BIOLOGICAL CLASSIFICATION MCQs for NEET UG BIOLOGY — Medium, Page 3

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How many statements are correct for Deuteromycetes?

(a) Only vegetative phases are known, no asexual and sexual phase

(b) Produces fruiting bodies

(c) Asexual spores are non-motile, exogenous in origin

(d) The mycelium is aseptate and branched

(e) A large number of members are decomposers of litter

View answer and explanation

Correct option: A

Two

Only two statements are correct for Deuteromycetes: (c) Asexual spores are non-motile, exogenous in origin and (e) A large number of members are decomposers of litter.

Solution approach

Correct Answer + Core Reason: Two. Only two statements are correct for Deuteromycetes.

Step-by-Step Solution:

Step 1: Analyze each statement:
(a) Only vegetative phases are known, no asexual and sexual phase - Incorrect. Asexual phase is known.
(b) Produces fruiting bodies - Incorrect. They generally don't produce fruiting bodies.
(c) Asexual spores are non-motile, exogenous in origin - Correct.
(d) The mycelium is aseptate and branched - Incorrect. The mycelium is septate and branched.
(e) A large number of members are decomposers of litter - Correct.

Step 2: Count the correct statements: (c) and (e) are correct.

Why other options are wrong:
- Option B: Three is incorrect because only two statements are correct.
- Option C: Four is incorrect because only two statements are correct.
- Option D: Five is incorrect because only two statements are correct.

Key Concepts:
- Characteristics of Deuteromycetes
- Asexual reproduction in fungi
- Mycelium structure

Memory Aids:
- Deuteromycetes are \"imperfect\" because their sexual stage is unknown, but asexual reproduction does occur.

Quick Tips:
- Carefully evaluate each statement based on the known characteristics of the fungal group.

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Concerning general characteristic of plants, which statement is correct
I. Some of these may be partially heterotrophic as in case of insectivorous plants line Venus fly trap
II. They have distinct nucleus, chloroplast and chitinous cell wall

View answer and explanation

Correct option: A

Only I

Option A (Only I) is correct because some plants are partially heterotrophic, like insectivorous plants, while plant cell walls are made of cellulose, not chitin.

Solution approach

Correct Answer + Core Reason: Only I. Insectivorous plants are partially heterotrophic, and plant cell walls are made of cellulose.

Step-by-Step Solution:

Step 1: Analyze statement I: Insectivorous plants like Venus fly traps are partially heterotrophic because they obtain nutrients from insects to supplement their photosynthetic activity.

Step 2: Analyze statement II: Plant cell walls are primarily composed of cellulose, not chitin. Chitin is a characteristic component of fungal cell walls.

Step 3: Determine the correct statement: Statement I is correct, and statement II is incorrect.

Step 4: Select the appropriate option.

Why other options are wrong:
- Option B: Statement II is incorrect as plant cell walls are made of cellulose, not chitin.
- Option C: Statement II is incorrect.
- Option D: Statement I is correct, so this option is incorrect.

Key Concepts:
- Plant nutrition (autotrophic and heterotrophic)
- Cell wall composition (cellulose vs. chitin)

Memory Aids:
- Remember that plants are primarily autotrophic but can be partially heterotrophic in some cases.
- Recall that cellulose is the main component of plant cell walls.

Quick Tips:
- Pay attention to the specific details in each statement.
- Remember the key differences between plant and fungal cell wall composition.

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Which is correct for the structure of cell wall of bacteria and fungi?

View answer and explanation

Correct option: C

Both are made up of N-acetylglucosamine

Option C (Both are made up of N-acetylglucosamine) is correct because N-acetylglucosamine is a component of both bacterial and fungal cell walls.

Solution approach

Correct Answer + Core Reason: Both are made up of N-acetylglucosamine because it is a component of peptidoglycan (bacteria) and chitin (fungi).

Step-by-Step Solution:

Step 1: Bacterial cell walls contain peptidoglycan, which includes N-acetylglucosamine (NAG).

Step 2: Fungal cell walls contain chitin, which is a polymer of N-acetylglucosamine (NAG).

Step 3: Conclusion: Both bacterial and fungal cell walls contain N-acetylglucosamine.

Why other options are wrong:
- Option A: Bacterial cell walls are not made of cellulose; plant cell walls are.
- Option B: Fungal cell walls do not contain mucopeptide (peptidoglycan).
- Option D: N-acetylglucosamine is a component of both bacterial and fungal cell walls.

Key Concepts:
- Bacterial cell walls
- Fungal cell walls
- N-acetylglucosamine (NAG)
- Peptidoglycan
- Chitin

Memory Aids:
- NAG = Bacteria & Fungi

Quick Tips:
- Remember the components of bacterial and fungal cell walls.

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Which of the given statement best describes the gametophytic in the alternation of generation life cycle?

View answer and explanation

Correct option: A

Generation that produces the gametes

Option A (Generation that produces the gametes) is correct because the gametophytic generation is the haploid phase in the alternation of generations life cycle that produces gametes through mitosis.

Solution approach

Correct Answer + Core Reason: Generation that produces the gametes. The gametophytic generation is the haploid phase that produces gametes.

Step-by-Step Solution:

Step 1: Recall the alternation of generations life cycle.

Step 2: The alternation of generations involves two distinct phases: the gametophytic and sporophytic generations.

Step 3: The gametophytic generation is haploid and produces gametes through mitosis.

Step 4: The sporophytic generation is diploid and produces spores through meiosis.

Why other options are wrong:
- Option B: Generation that produces the spores describes the sporophytic generation.
- Option C: Generation that has xylem and phloem describes the sporophytic generation in vascular plants.
- Option D: The diploid generation describes the sporophytic generation.

