butterflies and moths, the sex determination system is:

1. XX-XY, heterogametic male
2. XX-XO, heterogametic male
3. ZW-ZZ, heterogametic female
4. ZO-ZZ, heterogametic female

Correct Answer: 3

 

Explanation:

• Butterflies and moths (Lepidoptera) have a ZW-ZZsex determination system.
• Femalesare heterogametic (ZW) – produce two types of gametes (Z and W).
• Malesare homogametic (ZZ) – produce only Z gametes.
• This is opposite to the XX-XY system found in mammals (where males are heterogametic).
• Option 1:XX-XY (heterogametic male) – found in mammals and many insects like Drosophila.
• Option 2:XX-XO (heterogametic male) – found in some insects like grasshoppers.
• Option 4:ZO-ZZ – rare, found in some insects where females have only one Z.
• Thus, the correct answer is ZW-ZZ, heterogametic female → Option 3.

 

 

Question 2

Pollination by bats is found in:

1. Rafflesia
2. Colocasia
3. Anthocephalus
4. Ceratophyllum

Correct Answer: 3

 

Explanation:

• Chiropterophily– Pollination by bats.
• Anthocephalus(also known as Neolamarckia cadamba or Kadamba) is pollinated by bats. Its flowers are dull-colored, open at night, produce abundant nectar, and have a strong fragrance to attract bats.
• Rafflesia– Pollinated by carrion flies (mimics rotting flesh).
• Colocasia– Pollinated by beetles and flies (not bats).
• Ceratophyllum(hornwort) – Submerged aquatic plant; pollinated by water (hydrophily).
• Therefore, the correct answer is Anthocephalus→ Option 3.

 

 

Question 3

Which of the following is a polysaccharide?

1. Sucrose
2. Maltose
3. Starch
4. Lactose

Correct Answer: 3

 

Explanation:

• Polysaccharidesare long-chain polymers of monosaccharides (more than 10 units).
• Starch– A polysaccharide composed of glucose units (amylose and amylopectin). It is the storage carbohydrate in plants.
• Sucrose– Disaccharide (glucose + fructose).
• Maltose– Disaccharide (glucose + glucose).
• Lactose– Disaccharide (galactose + glucose).
• Therefore, the correct answer is Starch→ Option 3.

 

 

Question 4

The phenomenon of wobble hypothesis DOES NOT comply with which one of the following?

1. The multiple codes for a given amino acid
2. Possible suppression of point mutations in the third base of the codons
3. More easily removed deacylated RNA during protein synthesis
4. Fewer mRNA molecules than expected

Correct Answer: 3

 

Explanation:

• Wobble hypothesis(proposed by Francis Crick, 1966) explains how a single tRNA can recognize multiple codons due to flexibility (wobble) at the third base (3′ end) of the codon.
• Key features:
  • Explains degeneracyof genetic code (multiple codons for same amino acid) → Option 1 complies.
  • Allows suppression of point mutationsin the third base because the same tRNA can still bind → Option 2 complies.
  • Requires fewer tRNA moleculesthan the number of codons (not fewer mRNA molecules) → Option 4 is incorrectly phrased but “fewer mRNA molecules” is not a feature of wobble hypothesis.
• Option 3– “More easily removed deacylated RNA during protein synthesis” has no relation to the wobble hypothesis. Deacylated tRNA release is part of translation termination, not wobble.
• Therefore, the phenomenon of wobble hypothesis DOES NOT complywith Option 3 → Correct Answer: 3.

 

 

Question 5

Which one of the following is an example of a bacteria which is insecticidal?

1. Bacillus polymyxa
2. Yersinia pestis
3. Bacillus thuringiensis
4. Klebsiella oxytoca

Correct Answer: 3

 

Explanation:

• Bacillus thuringiensis(Bt) – A Gram-positive, spore-forming bacterium that produces crystal proteins (Cry toxins) that are insecticidal (especially against Lepidoptera, Coleoptera, Diptera). It is widely used as a biological pesticide and in Bt-transgenic crops.
• Bacillus polymyxa– Plant growth-promoting rhizobacterium; not primarily insecticidal.
• Yersinia pestis– Causes plague in humans (not insecticidal; transmitted by fleas).
• Klebsiella oxytoca– Opportunistic human pathogen; not insecticidal.
• Therefore, the correct answer is Bacillus thuringiensis→ Option 3.

