ICAR JRF Plant Science Practice Series Memory Based 2024 (Module 1) (41 – 60 MCQ) 

 

Question 41 

Given below are two statements:

• Statement I: Potato is essentially a crop of cool moist regions and grows best in a climate where cool nights alternate with warm days during the period of tuber formation.
• Statement II: Irrigation or rainfall at or just before the initiation of tuber formation will result in decrease of the number of tuber sets.

Options:

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

Correct Answer: 3

 

Explanation:

• Statement I is true:Potato is a cool-season crop. It grows best at temperatures of 15–20°C. Cool nights (10–15°C) and warm days (20–25°C) favor tuber formation. High temperatures (>30°C) inhibit tuberization and promote vegetative growth.
• Statement II is false:Irrigation or rainfall at or just before tuber initiation actually increases the number of tuber sets by providing adequate soil moisture for stolon development and tuber initiation. Water stress at this stage reduces tuber number. However, waterlogging should be avoided.
• Therefore, Statement I is correct but Statement II is false → Option 3.

 

 

Question 42 

Which disease greatly stimulated interest in the cause and control of plant disease?

1. Downy mildew of grapevine
2. Early blight of potato
3. Late blight of potato
4. Wheat rust disease

Correct Answer: 3

 

Explanation:

• Late blight of potato(caused by Phytophthora infestans) was responsible for the Irish Potato Famine (1845–1852) , which led to the death of over 1 million people and mass emigration.
• This catastrophic event greatly stimulated interestin understanding the cause and control of plant diseases.
• It led to the development of plant pathologyas a scientific discipline.
• Anton de Bary(1861) proved that Phytophthora infestans was the causal agent, establishing the germ theory of plant disease.
• Other options:
  • Downy mildew of grapevine– Led to the discovery of Bordeaux mixture (fungicide) but did not have the same historical impact on the field.
  • Early blight of potato– A less severe disease.
  • Wheat rust disease– Important but did not cause the same level of societal disruption as late blight.
• Therefore, the correct answer is Late blight of potato→ Option 3.

 

 

Question 43 

Which one of the following fungus is Perithecial ascomycetes?

1. Erysiphe
2. Uncinula
3. Podosphaera
4. Gibberella

Correct Answer: 4

 

Explanation:

• Perithecium(plural: perithecia) is a flask-shaped fruiting body (ascocarp) of certain ascomycetes, with an opening (ostiole) through which ascospores are released.
• Gibberella– A genus of ascomycete fungi that produces perithecia. It is the sexual stage (teleomorph) of Fusarium. Example: Gibberella fujikuroi (causes Bakanae disease of rice).
• Erysiphe, Uncinula, Podosphaera– These are powdery mildew fungi. They produce cleistothecia (closed, spherical fruiting bodies with no opening, not perithecia).
• Therefore, the correct answer is Gibberella→ Option 4.

 

 

Question 44 

Polycyclic diseases are usually caused by pathogens adapted:

1. on foliar plant parts
2. in soil
3. in rhizosphere
4. in phloem

Correct Answer: 1

 

Explanation:

• Polycyclic diseases– Diseases that have multiple infection cycles within a single growing season. The pathogen reproduces rapidly and produces multiple generations of inoculum.
• Foliar pathogens(e.g., rusts, powdery mildews, late blight, leaf spots) are typically polycyclic because they:
  • Produce abundant spores (e.g., urediniospores, conidia) that are easily dispersed by wind.
  • Have short latent periods (time between infection and spore production).
  • Can infect new leaves repeatedly during the season.
• Soilborne pathogens(e.g., Fusarium, Verticillium, Sclerotinia) are usually monocyclic (one infection cycle per season).
• Rhizosphere pathogens– Typically monocyclic or with limited cycles.
• Phloem pathogens(e.g., phytoplasmas, Candidatus Liberibacter) – Often have long latent periods and are not typically polycyclic.
• Therefore, polycyclic diseases are usually caused by pathogens adapted on foliar plant parts→ Option 1.

 

 

Question 45 

Options:

1. Crown gall
2. Die back
3. Leaf spot
4. Blight

Correct Answer: 1

 

Explanation:

• The question likely asks for a disease caused by Agrobacterium tumefaciensor a bacterial disease characterized by tumor formation.
• Crown gall– Caused by Agrobacterium tumefaciens (now Rhizobium radiobacter). It is characterized by the formation of galls (tumors) at the crown (root-stem junction) and on roots.
• Die back– Progressive death of shoots from the tip downward, caused by various fungi (e.g., Botryodiplodia, Phomopsis).
• Leaf spot– Caused by many fungi and bacteria; characterized by necrotic spots on leaves.
• Blight– Rapid, extensive necrosis of leaves, stems, or flowers (e.g., late blight, fire blight).
• Based on the answer key, Crown gallis correct → Option 1.

 

 

Question 46 

Which of the following vascular wilt is transmitted by beetles?

1. Bacterial wilt of tomato
2. Fungal wilt of banana
3. Bacterial wilt of banana
4. Fungal wilt of date palm

Correct Answer: 2

 

Explanation:

• Fungal wilt of banana– Caused by Fusarium oxysporum sp. cubense (Panama wilt). It is transmitted by beetles (e.g., banana weevil Cosmopolites sordidus, and other insects) that can carry fungal spores from infected to healthy plants. Also spread by infected suckers, soil, and water.
• Bacterial wilt of tomato– Caused by Ralstonia solanacearum. Primarily soilborne, transmitted through water, infected plant debris, and contaminated tools; not typically by beetles.
• Bacterial wilt of banana– Caused by Ralstonia solanacearum (Moko disease). Transmitted by insects (including beetles), but the question asks for “fungal wilt” specifically.
• Fungal wilt of date palm– Caused by Fusarium oxysporum sp. albedinis (Bayoud disease). Transmitted by soil and infected offshoots; beetle transmission is not primary.
• Therefore, the correct answer is Fungal wilt of banana→ Option 2.

 

 

Question 47 

Options:

1. Phytophthora capsici
2. Pythium ultimum
3. Phytophthora ramorum
4. Plasmodia viticola

Correct Answer: 4

 

Explanation:

• Plasmodia viticola– Likely a misspelling of Plasmopara viticola, the causal agent of downy mildew of grapevine (an Oomycete). The question likely asks for a downy mildew pathogen.
• Phytophthora capsici– Causes blight and root rot in peppers, cucurbits, and other crops.
• Pythium ultimum– Causes damping-off and root rot in many plants.
• Phytophthora ramorum– Causes sudden oak death and ramorum blight.
• Based on the answer key, Plasmodia viticola(presumably Plasmopara viticola) is correct → Option 4.

 

 

Question 48 

Which one of the following bacterial pathogenic genus is classified under Gram-negative aerobic and microaerophilic rods?

1. Acidovorax
2. Erwinia
3. Pantoea
4. Enterobacter

Correct Answer: 1

 

Explanation:

• Acidovorax– A genus of Gram-negative, aerobic Some species are microaerophilic. It includes plant pathogens (e.g., Acidovorax citrulli causes bacterial fruit blotch of watermelon).
• Erwinia– Gram-negative, facultatively anaerobic (not strictly aerobic/microaerophilic). Causes soft rot, fire blight.
• Pantoea– Gram-negative, facultatively anaerobic. Some species are plant pathogens (e.g., Pantoea agglomerans).
• Enterobacter– Gram-negative, facultatively anaerobic. Mostly opportunistic pathogens of humans, not typically plant pathogens.
• Therefore, the genus classified under Gram-negative aerobic and microaerophilic rodsis Acidovorax → Option 1.

 

 

Question 49 

Options:

1. Tospovirus
2. Tobravirus
3. Prunus necrotic ringspot virus
4. Papaya ringspot virus

Correct Answer: 2

 

Explanation:

• The question likely asks for a virus genus that is transmitted by nematodes.
• Tobravirus– A genus of plant viruses (e.g., Tobacco rattle virus, Pea early browning virus). They are transmitted by nematodes (trichodorid nematodes).
• Tospovirus– Transmitted by thrips (insects), not nematodes.
• Prunus necrotic ringspot virus– Transmitted by pollen, seeds, and grafting; not by nematodes.
• Papaya ringspot virus– Transmitted by aphids (insects), not nematodes.
• Therefore, the correct answer is Tobravirus→ Option 2.

 

 

Question 50 

In which host-patho system does the Gene-for-Gene hypothesis operate?

1. Host – Necrotrophic interaction
2. Host – Saprophytic interaction
3. Host – Biotrophic interaction
4. Host – Obligate Saprophytic interaction

Correct Answer: 3

 

Explanation:

• The Gene-for-Gene hypothesiswas proposed by H. Flor (1942) based on his studies of flax rust (caused by Melampsora lini).
• It operates in host-biotrophic pathogen interactions(specifically obligate biotrophs like rusts, powdery mildews, and downy mildews).
• Key concept:For each resistance gene in the host, there is a corresponding avirulence (Avr) gene in the pathogen. The interaction is specific: resistance occurs only when the host R gene matches the pathogen Avr gene.
• Biotrophic pathogensdepend on living host tissue for growth and reproduction. They have co-evolved with their hosts, leading to gene-for-gene relationships.
• Necrotrophic pathogens(Option 1) kill host tissue and generally do not follow gene-for-gene interactions (they follow inverse gene-for-gene or other models).
• Saprophytic interactions(Option 2 and 4) involve organisms that feed on dead organic matter; they are not pathogens and do not involve gene-for-gene relationships.
• Therefore, the correct answer is Host – Biotrophic interaction→ Option 3.

 

 

Question 51 

Statement I: Many viral diseases are transmitted by insect vectors.
Statement II: Few insects transmit bacterial diseases.

Options:

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

Correct Answer: 1

 

Explanation:

• Statement I is true:A large number of plant viruses are transmitted by insect vectors, including aphids, whiteflies, leafhoppers, planthoppers, thrips, and beetles. Insect transmission is the most common mode of virus spread in nature.
• Statement II is true:Few insects transmit bacterial diseases compared to viruses. However, some bacterial pathogens are transmitted by insects:
  • Erwinia amylovora(fire blight) – transmitted by bees, ants, and flies.
  • Ralstonia solanacearum(bacterial wilt) – transmitted by bees and other insects.
  • Xylella fastidiosa– transmitted by leafhoppers and sharpshooters.
  • Pantoea stewartii(Stewart’s wilt of maize) – transmitted by flea beetles (Chaetocnema pulicaria).
• The word “few” is accurate because bacterial transmission by insects is less common than viral transmission.
• Therefore, both statements are true → Option 1.

 

 

Question 52 

Statement I: Spatial and temporal transmission of plant pathogens are in general guided by asexual and sexual spores.
Statement II: Bacterial pathogens produce asexual cells.

Options:

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

Correct Answer: 1

 

Explanation:

• Statement I is true:Spatial transmission (spread from plant to plant) and temporal transmission (spread over time) of fungal and oomycete pathogens are primarily mediated by asexual spores (e.g., conidia, urediniospores, zoospores) for rapid spread and sexual spores (e.g., ascospores, oospores, basidiospores) for survival and long-distance dispersal.
• Statement II is true:Bacterial pathogens reproduce by binary fission (asexual reproduction), producing genetically identical asexual cells. They do not have sexual reproduction in the conventional sense (though some bacteria undergo genetic recombination through conjugation, transformation, and transduction, which are not sexual reproduction).
• Therefore, both statements are true → Option 1.

 

 

Question 53 

Statement I: Witches broom symptoms are caused by phytoplasmas.
Statement II: Nutrient deficiency also causes witches broom in crop plants.

Options:

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

Correct Answer: 1

 

Explanation:

• Statement I is true:Witches broom (abnormal proliferation of shoots with small leaves) is a classic symptom caused by phytoplasmas (wall-less bacteria, Mollicutes) in many crops (e.g., sesame, brinjal, groundnut, coconut).
• Statement II is true:Nutrient deficiencies (especially zinc deficiency and sometimes boron or copper deficiency) can also cause witches broom-like symptoms, characterized by stunted growth, shortened internodes, and proliferation of small shoots.
• Examples:Zinc deficiency in fruit trees (e.g., apple, pear, citrus) causes “little leaf” and rosetting, resembling witches broom.
• Therefore, both statements are true → Option 1.

 

 

Question 54 

Statement I: Xanthomonas causes disease on fibre plant parts.
Statement II: Xanthomonas can cause vascular wilt in rice.

Options:

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

Correct Answer: 3

 

Explanation:

• Statement I is true:Xanthomonas species cause diseases on fibre plant parts:
  • Xanthomonas citrimalvacearum – causes bacterial blight of cotton (affects bolls, stems, and leaves).
  • Xanthomonas campestrismalvacearum – affects cotton fibres.
• Statement II is false:Xanthomonas does not cause vascular wilt in rice. The vascular wilt of rice is caused by Ralstonia solanacearum (bacterial wilt). Xanthomonas oryzae oryzae causes bacterial leaf blight of rice (not vascular wilt; it causes leaf blight and can cause kresek symptoms in seedlings, but this is not true vascular wilt).
• Therefore, Statement I is correct but Statement II is false → Option 3.

 

 

Question 55 

Statement I: Leaf curl is one of the symptoms of viral disease.
Statement II: Fungi can also cause leaf curl symptoms.

Options:

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

Correct Answer: 1

 

Explanation:

• Statement I is true:Leaf curl is a classic symptom of several viral diseases:
  • Tomato leaf curl virus(Geminivirus) – causes upward curling of leaves.
  • Cotton leaf curl virus(Geminivirus) – causes leaf curling and enations.
  • Papaya leaf curl virus– causes downward curling of leaves.
• Statement II is true:Fungi can also cause leaf curl symptoms:
  • Peach leaf curl– caused by the fungus Taphrina deformans, which causes curling, puckering, and reddening of peach leaves.
  • Other Taphrinaspecies cause leaf curl on other trees (e.g., Taphrina ulmi on elm).
• Therefore, both statements are true → Option 1.

 

 

Question 56 

Match List I with List II:

List I	List II
A. Tumor	1. Xanthomonas
B. Wilt	2. Pectobacterium
C. Spot	3. Pantoea
D. Rot	4. Agrobacterium

Options:

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

Correct Answer: 1

 

Explanation:

• Tumor (A)– Caused by Agrobacterium (e.g., Agrobacterium tumefaciens causes crown gall). → A (IV)
• Wilt (B)– Caused by Pantoea (e.g., Pantoea stewartii causes Stewart’s wilt of maize). Also Ralstonia causes bacterial wilt, but Pantoea is in the list. → B (III)
• Spot (C)– Caused by Xanthomonas (e.g., Xanthomonas campestris vesicatoria causes bacterial leaf spot of tomato and pepper). → C (I)
• Rot (D)– Caused by Pectobacterium (e.g., Pectobacterium carotovorum causes soft rot). → D (II)
• Therefore, correct matching is A (IV), B (III), C (I), D (II)→ Option 1.

 

 

Question 57 

Match List I with List II:

List I	List II
A. White rust	I. Puccinia striiformis
B. Black rust	II. Albugo candida
C. Yellow rust	III. Cephalerous parasiticus
D. Red Rust	IV. Puccinia tritici

Options:

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

Correct Answer: 2

 

Explanation:

• White rust (A)– Caused by Albugo candida (an Oomycete, not a true rust fungus). → A (II)
• Black rust (B)– Caused by Puccinia tritici (also called Puccinia graminis sp. tritici), causes black stem rust of wheat. → B (IV)
• Yellow rust (C)– Caused by Puccinia striiformis (yellow/stripe rust of wheat). → C (I)
• Red rust (D)– Caused by Cephalerous parasiticus (red rust of tea). → D (III)
• Therefore, correct matching is A (II), B (IV), C (I), D (III)→ Option 2.

 

 

Question 58 

Naked asci can be observed in which ascomycete fungus?

1. Erysiphe
2. Claviceps
3. Taphrina
4. Uncinula

Correct Answer: 3

 

Explanation:

• Naked asciare asci that are not enclosed within a fruiting body (ascocarp) . They are exposed directly on the host surface.
• Taphrina– A genus of ascomycete fungi that causes leaf curl, witches broom, and other deformities. It produces naked asci (exposed asci) on the surface of infected plant tissues, with no ascocarp (perithecium, cleistothecium, or apothecium). Example: Taphrina deformans (peach leaf curl).
• Erysiphe– Produces cleistothecia (closed, spherical fruiting bodies), not naked asci.
• Claviceps– Produces perithecia embedded in the stroma (ergot body), not naked asci.
• Uncinula– A powdery mildew fungus that produces cleistothecia with appendages, not naked asci.
• Therefore, the correct answer is Taphrina→ Option 3.

 

 

Question 59 

Match List I with List II:

List I	List II
A. Fusarium oxysporum	I. Microsclerotia
B. Macrophomina phaseolina	II. Chlamydospores
C. Bipolaris sorokiniana	III. Thick-walled conidia
D. Ralstonia solanacearum	IV. Naked cells

Options:

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

Correct Answer: 3

 

Explanation:

• Fusarium oxysporum(A) – Produces chlamydospores (thick-walled, survival spores) that can survive in soil for many years. → A (II)
• Macrophomina phaseolina(B) – Produces microsclerotia (dark, compact masses of melanized hyphae) as survival structures in soil and plant debris. → B (I)
• Bipolaris sorokiniana(C) – Produces thick-walled conidia (also called dictyospores) that can survive under adverse conditions. → C (III)
• Ralstonia solanacearum(D) – A bacterial pathogen; produces naked cells (no spores, no specialized survival structures, but can survive in soil, water, and plant debris). → D (IV)
• Therefore, correct matching is A (II), B (I), C (III), D (IV)→ Option 3.

 

 

Question 61 

Match List I with List II (Pathogen and Mode of Entry):

List I	List II
A. Magnaporthe oryzae	I. Stomata
B. Puccinia tritici	II. Hydathodes
C. Xanthomonas axonopodis	III. Wound
D. Agrobacterium tumefaciens	IV. Direct penetration

Options:

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

Correct Answer: 1

 

Explanation:

• Magnaporthe oryzae(A) – Causes rice blast. Penetrates host tissue directly through the cuticle using an appressorium and penetration peg. → A (IV)
• Puccinia tritici(B) – Causes wheat rust. Enters through stomata (urediniospores germinate and enter via stomatal openings). → B (I)
• Xanthomonas axonopodis(C) – Causes citrus canker. Enters through hydathodes (natural openings at leaf margins) and also through stomata and wounds. → C (II)
• Agrobacterium tumefaciens(D) – Causes crown gall. Enters through wounds (requires wound site for infection). → D (III)
• Therefore, correct matching is A (IV), B (I), C (II), D (III)→ Option 1.

 

Question 62 

Match List I with List II (Fungicide and Chemical Group):

List I	List II
A. Dithiocarbamate	I. Metalaxyl
B. Oxathiin	II. Fosetyl-Al
C. Acylaniline	III. Mancozeb
D. Organophosphate	IV. Carboxin

Options:

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

Correct Answer: 3

 

Explanation:

• Dithiocarbamate (A)– A group of multi-site contact fungicides. Mancozeb is a dithiocarbamate (also maneb, zineb, etc.). → A (III)
• Oxathiin (B)– A group of systemic fungicides. Carboxin is an oxathiin fungicide (used for seed treatment against smuts and rusts). → B (IV)
• Acylaniline (C)– A group of systemic fungicides. Metalaxyl (an acylaniline) is effective against Oomycetes (downy mildews, late blight). → C (II)
• Organophosphate (D)– Fosetyl-Al (aluminum tris-O-ethylphosphonate) is an organophosphate fungicide effective against Oomycetes. → D (I)
• Therefore, correct matching is A (III), B (IV), C (II), D (I)→ Option 3.

 

 

Question 63 

Match List I with List II (Detection Method and Target Organism):

List I	List II
A. Baiting	I. Nematodes
B. Ooze test	II. Bacteria
C. ELISA	III. Fungi
D. Baermann-funnel technique	IV. Virus

Options:

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

Correct Answer: 2

 

Explanation:

• Baiting (A)– A method to isolate fungi (and Oomycetes) from soil or water using baits (e.g., seeds, plant parts). Used for detecting soilborne fungal pathogens. → A (III)
• Ooze test (B)– A simple test for bacterial pathogens (e.g., Ralstonia solanacearum). Bacterial ooze is observed when cut stems are suspended in water. → B (II)
• ELISA (C)– Enzyme-Linked Immunosorbent Assay. Widely used to detect viruses (also bacteria and fungi, but most common for viruses). → C (IV)
• Baermann-funnel technique (D)– A method for extracting nematodes from soil or plant tissue using a funnel, water, and heat. → D (I)
• Therefore, correct matching is A (III), B (II), C (IV), D (I)→ Option 2.

 

 

Question 64 

Two genes are linked or not can be ascertained by:

1. F-test
2. t-test
3. Chi-square test
4. Z-test

Correct Answer: 3

 

Explanation:

• Chi-square (χ²) test– Used to determine whether the observed phenotypic ratios in a genetic cross deviate significantly from expected Mendelian ratios.
• For linkage analysis: If two genes are independently assorting (not linked), the expected dihybrid test cross ratio is 1:1:1:1. The chi-square test is applied to the observed offspring numbers to see if they fit the independent assortment ratio. A significant deviation indicates linkage.
• F-test– Used for comparing variances (e.g., ANOVA).
• t-test– Used for comparing means of two groups.
• Z-test– Used for comparing proportions when sample size is large.
• Therefore, the correct answer is Chi-square test→ Option 3.

