ISC Class 12 Biology Sample Paper 2025 with Solutions

SECTION A - 20 MARKS

Question 1
Answer the following questions briefly.

(i) Sapna has been diagnosed with an infection of the reproductive tract caused by bacteria. She experiences burning sensation during urination, pain around her genitalia and observes pus containing discharge. Her doctor tells her that this infection has an incubation period of two to five days but can be cured.
What is the biological name of the causative agent of the disease Sapna is suffering from? [1 Mark]

Answer:
Neisseria gonorrhoeae

Teacher's Note:
a) The disease described is gonorrhea, which is a sexually transmitted bacterial infection.
b) Students must remember the exact binomial nomenclature rules when writing scientific names.

(ii) The number of chromosomes in the leaf cells of a male plant is 40 and in the stem cells of a female plant is 60. If they are artificially hybridised, what will be the number of chromosomes in the endosperm? [1 Mark]

Answer:
80

Teacher's Note:
a) Endosperm in angiosperms is triploid (\(3n\)). It is formed by triple fusion involving one male gamete (\(n\)) and two polar nuclei (\(2n\)).
b) The male plant leaf cell has \(2n = 40\), so its gamete has \(n = 20\). The female plant stem cell has \(2n = 60\), so each polar nucleus has \(n = 30\), making the endosperm \(n + 2n = 20 + 60 = 80\).

(iii) How many cycles of PCR are required to produce 250 molecules of DNA, starting with a single parental strand? [1 Mark]

Answer:
8 cycles

Teacher's Note:
a) The number of DNA molecules produced after \(n\) cycles of PCR is given by \(2^n\).
b) Since \(2^7 = 128\) and \(2^8 = 256\), 8 cycles are required to produce at least 250 molecules.

(iv) Which is the most commonly prescribed non-steroidal oral contraceptive pill in India? [1 Mark]

Answer:
Saheli

Teacher's Note:
a) Saheli is a non-steroidal oral contraceptive pill containing centchroman, developed by CDRI Lucknow.
b) It is taken once a week and has very few side effects.

(v) Tall pea plants having green pods were crossed with dwarf pea plants having yellow pods. Out of 80 plants, how many are likely to be tall plants in the F₂ generation? [1 Mark]

Answer:
60 plants

Teacher's Note:
a) Height is inherited independently of pod color following Mendelian dihybrid cross principles.
b) In the F₂ generation, the ratio of tall to dwarf plants is 3:1. Therefore, \(\frac{3}{4} \times 80 = 60\) plants will be tall.

(vi) Write a 6-nucleotide long palindromic sequence on a double stranded DNA, which was reverse transcribed by the following nucleotide sequence on a retroviral RNA.
3’-GUA- - - 5’ [1 Mark]

Answer:
5’- CATATG -3’
3’- GTATAC -5’

Teacher's Note:
a) A palindromic DNA sequence reads the same on both strands when orientation is read in the same direction (5' to 3').
b) CATATG is a well-known 6-base palindromic recognition sequence for the restriction enzyme NdeI.

(vii) In the given figure of mammalian spermatozoon, “X” is an organelle. Identify the organelle. [1 Mark]

[Figure: Diagram of a mammalian sperm showing head, middle piece, and tail, with label X pointing to the mitochondrion coiled around the axial filament in the middle piece.]

Answer:
Mitochondrion / Mitochondria

Teacher's Note:
a) Mitochondria provide energy (ATP) for the movement of the tail that is essential for sperm motility.
b) They form a spiral sheath in the middle piece of the spermatozoon.

(viii) Observe the relation between the first two words and then fill in the fourth word.
Histamine: Mast cells :: Antibody: ____________ [1 Mark]

Answer:
Plasma cells / B-Lymphocytes

Teacher's Note:
a) Mast cells secrete histamine during inflammatory and allergic reactions.
b) Plasma cells are differentiated B-lymphocytes that synthesize and secrete antibodies.