 

 

Question 6

Which of the following is an example of hesperidium?

1. Apple
2. Orange
3. Pear
4. Pomegranate

Correct Answer: 2

 

Explanation:

• Hesperidium– A type of berry with a tough, leathery rind (exocarp and mesocarp) and fleshy, juicy interior divided into segments (endocarp with juice sacs). It is characteristic of Citrus fruits.
• Orange– A classic example of hesperidium.
• Apple– Pome (accessory fruit from thalamus).
• Pear– Pome.
• Pomegranate– Balusta (a type of berry with hard rind and fleshy seeds).
• Therefore, the correct answer is Orange→ Option 2.

 

 

Question 7

If a male sterile line is crossed with heterozygous male fertile plant and the progeny segregated as 1:1 male fertile and male sterile plants, respectively then the male sterility is termed as:

1. Cytoplasmic male sterility
2. Genic male sterility
3. Cytoplasmic genic male sterility
4. Environment sensitive male sterility

Correct Answer: 2

 

Explanation:

• Genic male sterility (GMS)– Controlled by nuclear genes (recessive). When a male sterile line (ms ms) is crossed with a heterozygous male fertile plant (Ms ms), the progeny segregates in a 1:1 ratio (Ms ms fertile : ms ms sterile).
• Cytoplasmic male sterility (CMS)– Controlled by mitochondrial genes; all progeny are male sterile regardless of the male parent (no 1:1 segregation).
• Cytoplasmic genic male sterility (CGMS)– Requires both cytoplasmic and nuclear genes; restoration involves specific nuclear restorer genes (Rf).
• Environment sensitive male sterility– Expression depends on environmental conditions (e.g., temperature, photoperiod).
• Since the cross gave 1:1 segregation, it is genic male sterility→ Option 2.

 

 

Question 8

The tendency of genes to remain together in the same gamete during segregation and not show independent segregation is found in:

1. QTLs
2. Epistatic genes
3. Linked genes
4. Multiple alleles

Correct Answer: 3

 

Explanation:

• Linked genes– Genes located on the same chromosome tend to be inherited together (linkage) and do not show independent assortment (unless crossing over separates them).
• QTLs– Quantitative trait loci; not about linkage per se.
• Epistatic genes– Genes that interact (one masks the expression of another); not about physical linkage.
• Multiple alleles– More than two alleles of a gene; not about linkage.
• Therefore, the correct answer is Linked genes→ Option 3.

 

 

Question 9

A cell contains 40 chromatids at the beginning of meiosis. After the completion of meiosis, the cells would contain:

1. 20 chromosomes
2. 10 chromosomes
3. 40 chromosomes
4. 80 chromosomes

Correct Answer: 2

 

Explanation:

• At the beginning of meiosis(prophase I), DNA has already replicated. Each chromosome consists of two sister chromatids.
• If there are 40 chromatids, the number of chromosomes is 20(since 20 chromosomes × 2 chromatids each = 40 chromatids). The cell is diploid (2n = 20).
• After meiosis I(reductional division), two haploid cells are formed, each with 10 chromosomes (each chromosome still has 2 chromatids).
• After meiosis II(equational division), four haploid cells are formed, each with 10 chromosomes (each chromosome now has 1 chromatid).
• Therefore, after completion of meiosis, each cell would contain 10 chromosomes→ Option 2.

 

 

Question 10

In a type of apomixis known as adventive embryony, the embryos develop directly from:

1. Zygote
2. Synergids or antipodals in an embryo sac
3. Nucellus and integuments
4. Accessory embryo sacs in the ovule

Correct Answer: 3

 

Explanation:

• Adventive embryony(also called nucellar embryony) – A type of sporophytic apomixis where embryos develop directly from diploid cells of the nucellus or integuments (sporophytic tissue), not from the egg cell.
• Example:Citrus species (multiple embryos in a single seed – polyembryony).
• Option 1:Zygote – normal sexual embryogenesis.
• Option 2:Synergids/antipodals – parthenogenesis (gametophytic apomixis).
• Option 4:Accessory embryo sacs – rare, not adventive embryony.
• Therefore, the correct answer is Nucellus and integuments→ Option 3.

 

 

Question 11

Which plant is used for making cricket bat?

