The Directorate of Government Examinations (DGE), Tamil Nadu conducted the Tamil Nadu Class 12 Biology Public Exam 2026 on March 23, 2026. Tamil Nadu Board Class 12 Biology Public Question Paper with Solution PDF is now available for download.

The Tamil Nadu Class 12 Biology question paper covered important topics from genetics and evolution, molecular biology, reproduction, human health and diseases, biotechnology, ecology, and environmental biology. Students are advised to focus on understanding biological concepts, diagrams, scientific terminology, and application-based questions. The exam is conducted for a total of 100 marks, including theory and internal assessment.

Tamil Nadu Board 2026 Class 12th Biology Public Question Paper with Solution PDF

Tamil Nadu Class 12th Biology Public Question Paper 2026 Download PDF Check Solution

Question 1:

Explain the process of double fertilisation and its significance in angiosperms.

Correct Answer:
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Concept:
Double fertilisation is a unique feature of angiosperms in which two fertilisation events occur within the embryo sac.

Step 1: {\color{redFormation of male gametes.

The pollen grain contains two male gametes (sperm cells) which are delivered into the embryo sac through the pollen tube.

Step 2: {\color{redFirst fertilisation (Syngamy).

One male gamete fuses with the egg cell to form a diploid zygote (\(2n\)), which later develops into the embryo.

Step 3: {\color{redSecond fertilisation (Triple fusion).

The second male gamete fuses with two polar nuclei present in the central cell to form a triploid primary endosperm nucleus (\(3n\)).

Step 4: {\color{redDevelopment.


Zygote develops into the embryo
Primary endosperm nucleus develops into endosperm, which nourishes the embryo


Step 5: {\color{redSignificance.


Ensures efficient use of resources as endosperm develops only after fertilisation
Provides nourishment to the developing embryo
Leads to simultaneous development of embryo and endosperm
A characteristic feature distinguishing angiosperms from other plant groups


Step 6: {\color{redConclusion.

Thus, double fertilisation involves syngamy and triple fusion, resulting in the formation of both embryo and endosperm, making reproduction in angiosperms highly efficient. Quick Tip: Remember: Double fertilisation = 2 fusions → Zygote (2n) + Endosperm (3n).


Question 2:

Describe the structure and development of a mature 7-celled, 8-nucleated embryo sac.

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Concept:
The embryo sac is the female gametophyte of angiosperms. The most common type is the \textit{Polygonum type, which is 7-celled and 8-nucleated.

Step 1: {\color{redDevelopment (Megasporogenesis).

A diploid megaspore mother cell (MMC) undergoes meiosis to produce four haploid megaspores, out of which three degenerate and one functional megaspore remains.

Step 2: {\color{redMegagametogenesis.

The functional megaspore undergoes three successive mitotic divisions:

First division \(\rightarrow\) 2 nuclei
Second division \(\rightarrow\) 4 nuclei
Third division \(\rightarrow\) 8 nuclei

These nuclei arrange themselves within the embryo sac.

Step 3: {\color{redArrangement of nuclei.

The 8 nuclei are organized as follows:

Micropylar end: 3 cells (egg apparatus)

1 egg cell
2 synergids

Central cell: 2 polar nuclei
Chalazal end: 3 antipodal cells


Step 4: {\color{redCell formation.

Cell walls form around six nuclei, while two polar nuclei remain in the central cell:

Total cells = 7
Total nuclei = 8


Step 5: {\color{redStructure summary.


1 Egg cell
2 Synergids
3 Antipodal cells
1 Central cell (with 2 polar nuclei)


Step 6: {\color{redSignificance.

The embryo sac plays a crucial role in fertilisation:

Egg cell forms zygote
Polar nuclei participate in triple fusion
Synergids guide pollen tube


Step 7: {\color{redConclusion.

