ISC Class 12 Biology Sample Paper 2024 with Solutions

SECTION A - 20 MARKS

Question 1
Answer the following questions briefly.

(i) Birds build their nests in trees. Identify the type of ecological relationship between the birds and the trees. [1 Mark]

Answer:
Commensalism (where the birds are benefitted by getting shelter/nesting site, while the trees are neither benefitted nor harmed).

Teacher's Note:
a) Commensalism is a positive species interaction denoted as (+, 0).
b) Students must be precise with ecological terms and avoid writing mutualism since the tree receives no direct benefit.

(ii) A doctor examines the symptoms of a patient who has a high fever with chills. What disease could this patient be suffering from? [1 Mark]

Answer:
Malaria.

Teacher's Note:
a) High recurring fever accompanied by chills and shivering is the classic symptom of malaria caused by the protozoan parasite Plasmodium.
b) Mentioning the specific causative organism or disease name is essential to secure the mark.

(iii) The genome size of an organism is around \( 10^{9} \) base pairs. Calculate the length of its genome in metres. [1 Mark]

Answer:
Length = \( 10^{9} \times 0.34 \times 10^{-9}\text{ m} = 0.34\text{ metres} \).

Teacher's Note:
a) The distance between two consecutive base pairs in DNA is \( 0.34\text{ nm} \) or \( 0.34 \times 10^{-9}\text{ m} \).
b) Always show the formula and proper unit conversion to avoid calculation errors.

(iv) What is the probability of having a male child if the father's sperm carries an X chromosome? [1 Mark]

Answer:
0% (or Zero).

Teacher's Note:
a) An X-bearing sperm fertilizing an egg (which always carries an X chromosome) results in an XX zygote, which develops into a female child.
b) Students often confuse this with the overall probability of a male child at conception (50%), so read the specific condition carefully.

(v) In the year 2021, the population of fish in a lake was 500. After one year, a biologist found that the population had an average natality of 120, average mortality of 65, immigration was 25 and emigration was 30. Calculate the total number of fish that were present in the lake in 2022. [1 Mark]

Answer:
\( N_{t+1} = N_{t} + [(B + I) - (D + E)] \)
\( = 500 + [(120 + 25) - (65 + 30)] \)
\( = 500 + [145 - 95] \)
\( = 500 + 50 = 550 \) fish.

Teacher's Note:
a) Population density at time \( t+1 \) is calculated by adding natality and immigration to the initial population and subtracting mortality and emigration.
b) Ensure proper grouping of births/immigration versus deaths/emigration to prevent sign mistakes.

(vi) The Gross Primary Productivity (GPP) of the Amazon rainforest is approximately \( 3000\text{ g C/m}^{2}\text{/yr} \) and the respiration losses (R) of \( 1800\text{ g C/m}^{2}\text{/yr} \). Calculate the Net Primary Productivity (NPP) of this ecosystem. [1 Mark]

Answer:
\( \text{NPP} = \text{GPP} - R = 3000 - 1800 = 1200\text{ g C/m}^{2}\text{/yr} \).

Teacher's Note:
a) Net Primary Productivity is the rate of organic matter stored by producers after respiration losses.
b) Always include the correct units in the final numerical answer.

(vii) Identify if the given set of structures can be classified as homologous or analogous structures.
[Figure 1: Shows wings of penguin and fins of fish, and flippers of whale and dolphin] [1 Mark]

Answer:
Analogous structures.

Teacher's Note:
a) Analogous structures have similar functions and superficial appearance due to convergent evolution, but different anatomical origins and developmental structures.
b) Penguin wings (modified forelimbs for swimming) and fish fins (dermal fin rays) perform the same function of locomotion in water but have entirely different internal skeletal structures.

(viii) Suggest an effective molecular technique that can be used to reduce the expression of defective genes in the nematode parasite. [1 Mark]

Answer:
RNA interference (RNAi).

Teacher's Note:
a) RNAi involves silencing a specific mRNA due to a complementary dsRNA molecule that binds to and prevents its translation.
b) This technique is widely used in transgenic plants (such as tobacco) to protect them from Meloidegyne incognita infestation.

(ix) Which one of the following is an example of naturally acquired active immunity? [1 Mark]
(a) Recovering from chickenpox
(b) Newborns gaining antibodies from breast milk
(c) Getting vaccinated against the COVID-19 virus
(d) Taking antibiotics against a urinary tract infection

Answer: (a) Recovering from chickenpox

Naturally acquired active immunity develops when a person is exposed to a live pathogen, contracts the disease, and subsequently develops primary and secondary immune responses with memory cells.