1. Morus
2. Salix
3. Adina
4. Julgans

Correct Answer: 2

 

Explanation:

• Salix(Willow) – The wood of white willow (Salix alba) is used to make cricket bats. It is lightweight, strong, and has excellent shock-absorbing properties.
• Morus– Mulberry; used for making cricket balls? No, used for silk production and wood.
• Adina– Timber tree, not for cricket bats.
• Julgans(likely Juglans) – Walnut; used for furniture and gunstocks, not cricket bats.
• Therefore, the correct answer is Salix→ Option 2.

 

 

Question 12

While explaining cell organelles, in ribosomes, the suffix ‘s’ in ribosome unit indicates:

1. Sedimentation area
2. Sedimentation coefficient
3. Surface area
4. Structural protein

Correct Answer: 2

 

Explanation:

• The ‘S’in ribosome units (e.g., 70S, 80S) stands for Svedberg unit, which is the sedimentation coefficient (a measure of how fast a particle sediments in an ultracentrifuge).
• It is determined by the size, shape, and density of the particle.
• Higher S value indicates larger or more compact particles.
• Examples:Prokaryotic ribosomes are 70S (50S + 30S); eukaryotic ribosomes are 80S (60S + 40S).
• Therefore, the correct answer is Sedimentation coefficient→ Option 2.

 

 

Question 13

The cell organelles that act as factories which translate the genetic information from nucleus into usable protein molecules is:

1. mRNA
2. Endoplasmic reticulum
3. Ribosomes
4. Vesicles

Correct Answer: 3

 

Explanation:

• Ribosomes– The site of protein synthesis (translation) . They read mRNA (transcribed from nuclear DNA) and assemble amino acids into polypeptide chains.
• mRNA– Carries genetic information but does not translate it.
• Endoplasmic reticulum– Involved in protein processing and transport (rough ER has ribosomes attached).
• Vesicles– Transport proteins within the cell.
• Therefore, the correct answer is Ribosomes→ Option 3.

 

 

Question 14

The apomixis that occurs commonly in Citrus species is:

1. Gametophytic apomixis
2. Sporophytic apomixis
3. Diplospory
4. Apogamy

Correct Answer: 2

 

Explanation:

• Sporophytic apomixis– Embryos develop directly from somatic cells of the ovule (nucellus or integuments) without meiosis or gametophyte formation. This is also called adventive embryony or nucellar embryony.
• Citrusis a classic example of sporophytic apomixis (polyembryony – multiple embryos from nucellar cells).
• Gametophytic apomixis– Embryo develops from an unreduced egg cell (parthenogenesis) after aposporous or diplosporous embryo sac formation.
• Diplospory– A type of gametophytic apomixis where the embryo sac develops from the megaspore mother cell without reduction.
• Apogamy– Embryo develops from a cell other than the egg (e.g., synergid, antipodal) without fertilization.
• Therefore, the correct answer is Sporophytic apomixis→ Option 2.

 

 

Question 15

The multiple replicates of same chromosome holding together in a parallel fashion is called:

1. Endomitosis
2. Polyteny
3. Balbiani
4. Puffing

Correct Answer: 2

 

Explanation:

• Polyteny– The presence of giant chromosomes (polytene chromosomes) formed by multiple rounds of DNA replication without cell division (endomitosis). The replicated chromatids remain aligned in parallel, giving a banded appearance.
• Found in:Salivary glands of Drosophila, other dipterans.
• Endomitosis– DNA replication without nuclear division (leads to polyploidy, not necessarily polyteny).
• Balbiani– Refers to Balbiani rings (puffs) on polytene chromosomes.
• Puffing– Localized unwinding of polytene chromosomes indicating active transcription.
• Therefore, the correct answer is Polyteny→ Option 2.

 

 

Question 16

Heterobeltiosis is the superiority of F₁ over:

1. Mean parent
2. Popular check
3. Better parents
4. Commercial hybrids

Correct Answer: 3

 

Explanation:

• Heterobeltiosis– The superiority of the F₁ hybrid over the better parent (the higher-yielding or superior parental line).
• Formula:

Heterobeltiosis(%)=F₁ yield−Better parent yieldBetter parent yield×100Heterobeltiosis(%)=Better parent yieldF₁ yield−Better parent yield​×100

• Mean parent (mid-parent heterosis)– Superiority over the average of both parents.
• Popular check (standard heterosis)– Superiority over the standard commercial variety/hybrid.
• Commercial hybrids– Not a standard comparison for heterobeltiosis.
• Therefore, the correct answer is Better parents→ Option 3.

