Thursday, 13 March 2025

NEET 2024 Question paper with answer

Examination Pattern 2025

The Test pattern of NEET (UG) - 2025 comprises Physics, Chemistry, Biology (Botany & Zoology). Candidates are hereby informed that starting from NEET (UG)-2025 will revert back to pre-COVID pattern vide public notice dated 25.01.2025. The question paper will consist of 180 compulsory questions which must be attempted by the candidates in 180 minutes. These changes have been incorporated in the NEET (UG)-2025 test pattern. (Authority:- Vide DO letter no. 14023/19/2023/UGMEB. 

The pattern for the NEET (UG) - 2025 examination for admission in the Session 2025-26 is as follows:

Important Points to Note: To answer a question, the candidates need to choose one option corresponding to the correct answer or the most appropriate answer. Marking will be done as per following criteria: 

(i) Correct answer or the most appropriate answer: Four marks (+4) 

(ii) Any incorrect option marked will be given minus one mark (-1). 

(iii) Unanswered: No mark (0).

Medium of the Question Papers

Medium of the Question Papers Drawing from the National Education Policy (NEP), the NEET (UG) -2025 is being conducted in English, Hindi, Assamese, Bengali, Gujarati, Kannada, Malayalam, Marathi, Odia, Punjabi, Tamil, Telugu and Urdu. 3.5.1. Candidates can opt for a Question Paper in any one of the following 13 languages: 



 Let us see 2024 question paper (Botany Part) with explanation. If you get anything wrong, please let us notice.

Botany: Section - A (Q. No. 101 to 135)

101. Auxin is used by gardeners to prepare weed-free lawns. But no damage is caused to grass as auxin

(1) promote abscission of mature leaves only 

(2) does not affect mature monocotyledonous plants

(3) can help in cell division in grasses, to produce growth 

(4) promotes apical dominance 

Answer- The correct answer is: (2) does not affect mature monocotyledonous plants

[Auxins, such as 2,4-D (2,4-Dichlorophenoxyacetic acid), are commonly used as selective herbicides. They promote uncontrolled growth in broadleaf (dicot) weeds, leading to their death, while grasses (monocots) remain unaffected due to differences in their auxin metabolism and sensitivity].

102. Lecithin, a small molecular weight organic compound found in living tissues, is an example of

1) phospholipids

2) Glycerides 

3) Carbohydrates 

4) Amino acids 

Answer-- The correct answer is: (1) Phospholipids

[Lecithin is a type of phospholipid that contains a phosphate group and is a key component of cell membranes. It is commonly found in egg yolks, soybeans, and other biological tissues, playing a crucial role in membrane structure and function].


103. Match List I with List II

      List I                                                           List II

A. Two or more  alternative                           I. Back cross

forms of a gene

 B. Cross of F1 progeny with                        II. Ploidy 

homozygous receive parent

C. Cross of F progeny with                           III. Allele

any of the parents                       

 D. Number of   chromosome                        IV. Test cross 

set in plant 

 Choose the correct answer from the options given below:

(1) A-II,  B-I,   C-III, D-IV

(2) A-III, B-I,   C-I,   D-II

(3) A-IV, B-III, C-III, D-IV

(4) A-IV, B-III, C-III, D-IV

Answer-- The correct answer is: (2) A-III, B-I, C-IV, D-II

 You may read the following-

 A. Two or more alternative forms of a gene → III. Allele

(Alleles are different forms of a gene that occupy the same position on a chromosome.)

 B. Cross of F1 progeny with a homozygous recessive parent → IV. Test cross

(A test cross is used to determine the genotype of an organism by crossing it with a homozygous recessive parent.)

 C. Cross of F1 progeny with any of the parents → I. Back cross

(A back cross is when an F1 individual is crossed with one of its parental genotypes.)

 D. Number of chromosome sets in a plant → II. Ploidy

(Ploidy refers to the number of complete sets of chromosomes in a cell.)


104. Identify the set of correct statements:

A. The flowers of Vallisneria are colourful and produce nectar 

B. The flowers of waterlily are not pollinated by water 

C. In most of water -pollinated species, the pollen grains are protected from wetting.

D. Pollen grains of some hydrophytes are long and ribbon like 

E. In some hydrophytes, the pollen grains are carried passively inside water.

Choose the correct answer from the options given below:

1. A , B, C and D only 

2. A, C, D and E only 

3. B, C, D and E only 

4. C, D and E only 

 Answer-- The correct answer is: 3. B, C, D, and E only

Explanation:

A. The flowers of Vallisneria are colourful and produce nectar → ❌ Incorrect

Vallisneria is a water-pollinated (hydrophilous) plant. Its flowers are not colorful and do not produce nectar, as they rely on water for pollination.

 B. The flowers of water lily are not pollinated by water → ✅ Correct

Water lily (Nymphaea) is pollinated by insects, not water, even though it grows in aquatic environments.

 C. In most water-pollinated species, the pollen grains are protected from wetting → ✅ Correct

Water-pollinated plants have adaptations such as mucilage covering or being lightweight to prevent pollen from sinking or getting wet.

 D. Pollen grains of some hydrophytes are long and ribbon-like → ✅ Correct

In hydrophytes like Zostera (seagrass), pollen grains are long and ribbon-like to increase the chances of contact with the stigma.

 E. In some hydrophytes, the pollen grains are carried passively inside water → ✅ Correct

In submerged hydrophytes, pollen grains are transported by water currents passively.

Thus, the correct set of statements is B, C, D, and E (Option 3).


105. List of endangered species was released by

1) WWF.     2) FOAM.  3) IUCN.  4. GEAC

 Answer-The correct answer is: 3) IUCN (International Union for Conservation of Nature)

Explanation:

[The IUCN Red List of Threatened Species is the most comprehensive inventory of the conservation status of biological species. It classifies species into different categories such as Endangered (EN), Critically Endangered (CR), Vulnerable (VU), and Extinct (EX) based on their risk of extinction].

 106. What is the fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism?

A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism.

B. It may be integrated into the genome of the recipient

C. It may multiply and be inherited along with the host DNA

D.The alien piece of DNA is not an integral part of chromosome

E. It shows ability to replicate

Choose the correct answer from the option below:

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

Answer- The Correct Answer is: 2. B and C only

When a piece of DNA carrying a gene of interest is transferred into an alien organism, several outcomes are possible, but not all options provided are accurate. Let's analyze each option:

1.      : The piece of DNA would be able to multiply itself independently in the progeny cells of the organism.

o    Incorrect. A piece of DNA, even if it carries a gene of interest, cannot multiply independently without being integrated into the host genome or being part of a self-replicating plasmid.

2.      : It may be integrated into the genome of the recipient

o    Correct. The transferred DNA can be integrated into the host genome through homologous recombination or other integration mechanisms, allowing it to become a stable part of the host's genetic material.

3.      : It may multiply and be inherited along with the host DNA

o    Correct. If the DNA is integrated into the host genome or is part of a self-replicating plasmid, it can be replicated and inherited along with the host DNA during cell division.

4.      : The alien piece of DNA is not an integral part of chromosome

o    Incorrect in the context of integration. If the DNA is integrated into the host genome, it becomes an integral part of the chromosome.

5.      : It shows ability to replicate

o    Incorrect unless it is part of a self-replicating plasmid or integrated into the host genome. Naked DNA does not replicate on its own.


107. Which of the following are required for the dark reaction of photosynthesis

A.  Light

B. Chlorophyll

C.  CO2

D.   ATP

E.   NADPH

 Choose the correct answer from the option given below:

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

Answer- To answer this question, let's break down the requirements for the dark reaction (also known as the Calvin cycle) of photosynthesis:

Dark Reaction Requirements:

  1. CO2 (Carbon Dioxide): This is the primary carbon source that is fixed into organic compounds during the Calvin cycle.
  2. ATP (Adenosine Triphosphate): Energy from ATP is used to drive the reactions in the Calvin cycle.
  3. NADPH (Nicotinamide adenine dinucleotide phosphate): This is a reducing agent produced in the light-dependent reactions and used in the Calvin cycle to reduce CO2 into organic compounds.

Options Analysis:

  • A. Light: Not required directly in the dark reaction. Light is necessary for the light-dependent reactions, which produce ATP and NADPH used in the dark reaction.
  • B. Chlorophyll: Essential for absorbing light in the light-dependent reactions but not directly involved in the dark reaction.
  • C. CO2: Required for the dark reaction as it is the carbon source fixed into glucose.
  • D. ATP: Required as an energy source.
  • E. NADPH: Required as a reducing agent.

