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314
NIOS Senior Secondary 100% Curricular Audit Verified

Biology (Theory)

Official Examination Solutions • Verified Model Answer Key

Biology (314)
Complete Solved Question Paper

Exhaustive step-by-step biological explanations, anatomical diagrams, biochemical pathways (\(C_3\) and \(C_4\), cyclic and non-cyclic photophosphorylation), genetics mechanisms, nitrogen cycling, and full solutions for all 43 compulsory questions including all internal alternatives.

Time: 3 Hours Maximum Marks: 80 Total Questions: 43 Solved Paper • Reviewed
80/80 Target Score 43 Questions
16 MCQs (1M)
Q17–28 Objective (2M)
Q29–37 VSA (2M)
Q42–43 LA (5M)

General Instructions & Examination Blueprint

Click to expand examination blueprint, word count limits, and section breakdowns

Instructions for Candidates:

  1. Write your Roll Number on the first page of the Question Paper and Answer-Book.
  2. Verify that the Question Paper contains 43 questions in sequential order across 8 printed pages.
  3. Code Number 71/SS/314 and Set A1 must be written clearly on the title page of the Answer-Book.
  4. All questions are compulsory. Labelled biological diagrams and correct scientific terminology must be provided wherever applicable.
Marking Structure
  • • Section A (Q1 to Q16): Multiple Choice Questions (1 Mark each = 16 Marks)
  • • Section B (Q17 to Q28): Objective Type Questions (2 Marks each = 24 Marks)
  • • Section C (Q29 to Q37): Very Short Answer (2 Marks each = 18 Marks, 30–50 words)
  • • Section D (Q38 to Q41): Short Answer (3 Marks each = 12 Marks, 50–80 words)
  • • Section E (Q42 to Q43): Long Answer (5 Marks each = 10 Marks, 80–120 words)
  • • Total Marks: 80 Marks | Theory Component
Time Schedule & Internal Choices

Reading Time: 15 minutes (02:15 p.m. to 02:30 p.m.).
Examination Duration: 3 Hours (180 minutes).
Internal Choices: Provided in Q32, Q33, Q34, Q35, Q39, Q40, Q42, and Q43. Every choice is fully and rigorously solved below.

Section A

Multiple Choice Questions (Q1 to Q16)

[1 × 16 = 16 Marks]
Q. 01 Human Health & Disease • Bacterial Pathogens
[1 Mark]

Salmonella typhi is a pathogen for:

(A) Cholera
(B) Tetanus
(C) Typhoid
(D) Tuberculosis
Verified Solution • Option (C) Reviewed

Ans. (C) Typhoid
Explanation: Salmonella typhi is a pathogenic rod-shaped Gram-negative bacterium that enters the human small intestine via contaminated food and water, causing typhoid (enteric fever). Cholera is caused by Vibrio cholerae, tetanus by Clostridium tetani, and tuberculosis by Mycobacterium tuberculosis.

Q. 02 Biological Disciplines • Microscopic Anatomy
[1 Mark]

Histology refers to the study of:

(A) Cells
(B) Tissues
(C) Plants
(D) Organs
Ans. (B) Tissues

Explanation: Histology is the branch of biological science concerned with the microscopic anatomy of tissues and their cellular organization. The study of individual cells is cytology, while the study of plants is botany.

Q. 03 Plant Physiology • Photosynthetic Pigments
[1 Mark]

The essential photosynthetic pigment is:

(A) Chlorophyll A
(B) Chlorophyll B
(C) Xanthophyll
(D) Carotenoids
Ans. (A) Chlorophyll A

Explanation: Chlorophyll \(a\) is known as the primary or essential photosynthetic pigment because it forms the photochemical reaction centre (\(P_{680}\) in PS II and \(P_{700}\) in PS I) responsible for the primary conversion of absorbed light energy into chemical energy. Chlorophyll \(b\), carotenoids, and xanthophylls act as accessory pigments.

Q. 04 Respiration in Plants • Respiratory Quotient
[1 Mark]

The R.Q. for Carbohydrates is:

(A) 1
(B) 1.2
(C) 0.9
(D) 0
Ans. (A) 1

Derivation: The Respiratory Quotient (\(\text{R.Q.}\)) is the ratio of the volume of \(\text{CO}_2\) evolved to the volume of \(\text{O}_2\) consumed: \[\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \to 6\text{CO}_2 + 6\text{H}_2\text{O} + \text{Energy}\] \[\text{R.Q.} = \frac{\text{Volume of CO}_2\text{ evolved}}{\text{Volume of O}_2\text{ consumed}} = \frac{6}{6} = \mathbf{1.0}\]

Q. 05 Immunology • Acquired Immunity
[1 Mark]

A person recovered from measles in childhood does not get the disease again, this is due to:

(A) Innate immunity
(B) Active acquired immunity
(C) Passive immunity
(D) Non specific body defense
Ans. (B) Active acquired immunity

Explanation: During natural infection with the measles virus, the host's immune system actively produces specific antibodies and immunological memory cells (memory B-cells and memory T-cells). Upon subsequent re-exposure to the virus, these memory cells mount an immediate and heightened secondary immune response, granting lifelong active acquired immunity.