 

 

Question 17

The temperature, pressure and time combination usually employed in autoclaving is:

1. 125 °C, 15 lbs, 10-15 min
2. 128 °C, 20 lbs, 10-15 min
3. 125 °C, 20 lbs, 15-20 min
4. 121 °C, 15 lbs, 15-20 min

Correct Answer: 4

 

Explanation:

• Autoclaving– The standard method for moist heat sterilization under pressure.
• Standard conditions:121°C, 15 psi (pounds per square inch), 15–20 minutes.
• These conditions ensure the killing of all microorganisms, including bacterial endospores.
• Option 4matches the standard conditions.
• Other options have incorrect temperature or pressure values.
• Therefore, the correct answer is 121 °C, 15 lbs, 15-20 min→ Option 4.

 

 

Question 18

Assuming Hardy-Weinberg equilibrium, if the frequency of the two alleles at the gene being studied are 0.6 and 0.4, the genotypic frequency of heterozygotes will be:

1. 80
2. 48
3. 64
4. 16

Correct Answer: 2

 

Explanation:

• Hardy-Weinberg equilibrium formula: p² + 2pq + q² = 1
• p = frequency of dominant allele = 0.6
• q = frequency of recessive allele = 0.4
• Frequency of heterozygotes (Aa) = 2pq = 2 × 0.6 × 0.4 = 2 × 0.24 = 0.48
• Therefore, the correct answer is 48→ Option 2.

 

 

Question 19

The classical work of Beadle and Tatum on Neurospora crassa has proved that:

1. Replication of DNA is semiconservative
2. Viruses have genetic material
3. One gene is responsible for synthesis of one enzyme
4. Plant cells contain prokaryotic genome

Correct Answer: 3

 

Explanation:

• Beadle and Tatum (1941)– Worked on Neurospora crassa (bread mold) using induced mutations and nutritional mutants.
• They proposed the “one gene–one enzyme” hypothesis(later modified to one gene–one polypeptide).
• They showed that each mutation affected a single enzyme in a metabolic pathway.
• Option 1:Semiconservative replication – Meselson and Stahl (1958).
• Option 2:Viruses have genetic material – Hershey and Chase (1952).
• Option 4:Plant cells contain prokaryotic genome – Not relevant.
• Therefore, the correct answer is One gene is responsible for synthesis of one enzyme→ Option 3.

 

 

Question 20

Options:

1. Parallel
2. Antiparallel
3. Forward
4. Alternate

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates Antiparallel→ Option 2.
• This likely asks: The two strands of DNA are oriented in which direction?
• DNA double helix consists of two strands that run in antiparalleldirections (one 5’→3′, the other 3’→5′).
• Therefore, the correct answer is Antiparallel→ Option 2.

 

 

Question 21

During the process of lint development in cotton, how many days after fertilization does the lint gain maturity?

1. 45-50 days
2. 70-75 days
3. 100-105 days
4. 75-80 days

Correct Answer: 4

 

Explanation:

• In cotton, lint(cotton fibre) develops from epidermal cells of the seed coat.
• Lint development begins soon after fertilization and continues for about 75-80 daysafter fertilization until boll opening.
• Mature lint is fully elongated and thickened with secondary cell wall cellulose.
• Therefore, the correct answer is 75-80 days→ Option 4.

 

 

Question 22

The unit of radiation dose is called Gray (Gy) and 1 Gray equals:

1. One Jule/gram of the irradiated object
2. One Jule/kg of the irradiated object
3. 100 J/kg of the irradiated object
4. 1000 J/kg of the irradiated object

Correct Answer: 2

 

Explanation:

• Gray (Gy)– The SI unit of absorbed radiation dose.
• 1 Gy = 1 Joule per kilogram (J/kg)of the irradiated material.
• Option 1:J/gram – That would be 1000 Gy (incorrect).
• Option 3:100 J/kg = 100 Gy (incorrect).
• Option 4:1000 J/kg = 1000 Gy (incorrect).
• Therefore, the correct answer is One Jule/kg of the irradiated object→ Option 2.

