ISC Biology 2026 — Questions and Answers

Section A — 20 Marks

Question 1 — Short Answer Questions

i) Zara is suffering from cough and has bluish lips. Her mucus is mixed with blood. What is the biological name of the causative organism responsible for Zara's condition?

Answer: The symptoms suggest severe Pneumonia. The biological name of the causative organism is Streptococcus pneumoniae (or Haemophilus influenzae).

ii) Observe the relation between the first two words and complete the analogy:

Dengue : Aedes :: Plague : _________

Answer: Xenopsylla cheopis (the Rat flea)

iii) If the number of chromosomes in the cells of the calyx is 16, how many chromosomes will be formed in the endosperm after double fertilisation?

• Calyx cells are diploid (2n = 16), so n = 8
• Endosperm is triploid (3n) after double fertilisation
• Chromosomes in endosperm: 3 x 8 = 24

Answer: 24 chromosomes

iv) A scientist isolates a second restriction endonuclease from the strain DC3 of Helicobacter pyogenes. What would the scientist name it using the standard naming technique?

Following standard nomenclature:

Answer: HpyDC3II

v) Meenu is suffering from a genetic disorder in which phenylpyruvic acid and its metabolites accumulate in blood. It causes impairment of nervous tissues. Which enzyme deficiency has caused this disorder in Meenu?

Answer: Meenu is suffering from Phenylketonuria (PKU).

Enzyme deficiency: Phenylalanine hydroxylase

vi) There were 770 frogs in a pond. 70 of them died within a month. Calculate the death rate of the population.

Answer:

Death Rate = Number of deaths / Initial population
           = 70 / 770
           = 1/11
           ≈ 0.091 individuals per frog per month

vii) A farmer notices that the leaves of his cabbage plants are curling up because of heavy infestation of aphids. Suggest one eco-friendly method to control their spread.

Answer: Introduce natural predators such as Ladybirds (ladybug beetles) or Lacewings to the cabbage patch. Alternatively, apply neem oil spray as a biological pesticide.

viii) In a particular plant with bilobed dithecous anther, there were 50 Pollen Mother Cells. Only 25% could develop their pollen tubes. Calculate the number of pollen grains that would proceed for fertilisation.

Each PMC produces 4 pollen grains.
Total pollen grains = 50 x 4 = 200
Only 25% develop pollen tubes = 25% of 200 = 50

Answer: 50 pollen grains

ix) Choose the components required for the proper functioning of DNA polymerase enzyme.

• (P) DNA template
• (Q) A DNA primer
• (R) An RNA primer
• (S) Four different dNTPs

(a) Only (P), (Q) and (R)

(b) Only (P), (R) and (S)

(c) Only (Q), (R) and (S)

(d) Only (P), (Q) and (S)

Answer: (d) Only (P), (Q) and (S)

DNA polymerase requires a DNA template, a DNA primer, and all four dNTPs. In lab settings, DNA primers are used for stability. RNA primers (R) are used in vivo but are not a direct component of DNA polymerase itself.

x) The left hind limb of a person was swollen like an elephant's limb. Which one of the following was MOST likely the cause of swelling?

• (a) Accumulation of uric acid
• (b) Blockage of blood vessels
• (c) Blockage of lymph vessels
• (d) Hypertrophy of skeletal muscles

Answer: (c) Blockage of lymph vessels

This condition is Filariasis (Elephantiasis), caused by blockage of the lymphatic system by filarial worms (Wuchereria bancrofti).

xi) Assertion and Reason — RNAi Technique

• Assertion: RNAi technique is applied to plants and animals to protect them from pest infestation.
• Reason: This technique inhibits the expression of certain genes in pests.

(a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.

(b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.

(c) Assertion is true and Reason is false.

(d) Both Assertion and Reason are false.

Answer: (a)

RNAi silences specific mRNA sequences, inhibiting the expression of essential genes in pests (e.g., Meloidogyne incognita), thereby protecting the plant.

xii) Assertion and Reason — Klinefelter Syndrome

• Assertion: Individuals with Klinefelter syndrome lose their secondary sexual characters and develop gynaecomastia.
• Reason: The presence of the XYY chromosomal pattern leads to underdeveloped testes and reduced testosterone levels in affected males.

Answer: (c) Assertion is true and Reason is false.

Klinefelter syndrome is characterised by the XXY chromosomal pattern, not XYY. XYY is a different condition (Jacob's syndrome).

xiii) In a garden pea plant, the flowers are in axial (A) position. Find out the proportion of flowers in terminal position (a), if a cross is made between two heterozygous plants.

Cross: Aa x Aa

Ratio = 3 Axial : 1 Terminal

Answer: Proportion of terminal flowers (aa) = 1/4 or 25%

xiv) Name the autoimmune disorder that affects the skeletal muscles in humans.