(ix) Which of the following sequences of mRNA will NOT translate completely? [1 Mark]
(a) 5' - AUG UUC AGC UCG UGA - 3'
(b) 5'- AUG AAC UAA CCA CUC - 3'
(c) 5' - AUG UUA CUC GCG UAA - 3'
(d) 5' - AUG CCA UAC GAC UAG - 3'

Answer: (b) 5'- AUG AAC UAA CCA CUC - 3'

Translation stops prematurely at the internal stop codon UAA, preventing complete translation.

Teacher's Note:
a) UAA, UAG, and UGA are stop codons that terminate translation.
b) Option (b) contains UAA right after the second codon, halting translation before the entire sequence is read.

(x) Which one of the following choices reflects the correct number of chromosomes in the zygote and in the second polar body of Drosophila melanogaster? [1 Mark]

(a) I
(b) II
(c) III
(d) IV

Answer: (c) III

Zygote has \(2n = 8\) and second polar body (or meiotic product/haploid cell) has \(n = 4\).

Teacher's Note:
a) Drosophila melanogaster has a diploid chromosome number (\(2n\)) of 8.
b) The zygote is diploid (\(2n = 8\)) and polar bodies formed during meiosis are haploid (\(n = 4\)).

(xi) Given below are two statements marked Assertion and Reason. Read both the statements carefully and choose the correct option.
Assertion: Earthworm is called a detritivore.
Reason: It breaks down the water-soluble inorganic nutrients, which percolate down into the soil. [1 Mark]
(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: (c) Assertion is true and Reason is false.

Earthworms break down detritus into smaller particles (fragmentation), not water-soluble inorganic nutrients (which is catabolism).

Teacher's Note:
a) Earthworms feed on detritus and are termed detritivores.
b) The process of water-soluble inorganic nutrients going down into the soil is leaching, carried out by bacterial and fungal enzymes during catabolism.

(xii) Given below are two statements marked Assertion and Reason. Read both the statements carefully and choose the correct option.
Assertion: E. coli was transformed by inserting the foreign DNA by using PvuI in pBR322.
Reason: It was done to make the transformed cells survive in the culture medium containing ampicillin. [1 Mark]
(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: (c) Assertion is true and Reason is false.

Insertion of foreign DNA at the PvuI site inactivates the ampicillin resistance gene, making the transformed cells sensitive to ampicillin.

Teacher's Note:
a) PvuI is located within the amp^R gene of pBR322.
b) Insertion of foreign DNA causes insertional inactivation of ampicillin resistance, so transformed cells cannot survive on ampicillin plates.

(xiii) What is the advantage of growing apomictic seeds in crop improvement? [1 Mark]

Answer:
It helps in retaining parental characters without segregation across generations.

Teacher's Note:
a) Apomixis is asexual reproduction mimicking sexual reproduction.
b) Hybrid seeds produced through apomixis do not show segregation of traits, allowing farmers to use hybrid seeds year after year without loss of vigour.

(xiv) A farmer has two fields in which he wants to grow a cereal crop and a legume. He cannot afford to spend a lot of money on chemical fertilisers to increase the fertility of the soil.
Suggest a way to help him resolve the issue. [1 Mark]

Answer:
By recommending him to use cyanobacteria and / or Nitrogen-fixing bacteria (or crop rotation with legumes).

Teacher's Note:
a) Biological nitrogen fixation by Rhizobium in legumes enriches soil nitrogen naturally.
b) Biofertilisers like Anabaena, Nostoc, and Azotobacter reduce the need for costly chemical fertilisers.

(xv) Answer the following questions. [2 Marks]
(a) Expand the abbreviation MALT. [1 Mark]

Answer:
Mucosal Associated Lymphoid Tissues

Teacher's Note:
a) MALT constitutes about 50 percent of the lymphoid tissue in the human body.
b) It is located within the lining of major tracts such as respiratory, digestive, and urogenital tracts.

(b) David is a molecular biologist. He uses a technique to know the amino acids occupy the first and the last positions in the Chain – A and Chain – B of insulin. Name the scientist whose contribution made it possible for David to use this technique. [1 Mark]

Answer:
Sanger

Teacher's Note:
a) Frederick Sanger developed the protein sequencing technique (Sanger method for amino acid sequencing).
b) He determined the complete amino acid sequence of insulin.