Correct Options:

  • C. CO2
  • D. ATP
  • E. NADPH

The correct option for this answer is 2. C, D and E only

These are the components directly required for the dark reaction of photosynthesis.


108. The type of conservation in which the threatened species are taken out from their natural habitat and placed in species setting where they can be protected and given special care is called:

  1. Biodiversity conservation
  2. Semi-conservative method
  3. Sustainable development
  4. In-situ conservation

Answer- 

Analysis: No correct options are here, but we have to choose the answer. Lets see our analysis-

The correct option for the type of conservation described is not listed directly as described, but based on the definition provided, it aligns more closely with ex-situ conservation. However, since "ex-situ conservation" is not an option, let's clarify the terms:

  • Ex-situ conservation involves removing threatened species from their natural habitat and placing them in controlled environments like zoos, botanical gardens, or seed banks for protection and care. This matches the description given.
  • In-situ conservation (Option 4) involves protecting species in their natural habitats.
  • Biodiversity conservation (Option 1) is a broader term that encompasses various methods to preserve biodiversity.
  • Sustainable development (Option 3) refers to meeting the needs of the present without compromising the ability of future generations to meet their own needs, which is not directly related to the specific conservation method described.
  • Semi-conservative method (Option 2) is not a recognized term in conservation biology.

Given the options, none directly states "ex-situ conservation," but based on the description, the closest correct concept is ex-situ conservation, which is not listed. However, if you must choose from the provided options, none accurately describe the method as defined. But if you're looking for a term related to conservation in a controlled environment, you would typically look for "ex-situ conservation," which is not an option here.

If you had to choose the least incorrect option based on the context of conservation, Option 1: Biodiversity conservation might seem like a broad category that could encompass ex-situ methods, but it's not a precise match.

 109. Given below are two statements:

Statement I: Bt toxin toxins are insect group specific and coded by a gene cry IAc

Statement II: Bt toxin exists as inactive protoxin in B. thuringnesis. However, after ingestion by the insect the inactive protoxin gets converted into active form due to acidic pH of the insect gut.

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

  1. Both statement I and satement II are false
  2. Statement I is true but statement II is false
  3. Statement I is false but statement II is true
  4. Both statement I and satement II are true

 Answer-

Let's analyze each statement logically: Read carefully following analysis and grasp the concept for future.

Statement I:

"Bt toxin toxins are insect group specific and coded by a gene cry IAc."

  • This statement is partially incorrect.
    • Insect group specificity: True. Different Bt toxins target specific insect groups (e.g., Lepidoptera, Coleoptera, Diptera).
    • Gene specificity: The statement implies that all Bt toxins are coded by cry IAc, which is false.
      • cry IAc is just one of many genes encoding Bt toxins. Other genes like cry1Ab, cry3Aa, and cry4Ba also exist, each targeting different insects.
      • cry IAc is specific to certain lepidopteran pests.

Statement II:

"Bt toxin exists as inactive protoxin in B. thuringiensis. However, after ingestion by the insect the inactive protoxin gets converted into active form due to acidic pH of the insect gut."

  • This statement is false.
    • Existence as inactive protoxin: True. Bt toxins are produced as inactive protoxins in B. thuringiensis.
    • Activation mechanism: False. The activation occurs in the alkaline midgut (pH 9–12) of susceptible insects, not due to acidic pH.
      • The alkaline environment dissolves the crystalline protoxin, and midgut proteases cleave it into the active toxin.

Conclusion:

  • Statement I is incorrect because it inaccurately suggests all Bt toxins are encoded by cry IAc.
  • Statement II is incorrect because it incorrectly states that the protoxin is activated by acidic pH.

Final Answer: Option 1 (Both statement I and statement II are false) is correct.

 

110. A transcription unit in DNA is defined primarily by the three regions in DNA and these are with respect to upstream and down stream ends

  1. Structural gene, Transposons, Operator gene
  2. Inducer, Repressor gene and Terminator
  3. Promoter, Structural gene, Terminator
  4. Repressor, Operator gene, Structural gene

Answer- The correct answer is: 3. Promoter, Structural gene, Terminator

Logical Explanation:

A transcription unit in DNA is defined by three key regions that are essential for the process of transcription, with respect to upstream and downstream ends:

1.      Promoter:

o    The promoter is located upstream of the structural gene.

o    It is the site where RNA polymerase binds to initiate transcription.

o    It determines the start point and direction of transcription.

2.      Structural Gene:

o    This region contains the coding sequence that is transcribed into RNA.

o    It provides the template for RNA synthesis.

3.      Terminator:

o    The terminator is located downstream of the structural gene.

o    It signals the end of transcription, causing RNA polymerase to stop and release the newly synthesized RNA.

Why Other Options Are Incorrect:

1.      Structural gene, Transposons, Operator gene:

o    Transposons and operator genes are not part of a transcription unit. Operators are regulatory elements in operons, and transposons are mobile genetic elements.

2.      Inducer, Repressor gene, Terminator:

o    Inducers and repressors are regulatory molecules or genes but are not structural components of a transcription unit.

3.      Repressor, Operator gene, Structural gene:

o    Repressors and operators are part of regulatory mechanisms (e.g., in operons) but are not components of a transcription unit.

Thus, Promoter, Structural gene, Terminator (Option 3) correctly defines the primary regions of a transcription unit.


111. In the given figure, which component has thin outer walls and highly thickened inner walls?

  1. D             (2) A          (3) B           (4) C

Answer- The correct answer is (4) C


112. Hind II always cuts DNA molecules at a particular point called recognition sequence and it consists of:

  1. 6 bp      (2) 4 bp      (3) 10 bp       (4) 8 bp

The correct answer is (1) 6 bp

Logical Explanation:

HindII, the first discovered restriction enzyme of the Type II class, specifically recognizes and cuts DNA at a 6 base pair (bp) palindromic sequence. This recognition site is unique to HindII and ensures precise cleavage within the DNA molecule.

Supporting Evidence:

1.      Recognition Sequence Length:

o Type II restriction enzymes, including HindII, generally recognize short DNA sequences that are 4–8 base pairs (bp) in length. HindII specifically recognizes a 6 bp palindromic sequence.

2.      Mechanism of Action:

o HindII binds to its recognition site and cleaves the DNA precisely within this 6 bp sequence. This specificity makes it a valuable tool in molecular biology for DNA manipulation.

Why Other Options Are Incorrect:

·         4 bp (Option 2): Some restriction enzymes recognize 4 bp sequences, but HindII specifically recognizes a 6 bp sequence.

·         10 bp (Option 3) and 8 bp (Option 4): These lengths are outside the range of HindII's recognition sequence.

Thus, the correct answer is 6 bp.

 

113. Identify the type of flowers based on the position of calyx, corolla and androecium with respect to the ovary from the given figures (a) and (b)

  1. (a) Hypogynous; (b) Epigynous
  2. (a) Perigynous; (b) Epigynous
  3. (a) Perigynous; (b) Perigynous
  4. (a) Epigynous; (b) Hypogynous

Answer- The correct answer is (3) (a) Perigynous; (b) Perigynous


114. Which of the following is an example of actinomorphic flower?

  1. Cassia    (2) Pisum       (3) Sesbania      (4) Datura

Answer- The correct answer is (4) Datura

 115. Which one of the following is not a criterion for classification of fungi?

  1. Mode of nutrition  (2) Mode of spore formation  (3) Fruiting body  (4) Morphology of mycelium

Answer- The correct answer is (1) Mode of nutrition

 

116. The equation of Verhulst-Pearl logistic growth is dN/dt=rN[K-N/K]

From this equation, K indicates:

  1. Biotic potential              (2) Carrying capacity   

     (3) Population density         (4) Intrinsic rate of natural increase

Answer- The correct answer is (2) Carrying capacity


117. Which one of the following can be explained on the basis of Mendel’s Law of Dominance?

A.                Out of one pair of factors one is dominant and the other is recessive.

B.                 Alleles do not show any expression and both the characters appear as such in F2 generation

C.                 Factors occur in pairs in normal diploid plants

D.                The discrete unit controlling a particular character is called factor

E.                 The expression of only one of the parental characters is found in a monohybrid cross

 Choose the correct answer from the option given below:

(1) A, C, D and E only

(2) B, C and D only

(3) A, B, C, D and E

(4) A, B and C only

Answer- The correct answer is  (1) A, C, D and E only

More reading and concept behind this question

Analysis of Mendel’s Law of Dominance

To determine which statements align with Mendel’s Law of Dominance, let’s evaluate each option based on the law’s principles:

Key Points of the Law of Dominance:

  1. In a heterozygous pair, one allele (dominant) masks the expression of the recessive allele.
  2. Only the dominant trait appears in the F1 generation of a monohybrid cross.
  3. Recessive traits reappear in the F2 generation when homozygous.
Evaluation of Options:
  1. A: Out of one pair of factors, one is dominant and the other is recessive.
     ✅ Directly aligns with the law. Dominance implies one allele suppresses the other.