Q. 06 Circulatory System • Cardiac Regulation
[1 Mark]

Use of an artificial pacemaker helps the patient to:

(A) Regulate body temperature
(B) Regularise heart beat
(C) Helps in respiration
(D) Helps in recording heart beat
Ans. (B) Regularise heart beat

Explanation: An artificial pacemaker is an electronic device surgically implanted when the natural cardiac pacemaker—the Sinoatrial node (SA node)—becomes defective or fails to initiate rhythmic electrical impulses. It delivers electrical impulses to the myocardium to regularise and maintain a normal heartbeat rhythm.

Q. 07 Plant Anatomy • Stem Cross-Section
[1 Mark]

This T.S. of stem is of ________ because vascular bundles are in ________ arrangement:

(A) dicot, ring
(B) dicot, scattered
(C) monocot, ring like
(D) monocot, scattered
Ans. (A) dicot, ring [or (D) monocot, scattered depending on diagram specimen]

Diagnostic Botanical Rule: In standard plant anatomy:
• In dicotyledonous stems (e.g. Sunflower), vascular bundles are conjoint, collateral, open, and arranged in a definite ring around the central pith.
• In monocotyledonous stems (e.g. Maize), vascular bundles are closed and scattered throughout the ground tissue.
Option (A) accurately matches the diagnostic pair: dicot, ring.

Q. 08 Reproduction in Plants • Embryo Sac Structure
[1 Mark]

An unfertilised female gametophyte of an angiosperm has:

(A) 8 cells, 7 nuclei
(B) 7 cells, 7 nuclei
(C) 7 cells, 8 nuclei
(D) 8 cells, 8 nuclei
Ans. (C) 7 cells, 8 nuclei

Explanation: A typical mature polygonum-type female gametophyte (embryo sac) undergoes three rounds of free-nuclear mitotic divisions resulting in 8 nuclei. Cytokinesis organizes them into 7 distinct cells:
• 3 antipodal cells at the chalazal end (3 cells, 3 nuclei)
• 1 egg apparatus at the micropylar end containing 2 synergids and 1 egg cell (3 cells, 3 nuclei)
• 1 large central cell containing 2 polar nuclei (1 cell, 2 nuclei)
Total = 7 cells and 8 nuclei.

Q. 09 Genetics • Allelic Expression
[1 Mark]

The alleles that express in homozygous condition are:

(A) dominant, codominant
(B) dominant, recessive
(C) recessive, dominant
(D) incompletely dominant
Ans. (B) dominant, recessive [or (C)]

Explanation: Both dominant alleles (e.g. \(TT\)) and recessive alleles (e.g. \(tt\)) express phenotypically in the homozygous condition. While a dominant allele also expresses in the heterozygous state (\(Tt\)), a recessive allele expresses exclusively when present in the homozygous state.

Q. 10 Plant Reproduction • Microsporangium Wall Layers
[1 Mark]

The correct order of layers of a young anther from outer to inner side is:

(A) endothecium, middle layer, epidermis, tapetum
(B) epidermis, endothecium, middle layer, tapetum
(C) tapetum, middle layer, endothecium, epidermis
(D) epidermis, tapetum, endothecium, middle layer
Ans. (B) epidermis, endothecium, middle layer, tapetum

Explanation: In a transverse section of an anther, the microsporangium wall consists of four concentric layers from periphery to interior:
1. Epidermis (outer single protective layer)
2. Endothecium (fibrous hygroscopic sub-epidermal layer assisting in dehiscence)
3. Middle layers (1–3 ephemeral parenchymatous layers)
4. Tapetum (innermost nutritive layer that nourishes developing microspores).

Q. 11 Human Physiology • Health & Lifestyle Disorders
[1 Mark]

Obese person may get:

(A) hypertension
(B) atherosclerosis
(C) heart attack
(D) all of these
Ans. (D) all of these

Explanation: Excess adipose tissue deposits cause hypercholesterolemia leading to plaque accumulation in arterial walls (atherosclerosis). This narrows the arterial lumen and stiffens blood vessels, escalating blood pressure (hypertension) and substantially elevating the risk of myocardial infarction (heart attack).

Q. 12 Ecology • Ecosystem Boundaries
[1 Mark]

A transitional zone between two ecosystems is:

(A) ecotone
(B) biome
(C) wetland
(D) biosphere
Ans. (A) ecotone

Explanation: An ecotone is a zone of junction or transition between two diverse ecological communities or biomes (e.g. mangrove forest between terrestrial and marine ecosystems, or marshland between dry land and water body). It often exhibits greater species diversity, known as the edge effect.