(xvi) In a flower, the megaspore mother cell formed four megaspores without undergoing meiosis. One of the four megaspores developed into the embryo sac. What would be the ploidy level of the antipodal cells in this embryo sac? [1 Mark]

Answer:
Diploid (2n)

Teacher's Note:
a) Since meiosis did not occur, the megaspore mother cell and consequently the megaspore remain diploid (\(2n\)).
b) Development of the embryo sac from this diploid megaspore (apospory/diplospory) results in all cells of the embryo sac being diploid.

(xvii) The diagram given below represents a specific stage of human embryonic development. Identify the stage. [1 Mark]

[Figure: Diagram of a solid ball of 16 cells (blastomeres) representing an early human embryo.]

Answer:
Morula

Teacher's Note:
a) The embryo with 8 to 16 blastomeres is called a morula.
b) It continues to divide and transforms into blastocyst as it moves into the uterus.

(xviii) Give a reason for each of the following. [2 Marks]
(a) The diagram given below represents vector pBr322 which was modified by a research scholar Dr. Rayon, but the vector was rejected by his guide. [1 Mark]

[Figure: Diagram of plasmid vector pBR322 showing multiple restriction sites clustered within the marker genes or origin, notably showing more than one restriction site for the same restriction enzyme.]

Answer:
More than one restriction site for the same restriction enzyme.

Teacher's Note:
a) Vectors should ideally have single recognition sites for commonly used restriction enzymes.
b) Multiple sites for the same enzyme would result in several fragments, complicating gene cloning.

(b) The continuous inbreeding of crops may lead to reduced fertility and productivity. However, the self-pollinated crops do not show the ill-effects of inbreeding. [1 Mark]

Answer:
Because the weaker alleles become homozygous and exhibit their harmful effects. So, such plants die and get eliminated out of population. In this way, the population becomes free of such genes.

Teacher's Note:
a) Continuous inbreeding in normally cross-pollinated crops leads to inbreeding depression due to the accumulation of homozygous recessive deleterious alleles.
b) Naturally self-pollinated crops have already purged most harmful recessive alleles through generations of natural and artificial selection.

SECTION B - 14 MARKS

Question 2 [2 Marks]

(i) During embryogenesis in dicots, the zygote divides into a basal cell and a terminal cell. The basal cell divides repeatedly to produce a structure called suspensor. Carefully observe the image given below and describe the function of cell – A and cell – B of the suspensor. [2 Marks]

[Figure: Diagram of a dicot embryo development showing a filamentous suspensor with terminal cell A (haustorial cell at the top) and basal cell B (hypophysis at the junction with the globular embryo).]

Answer:
Cell - A: The haustorial cell absorbs and transfers nutrients from the endosperm to the proembryo.
Cell - B: Hypophysis forms the radicle (embryonic root).

Teacher's Note:
a) The suspensor pushes the proembryo deep into the endosperm for nourishment.
b) The terminal cell of the suspensor adjacent to the embryo is the hypophysis, which gives rise to the root tip and root cap.

OR

(ii) Given below are the features of Seed A and Seed B. Study them carefully and answer the question that follows.
SEED A: Papery endosperm, swollen cotyledon, perisperm develops from nucellus, suspensor/haustoria substitute endosperm
SEED B: Swollen endosperm, papery cotyledons or absent, perisperm does not develop from nucellus, suspensor/haustoria do not substitute endosperm.
Identify the type of Seed A and Seed B. Cite one example of each of Seed A and Seed B. [2 Marks]

Answer:
Seed A: Non-endospermic / Exalbuminous seed. Example: Pea / Bean / Gram.
Seed B: Endospermic / Albuminous seed. Example: Castor / Maize / Wheat.

Teacher's Note:
a) Non-endospermic seeds consume endosperm completely during embryo development, storing food in cotyledons.
b) Endospermic seeds retain endosperm in the mature seed to nourish the seedling during germination.

Question 3 [2 Marks]
Name the pollinating agent of a flowering plant with large and coloured flowers with sticky stigma. State one characteristic feature of the pollen grains produced in such flowers. [2 Marks]

Answer:
1. Pollinating agent: Insects (Entomophily).
2. Characteristic feature of pollen grains: Pollen grains are sticky due to a yellowish, sticky oily layer called pollen kit, or have spiny surfaces.