  2. B: Alleles do not show any expression, and both characters appear as such in F2.
     ❌ Incorrect. The dominant allele is expressed in F1, while the recessive reappears only in homozygous F2 individuals.

  3. C: Factors occur in pairs in normal diploid plants.
     ✅ Mendel’s model assumes paired factors (alleles), forming the basis for dominance interactions.

  4. D: The discrete unit controlling a character is called a factor.
     ✅ Mendel’s concept of "factors" (now genes) as discrete units underpins his laws, including dominance.

  5. E: The expression of only one parental character is found in a monohybrid cross.
     ✅ The F1 generation exhibits only the dominant trait, a core tenet of the law.

118. Match List I with List II

 List I                                  List II

A.  Rhizopus                I. Mushroom

B.  Ustilago                 II. St fungus

C.  Puccinia                III. Bread mould

D.  Agaricus               IV. Rust fungus

Choose the correct answer from the options given

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

Answer- Correct Option: (4) A-III, B-II, C-IV, D-I

Matching List I with List II:

  1. Rhizopus → III. Bread mould

  2. Rhizopus species (e.g., R. stolonifer) are saprophytic fungi commonly known as bread molds, thriving on decaying organic matter like bread.
  3. Ustilago → II. Smut fungus
    • Ustilago (e.g., U. maydis) are parasitic fungi causing smut diseases in crops like maize, forming tumors and teliospores
  4. Puccinia → IV. Rust fungus

    • Puccinia species (e.g., P. graminis) are plant pathogens causing rust diseases, characterized by reddish-brown pustules on leaves

  5. Agaricus → I. Mushroom

    • Agaricus includes edible species like the button mushroom (A. bisporus), a widely cultivated mushroom

Explanation:

  • Rhizopus-III and Agaricus-I are direct matches based on their ecological roles and common names.

  • Ustilago-II and Puccinia-IV align with their pathogenic classifications (smut and rust fungi, respectively).

The correct sequence is A-III, B-II, C-IV, D-I, making Option 4 the accurate choice.

119. Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of :

  1. Feedback inhibition
  2. Competitive inhibition
  3. Enzyme activation
  4. Cofactor inhibition

Answer- The Correct Answer is: Competitive inhibition

Explanation:

Malonate's inhibition of succinate dehydrogenase (SDH) is a classic example of competitive inhibition due to:

  1. Structural Similarity:

    • Malonate closely resembles succinate, the enzyme's natural substrate.

    • Both molecules share a similar molecular structure, enabling malonate to bind reversibly to SDH’s active site.

  2. Mechanism:

    • Malonate competes with succinate for binding at the enzyme's active site.

    • This blocks the substrate (succinate) from accessing the catalytic site, reducing SDH activity.

  3. Reversibility:

    • Inhibition can be reversed by increasing succinate concentration, a hallmark of competitive inhibition.

Why Not Other Options?

  • Feedback inhibition: Involves end-products regulating earlier enzymes in a pathway (not applicable here).

  • Enzyme activation: Refers to processes increasing enzyme activity (opposite of inhibition).

  • Cofactor inhibition: Targets non-protein components (e.g., metal ions) essential for enzyme function, which is unrelated to malonate’s action.

Thus, competitive inhibition is the correct classification


120. Formation of interfascicular cambium from fully developed parenchyma cells is an example for 

  1. Redifferentiation
  2. Dedifferentiation
  3. Maturation
  4. Differentiation 

Answer- The Correct Answer is: Dedifferentiation

Explanation:

The formation of interfascicular cambium from fully developed parenchyma cells is an example of dedifferentiation. This process involves differentiated cells regaining their ability to divide and form new meristematic tissue.

  1. Definition of Dedifferentiation:

    • Dedifferentiation is the process where specialized or mature cells lose their specific functions and revert to a more primitive, undifferentiated state, allowing them to divide and form new tissues.
  1. Context in Cambium Formation:

    • In the case of interfascicular cambium, fully developed parenchyma cells undergo dedifferentiation to form cambial tissue, which is capable of further division and contributes to secondary growth in plants.

Why Not Other Options?

  • Redifferentiation: Refers to the process where dedifferentiated cells become specialized again, which is not applicable here as we are discussing the initial formation of cambium.

  • Maturation: This term describes the process of cells reaching their final functional state, which does not involve cell division.

  • Differentiation: This is the process by which unspecialized cells become specialized, opposite to what occurs in dedifferentiation.

Thus, the correct answer is dedifferentiation, as it accurately describes the transformation of parenchyma cells into cambial tissue capable of further development.


121. A pink flowered Snapdragon plant was crossed with a red flowered Snapdragon plant. What type of phenotype/s is/are expected in the progeny?

  1. Red flowered as well as pink flowered plants
  2. Only pink flowered plants
  3. Red, Pink as well as white flowered plants
  4. Only red flowered plants

Answer- The correct answer is 1. Red flowered as well as pink flowered plants

Explanation:

Snapdragon flower color follows incomplete dominance, where neither allele is fully dominant. Instead, the heterozygous genotype produces a blended phenotype.

  • Red-flowered plant: Homozygous dominant (RR).

  • Pink-flowered plant: Heterozygous (RW), resulting from blending red (R) and white (W) alleles.

Cross: RW (pink) × RR (red)

Parental Gametes:

  • Pink parent (RW): Can donate R or W.

  • Red parent (RR): Can only donate R.
Punnett Square:

R

R

R

RR (Red)

RR (Red)

W

RW (Pink)

RW (Pink)

Progeny Genotypes & Phenotypes:

  • 50% RR: Red flowers (homozygous dominant).

    50% RW: Pink flowers (heterozygous blend).

No WW (white) offspring are possible because the red parent (RR) cannot donate a W allele.

Why Other Options Are Incorrect:

  1. Only pink: Requires all offspring to be heterozygous (RW), but the red parent (RR) ensures 50% RR (red).

  2. Red, pink, white: White (WW) requires both parents to donate W, but the red parent (RR) has no W allele.

  3. Only red: Ignores the W allele from the pink parent (RW).

Thus, the correct answer is red and pink phenotypes.


122. In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). In order to find out the genotype of the black seed plant, with which of the following genotype will cross it?

  1. bb            (2) Bb         (3) BB/Bb           (4) BB

Answer- The correct answer is (1) bb

Explanation:

To determine the genotype of a black-seed plant (BB or Bb), a test cross is performed by crossing it with a recessive homozygous plant (bb). Here’s why:

Case 1: If the black plant is BB

  • Cross: BB (black) × bb (white)

  • Progeny: All Bb (black seeds).

  • Result: No white seeds appear.

Case 2: If the black plant is Bb

  • Cross: Bb (black) × bb (white)

  • Progeny: 50% Bb (black) and 50% bb (white).

  • Result: Half the offspring will show white seeds.

Why Other Options Fail:

  • (2) Bb: Crossing with another heterozygous plant (Bb) produces a 3:1 phenotypic ratio (black:white), which doesn’t conclusively reveal the parent’s genotype.

  • (3) BB/Bb: Using a dominant plant (BB or Bb) won’t expose hidden recessive alleles.

  • (4) BB: A homozygous dominant parent (BB) will only produce black offspring, regardless of the test plant’s genotype.

Key Concept: A test cross with bb is the only way to distinguish between homozygous (BB) and heterozygous (Bb) genotypes by observing recessive phenotypes (white seeds) in the offspring.

 

123. Match List I with List II

            List I                                                      List II

A.                Clostridium butylicum                        I. Ethanol

B.                 Saccharomyces cerevisiae                  II. Streptokinase

C.                 Trichoderma polysporum                   III. Butyric acid

D.                Streptococcus sp.                                IV. Cyclosporin-A

Choose the correct answer from the options given below

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

Answer- The correct answer is (2): A-III, B-I, C-IV, D-II

[Explanation:

Matching List I (Organisms) with List II (Products):

  1. Clostridium butylicum (A) → III. Butyric acid

    • Clostridia species like C. acetobutylicum and C. butylicum are known for producing butyric acid as a primary fermentation product. Search results13 highlight their role in acidogenesis and solvent production.