Thus, a mature embryo sac is 7-celled and 8-nucleated, formed through orderly mitotic divisions and nuclear organization. Quick Tip: Remember: 7 cells, 8 nuclei → Egg (1) + Synergids (2) + Antipodals (3) + Central cell (2 nuclei).


Question 3:

Explain the basic techniques and applications of plant tissue culture.

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Concept:
Plant tissue culture is the technique of growing plant cells, tissues, or organs under sterile and controlled conditions on a nutrient medium.

Step 1: {\color{redBasic principle.

It is based on \textit{totipotency, the ability of a single plant cell to develop into a complete plant.

Step 2: {\color{redBasic techniques.


Explant selection: A small piece of plant tissue (leaf, stem, root, meristem) is selected.
Surface sterilization: Explant is sterilized using chemicals (e.g., HgCl\(_2\), alcohol) to remove contaminants.
Inoculation: Explant is placed on a sterile nutrient medium (like MS medium).
Incubation: Cultures are kept under controlled temperature, light, and humidity.
Callus formation: Undifferentiated mass of cells develops.
Organogenesis/Embryogenesis: Differentiation into shoots and roots.
Hardening: Gradual acclimatization of plantlets to natural conditions before field transfer.


Step 3: {\color{redTypes of culture.


Callus culture
Suspension culture
Meristem culture
Anther/pollen culture
Embryo culture


Step 4: {\color{redApplications.


Micropropagation: Rapid multiplication of plants
Disease-free plants: Production via meristem culture
Crop improvement: Development of hybrids and transgenic plants
Germplasm conservation: Preservation of rare and endangered species
Secondary metabolite production: Alkaloids, hormones, medicines
Haploid production: Useful in plant breeding


Step 5: {\color{redAdvantages.


Large-scale production in short time
Uniform and true-to-type plants
Year-round production independent of season


Step 6: {\color{redConclusion.

Thus, plant tissue culture is a powerful biotechnological tool for plant propagation, improvement, and conservation. Quick Tip: Remember: Tissue culture = Totipotency + Sterile conditions + Nutrient medium.


Question 4:

Write an essay on the steps involved in recombinant DNA (rDNA) technology.

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Concept:
Recombinant DNA (rDNA) technology involves the artificial combination of DNA from different sources to create new genetic combinations for useful purposes.

Step 1: {\color{redIsolation of genetic material.

DNA is isolated from the donor organism using suitable methods. Since DNA is enclosed within cells, cell walls and membranes are broken using enzymes or chemicals.

Step 2: {\color{redCutting of DNA (Restriction digestion).

The isolated DNA is cut into fragments using restriction enzymes (restriction endonucleases), which recognize specific nucleotide sequences.

Step 3: {\color{redSelection of vector.

A suitable vector (such as plasmid, bacteriophage, or cosmid) is selected to carry the desired DNA fragment into a host cell.

Step 4: {\color{redLigation of DNA fragments.

The desired DNA fragment is inserted into the vector using the enzyme DNA ligase, forming recombinant DNA.

Step 5: {\color{redTransformation.

The recombinant DNA is introduced into a host cell (commonly bacteria like \textit{E. coli) through transformation.

Step 6: {\color{redSelection and screening.

Transformed cells are selected using selectable markers (such as antibiotic resistance). Screening ensures that the cells contain the desired recombinant DNA.

Step 7: {\color{redCloning and expression.

The host cells multiply, producing multiple copies (clones) of the recombinant DNA. The inserted gene may also be expressed to produce a desired protein.

Step 8: {\color{redDownstream processing.

The final product (such as a protein or enzyme) is extracted, purified, and processed for commercial or medical use.

Step 9: {\color{redApplications.


Production of insulin and vaccines
Gene therapy
Development of genetically modified crops
Industrial enzyme production


Step 10: {\color{redConclusion.

Thus, rDNA technology is a powerful tool in biotechnology that enables genetic manipulation for medical, agricultural, and industrial advancements. Quick Tip: Remember: rDNA steps → Isolation → Cutting → Ligation → Transformation → Selection → Expression.