Key Concepts:
- Alternation of generations
- Gametophytic generation
- Sporophytic generation
- Haploid
- Diploid

Memory Aids:
- Gametophyte = Gametes.

Quick Tips:
- Know the roles of the gametophytic and sporophytic generations in the plant life cycle.

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Analyse the following statements regarding cyanobacteria and identify the correct option given below
I. The cyanobacteria are unicellular, colonial or filamentous, marine or terrestrial algae
II. The colonies of cyanobacteria are generally surrounded by gelatinous sheath
Codes

View answer and explanation

Correct option: C

I and II

Option C is correct because both statements accurately describe cyanobacteria: they can be unicellular, colonial, or filamentous, and their colonies are often surrounded by a gelatinous sheath.

Solution approach

Correct Answer + Core Reason: I and II. Both statements accurately describe the characteristics of cyanobacteria.

Step-by-Step Solution:
Step 1: Analyze the given statements regarding cyanobacteria.
Step 2: Recall the characteristics of cyanobacteria:
- They are prokaryotic organisms.
- They can be unicellular, colonial, or filamentous.
- They can be found in marine or terrestrial environments.
- Colonies are often surrounded by a gelatinous sheath.
Step 3: Evaluate each statement:\nI. The cyanobacteria are unicellular, colonial or filamentous, marine or terrestrial algae: This statement is correct.\nII. The colonies of cyanobacteria are generally surrounded by gelatinous sheath: This statement is also correct.

Why other options are wrong:
- Option A: Only I - Statement II is also correct.
- Option B: Only II - Statement I is also correct.
- Option D: None of these - Both statements are correct.

Key Concepts:
- Characteristics of Cyanobacteria
- Prokaryotic Organisms

Memory Aids:
- Think of cyanobacteria as \"blue-green algae\" that can form colonies with a slimy covering.

Quick Tips:
- Remember the key features of prokaryotes and how they differ from eukaryotes.

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Viruses have

View answer and explanation

Correct option: C

Both living and non-living characteristics

Option C (Both living and non-living characteristics) is correct because viruses exhibit characteristics of both living and non-living entities depending on whether they are inside or outside a host cell.

Solution approach

Correct Answer + Core Reason: Both living and non-living characteristics. Viruses possess both living and non-living characteristics.

Step-by-Step Solution:

Step 1: Understand the nature of viruses. Viruses are unique entities that exist at the boundary between living and non-living things.

Step 2: Identify living characteristics. Inside a host cell, viruses can replicate, mutate, and evolve, which are characteristics of living organisms.

Step 3: Identify non-living characteristics. Outside a host cell, viruses are inert and cannot reproduce or carry out metabolic processes on their own. They exist as particles.

Step 4: Combine the characteristics. Since viruses exhibit both living and non-living characteristics, the correct answer is that they have both.

Why other options are wrong:
- Option A: Viruses are not exclusively living because they are inert outside of a host cell.
- Option B: Viruses are not exclusively non-living because they can replicate and evolve inside a host cell.
- Option D: While viruses are parasitic, this doesn't encompass their full range of characteristics, including their non-living aspects.

Key Concepts:
- Characteristics of viruses
- Living vs. Non-living
- Viral replication

Memory Aids:
- Remember that viruses are \"in-between\" living and non-living.

Quick Tips:
- Consider the environment (inside or outside a host) when determining whether a virus is exhibiting living or non-living characteristics.

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The non-living characteristic of viruses is

View answer and explanation

Correct option: D

Ability to be crystallize

Option D (Ability to be crystallize) is correct because crystallization is a property of non-living matter, and viruses can be crystallized outside of a host cell.

Solution approach

Correct Answer + Core Reason: Ability to be crystallize. Crystallization is a property of non-living matter.

Step-by-Step Solution:

Step 1: Viruses can be crystallized outside of a host cell, which is a characteristic of non-living matter.

Step 2: This ability to form crystals indicates that they lack the complex organization and metabolic activity of living cells when outside a host.

Why other options are wrong:
- Option A: Ability to multiply only inside the host is a living characteristic.
- Option B: Ability to cause diseases in the host is a living characteristic (related to their parasitic nature).
- Option C: Ability to undergo mutation is a living characteristic.

Key Concepts:
- Characteristics of viruses
- Living vs. non-living
- Crystallization

Memory Aids:
- Crystallization = Non-living.

Quick Tips:
- Remember that the ability to crystallize is a key non-living characteristic of viruses.

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‘Ordines Anomali’ of Bentham and Hooker includes

View answer and explanation

Correct option: D

A few orders which could not be placed satisfactorily in the classification

Option D (A few orders which could not be placed satisfactorily in the classification) is correct because 'Ordines Anomali' in Bentham and Hooker's classification included orders that didn't fit neatly into their system.

Solution approach

Correct Answer + Core Reason: A few orders which could not be placed satisfactorily in the classification. 'Ordines Anomali' were orders that didn't fit well in Bentham & Hooker's system.

Step-by-Step Solution:

Step 1: Understand the Bentham and Hooker classification system.

Step 2: Bentham and Hooker's system is a natural system of classification based on observable characteristics.

Step 3: 'Ordines Anomali' refers to a group of plant orders that could not be easily placed within the main classification due to unusual or unclear characteristics.

Step 4: These orders were considered anomalous or irregular.

Why other options are wrong:
- Option A: Seed plants showing abnormal forms of growth and development - While some plants in Ordines Anomali might have shown abnormal growth, this isn't the primary reason for their inclusion.
- Option B: Plants represented only in fossil state - Fossil plants were not the sole constituents of Ordines Anomali.
- Option C: Plants described in the literature but which Bentham and Hooker did not see in original - While they might not have seen all plants, the primary reason was their difficulty in fitting the classification.

