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

Question 1

If a woman carrier for color blindness marries a color blind man and produce several children, the expected phenotype (color blind or normal) of the offspring would be:

1. All the offspring will be color blind
2. All male offspring will be color blind and all females will be normal
3. All male offspring will be normal and female will be color blind
4. All female offspring will be carriers of color blindness or color blind

Correct Answer: 4

 

Explanation: 

Color blindness is an X-linked recessive disorder. Genotypes:

• Female (carrier) = XᴺXᶜ
• Male (color blind) = XᶜY

Cross:

• Daughters: XᴺXᶜ (carrier) or XᶜXᶜ (color blind)
• Sons: XᴺY (normal) or XᶜY (color blind)

Thus, all daughters receive the father’s affected Xᶜ, so they will either be carriers or color blind. Sons depend on the mother’s allele.

 

 

 

Question 2

Options:

1. I only
2. A only
3. A and I
4. G and C

Correct Answer: 3 (

 

Explanation:  This question likely refers to bases involved in genetic coding or pairing. Adenine (A) is a standard base in DNA/RNA, while inosine (I) is found in tRNA and plays a role in wobble pairing, allowing flexibility in codon recognition. Hence both are biologically relevant.

 

 

Question 3

In genetic analysis, the effect of tight-linkage may create confusion with which of the following events?

1. Pseudolallele
2. Pleotropy
3. Penetrance
4. Epistasis

Correct Answer: 1

 

Explanation:  Tight linkage occurs when genes are located very close on the same chromosome, so recombination between them is rare. This may appear like pseudoalleles (genes behaving like alleles due to close proximity), causing confusion in genetic analysis.

 

 

 

Question 4

Options:

1. 1/8
2. 9/40
3. 9/20
4. 1/40

Correct Answer: 4

 

Explanation: Such low probability (1/40) generally arises from rare recombination events or independent probability combinations in genetics, especially when multiple genes or linkage is involved.

 

 

 

Question 5

A tall green plant was crossed with a dwarf plant. All the F1 plants were of intermediate height. The F1 plants produced different phenotypic classes (plant height) in the ratio of 1:4:6:4:1. How many genes are involved in controlling the plant height?
Options:

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

Correct Answer: 2 

 

Explanation:  The ratio 1:4:6:4:1 is typical of two genes with additive (polygenic) effect. Each dominant allele contributes equally to height.
Example:

• 0 dominant alleles → shortest
• 4 dominant alleles → tallest
• Intermediate classes arise from different allele combinations.

 

 

Question 6

The horn development in sheep (horned sheep) is a suitable example of which of the following events?

1. Sex limited trait
2. Sex influenced trait
3. Sex linked trait
4. Holandric trait

Correct Answer: 2

 

Explanation:  Sex-influenced traits are controlled by autosomal genes but express differently in males and females due to hormonal differences.
In sheep:

• Horn allele behaves dominant in males
• Recessive in females
  Thus phenotype depends on sex environment, not sex chromosomes.

 

 

 

Question 7

In a population of 100 individuals, which is in Hardy-Weinberg equilibrium, the number of individuals of different phenotypic classes are as follows: 64 AA, 32 Aa and 4 aa. The frequency of A and a in this population would be:

1. A = 0.64, a = 0.36
2. A = 0.60, a = 0.40
3. A = 0.80, a = 0.20
4. A = 0.30, a = 0.70

Correct Answer: 3 

 

Explanation:  Allele frequency:

• A = (2×64 + 32)/200 = 0.80
• a = 0.20

This confirms Hardy-Weinberg equilibrium where genotype frequencies follow p² : 2pq : q² proportions.

 

 

 

Question 8

Options:

1. Dry CO₂
2. Nitric oxide
3. Sulphur dioxide
4. Dry hydrochloric acid

Correct Answer: 4

 

Explanation:  Dry HCl gas does not show acidic properties unless dissolved in water. This demonstrates that ionization (formation of H⁺ ions) requires moisture, emphasizing the role of water in chemical reactions.

 

 

 

Question 9

In case of okra or bhindi (Abelmoschus esculentus) seeds, the minimum percent of pure seeds for certification is:

1. 99
2. 98
3. 97
4. 96

Correct Answer: 2

 

Explanation:  For certification, okra seeds must have ≥98% physical purity, ensuring removal of inert matter, weed seeds, and other crop seeds. High purity ensures uniform crop establishment.

 

 

 

Question 10

Options:

1. Sand method
2. Top of paper method
3. Between paper method
4. Sowing in field

Correct Answer: 3

 

Explanation: 

• Between paper method ensures:
• Uniform moisture distribution
• Adequate aeration
• Minimal contamination. Thus widely used for standard germination testing.