Teacher's Note:
a) Insect-pollinated flowers are large, colorful, fragrant, and rich in nectar to attract insect visitors.
b) The sticky pollen grains easily adhere to the bodies of visiting insects for cross-pollination.

Question 4 [2 Marks]
Read the passage given below carefully and answer the questions that follow.
The bacterium Bacillus thuringiensis serovar israelensis (Bti) is commercially prepared in various formulations such as liquid, water dispersible granules, powders, and pellets. It is used as a larvicide all over the world due to its ability to produce a toxic protein that primarily targets the larvae of mosquitoes.
In Sweden, Bti has been applied on a large scale in the form of commercially available granular formulation Vecto Bac G (Valent BioScience, USA). The applications have taken place in the lower Dalalven River Area to control mass outbreaks of the floodplain mosquito Aedes sticticus.
(Source: Environmental Evidence Journal, page 26, November 2023)
(i) Why does the toxic protein kill the insects but not the bacterium? [1 Mark]
(ii) Can the formulation Vecto Bac G be used to eliminate the species of mosquito referred to above? Why? [1 Mark]

Answer:
(i) The protein remains insoluble at low pH in the bacterial cell, but in the alkaline gut of insect larvae it becomes soluble and damages the epithelial cells.
(ii) No, because the biological toxin is species-specific and controls mosquito populations rather than totally eliminating the species from the ecosystem.

Teacher's Note:
a) Bt toxin exists as an inactive protoxin inside the bacterium, avoiding self-toxicity.
b) Biocontrol agents suppress pest populations below economic injury levels rather than causing complete extinction.

Question 5 [2 Marks]
What happens to an inferior competitor if:
(i) a superior competitor is present in the same environment? [1 Mark]
(ii) the superior competitor is removed from the environment? [1 Mark]

Answer:
(i) The inferior competitor will be eliminated as per Gause's Competitive Exclusion Principle.
(ii) The inferior competitor expands its territory and population density through a process called Competitive Release.

Teacher's Note:
a) Gause's principle states that two closely related species competing for the same resources cannot coexist indefinitely if resources are limiting.
b) Competitive release occurs when the dominant competitor is removed, allowing the weaker species to exploit resources freely.

Question 6 [2 Marks]
The diagram given below shows the early embryonic stages of fish, reptiles, birds and humans. Though adult reptiles, birds and mammals breathe air, their embryos possess gills. How does this fact support evolution? [2 Marks]

[Figure: Comparative embryology diagrams showing early developmental stages of fish, reptile, bird, and human embryos with labeled gill slits and tails.]

Answer:
1. The presence of gill slits in embryos of land vertebrates indicates that terrestrial vertebrates have evolved from aquatic ancestors.
2. This supports Ernst Haeckel's Biogenetic Law, summarized as "ontogeny recapitulates phylogeny".

Teacher's Note:
a) Embryological support for evolution was proposed by Ernst Haeckel, showing common ancestry.
b) Features absent in adults often appear during embryonic development due to evolutionary conservation.

Question 7 [2 Marks]
Name the technique used to detect the presence of HIV in the body of an individual. Justify the principle associated with this technique. [2 Marks]

Answer:
1. Technique: ELISA (Enzyme-Linked Immunosorbent Assay).
2. Principle: It is based on antigen-antibody interactions where the presence of specific viral antigens or anti-viral antibodies in the patient's serum is detected using enzyme-linked secondary antibodies.

Teacher's Note:
a) ELISA is widely used as a preliminary diagnostic screening test for HIV.
b) The enzyme linked to the antibody converts a colorless substrate into a colored product, indicating a positive test.

Question 8 [2 Marks]
Name any two factors responsible for the loss of biodiversity in a geographical region. [2 Marks]

Answer:
1. Habitat loss and fragmentation.
2. Alien species invasions (or Over-exploitation / Co-extinctions).

Teacher's Note:
a) These are part of the "Evil Quartet" causing rapid biodiversity decline globally.
b) Habitat destruction is recognized as the primary driver of extinction for plants and animals.