  2. Saccharomyces cerevisiae (B) → I. Ethanol

    • Saccharomyces cerevisiae (yeast) is widely used in ethanol production via alcoholic fermentation. This is a well-established biological process not explicitly covered in the search results but confirmed by general knowledge.

  3. Trichoderma polysporum (C) → IV. Cyclosporin-A

    • Trichoderma polysporum is a fungus that produces cyclosporin-A, an immunosuppressant drug. This is a known application of the species.

  4. Streptococcus sp. (D) → II. Streptokinase
Streptococcus species produce streptokinase, an enzyme used clinically to dissolve blood clots. This matches standard microbiological knowledge.

Why Other Options Are Incorrect:
  • Option A: Incorrectly links Clostridium butylicum to streptokinase (unrelated).

  • Option C: Assigns cyclosporin-A to Streptococcus (illogical; cyclosporin is fungal).

  • Option D: Misplaces ethanol production with Trichoderma (ethanol is yeast-associated)].


124. How many molecules of ATP and NADPH required for every molecule of CO2 fixed in the Calvin cycle.

  1. 2 molecules of ATP and 2 molecules of NADPH
  2. 3 molecules of ATP and 3 molecules of NADPH
  3. 3 molecules of ATP and 2 molecules of NADPH
  4. 2 molecules of ATP and 3 molecules of NADPH

Answer-  The correct answer is 3. 3 molecules of ATP and 2 molecules of NADPH

[Explanation: For every molecule of CO₂ fixed in the Calvin cycle, the following energy molecules are required:

·         ATP: 3 molecules

·         NADPH: 2 molecules

Logical Explanation:

The Calvin cycle consists of three main stages:

1.Carbon Fixation: CO₂ is fixed to ribulose-1,5-bisphosphate (RuBP) by the enzyme RuBisCO. This step does not require ATP or NADPH directly.

2.Reduction: The intermediate 3-phosphoglycerate (3-PGA) is converted into glyceraldehyde-3-phosphate (G3P). This step requires 2 NADPH and 2 ATP per CO₂ molecule.

3. Regeneration of RuBP: RuBP is regenerated from G3P, which requires 1 ATP per CO₂ molecule.

Thus, for every CO₂ molecule fixed:

·         ATP requirement: 2 ATP for reduction + 1 ATP for regeneration = 3 ATP

·         NADPH requirement: 2 NADPH for reduction = 2 NADPH

This energy input ensures the Calvin cycle can continuously fix carbon dioxide and produce carbohydrates].

 

125. The capacity to generate a whole plant from any cell of the plant is called:

  1. Micropropagation
  2. Differentiation
  3. Somatic hybridization
  4. Totipotency

Answer- The correct answer is (4) Totipotency

[Explanation:

Totipotency refers to the ability of a single cell to divide and produce all the differentiated cells in an organism, including the capacity to regenerate into a complete organism. In plants, this means that theoretically, any living cell can grow back into a full plant under the right conditions. This concept is fundamental to plant tissue culture and regeneration techniques, where cells can be induced to dedifferentiate and then redifferentiate into various plant tissues, ultimately forming a whole plant.

Other Options Explained:

  1. Micropropagation: This is a technique used to propagate plants in vitro, often involving the use of totipotent cells. However, it is not the term for the capacity itself but rather a method that exploits totipotency. 
  1. Differentiation: This is the process by which cells become specialized to perform specific functions. While differentiation is crucial in plant development, it is the opposite of what is required for a cell to regenerate into a whole plant (which involves dedifferentiation followed by redifferentiation)].

  2. Somatic Hybridization: This involves the fusion of somatic cells from two different plants to create a new hybrid plant. It does not directly relate to the ability of a single cell to regenerate into a whole plant].

Why Totipotency Fits:

Totipotency is the correct term because it specifically describes the potential of a single plant cell to develop into a complete organism, which is the basis for techniques like somatic embryogenesis and plant tissue culture].


126. Tropical regions shows greatest level of species richness because

A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.

B. Tropical environments are more seasonal.

C. More solar energy is available in tropics.

D. Constant environments promote niche specialization.

E. Tropical environments are constant and predictable.

Choose the correct answer from the options given below:

(1) A and B only,     (2) A, B and E only

(3) A, B and D only,    (4) A, C, D and E only

Answer- The correct answer is: (4) A, C, D, and E only

Logical Explanation:

Tropical regions exhibit the greatest level of species richness due to several factors:

  1. A. Tropical latitudes have remained relatively undisturbed for millions of years, hence more time was available for species diversification.
    • True. The stability of tropical regions over geological time scales has allowed for prolonged evolutionary processes, leading to increased species diversity.
  2. B. Tropical environments are more seasonal.
    • False. Tropical environments are generally less seasonal compared to temperate regions. This lack of strong seasonal variation contributes to higher species richness.
  3. C. More solar energy is available in tropics.
    • True. The tropics receive more consistent and abundant solar energy throughout the year, supporting higher productivity and biodiversity.
  4. D. Constant environments promote niche specialization.
    • True. The relatively stable conditions in tropical regions allow for finer niche partitioning among species, enhancing diversity.
  5. E. Tropical environments are constant and predictable.
    • True. This stability allows species to adapt to specific ecological niches, contributing to higher species richness.

Why the Correct Answer Includes A, C, D, and E:

  • A provides historical stability for diversification.
  • C explains the energy availability that supports high productivity.
  • D highlights how constant conditions allow for niche specialization.
  • E emphasizes the predictability of tropical environments, which supports biodiversity.

Thus, the correct answer is (4) A, C, D, and E only.


127. Match List I with List II

List I                                 List II

A. Nucleolus                  I. Site of formation of glycolipid

B. Centriole                    II. Organization like the cartwheel

C. Leucoplasts               III. Site of active ribosomal RNA synthesis

D. Golgi apparatus        IV. For storing nutrients

Choose the correct answer from the options given below:

(1) A-II, B-III, C-I, D-IV

(2) A-III. B-IV. C-II, D-I

(3) A-I, B-II, C-III, D-IV

(4) A-III, B-II, C-IV, D-I

Answer- The correct answer is: (4) A-III, B-II, C-IV, D-I

Explanation:

1.      A. Nucleolus - III. Site of active ribosomal RNA synthesis

o    The nucleolus is the site where ribosomal RNA (rRNA) is synthesized and ribosome subunits are assembled. It plays a critical role in protein synthesis by producing the components required for ribosomes.

2.      B. Centriole - II. Organization like the cartwheel

o    Centrioles are cylindrical structures with a "cartwheel" organization of microtubules. They play a role in cell division by forming spindle fibers and organizing the centrosome.

3.      C. Leucoplasts - IV. For storing nutrients

o    Leucoplasts are non-pigmented plastids found in plant cells that primarily store nutrients such as starch, proteins, or lipids.

4.      D. Golgi apparatus - I. Site of formation of glycolipid

o    The Golgi apparatus is involved in the modification, sorting, and packaging of macromolecules like glycoproteins and glycolipids for secretion or use within the cell.

Final Matching:

 A → III,          B → II,          C → IV,           D → I

Thus, the correct option is (4) A-III, B-II, C-IV, D-I.


128. Identify the part of the seed from the given figure which is destined to form root when the seed germinates.

(1) A   (2) B    (3) C  (4) D

Answer- The correct answer is (3) C

Reasoning:

·   In a seed embryo, the embryonic root is called the radicle. When the seed germinates, the radicle is the first part to emerge and develops into the primary root.

·   In the diagram, "C" appears to point to the lowermost part of the embryo, which is typically where the radicle is located.

·   Therefore, (3) C is the most likely answer.


129. Spindle fibers attach to kinetochores of chromosomes during

(1) Metaphase     (2) Anaphase    (3) Telophase    (4) Prophase

Answer- The correct answer is: (1) Metaphase

Explanation:

Here's the breakdown:

1.      Kinetochores: These are protein structures on chromosomes where spindle fibers attach.

2.      Spindle fibers: These are microtubules that help separate chromosomes during cell division.

During the different phases of mitosis:

·   Prophase: Chromosomes condense, and the spindle apparatus begins to form, but spindle fibers do not yet attach to kinetochores.

· Metaphase: This is the stage when the spindle fibers attach to the kinetochores of the chromosomes. The chromosomes then align along the metaphase plate.

·    Anaphase: The sister chromatids separate and move to opposite poles of the cell, guided by the spindle fibers.

·      Telophase: The chromosomes arrive at the poles, and the nuclear envelope reforms.

Thus, spindle fibers attach to the kinetochores of chromosomes during Metaphase.