Answer: Myasthenia gravis

It affects the neuromuscular junction and leads to progressive weakness of skeletal muscles.

xv) Answer the following:

(a) Expand the abbreviation IUI.

IUI = Intrauterine Insemination

(b) The enzyme polynucleotide phosphorylase helps the in-vitro synthesis of RNA. Name the scientist who discovered this enzyme.

Answer: Severo Ochoa

xvi) Shama complained of severe itching where ring-like patches appeared on her skin. Write the biological name of the causative agent.

The condition is Ringworm (Dermatophytosis).

Causative agents: Fungi belonging to the genera Microsporum, Trichophyton, or Epidermophyton

xvii) Name the chemical messenger that activates Natural Killer T-cells and macrophages.

Answer: Interferons (specifically Interferon-gamma) / Cytokines

xviii) Give a reason for each of the following:

(a) Kanika was suffering from SCID and needed periodic infusion of transformed lymphocytes.

Lymphocytes are not immortal. The transformed lymphocytes carrying the functional ADA gene eventually die, so the patient requires regular infusions to maintain an active immune system.

(b) In HIV infected persons, the immune system is compromised.

HIV specifically attacks and destroys Helper T-cells (CD4+ T-lymphocytes), which are essential for coordinating the entire immune response, thereby leaving the body unable to fight infections.

Section B — 14 Marks

Question 2 — DNA Fingerprinting Analysis

i) Which suspect's DNA sample matches the sample collected from the crime scene?

Answer: Suspect 2 (S2)

Comparing the banding pattern in column C (Crime scene) with all suspects, S2 shows an identical pattern at all levels (Levels 1, 2, 5, and 7 match the dark bands).

ii) Mention the principle on which DNA profiling is based.

DNA profiling is based on DNA Polymorphism, specifically variations in VNTRs (Variable Number of Tandem Repeats) in the non-coding (satellite) sequences of DNA. These repeats differ between individuals, producing unique banding patterns.

Question 3 — Pollination Argument

In a group discussion, Shubham argued that Cleistogamy was a type of Xenogamy whereas Ameena was of the opinion that it was a kind of Autogamy. Whose argument is correct? Justify.

Answer: Ameena is correct.

Justification: Cleistogamy occurs in flowers that never open. Since the pollen must come from the same flower, cross-pollination is impossible. It is therefore a strictly specialised form of Autogamy (self-pollination), not Xenogamy (cross-pollination between genetically different individuals).

Question 4 — Energy Flow / Growth Rate

i) The amount of energy at the fourth trophic level in the food chain (Grass → Zebra → Crocodile → Vulture) is 4 J. What will be the amount of energy available at the sunlight and transducer level?

Applying Lindeman's 10% Law (only 10% energy transfers between trophic levels):

Vulture       =     4 J   (4th trophic level)
Crocodile     =    40 J   (3rd trophic level)
Zebra         =   400 J   (2nd trophic level)
Grass         = 4,000 J   (1st trophic / Transducer level)

Plants fix only 1% of sunlight:
Sunlight = 4,000 x 100 = 4,00,000 J

Answer:

• Transducer (Grass) level = 4,000 J
• Sunlight level = 4,00,000 J

ii) OR — Roselin had an aquarium with 10 fish. In a year, a fish gave birth to 7 fish. 3 adult fish died. Father purchased 2 more pairs. Later, Roselin gifted 4 fish to her friend. Find out the growth rate of fish.

Growth Rate = (Births + Immigration) - (Deaths + Emigration)
            = (7 + 4) - (3 + 4)
            = 11 - 7
            = 4

Answer: Growth Rate = 4

Question 5 — Bioreactor Vessels

Anthony has two large vessels — one with a sparger and the other with a stirrer. Which vessel will be more advantageous? Why?

Answer: The Sparged Stirred-tank Bioreactor (with sparger) is more advantageous.

A sparger bubbles air directly through the culture medium, dramatically increasing the oxygen transfer surface area and ensuring far better oxygen availability compared to a simple stirrer. This leads to higher cell density and improved productivity.

Question 6 — Decline in Migratory Birds

Based on the provided table, interpret two reasons for the change in the bird population.

1. Habitat Loss: The Aquatic Area decreased from 350 ha to 220 ha, reducing the space available for foraging, roosting, and nesting of migratory birds.

2. Food Scarcity: The percentage of Aquatic Plants dropped from 85% to 45%, indicating severe depletion of the primary food source for many migratory species.

Question 7 — Ecological Interaction (Paramecium)

Paramecium caudatum and Paramecium aurelia were grown together; P. caudatum got eliminated. Identify and explain the ecological interaction based on Gause's principle.