Key Concepts:
- Bentham and Hooker's classification system
- Natural system of classification
- Ordines Anomali

Memory Aids:
- Anomali = Anomalous = Irregular

Quick Tips:
- Remember that 'Ordines Anomali' refers to plants that were difficult to classify.

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In which of the following patterns of viral replication, viruses enter a cell, replicate and then cause the cell to burst, releasing new viruses?

View answer and explanation

Correct option: A

Lytic

Option A (Lytic) is correct because the lytic cycle involves viral replication and cell lysis.

Solution approach

Correct Answer + Core Reason: Lytic. In the lytic cycle, viruses replicate and cause the cell to burst, releasing new viruses.

Step-by-Step Solution:

Step 1: The question describes a viral replication pattern where the cell bursts and releases new viruses.

Step 2: Consider the options:
- Lytic: Viral replication and cell lysis.
- Lysogenic: Viral DNA integrates into the host DNA without immediate cell lysis.
- Repreogenic: Not a recognized term in virology.

Step 3: The lytic cycle is characterized by viral replication, assembly, and release through cell lysis.

Why other options are wrong:
- Option B: In the lysogenic cycle, the viral DNA integrates into the host cell's DNA and does not immediately cause cell lysis.
- Option C: Repreogenic is not a recognized term in virology.
- Option D: The lysogenic cycle does not cause immediate cell lysis.

Key Concepts:
- Lytic cycle
- Lysogenic cycle
- Viral replication

Memory Aids:
- Lytic = Lysis (bursting)

Quick Tips:
- Differentiate between the lytic and lysogenic cycles.

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Red tide in warms coastal water is caused due to the rapid multiplications of

View answer and explanation

Correct option: C

Gonyaulax

Option C (Gonyaulax) is correct because red tide is caused by the rapid multiplication of Gonyaulax, a type of dinoflagellate.

Solution approach

Correct Answer + Core Reason: Gonyaulax is the correct answer because it is the dinoflagellate responsible for red tides.

Step-by-Step Solution:

Step 1: Understand the phenomenon of red tide. Red tide is a harmful algal bloom that occurs in coastal waters.

Step 2: Identify the causative organism. Red tides are primarily caused by dinoflagellates, particularly Gonyaulax.

Step 3: Consider the given options. Among the options, Gonyaulax is a dinoflagellate known to cause red tides.

Step 4: Conclude that Gonyaulax is the correct answer.

Why other options are wrong:
- Option A: Euglena is a genus of single-celled flagellate eukaryotes, not typically associated with red tides.
- Option B: Diatoms are a major group of algae, but they are not the primary cause of red tides. While some diatoms can contribute to harmful algal blooms, Gonyaulax is more commonly associated with red tides.
- Option D: Paramecium is a genus of unicellular ciliated protozoa, not involved in red tides.

Key Concepts:
- Red tide
- Dinoflagellates
- Harmful algal blooms

Memory Aids:
- Remember that \"Gonyaulax gives you a gut ache\" (associating it with the negative effects of red tide).

Quick Tips:
- Focus on the organisms known to cause harmful algal blooms when answering questions about red tides.

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Chromosomes in a bacterial cell can be 1-3 in number and

View answer and explanation

Correct option: C

Are always circular

Option C (Are always circular) is correct because bacterial chromosomes are typically circular and not enclosed within a nucleus.

Solution approach

Correct Answer + Core Reason: Are always circular. Bacterial chromosomes are typically circular.

Step-by-Step Solution:

Step 1: Recall the structure of bacterial chromosomes.

Step 2: Bacterial chromosomes are typically circular, double-stranded DNA molecules.

Step 3: They are not enclosed within a nucleus.

Step 4: Therefore, the chromosomes in a bacterial cell are always circular.

Why other options are wrong:
- Option A: Bacterial chromosomes are generally circular, not linear.
- Option B: Bacterial chromosomes are generally circular, not both circular and linear within the same cell.
- Option D: Bacterial chromosomes are generally circular, not linear.

Key Concepts:
- Bacterial chromosome structure
- Circular DNA

Memory Aids:
- Bacteria = Circular chromosome

Quick Tips:
- Remember that bacterial DNA is typically a single, circular chromosome.

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Assertion: Infectious agents causing mad cow disease can not show genetic integrity

Reason: PSTV caused by viriods.

View answer and explanation

Correct option: A

If both Assertion Reason are true and Reason in the correct explanation of Assertion.

Option A (If both Assertion Reason are true and Reason in the correct explanation of Assertion) is correct because prions, the infectious agents causing mad cow disease, lack nucleic acids and therefore cannot show genetic integrity, and PSTV caused by viroids are infectious agents that also lack genetic integrity.

Solution approach

Correct Answer + Core Reason: If both Assertion Reason are true and Reason in the correct explanation of Assertion. Prions lack nucleic acids and therefore genetic integrity, and PSTV caused by viroids also lack genetic integrity.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Infectious agents causing mad cow disease (prions) are proteins and do not contain nucleic acids. Therefore, they cannot show genetic integrity. This statement is true.

Step 2: Analyze the Reason: PSTV (Potato Spindle Tuber Viroid) is caused by viroids, which are infectious agents consisting of only RNA (no protein). This is an example of an infectious agent that does not have genetic integrity in the traditional sense (DNA). This statement is true.

Step 3: The reason explains why the assertion is true because both prions and viroids lack the traditional genetic material (DNA) and cannot show genetic integrity in the same way as organisms with DNA.