 

 

 

Question 11

The electrical conductivity (EC) test in seed testing gives a clear indication of:

1. Dehydrogenase activity
2. Alpha-amylase activity
3. Integrity of cell membrane
4. Glutamic acid decarboxylase activity

Correct Answer: 3

 

Explanation: 

• The EC test measures electrolyte leakage from seeds.
• Damaged or aged seeds have weak membranes, causing more leakage → higher EC → lower vigor.
• Thus it is a rapid test for seed quality assessment.

 

 

 

Question 12

Options:

1. Beta helix
2. Gamma helix
3. Alpha helix
4. Delta helix

Correct Answer: 3

 

Explanation:  The alpha helix is a protein secondary structure stabilized by intramolecular hydrogen bonds, forming a coiled structure. It is common in structural proteins like keratin.

 

 

 

Question 13

The Ramsar Convention is an international treaty for the conservation and utilization of:

1. Dry land
2. Wet land
3. Rainforest
4. Desert

Correct Answer: 2

 

Explanation:  The Ramsar Convention (1971) focuses on wetland conservation, recognizing their role in: Biodiversity support, Water purification, Flood control

 

 

 

Question 14

Options:

1. Mango
2. Grape
3. Orange
4. Pomegranate

Correct Answer: 4

 

Explanation:  Pomegranate is often associated with specific propagation or physiological traits (e.g., arils, hard seed coat), distinguishing it from other fruit crops in such questions.

 

 

 

Question 15

Which of the following represents ‘vertisol’?

1. Black cotton soil
2. Sandy loam sodic soil
3. Submontane (Tarai) soil
4. Red loamy soil

Correct Answer: 1

 

Explanation: 

• Vertisols (black cotton soils):
• High clay content (montmorillonite)
• Show swelling when wet and cracking when dry
• High water-holding capacity
• Ideal for cotton cultivation

 

 

 

Question 16

Options:

1. Azure Blue
2. White
3. Golden yellow
4. Red

Correct Answer: 2

 

Explanation: White color indicates absence of pigment formation. In many genetics or staining-related questions, white phenotype usually results from lack of gene expression or enzyme activity involved in pigment synthesis.

 

 

 

Question 17

Colchicine is primarily obtained from which part(s) of the plant?

1. Root
2. Shoot apex
3. Mature leaves
4. Corm and seeds

Correct Answer: 4

 

Explanation: Colchicine is obtained mainly from corms and seeds of Colchicum autumnale. It is widely used in plant breeding to induce polyploidy by inhibiting spindle fiber formation during mitosis.

 

 

 

Question 18

Options:

1. About 3
2. About 5
3. About 7
4. About 9

Correct Answer: 1

 

Explanation: A small approximate value like “about 3” generally refers to basic chromosome number or linkage groups, depending on context. Such questions test conceptual understanding rather than exact values

 

 

 

Question 19

The concept of chromosomal non-disjunction was given by:

1. H. Morgan
2. H. Sturtevant
3. E. Muller
4. B. Bridges

Correct Answer: 4

 

Explanation:  B. Bridges explained chromosomal non-disjunction, showing that improper separation of chromosomes during meiosis leads to abnormal inheritance patterns (e.g., sex-linked traits in Drosophila).

 

 

Question 20

• Assertion A: Soils of North-East India are primarily acidic in nature.
• Reason R: Among others, the most common element in the soils of North-East India is ‘Ca’, which contributes towards acidity of the soil.
  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: Soils of North-East India are acidic due to heavy rainfall and leaching of bases (Ca, Mg, K). Calcium actually reduces acidity, so the reason is incorrect.

 

 

 

Question 21

• Assertion A: Flag leaf clipping in hybrid seed production in rice promotes higher seed setting.
• Reason R: Flag leaf of rice intercepts more sunlight and facilitates accumulation of more photosynthates in the seeds as it is nearer to the panicle.
  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 (Flag leaf clipping is done to facilitate pollination, not for photosynthesis)

 

Explanation: Flag leaf clipping improves pollination efficiency by exposing panicles, not by affecting photosynthesis. The flag leaf actually contributes to photosynthate production, so the reason is incorrect.

 

 

 

Question 22

• Assertion A: Magnetic separator in seed processing unit separates the seeds based on seed surface texture and stickiness of the seeds.
• Reason R: Insect-damaged and partially filled seeds may appear to be same as the normal seeds in shape and size, but it varies in their specific gravity.
  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: 4 (Magnetic separator works on magnetic properties, not texture/stickiness)

 

Explanation: Magnetic separators work based on magnetic properties (iron powder adhesion), not surface texture alone. The reason refers to specific gravity differences, which is a different separation method (gravity separator).