Q. 13 Population Ecology • Population Density Growth
[1 Mark]

A human population density in an area is 100 and after one year, it increases to 110. The factors responsible are:

(A) Immigration, Natality
(B) Immigration, Mortality
(C) Emigration, Natality
(D) Emigration, Mortality
Ans. (A) Immigration, Natality

Derivation: Population change formula: \(N_{t+1} = N_t + [(B + I) - (D + E)]\).
• Natality (\(B\)) and Immigration (\(I\)) increase population density.
• Mortality (\(D\)) and Emigration (\(E\)) decrease population density.
Because the density grew from \(100 \to 110\), the positive drivers responsible are Natality and Immigration.

Q. 14 Biological Classification • Microorganisms
[1 Mark]

Saccharomyces cerevisiae belongs to:

(A) algae
(B) fungi
(C) bacteria
(D) virus
Ans. (B) fungi

Explanation: Saccharomyces cerevisiae (commonly known as Baker's yeast or Brewer's yeast) is a unicellular, eukaryotic microorganism belonging to Kingdom Fungi (Class Ascomycetes).

Q. 15 Biotechnology • Fermentation Engineering
[1 Mark]

Large tanks used to carry out fermentation process are:

(A) Stirrer
(B) Bioreactor
(C) Autoclave
(D) Motor
Ans. (B) Bioreactor (Fermenter)

Explanation: A bioreactor (or fermenter) is a large closed vessel (\(100-10,000\text{ litres}\)) engineered to biologically convert raw materials into specific products by microbial cells under controlled optimum conditions (temperature, pH, oxygen, and nutrient aeration).

Q. 16 Innate Immunity • Physical Barriers
[1 Mark]

Skin serves as 1st line of defence because:

(A) it prevents the entry of pathogens into the body
(B) it is the outer tough layer
(C) it is the sense organ
(D) none of the above
Ans. (A) it prevents the entry of pathogens into the body

Explanation: The intact stratum corneum of human skin forms a mechanical and anatomical physical barrier that prevents microbial pathogens from entering underlying tissues, functioning as the body's primary line of non-specific innate defense.

Section B

Objective Type Questions (Q17 to Q28)

[2 × 12 = 24 Marks]
Q. 17 [1 × 2 = 2 Marks]

Fill in the blanks:
"In human, since mitochondria come into ________ from the egg, inheritance of mitochondrial DNA is said to be a case of ________ inheritance."

1st Blank: zygote (or embryo / offspring)

2nd Blank: maternal (or cytoplasmic / extra-nuclear)

Scientific Context: During fertilization, only the sperm nucleus enters the ovum while sperm mitochondria located in the neck/middle piece degenerate. All mitochondria in the resulting zygote originate solely from the maternal ooplasm.

Q. 18 [1 + 1 = 2 Marks]

Identify A and B in the human endocrine diagram:

B A

• A: Uterus / Ovary (Female reproductive organ situated in the pelvic cavity)

• B: Adrenal Glands / Kidneys (Endocrine glands positioned retroperitoneally above each kidney)

Q. 19 [1 + 1 = 2 Marks]

In the Central Dogma schematic: \(\text{DNA} \xrightarrow{\mathbf{A}} \text{mRNA} \xrightarrow{\mathbf{B}} \text{Protein}\). Name \(\mathbf{A}\) and \(\mathbf{B}\):

• \(\mathbf{A}\): Transcription (Enzymatic synthesis of mRNA from a DNA template directed by RNA Polymerase)

• \(\mathbf{B}\): Translation (Decoding genetic codons on mRNA into an amino acid polypeptide chain on ribosomes)

Q. 20 [1 + 1 = 2 Marks]

Match the plant families with their respective representative species:

Column - I (Family) Column - II (Plant Species)
(1) Malvaceae(A) Hibiscus rosa-sinensis
(2) Fabaceae(B) Pisum sativum
(C) Oryza sativa
Matching Pairs:
  • • (1) Malvaceae \(\to\) (A) Hibiscus rosa-sinensis (China rose / Shoe-flower)
  • • (2) Fabaceae \(\to\) (B) Pisum sativum (Garden pea)
  • (Note: Oryza sativa belongs to family Poaceae / Gramineae).
Q. 21 [1 + 1 = 2 Marks]

In the endocrine regulatory pathway of thyroid hormone:
What do the solid line and broken line represent?

Hypothalamus Pituitary Thyroid Thyroxin in Blood (-)

• Solid Line: Represents Stimulatory / Positive Action (Forward Endocrine Stimulation). Hypothalamus secretes TRH to stimulate the Pituitary, which in turn releases TSH to stimulate the Thyroid gland to secrete thyroxin.

• Broken Line: Represents Inhibitory Feedback (Negative Feedback Mechanism). Elevated circulating levels of thyroxin act back on the anterior pituitary and hypothalamus to suppress further release of TSH and TRH, maintaining homeostatic hormonal equilibrium.

Q. 22 [1 + 1 = 2 Marks]

Write True (T) or False (F):

  1. The edible part of coconut is endocarp.
  2. Apple is having fleshy thalamus.

(a) False (F) (The edible portion of coconut is the cellular and liquid endosperm; the endocarp is the hard stony inner shell that is inedible).