SECTION C - 21 MARKS

Question 9 [3 Marks]
(i) What are ZIFT and GIFT? [1½ Marks]
(ii) All STDs do not affect the genital organs. Justify the statement. [1½ Marks]

Answer:
(i) ZIFT (Zygote Intra-Fallopian Transfer): Transfer of a zygote or early embryo (upto 8 blastomeres) into the fallopian tube.
GIFT (Gamete Intra-Fallopian Transfer): Transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one.
(ii) Certain Sexually Transmitted Diseases (STDs) like AIDS and Hepatitis-B do not directly affect the genital organs because HIV attacks the immune system (helper T cells) and Hepatitis-B primarily infects and damages the liver.

Teacher's Note:
a) Assisted Reproductive Technologies (ART) like ZIFT and GIFT help infertile couples conceive.
b) While most STDs like gonorrhea and syphilis cause localized genital symptoms, blood-borne pathogens like HIV and HBV affect systemic organs.

Question 10 [3 Marks]
(i) The DNA molecules of the same size were extracted from E. coli and Plasmodium vivax. It was discovered that both the DNA molecules had one target site each for the restriction enzyme Hind II. After being digested with Hind II, the DNA fragments were subjected to gel electrophoresis.
With reference to the diagram given below, identify the lanes that represent the DNA fragments of E. coli and Plasmodium vivax respectively. Justify your answer with a reason for each. [1½ Marks]

[Figure: Gel electrophoresis plate showing Lane A with a single DNA band and Lane B with two DNA bands separated by size.]

Answer:
Lane A - DNA of E. coli
Lane B - DNA of Plasmodium vivax
Reason: E. coli possesses circular DNA, which when cut at one site yields a single linear fragment. Plasmodium vivax possesses linear DNA, which when cut at one target site yields two fragments.

Teacher's Note:
a) Restriction endonucleases cleave phosphodiester bonds at specific palindrome sequences.
b) Cutting a circular DNA molecule once produces 1 linear piece, while cutting a linear DNA molecule once produces 2 pieces.

OR

(ii) The diagram given below represents the image of the cloning vector pUC 18. The gene of interest is inserted and ligated within the gene lacZ. The recombinant DNA is introduced in the host bacterial cell. Explain the method that would help in selection of recombinant colonies from non-recombinant colonies. [1½ Marks]

[Figure: Diagram of plasmid cloning vector pUC18 showing ori, ampR, PvuI, SalI, and lacZ gene region.]

Answer:
1. Selection method: Blue-white selection.
2. Mechanism: The lacZ gene encodes for beta-galactosidase. Insertion of foreign DNA inside lacZ causes insertional inactivation of the enzyme.
3. Non-recombinant colonies produce active beta-galactosidase and turn blue in the presence of a chromogenic substrate, whereas recombinant colonies remain colorless (white).

Teacher's Note:
a) Insertional inactivation is a reliable technique for identifying recombinant clones.
b) White colonies represent successful recombinants containing the foreign DNA insert.

Question 11 [3 Marks]
Samson and Dorothy surveyed two islands, 'A' and 'B'. They recorded the relevant data in the following format in their research folder:

(i) Derive the species-richness for each island. [1½ Marks]
(ii) Which island exhibits greater biodiversity? Support your answer with a reason. [1½ Marks]

Answer:
(i) Using species-area relationship formula \(S = C A^Z\):
For Island A: \(S_A = 20 \times (45 \times 10^3)^1 = 900 \times 10^3 = 9,00,000\)
For Island B: \(S_B = 10 \times (12 \times 10^5)^1 = 120 \times 10^5 = 1,20,00,000\)
(ii) Island B exhibits greater biodiversity because its species richness (\(S_B = 1.2 \times 10^7\)) is significantly higher than that of Island A (\(S_A = 9 \times 10^5\)), supported by a larger area and favorable environmental conditions.

Teacher's Note:
a) Alexander von Humboldt observed that within a region, species richness increases with increasing explored area up to a limit.
b) The equation is logarithmic as \(\log S = \log C + Z \log A\).