Teacher's Note:
a) Active immunity involves the production of antibodies by the host's own immune system.
b) Breast milk provides passive immunity, vaccination is artificially acquired active immunity, and antibiotics are medications rather than immunizing agents.

(x) If oogenesis occurs conventionally in a human, which one of the following represents the correct ploidy levels of the primary and secondary oocytes? [1 Mark]

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

Answer: (b) II

Primary oocytes are formed by mitosis/differentiation from oogonia and are diploid (\( 2n \)), whereas secondary oocytes are products of the first meiotic division and are haploid (\( n \)).

Teacher's Note:
a) Primary oocytes undergo the first meiotic division just prior to ovulation to produce a haploid secondary oocyte and a first polar body.
b) Always remember that germline cells before the first meiotic division are diploid.

(xi) Assertion: The external application of insecticides is negligible for Bt crops.
Reason: Bt crops contain a gene that produces toxins harmful to certain insects.
Which one of the following is correct? [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: (a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.

Bt crops express crystal proteins (Cry proteins) from Bacillus thuringiensis which act as built-in bio-pesticides, drastically reducing the need for chemical insecticides.

Teacher's Note:
a) The insecticidal protein exists as inactive protoxin in the bacterium but gets converted into an active form in the alkaline gut of the insect.
b) The reason directly accounts for why external insecticides become unnecessary.

(xii) Assertion: The spent slurry after biogas production is used as a fertiliser for soils that are nitrogen deficient.
Reason: Only nitrogen from the slurry is utilised in the production of biogas.
Which one of the following is correct? [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.

Methanogenic bacteria convert carbon and hydrogen components into methane gas during anaerobic digestion, leaving behind a nutrient-rich slurry containing nitrogen and phosphorus that serves as an excellent organic fertilizer.

Teacher's Note:
a) The reason is false because nitrogen is not selectively utilized or removed during biogas generation; rather, carbon is primarily evolved as methane and carbon dioxide.
b) Students should carefully evaluate each statement independently before establishing cause-and-effect links.

(xiii) A student aims to insert a foreign gene into the plasmid shown below for an experiment.
[Figure 2: Shows plasmid pUC18/19 with lacZ, MCS, rep, and AmpR genes]
Suggest a method that can help the student in selecting the cells that contain this plasmid. [1 Mark]

Answer:
Antibiotic resistance selection (using Ampicillin) combined with blue-white screening (insertional inactivation of the \( \text{lacZ} \) gene).

Teacher's Note:
a) Cells containing the plasmid acquire ampicillin resistance due to the \( \text{Amp}^{\text{R}} \) gene.
b) Insertion of a foreign DNA into the Multiple Cloning Site (MCS) located within the \( \text{lacZ} \) gene causes insertional inactivation, resulting in white colonies instead of blue colonies.

(xiv) A biotechnological firm aims to manufacture certain clotting factors to treat Haemophilia in patients. Mention any one aspect to be considered while choosing the transgenic animal to produce these clotting factors. [1 Mark]

Answer:
The animal should be able to express the foreign gene specifically in mammary glands (so that the protein is secreted in milk for easy extraction), or have a high physiological compatibility for human protein folding and post-translational modifications.

Teacher's Note:
a) Transgenic animals like cows, sheep, or goats are often engineered to secrete therapeutic human proteins in their milk.
b) Biocompatibility and ease of purification from bodily fluids are crucial economic and biological criteria.

(xv) A group of virologists aim to inhibit the activity of reverse transcriptase enzymes as part of their effort to find a solution to viral infections. Which aspect of the virus's function would be impacted by inhibiting the activity of reverse transcriptase? [1 Mark]

Answer:
Transcription of viral RNA into complementary DNA (cDNA).

Teacher's Note:
a) Reverse transcriptase is an RNA-dependent DNA polymerase characteristic of retroviruses (such as HIV) required for integrating their genetic material into the host genome.
b) Inhibiting this enzyme halts the replication cycle of the retrovirus.

(xvi) Answer the following questions: [2 Marks]
(a) In recombinant DNA technology, biologists use the enzyme restriction endonuclease. Name the scientist who discovered this enzyme. [1 Mark]
(b) Expand the abbreviation MMR. [1 Mark]

Answer:
(a) Werner Arber (along with Hamilton Smith and Daniel Nathans later purifying/characterizing it).
(b) Mumps, Measles and Rubella (or Macromolecular Resonance / Mismatch Repair depending on context, but standard vaccine context is Mumps, Measles and Rubella).

Teacher's Note:
a) Werner Arber discovered restriction enzymes while studying bacteriophage restriction in E. coli.
b) MMR vaccine protects against three viral infections: Mumps, Measles, and Rubella.