(b) True (T) (Apple is a false fruit or pome where the fleshy, edible portion develops from the swollen thalamus surrounding the central ovary core).

Q. 23 [1 + 1 = 2 Marks]

Answer the following questions regarding diagnostic prenatal testing (Amniocentesis):

  1. Name the technique.
  2. What is its significance?

(a) Amniocentesis (or Gel Electrophoresis / Micropropagation depending on curriculum diagram).

(b) Significance:
• It allows prenatal detection of chromosomal abnormalities (e.g. Down syndrome, Turner syndrome, Klinefelter syndrome) and metabolic genetic disorders in the developing fetus.
• It assesses fetal health and lung maturity before delivery.

Q. 24 [1 + 1 = 2 Marks]

Choose the two correct words to fill in the blanks [nodules, nodes, nodulins]:

  1. In legumes, nitrogen fixation occurs in specialized bodies called ________.
  2. ________ is the special protein formed as a result of symbiosis in legumes.

(a) nodules (Root nodules housing symbiotic Rhizobium bacteria)

(b) Nodulins (Special plant-encoded proteins specifically induced during nodule organogenesis and symbiosis)

Q. 25 [1 + 1 = 2 Marks]

Complete the diagram of the multipolar neuron by correctly labelling the indicated parts (including Axon and Dendrites):

N Dendrites Nucleus Cyton (Soma) Myelin Sheath Node of Ranvier Axon

• Dendrites: Short, profusely branched protoplasmic projections that receive electrical impulses and transmit them toward the cyton (soma).

• Axon: Long, single, cylindrical process that conducts nerve impulses away from the cell body toward synaptic terminals.

Q. 26 [1 + 1 = 2 Marks]

Regarding the nutritional deficiency disorder shown in the child (distended pot belly, limb edema, wasted muscles):

  1. Name the disorder in this child's body.
  2. What is the reason?

(a) Kwashiorkor [or Protein-Energy Malnutrition (PEM) / Cretinism].

(b) Reason: Severe deficiency of proteins in the diet despite adequate or marginal caloric/carbohydrate intake. Hypoalbuminemia reduces blood oncotic pressure, resulting in fluid accumulation (edema) in tissues and the abdominal cavity (ascites / pot-belly).

Q. 27 [2 Marks]

After organ transplantation, why should the patient be administered with immunosuppressants?

• Graft Rejection Prevention: The recipient's immune system recognizes cell-surface major histocompatibility complex (MHC/HLA) markers on the transplanted organ as non-self foreign antigens.

• Role of Cell-Mediated Immunity: T-lymphocytes (cytotoxic T-cells) initiate a cell-mediated immune response to attack and destroy the foreign donor tissue (graft rejection). Immunosuppressive drugs (such as Cyclosporin A) inhibit T-cell activation, preventing graft rejection and ensuring survival of the transplant.

Q. 28 [1 + 1 = 2 Marks]

Fill in the blanks with the appropriate words given in bracket [wings, beak, analogous, homologous]:
"The ________ of bird and bat are an example of ________ organs."

1st Blank: wings

2nd Blank: analogous

Evolutionary Explanation: Bird wings (feathers attached to forelimb skeleton) and bat wings (membranous patagium stretched between elongated digits) perform identical flight functions but possess fundamentally different structural origins, demonstrating convergent evolution.

Section C

Very Short Answer Questions (Q29 to Q37)

[2 × 9 = 18 Marks • 30–50 Words]
Q. 29 [2 Marks]

Construct a pyramid of numbers in a grassland with 4 trophic levels.

T4: Top Carnivores (Hawks) T3: Carnivores (Frogs / Birds) T2: Herbivores (Grasshoppers) T1: Producers (Grasses)

In a grassland ecosystem, the pyramid of numbers is upright because the number of individuals progressively decreases at successive higher trophic levels:

  • 1. T1 (Producers): Grasses / vegetation (numerically most abundant).
  • 2. T2 (Primary Consumers): Herbivores such as grasshoppers and rabbits.
  • 3. T3 (Secondary Consumers): Primary carnivores such as frogs and lizards.
  • 4. T4 (Tertiary Consumers): Top carnivores such as snakes and hawks (fewest in number).
Q. 30 [2 Marks]

With reference to class Reptilia, identify the two wrong statements:

  1. Body is covered by hairs
  2. Has paired pentadactyl limbs with clawed digits
  3. Respiration is by lungs
  4. Heart is two chambered
Identified Wrong Statements:

1. Statement (a): "Body is covered by hairs"
Correction: Reptilian skin is dry, non-glandular, and covered by epidermal scales or scutes; hair is a diagnostic characteristic exclusive to mammals.

2. Statement (d): "Heart is two chambered"
Correction: Reptiles possess a 3-chambered heart with an incompletely partitioned ventricle (and fully 4-chambered in crocodiles); a 2-chambered venous heart is found in fishes.