Question 12 [3 Marks]
The black colour on the beak of finches dominates over the yellow colour. There are 210 individuals with the genotype DD, 245 individuals with the genotype Dd and 45 individuals with the genotype dd. Deduce the frequency of individuals with dominant, heterozygous, and recessive traits. [3 Marks]

Answer:
Total individuals = \(210 + 245 + 45 = 500\)
1. Frequency of homozygous dominant genotype (\(DD\), \(p^2\)) = \(210 \div 500 = 0.42\)
2. Frequency of heterozygous genotype (\(Dd\), \(2pq\)) = \(245 \div 500 = 0.49\)
3. Frequency of homozygous recessive genotype (\(dd\), \(q^2\)) = \(45 \div 500 = 0.09\)
*(Note: Official key lists \(p^2 = 0.49\), \(2pq = 0.42\), \(q^2 = 0.09\) based on gene frequencies \(p = 0.7, q = 0.3\).)*

Teacher's Note:
a) Hardy-Weinberg equilibrium relates allele frequencies to genotype frequencies as \(p^2 + 2pq + q^2 = 1\).
b) Always calculate the total population first before determining individual genotype fractions.

Question 13 [3 Marks]
Robert was suffering from chronic renal failure. At the doctors' recommendation of transplantation of kidney received from a healthy donor, Robert underwent kidney transplant. However, after two weeks, the transplanted kidney was rejected by the immune system of Robert.
(i) Identify and define the type of immune response that is responsible for the rejection of the grafted organ. [1½ Marks]
(ii) Suggest a clinical method by which the rejection of the transplanted organ can be prevented. [1½ Marks]

Answer:
(i) Cell-Mediated Immunity (CMI): It is the type of immune response mediated by T-lymphocytes (T-cells) that recognizes tissue grafts as foreign and destroys them.
(ii) Administration of immunosuppressant drugs (such as cyclosporin) to suppress the patient's immune response.

Teacher's Note:
a) The body's immune system can distinguish between self and non-self, leading to graft rejection.
b) Patients receiving organ transplants must take immunosuppressants for the rest of their lives to prevent rejection.

Question 14 [3 Marks]
The red panda has been listed as an endangered species on the IUCN Red List since 2015. Regional captive breeding programmes have been established in the zoos around the world to protect the red panda from extinction.
(i) Classify the biodiversity conservation programme referred to above. [1 Mark]
(ii) Mention any other two methods of conservation which belong to the same category. [2 Marks]

Answer:
(i) Ex-situ conservation.
(ii) Cryopreservation and tissue culture (or gene banks / botanical gardens / zoological parks).

Teacher's Note:
a) Ex-situ conservation involves protecting endangered species outside their natural habitats in specialized facilities.
b) In-situ conservation protects species within their natural habitats (e.g., national parks, biosphere reserves).

Question 15 [3 Marks]
(i) Draw a well-labelled diagram of tRNA. [3 Marks]

[Figure: Cloverleaf model of tRNA showing amino acid attachment site at 3' end, TPsiC loop, D loop, anticodon loop with anticodon, and variable extra arm.]

Answer:
1. Proper cloverleaf shape with 5'-arm at a lower level than 3'-arm.
2. Three characteristic loops: D-loop, Anticodon loop, and TpsiC loop with proper orientation with reference to 3' and 5' ends.

Teacher's Note:
a) tRNA acts as an adapter molecule that reads genetic code and brings specific amino acids during translation.
b) The 3' end carries the CCA-amino acid attachment site.

OR

(ii) Draw a well-labelled diagram of nucleosome. [3 Marks]

[Figure: Diagram of a nucleosome core particle showing histone octamer (H2A, H2B, H3, H4), core DNA wrapped around it, linker DNA, and H1 histone protein sealing the loop.]

Answer:
1. Depiction of histone octamer core containing two molecules each of H2A, H2B, H3, and H4.
2. Wrapping of negatively charged DNA around the positively charged histone core.
3. Labeling of linker DNA and H1 histone protein.

Teacher's Note:
a) Packaging of DNA into nucleosomes allows long DNA molecules to fit inside eukaryotic nuclei.
b) A typical nucleosome contains 200 base pairs of DNA helix.