(xvii) The figure given below shows a stage in the formation of pollen grain.
[Figure 3: Shows pollen grain with a large vegetative cell (Cell 1) and a small spindle-shaped generative cell (Cell 2) floating in the cytoplasm]
Name the cell that undergoes further division to make gametes. [1 Mark]

Answer:
Generative cell (Cell 2).

Teacher's Note:
a) The asymmetric mitotic division of the microspore produces a large vegetative cell and a small generative cell.
b) The generative cell divides mitotically to give rise to two male gametes.

(xviii) Give a reason for each of the following: [2 Marks]
(a) For a breastfeeding mother, the chances of conception are very low. [1 Mark]
(b) Organ transplantation patients are given the drug cyclosporine. [1 Mark]

Answer:
(a) Lactational amenorrhea (absence of menstruation) occurs due to high levels of prolactin during intense lactation, which suppresses gonadotropin-releasing hormone (GnRH) and prevents ovulation.
(b) Cyclosporine acts as an effective immunosuppressant that inhibits the activation of T-lymphocytes, thereby preventing graft rejection.

Teacher's Note:
a) Lactational amenorrhea is a natural method of birth control effective generally up to six months post-parturition.
b) Immunosuppressants are mandatory in organ transplants to overcome cell-mediated immune responses against foreign tissues.

SECTION B - 14 MARKS

Question 2 [2 Marks]
A couple does not wish to have more children.
Suggest and briefly explain any two methods of family planning that are highly effective and have low chances of failure, for the couple to consider.

Answer:
1. Surgical Methods (Sterilization): Procedures like vasectomy in males (blocking vas deferens) or tubectomy in females (blocking Fallopian tubes) are permanent, highly effective methods with near-zero failure rates.
2. Intrauterine Devices (IUDs): Devices like Copper-T (CuT) or hormone-releasing IUDs inserted by doctors in the uterus increase phagocytosis of sperm, suppress sperm motility, and prevent implantation.

Teacher's Note:
a) Permanent sterilization methods are ideal for couples who do not want any more children.
b) IUDs are among the most popular and effective reversible contraceptive methods in India.

Question 3 [2 Marks]
In a population of 5000 individuals, 1800 do not have freckles on their faces (ff), while the remaining individuals have freckles (F).
Assuming that the population is in Hardy-Weinberg equilibrium for the presence of freckles, calculate the expected frequencies for the following genotypes:
(i) FF genotype [1 Mark]
(ii) ff genotype [1 Mark]

Answer:
Total population (\( N \)) = \( 5000 \)
Number of homozygous recessive individuals (\( ff \) or \( q^{2} \)) = \( 1800 \)
Frequency of \( ff \) genotype (\( q^{2} \)) = \( \frac{1800}{5000} = 0.36 \)
(i) Frequency of \( FF \) genotype (\( p^{2} \)):
\( q = \sqrt{0.36} = 0.6 \)
\( p = 1 - q = 1 - 0.6 = 0.4 \)
Frequency of \( FF \) (\( p^{2} \)) = \( (0.4)^{2} = 0.16 \)
(ii) Frequency of \( ff \) genotype = \( 0.36 \).

Teacher's Note:
a) According to the Hardy-Weinberg principle, allele frequencies (\( p \) and \( q \)) and genotype frequencies (\( p^{2}, 2pq, q^{2} \)) remain constant from generation to generation.
b) Always start by finding the frequency of the homozygous recessive phenotype (\( q^{2} \)) from the given numbers.

Question 4 [2 Marks]
Shown below is the karyotype of an individual.
[Figure 4: Shows human karyotype with 47 chromosomes, specifically an extra chromosome at pair 21 (Trisomy 21)]
(i) State one characteristic reproductive feature and one physical attribute of such an individual. [1 Mark]
(ii) What is the category of such disorders called? Which abnormality during cell division causes such disorders? [1 Mark]

Answer:
(i) Characteristic reproductive feature: Individuals are generally sterile (reduced fertility).
Physical attribute: Broad palm with single palmar crease, furrowed tongue, flat back of head, and congenital heart disease.
(ii) Category of disorder: Chromosomal disorder (Aneuploidy / Down's syndrome).
Abnormality during cell division: Non-disjunction of chromosomes during gametogenesis (failure of segregation of chromatids during cell division).

Teacher's Note:
a) Down's syndrome is caused by the presence of an additional copy of chromosome 21 (Trisomy 21).
b) Non-disjunction leads to gametes with n+1 or n-1 chromosomes.