Q. 31 [2 Marks]

Construct a flow chart using these words:
[Xylem, Complex permanent tissue, simple permanent tissue, plant tissue, meristem, parenchyma]

Plant Tissue Meristem Permanent Tissue Simple Perm. Tissue Complex Perm. Tissue Parenchyma Xylem
Hierarchical Scheme:

Plant Tissue \(\to\) Meristem & Permanent Tissue.
Permanent Tissue \(\to\) Simple permanent tissue (Example: Parenchyma) & Complex permanent tissue (Example: Xylem).

Q. 32 (Choice I) [2 Marks]

Write any two biotic and two abiotic components of a pond ecosystem.

Solution to Choice I:

• Biotic Components (Living):
1. Producers: Phytoplankton, algae, and submerged/floating macrophytes (e.g. Hydrilla, Nymphaea).
2. Consumers: Zooplankton, aquatic insects, crustaceans, and fishes.

• Abiotic Components (Non-living):
1. Water containing dissolved oxygen, minerals, and carbon dioxide.
2. Solar radiation (light penetration), temperature, and bottom pond soil/silt.

OR Alternative

Explain the term commensalism with an example.

Solution to Choice II:

• Definition: Commensalism is an interspecific ecological interaction between two organisms of different species in which one species benefits (\(+\)) while the other remains completely unaffected, deriving neither benefit nor harm (\(0\)).

• Example: An epiphytic orchid growing on the bark of a mango tree. The orchid derives physical support, space, and sunlight without extracting water or sap from the mango tree, leaving the tree unharmed.

Q. 33 (Choice I) [1 + 1 = 2 Marks]

Name the microbial source used to extract the following antibiotics:

  1. Tetracycline
  2. Streptomycin
Solution to Choice I:

(a) Tetracycline: Extracted from actinomycete bacterium Streptomyces aureofaciens (or Streptomyces rimosus).

(b) Streptomycin: Extracted from actinomycete bacterium Streptomyces griseus.

OR Alternative

Write any two advantages of biogas.

Solution to Choice II:

1. Clean, Non-Polluting Fuel: Biogas (composed of \(50-70\%\) methane) burns with a high calorific value without smoke or residue, preventing household respiratory ailments and deforestation.

2. Enriched Organic Manure: The spent fermented slurry discharged from the biogas digester is rich in nitrogen and phosphorus, serving as high-quality organic biofertilizer for agricultural crops.

Q. 34 (Choice I) [1 + 1 = 2 Marks]

Write the function of the following carbohydrates:

  1. Glucose
  2. Cellulose
Solution to Choice I:

(a) Glucose: Serves as the universal cellular fuel and primary respiratory substrate oxidized during glycolysis and the Krebs cycle to synthesize ATP energy for metabolic activities.

(b) Cellulose: A structural polysaccharide forming microfibrils that provide tensile strength, mechanical rigidity, and structural integrity to plant cell walls.

OR Alternative

Write any two functions of roughage (dietary fibre).

Solution to Choice II:

1. Promotes Healthy Bowel Movement: Roughage adds bulk to food bolus and retains water, facilitating smooth intestinal peristalsis and preventing chronic constipation.

2. Lowers Cholesterol and Blood Sugar Spikes: Soluble dietary fibres bind to bile acids, aiding in cholesterol excretion and slowing glucose absorption in the small intestine.

Q. 35 (Choice I) [2 Marks]

AUG has dual function, which are they?

Solution to Choice I:

1. Initiation Codon: It acts as the initiation codon on mRNA, signalling the starting point of translation (protein synthesis) on the ribosome.

2. Codes for Methionine: It encodes the specific amino acid Methionine (Met) in eukaryotes (and \(N\)-formylmethionine in prokaryotes).

OR Alternative

Write the anticodon for mRNA codons: (i) \(\text{AGU}\) and (ii) \(\text{UGC}\).

Solution to Choice II:

(i) For codon \(5'-\text{AGU}-3'\): Complementary anticodon is \(3'-\text{UCA}-5'\) (or \(5'-\text{ACU}-3'\)).

(ii) For codon \(5'-\text{UGC}-3'\): Complementary anticodon is \(3'-\text{ACG}-5'\) (or \(5'-\text{GCA}-3'\)).

Q. 36 [2 Marks]

Draw the diagram to show the double helical structure of a DNA molecule.

A = T G ≡ C T = A C ≡ G 5' 3' 3' 5' Pitch: 3.4 nm Dia: 2.0 nm

Watson-Crick B-DNA Model Features:

  • Two polynucleotide chains run in antiparallel directions (\(5' \to 3'\) and \(3' \to 5'\)).
  • Sugar-phosphate backbones form the exterior; nitrogenous bases project inward.
  • Purines pair with pyrimidines: \(\text{Adenine} = \text{Thymine}\) (\(2\text{ H-bonds}\)) and \(\text{Guanine} \equiv \text{Cytosine}\) (\(3\text{ H-bonds}\)).
  • Diameter is \(2.0\text{ nm}\) (\(20\text{ \AA}\)); each complete helical turn has a pitch of \(3.4\text{ nm}\) containing 10 base pairs.
Q. 37 [2 Marks]

What is PEM? Give two reasons for PEM.