130. Given below are two statements:

Statement I: Chromosomes become gradually visible under light microscope during leptotene stage

Statement II: The beginning of diplotene stage is recognized by dissolution of synaptonemal complex

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

(1) Both Statement I and Statement II are false

(2) Statement I is true but Statement II id false

(3) Statement I is false but Statement II is true

(4) Both Statement I and Statement II are true

Answer- The correct answer is (4) Both Statement I and Statement II are true

Here's a more detailed explanation:

·         MMeiosis:

This is a type of cell division that reduces the number of chromosomes in a cell, resulting in the production of gametes (sperm and egg cells). 

·         PProphase I:

The first phase of meiosis I, which is further divided into several substages, including leptotene. 

·         LLeptotene:

During this stage, the chromosomes, which were previously in a diffuse, thread-like form (chromatin), start to condense and become visible under a light microscope. 

·         CChromatin Condensation:

As the cell prepares for division, the chromatin within the nucleus coils and condenses, becoming increasingly compact and visible as distinct chromosomes. 

·         SSister Chromatids:

Each chromosome at this stage consists of two identical sister chromatids joined at the centromere. 


131. Given below are two statements:

Statement I: Parenchyma is living but collenchymas is dead tissue

Statement II: Gymnosperms lack xylem vessels but presence of xylem vessels is the characteristic of angiosperms.

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

(1) Both Statement I and Statement II are false

(2) Statement I is true and Statement II is false

(3) Statement I is false and Statement II is true

(4) Both Statement I and Statement II are true

 Answer- The correct answer is: (3) Statement I is false and Statement II is true

Justification:

·  Statement I: Parenchyma is living but collenchymas is dead tissue.

o This statement is false. Parenchyma cells are indeed living and have thin cell walls. However, collenchyma cells are also living cells, although they have unevenly thickened cell walls that provide support to young plant parts.

· Statement II: Gymnosperms lack xylem vessels but the presence of xylem vessels is the characteristic of angiosperms.

o    This statement is true. Gymnosperms typically lack xylem vessels; their water transport is mainly through tracheids. Angiosperms, on the other hand, have xylem vessels, which are more efficient in water transport. This is a distinguishing characteristic between these two groups of plants.

Thus, Statement I is incorrect, and Statement II is correct, making option (3) the right choice.


132. These are regarded as major causes of biodiversity loss:

A. Over exploitation,   B. Co-extinction,    C. Mutation,                 

D. Habitat loss and fragmentation, E. Migration

Choose the correct option:

(1) A, B, C and D only,  (2) A, B and E only

(3) A, B and D only, (4) A, C and D only

Answer- The correct answer is:(3) A, B, and D only

Justification:

Let's analyze each option:

·    A. Overexploitation: This is a major cause of biodiversity loss. Overuse of resources can lead to the depletion or extinction of species.

·  B. Co-extinction: This is a significant factor in biodiversity loss. When one species goes extinct, other species that are dependent on it (e.g., parasites, mutualists) may also face extinction.

·  C. Mutation: While mutation is a fundamental process in evolution, it is not a primary cause of biodiversity loss in the current ecological context. Mutations occur naturally and are a source of genetic variation, but they don't directly drive large-scale species extinctions.

·   D. Habitat loss and fragmentation: This is a leading cause of biodiversity loss. Destruction or division of habitats reduces the space and resources available for species to survive, often leading to population declines and extinctions.

·  E. Migration: Migration, a natural process of species relocation, is not a primary cause of biodiversity loss itself. It can be affected by habitat loss and climate change, but it is not a direct driver of species extinction.

Therefore, A, B, and D are the most significant and directly linked causes of biodiversity loss among the options provided.
The correct answer is (3) A, B, and D only.


133. The lactose present in the growth medium of bacteria is transported to the cell by the action of:

(1) Acetylas,   (2) Permease,     (3) Polymerase,    (4) Beta-galactosidase

Answer- The correct answer is: (2) Permease

Justification:

1.      Permease: This is a specific membrane protein that facilitates the transport of lactose across the bacterial cell membrane. In the lac operon system, lactose permease is essential for bringing lactose into the cell, where it can then be metabolized.

2.      Other options:

o  Acetylase: This is an enzyme involved in acetylation reactions and not directly involved in lactose transport.

o Polymerase: Polymerases are enzymes involved in DNA or RNA synthesis and are not responsible for transporting lactose.

o Beta-galactosidase: This enzyme is responsible for breaking down lactose into glucose and galactose inside the cell, but it doesn't transport lactose into the cell.

Thus, permease is the correct enzyme responsible for transporting lactose into the bacterial cell.


134. Bulliform cells are responsible for

(1) Protecting the plant from salt stress

(2) Increase photosynthesis in monocots

(3) Providing large spaces for storage of sugars

(4) Inward curling of leaves in monocots.

 Answer- The correct answer is: (4) Inward curling of leaves in monocots.

Justification:

·    Bulliform cells are large, empty, bubble-shaped epidermal cells that occur in groups on the upper surface of leaves, particularly in monocots. They are involved in the mechanism that causes leaves to curl inward to minimize water loss during drought stress.

·   When water is scarce, bulliform cells lose turgidity (water pressure), causing the leaf to fold or curl inward, reducing the surface area exposed to the environment and thus decreasing transpiration.

Here's why the other options are incorrect:

(1) Protecting the plant from salt stress: While some plants have adaptations to deal with salt stress, bulliform cells are not directly involved in this process. Halophytes, for example, have other mechanisms for salt tolerance.

 (2) Increase photosynthesis in monocots: Bulliform cells do not directly increase photosynthesis. Photosynthesis primarily occurs in mesophyll cells.

(3) Providing large spaces for storage of sugars: While some plant cells are specialized for storage, this is not the primary function of bulliform cells.

Therefore, the correct answer is that bulliform cells are responsible for the inward curling of leaves in monocots during water stress.


135. The cofactor of the enzyme carboxylpeptidase is:

(1) Niacin           (2) Flavin             (3) Haem           (4) Zinc

Answer- The correct answer is (4) Zinc

Explanation

Carboxypeptidase is a proteolytic enzyme that requires zinc as an inorganic cofactor for its activity. Cofactors are non-protein molecules essential for enzyme function, categorized as:

·   Inorganic ions (e.g., zinc, magnesium)

·   Organic coenzymes (e.g., NAD, FAD derived from vitamins).

Zinc acts by binding to the enzyme's active site, stabilizing its structure, and facilitating substrate binding or catalysis12. The other options are organic coenzymes:

·    Niacin (option 1) is a precursor for NAD, involved in redox reactions.

·    Flavin (option 2) forms FMN/FAD, also used in redox processes.

·    Haem (option 3) is an iron-containing prosthetic group in proteins like hemoglobin.

Since carboxypeptidase relies on a metal ion cofactor rather than an organic coenzyme, zinc is the correct answer.

 Botany: Section – B (Q. No. 136 to 150)

136. Read the following statements and choose the set of correct statements:

In the members of Phaeophyceae,

A. Asexual reproduction occurs usually by biflagellate zoospores.

B. Sexual reproduction is by oogamous method only.

C. Stored food is in the form of carbohydrates which is either mannitol or laminarian

D. The majorpigments found are chlorophyll a, c and carotenoids and xanthophylls.

E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin

Choose the correct answer from the options given below:

(1) B, C, D and E only

(2) A, C, D and E only

(3) A, B, C and E only

(4) A, B, C and D only

Answer- The correct answer is (2) A, C, D and E only

Read carefully following explanation to enhance your understanding.

Phaeophyceae, or brown algae, exhibit specific characteristics related to their reproduction, pigments, and cell structure. Let's analyze each statement to determine the correct set:

A. Asexual reproduction occurs usually by biflagellate zoospores.
Correct. Phaeophyceae reproduce asexually via biflagellate zoospores, which have one anterior tinsel and one posterior whiplash flagellum.

B. Sexual reproduction is by oogamous method only.
Incorrect. Sexual reproduction in brown algae can be isogamous, anisogamous, or oogamous. The term "only" makes this statement false.

C. Stored food is in the form of carbohydrates, either mannitol or laminarin.
Correct. Laminarin (a β-1,3-glucan) and mannitol are the primary storage carbohydrates.

D. Major pigments include chlorophyll a, c, carotenoids, and xanthophylls.
Correct. Pigments are chlorophyll a, c, fucoxanthin (a xanthophyll), and other carotenoids. While xanthophylls are a subset of carotenoids, the statement is accurate.

E. Vegetative cells have a cellulosic wall with an outer gelatinous algin coating.
Correct. Cell walls contain cellulose and are coated with algin, a gelatinous polysaccharide.

