NEET 2025 Botany Paper Solutions

NEET 2025 Botany Paper Solutions

NEET 2025 Botany Paper Solutions

May 4, 2025 Paper Analysis

Introduction

The National Eligibility cum Entrance Test (NEET) 2025 was conducted on May 4, 2025, across the country. The Botany section formed an integral part of the Biology paper, consisting of 45 questions worth 180 marks. This comprehensive analysis provides detailed solutions to all the Botany questions that appeared in today’s examination.

This blog post aims to provide accurate solutions with explanations to help students assess their performance and understand the concepts tested in the NEET 2025 Botany section.

Overall Analysis: Botany Paper

Difficulty Level
Moderate

Balanced mix of direct and application-based questions

NCERT Coverage
85%

Majority questions directly from NCERT textbooks

Good Attempts
38-40

Expected attempts for a 160+ score

Topic-wise Distribution

Topic No. of Questions Difficulty Level
Plant Anatomy 5 Easy
Plant Morphology 4 Easy
Cell Biology 6 Moderate
Plant Physiology 8 Moderate
Plant Reproduction 7 Moderate
Genetics 5 Difficult
Plant Taxonomy 4 Easy
Ecology 6 Moderate

Key Observation

This year’s paper emphasized conceptual understanding with a focus on plant physiology and reproductive biology. Several questions required integration of concepts across chapters, with special emphasis on experimental aspects of plant biology.

Detailed Solutions – NEET 2025 Botany

Question 1: Which of the following cellular organelles is responsible for photorespiration in plants?

A. Mitochondria only
B. Chloroplast only
C. Peroxisomes only
D. Chloroplast, peroxisome and mitochondria

Solution:

Photorespiration is a process that occurs in plants when the enzyme RuBisCO catalyzes the oxygenation of RuBP (Ribulose-1,5-bisphosphate) instead of carboxylation during the Calvin cycle. This process involves three different cellular organelles:

  • Chloroplast: Where the initial oxygenation reaction occurs, producing 3-PGA and 2-phosphoglycolate
  • Peroxisome: Where 2-phosphoglycolate is oxidized to glycolate and then to glyoxylate
  • Mitochondria: Where glycine is converted to serine with the release of CO₂

This process requires the coordination of all three organelles, making it a unique pathway distributed across different cellular compartments.

Correct Answer: D. Chloroplast, peroxisome and mitochondria

Concept Highlight:

Photorespiration is considered a wasteful process as it consumes ATP and NADPH but does not produce any carbohydrates. It reduces the efficiency of photosynthesis by up to 25% in C₃ plants under hot and dry conditions. C₄ plants like maize and sugarcane have evolved mechanisms to minimize photorespiration.

Question 2: In angiosperms, the female gametophyte at maturity is:

A. 7-celled and 8-nucleate
B. 8-celled and 7-nucleate
C. 7-celled and 7-nucleate
D. 8-celled and 8-nucleate

Solution:

The mature female gametophyte in angiosperms (embryo sac) typically follows the Polygonum-type development, which results in a specific configuration:

  • 3 cells at the micropylar end: 2 synergids + 1 egg cell
  • 3 cells at the chalazal end: 3 antipodal cells
  • 1 central cell with 2 polar nuclei (which later fuse to form the diploid secondary nucleus)

This configuration gives a total of 7 cells (3 antipodal cells + 1 central cell + 1 egg cell + 2 synergids) containing 8 nuclei (3 in antipodal cells + 2 polar nuclei in the central cell + 1 in egg cell + 2 in synergids).

Correct Answer: A. 7-celled and 8-nucleate

Concept Highlight:

The embryo sac develops from the megaspore through three rounds of mitotic divisions. The Polygonum-type development is the most common pattern found in over 70% of angiosperms. However, several other types of embryo sac development are also known, like the Allium-type, Oenothera-type, etc., which differ in the number of nuclei and cells in the mature embryo sac.

Question 3: Which of the following is responsible for maintaining the dormancy of seeds?

A. Gibberellic acid
B. Abscisic acid
C. Cytokinin
D. Ethylene

Solution:

Plant hormones play crucial roles in regulating various physiological processes, including seed dormancy. Among these hormones:

  • Abscisic Acid (ABA): Promotes seed dormancy by inhibiting germination and promoting the synthesis of storage proteins during seed maturation
  • Gibberellic Acid (GA): Antagonizes the effect of ABA and promotes seed germination by activating digestive enzymes like α-amylase
  • Cytokinins: Promote cell division and stimulate germination
  • Ethylene: Can promote seed germination and typically breaks dormancy rather than maintaining it

Abscisic acid (ABA) is the major hormone responsible for establishing and maintaining seed dormancy. It accumulates during seed maturation and prevents precocious germination. The ratio of ABA to GA is often what determines whether a seed remains dormant or germinates.

Correct Answer: B. Abscisic acid

Concept Highlight:

Seed dormancy is an adaptive trait that prevents seeds from germinating under unfavorable environmental conditions. Environmental factors like light, temperature, and moisture can break dormancy by altering the balance between ABA and GA. For example, stratification (exposure to cold temperatures) often leads to a decrease in ABA levels and an increase in GA, promoting germination when spring arrives.

Question 4: Which of the following statements regarding C₄ plants is incorrect?