Identification: Competitive Exclusion (Interspecific Competition)

Explanation: According to Gause's Competitive Exclusion Principle, two closely related species competing for the same limiting resources cannot co-exist indefinitely. The competitively superior species (P. aurelia) eventually eliminates the inferior one (P. caudatum) from the habitat.

Question 8 — Plasmid without Selectable Marker

i) Predict the outcome of the experiment (plasmid without a selectable marker used as a vector).

It will be impossible to distinguish between transformants (cells that took up the plasmid) and non-transformants. Without a selectable marker (e.g., an antibiotic resistance gene), all cells grow or die equally, making selection of recombinant cells impossible.

ii) Name an artificial chromosome used as a vector in Penicillium notatum.

Answer: YAC (Yeast Artificial Chromosome)

YAC is used as a vector for cloning large DNA fragments in fungal hosts like Penicillium.

Section C — 21 Marks

Question 9 — Genetically Engineered Insulin

i) How is mature insulin different from pro-insulin?

ii) Name the company that produced genetically engineered insulin for the first time.

Answer: Eli Lilly (American company, 1983)

iii) Why is functional insulin produced by rDNA technology better than the ones produced earlier?

rDNA-produced insulin is identical to human insulin. It does not cause allergic reactions or local immune responses, which were common with insulin extracted from the pancreas of slaughtered cattle and pigs (bovine/porcine insulin).

Question 10 — Species–Area Relationship

i) What pattern do you observe in the data (Island Area vs. No. of mammalian species)?

The data shows a Species–Area Relationship: as the area of the island increases, the species richness (number of mammalian species) also increases, following a consistent pattern.

ii) Name the scientist who proposed this concept.

Answer: Alexander von Humboldt

iii) Write the mathematical expression for this relationship.

log S = log C + Z log A

Where:
  S = Species richness
  A = Area
  Z = Slope of the line (regression coefficient), typically 0.1 to 0.2
  C = Y-intercept (a constant)

Question 11 — Sex Determination in Honeybees

i) Draw a flowchart to show the process of sex determination in honeybees.

Sex determination in honeybees is based on the Haplodiploid system:

Queen (Female, 2n)
        |
        | Meiosis
        v
      Eggs (n)
        |
  +-----+---------------------------------+
  | Fertilised by Sperm (n) from Drone    | Not fertilised (Parthenogenesis)
  v                                       v
Zygote (2n)                          Male Drone (n)
  |
  v
Female (Queen or Worker, 2n)

ii) If there are 16 chromosomes in the somatic cells of a male honeybee, how many chromosomes are present in spermatozoa?

Male honeybees are haploid (n). They produce sperm by mitosis (not meiosis).

Answer: 16 chromosomes (same as somatic cells, since they are already haploid)

Question 12 — Cannabinoid Drug

i) Mention the scientific name of the source plant (Cannabinoid / THC).

Answer: Cannabis sativa

ii) State any one effect of this drug on the human body.

It primarily affects the cardiovascular system — causes increased heart rate (tachycardia). It also acts on the CNS, causing euphoria, altered sensory perception, and impaired coordination.

iii) What is meant by withdrawal symptom?

Withdrawal symptoms are the unpleasant physical or psychological reactions (such as anxiety, shakiness, nausea, or sweating) that occur when an individual abruptly stops or reduces the intake of an addictive drug after a period of regular use.

Question 13 — Embryo Development (Dicot)

A researcher compares Embryo X (equal division) and Embryo Y (unequal division — small apical cell + large basal cell).

i) Which embryo is likely to develop into a normal dicot embryo?

Answer: Embryo Y

In dicots, the first division of the zygote is characteristically unequal, producing a small apical cell and a large basal cell. Embryo Y matches this pattern.

ii) What is the fate of:

(a) Apical cell: Undergoes further divisions to form the embryo proper (plumule, radicle, and cotyledons).

(b) Basal cell: Undergoes transverse divisions to form the suspensor, which anchors the embryo and pushes it into the endosperm for nutrition.

Question 14 — Immune Response

i) Study the antibody level graph:

(a) What response types do peaks A and B represent?

• Peak A = Primary Immune Response — lower intensity, longer lag period; occurs on first encounter with the antigen.
• Peak B = Secondary (Anamnestic) Immune Response — highly intensified and faster; occurs due to immunological memory cells.

(b) Name the antibody at the highest level in the 7th week.

Answer: IgG

While IgM appears first in the primary response, IgG is the predominant antibody produced during the secondary response and maintains high levels for a longer duration.

ii) OR — Draw a flowchart to outline the erythrocytic cycle of Plasmodium in humans.