Question 5:

Explain Bentham and Hooker's system of classification.

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Concept:
Bentham and Hooker proposed a natural system of classification for flowering plants in their book \textit{Genera Plantarum (1862–1883). This system is based on observable morphological characters and is widely used for practical plant identification.

Step 1: {\color{redBasis of classification.

The system is based on:

Morphological features of plants
Floral characters
Natural affinities among plants


Step 2: {\color{redMajor divisions.

They divided flowering plants (Phanerogams) into three major classes:

Dicotyledons
Gymnosperms
Monocotyledons


Step 3: {\color{redClassification of Dicotyledons.

Dicots are further divided into three subclasses:

Polypetalae – petals free
Gamopetalae – petals fused
Monochlamydeae – perianth absent or undifferentiated


Step 4: {\color{redClassification of Monocotyledons.

Monocots are grouped into several series based on floral characteristics such as perianth and ovary position.

Step 5: {\color{redGymnosperms.

Placed between dicots and monocots, representing plants with naked seeds.

Step 6: {\color{redMerits.


Based on natural relationships
Useful for identification of plants
Widely accepted and used in herbaria


Step 7: {\color{redDemerits.


Does not consider evolutionary relationships
Gymnosperms placed between dicots and monocots artificially
Monochlamydeae is a heterogeneous group


Step 8: {\color{redConclusion.

Bentham and Hooker's system is a natural and practical classification system, highly valuable for plant identification despite lacking evolutionary basis. Quick Tip: Remember: Bentham \& Hooker = Natural system based on morphology.


Question 6:

Describe the mechanism of crossing over and its significance.

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Concept:
Crossing over is the exchange of genetic material between non-sister chromatids of homologous chromosomes during meiosis (prophase I), leading to genetic recombination.

Step 1: {\color{redOccurrence.

Crossing over occurs during the pachytene stage of prophase I of meiosis after homologous chromosomes have paired (synapsis).

Step 2: {\color{redSynapsis and bivalent formation.

Homologous chromosomes pair closely to form a bivalent or tetrad consisting of four chromatids.

Step 3: {\color{redBreakage and exchange.

Non-sister chromatids break at corresponding points and exchange segments of DNA.

Step 4: {\color{redFormation of chiasmata.

The points where exchange occurs are called chiasmata, which become visible during diplotene stage.

Step 5: {\color{redSeparation.

After exchange, chromatids separate but remain attached at chiasmata until they move apart during later stages of meiosis.

Step 6: {\color{redResult.

This leads to recombination of genes, producing new combinations of alleles in gametes.

Step 7: {\color{redSignificance.


Genetic variation: Produces new gene combinations
Evolution: Provides raw material for natural selection
Linkage mapping: Helps in determining gene distances on chromosomes
Proper segregation: Ensures correct separation of homologous chromosomes


Step 8: {\color{redConclusion.

Thus, crossing over is a vital process in meiosis that enhances genetic diversity and ensures proper chromosome behavior. Quick Tip: Remember: Crossing over = Exchange of segments → Genetic variation.


Question 7:

Describe the hormonal control of the human menstrual cycle.

Correct Answer:
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Concept:
The menstrual cycle is regulated by a complex interaction of hormones secreted by the hypothalamus, pituitary gland, and ovaries.

Step 1: {\color{redRole of hypothalamus.

The hypothalamus secretes Gonadotropin-Releasing Hormone (GnRH), which stimulates the anterior pituitary gland.

Step 2: {\color{redRole of pituitary hormones.

The anterior pituitary releases:

Follicle Stimulating Hormone (FSH) – stimulates growth of ovarian follicles
Luteinizing Hormone (LH) – triggers ovulation and formation of corpus luteum


Step 3: {\color{redFollicular phase (Day 1–14).