Why other options are wrong:
- Option B: If both Assertion and Reason are true but Reason is not the correct explanation of Assertion - The reason correctly explains the assertion.
- Option C: If Assertion is true but Reason is false - The reason is true.
- Option D: If both Assertion and Reason are false - Both are true.

Key Concepts:
- Prions
- Viroids
- Genetic integrity
- Mad cow disease
- PSTV

Memory Aids:
- Remember that prions are proteins, and viroids are RNA.

Quick Tips:
- Know the composition of prions and viroids.

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Consider the following features
(a) Lack of cell wall and cannot pass through bacterial filter.
(b) Can survive without oxygen and are free living eukaryotes.
(c) Reproduce mainly by fission
(d) Insensitive to penicillin.
Select the option which includes incorrect feature w.r.t. highly pleomorphic and smallest organisms.

View answer and explanation

Correct option: A

(a), (b) and (c)

The correct answer is A ((a), (b) and (c)) because Mycoplasmas have no cell wall, cannot pass through bacterial filters due to their size, can survive without oxygen, and reproduce mainly by fission. Therefore, statements (a), (b), and (c) are incorrect when referring to Mycoplasma.

Solution approach

Correct Answer + Core Reason: (a), (b) and (c). Mycoplasmas lack a cell wall, are small enough to pass through bacterial filters, and are not free-living eukaryotes.

Step-by-Step Solution:
Step 1: Understand the question. The question asks for incorrect features related to Mycoplasma.
Step 2: Recall the characteristics of Mycoplasma. They lack a cell wall, are small, can survive without oxygen, and are insensitive to penicillin.
Step 3: Evaluate statement (a). Lack of cell wall and cannot pass through bacterial filter. Mycoplasmas lack a cell wall, but they *can* pass through bacterial filters due to their small size. So, this statement is incorrect.
Step 4: Evaluate statement (b). Can survive without oxygen and are free-living eukaryotes. Mycoplasmas can survive without oxygen, but they are prokaryotes, not eukaryotes. So, this statement is incorrect.
Step 5: Evaluate statement (c). Reproduce mainly by fission. Mycoplasmas reproduce mainly by fission, so this statement is incorrect in the context of the question asking for *incorrect* features.
Step 6: Evaluate statement (d). Insensitive to penicillin. Mycoplasmas are insensitive to penicillin because they lack a cell wall, which is the target of penicillin. So, this statement is correct.
Step 7: Identify the incorrect features. Statements (a), (b), and (c) are incorrect.

Why other options are wrong:
- Option B: (a) and (b) only. This is incorrect because statement (c) is also incorrect in the context of the question.
- Option C: (c) only. This is incorrect because statements (a) and (b) are also incorrect.
- Option D: (d) only. This is incorrect because statement (d) is correct, not incorrect.

Key Concepts:
- Mycoplasma characteristics (lack of cell wall, small size, antibiotic resistance)
- Prokaryotic vs. eukaryotic cells
- Bacterial reproduction (fission)

Memory Aids:
- Remember that Mycoplasmas are bacteria that lack a cell wall and are very small.

Quick Tips:
- When answering questions about microorganisms, pay close attention to their defining characteristics.

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What is the total number of amino acids when
the capsid of TMV contains 2130 capsomers?

View answer and explanation

Correct option: A

336540

Option A (336540) is correct because each capsomer in TMV contains 158 amino acids, and multiplying this by the number of capsomers gives the total number of amino acids.

Solution approach

Correct Answer + Core Reason: 336540. Each capsomer contains 158 amino acids. Multiply the number of capsomers by the number of amino acids per capsomer.

Step-by-Step Solution:

Step 1: Identify the number of capsomers: 2130.

Step 2: Recall the number of amino acids per capsomer in TMV: 158.

Step 3: Multiply the number of capsomers by the number of amino acids per capsomer: 2130 * 158.

Step 4: Calculate: 2130 * 158 = 336540.

Why other options are wrong:
- Option B: 162622 - This number is incorrect and doesn't represent a valid calculation based on the given information.
- Option C: 948 - This number is incorrect and doesn't represent a valid calculation based on the given information.
- Option D: 387921 - This number is incorrect and doesn't represent a valid calculation based on the given information.

Key Concepts:
- TMV structure
- Capsomers and amino acids

Memory Aids:
- Remember that each TMV capsomer has 158 amino acids.

Quick Tips:
- Pay attention to the units and what each number represents.

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A pathogen accumulates its secretion in the
xylem vessels of a plant, whose seed epidermal
hairs are of great economic importance.

Identify the pathogen.

View answer and explanation

Correct option: A

Fusarium oxysporum

Fusarium oxysporum is the correct answer because it is a xylem-inhabiting pathogen that can affect plants with economically important seed epidermal hairs (e.g., cotton).

Solution approach

Correct Answer + Core Reason: Fusarium oxysporum. This pathogen accumulates in xylem vessels and can affect plants with economically important seed epidermal hairs.

Step-by-Step Solution:

Step 1: Analyze the given information: The pathogen accumulates in xylem vessels and affects plants with economically important seed epidermal hairs.

Step 2: Consider the characteristics of each pathogen:
- Fusarium oxysporum: A vascular wilt fungus that colonizes the xylem vessels, causing wilting and death. It can affect cotton, which has economically important seed epidermal hairs (cotton fibers).
- Xanthomonas citri: Causes citrus canker, affecting citrus plants.
- Phytophthora infestans: Causes late blight of potato and tomato.
- Puccinia purpurea: Causes sorghum rust.

Step 3: Determine the pathogen that matches the description: Fusarium oxysporum is the most likely candidate because it is a xylem-inhabiting pathogen that can affect plants with economically important seed epidermal hairs (e.g., cotton).