SECTION D - 15 MARKS

Question 16 [5 Marks]
(i) Answer the following questions.
(a) Why are the biocontrol agents preferred over the chemical pesticides? [2 Marks]
(b) Explain the role of any two biocontrol agents by mentioning their target pests. [3 Marks]

Answer:
(a) Biocontrol agents do not pollute the environment, are eco-friendly, and are highly specific in nature, thus sparing non-target beneficial organisms.
(b) 1. Baculoviruses (Nucleopolyhedrovirus): Specifically attack insect pests and arthropods.
2. Bacillus thuringiensis: Produces toxic proteins that specifically destroy cotton bollworms and corn borers.

Teacher's Note:
a) Biological control relies on natural predation rather than toxic chemical residues.
b) Specificity ensures that only harmful pests are eradicated without disturbing local biodiversity.

OR

(ii) Answer the following questions.
(a) Why are some molecules called bioactive molecules? [2 Marks]
(b) Mention the respective source and state the specific use of any two bioactive molecules. [3 Marks]

Answer:
(a) They are called bioactive molecules because they function by modulating metabolic pathways in living organisms.
(b) 1. Streptokinase: Source - Streptococcus; Use - Intravascular clot buster for patients with myocardial infarction.
2. Cyclosporin A: Source - Trichoderma polysporum; Use - Immunosuppressive agent used in organ transplant patients.

Teacher's Note:
a) Bioactive molecules are often derived from microbial fermentation.
b) They play crucial roles in modern medicine as therapeutic drugs.

Question 17 [5 Marks]
In a forest ecosystem, a large population of insect feeds upon a banyan tree. Several small birds feed upon these insects. The small birds are fed upon by big-sized birds.
(i) With respect to this ecosystem, draw a pyramid each of biomass and a pyramid of number. [3 Marks]
(ii) If 20,000 Kcal energy is available at the level of insects, calculate the amount of energy available at the level of big-sized birds. [2 Marks]

Answer:
(i) Pyramid of Number: Inverted pyramid (Tree -> Insects -> Small birds -> Big birds).
Pyramid of Biomass: Upright pyramid in terms of producers having the largest biomass, tapering towards top carnivores (or inverted tree biomass if drawn strictly as tree support, but standard forest key shows upright biomass pyramid).
(ii) Energy calculation based on 10% law:
Energy at insect level = 20,000 Kcal.
Energy at small bird level = \(10\% \text{ of } 20,000 = 2,000 \text{ Kcal}\).
Energy at big-sized bird level = \(10\% \text{ of } 2,000 = 200 \text{ Kcal}\).

Teacher's Note:
a) Ecological pyramids represent trophic structure and function.
b) According to Lindeman's 10% law, only 10% of energy is transferred from one trophic level to the next.

Question 18 [5 Marks]
(i) Consider the following information and answer the question that follows
- Reshma's mother is normal, but her father is suffering from PKU.
- Reshma is suffering from PKU. She has two younger brothers who are identical twins; both are suffering from PKU. Reshma's two elder sisters are normal.
- Reshma marries Robert, who is normal. Reshma gives birth to a son named Jason who is normal. After some time, Reshma and Robert have two more children: a daughter with symptoms of PKU and a son without any symptoms of PKU.
Make a single pedigree chart, using the above information, to show the pattern of inheritance of phenylketonuria (PKU) in the family of Reshma. [3 Marks]
(ii) Enumerate the cause and the characteristic symptom of PKU. [2 Marks]

Answer:
(i) Pedigree chart showing affected father married to normal mother, producing affected daughter Reshma, normal sisters, and affected twin brothers. Reshma (affected) married to Robert (normal) having children: Jason (normal), an affected daughter, and a normal son.
(ii) Cause: Deficiency of the enzyme phenylalanine hydroxylase due to an autosomal recessive gene mutation.
Symptom: Retarded mental growth (mental retardation) and light skin pigmentation.

Teacher's Note:
a) Phenylketonuria (PKU) is an inborn error of metabolism inherited as an autosomal recessive trait.
b) The absence of phenylalanine hydroxylase leads to the accumulation of phenylalanine in the brain, causing severe intellectual disability.