Merozoites released from liver cells
        |
        v
Merozoites infect Red Blood Cells (RBCs)
        |
        v
Trophozoite stage (signet ring stage)
        |
        v
Erythrocytic Schizont (matures inside RBC)
        |
        v
Schizont ruptures:
  - Releases new Merozoites
  - Releases Haemozoin (toxin causing recurring fever and chills)
        |
   +----+----------------------------+
   |                                 |
   v                                 v
Infects new RBCs              Differentiates into Gametocytes
(cycle repeats)         (taken up by Anopheles mosquito)

Question 15 — Spermiogenesis

Structure A: Spermatid | Structure B: Spermatozoon (Sperm)

i) Identify the process which converts Structure A (Spermatid) to Structure B (Spermatozoon).

Answer: Spermiogenesis

ii) State any two modifications observable in Structure B:

1. Formation of Tail (Flagellum): Development of a long flagellum from the centriole, providing motility.
2. Acrosome Formation: The Golgi apparatus modifies into a cap-like acrosome at the head, containing hydrolytic enzymes (e.g., hyaluronidase) required for fertilisation.
3. Condensation of Nucleus: The nucleus becomes compact and streamlined for efficient swimming.
4. Mitochondrial Arrangement: Mitochondria spiral around the middle piece to form the mitochondrial sheath, providing ATP energy for movement.

(Any two of the above are acceptable)

Section D — 15 Marks

Question 16 — Locust Population Growth Curve

i) Construct a growth curve. What type of curve is obtained?

By plotting Number of Locusts (Y-axis) against Day (X-axis):

• The population rises slowly at first (Lag phase)
• Then increases rapidly (Log / Exponential phase)
• Then crashes sharply after Day 30 (resource exhaustion / environmental resistance)

Type of curve: J-shaped growth curve (Exponential growth)

ii) Write a mathematical expression for the growth curve obtained.

dN/dt = rN

Where:
  N = Population size
  r = Intrinsic rate of natural increase (biotic potential)
  t = Time

Question 17 — Transcription Unit

i) Identify the parts marked 'E' and 'F'.

• E = Promoter — the region where RNA polymerase binds to initiate transcription.
• F = hnRNA (Heterogeneous nuclear RNA) / Primary transcript — the newly synthesised RNA strand before processing.

ii) Explain the post-transcriptional modifications that occur in part 'F' (hnRNA).

In eukaryotes, hnRNA undergoes three major post-transcriptional modifications to become functional mRNA:

1. Splicing: Removal of non-coding sequences (introns) and joining of coding sequences (exons). Catalysed by spliceosomes.

2. 5' Capping: Addition of an unusual nucleotide — methyl guanosine triphosphate (7-methyl guanosine) — at the 5'-end. Protects mRNA from degradation and aids ribosome recognition.

3. 3' Poly-A Tailing: Addition of a stretch of adenylate (poly-A) residues at the 3'-end. Enhances mRNA stability and assists in export from the nucleus.

Question 18 — Artificial Hybridisation / Oogenesis

i) Outline three major steps in the process of artificial hybridisation.

1. Emasculation: Removal of anthers from the bisexual flower bud of the female parent plant before they dehisce (shed pollen). Not required for unisexual (pistillate) flowers.

2. Bagging: Covering the emasculated (or pistillate) flower with a butter-paper bag to prevent contamination of the stigma with unwanted pollen.

3. Rebagging: After the stigma matures, desired pollen grains are dusted onto it, and the flower is bagged again to prevent further contamination and to allow the fruit to develop.

ii) Discuss the role of pollen–pistil interaction in ensuring successful hybrid formation.

Pollen–pistil interaction acts as a molecular checkpoint — a biochemical dialogue between the pollen and the pistil:

• The pistil recognises whether incoming pollen is compatible (correct species/variety) or incompatible (wrong type/self-pollen in self-incompatible plants).
• Only compatible pollen is permitted to germinate and grow a pollen tube through the style to reach the ovary.
• This ensures fertilisation occurs exclusively with the intended male gamete, guaranteeing true and successful hybrid formation.

iii) OR — Draw a flowchart to represent the process of oogenesis.

Oogonia (2n)   [Fetal life]
      |
      | Mitosis + Differentiation
      v
Primary Oocyte (2n)
[Enters Meiosis I; arrested at Prophase I until puberty]
      |
      | Puberty: Completion of Meiosis I (unequal division)
      v
Secondary Oocyte (n)  +  First Polar Body (n)
[Arrested at Metaphase II; released at Ovulation]
      |
      | Fertilisation by Sperm: Completion of Meiosis II
      v
Ootid / Ovum (n)  +  Second Polar Body (n)

iv) Differentiate between oogenesis and spermatogenesis: