Question 81.

Which of the following plant disease epidemics is the best example of an introduced pathogen causing near-extinction of a native tree species due to lack of co-evolved resistance?

1. Late blight of potato
2. Downy mildew of grapevine
3. Chestnut blight
4. Southern corn leaf blight

Correct Answer: 3

 

Explanation:

• Chestnut blightis the classic textbook example of an introduced pathogen causing a catastrophic epidemic.
• Causal agent:Cryphonectria parasitica (formerly Endothia parasitica), an ascomycete fungus.
• Origin and introduction:The pathogen was introduced from Asia (China/Japan) to North America in the early 1900s via infected nursery stock.
• Impact:It devastated the American chestnut (Castanea dentata), which was a dominant forest tree in eastern North America. Over 3–4 billion trees were killed, and the species was driven to functional extinction (few living trees, no reproduction).
• Comparison with other options:
  • Late blight of potato (1)– Caused by Phytophthora infestans; led to the Irish famine (1840s). While severe, it did not drive the potato species to extinction, and the pathogen was not introduced to a completely naive host (potato was already cultivated).
  • Downy mildew of grapevine (2)– Caused by Plasmopara viticola, introduced from North America to Europe in the 1870s. It was controlled by grafting European grapevines onto resistant American rootstocks, so it did not cause near-extinction.
  • Southern corn leaf blight (4)– Caused by Bipolaris maydis race T; epidemic in the USA in 1970. It was due to a race shift in an existing pathogen (not a new introduction) and was controlled by changing hybrid varieties.
• Key concept:Introduced pathogens can cause catastrophic epidemics when the host population has no co-evolved resistance, leading to functional extinction.
• Thus, the correct answer is Chestnut blight → Option 3.

 

 

Question 82

Light pathway of a typical microscope consists of several items:
A. Light source to Mirror
B. Field Diaphragm to Condenser
C. Stage to Specimen
D. Objective to Body
E. Ocular to Eye/Camera

Options:

1. Light path is A to C to B to D to E
2. Light path is A to D to C to B to E
3. Light path is A to B to C to D to E
4. Light path is A to E to C to B to D

Correct Answer: 3

 

Explanation: The correct light path in a compound light microscopeis:

Light source → Mirror → Field Diaphragm → Condenser → Specimen (on stage) → Objective lens → Body tube → Ocular lens → Eye/Camera

• Step-by-step:
  • Light source to Mirror– Light from the illuminator or external source reflects off the mirror (or passes directly in modern microscopes).
  • Field Diaphragm to Condenser– Light passes through the field diaphragm (controls light intensity/field of view) and then through the condenser (focuses light onto the specimen).
  • Stage to Specimen– Light passes through the specimen placed on the stage.
  • Objective to Body– The objective lens magnifies the image, and the light travels up the body tube.
  • Ocular to Eye/Camera– The ocular (eyepiece) further magnifies the image and directs it to the eye or camera.
  • Therefore, the correct sequence is A → B → C → D → E→ Option 3.
  • Key concept: Light must pass through the specimen before entering the objective lens.

 

 

Question 83

Statement I: Horizontal and Vertical Resistance defend crop plants against pathogens.
Statement II: Horizontal Resistance is more frequently exploited in resistance breeding programmes than Vertical Resistance.

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:

• Statement I is true:Both vertical resistance (race-specific, controlled by major genes) and horizontal resistance (race-nonspecific, controlled by polygenes) are used to defend plants against pathogens.
• Statement II is false:Vertical resistance is more frequently exploited in resistance breeding programmes because:
  • It is easier to select(single gene control, clear segregation patterns).
  • It provides complete (high-level) resistance.
  • It is more straightforward to transfervia backcrossing.
  • Horizontal resistanceis more durable but is quantitative (polygenic), harder to select, and often provides only partial resistance, making it less attractive for many breeding programmes.

Feature	Vertical Resistance	Horizontal Resistance
Genetic control	Major genes (oligogenic)	Polygenic (quantitative)
Race-specificity	Yes (race-specific)	No (race-nonspecific)
Level of resistance	High (complete)	Partial (incomplete)
Durability	Low (easily broken)	High (durable)
Frequency in breeding	More common	Less common

Therefore, Statement I is true but Statement II is false → Option 3.

 

 

 

Question 84

Plant pathogens can be detected and identified using ELISA and PCR.
A. ELISA based detection is more useful for viral pathogen detection than bacterial pathogens
B. Fungal diseases are more amenable for ELISA based detection than plant viruses
C. Bacterial pathogens are more amenable for PCR based detection than plant viruses
D. Bacterial pathogens are more amenable for PCR based detection than fungal pathogens
E. Phytoplasma is more amenable for PCR based detection than ELISA

Options:

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

Correct Answer: 4

 

Explanation:

• A is correct:ELISA (Enzyme-Linked Immunosorbent Assay) is more useful for viral detection because:
  • Viruses have protein coats (capsids)that serve as excellent antigens.
  • Specific antibodies can be produced against viral coat proteins.
  • ELISA is rapid, sensitive, and can process many samples.
  • Bacterial detection by ELISA is possible but less common due to cross-reactivity and lower sensitivity compared to culture or PCR.
• B is incorrect:Fungal diseases are less amenable to ELISA than plant viruses because:
  • Fungi have complex cell walls and variable antigen expression.
  • Antibody production is more difficult and often less specific.
  • PCR is preferred for fungal detection.
• C is incorrect:Bacterial pathogens are not more amenable to PCR than plant viruses. Both are amenable to PCR, but the statement implies a comparison that is not accurate.
• D is incorrect:Bacterial pathogens are not necessarily more amenable to PCR than fungal pathogens. Both are routinely detected by PCR; fungi may require more complex DNA extraction, but PCR works well for both.
• E is correct:Phytoplasma is more amenable to PCR than ELISA because:
  • Phytoplasmas are unculturablein vitro, making antibody production difficult.
  • They are present in low titers in phloem.
  • PCR (especially nested PCR) is the gold standardfor phytoplasma detection and identification.

Method	Best for	Principle
ELISA	Viruses (antigen-antibody reaction)	Immunological
PCR	Bacteria, fungi, phytoplasma, viruses	DNA amplification

Therefore, the correct statements are A and E → Option 4.

 

 

 

Question 85

Wilt diseases are caused by:

1. Heat
2. Drought
3. Bacteria and Fungi
4. Viroids

Correct Answer: 3

 

Explanation:

• Wilt diseasesare primarily caused by bacterial and fungal pathogens that invade the vascular system (xylem) of plants.
• Bacterial wilt agents:Ralstonia solanacearum, Clavibacter michiganensis, Erwinia tracheiphila.
• Fungal wilt agents:Fusarium oxysporum (Fusarium wilt), Verticillium dahliae (Verticillium wilt), Ceratocystis fimbriata (Oak wilt).
• Mechanism:Pathogens block the xylem vessels through:
  • Mycelial growth(fungi)
  • Exopolysaccharide (EPS) production(bacteria)
  • Ty lose formation(host response)
  • Vascular discolorationoccurs due to phenolic compounds.
• Symptoms:Wilting despite adequate soil moisture, leaves remain green initially (no yellowing), vascular browning.
• Heat and drought(Options 1 and 2) cause physiological wilting (due to water deficit), not “wilt diseases” in the pathological sense.
• Viroids(Option 4) cause stunting, leaf distortion, and fruit deformation, but not typically vascular wilts.
• Therefore, wilt diseases are caused by bacteria and fungi → Option 3.

 

 

Question 86

Assertion A: Foliar plant pathogens produce pigmented and colored spores.
Reason R: The dark pigmentation helps the pathogens to adapt on foliar parts to withstand high radiations.

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:

• Assertion (A) is true:Many foliar plant pathogens produce pigmented (melanized) spores (e.g., Alternaria, Helminthosporium, Puccinia).
• Reason (R) is true and correctly explains A:The dark pigmentation (melanin) provides protection against:
  • UV radiation– melanin absorbs and dissipates harmful UV light.
  • Desiccation– helps retain moisture.
  • Extreme temperatures– provides thermal stability.
  • Microbial degradation– melanin is resistant to enzymatic breakdown.
• Foliar pathogens are exposed to direct sunlight, high temperatures, and variable humidity. Melanized spores have a survival advantageon leaf surfaces.
• Therefore, both A and R are true, and R correctly explains A→ Option 1.

 

 

Question 87

Match List I with List II:

List I	List II
A. Chlorosis	I. Fire blight of apple
B. Hyperplasia	II. Mosaic disease of tobacco
C. Epinasty	III. Bacterial wilt of tomato
D. Etiolation	IV. Crown gall of rose
E. Necrosis	V. Bakanae disease of rice

Options:

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

Correct Answer: 2

 

Explanation:

• Chlorosis (A)– Yellowing of leaves due to loss of chlorophyll. Associated with mosaic disease of tobacco (virus causes chlorotic mottling) → A-II
• Hyperplasia (B)– Excessive cell division leading to abnormal growth (galls, tumors). Associated with crown gall of rose (caused by Agrobacterium tumefaciens) → B-IV
• Epinasty (C)– Downward bending of leaves due to uneven growth or hormonal imbalance. Associated with bacterial wilt of tomato (caused by Ralstonia solanacearum; ethylene production causes epinasty) → C-III
• Etiolation (D)– Pale, elongated growth due to lack of light. Associated with Bakanae disease of rice (caused by Gibberella fujikuroi; produces gibberellins causing etiolation-like symptoms) → D-V
• Necrosis (E)– Death of plant tissues (spots, blights). Associated with fire blight of apple (caused by Erwinia amylovora; necrotic blighting of blossoms, shoots) → E-I
• Therefore, correct matching is A-II, B-IV, C-III, D-V, E-I → Option 2.

 

 

Question 88

Match List I with List II:

List I	List II
A. Anton de Bary	I. Gene for Gene hypothesis
B. M.W. Beijerinck	II. Late blight of potato
C. T.J. Burrill	III. Bacterial disease of plants
D. E.J. Butler	IV. Viruses
E. H.J. Flor	V. Imperial Mycologist

Options:

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

Correct Answer: 2

 

Explanation:

• Anton de Bary (A)– German botanist, Father of Plant Pathology. Proved that Phytophthora infestans causes late blight of potato (1861) → A-II
• Martinus Beijerinck (B)– Dutch microbiologist. Coined the term “virus” (contagium vivum fluidum) and discovered tobacco mosaic virus (TMV) as the first virus → B-IV
• J. Burrill (C)– American botanist. First to demonstrate that bacteria cause plant diseases (fire blight of pear, 1878) → C-III
• J. Butler (D)– British mycologist. Served as the Imperial Mycologist in India (1905–1921). Known as “Father of Indian Mycology and Plant Pathology” → D-V
• J. Flor (E)– American plant pathologist. Proposed the Gene-for-Gene hypothesis (1942) based on studies of flax rust → E-I
• Therefore, correct matching is A-II, B-IV, C-III, D-V, E-I → Option 2.

 

 

Question 89

Rust diseases are caused by:
A. Puccinia
B. Albugo
C. Cephalerous
D. Uromyces
E. Gymnosporangium

Options:

1. Pucciniacauses rust only on wheat
2. Albugocauses rust on mustards
3. Cephalerouscauses rust on barley
4. Uromycesand Gymnosporangium cause rust on tomato

Correct Answer: 3

 

Explanation:

• True rust fungibelong to the order Pucciniales (Basidiomycetes). Genera include Puccinia, Uromyces, Gymnosporangium, Hemileia, etc.
• Albugo(Option 2) causes white rust (not true rust); it is an Oomycete (not a true fungus).
• Option 1 is false:Puccinia causes rust on many cereals and grasses, not only on wheat (e.g., Puccinia graminis on wheat, Puccinia striiformis on wheat, Puccinia sorghi on maize, Puccinia recondita on rye).
• Option 2 is false:Albugo causes white rust (not true rust) on mustards and other crucifers.
• Option 3 is true:Cephalerous (likely a misspelling of Cephalosporium or a less common rust genus) – but in the context of this question, this is the statement that is considered correct. (Note: Some rust fungi like Puccinia hordei cause barley rust.)
• Option 4 is false:Uromyces and Gymnosporangium do not cause rust on tomato. Uromyces causes rust on legumes; Gymnosporangium causes rust on juniper and apple/pear (cedar-apple rust).
• Therefore, the correct statement among the options is Option 3.

 

 

Question 90

Panama wilt of banana caused by Fusarium oxysporum f. sp. cubense is an internationally known plant disease.
A. Gros Michel variety of banana is susceptible to race 1
B. Gros Michel variety of banana is resistant to race 1
C. Cavendish variety of banana is susceptible to race 4
D. Cavendish variety of banana is resistant to race 4
E. Both Gros Michel and Cavendish are susceptible to race 1

Options:

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

Correct Answer: 4

 

Explanation:

• Panama wilt(Fusarium wilt of banana) is caused by the soilborne fungus Fusarium oxysporum sp. cubense (Foc).
• Race 1– Infects and destroys the Gros Michel variety, which was the main export banana in the early 20th century → A is true, B is false.
• Cavendishvariety is resistant to Race 1 → E is false.
• Tropical Race 4 (TR4)– A more aggressive strain that infects the Cavendish variety, which replaced Gros Michel. TR4 is spreading globally and threatens the banana industry → C is true, D is false.
• Therefore:
  • A is correct (Gros Michel susceptible to race 1)
  • C is correct (Cavendish susceptible to race 4)
  • A and C are correct → Option 4.

Variety	Race 1	Tropical Race 4
Gros Michel	Susceptible	Susceptible
Cavendish	Resistant	Susceptible

 

 

Question 91

Assertion A: Barberry is an alternate host for Puccinia graminis f. sp. tritici.
Reason R: Barberry plays a key role in rust epidemics in India.

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:

• Assertion (A) is true:Barberry (Berberis vulgaris and other Berberis species) is the alternate (secondary) host for the black stem rust fungus Puccinia graminis sp. tritici. The fungus completes its life cycle on barberry, where sexual reproduction (pycnia and aecia) occurs.
• Reason (R) is false:Barberry does not play a key role in rust epidemics in India because:
  • Barberry species are rare or absentin the major wheat-growing regions of India (Indo-Gangetic plains).
  • The pathogen overwinters as urediniosporeson wheat stubble and volunteer plants, or spreads via air currents from the hills.
  • Sexual reproduction on barberry is not required for epidemic development in India due to the mild winter climate.
  • In contrast, in regions where barberry is common (e.g., parts of the USA, Europe), it plays a significant role in the disease cycle.
• Therefore, A is true but R is false → Option 3.

 

 

Question 92

Match List I with List II (Culture Collections and Locations):

List I	List II
A. ITCC	I. Belgium
B. MTCC	II. England
C. NCPPB	III. New Delhi
D. ATCC	IV. Chandigarh
E. LMG	V. USA

Options:

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

Correct Answer: 1

 

Explanation:

• ITCC– Indian Type Culture Collection (IARI, New Delhi) → A-III
• MTCC– Microbial Type Culture Collection (IMTECH, Chandigarh) → B-IV
• NCPPB– National Collection of Plant Pathogenic Bacteria (England, UK) → C-II
• ATCC– American Type Culture Collection (USA) → D-V
• LMG– Laboratorium voor Microbiologie Gent (Belgium) → E-I
• Therefore, correct matching is A-III, B-IV, C-II, D-V, E-I → Option 1.

 

 

Question 93

Statement I: Plant viruses are transmitted by insects.
Statement II: Only sucking insects are capable of transmitting plant viruses.

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:

• Statement I is true:Many plant viruses are transmitted by insects (vectors). Major insect vectors include:
  • Aphids(sucking) – transmit many viruses (e.g., Potato virus Y, Cucumber mosaic virus)
  • Whiteflies(sucking) – transmit geminiviruses (e.g., Tomato yellow leaf curl virus)
  • Leafhoppers(sucking) – transmit phytoplasma and some viruses
  • Thrips(piercing-sucking) – transmit tospoviruses
• Statement II is false:Not only sucking insects transmit plant viruses. Some chewing insects also transmit viruses:
  • Beetles(chewing) – transmit several plant viruses (e.g., Cowpea mosaic virus, Beetle-transmitted tymoviruses).
  • Grasshoppers– can transmit some viruses mechanically.
  • Other chewing insects– may also act as vectors for certain viruses.
• Therefore, Statement I is true but Statement II is false → Option 3.

 

 

Question 94

Motile infective propagules are produced by:
A. Pythium
B. Phytophthora
C. Xanthomonas
D. Ralstonia
E. Nematodes

Options:

1. Pythiumand Phytophthora possess pili
2. Pythiumand Phytophthora possess fimbriae
3. Nematodes have flagella
4. Xanthomonasand Ralstonia possess flagella

Correct Answer: 4

Explanation:

• Xanthomonasand Ralstonia are motile bacteria that possess flagella:
  • Xanthomonas– single polar flagellum (motile).
  • Ralstonia solanacearum– multitrichous (multiple flagella), motile.
  • Flagellar motility helps these bacteria move towards host roots (chemotaxis) and colonize xylem vessels.
• Pythiumand Phytophthora (Options 1 and 2) produce motile zoospores (biflagellate: one tinsel, one whiplash). They do not possess pili or fimbriae – pili/fimbriae are bacterial structures. Zoospores are unique to Oomycetes and some fungi.
• Nematodes(Option 3) do not have flagella; they move by undulatory movement (body muscles). Some nematodes are vectors for viruses, but they are not “motile infective propagules” in the same sense as zoospores or bacteria.

Therefore, the correct statement is: Xanthomonas and Ralstonia possess flagella → Option 4.

 

 

Question 95

Assertion A: Polycyclic diseases are known to cause plant disease epidemics.
Reason R: Polycyclic disease inciting pathogens produce multiplicity of infections in short time.

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:

• Assertion (A) is true:Polycyclic diseases are indeed known to cause epidemics because they involve multiple infection cycles within a single growing season.
• Reason (R) is true and correctly explains A:Polycyclic pathogens (e.g., rusts, mildews, late blight) produce multiple generations of inoculum (urediniospores, conidia, zoospores) in a short period. Each infection cycle produces new inoculum, leading to exponential (compound interest) increase in disease.
• Comparison with monocyclic diseases:
  • Monocyclic diseases– one infection cycle per season (e.g., soilborne diseases like Fusarium wilt, smuts). Disease increase is linear (simple interest).
  • Polycyclic diseases– multiple cycles per season → exponential increase → epidemics.
• R explains A:The ability to produce “multiplicity of infections in short time” (R) is precisely why polycyclic diseases cause epidemics (A). Therefore, R is the correct explanation of A.
• Thus, both A and R are true, and R is the correct explanation of A → Option 1.

 

 

 

Question 96

Options:

1. Fungal Biotrophs
2. Bacterial Biotrophs
3. Fungal Hemibiotrophs
4. Fungal Necrotrophs

Correct Answer: 1

 

Explanation:

• Biotrophic fungi are pathogens that derive nutrients from living host cells without immediately killing them.
• They have high host specificity and often form specialized feeding structures (e.g., haustoria in rusts and powdery mildews).
• Examples: Rusts (Puccinia spp.), powdery mildews (Blumeria graminis), smuts (Ustilago spp.).
• Key characteristics:
  • Require living host tissue throughout their life cycle.
  • Cannot be cultured on artificial media (obligate biotrophs).
  • Co-evolved with host plants, often showing gene-for-gene interactions.
• Comparison with other trophic types:

Trophic type	Host interaction	Example
Biotroph	Live host (does not kill)	Rusts, powdery mildews
Necrotroph	Kills host, feeds on dead tissue	Sclerotinia, Botrytis
Hemibiotroph	Biotrophic initially, then necrotrophic	Magnaporthe oryzae, Phytophthora infestans
Saprotroph	Dead organic matter	Trichoderma, Aspergillus

Therefore, the correct answer is Fungal Biotrophs → Option 1.

 

 

Question 97

Arrange the following organelles starting from the center to periphery of the cell:
A. Golgi apparatus
B. Nucleolus
C. Rough endoplasmic reticulum
D. Cell wall
E. Plasma membrane

Options:

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

Correct Answer: 3

Explanation:

• The correct order from the center to the periphery of a typical plant cell is: Nucleolus → Golgi apparatus → Rough endoplasmic reticulum → Plasma membrane → Cell wall

• Step-by-step:
  • Nucleolus (B) – Located inside the nucleus (center of the cell). Responsible for ribosome synthesis.
  • Golgi apparatus (A) – Located near the nucleus. Functions in modification, sorting, and packaging of proteins.
  • Rough endoplasmic reticulum (C) – Extends from the nuclear envelope outward. Studded with ribosomes; involved in protein synthesis.
  • Plasma membrane (E) – The boundary of the cytoplasm, located inside the cell wall.
  • Cell wall (D) – The outermost layer in plant cells (not present in animal cells).
• Therefore, the correct sequence is B → A → C → E → D → Option 3.

 

 

Question 98

Options:

1. Lysosome
2. Mitochondria
3. Chloroplast
4. Nucleus

Correct Answer: 4

 

Explanation:

• The nucleus is the control center of the cell because:
  • It contains the genetic material (DNA) in the form of chromosomes.
  • It regulates gene expression and controls all cellular activities including growth, metabolism, and reproduction.
  • It directs cell division (mitosis and meiosis).
  • It coordinates the synthesis of proteins via mRNA transcription.
• Other organelles:
  • Lysosomes – Digestive organelles; break down waste materials.
  • Mitochondria – Powerhouse of the cell; produce ATP via cellular respiration.
  • Chloroplasts – Site of photosynthesis in plants; convert light energy into chemical energy.
• While mitochondria and chloroplasts are essential, the nucleus is the primary control center.

Therefore, the correct answer is Nucleus → Option 4.

 

 

Question 99

Which of the following statements are true while comparing prokaryotes and eukaryotes?
A. Prokaryotes have 70S ribosomes, eukaryotes have 80S ribosomes.
B. Nuclear material is linear in both prokaryotes and eukaryotes.
C. Meiosis is the characteristic of eukaryotes.
D. Genetic exchange occurs in both prokaryotes and eukaryotes.
E. Nuclear membrane is present in both prokaryotes and eukaryotes.

Options:

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

Correct Answer: 1

 

Explanation:

• A is true: Prokaryotes have 70S ribosomes (smaller), while eukaryotes have 80S ribosomes (larger). This difference is targeted by certain antibiotics (e.g., tetracycline affects 70S).
• B is false: Nuclear material in prokaryotes is circular (single circular chromosome), while in eukaryotes it is linear (multiple linear chromosomes).
• C is true: Meiosis (reductional division) is characteristic only of eukaryotes. Prokaryotes reproduce asexually by binary fission; they do not undergo meiosis.
• D is true: Genetic exchange occurs in both:
  • Prokaryotes – conjugation, transformation, transduction.
  • Eukaryotes – sexual reproduction (crossing over, independent assortment), horizontal gene transfer (rare).
• E is false: A nuclear membrane is present only in eukaryotes (true nucleus). Prokaryotes lack a nuclear membrane; their genetic material is in the nucleoid region.

Therefore, the true statements are A, C, and D → Option 1.

 

 

Question 100

Which of the following statements are true for Koch’s postulates?
A. The microorganisms must be present in every case of the disease.
B. The microorganisms may not be present in every case of the disease.
C. The microorganisms must be isolated from the diseased host and grown in pure culture.
D. The specific disease must be reproduced on inoculation of a pure culture of microorganisms in the healthy susceptible host.
E. The microorganism must be recoverable once again from the inoculated susceptible host.

Options:

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

Correct Answer: 3

 

Explanation:

• Koch’s postulates (developed by Robert Koch, 1884) are the standard criteria for establishing a causal relationship between a microorganism and a disease.
• The four original postulates are:
  1. The microorganism must be present in every case of the disease (A is true).
  2. It must be isolated from the diseased host and grown in pure culture (C is true).
  3. The pure culture must reproduce the disease when inoculated into a healthy, susceptible host (D is true).
  4. The microorganism must be re-isolated from the experimentally infected host and shown to be identical to the original (E is true).
• B is false: The microorganism must be present in every case (not “may not be present”).
• Therefore, the true statements are A, C, D, and E → Option 3.
• Note: Koch’s postulates have limitations (e.g., viruses, unculturable organisms, asymptomatic carriers), but the question asks for the traditionally true statements.

 

 

Question 101

Which of the following require oxygen for aerobic respiration but can also grow under anaerobic conditions?

1. Strict anaerobic bacteria
2. Micro-aerophilic bacteria
3. Facultative anaerobic bacteria
4. Aerobic bacteria

Correct Answer: 3

 

Explanation:

• Facultative anaerobic bacteria (Option 3) are organisms that:
  • Prefer oxygen for aerobic respiration (producing more ATP).
  • Can grow without oxygen by switching to fermentation or anaerobic respiration (producing less ATP).
• Examples: Escherichia coli, Staphylococcus aureus, Salmonella.
• Comparison with other types:

Type	Oxygen requirement	Example
Strict (obligate) aerobe	Requires oxygen	Bacillus subtilis
Strict (obligate) anaerobe	Cannot tolerate oxygen	Clostridium botulinum
Microaerophile	Requires low oxygen (2-10%)	Campylobacter jejuni
Facultative anaerobe	Can grow with or without oxygen	E. coli

Therefore, the correct answer is Facultative anaerobic bacteria → Option 3.

 

 

Question 102

Match List I with List II:

List I	List II
A. Dipicolinic acid	I. Filter sterilization
B. Mycoplasmas	II. Endospore
C. Heat labile liquids	III. Mollicutes
D. Steam under pressure	IV. Autoclave

Options:

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

Correct Answer: 2

 

Explanation:

• Dipicolinic acid (A) – Found in bacterial endospores (e.g., Bacillus, Clostridium). Contributes to heat resistance and stability of the spore → A-II
• Mycoplasmas (B) – Belong to the class Mollicutes (meaning “soft skin”). They lack a cell wall and are pleomorphic → B-III
• Heat labile liquids (C) – Liquids that are sensitive to heat (e.g., antibiotics, vitamins, serum). Sterilized by filter sterilization (membrane filters, 0.22 µm) → C-I
• Steam under pressure (D) – Used in an autoclave (121°C, 15 psi, 15-20 min) for sterilization of heat-stable materials → D-IV
• Therefore, correct matching is A-II, B-III, C-I, D-IV → Option 2.

 

 

Question 103

Options:

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

Correct Answer: 3

 

Explanation:

• The specific question text is missing from the document, but the answer key indicates Option 3 (A, B, C, D, and E) is correct.
• This suggests that all given statements are true, or all are relevant to the question asked.
• Without the full question, a detailed explanation cannot be provided. However, the answer key confirms Option 3.

 

 

Question 104

Statement I: A phototrophic microorganism can derive all carbon requirements from the principal carbon source.
Statement II: An auxotrophic microorganism requires one or more growth factors in addition to the principal carbon source.

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:

• Statement I is true: A phototrophic microorganism uses light as its energy source. It can derive its carbon requirements from:
  • CO₂ (photoautotrophs – fix carbon via photosynthesis).
  • Organic carbon sources (photoheterotrophs – require organic compounds).
  • In both cases, it can obtain carbon from the principal carbon source (CO₂ or organic carbon).
• Statement II is true: An auxotrophic microorganism has a mutation in a biosynthetic pathway and cannot synthesize certain essential compounds (e.g., amino acids, vitamins). It therefore requires one or more growth factors (supplements) in addition to the principal carbon source.
• Example: E. coli auxotroph for tryptophan cannot grow on minimal medium without added tryptophan.
• Therefore, both statements are correct → Option 1.

 

 

Question 105

Arrange the following events of phage replication in correct sequence:
A. Replication and synthesis
B. Penetration
C. Adsorption
D. Lysis and release
E. Assembly of phage particles

Options:

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

Correct Answer: 3

 

Explanation:

• The correct sequence of the lytic cycle of a bacteriophage is:

Adsorption → Penetration → Replication and synthesis → Assembly → Lysis and release

• Step-by-step:
  • C. Adsorption – The phage attaches to specific receptors on the bacterial cell surface.
  • B. Penetration – The phage injects its genetic material (DNA or RNA) into the bacterial cell.
  • A. Replication and synthesis – Phage nucleic acid is replicated, and phage proteins (capsid, enzymes) are synthesized using host machinery.
  • E. Assembly – New phage particles are assembled from the synthesized components.
  • D. Lysis and release – The bacterial cell lyses (breaks open), releasing new phages to infect other cells.
• Therefore, the correct sequence is C → B → A → E → D → Option 3.

 

 

Question 106

Assertion A: In lysogenic cycle, the viral nucleic acid does not hamper the functions of host bacterium.
Reason R: In lysogenic cycle, the nucleic acid of phage gets incorporated into the host DNA and becomes a prophage, acting as a gene.

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:

• Assertion (A) is true: In the lysogenic cycle, the viral nucleic acid (prophage) is integrated into the host chromosome and replicates passively along with the host DNA. It does not disrupt normal host functions or kill the host immediately.
• Reason (R) is true and correctly explains A: The prophage acts as a set of additional genes within the host genome. It does not interfere with host metabolism because viral genes for replication and lysis are repressed (e.g., by lambda repressor protein). The host continues to grow and divide normally.
• Key concept: The lysogenic cycle allows the phage to coexist with the host. Under stress conditions, the prophage may be induced to enter the lytic cycle.
• Therefore, both A and R are true, and R correctly explains A → Option 1.

 

 

Question 107

Statement I: Operon models of gene regulation are applicable to prokaryotes only.
Statement II: Eukaryotic DNA contains intervening pieces of DNA called Introns.

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:

• Statement I is true: The operon model (proposed by Jacob and Monod, 1961) describes a cluster of genes transcribed as a single polycistronic mRNA under the control of a single promoter and operator. This model is applicable primarily to prokaryotes (bacteria and archaea). Eukaryotes have more complex gene regulation mechanisms (e.g., individual promoters for each gene, enhancers, transcription factors, chromatin remodeling).
• Statement II is true: Eukaryotic DNA contains introns (non-coding intervening sequences) that are transcribed into pre-mRNA and later spliced out during RNA processing. Prokaryotic genes generally lack introns (though some archaeal and rare bacterial genes may have introns in tRNA or rRNA).
• Therefore, both statements are correct → Option 1.

 

 

Question 108

Match List I with List II:

List I	List II
A. Pseudomonas	I. Gram negative bacteria
B. Cell Outer Membrane	II. Bacteria
C. Algae-fungus association	III. Induction of Nod genes
D. Flavonoids	IV. Lichens

Options:

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

Correct Answer: 2

 

Explanation:

• Pseudomonas (A) – A genus of bacteria (gram-negative, rod-shaped) → A-II
• Cell Outer Membrane (B) – A characteristic feature of gram-negative bacteria (present in addition to the inner/cytoplasmic membrane) → B-I
• Algae-fungus association (C) – A symbiotic association between an alga (or cyanobacterium) and a fungus forms a lichen → C-IV
• Flavonoids (D) – Plant secondary metabolites released by legume roots that act as signals for induction of Nod (nodulation) genes in Rhizobium bacteria, initiating root nodule formation → D-III
• Therefore, correct matching is A-II, B-I, C-IV, D-III → Option 2.

 

 

Question 109

Match List I with List II:

List I	List II
A. Sesbania rostrata	I. Azotobacter
B. Vermi-composting	II. Anabaena
C. Heterocyst	III. Green manure
D. Cyst	IV. Earthworm

Options:

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

Correct Answer: 4

 

Explanation:

• Sesbania rostrata (A) – A leguminous plant that forms stem nodules and is used as a green manure (biofertilizer) to enrich soil nitrogen → A-III
• Vermi-composting (B) – The process of using earthworms to convert organic waste into nutrient-rich compost (vermicompost) → B-IV
• Heterocyst (C) – A specialized, thick-walled cell found in filamentous cyanobacteria like Anabaena. Heterocysts are the site of nitrogen fixation (contain nitrogenase) → C-II
• Cyst (D) – A dormant, thick-walled resting structure formed by certain bacteria like Azotobacter to survive unfavorable conditions → D-I
• Therefore, correct matching is A-III, B-IV, C-II, D-I → Option 4.

 

 

Question 110

Which of the following processes cause losses of nitrogen from cultivated soils?
A. Leaching
B. Volatilization
C. Biological nitrogen fixation
D. Denitrification
E. Green manuring

Options:

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

Correct Answer: 3

 

Explanation:

• Nitrogen loss from soil occurs through processes that remove nitrogen from the plant-available pool.
• A. Leaching – Loss of nitrate (NO₃⁻) (highly soluble) through percolating water beyond the root zone. This causes nitrogen loss and groundwater pollution.
• B. Volatilization – Loss of ammonia (NH₃) gas from soil surface, especially from urea or ammonium-based fertilizers under alkaline conditions or high temperatures.
• C. Biological nitrogen fixation – Adds nitrogen to the soil (converts atmospheric N₂ to ammonia). This is a gain, not a loss.
• D. Denitrification – Conversion of nitrate (NO₃⁻) to gaseous forms (N₂, N₂O) by denitrifying bacteria (e.g., Pseudomonas) under anaerobic conditions (waterlogged soils). This is a major loss pathway.
• E. Green manuring – Incorporation of green plant biomass into soil to add nitrogen and organic matter. This is a gain, not a loss.
• Therefore, the processes causing nitrogen loss are Leaching (A), Volatilization (B), and Denitrification (D) → Option 3 (A, B and D only).

 

 

Question 111

Statement I: Nitrification is the biological oxidation of ammonia to nitrates, in two steps.
Statement II: Members of genus Nitrobacter are involved in the oxidation of nitrites to nitrates.

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:

• Statement I is true: Nitrification is the biological oxidation of ammonia (NH₃) to nitrate (NO₃⁻) via two sequential steps carried out by different groups of chemoautotrophic bacteria.
• Step 1: Nitrosomonas (and other ammonia-oxidizing bacteria) convert ammonia to nitrite (NO₂⁻).
• Step 2: Nitrobacter (and other nitrite-oxidizing bacteria) convert nitrite to nitrate (NO₃⁻).
• Statement II is true: Members of the genus Nitrobacter are responsible for the second step of nitrification: oxidation of nitrite (NO₂⁻) to nitrate (NO₃⁻).
• Therefore, both statements are correct → Option 1.

 

 

 

Question 112

Assertion A: Rotation with legume crops can increase the nitrogen levels in the soil, thus improving soil fertility.
Reason R: Legume crops harbor nitrogen-fixing bacteria in the root nodules that can convert inert nitrogen from the atmosphere to plant-usable form.

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:

• Assertion (A) is true: Rotating crops with legumes increases soil nitrogen levels because legumes fix atmospheric nitrogen, enriching the soil for subsequent crops (e.g., wheat, maize). This reduces the need for synthetic nitrogen fertilizers.
• Reason (R) is true and correctly explains A: Legumes form a symbiotic association with rhizobia bacteria (e.g., Rhizobium, Bradyrhizobium, Sinorhizobium) in root nodules. These bacteria possess the nitrogenase enzyme that converts inert atmospheric N₂ into ammonia (NH₃), which is then assimilated into amino acids and other organic compounds. When legume residues decompose, this fixed nitrogen becomes available to the soil.
• Key concept: Biological nitrogen fixation (BNF) is a natural, eco-friendly way to improve soil fertility. The reason directly explains why legume rotation increases soil nitrogen.
• Therefore, both A and R are true, and R is the correct explanation of A → Option 1.

 

 

Question 113

Which of the following can be used for surface sterilization of plant material?

1. Sodium chloride
2. Sodium hydroxide
3. Sodium sulphate
4. Sodium hypochlorite

Correct Answer: 4

 

Explanation:

• Sodium hypochlorite (NaOCl) is the most commonly used surface sterilant for plant explants (seeds, leaves, stem segments, meristems) in plant tissue culture.
• Mechanism: NaOCl acts as a strong oxidizing agent. It releases chlorine radicals that oxidize and denature proteins, lipids, and nucleic acids of surface microbes (bacteria, fungi, spores) leading to their death.
• Advantages:
  • Effective at low concentrations (0.5–2% for seeds, 0.5–1% for vegetative tissues).
  • Does not penetrate deeply into plant tissues when exposure time is optimized (typically 5–20 minutes).
  • Easily removed by rinsing with sterile distilled water.
• Other options:
  • Sodium chloride (NaCl) – Table salt; has weak antiseptic properties but not used for surface sterilization in tissue culture.
  • Sodium hydroxide (NaOH) – Highly caustic; kills plant tissues, not suitable for surface sterilization of live plant material.
  • Sodium sulphate (Na₂SO₄) – A salt; has no antimicrobial activity.
• Other common surface sterilants: Ethanol (70%), mercuric chloride (HgCl₂), calcium hypochlorite, hydrogen peroxide.
• Therefore, the correct answer is Sodium hypochlorite → Option 4.

 

 

Question 114

Options:

1. Nostoc
2. Anabaena
3. Azotobacter
4. Calothrix

Correct Answer: 3

 

Explanation:

• The question likely asks for a free-living (non-symbiotic) nitrogen-fixing bacterium (common topic in microbiology/biofertilizers).
• Azotobacter (Option 3) is a free-living, aerobic, heterotrophic nitrogen-fixing bacterium found in soil. It fixes atmospheric N₂ independently without forming symbiosis with plants.
• Nostoc (Option 1) – A filamentous cyanobacterium (blue-green alga) that fixes nitrogen in heterocysts. It can be free-living or form symbiosis (e.g., with cycads, Anthoceros).
• Anabaena (Option 2) – A filamentous cyanobacterium that fixes nitrogen in heterocysts. It can be free-living or form symbiosis (e.g., in the water fern Azolla).
• Calothrix (Option 4) – A filamentous cyanobacterium with basal heterocysts; fixes nitrogen and can be free-living or epiphytic.
• Key distinction: Azotobacter is a bacterium (not a cyanobacterium) and is strictly free-living. The others are cyanobacteria (also free-living or symbiotic).
• Therefore, if the question expects the free-living bacterial nitrogen fixer, the answer is Azotobacter → Option 3.

 

 

Question 115

Which of the following molecules is known as the energy currency of the cell?

1. ATP
2. AMP
3. NADH
4. FADH₂

Correct Answer: 1

 

Explanation:

• ATP (Adenosine Triphosphate) is universally known as the energy currency of the cell because:
  • It stores and transfers chemical energy for cellular processes.
  • Hydrolysis of ATP to ADP + Pi (or AMP + PPi) releases energy (≈7.3 kcal/mol under standard conditions).
  • It is used in metabolism (biosynthesis), transport (active transport), mechanical work (muscle contraction, cell motility), signal transduction, and cellular respiration.
• AMP (Adenosine Monophosphate) – Has only one phosphate group; contains less energy than ATP. It is a product of ATP breakdown and can be recycled to ADP/ATP.
• NADH (Nicotinamide Adenine Dinucleotide + Hydrogen) – An electron carrier (reducing agent) that donates electrons to the electron transport chain to produce ATP. It is not the direct energy currency.
• FADH₂ (Flavin Adenine Dinucleotide + Hydrogen) – Another electron carrier that donates electrons to the ETC, yielding fewer ATP than NADH.
• Therefore, the correct answer is ATP → Option 1.

 

 

Question 116

Which of the following are NOT chain initiation codons?
A. UAA
B. UAG
C. UGA
D. AUG
E. GUG

Options:

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

Correct Answer: 3

 

Explanation:

• Initiation codons (start codons) signal the start of translation:
  • AUG (codes for methionine in eukaryotes and formylmethionine in prokaryotes) is the primary start codon.
  • GUG (codes for valine) can also serve as an alternative start codon in some prokaryotes (e.g., E. coli), though less common.
• Stop codons (termination codons) signal the end of translation:
  • UAA (ochre)
  • UAG (amber)
  • UGA (opal/umber)
• NOT chain initiation codons = stop codons = UAA, UAG, UGA → A, B, and C.
• D (AUG) and E (GUG) are initiation codons (AUG universally, GUG in some prokaryotes).
• Therefore, the correct answer is A, B and C only → Option 3.

 

 

Question 117

Ideal C:N ratio of the mature compost is:

1. <20:1
2. 35:1
3. 60:1
4. 100:1

Correct Answer: 1

 

Explanation:

• C:N ratio (Carbon to Nitrogen ratio) is a critical factor in composting.
• Ideal C:N ratio for mature compost is <20:1 (typically 10:1 to 20:1).
• Why?
  • Microorganisms (bacteria, fungi) require both carbon (energy source) and nitrogen (protein synthesis) for decomposition.
  • At the start of composting, an ideal initial C:N ratio is 25:1 to 30:1 (balanced).
  • As composting progresses, carbon is lost as CO₂ (respiration), while nitrogen is conserved (some loss as NH₃). Therefore, the C:N ratio decreases.
  • Mature compost with a C:N ratio <20:1 indicates:
    • Stability (minimal further decomposition).
    • No nitrogen immobilization (nitrogen is available for plants).
    • Safe for plant application (no phytotoxicity).
• High C:N ratio (>30:1) → Slow decomposition, nitrogen immobilization (microbes compete with plants for N).
• Low C:N ratio (<20:1) → Rapid decomposition, risk of ammonia volatilization if too low, but mature compost should be <20:1.
• Therefore, the correct answer is <20:1 → Option 1.

 

 

Question 118

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: 4

 

Explanation:

• The specific question text is missing from the document, but the answer key indicates Option 4 (Statement I is incorrect but Statement II is correct).
• Without the full question, a detailed explanation cannot be provided. However, the answer key confirms Option 4.

 

 

Question 119

Statement I: National Agriculturally Important Microbial Culture Collection (NAIMCC) is located in Mau Nath Bhanjan, Uttar Pradesh.
Statement II: NAIMCC has acquired the status of International Depository Authority under Budapest Treaty in 2020.

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:

• Statement I is true: The National Agriculturally Important Microbial Culture Collection (NAIMCC) is located at Mau Nath Bhanjan (Mau district), Uttar Pradesh. It is under the ICAR-National Bureau of Agriculturally Important Microorganisms (NBAIM).
• Statement II is true: In 2020, NAIMCC was granted the status of an International Depository Authority (IDA) under the Budapest Treaty on the International Recognition of the Deposit of Microorganisms for the Purposes of Patent Procedure (1977, amended 1980). This allows NAIMCC to accept and preserve microbial strains for patent purposes.
• Functions of NAIMCC:
  • Collection, preservation, and supply of agriculturally important microorganisms (bacteria, fungi, actinomycetes, etc.).
  • Long-term preservation using techniques like lyophilization, cryopreservation, and mineral oil overlay.
  • Microbial identification and characterization.
• Therefore, both statements are correct → Option 1.

 

Question 120 

Match List I with List II (Bureaus and Locations):

List I	List II
A. National Bureau of Plant Genetic Resources (NBPGR)	I. Karnal
B. National Bureau of Animal Genetic Resources (NBAGR)	II. Lucknow
C. National Bureau of Fish Genetic Resources (NBFGR)	III. Delhi
D. National Bureau of Soil Survey and Land Use Planning (NBSSLUP)	IV. Nagpur

Options:

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

Correct Answer: 4

 

Explanation:

• NBPGR (A) – National Bureau of Plant Genetic Resources is located in New Delhi → A-III (Delhi)
• NBAGR (B) – National Bureau of Animal Genetic Resources is located in Karnal, Haryana → B-I (Karnal)
• NBFGR (C) – National Bureau of Fish Genetic Resources is located in Lucknow, Uttar Pradesh → C-II (Lucknow)
• NBSSLUP (D) – National Bureau of Soil Survey and Land Use Planning is located in Nagpur, Maharashtra → D-IV (Nagpur)
• Therefore, correct matching is A-III, B-I, C-II, D-IV → Option 4.

Bureau	Location
NBPGR	New Delhi (Delhi)
NBAGR	Karnal (Haryana)
NBFGR	Lucknow (Uttar Pradesh)
NBSSLUP	Nagpur (Maharashtra)