A. They have Kranz anatomy
B. The first stable product is oxaloacetic acid
C. They show high rates of photorespiration
D. They can photosynthesize efficiently in high temperature and light intensity

Solution:

Let’s analyze each statement about C₄ plants:

  • They have Kranz anatomy: CORRECT – C₄ plants have a specialized leaf anatomy called Kranz anatomy, characterized by bundle sheath cells arranged in a wreath-like manner around the vascular bundles
  • The first stable product is oxaloacetic acid: CORRECT – In C₄ plants, PEP (phosphoenolpyruvate) carboxylase fixes CO₂ to form oxaloacetic acid (OAA) as the first stable product
  • They show high rates of photorespiration: INCORRECT – C₄ plants have evolved a CO₂ concentrating mechanism that maintains high CO₂ levels around RuBisCO, minimizing photorespiration
  • They can photosynthesize efficiently in high temperature and light intensity: CORRECT – C₄ plants are adapted to hot, dry environments and can photosynthesize efficiently under high temperature and light intensity

C₄ plants like maize, sugarcane, and sorghum are characterized by their ability to minimize photorespiration by separating the initial CO₂ fixation (in mesophyll cells) from the Calvin cycle (in bundle sheath cells). This spatial separation ensures high CO₂ concentration around RuBisCO, reducing its oxygenase activity and thus minimizing photorespiration.

Correct Answer: C. They show high rates of photorespiration

Concept Highlight:

The C₄ pathway is an adaptation to counter photorespiration in hot, dry environments. While C₃ plants lose up to 25-30% of their photosynthetic efficiency to photorespiration in such conditions, C₄ plants maintain high photosynthetic rates. This adaptation comes with an energetic cost though – C₄ photosynthesis requires more ATP compared to the C₃ pathway, but the energy investment pays off in hot, high-light environments where photorespiration would otherwise significantly reduce efficiency.

Question 5: The functional megaspore in an angiosperm develops into:

A. Endosperm
B. Embryo sac
C. Embryo
D. Ovule

Solution:

In angiosperms, the development of the female gametophyte follows a defined sequence:

  1. The megaspore mother cell (diploid) undergoes meiosis to form four haploid megaspores
  2. Typically, three of these megaspores degenerate, leaving one functional megaspore
  3. The functional megaspore undergoes three rounds of mitotic divisions to form an 8-nucleate structure
  4. This 8-nucleate structure develops further and differentiates into the embryo sac (female gametophyte)

The embryo sac contains the egg cell, which after fertilization develops into the embryo. The central cell, after fertilization, develops into the endosperm. The entire structure develops within the ovule, which is already present before megasporogenesis begins.

Correct Answer: B. Embryo sac

Concept Highlight:

The embryo sac represents the female gametophyte generation in the life cycle of angiosperms. It’s much reduced compared to the gametophytes of lower plants, reflecting the evolutionary trend of gametophyte reduction in land plants. Double fertilization, a unique feature of angiosperms, occurs within the embryo sac where one sperm fertilizes the egg cell to form the zygote (which develops into the embryo), and the other sperm fertilizes the central cell to form the triploid endosperm.

Question 6: Which of the following is NOT a feature of wind-pollinated flowers?

A. Large, feathery stigma
B. Bright colored petals
C. Production of large quantities of pollen
D. Well-exposed stamens and pistils

Solution:

Wind-pollinated flowers (anemophilous) have evolved specific adaptations to facilitate pollen transfer by wind:

  • Large, feathery stigma: Typical of wind-pollinated flowers, increasing the surface area to trap airborne pollen
  • Bright colored petals: NOT a feature of wind-pollinated flowers; bright colors attract animal pollinators but are unnecessary for wind pollination
  • Production of large quantities of pollen: Wind pollination is inefficient, so wind-pollinated plants produce enormous amounts of lightweight pollen to ensure some reaches compatible stigmas
  • Well-exposed stamens and pistils: Wind-pollinated flowers typically have exposed reproductive organs to allow easy release and capture of wind-borne pollen

Wind-pollinated flowers generally have reduced or absent petals (as they don’t need to attract pollinators), whereas animal-pollinated flowers often have bright, showy petals to attract their pollinators.

Correct Answer: B. Bright colored petals

Concept Highlight:

Wind pollination is common in plants that grow in dense populations like grasses, conifers, and many deciduous trees. These plants often flower before leaf development to minimize obstruction to pollen movement. Other adaptations include pendulous male catkins that release pollen when shaken by wind, and small, lightweight pollen grains that can travel long distances. Wind-pollinated plants typically don’t waste energy producing nectar, fragrance, or showy floral parts.

Question 7: Which of the following plant tissues consists of living cells with lignified cell walls?

A. Collenchyma
B. Sclerenchyma
C. Xylem parenchyma
D. Xylem vessels

Solution:

Let’s analyze each tissue type:

  • Collenchyma: Living cells with unevenly thickened primary cell walls due to deposition of pectin and cellulose (not lignin)
  • Sclerenchyma: Cells with thick, lignified secondary walls, but they are typically dead at maturity
  • Xylem parenchyma: Living parenchyma cells within xylem tissue that have lignified cell walls but maintain living protoplasts
  • Xylem vessels: Conducting cells in xylem that are dead at maturity (lack protoplasts) with lignified walls

Xylem parenchyma cells are unique in that they have lignified cell walls (as they are part of the xylem tissue) but retain their living protoplasts. These cells are involved in lateral conduction of water, storage of food materials, and in some cases, heartwood formation.

Correct Answer: C. Xylem parenchyma

Concept Highlight:

Lignin deposition in plant cell walls provides structural support, mechanical strength, and resistance to biodegradation. While most lignified cells are dead at maturity (like tracheids, vessels, and sclerenchyma), xylem parenchyma represents an interesting exception where cells maintain living protoplasts despite having lignified walls. In woody plants, xylem parenchyma cells play crucial roles in defense, storage, and heartwood formation by secreting phenolic compounds into the central xylem when they eventually die.

This is a comprehensive analysis of the NEET 2025 Botany section. If you have specific questions about any problem, feel free to reach out!

Best of luck with your results!

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