• Definition: PEM stands for Protein-Energy Malnutrition, an umbrella term for a spectrum of nutritional deficiency disorders (such as Kwashiorkor and Marasmus) affecting primarily infants and young children.

• Two Main Reasons:
1. Inadequate Nutritional Intake: Chronic deficiency of protein and calories in weaning food due to poverty, lack of balanced dietary resources, and maternal malnutrition.
2. Recurrent Gastrointestinal Infections: Frequent bouts of diarrhea and enteritis impair intestinal absorption of nutrients, rapidly precipitating acute wasting and edema.

Section D

Short Answer Questions (Q38 to Q41)

[3 × 4 = 12 Marks • 50–80 Words]
Q. 38 [3 Marks]

Write the schematic representation of the nitrogen cycle.

Atmospheric N₂ (78%) Biological Fixation Electrical / Industrial Ammonia (NH₃) Dead Biomass Ammonification Nitrosomonas Nitrite (NO₂⁻) Nitrobacter Nitrate (NO₃⁻) Denitrification (Pseudomonas) Plant Uptake / Assimilation

Key Stages of the Nitrogen Cycle:

  1. Nitrogen Fixation: Conversion of atmospheric \(\text{N}_2\) into \(\text{NH}_3\) by biological fixers (Rhizobium, Azotobacter, Anabaena) or industrial processes.
  2. Nitrification: Two-step biological oxidation of ammonia into nitrite by Nitrosomonas, followed by oxidation to nitrate by Nitrobacter: \[2\text{NH}_3 + 3\text{O}_2 \to 2\text{NO}_2^- + 2\text{H}^+ + 2\text{H}_2\text{O}; \quad 2\text{NO}_2^- + \text{O}_2 \to 2\text{NO}_3^-\]
  3. Assimilation & Ammonification: Plants absorb \(\text{NO}_3^-\); decomposing organic remains are converted back to ammonia by ammonifying bacteria.
  4. Denitrification: Anaerobic reduction of nitrate (\(\text{NO}_3^-\)) back into inert \(\text{N}_2\) gas by Pseudomonas and Thiobacillus.
Q. 39 (Choice I) [3 Marks]

Name the three subunits of a nucleotide.

Solution to Choice I:

Every nucleotide monomer consists of three distinct chemical components:

  1. Pentose Sugar: A five-carbon ring sugar—either \(\beta\)-D-ribose (in RNA nucleotides) or \(\beta\)-D-2'-deoxyribose (in DNA nucleotides).
  2. Nitrogenous Base: Heterocyclic nitrogen ring attached via an \(N\)-glycosidic bond to carbon-1' of the sugar. Can be a Purine (Adenine, Guanine) or a Pyrimidine (Cytosine, Thymine, Uracil).
  3. Phosphate Group: A phosphoric acid (\(\text{H}_3\text{PO}_4\)) group esterified via a phosphoester linkage to the 5'-hydroxyl (\(-\text{OH}\)) of the pentose sugar.
OR Alternative

Write any three differences between DNA and RNA.

Solution to Choice II:
FeatureDNA (Deoxyribonucleic Acid)RNA (Ribonucleic Acid)
1. Pentose SugarContains 2'-deoxyribose sugar (lacks an oxygen at C-2').Contains ribose sugar (has a reactive \(-OH\) group at C-2').
2. Pyrimidine BaseContains Thymine (T) along with Cytosine.Contains Uracil (U) in place of Thymine.
3. Structure & StabilityDouble-stranded (\(ds\)); chemically stable and less reactive; primary hereditary material.Single-stranded (\(ss\)); chemically more labile and catalytic; acts mainly as a messenger/adapter.
Q. 40 (Choice I) [1 + 1 + 1 = 3 Marks]

Give scientific reasons:

  1. Testes are extra-abdominal.
  2. In aphids, we can see only females.
  3. A mother should feed her new born baby just after birth.
Solution to Choice I:

(a) Extra-abdominal Testes: Human spermatogenesis requires an optimal temperature \(2-2.5^\circ\text{C}\) lower than the core body temperature (\(37^\circ\text{C}\)). The scrotum hangs outside the pelvic cavity, acting as a physical thermoregulator.

(b) Female Aphids: During favorable spring and summer conditions, aphids reproduce through parthenogenesis (thelytoky), wherein unfertilized diploid eggs develop directly into generations of exclusively female offspring without mating.

(c) Feeding Newborn Immediately: Immediate suckling delivers early breast milk rich in essential antibodies (IgA), establishes maternal-infant bonding, and releases maternal oxytocin to accelerate uterine involution.

OR Alternative

What is colostrum? Why is it essential for a newly born baby?

Solution to Choice II:

• Definition: Colostrum is the thick, yellowish initial milk produced and secreted by the maternal mammary glands during the first \(2-4\text{ days}\) postpartum following parturition.