FSH stimulates follicle development
Growing follicles secrete estrogen
Estrogen promotes thickening of endometrium


Step 4: {\color{redOvulation (Around Day 14).


High estrogen levels trigger LH surge
LH surge causes release of ovum (ovulation)


Step 5: {\color{redLuteal phase (Day 15–28).


Corpus luteum forms and secretes progesterone
Progesterone maintains and prepares endometrium for implantation


Step 6: {\color{redMenstruation.


If fertilization does not occur, corpus luteum degenerates
Estrogen and progesterone levels fall
Endometrial lining sheds → menstruation


Step 7: {\color{redFeedback regulation.


Estrogen and progesterone exert feedback control on hypothalamus and pituitary
Regulates secretion of GnRH, FSH, and LH


Step 8: {\color{redConclusion.

Thus, the menstrual cycle is hormonally regulated through coordinated actions of GnRH, FSH, LH, estrogen, and progesterone. Quick Tip: Remember: FSH → Follicle, LH → Ovulation, Progesterone → Maintains uterus.


Question 8:

Differentiate between active and passive immunity with suitable examples.

Correct Answer:
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Concept:
Immunity is the ability of the body to resist infection. It is broadly classified into active and passive immunity based on how antibodies are acquired.

Step 1: {\color{redActive immunity.

Active immunity is developed when the body produces its own antibodies in response to an antigen.

Source: Own immune system
Onset: Slow (takes time to develop)
Duration: Long-lasting (often lifelong)
Memory: Immunological memory present
Examples: Natural infection (e.g., measles), vaccination


Step 2: {\color{redPassive immunity.

Passive immunity is acquired by receiving ready-made antibodies from another source.

Source: External antibodies
Onset: Immediate protection
Duration: Short-lived
Memory: No immunological memory
Examples: Maternal antibodies (through placenta or milk), antiserum injections


Step 3: {\color{redComparison table.
\[ \begin{array}{|c|c|c|} \hline \textbf{Feature} & \textbf{Active Immunity} & \textbf{Passive Immunity}
\hline Source & Own body & External source
\hline Onset & Slow & Immediate
\hline Duration & Long-lasting & Short-term
\hline Memory & Present & Absent
\hline Example & Vaccination & Maternal antibodies
\hline \end{array} \]

Step 4: {\color{redConclusion.

Thus, active immunity provides long-term protection through antibody production and memory, while passive immunity offers immediate but temporary protection. Quick Tip: Remember: Active = Own antibodies (long-term), Passive = Ready-made antibodies (short-term).


Question 9:

Explain the semi-conservative mode of DNA replication.

Correct Answer:
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Concept:
The semi-conservative mode of DNA replication means that after replication, each newly formed DNA molecule contains one parental (old) strand and one newly synthesized strand.

Step 1: {\color{redMeaning of semi-conservative replication.

During DNA replication, the two strands of the parent DNA separate and each strand acts as a template for the synthesis of a new complementary strand. As a result, each daughter DNA molecule conserves one original strand and contains one new strand.

Step 2: {\color{redUnwinding of DNA.

The double helix unwinds with the help of enzymes such as helicase. Hydrogen bonds between the nitrogenous bases are broken, and the two strands separate.

Step 3: {\color{redTemplate function of parental strands.

Each separated parental strand serves as a template. Free nucleotides from the nucleoplasm pair with the exposed bases according to the base-pairing rule: \[ A \leftrightarrow T, \qquad G \leftrightarrow C \]

Step 4: {\color{redFormation of new strands.

DNA polymerase joins the incoming nucleotides and synthesizes new complementary strands along each parental strand.

Step 5: {\color{redResult of replication.

At the end of replication, two identical DNA molecules are produced. Each consists of:

one old (parental) strand
one new (daughter) strand


Step 6: {\color{redExperimental proof.