 

 

Question 23

The hybrid progeny between a single cross and an inbred line is referred to as:

1. Three way cross hybrid
2. Double top cross hybrid
3. Double cross hybrid
4. Two way cross hybrid

Correct Answer: 1

 

Explanation:

• Three way cross hybrid– Produced by crossing a single cross hybrid (A × B) with an inbred line (C) → (A × B) × C.
• Single cross hybrid– Cross between two inbred lines (A × B) – also called two-way cross.
• Double cross hybrid– Cross between two single cross hybrids (A × B) × (C × D).
• Double top cross hybrid– Cross between a single cross hybrid and an open pollinated variety (OPV).
• Therefore, the correct answer is Three way cross hybrid→ Option 1.

 

 

Question 24

The process of breeding food crops that are rich in bioavailable micronutrients is referred to as:

1. Molecular breeding
2. Speed breeding
3. Quality breeding
4. Biofortification

Correct Answer: 4

 

Explanation:

• Biofortification– The process of breeding or developing crop varieties that have enhanced levels of bioavailable micronutrients (e.g., iron, zinc, vitamin A, iodine) in their edible parts.
• Examples:Golden rice (vitamin A), iron-rich pearl millet, zinc-rich wheat.
• Molecular breeding– Use of DNA markers in breeding.
• Speed breeding– Accelerating the breeding cycle using controlled environments.
• Quality breeding– Breeding for quality traits (e.g., protein, oil content).
• Therefore, the correct answer is Biofortification→ Option 4.

 

 

Question 25

A set of germplasm accession derived from base collection to represent the genetic diversity in the whole collection is:

1. Active collection
2. Reference set
3. Working collection
4. Core collection

Correct Answer: 4

 

Explanation:

• Core collection– A subset of the entire germplasm collection (base collection) that captures the maximum genetic diversity (usually about 10% of the total accessions) with minimum redundancy. It is used for efficient evaluation and utilization.
• Active collection– Medium-term storage (2-10 years) used for distribution and research.
• Reference set– A specific subset for comparative studies.
• Working collection– Actively used in breeding programs.
• Therefore, the correct answer is Core collection→ Option 4.

 

 

Question 26

Options:

1. RNAi technology
2. PCR technique
3. DNA sequencing technique
4. DNA walking

Correct Answer: 1

 

Explanation:

• The specific question text is missing, but the answer key indicates RNAi technology→ Option 1.
• RNAi (RNA interference)– A gene silencing technique used to study gene function and develop pest-resistant crops (e.g., against viruses, insects, nematodes).
• Therefore, the correct answer is RNAi technology→ Option 1.

 

 

Question 27

In recombinant DNA technology, which enzyme is used to join fragments that have been cut by restriction enzymes?

1. EcoRI
2. DNA polymerase
3. DNA methylase
4. DNA ligase

Correct Answer: 4

 

Explanation:

• DNA ligase– The enzyme that catalyzes the formation of phosphodiester bonds between adjacent DNA fragments (joins DNA fragments).
• It is used to ligate(join) restriction fragments into vectors (plasmids) in recombinant DNA technology.
• EcoRI– A restriction enzyme (cuts DNA, does not join).
• DNA polymerase– Synthesizes new DNA strands.
• DNA methylase– Adds methyl groups to DNA (modification).
• Therefore, the correct answer is DNA ligase→ Option 4.

 

 

Question 28

Monocot seed sowing epigeal germination is found in:

1. Rice
2. Wheat
3. Onion
4. Maize

Correct Answer: 3

 

Explanation:

• Epigeal germination– The cotyledons are brought above the soil surface due to elongation of the hypocotyl. The cotyledons become photosynthetic.
• Onion(Allium cepa) – A monocot that shows epigeal germination.
• Rice, wheat, maize– Monocots that show hypogeal germination (cotyledons remain below ground; epicotyl elongates).
• Therefore, the correct answer is Onion→ Option 3.

 

 

Question 29

Which bacteria converts nitrite into nitrate form?

1. Nitrosomonas
2. Nitrosococcus
3. Nitrobacter
4. Rhizobium

Correct Answer: 3

 

Explanation:

• Nitrification– Two-step process:
  1. Ammonia (NH₃) → Nitrite (NO₂⁻)– carried out by Nitrosomonas, Nitrosospira, Nitrosococcus.
  2. Nitrite (NO₂⁻) → Nitrate (NO₃⁻)– carried out by Nitrobacter, Nitrospira.
• Rhizobium– Symbiotic nitrogen fixer (N₂ → NH₃).
• Therefore, the correct answer is Nitrobacter→ Option 3.