Why other options are wrong:
- Option B: Xanthomonas citri affects citrus plants, not plants with seed epidermal hairs.
- Option C: Phytophthora infestans affects potato and tomato, not plants with seed epidermal hairs.
- Option D: Puccinia purpurea affects sorghum, not plants with seed epidermal hairs.

Key Concepts:
- Plant pathogens
- Vascular wilt fungi
- Xylem colonization

Memory Aids:
- Fusarium = vascular wilt.

Quick Tips:
- Consider the host range and mode of infection of each pathogen.

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The plant cell have an eukaryotic structure with prominent …A… and cell wall is made up of …B…. Identify the correct options for A and B to complete the given statement

View answer and explanation

Correct option: A

A-chloroplast; B-cellulose

Option A (A-chloroplast; B-cellulose) is correct because plant cells have a eukaryotic structure with prominent chloroplasts, and their cell wall is made up of cellulose.

Solution approach

Correct Answer + Core Reason: A-chloroplast; B-cellulose. Plant cells contain chloroplasts for photosynthesis and have a cell wall made of cellulose.

Step-by-Step Solution:

Step 1: Recall the structure of plant cells.

Step 2: Plant cells are eukaryotic and contain organelles such as chloroplasts, which are responsible for photosynthesis.

Step 3: The cell wall of plant cells is primarily composed of cellulose, a polysaccharide.

Why other options are wrong:
- Option B: Plant cells have a nucleus, but their cell wall is not made of chitin (chitin is found in fungal cell walls).
- Option C: Plant cells have chloroplasts, but their cell wall is not primarily made of lignin (lignin provides rigidity but is not the main component).
- Option D: Plant cells have a nucleus, and their cell wall is made of polysaccharide (cellulose), but the more specific answer is cellulose.

Key Concepts:
- Plant cell structure
- Chloroplasts
- Cell wall composition
- Cellulose

Memory Aids:
- Plant cells = Chloroplasts + Cellulose.

Quick Tips:
- Know the key components of plant cells and their functions.

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In Basidiomycetes, the vegetative reproduction takes place by

View answer and explanation

Correct option: D

Fragmentation

Option D (Fragmentation) is correct because vegetative reproduction in Basidiomycetes commonly occurs through fragmentation of the mycelium.

Solution approach

Correct Answer + Core Reason: Fragmentation. It is a common method of vegetative reproduction in Basidiomycetes.

Step-by-Step Solution:

Step 1: Define Vegetative Reproduction: Vegetative reproduction is a form of asexual reproduction in plants and fungi where new individuals arise from portions of the parent organism.

Step 2: Basidiomycetes: In Basidiomycetes, vegetative reproduction often occurs through fragmentation, where the mycelium (the vegetative part of the fungus) breaks into fragments, and each fragment can grow into a new individual.

Step 3: Endospore: Endospores are typically formed by bacteria, not Basidiomycetes.

Step 4: Conidia: Conidia are asexual spores formed by some fungi, but fragmentation is more common in Basidiomycetes.

Step 5: Akinetes: Akinetes are resting cells formed by some bacteria and algae, not Basidiomycetes.

Step 6: Conclusion: Therefore, fragmentation is the primary method of vegetative reproduction in Basidiomycetes.

Why other options are wrong:
- Option A: Endospores are bacterial, not fungal.
- Option B: Conidia are asexual spores, but fragmentation is more common in Basidiomycetes.
- Option C: Akinetes are bacterial/algal, not fungal.

Key Concepts:
- Vegetative Reproduction
- Fragmentation
- Basidiomycetes

Memory Aids:
- "Basidiomycetes Break Apart" (Fragmentation in Basidiomycetes).

Quick Tips:
- Know the different methods of asexual reproduction in fungi.

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Regarding sexual reproduction in fungi. Which of the following statement is correct?

View answer and explanation

Correct option: A

Plasmogamy followed by karyogamy

Option A (Plasmogamy followed by karyogamy) is correct because it represents the correct sequence of events in fungal sexual reproduction.

Solution approach

Correct Answer + Core Reason: Plasmogamy followed by karyogamy. This is the correct sequence of events in sexual reproduction in fungi.

Step-by-Step Solution:

Step 1: Define the terms:
- Plasmogamy: Fusion of protoplasts of two cells.
- Karyogamy: Fusion of two nuclei.

Step 2: Understand the sequence: In fungi, sexual reproduction involves the following steps:
1. Plasmogamy: The fusion of the cytoplasm of two parent cells.
2. Karyogamy: The fusion of the two nuclei within the same cell.

Step 3: Conclusion: The correct sequence is Plasmogamy, then Karyogamy.

Why other options are wrong:
- Option B: Karyogamy occurs after plasmogamy, not before.
- Option C: Karyogamy and plasmogamy do not occur together; they are sequential events.
- Option D: Sexual reproduction is present in most fungi, so it is not absent.

Key Concepts:\nSexual reproduction in fungi, plasmogamy, karyogamy.

Memory Aids:\nPK - Plasmogamy then Karyogamy.

Quick Tips:\nRemember the order of nuclear events in fungal sexual reproduction.

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The smallest free-living organism is

View answer and explanation

Correct option: B

Mycoplasma

Option B (Mycoplasma) is correct because Mycoplasmas are the smallest known free-living organisms, lacking a cell wall.

Solution approach

Correct Answer + Core Reason: Mycoplasma. Mycoplasmas are the smallest free-living organisms.

Step-by-Step Solution:

Step 1: Recall the size and characteristics of different microorganisms.

Step 2: Mycoplasmas are the smallest known free-living organisms and lack a cell wall.

Why other options are wrong:
- Option A: Viruses are smaller than mycoplasmas, but they are not free-living; they require a host cell to replicate.
- Option C: Diatoms are unicellular algae and are larger than mycoplasmas.
- Option D: Cyanobacteria are photosynthetic bacteria and are larger than mycoplasmas.

Key Concepts:
- Mycoplasmas
- Size of microorganisms

Memory Aids:
- Mycoplasma = Smallest

Quick Tips:
- Remember that mycoplasmas lack a cell wall and are the smallest free-living organisms.

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Slime moulds are dependent on

View answer and explanation

Correct option: B

Dead and decaying organic matter

The correct answer is B (Dead and decaying organic matter) because slime molds are saprophytic organisms that feed on dead and decaying organic matter.

Solution approach

Correct Answer + Core Reason: Slime moulds are dependent on dead and decaying organic matter.

Step-by-Step Solution:

Step 1: Understand the nutritional mode of slime molds.

Step 2: Slime molds are saprophytes, meaning they obtain nutrients from dead and decaying organic matter.

Step 3: They play an important role in decomposition and nutrient cycling in ecosystems.

Step 4: They engulf bacteria, fungi, and other microorganisms found in decaying matter.

Why other options are wrong:
- Option A (Water plants): Slime molds are not typically associated with water plants as a primary food source.
- Option C (Plants): Slime molds are not parasitic on living plants; they feed on decaying matter.
- Option D (Weeds): Slime molds are not specifically dependent on weeds; they feed on a variety of decaying organic materials.

Key Concepts:
- Saprophytes
- Slime mold nutrition
- Decomposition

Memory Aids:
- Saprophyte = decaying matter

Quick Tips:
- Remember that saprophytes are decomposers.

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Organisms having cellular body organisation and with ability to oxidise various inorganic substances and use the released energy for their ATP production are
A. Nitrifying bacteria of Kingdom Monera in Linnaeus's system.
B. Sulphur bacteria belonging to Kingdom Protista in Haeckel's system.
C. Holophytic protistans.
D. Non-pigmented iron bacteria of Kingdom Monera in Whittaker's system.

View answer and explanation

Correct option: D

B and D are correct

Option D (B and D are correct) is correct because sulphur bacteria in Kingdom Protista (Haeckel's system) and non-pigmented iron bacteria in Kingdom Monera (Whittaker's system) can oxidize inorganic substances for ATP production.

Solution approach

Correct Answer + Core Reason: B and D are correct. Sulphur bacteria and non-pigmented iron bacteria fit the description of organisms oxidizing inorganic substances for ATP production.

Step-by-Step Solution:
Step 1: Understand the question. The question asks about organisms that can oxidize inorganic substances for ATP production.
Step 2: Evaluate option A. Nitrifying bacteria of Kingdom Monera in Linnaeus's system can oxidize inorganic substances. However, Linnaeus's system is outdated.
Step 3: Evaluate option B. Sulphur bacteria belonging to Kingdom Protista in Haeckel's system can oxidize inorganic substances.
Step 4: Evaluate option C. Holophytic protistans are photosynthetic and do not oxidize inorganic substances for energy.
Step 5: Evaluate option D. Non-pigmented iron bacteria of Kingdom Monera in Whittaker's system can oxidize inorganic substances.
Step 6: Determine the correct combination. Options B and D are correct.

Why other options are wrong:
- Option A: A, B and D are correct. Option A includes nitrifying bacteria in Linnaeus's system, which is outdated.
- Option C: C and D are correct. Option C includes holophytic protistans, which are photosynthetic and don't oxidize inorganic substances for energy.

Key Concepts:
- Chemosynthesis
- Oxidation of inorganic substances
- Bacterial metabolism
- Kingdom classifications (Linnaeus, Haeckel, Whittaker)

Memory Aids:
- Remember that chemosynthetic bacteria obtain energy by oxidizing inorganic compounds.

Quick Tips:
- When answering questions about microbial metabolism, consider the different ways organisms obtain energy.

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Limitations of two kingdom classification is?

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the two-kingdom classification failed to differentiate between unicellular and multicellular organisms, photosynthetic and non-photosynthetic organisms, and prokaryotic and eukaryotic cells.

Solution approach

Correct Answer + Core Reason: All of these. The two-kingdom classification system had several limitations, including the inability to distinguish between different types of organisms.

Step-by-Step Solution:

Step 1: Recall the limitations of the two-kingdom classification system.

Step 2: Consider the placement of unicellular and multicellular organisms together (e.g., Chlamydomonas and Spirogyra).

Step 3: Consider the placement of photosynthetic and non-photosynthetic organisms together (e.g., algae and fungi).

Step 4: Consider the placement of prokaryotic and eukaryotic cells together (e.g., bacteria and other cells).

Step 5: Recognize that all the given options are valid limitations of the two-kingdom classification.

Why other options are wrong:
- Option A: This is a valid limitation, but not the only one.
- Option B: This is a valid limitation, but not the only one.
- Option C: This is a valid limitation, but not the only one.

Key Concepts:
- Two-kingdom classification system
- Limitations of early classification systems
- Prokaryotes vs. Eukaryotes
- Unicellular vs. Multicellular
- Photosynthetic vs. Non-photosynthetic

Memory Aids:
- Remember that the two-kingdom system was very basic and couldn't account for the diversity of life.

Quick Tips:
- When asked about limitations, consider the broad categories that the classification system fails to address.