The semi-conservative nature of DNA replication was demonstrated by Meselson and Stahl using E. coli. They grew bacteria in heavy nitrogen (\(^{15N\)) and then shifted them to light nitrogen (\(^{14}N\)). The results proved that each new DNA molecule had one old and one new strand.

Step 7: {\color{redSignificance.


Ensures accurate transmission of genetic information
Maintains continuity of genetic material from one generation to the next
Reduces errors during DNA duplication


Step 8: {\color{redConclusion.

Thus, in semi-conservative replication, each daughter DNA molecule contains one parental strand and one newly synthesized strand, ensuring faithful copying of genetic material. Quick Tip: Remember: Semi-conservative = Half old + Half new in each daughter DNA molecule.


Question 10:

Describe the mechanism of urine formation, including filtration, reabsorption, and secretion.

Correct Answer:
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Concept:
Urine formation is a vital process carried out in the nephrons of the kidneys. It involves three main steps: glomerular filtration, tubular reabsorption, and tubular secretion.

Step 1: {\color{redGlomerular filtration.


Occurs in the glomerulus and Bowman's capsule
Blood is filtered under high pressure
Water, glucose, salts, amino acids, and urea pass into the filtrate
Large molecules like proteins and blood cells are retained in blood


Step 2: {\color{redTubular reabsorption.


Takes place mainly in the proximal convoluted tubule (PCT), loop of Henle, and distal convoluted tubule (DCT)
Useful substances such as glucose, amino acids, ions, and most water are reabsorbed back into the blood
Reabsorption may be active (requiring energy) or passive


Step 3: {\color{redTubular secretion.


Occurs mainly in DCT and collecting duct
Waste substances like hydrogen ions, potassium ions, ammonia, and drugs are secreted from blood into the filtrate
Helps in maintaining pH and ionic balance


Step 4: {\color{redConcentration of urine.


Loop of Henle and collecting duct regulate water reabsorption
Antidiuretic hormone (ADH) controls water permeability
Produces concentrated or dilute urine depending on body needs


Step 5: {\color{redFinal urine formation.

The remaining filtrate, containing urea, excess salts, and water, is excreted as urine through ureters to the urinary bladder.

Step 6: {\color{redConclusion.

Thus, urine formation involves filtration of blood, selective reabsorption of useful substances, and secretion of wastes, ensuring proper excretion and homeostasis. Quick Tip: Remember: Urine formation = Filtration → Reabsorption → Secretion.


Question 11:

Explain the production of recombinant insulin and its advantages over conventional insulin.

Correct Answer:
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Concept:
Recombinant insulin is produced using recombinant DNA (rDNA) technology by inserting the human insulin gene into microorganisms such as \textit{E. coli.

Step 1: {\color{redIsolation of insulin gene.

The gene responsible for insulin production is identified and isolated from human DNA.

Step 2: {\color{redInsertion into vector.

The insulin gene is inserted into a plasmid vector using restriction enzymes and DNA ligase to form recombinant DNA.

Step 3: {\color{redTransformation.

The recombinant plasmid is introduced into bacterial cells (commonly \textit{E. coli).

Step 4: {\color{redExpression of insulin.

The bacteria multiply and express the insulin gene, producing insulin protein (initially as separate A and B chains or proinsulin).

Step 5: {\color{redProcessing and purification.

The insulin chains are purified and combined to form functional insulin, which is then processed for medical use.

Step 6: {\color{redAdvantages over conventional insulin.


High purity: Free from animal impurities
Reduced allergic reactions: More compatible with human body
Large-scale production: Easily produced in bulk
Ethical advantage: No need to extract from animal pancreas
Consistent quality: Uniform and reliable product


Step 7: {\color{redConclusion.

Thus, recombinant insulin is produced through genetic engineering and offers safer, more efficient, and ethical treatment for diabetes compared to conventional insulin. Quick Tip: Remember: Recombinant insulin = Human gene + Bacteria → Safe insulin.

TN Board 2026 Class 12 Preparation