 

 

Question 30

Which of the following represents genome and chromosome number of the oilseed Brassica napus sub. sp. napus?

1. AACC, 38
2. BBCC, 34
3. CC, 18
4. AA, 20

Correct Answer: 1

 

Explanation:

• Brassica napus(rapeseed/canola) is an allotetraploid (amphidiploid) derived from hybridization between Brassica rapa (AA, 2n=20) and Brassica oleracea (CC, 2n=18).
• Genome designation: AACC(2n = 20 + 18 = 38).
• Option 1:AACC, 38 – Correct.
• Option 2:BBCC, 34 – Brassica juncea (mustard) genome.
• Option 3:CC, 18 –  oleracea (cabbage, cauliflower).
• Option 4:AA, 20 –  rapa (turnip, Chinese cabbage).
• Therefore, the correct answer is AACC, 38→ Option 1.

 

 

Question 31

Entry of the pollen tube through micropyle is known as:

1. Mesogamy
2. Microgamy
3. Porogamy
4. Chalazogamy

Correct Answer: 3

 

Explanation:

• Porogamy– The pollen tube enters the ovule through the micropyle. This is the most common type of fertilization in angiosperms.
• Mesogamy– Pollen tube enters through the integuments (not micropyle or chalaza).
• Chalazogamy– Pollen tube enters through the chalaza (e.g., in Casuarina).
• Microgamy– Not a standard term for pollen tube entry.
• Therefore, the correct answer is Porogamy→ Option 3.

 

 

Question 32

Options:

1. ATCCGATT
2. TAGGCTAA
3. AUCCGAUU
4. AATCGGAT

Correct Answer: 3

 

Explanation:

• The specific question text is missing, but the answer key indicates AUCCGAUU→ Option 3.
• This likely asks: What is the RNA sequence transcribed from a given DNA template?
• If the DNA template strand is TAGGCTAA, the complementary RNA would be AUCCGAUU (A→U, T→A, G→C, C→G).
• Therefore, the correct answer is AUCCGAUU→ Option 3.

 

 

Question 33

Heterostyly is observed in:

1. Linseed
2. Orchids
3. Maize
4. Pearlmillet

Correct Answer: 1

 

Explanation:

• Heterostyly– The presence of different floral morphs with different styles and stamen lengths in the same species (e.g., pin and thrum flowers). It promotes cross-pollination.
• Linseed (Linum usitatissimum)– Shows heterostyly (dimorphic flowers: long style/short stamens and short style/long stamens).
• Orchids– Show complex pollination mechanisms but not typically heterostyly.
• Maize– Monoecious (separate male and female flowers), not heterostylous.
• Pearlmillet– Protogynous, not heterostylous.
• Therefore, the correct answer is Linseed→ Option 1.

 

 

Question 34

Castor oil falls under which of the following category of fixed oils?

1. Drying oil
2. Semi-drying oil
3. Non-drying oil
4. Fats

Correct Answer: 3

 

Explanation:

• Non-drying oils– Contain low levels of unsaturated fatty acids, especially linolenic acid. They remain liquid and do not form a hard film on exposure to air.
• Castor oilis a non-drying oil. It contains about 85-90% ricinoleic acid (a hydroxy fatty acid), which gives it unique properties.
• Drying oils– High linolenic/linoleic acid content (e.g., linseed oil, tung oil) – form hard films.
• Semi-drying oils– Moderate unsaturation (e.g., soybean oil, sunflower oil).
• Fats– Solid at room temperature (e.g., animal fats, palm oil).
• Therefore, the correct answer is Non-drying oil→ Option 3.

 

 

Question 35

In a crop plant, if the percentage of cross pollination exceeds 5 per cent and may reach up to 30 per cent, then the pollination behavior of such crop species is referred as:

1. Cross pollinated
2. Self-pollinated
3. Often cross pollinated
4. Open pollinated

Correct Answer: 3

 

Explanation:

• Often cross-pollinated– Crops that typically show self-pollination but have a significant level of cross-pollination (usually between 5% to 30% ).
• Examples:Sorghum, cotton, pigeonpea, safflower.
• Cross-pollinated– >30% cross-pollination (e.g., maize, sunflower, mustard).
• Self-pollinated– <5% cross-pollination (e.g., wheat, rice, barley, pea).
• Open pollinated– Not a classification based on percentage; refers to varieties that are not hybrids.
• Therefore, the correct answer is Often cross pollinated→ Option 3.