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Consider the following statements
I. Biological classification is the scientific ordering of organisms in a hierarchial series of groups on the basis of their relationships, i.e., morphological, evolutionary and others
II. Whittaker classified organisms on the basis of autotrophic and heterotrophic mode of nutrition
III. In five kingdom system of classification, living organisms can be divided into prokaryotic and eukaryotic cells on the basis of cell structure
Which of the statements given above are correct?

View answer and explanation

Correct option: D

I, II and III

Option D (I, II and III) is correct because all three statements accurately describe biological classification, Whittaker's criteria, and the division of organisms based on cell structure.

Solution approach

Correct Answer + Core Reason: I, II and III. All three statements are correct regarding biological classification and Whittaker's system.

Step-by-Step Solution:

Step 1: Evaluate each statement:
- I. Biological classification is the scientific ordering of organisms in a hierarchial series of groups on the basis of their relationships, i.e., morphological, evolutionary and others: This is a correct definition of biological classification.
- II. Whittaker classified organisms on the basis of autotrophic and heterotrophic mode of nutrition: This is correct; mode of nutrition was one of Whittaker's criteria.
- III. In five kingdom system of classification, living organisms can be divided into prokaryotic and eukaryotic cells on the basis of cell structure: This is also correct; cell structure is a key criterion.

Step 2: Since all three statements are correct, the answer is I, II, and III.

Why other options are wrong:
- Option A: I and II is incorrect because statement III is also correct.
- Option B: I and III is incorrect because statement II is also correct.
- Option C: II and III is incorrect because statement I is also correct.

Key Concepts:
- Biological classification
- Whittaker's five kingdom classification
- Prokaryotic vs. eukaryotic cells

Memory Aids:
- Remember the key criteria for classification.

Quick Tips:
- Understand the principles of biological classification.

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Select incorrect matching

View answer and explanation

Correct option: D

Thermoacidophiles - Heterotrophs and pathogenic

Option D (Thermoacidophiles - Heterotrophs and pathogenic) is the incorrect matching because while some thermoacidophiles can be heterotrophic, they are not all pathogenic.

Solution approach

Correct Answer + Core Reason: Thermoacidophiles - Heterotrophs and pathogenic. Not all thermoacidophiles are pathogenic; some are autotrophic.

Step-by-Step Solution:
Step 1: Understand the question. The question asks to select the incorrect matching.
Step 2: Evaluate option A. Methanogens - Obligate anaerobes. This is correct; methanogens are obligate anaerobes.
Step 3: Evaluate option B. Whittaker’s system of Classification - Brought together Chlorella with Paramoecium and Amoeba in Protista. This is correct; Whittaker's system placed these organisms in Protista.
Step 4: Evaluate option C. Recycling nutrients - Chemosynthetic bacteria. This is correct; chemosynthetic bacteria play a role in recycling nutrients.
Step 5: Evaluate option D. Thermoacidophiles - Heterotrophs and pathogenic. This is incorrect; while some thermoacidophiles are heterotrophic, not all are pathogenic.

Why other options are wrong:
- Option A: Methanogens - Obligate anaerobes. This is a correct matching.
- Option B: Whittaker’s system of Classification - Brought together Chlorella with Paramoecium and Amoeba in Protista. This is a correct matching.
- Option C: Recycling nutrients - Chemosynthetic bacteria. This is a correct matching.

Key Concepts:
- Methanogens
- Whittaker's classification system
- Chemosynthetic bacteria
- Thermoacidophiles

Memory Aids:
- Remember that methanogens are obligate anaerobes and that Whittaker's system grouped certain organisms into Protista.

Quick Tips:
- When evaluating matching questions, consider the defining characteristics of each term.

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Identify from the following examples, a fungus, which is of medicinal importance

View answer and explanation

Correct option: C

Penicillium

Option C (Penicillium) is correct because Penicillium is used to produce the antibiotic penicillin, making it of medicinal importance.

Solution approach

Correct Answer + Core Reason: Penicillium. It is used to produce the antibiotic penicillin.

Step-by-Step Solution:

Step 1: Identify Medicinal Importance: Medicinal importance refers to the use of a substance in treating or preventing diseases.

Step 2: Penicillium: Penicillium is a genus of fungi, and certain species (e.g., Penicillium notatum, Penicillium chrysogenum) are used to produce the antibiotic penicillin.

Step 3: Agaricus: Agaricus is an edible mushroom but does not have significant medicinal applications.

Step 4: Saccharomyces: Saccharomyces (yeast) is used in baking and brewing but is not primarily known for medicinal properties, although it is used in some probiotic supplements.

Step 5: Cercospora: Cercospora is a plant pathogen and does not have medicinal importance.

Step 6: Conclusion: Therefore, Penicillium is the fungus with medicinal importance due to its production of penicillin.

Why other options are wrong:
- Option A: Agaricus is edible but not medicinally important.
- Option B: Saccharomyces is used in food production, not primarily medicinal.
- Option D: Cercospora is a plant pathogen, not medicinal.

Key Concepts:
- Antibiotics
- Penicillium
- Medicinal Fungi

Memory Aids:
- "Penicillin from Penicillium".

Quick Tips:
- Remember the specific uses of common fungi.

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Thermococcus,Methanococcus and Methanobacterium are

View answer and explanation

Correct option: C

Archaebacteria with negatively supercoiled DNA as eukaryotes but lacking histones

Option C (Archaebacteria with negatively supercoiled DNA as eukaryotes but lacking histones) is correct because these organisms are Archaebacteria, have negatively supercoiled DNA, and lack histones.

Solution approach

Correct Answer + Core Reason: Archaebacteria with negatively supercoiled DNA as eukaryotes but lacking histones because Thermococcus, Methanococcus, and Methanobacterium are Archaebacteria with unique DNA structure.

Step-by-Step Solution:

Step 1: Thermococcus, Methanococcus, and Methanobacterium are classified as Archaebacteria.

Step 2: Archaebacteria have negatively supercoiled DNA, similar to eukaryotes, but they lack histones, which are proteins that package DNA in eukaryotes.

Step 3: Conclusion: The correct description is Archaebacteria with negatively supercoiled DNA but lacking histones.

Why other options are wrong:
- Option A: Incorrect because Archaebacteria lack eukaryotic histone homologues.
- Option B: Incorrect because Bacteria do not have cytoskeleton like eukaryotes.
- Option D: Incorrect because Bacteria do not have positively coiled DNA, cytoskeleton, or mitochondria.

Key Concepts:
- Archaebacteria
- DNA structure
- Supercoiling
- Histones

Memory Aids:
- Archaebacteria = Unique DNA, No Histones

Quick Tips:
- Remember the key differences between Archaebacteria, Bacteria, and Eukaryotes.

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Viruses contain

View answer and explanation

Correct option: C

Either DNA or RNA

Option C (Either DNA or RNA) is correct because viruses contain either DNA or RNA as their genetic material, but not both.

Solution approach

Correct Answer + Core Reason: Either DNA or RNA. Viruses contain either DNA or RNA as their genetic material, but not both.

Step-by-Step Solution:

Step 1: Recall the structure of viruses.

Step 2: Viruses consist of a nucleic acid core (DNA or RNA) surrounded by a protein coat (capsid).

Step 3: The nucleic acid core can be either DNA or RNA, but not both in the same virus.

Why other options are wrong:
- Option A: Some viruses contain RNA, but not all viruses contain only RNA. Some contain DNA.
- Option B: Some viruses contain DNA, but not all viruses contain only DNA. Some contain RNA.
- Option D: Viruses contain either DNA or RNA, so this option is incorrect.

Key Concepts:
- Virus structure
- Nucleic acids in viruses
- DNA
- RNA

Memory Aids:
- Viruses = DNA or RNA + Protein coat.

Quick Tips:
- Remember that viruses can have either DNA or RNA, but not both.

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Which of the following statement is correct?

View answer and explanation

Correct option: A

Viruses are obligate parasites

Option A (Viruses are obligate parasites) is correct because viruses can only replicate inside a host cell, making them obligate parasites.

Solution approach

Correct Answer + Core Reason: Viruses are obligate parasites. They require a host cell to replicate.

Step-by-Step Solution:

Step 1: Obligate parasites are organisms that cannot complete their life cycle without exploiting a suitable host.

Step 2: Viruses are obligate intracellular parasites because they lack the cellular machinery to replicate on their own and must use the host cell's resources.

Why other options are wrong:
- Option B: Not all fungi are pathogenic; many are beneficial or harmless.
- Option C: All algae are eukaryotes is correct, but it's not the *most* correct statement in the context of the question.
- Option D: Not all bacteria are harmful to mankind; many are beneficial or harmless.

Key Concepts:
- Obligate parasites
- Viral replication
- Characteristics of fungi, algae, and bacteria

Memory Aids:
- Viruses = Obligate Parasites.

Quick Tips:
- Remember the definition of obligate parasites and how it applies to viruses.

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Which is false for nutrition in Amoeba?

View answer and explanation

Correct option: D

Photoautotroph

Option D (Photoautotroph) is correct because Amoeba is heterotrophic and obtains nutrition by holozoic nutrition, not by photosynthesis.

Solution approach

Correct Answer + Core Reason: Photoautotroph. Amoeba is heterotrophic, not photoautotrophic.

Step-by-Step Solution:

Step 1: Analyze option A: Amoeba is omnivorous, feeding on various small organisms.

Step 2: Analyze option B: Amoeba uses pseudopodia to engulf food.

Step 3: Analyze option C: Amoeba exhibits holozoic nutrition (ingestion, digestion, absorption, assimilation, egestion).

Step 4: Analyze option D: Amoeba is not a photoautotroph; it cannot produce its own food through photosynthesis.

Step 5: Conclusion: Option D is false for nutrition in Amoeba.

Why other options are wrong:
- Option A: This is a correct statement about Amoeba.
- Option B: This is a correct statement about Amoeba.
- Option C: This is a correct statement about Amoeba.

Key Concepts:
- Nutrition in Amoeba
- Heterotrophic vs. Autotrophic
- Holozoic nutrition

Memory Aids:
- Amoeba = heterotroph

Quick Tips:
- Remember the mode of nutrition in different organisms.

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Identify the odd one out from the given options.

View answer and explanation

Correct option: A

Ustilago

Ustilago is the odd one out because it belongs to Basidiomycetes, while the others are Deuteromycetes.

Solution approach

Correct Answer + Core Reason: Ustilago. Ustilago belongs to Basidiomycetes, while Alternaria, Colletotrichum, and Trichoderma belong to Deuteromycetes.

Step-by-Step Solution:

Step 1: Recall the classification of each fungus.

Step 2: Alternaria, Colletotrichum, and Trichoderma are all members of Deuteromycetes (Fungi Imperfecti).

Step 3: Ustilago is a member of Basidiomycetes.

Why other options are wrong:
- Option B: Alternaria is a Deuteromycete.
- Option C: Colletotrichum is a Deuteromycete.
- Option D: Trichoderma is a Deuteromycete.

Key Concepts:
- Classification of fungi
- Deuteromycetes (Fungi Imperfecti)
- Basidiomycetes

Memory Aids:
- Deuteromycetes are \"imperfect\" because their sexual stage is unknown.

Quick Tips:
- Know the defining characteristics of different fungal classes.

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