Correct Answer: Option (2) A, C, D, and E only
Justification:

·         B is excluded because sexual reproduction in Phaeophyceae is not exclusively oogamous.

·         A, C, D, and E align with the characteristics confirmed by multiple sources.


137. Match List I with List II

  List I                                                   List II

A. Robert May                                    I. Species-Area relationship

B. Alexander von Humboldt              II. Long term ecosystem experiment using out door plots

C. Paul Ehrlich                                   III. Global species diversity at about 7 million

D. David Tilman                                IV. River popper hypothesis

Choose the correct answer from the options given below:

(1) A-III, B-I, C-IV, D-II

(2) A-I, B-III, C-II, D-IV

(3) A-III, B-IV, C-II, D-I

(4) A-II, B-III, C-I, D-IV

Answer- The correct answer is (1) A-III, B-I, C-IV, D-II

Read the following and find the answer

To match the names in List I with their corresponding contributions in List II, let us analyze each individual:

Analysis:

1. Robert May: Known for his work on the relationship between species diversity and ecosystem stability. He also contributed to the understanding of species-area relationships, which describe how species richness increases with area siz

     o  Match: I. Species-Area Relationship

2. Alexander von Humboldt: He is credited with pioneering the concept of the species-area relationship, where larger areas support more species due to greater habitat diversity

     o  Match: I. Species-Area Relationship

3. Paul Ehrlich: Proposed the Rivet Popper Hypothesis, which compares species in an ecosystem to rivets in an airplane wing, emphasizing the importance of each species in maintaining ecosystem stability

      o  Match: IV. Rivet Popper Hypothesis

4. David Tilman: Conducted long-term ecosystem experiments using outdoor plots to study biodiversity and its impact on ecosystem productivity and stability

      o    Match: II. Long-term ecosystem experiment using outdoor plots

Justification:

·         Robert May (A) is associated with global species diversity estimates (~7 million).

·         Alexander von Humboldt (B) is linked to the species-area relationship.

·         Paul Ehrlich (C) developed the Rivet Popper hypothesis.

·         David Tilman (D) is known for long-term biodiversity experiments using outdoor plots.


138. Given below are two statements:

Statement I: In C3 plants, some O2 binds to RuBisCO, hence CO2 fixation is decreased.

Statement II: In C4 plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.

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

(1) Both Statement I and Statement II are false

(2) Statement I is true but Statement II are false

(3) Statement I is false but Statement II is true

(4) Both Statement I and Statement II are true

Answer-  The correct answer is (4) Both Statement I and Statement II are true

Logical analysis-

Statement I: In C3 plants, some O2 binds to RuBisCO, hence CO2 fixation is decreased.

·     Explanation: In C3 plants, RuBisCO (Ribulose-1,5-Bisphosphate Carboxylase/Oxygenase) is the enzyme responsible for fixing CO2 into organic compounds during the Calvin cycle. However, RuBisCO can also bind to O2, leading to photorespiration. When O2 binds to RuBisCO, it initiates a process that results in the loss of CO2 and energy, thereby decreasing the efficiency of CO2 fixation. Thus, Statement I is true.

Statement II: In C4 plants, mesophyll cells show very little photorespiration while bundle sheath cells do not show photorespiration.

·  Explanation: C4 plants have evolved a mechanism to reduce photorespiration by spatially separating CO2 fixation and the Calvin cycle. In C4 plants, CO2 is first fixed into a four-carbon molecule in the mesophyll cells, which then moves to the bundle sheath cells where the Calvin cycle occurs. This process concentrates CO2 in the bundle sheath cells, reducing the likelihood of O2 binding to RuBisCO and thus minimizing photorespiration in these cells. Mesophyll cells, however, do not perform the Calvin cycle and therefore do not exhibit photorespiration related to RuBisCO activity in the same way as C3 plants. Thus, Statement II is true.

Justification:

·   Statement I is true because in C3 plants, O2 binding to RuBisCO leads to photorespiration, which decreases CO2 fixation efficiency.

·   Statement II is true because C4 plants minimize photorespiration by concentrating CO2 in bundle sheath cells, where the Calvin cycle occurs, and mesophyll cells do not show significant photorespiration due to their role in initial CO2 fixation.


139. The DNA present in chloroplast is:

(1) Circular, double stranded

(2) Linear, single stranded

(3) Circular, single stranded

(4) Linear, double stranded

Answer- The correct answer is (1) Circular, double stranded

Read these facts carefully and build subject concept along with the answer

Chloroplast DNA Characteristics:

·         Structure: Chloroplast DNA is typically circular in structure, similar to bacterial DNA. This circular structure is a common feature among many organelle DNAs.

·         Strand Type: Chloroplast DNA is double-stranded, meaning it consists of two complementary strands of nucleotides.

Justification:

 Option (1) Circular, double-stranded: This is the correct description of chloroplast DNA. It is circular, similar to bacterial DNA, and double-stranded, allowing for replication and transcription processes.

Option (2) Linear, single-stranded: Incorrect. Chloroplast DNA is not linear like some viral genomes, nor is it single-stranded.

Option (3) Circular, single-stranded: Incorrect. While chloroplast DNA is circular, it is not single-stranded.

 Option (4) Linear, double-stranded: Incorrect. Chloroplast DNA is not linear.

Justification:

Chloroplast DNA is circular and double-stranded, reflecting its evolutionary origin from cyanobacteria, which also have circular DNA. This structure supports the replication and transcription processes necessary for chloroplast function.


140. In an ecosystem if the Net Primary Productivity (NPP) of first trophic level is


Answer- The correct answer is :(1) x(kcal m-2)/y-1

Read the following explanation and learn the facts behind it

Definitions:

·Gross Primary Productivity (GPP): The total amount of organic matter produced by photosynthesis in an ecosystem, including the energy used by the plants themselves for respiration.

·Net Primary Productivity (NPP): The amount of organic matter left after the plants have used some of the GPP for their own respiration. It is essentially GPP minus the energy used by plants for respiration.

Relationship Between GPP and NPP:

The relationship between GPP and NPP can be expressed as:
NPP=GPP−R
where R is the respiration of the plants. Typically, NPP is about half of GPP, but this ratio can vary.

Trophic Levels and Energy Transfer:

Energy is transferred from one trophic level to the next with a significant loss at each step, typically following the "10% rule," where about 10% of the energy from one trophic level is transferred to the next.

Given Information:

·  The NPP of the first trophic level (primary producers) is 100x (kcal m−2) yr−1

Calculating GPP of the First Trophic Level:

If we assume that NPP is roughly half of GPP (though this can vary), then:
GPP=2×NPP=2×100x=200x (kcal m−2) yr−1

Energy Transfer to Higher Trophic Levels:

Applying the 10% rule:

·From First to Second Trophic Level: 10%×200x=20x (kcal m−2) yr−1  (GPP of the second trophic level, assuming it's all used by the next level).

·From Second to Third Trophic Level: 10%×20x=2x (kcal m−2) yr−1 (GPP of the third trophic level).

However, the question asks for the GPP of the third trophic level, which would be the energy available for that level to produce biomass, not necessarily what is transferred to the next level. Since the GPP of a trophic level includes the energy used by organisms in that level for their own metabolic processes, the GPP of the third trophic level would be slightly higher than the energy transferred to it, but in this simplified model, we focus on the energy transferred.

Justification:

· The energy transferred from the second to the third trophic level is 2x, but considering the simplified nature of the question and typical assumptions, the closest match is x (kcal m-2)/yr-1, acknowledging that the exact calculation might slightly differ based on specific ecosystem efficiencies and assumptions about energy transfer. However, given the options provided and the typical understanding of energy transfer in ecosystems, the closest match is x.

Note: The exact calculation provided suggests 2x for the energy transferred to the third trophic level, but none of the options directly match this. The closest interpretation under typical assumptions and given the options is x, acknowledging that the question might be simplifying or assuming different efficiencies in energy transfer.


141. Which of the following are fused in somatic hybridization involving two varieties of plants?

(1) Somatic embryos,  (2) Protoplasts

(3) Pollens,  (4) Callus

Answer- The Correct Answer is (2) Protoplasts

Explanation:

Somatic hybridization involves the fusion of protoplasts, which are plant cells that have had their cell walls removed. This process allows for the combination of genetic material from two different plant varieties, species, or genera to create a hybrid plant with desirable traits.

Key Points:

1.      Protoplasts in Somatic Hybridization:

o Protoplasts are isolated by enzymatic digestion of the cell wall using enzymes like cellulase and pectinase.

o  These protoplasts can then be fused using chemical agents (like polyethylene glycol, PEG) or physical methods (like electrofusion).