 

 

Question 65 

Foliar plant pathogens are pigmented in nature and cause crop losses in major staple crops in the world as:
A. Pigments help them to tolerate high radiation
B. They cause diseases only on monocots
C. They cause diseases only on dicots
D. They move from one plant to another plant with the help of wind
E. They cause polycyclic diseases

Options:

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

Correct Answer: 4

 

Explanation:

• A is correct:Pigments (especially melanin) in foliar pathogens help them tolerate high radiation (UV protection) and desiccation.
• B is incorrect:Foliar pathogens cause diseases on both monocots and dicots, not only monocots.
• C is incorrect:Foliar pathogens cause diseases on both monocots and dicots, not only dicots.
• D is correct:Many foliar pathogens produce spores that are dispersed by wind (e.g., urediniospores of rusts, conidia of powdery mildews), moving from plant to plant.
• E is correct:Foliar pathogens typically cause polycyclic diseases (multiple infection cycles per season), leading to rapid epidemic development.
• Therefore, the correct statements are A, D, and E→ Option 4.

 

 

Question 66 

Certain plant pathogens do not live at the site of the symptom or symptomate plant niche:
A. They only transfer virulent factors to host
B. Plant attracts the virulence factors by its chemical signals
C. This kind of pathogen survive in the rhizosphere of the infected plant
D. Virulence factors can be cultured on media
E. Infected plant can be recovered with time

Options:

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

Correct Answer: 4

 

Explanation:

• This question describes pathogens that do not colonize the symptomatic tissue but deliver effectors/virulence factorsinto the host (e.g., some bacteria, fungi, or nematodes).
• A is correct:They only transfer virulence factors (effectors) to the host, often through specialized secretion systems (e.g., Type III secretion system in bacteria).
• B is correct:The plant attracts the virulence factors by its chemical signals (e.g., root exudates, flavonoids) that trigger pathogen gene expression.
• C is correct:This kind of pathogen can survive in the rhizosphere of the infected plant (e.g., Ralstonia solanacearum survives in soil and rhizosphere).
• D is incorrect:Virulence factors themselves are proteins or small molecules that may not be culturable on media; the pathogen can be cultured, but the virulence factors are not independently culturable.
• E is incorrect:The infected plant may not fully recover; symptoms often persist or worsen.
• Therefore, the correct statements are A, B, and C→ Option 4.

 

 

Question 67 

Rishitin is a phytoalexin produced by:

1. Cotton
2. Potato
3. Pepper
4. Pea

Correct Answer: 2

 

Explanation:

• Rishitinis a phytoalexin (antimicrobial compound produced by plants in response to pathogen attack).
• It is produced by potato(Solanum tuberosum) in response to infection by Phytophthora infestans (late blight) and other pathogens.
• Rishitin is a sesquiterpenoid phytoalexin that accumulates in potato tubers and leaves following pathogen infection or elicitor treatment.
• Other phytoalexins:
  • Cotton– Gossypol.
  • Pepper– Capsidiol.
  • Pea– Pisatin.
• Therefore, the correct answer is Potato→ Option 2.

 

 

Question 68

Given below are two statements, one is labelled as Assertion (A) and the other as Reason (R).

• Assertion (A): Vertical resistance is widely used in crop improvement programmes, but it is often short-lived in the field.
• Reason (R): Vertical resistance is controlled by a single (monogenic) gene and can be easily broken when the pathogen evolves a new race.

In light of the above statements, choose the correct answer from the options given below:

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:

• Assertion (A) is true:Vertical resistance (also called race-specific resistance) is widely used in crop breeding because it provides complete (high-level) resistance and is easy to select. However, it is often short-lived in the field because the pathogen population can evolve new races that overcome the resistance.
• Reason (R) is true and correctly explains (A):Vertical resistance is controlled by monogenic (single gene) inheritance (major R genes). The pathogen can overcome this resistance by a simple mutation in the corresponding avirulence (Avr) gene, leading to the evolution of a new race that breaks the resistance. This is the gene-for-gene interaction. The monogenic nature makes it vulnerable to breakdown.
• Key concept:In contrast, horizontal resistance (polygenic, race-nonspecific) is more durable but harder to select and provides only partial resistance.
• Therefore, both (A) and (R) are true, and R is the correct explanation of A→ Option 1.

 

 

 

Question 69 

Options:

1. Positive
2. Negative
3. Both positive and negative
4. Neutral

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates Negative→ Option 2.
• This likely refers to a concept such as negative correlation(e.g., between grain yield and protein content in cereals), negative feedback (e.g., in plant hormone regulation), or negative staining (in microbiology).

 

 

Question 70 

Cell theory was put forward by:

1. Singh & Nicolsan
2. Leeuwenhoek
3. Schleiden & Schwann
4. Calvin & Malvin

Correct Answer: 3

 

Explanation:

• Cell theorystates that:
  1. All living organisms are composed of one or more cells.
  2. The cell is the basic structural and functional unit of life.
  3. All cells arise from pre-existing cells.
• Matthias Schleiden(1838) – Proposed that all plants are made of cells.
• Theodor Schwann(1839) – Extended this to animals and formally proposed the cell theory.
• Rudolf Virchow(1855) – Added “Omnis cellula e cellula” (all cells come from cells).
• Antonie van Leeuwenhoek– Improved the microscope and observed microorganisms, but did not propose cell theory.
• Therefore, cell theory was put forward by Schleiden & Schwann→ Option 3.

 

 

Question 71 

Options:

1. Acetyl CoA
2. Pyruvic acid
3. Glucose-1-Phosphate
4. Fructose-1-Phosphate

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates Pyruvic acid→ Option 2.
• This likely asks for the end product of glycolysis(pyruvic acid) or a key intermediate in cellular respiration.
• Pyruvic acid (pyruvate)is produced from glucose via glycolysis and is then converted to acetyl-CoA (which enters the Krebs cycle) or fermented to lactate/ethanol.

 

 

Question 72 

Chief functions of leaves are:

1. Gaseous exchange and photosynthesis
2. Respiration and photosynthesis
3. Respiration and Digestion
4. Respiration and gaseous exchange

Correct Answer: 1

 

Explanation:

• Photosynthesis– Leaves contain chlorophyll in chloroplasts and are the primary sites of photosynthesis (conversion of light energy into chemical energy, producing glucose and oxygen).
• Gaseous exchange– Leaves exchange gases (CO₂ uptake for photosynthesis, O₂ release, and water vapor loss through transpiration) via stomata.
• Respiration(Options 2, 3, 4) occurs in all living plant cells, including leaves, but it is not the “chief” or unique function of leaves. Digestion does not occur in leaves.
• Therefore, the chief functions of leaves are gaseous exchange and photosynthesis→ Option 1.

 

 

Question 73 

Options:

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

Correct Answer: 4

 

Explanation:

• The specific question text is missing, but the answer key indicates Sucrose→ Option 4.
• This likely asks: The primary form of carbohydrate transported in plants is?
• Sucroseis the main transport sugar in plants (via phloem). It is a disaccharide composed of glucose and fructose.
• Glucose– Transported in small amounts; mostly used locally or stored as starch.
• Maltose– A disaccharide of two glucose units; not a major transport form.
• Starch– Storage polysaccharide, not transported.
• Therefore, the correct answer is Sucrose→ Option 4.

 

 

Question 74 

The first product to be tagged with GI in India is:

1. Assam Tea
2. Nilgiri Tea
3. Darjeeling tea
4. Organic tea

Correct Answer: 3

 

Explanation:

• Geographical Indication (GI)tag is given to products that have a specific geographical origin and possess qualities or reputation due to that origin.
• Darjeeling teawas the first product in India to receive a GI tag (in 2004-2005). It is protected under the Geographical Indications of Goods (Registration and Protection) Act, 1999.
• GI registration for Darjeeling tea was granted to the Tea Board of India.
• Assam Teaand Nilgiri Tea also have GI tags, but they were registered later.
• Therefore, the correct answer is Darjeeling tea→ Option 3.

 

 

Question 75 

Options:

1. 200 m
2. 300 m
3. 100 m
4. 50 m

Correct Answer: 1

 

Explanation:

• The specific question text is missing, but the answer key indicates 200 m→ Option 1.
• This likely asks for the isolation distancerequired for seed certification of a particular crop (e.g., for certified seed production of cross-pollinated crops like maize or sunflower, isolation distance is often 200 m or more).
• In seed certification, isolation distances vary by crop and seed class:
  • 200 m– Common for certified seeds of cross-pollinated crops (e.g., maize, sunflower, mustard).
  • 100 m, 50 m– Used for self-pollinated crops or foundation seeds.
• Therefore, the correct answer is 200 m→ Option 1.

 

 

Question 76 

The seed dormancy can be detected by:

1. Vigor test
2. Germination test
3. Tetrazolium test
4. Purity test

Correct Answer: 2

 

Explanation:

• Seed dormancyis the inability of a viable seed to germinate under favorable conditions (water, temperature, oxygen, light).
• Germination test– Seeds are placed under optimal germination conditions. If they fail to germinate despite being viable (confirmed by tetrazolium test), dormancy is detected. The germination test directly reveals dormancy as low germination percentage.
• Vigor test– Measures seed strength and performance under stress, not directly dormancy.
• Tetrazolium test– A biochemical test for seed viability (alive or dead), but it does not detect dormancy (dormant seeds are viable but will stain in TZ test).
• Purity test– Determines physical purity (presence of inert matter, weed seeds, other crop seeds), not dormancy.
• Therefore, seed dormancy is detected by germination test→ Option 2.

 

 

Question 77 

Options:

1. Developed by a breeder in a government organization
2. Registered with the PPV & FR Authority
3. Cultivated in a large area
4. Released by any Variety Release Committee

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates Registered with the PPV & FR Authority→ Option 2.
• This likely asks: What is required for a plant variety to be legally protected under the PPV&FR Act?
• Registration with the PPV&FR Authority(Protection of Plant Varieties and Farmers’ Rights Authority) is necessary to obtain plant variety protection (breeder’s rights) in India.
• Option 1– Not sufficient; many varieties are developed but not protected.
• Option 3– Cultivation area does not confer legal protection.
• Option 4– Release by a variety release committee is for notification (not for IP protection).
• Therefore, the correct answer is Registered with the PPV & FR Authority→ Option 2.

 

 

Question 78 

The tag for Breeders Seed is issued by:

1. State Seed Certification Agency
2. Central Seed Certification Agency
3. Concerned Breeder of the variety
4. Independent organization operated under the state government

Correct Answer: 3

 

Explanation:

• Breeder seedis the first stage of seed multiplication. It is produced under the direct supervision of the concerned breeder (or the institute/University that developed the variety).
• The tag for breeder seed is issued by the breeder(or the breeder’s institution), not by a seed certification agency.
• Seed certification tags(Foundation, Certified) are issued by the State Seed Certification Agency (SSCA) or Central Seed Certification Agency (CSCA).
• Therefore, the correct answer is Concerned Breeder of the variety→ Option 3.

 

 

Question 79 

Options:

1. 2 years
2. 3 years
3. 4 years
4. 5 years

Correct Answer: 2

 

Explanation:

• The specific question text is missing, but the answer key indicates 3 years→ Option 2.
• This likely asks: The validity period of a certified seed tag?or The minimum duration for a variety trial?
• In seed certification, the certified seed tagis typically valid for one year (or until the next harvest), not 3 years. This may refer to the period for which seed certification standards apply or the period for which a registered variety is protected.
• Under the PPV&FR Act, a registered variety is protected for 15 years(annuals) or 18 years (trees/vines). So 3 years is not matching.
• Alternatively, this could refer to the period of seed viabilityfor certain crops under optimal storage (e.g., 3 years for many cereal seeds).
• Given the answer key, the correct answer is 3 years→ Option 2.

 

 

Question 80 

In which of the following seed storage conditions, the longevity of the seeds would be maximum?

1. Normal ambient storage
2. Moisture proof storage
3. Dehumidified with low temperature (5-8 °C) storage
4. Dehumidified storage only

Correct Answer: 3

 

Explanation:

• Seed longevity is maximized when temperature and seed moisture content are reduced(according to the Harrington’s rule and the seed viability equation).
• Normal ambient storage(Option 1) – High temperature and humidity, leads to rapid seed deterioration.
• Moisture proof storage(Option 2) – Prevents moisture ingress but does not lower temperature; seeds still respire.
• Dehumidified with low temperature (5-8 °C) storage(Option 3) – Low temperature + low humidity (dehumidified) significantly reduces metabolic activity and microbial growth, maximizing seed longevity.
• Dehumidified storage only(Option 4) – Reduces moisture but temperature remains high; seeds will still age faster than at low temperature.
• Therefore, the best condition for maximum seed longevity is dehumidified with low temperature (5-8 °C) storage→ Option 3.