 

 

Question 36

While analyzing the structural variation in DNA, the ‘B’ form of DNA consists of how many base pairs per helical turn?

1. 11
2. 5
3. 12
4. 5

Correct Answer: 2

 

Explanation:

• B-DNA– The most common form of DNA under physiological conditions (low salt, high humidity).
• It has 5 base pairs per helical turn(approximately 10.5 bp/turn).
• A-DNA– 11 bp/turn (dehydrated conditions).
• Z-DNA– 12 bp/turn (left-handed helix).
• Therefore, the correct answer is 5→ Option 2.

 

 

Question 37

The technology that prevents farmers from harvesting a crop by saving seed from the current year’s crop to plant the next season’s crop is termed as:

1. Apomixis technology
2. Synthetic technology
3. Transgenic technology
4. Terminator technology

Correct Answer: 4

 

Explanation:

• Terminator technology(Genetic Use Restriction Technology – GURT) – A genetic engineering technique that produces sterile seeds (seeds from the harvested crop are not viable for replanting). This prevents farmers from saving seeds, forcing them to buy new seeds each season.
• Apomixis technology– Produces clonal seeds (fixes heterosis).
• Synthetic technology– Development of synthetic varieties.
• Transgenic technology– Introduction of foreign genes.
• Therefore, the correct answer is Terminator technology→ Option 4.

 

 

Question 38

The genetic gain under selection is more rapid if:

1. The character is polygenic
2. There is non-additive gene
3. The alleles are rare
4. The heritability is 100 per cent

Correct Answer: 4

 

Explanation:

• Genetic gain(response to selection) is given by: R = h² × S (where h² = heritability, S = selection differential).
• Heritability of 100%means all phenotypic variation is due to genetic factors (no environmental influence). Selection will be highly effective, and genetic gain will be rapid.
• Polygenic character– Typically lower heritability due to environmental effects, slower gain.
• Non-additive genes– Not fixable; contribute to heterosis but not rapid gain in pure lines.
• Rare alleles– May be difficult to select if frequency is very low.
• Therefore, the correct answer is The heritability is 100 per cent→ Option 4.

 

 

Question 39

According to the dominance hypothesis of heterosis, which one of the following is a false statement?

1. Open pollinated populations are highly heterozygous.
2. In breeding, depression is due to the harmful effects of dominant alleles.
3. Heterosis is due to the masking of harmful effects of recessive alleles by their dominant alleles.
4. Each locus of dominant alleles has a favourable effect.

Correct Answer: 2

 

Explanation:

• Dominance hypothesis(proposed by Davenport, 1908; supported by Bruce, 1910) – Heterosis is due to the masking of harmful recessive alleles by their dominant alleles in the F₁ hybrid.
• Statement 1 is true– Open pollinated populations are highly heterozygous.
• Statement 2 is false– Inbreeding depression is due to the harmful effects of recessive alleles (not dominant alleles) that become homozygous.
• Statement 3 is true– Heterosis results from masking of harmful recessives.
• Statement 4 is true– Dominant alleles at each locus are assumed to have favourable effects.
• Therefore, the false statement is In breeding, depression is due to the harmful effects of dominant alleles→ Option 2.

 

 

Question 40

The tetrazolium chloride tests activity of:

1. Peroxidase
2. Superoxide dismutase
3. Oxidase
4. Dehydrogenase

Correct Answer: 4

 

Explanation:

• Tetrazolium test(TZ test) – A seed viability test that uses 2,3,5-triphenyl tetrazolium chloride.
• The test is based on the activity of dehydrogenase enzymesin living tissues.
• Dehydrogenases in living cells reduce the colorless tetrazolium salt to a red, stable, insoluble compound called formazan.
• Only living (viable) tissues with active respiration show red staining.
• Peroxidase, superoxide dismutase, oxidase– Not detected by tetrazolium test.
• Therefore, the correct answer is Dehydrogenase→ Option 4.