• Essential Significance:
1. Passive Immunization: It contains exceptionally high titers of Secretory Immunoglobulin A (\(\text{IgA}\)), lactoferrin, and lysozyme that coat the newborn's immature gut lining, providing immediate passive immunity against diarrheal and respiratory pathogens.
2. Clears Meconium & Nutritive: It contains mild laxative properties that facilitate the expulsion of meconium (the infant's first stool, clearing fetal bilirubin) and provides concentrated proteins, vitamins, and minerals.

Q. 41 [1 + 2 = 3 Marks]

What is placenta? Give any two of its functions.

• Definition: The placenta is a temporary, specialized organic structural and physiological disc-like connection established between the chorionic villi of the developing embryo/fetus and the maternal uterine endometrium during pregnancy.

• Two Major Functions:
1. Physiological Exchange (Nutrition, Respiration & Excretion): Mediates the diffusion of oxygen (\(\text{O}_2\)), glucose, amino acids, and antibodies from maternal blood to the fetus, while concurrently transferring fetal metabolic wastes (\(\text{CO}_2\), urea, creatinine) into maternal circulation for maternal elimination.
2. Endocrine Function: Acts as a temporary endocrine gland synthesizing essential pregnancy hormones, including human Chorionic Gonadotropin (hCG), human Placental Lactogen (hPL), progesterone, and estrogens to sustain gestation.

Section E

Long Answer Questions (Q42 to Q43)

[5 × 2 = 10 Marks • 80–120 Words]
Q. 42 (Choice I) [1 + 4 = 5 Marks]

What is photophosphorylation? Write the differences between cyclic and non-cyclic photophosphorylation.

Solution to Choice I:
1. Definition of Photophosphorylation (1 Mark)

Photophosphorylation is the light-driven synthesis of ATP from ADP and inorganic phosphate (\(\text{P}_i\)) within the thylakoid membranes of chloroplasts using photon energy harvested during the photochemical phase of photosynthesis: \[\text{ADP} + \text{P}_i \xrightarrow{\text{Light energy, Chloroplast}} \text{ATP}\]

2. Differences: Cyclic vs Non-Cyclic Photophosphorylation (4 Marks)
Feature Cyclic Photophosphorylation Non-Cyclic Photophosphorylation
Photosystems Involved Involves only Photosystem I (PS I, \(P_{700}\)). Involves both PS II (\(P_{680}\)) and PS I (\(P_{700}\)) operating cooperatively.
Electron Trajectory Electrons expelled from \(P_{700}\) pass through carriers (Fd \(\to\) Cyt \(b_6f\) \(\to\) PC) and return back to PS I in a closed circuit. Electrons follow a unidirectional open Z-scheme from \(\text{H}_2\text{O} \to \text{PS II} \to \text{PS I} \to \text{NADP}^+\).
Photolysis of Water Does NOT occur. Occurs on the inner luminal side of PS II (\(2\text{H}_2\text{O} \to 4\text{H}^+ + 4e^- + \text{O}_2\)).
Oxygen (\(\text{O}_2\)) Evolution No oxygen is evolved. Oxygen gas (\(\text{O}_2\)) is released as a vital byproduct.
End Products Synthesizes ATP only. Synthesizes both \(\text{ATP}\) and \(\text{NADPH} + \text{H}^+\) (assimilatory power).
OR Alternative

Name the scientists who discovered the \(C_3\) cycle in plants. Write any four differences between \(C_3\) plants and \(C_4\) plants.

Solution to Choice II:
1. Discovery of \(C_3\) Cycle (1 Mark)

The \(C_3\) cycle (Calvin Cycle / Reductive Pentose Phosphate Pathway) was elucidated by Melvin Calvin, along with James Bassham and Andrew Benson, using radioactive \(^{14}\text{C}\) in green algal cultures (*Chlorella* and *Scenedesmus*). Melvin Calvin was awarded the Nobel Prize in Chemistry in 1961.

2. Four Differences: \(C_3\) Plants vs \(C_4\) Plants (4 Marks)
Characteristic \(C_3\) Plants \(C_4\) Plants
1. First Stable Product A 3-carbon compound: 3-Phosphoglyceric acid (3-PGA). A 4-carbon dicarboxylic acid: Oxaloacetic acid (OAA).
2. Primary \(\text{CO}_2\) Acceptor RuBP (Ribulose-1,5-bisphosphate, a 5-carbon ketose sugar) catalyzed by RuBisCO. PEP (Phosphoenolpyruvate, a 3-carbon acid) catalyzed by PEP carboxylase.
3. Leaf Anatomy Possess standard mesophyll; lack Kranz anatomy. Possess Kranz anatomy (dimorphic chloroplasts in bundle-sheath and mesophyll cells).
4. Photorespiration (\(C_2\) Cycle) High photorespiratory loss (up to \(25-40\%\) of fixed carbon is lost in hot, bright conditions). Photorespiration is negligible or completely absent due to high internal \(\text{CO}_2\) concentration around RuBisCO.
5. Temperature Optimum Lower optimum temperature range: \(20^\circ\text{C} - 25^\circ\text{C}\). Higher optimum temperature range: \(35^\circ\text{C} - 45^\circ\text{C}\) (adapted to tropical and arid climates).
Q. 43 (Choice I) [2 + 3 = 5 Marks]

(a) Draw a neat and well-labelled diagram of a human nephron showing its different functional segments.
(b) Describe the three essential physiological steps involved in the process of urine formation in humans.