2.      Fusion Process:

o  The fusion of protoplasts results in a heterokaryon (a cell containing nuclei from both parent cells).

o  The fused protoplast is then cultured to regenerate a new cell wall, divide, and develop into a callus, which can be grown into a hybrid plant.

3.      Why Not Other Options?:

o  (1) Somatic embryos: These are formed later during tissue culture but are not involved in the fusion process.

o  (3) Pollens: Pollens are involved in sexual reproduction, not somatic hybridization.

o  (4) Callus: Callus is an undifferentiated mass of cells formed after protoplast fusion but is not directly involved in the fusion process.

The correct answer is (2) Protoplasts, as they are the cellular components fused during somatic hybridization to create somatic hybrids.


142. Match List I with List II

List I                                                           List II

A. Citric acid cycle                                  I. Cytoplasm

B. Glycolysis                                          II. Mitochondrial matrix

C. Electron Transport system                III. Intermembrane space of mitochondria

D. Proton gradient                                 IV. Inner mitochondrial membrane

Choose the correct answer from the options given below:

(1) A-II, B-I, C-IV, D-III

(2) A-III, B-IV, C-I, D-II

(3) A-IV, B-III, C-II, D-I

(4) A-I, B-II, C-III, D-V

Answer-  The correct answer is (1) A-II, B-I, C-IV, D-III

Explanation:

Let us analyze the locations of the cellular processes listed in List I and match them with their corresponding locations in List II:

1.      A. Citric Acid Cycle (Krebs Cycle)

o    The citric acid cycle occurs in the mitochondrial matrix, where acetyl-CoA is oxidized to produce NADH, FADH2, and ATP/GTP.

o    Match: II. Mitochondrial matrix.

2.      B. Glycolysis

o    Glycolysis is the first step of cellular respiration and occurs in the cytoplasm, where glucose is broken down into pyruvate.

o    Match: I. Cytoplasm.

3.      C. Electron Transport System (ETS)

o    The ETS is located in the inner mitochondrial membrane, where electrons are transferred through complexes to generate a proton gradient.

o    Match: IV. Inner mitochondrial membrane.

4.      D. Proton Gradient

o    The proton gradient is established in the intermembrane space of mitochondria, as protons are pumped from the matrix into this space during electron transport.

o    Match: III. Intermembrane space of mitochondria.

The correct matches are:

·         A-II: Citric Acid Cycle → Mitochondrial Matrix

·         B-I: Glycolysis → Cytoplasm

·         C-IV: Electron Transport System → Inner Mitochondrial Membrane

·         D-III: Proton Gradient → Intermembrane Space


143. Match List I with List II

List I                                                               List II

A. Frederick Griffith                                    I. Genetic code

B. Francisco Jacob & Jacque Monod           II. Semi-conservative mode of DNA replication

C. Har Gobind Khorana                               III. Transformation

D. Meselson & Stahl                                    IV. Lac operon

Choose the correct answer from the options given below:

(1) A-III, B-IV, C-I, D-II

(2) A-II, B-III, C-IV, D-I

(3) A-IV, B-I, C-II, D-III

(4) A-III, B-II, C-I, D-IV

Answer- The correct answer is (1) A-III, B-IV, C-I, D-II

Explanation:

Let us analyze each scientist and their corresponding discovery or contribution:

1.      Frederick Griffith (A)

o    Griffith's experiment in 1928 demonstrated the phenomenon of bacterial transformation, where genetic material from dead virulent bacteria transformed non-virulent bacteria into virulent ones. This is referred to as the "transforming principle".

o    Match: III. Transformation

2.      Francisco Jacob & Jacques Monod (B)

o    Jacob and Monod proposed the Lac operon model, which explains the regulation of gene expression in prokaryotes through an operon system.

o    Match: IV. Lac operon

3.      Har Gobind Khorana (C)

o    Khorana played a key role in deciphering the genetic code, determining how nucleotide sequences specify amino acids during protein synthesis.

o    Match: I. Genetic code

4.      Meselson & Stahl (D)

o    Meselson and Stahl conducted the famous experiment that demonstrated the semi-conservative mode of DNA replication, where each daughter DNA molecule consists of one parental strand and one newly synthesized strand.

o    Match: II. Semi-conservative mode of DNA replication

Final Matches:

·         A-III: Frederick Griffith → Transformation

·         B-IV: Francisco Jacob & Jacques Monod → Lac operon

·         C-I: Har Gobind Khorana → Genetic code

·         D-II: Meselson & Stahl → Semi-conservative mode of DNA replication


144. Match List I with List II

List I                                            List II

(Type of Stamens)                       (Example)

A. Monadelphous                       I. Citrus

B. Diadelphous                           II. Pea

C. Polydelphous                          III. Lily

D. Epiphyllous                             IV. China-rose

Choose the correct answer from the options given below:

(1) A-IV, B-I, C-II, D-III

(2) A-I, B-II, C-IV, D-III

(3) A-III, B-I, C-IV, D-II

(4) A-IV, B-II, C-I, D-III

Answer-  The correct answer is (4) A-IV, B-II, C-I, D-III

Explanation:

Let us match the types of stamens in List I with their corresponding examples in List II:

1.      A. Monadelphous

o    In this type, all the filaments are united into a single bundle, while the anthers remain free.

o    Example: China-rose (Hibiscus) belongs to the Malvaceae family and exhibits monadelphous stamens.

o    Match: IV. China-rose

2.      B. Diadelphous

o    In this type, the filaments are united into two bundles, while the anthers remain free.

o    Example: Pea (Pisum sativum) exhibits diadelphous stamens, where 9 stamens form one bundle and 1 remains separate.

o    Match: II. Pea

3.      C. Polydelphous

o    In this type, the filaments are united into more than two bundles, while the anthers remain free.

o    Example: Citrus exhibits polydelphous stamens, where the filaments form multiple groups.

o    Match: I. Citrus

4.      D. Epiphyllous

o    In this type, the stamens are attached to the tepals (undifferentiated perianth).

o    Example: Lily exhibits epiphyllous stamens attached to its tepals.

o    Match: III. Lily

Final Matches:

·         A-IV: Monadelphous → China-rose

·         B-II: Diadelphous → Pea

·         C-I: Polydelphous → Citrus

·         D-III: Epiphyllous → Lily


145. Identify the correct description about the given figure:


(1) Water pollinated flowers showing stamens with mucilaginous covering.

(2) Cleistogamous flowers showing autogamy.

(3) Compact inflorescence showing complete autpgamy

(4) Wind pollination plant inflorescence showing flowers with well exposed stamens.           

Answer- The correct answer is (4) Wind pollination plant inflorescence showing flowers with well-exposed stamens.

Explanation:

The question refers to identifying the correct description of a plant structure based on the figure. Let us analyze each option in light of the characteristics of pollination mechanisms:

1.  Option (1): Water-pollinated flowers showing stamens with mucilaginous covering

o    Water-pollinated flowers (hydrophily) typically have pollen grains protected by a mucilaginous covering to prevent wetting. However, this is specific to aquatic plants like Vallisneria and Hydrilla, and there is no mention of such adaptations in the question.

o    Incorrect.

2.  Option (2): Cleistogamous flowers showing autogamy

o    Cleistogamous flowers are closed flowers that do not open and enforce self-pollination (autogamy). These flowers lack exposed stamens as they do not rely on external pollination agents.

o    Incorrect.

3.  Option (3): Compact inflorescence showing complete autogamy

o    Compact inflorescences may promote self-pollination, but this does not align with the description of exposed stamens, which are typically associated with wind pollination rather than autogamy.

o    Incorrect.

4.   Option (4): Wind pollination plant inflorescence showing flowers with well-exposed stamens

o    Wind-pollinated plants (anemophily) have adaptations such as:

§  Well-exposed stamens for easy pollen dispersal by wind.

§  Large feathery stigmas to trap airborne pollen.

§  Light, non-sticky pollen grains for efficient wind transport.

o    Examples include grasses, maize, and other plants with compact inflorescences adapted for wind pollination. This matches the description given in the question.

o    Correct.

The correct answer is (4) Wind pollination plant inflorescence showing flowers with well-exposed stamens, as this aligns with the adaptations seen in anemophilous plants like grasses and maize.


146. Match List I with List II

List I                                 List II

A. GLUT-4                      I. Hormone

B. Insulin                          II. Enzyme

C. Trypsin                        III. Intercellular ground substance

D. Collagen                       IV. Enables glucose transport into cells

Choose the correct answer from the options given below:

(1) A-I, B-II, C-III, D-IV

(2) A-II, B-III, C-IV, D-I

(3) A-III, B-IV, C-I, D-II

(4) A-IV, B-I, C-II, D-III

Answer-The correct answer is (4) A-IV, B-I, C-II, D-III

Explanation:

Let us analyze each item in List I and match it with its corresponding description in List II:

1.      A. GLUT-4

o    GLUT-4 is a glucose transporter protein that facilitates the uptake of glucose into cells, particularly in adipose tissue and striated muscle. It is regulated by insulin and enables glucose transport into cells.

o    Match: IV. Enables glucose transport into cells

2.      B. Insulin

o    Insulin is a hormone secreted by the pancreas that regulates blood glucose levels by promoting the uptake of glucose into cells through mechanisms like GLUT-4 activation.

o    Match: I. Hormone

3.      C. Trypsin

o    Trypsin is an enzyme involved in protein digestion. It breaks down proteins into smaller peptides in the small intestine.

o    Match: II. Enzyme

4.      D. Collagen

o    Collagen is a structural protein found in connective tissues and serves as an intercellular ground substance, providing strength and support to tissues like skin, bones, and cartilage.

o    Match: III. Intercellular ground substance

Final Matches:

·         A-IV: GLUT-4 → Enables glucose transport into cells

·         B-I: Insulin → Hormone

·         C-II: Trypsin → Enzyme

·         D-III: Collagen → Intercellular ground substance


147. Identify the step in tricarboxylic acid cycle, which does not involve oxidation of substrate.

   to

(1) Succinic acid         Malic acid

   to(2) Succinyl – Co-A         Succinic acid

  to(3) Isocitrate          α-ketoglutaric acid

  to(4) Malic acid         Oxaloacetic acid

Answer-The  correct answer is (2) Succinyl-CoA → Succinic acid

Explanation:

The Tricarboxylic Acid (TCA) cycle, also known as the citric acid cycle or Krebs cycle, involves multiple steps where substrates are oxidized to generate energy in the form of NADH, FADH₂, and ATP. However, not all steps involve oxidation. Let us analyze each option:

1.      Succinic acid → Malic acid

o    This step involves the oxidation of succinate to fumarate and then to malate. FAD is reduced to FADH₂ during this process.

o    Involves oxidation, so this option is incorrect.

2.      Succinyl-CoA → Succinic acid

o    In this step, Succinyl-CoA is converted to succinate by replacing CoA with a phosphate group, which is then transferred to ADP (or GDP) to form ATP (or GTP). This is a substrate-level phosphorylation step and does not involve oxidation.

o    Does not involve oxidation, so this is the correct answer.

3.      Isocitrate → α-ketoglutaric acid

o    This step involves the oxidation of isocitrate to α-ketoglutarate, with NAD⁺ being reduced to NADH and CO₂ being released.

o    Involves oxidation, so this option is incorrect.

4.      Malic acid → Oxaloacetic acid

o    This step involves the oxidation of malate to oxaloacetate, with NAD⁺ being reduced to NADH.

o    Involves oxidation, so this option is incorrect.

The correct answer is (2) Succinyl-CoA → Succinic acid, as it is the only step in the TCA cycle that does not involve oxidation but instead involves substrate-level phosphorylation.


148. Spraying sugarcane crop with which of the following plant growth regulators, increases the length of stem, thus, increasing the yield?

(1) Gibberellin

(2) Cytokinin

(3) Abscisic acid

(4) Auxin

Answer-The correct answer is (1) Gibberellin

Explanation:

Gibberellins (GAs) are plant growth regulators known for their role in promoting stem elongation and internodal growth. Spraying sugarcane crops with gibberellins increases the length of the stem, which directly enhances the yield by increasing the biomass and sucrose accumulation in the elongated internodes.

Key Points:

1.      Role of Gibberellins:

o    Gibberellins regulate stem elongation by promoting both cell division and cell elongation.

o    They reduce the activity of DELLA nuclear repressor proteins, enabling the expression of genes involved in growth and development.

o    Studies have shown that applying gibberellins (GA3) increases internode elongation in sugarcane, leading to higher yields.

2.      Other Plant Growth Regulators:

o    Cytokinin: Primarily involved in cell division and delaying senescence but does not significantly influence stem elongation.

o    Abscisic Acid (ABA): Known for its role in stress responses and inhibiting growth, not promoting stem elongation.

o    Auxin: While it plays a role in cell elongation, its effect on internode elongation is secondary compared to gibberellins.

3.      Scientific Evidence:

o    Research indicates that spraying gibberellins increases endogenous GA levels, which stimulate internode elongation and improve sugarcane yield.

Spraying sugarcane crops with gibberellins effectively increases stem length, enhancing yield. Thus, the correct answer is (1) Gibberellin.


149. Match List I with List II

List I                          List II

A. Rose                    I. Twisted aestivation

B. Pea                      II. Perigynous flower

C. Cotton                 III. Drupe

D. Mango                  IV. Marginal placentation

Choose the correct answer from the options given below:

(1) A-I, B-II, C-III, D-IV

(2) A-IV, B-III, C-II, D-I

(3) A-II, B-III, C-IV, D-I

(4) A-II, B-IV, C-I, D-III

Answer-The correct answer is (4) A-II, B-IV, C-I, D-III

Explanation:

Let us match the items in List I with their corresponding descriptions in List II based on their botanical characteristics:

1.      A. Rose → Perigynous flower (II)

o    Roses are perigynous flowers, where the ovary is partially embedded in the floral cup (hypanthium), and other floral parts (sepals, petals, and stamens) are attached at the same level around the ovary.

o    Match: II. Perigynous flower

2.      B. Pea → Marginal placentation (IV)

o    In peas, the ovules are attached along the margin of a single carpel, a characteristic of marginal placentation.

o    Match: IV. Marginal placentation

3.      C. Cotton → Twisted aestivation (I)

o    Cotton exhibits twisted aestivation, where one petal overlaps the next petal in a twisted manner in the bud stage.

o    Match: I. Twisted aestivation

4.      D. Mango → Drupe (III)

o    Mango is a classic example of a drupe, a type of fleshy fruit with a single seed enclosed in a hard endocarp (stone).

o    Match: III. Drupe

Final Matches:

·         A-II: Rose → Perigynous flower

·         B-IV: Pea → Marginal placentation

·         C-I: Cotton → Twisted aestivation

·         D-III: Mango → Drupe


150. Which of the following statement is correct regarding the process of replication in E. coli?

(1) The DNA dependent RNA polymerase catalyses polymerization in one direction, that is

 5’     3’.

(2) The DNA dependent DNA polymerase catalyses polymerization in 5’   to  3’ as well as 3’  to   5’ direction

  (3) The DNA dependent DNA polymerase catalyses polymerization in 5’    3’direction.

(4) The DNA dependent DNA polymerase catalyses polymerization in one direction that is 3’ to    5’

Answer-The correct answer is (3) The DNA-dependent DNA polymerase catalyzes polymerization in the 5' → 3' direction.

Explanation:

DNA replication in E. coli involves several enzymes, with DNA-dependent DNA polymerase playing a central role. Let us analyze the given options based on the process of replication:

1.      DNA Polymerization Direction:

o    DNA polymerases, including the main enzyme in E. coli (DNA polymerase III), can only add nucleotides to the 3' end of a growing DNA strand.

o    This means that DNA synthesis always occurs in the 5' → 3' direction, as new nucleotides are added to the free hydroxyl group (-OH) at the 3' end of the growing strand.

2.      Template Reading Direction:

o    To synthesize DNA in the 5' → 3' direction, the template strand is read in the 3' → 5' direction.

3.      Analysis of Options:

o    Option (1): Incorrect. RNA polymerase is not involved in DNA replication; it is involved in transcription.

o    Option (2): Incorrect. DNA-dependent DNA polymerase does not catalyze polymerization in both directions; it works only in the 5' → 3' direction.

o    Option (3): Correct. The DNA-dependent DNA polymerase catalyzes polymerization exclusively in the 5' → 3' direction, which aligns with how nucleotide chains are elongated during replication.

o    Option (4): Incorrect. Polymerization does not occur in the 3' → 5' direction.

4.      Supporting Evidence:

o    During replication, one strand (the leading strand) is synthesized continuously in the 5' → 3' direction toward the replication fork.

o    The other strand (the lagging strand) is synthesized discontinuously as Okazaki fragments, also in the 5' → 3' direction, but away from the fork.

The correct answer is (3) The DNA-dependent DNA polymerase catalyzes polymerization in the 5' → 3' direction, as this is a fundamental property of all known DNA polymerases involved in replication.





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