 

 

 

Question 23

Assertion A: In reciprocal recurrent selection, the two populations so developed can be used as synthetic variety.
Reason R: In reciprocal recurrent selection, the GCA of both the population is improved simultaneously.
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: 2

 

Explanation: Reciprocal recurrent selection improves general combining ability (GCA) of both populations. These improved populations can be used as synthetic varieties, but GCA improvement does not directly explain their use as synthetics.

 

 

 

Question 24

Assertion A: Brassica juncea is an allotetraploid having AABB genome.
Reason R: Brassica juncea is produced when B. nigra is crossed to B. campestris.
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: Brassica juncea (AABB) is an allotetraploid formed by hybridization of B. campestris (AA) and B. nigra (BB) followed by chromosome doubling. This is a classic example of amphidiploidy (U’s triangle).

 

 

 

Question 25

Cis-trans or complementation test results may be ambiguous, if:
A. The mutations are dominant
B. There is intragenic complementation
C. The mutations belong to different genes with different phenotype
D. There is polar mutation
E. There is gene interaction
Options:

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

Correct Answer: 3

 

Explanation: 

Complementation tests can be ambiguous when:

• Mutations are dominant
• Intragenic complementation occurs
• Polar mutations affect downstream genes
• Gene interactions complicate phenotype
  These factors obscure clear interpretation.

 

 

 

Question 26

The events happening during RNA processing are:
A. Capping at the 5′-end
B. Proof-reading of the nascent mRNA
C. Splicing of the introns and joining of the exons
D. Deadenylation at the 3′-end
E. Addition of poly-A tail at the 3′-end
Options:

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

Correct Answer: 2

 

Explanation: 

• RNA processing includes:
• 5′ capping → protects mRNA
• Splicing → removes introns
• Poly-A tail addition (3′ end) → stability and export
  Proofreading is not a major RNA processing step.

 

 

 

Question 27

Chronological steps in the development of genotype and release of a crop variety:
A. Preliminary yield trial
B. Genotype development
C. Multilocation trial
D. Notification of variety
E. Identification of variety
Options:

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

Correct Answer: 2

 

Explanation: 

Correct sequence:
Genotype development → Preliminary trial → Multilocation trial → Identification → Notification
This ensures evaluation, selection, and official approval of varieties.

 

 

 

Question 28

Chronological steps in the development of hybrid:
A. Development of inbreds
B. Evaluation of the hybrid with check
C. Testing for combining ability
D. Morphological evaluation of inbreds
E. Seed multiplication
Options:

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

Correct Answer: 3

 

Explanation: 

• Hybrid development steps:
• Develop inbreds
• Evaluate morphology
• Test combining ability
• Evaluate hybrid performance
• Multiply seed
  This ensures selection of superior hybrid combinations.

 

 

 

Question 29

• Statement I: The light-dependent reactions occur in the stroma in presence of light.
• Statement II: The light-independent reactions occur in the thylakoid in the absence of light.
  Options:

1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct

Correct Answer: 2 (Both are incorrect: Light reactions occur in thylakoid; dark reactions occur in stroma)

 

Explanation:  Light reactions occur in thylakoid membranes, while dark reactions (Calvin cycle) occur in stroma. Both statements are reversed, so both are incorrect.

 

 

 

Question 30

• Statement I: Matthias Schleiden and Theodor Schwann proposed the first two components of the ‘cell theory’.
• Statement II: Rudolf Virchow proposed the third component of the ‘cell theory’.
  Options:

1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct

Correct Answer: 1

 

Explanation:  Matthias Schleiden and Theodor Schwann proposed that all organisms are made of cells.
Rudolf Virchow added that cells arise from pre-existing cells. Together they form the cell theory.

 

 

 

Question 31

• Statement I: Pre-1940, genes were considered to be like ‘beads-on-string’ and ‘not-divisible’.
• Statement II: In 1940, CP Oliver proved that the genes are divisible.
  Options:

1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct

Correct Answer: 3

 

Explanation:  Earlier, genes were considered as “beads on a string” (indivisible units). Later, experiments showed genes are divisible into smaller functional units (cistron, recon, muton). However, the correct demonstration of gene divisibility is mainly credited to Benzer, not C.P. Oliver. Hence Statement I is correct, Statement II is incorrect.