Solution to Choice I:
Part (a): Structural Anatomy of a Nephron (2 Marks)

The nephron is the microscopic structural and functional unit of the human kidney. Each kidney accommodates approximately \(1.0 - 1.2\text{ million}\) nephrons consisting of two main components: the Renal Corpuscle (Malpighian Body) and the Renal Tubule.

Afferent Efferent Bowman's Capsule PCT Descending Limb Ascending Limb Loop of Henle DCT Collecting Duct
Part (b): Three Essential Steps in Urine Formation (3 Marks)
  1. 1. Glomerular Ultrafiltration: Occurs across the three-layered filtration barrier (capillary endothelium, basement membrane, and podocyte slit pores). Blood enters via the wider afferent arteriole under net filtration pressure (\(\text{NFP} \approx 10-15\text{ mmHg}\)). Water, glucose, amino acids, urea, and electrolytes filter into Bowman's space forming primary urine (glomerular filtrate) at a rate of \(125\text{ mL/min}\) (\(180\text{ litres/day}\)). Plasma proteins and blood cells are retained.
  2. 2. Selective Tubular Reabsorption: Over \(99\%\) of the glomerular filtrate volume is reabsorbed back into peritubular capillaries:
    • PCT: Reabsorbs nearly \(100\%\) of essential nutrients (glucose, amino acids) by active transport, and \(70-80\%\) of electrolytes and water by obligate osmosis.
    • Henle's Loop: Countercurrent multiplier mechanism; descending limb is permeable to water but impermeable to salts, whereas ascending limb actively reabsorbs \(\text{Na}^+\) and \(\text{Cl}^-\) but is impermeable to water.
    • DCT & Collecting Duct: Facultative reabsorption of water and sodium controlled hormonally by ADH (Vasopressin) and Aldosterone.
  3. 3. Tubular Secretion: Tubular epithelial cells actively transport metabolic end-products and exogenous substances—principally \(\text{H}^+\), \(\text{K}^+\), \(\text{NH}_4^+\), creatinine, and drug metabolites—from the peritubular capillary blood into the tubular lumen. This active process maintains homeostatic ionic balance and normal blood \(\text{pH}\) (\(7.35 - 7.45\)).
OR Alternative

(a) Explain the hormonal regulation and ovarian changes occurring during the phases of the human menstrual cycle.
(b) State three major physiological differences between Spermatogenesis and Oogenesis.

Solution to Choice II:
Part (a): Menstrual Cycle Phases & Hormonal Orchestration (3 Marks)

The human female reproductive cycle spans approximately 28 days and comprises four distinct sequential phases:

  • • 1. Menstrual Phase (Days 1–5): Triggered by the abrupt decline of progesterone and estrogen due to degeneration of the corpus luteum. The functional endometrial layer sloughs off along with unfertilized ovum, producing \(40-80\text{ mL}\) of menstrual blood.
  • • 2. Follicular / Proliferative Phase (Days 6–13): Anterior pituitary hormone FSH (Follicle Stimulating Hormone) stimulates maturation of primary follicles into a mature Graafian follicle. Developing granulosa cells secrete estrogen, which induces mitotic repair and thickening of the uterine endometrium.
  • • 3. Ovulatory Phase (Day 14): Peak levels of estrogen trigger a sharp surge in anterior pituitary LH (Luteinizing Hormone Surge). This causes the mature Graafian follicle to rupture and release the secondary oocyte into the fallopian tube (ovulation).
  • • 4. Luteal / Secretory Phase (Days 15–28): Under LH influence, the ruptured follicle transforms into an endocrine gland, the Corpus Luteum, which secretes abundant progesterone. Progesterone renders the endometrium vascular, glandular, and receptive for blastocyst implantation. In the absence of fertilization, it degenerates into the scar-like corpus albicans.
Part (b): Comparison: Spermatogenesis vs Oogenesis (2 Marks)
Parameter Spermatogenesis Oogenesis
1. Site of Occurrence Occurs continuously in the seminiferous tubules of the male testes. Occurs within the cortical follicles of the female ovaries.
2. Gamete Yield per Precursor 1 primary spermatocyte undergoes equal cytokinesis to yield 4 functional spermatozoa (\(n\)). 1 primary oocyte undergoes unequal cytokinesis to yield 1 functional ovum (\(n\)) and 2–3 non-functional polar bodies.
3. Temporal Pattern Initiates at puberty and continues uninterrupted throughout adult life into old age. Initiates prenatally during embryonic development, pauses at Diplotene of Prophase I, resumes at puberty, and permanently ceases at menopause (\(\approx 45-50\text{ years}\)).