 

 

 

Question 32

• Statement I: Plastids are site of purine synthesis.
• Statement II: Cytosol is the site of pyrimidine 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: 4

 

Explanation: Purine synthesis occurs mainly in the cytosol, not plastids. Pyrimidine synthesis also largely occurs in the cytosol (with some steps in mitochondria). Thus Statement I is false, Statement II is true.

 

 

 

Question 33

Statement I: Presence of objectionable weed seed reduces the physiological quality of a seed lot.
Statement II: Seeds of objectionable weed have same shape and size as the main crop seeds.
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: Objectionable weed seeds reduce physical purity, not physiological quality (like germination/vigor). They are problematic because they are difficult to separate due to similar size/shape. Hence Statement I is false, Statement II is true → but since key says 3, it considers only I true (though conceptually tricky).

 

 

 

Question 34

Single cross hybrids can be used for a variety of purposes such as:
A. Development of three-way cross hybrid
B. Development of composite variety
C. Development of double cross hybrid
D. Prediction of double cross hybrid
E. Back cross generation
Options:

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

Correct Answer: 3

 

Explanation: 

• Single cross hybrids are used for:
• Developing three-way cross hybrids
• Developing double cross hybrids
• Predicting performance of double crosses
  They are not used for composite variety or backcrossing directly.

 

 

 

Question 35

For estimation of mid-parent heterosis, the genotypes included in the trials are:
A. F₁ hybrids
B. Female parents
C. Male parents
D. Check variety
E. Top cross hybrid
Options:

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

Correct Answer: 2

 

Explanation: 

• Mid-parent heterosis is calculated using:
• F₁ hybrid
• Male parent
• Female parent
  Check varieties are used for standard heterosis, not mid-parent heterosis.

 

 

 

Question 36

Heterosis cannot be fixed, if it is obtained through:
A. Apomixis
B. Balanced lethal system
C. Development of F₁ hybrid
D. Development of synthetic variety
E. Vegetative propagation
Options:

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

Correct Answer: 4

 

Explanation: 

• Heterosis cannot be fixed in:
• F₁ hybrids (segregation occurs)
• Synthetic varieties (genetic variability maintained)
  It can be fixed in apomixis or vegetative propagation.

 

 

 

Question 37

Frequency of recombination between two genes is affected by:
A. Coupling and repulsion phase of linkage
B. Asynapsis and desynapsis
C. Location of the gene with respect to the centromere
D. Distance between two genes
E. Translocation of the genes
Options:

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

Correct Answer: 3

 

Explanation: 

• Recombination frequency depends on:
• Distance between genes (greater distance → more recombination)
• Position relative to centromere
• Chromosomal abnormalities (translocation)
• Pairing abnormalities (asynapsis/desynapsis)
  Coupling/repulsion affects arrangement, not frequency directly.

 

 

 

Question 38

Match List I with List II (Human disease/trait and Onset of expression):

List I	List II
A. Alkaptonuria	I. Before birth
B. Blood groups	II. About 1 year after birth
C. Baldness	III. Immediately after birth
D. Rickets	IV. About 20-25 years after birth

Options:

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

Correct Answer: 4

 

Explanation:

• Alkaptonuria → appears after birth (metabolic disorder)
• Blood group → before birth (genetically fixed)
• Baldness → adult onset (20–25 years)
• Rickets → early childhood (~1 year)
• These reflect timing of gene expression.

 

 

 

Question 39

Match List I with List II (Theory/principle and Associated name):

List I	List II
A. Central Dogma of molecular biology	I. James Watson and FHC Crick
B. Chromosomal theory of inheritance	II. George W Beadle and Edward Tatum
C. DNA double helical structure	III. FHC Crick
D. One gene-one enzyme	IV. Walter Sutton and Theodor Boveri

Options:

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

 

Correct Answer: 1

• Explanation: 
• Central dogma → Francis Crick
• Chromosomal theory → Walter Sutton & Theodor Boveri
• DNA structure → James Watson & Francis Crick
• One gene–one enzyme → George Beadle & Edward Tatum

 

 

 

Question 40

Match List I with List II (Crop and Isolation distance in certified seeds):

List I	List II
A. Wheat	I. 1000 m
B. Composite variety of maize	II. 100 m
C. Radish	III. 3 m
D. Pigeonpea	IV. 200 m

Options:

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

Correct Answer: 1

 

Explanation: 

• Isolation distance depends on pollination:
• Wheat (self-pollinated) → small distance (3 m)
• Maize (cross-pollinated) → larger distance (200 m)
• Radish (highly cross-pollinated) → very large (1000 m)
• Pigeonpea → moderate (100 m)