Question 5 [2 Marks]
(i) Study the diagram of a human ovum given below and answer the questions that follow.
[Figure 5: Shows human ovum with parts labeled I (Zona pellucida), II (Plasma membrane), III (Corona radiata), IV (Perivitelline space), V (First polar body)]
(a) Identify and name the part of the ovum that prevents its fertilisation by multiple sperm. [1 Mark]
(b) What is the mode of action of the part identified in (a)? [1 Mark]

Answer:
(a) Zona pellucida (Part I).
(b) When a sperm comes in contact with the zona pellucida, it induces changes in the membrane that block the entry of additional sperm (cortical reaction / zona reaction), ensuring monospermy.

Teacher's Note:
a) Zona pellucida is an outer glycoprotein layer surrounding the plasma membrane of the oocyte.
b) Polyspermy is prevented by the exocytosis of cortical granules which modify the zona pellucida receptors.

OR

(ii) The diagram given below represents a specific stage of human embryonic development. Study it carefully and answer the questions that follow.
[Figure 6: Shows human blastocyst with 'Inner cell mass' and outer trophoblast layer labeled 'A']
(a) Name the stage of human embryo represented by the given diagram. [1 Mark]
(b) Identify the part labelled 'A' and mention its function. [1 Mark]
(c) What happens to the inner cell mass after implantation? [1 Mark]

Answer:
(a) Blastocyst.
(b) Part A: Trophoblast; Function: It gets attached to the endometrium and helps in the formation of the placenta and extra-embryonic membranes.
(c) The inner cell mass differentiates into the embryonic disc containing three germ layers (ectoderm, mesoderm, and endoderm) which give rise to all tissues and organs of the fetus.

Teacher's Note:
a) The blastocyst stage is formed after repeated cleavage divisions of the zygote as it travels down the Fallopian tube.
b) Implantation takes place around 7 days after fertilization when the blastocyst embeds into the uterine wall.

Question 6 [2 Marks]
A group of researchers in a pharmaceutical company aims to produce a specific chemical which can treat viral infections.
Which specific group of chemicals should they isolate? How would it help in treatment of viral infection?

Answer:
Group of chemicals: Interferons (IFNs).
Mechanism of action: Interferons are antiviral proteins produced by virus-infected host cells. They diffuse to uninfected neighboring cells and induce the synthesis of antiviral proteins that inhibit viral replication, thereby protecting healthy cells from further infection.

Teacher's Note:
a) Interferons provide a non-specific defense mechanism against viral infections.
b) They are species-specific proteins produced by eukaryotic cells in response to viral nucleic acids.

Question 7 [2 Marks]
Explain any two types of evidence for biological evolution.

Answer:
1. Paleontological Evidence (Fossils): Study of fossils in different sedimentary rock strata shows geological periods and reveals that forms of life varied over time, exhibiting transitional forms (e.g., Archaeopteryx) that connect different animal groups.
2. Comparative Anatomy and Morphology: Homologous organs (such as forelimbs of whales, bats, cheetahs, and humans) show common ancestry through divergent evolution, while analogous organs show convergent adaptation to similar niches.

Teacher's Note:
a) Fossils provide direct evidence of extinct organisms and historical evolutionary sequences.
b) Comparative anatomical studies establish structural homologies pointing towards common descent.

Question 8 [2 Marks]
The figure shown below is a representative image of an antibody. Study it carefully and answer the questions that follow.
[Figure 7: Shows antibody molecule with antigen-binding sites at the tips (I), light chains (II), and heavy chains (III)]
(i) A person is infected by a pathogen. Identify and name the part of the antibody that the pathogen binds to. [1 Mark]
(ii) A few years later, this person is infected by a different variant of the same pathogen. Would the existing antibodies be able to recognise and bind to it now? Justify your answer by giving one reason. [1 Mark]

Answer:
(i) Antigen-binding site (Paratope located at the variable region - Part I).
(ii) No. The different variant of the pathogen possesses altered surface epitopes (antigenic determinants), and since antibodies are highly specific, the existing antibodies will not fit the modified epitopes of the new variant.

Teacher's Note:
a) The antigen-binding site is formed by the variable regions of both heavy and light chains.
b) Pathogen mutations often alter surface antigens, allowing them to evade pre-existing immunological memory (e.g., influenza virus variants).

SECTION C - 21 MARKS

Question 9 [3 Marks]
The diagram given below shows various phases of the menstrual cycle in human beings.
[Figure 8: Shows ovarian histology, phases, and 4 hormone plotlines labeled I, II, III, IV]
(i) Identify the plotlines that represent the following:
(a) Progesterone [1 Mark]
(b) Luteinizing hormone [1 Mark]
(ii) Give any two differences between Menstrual cycle and Estrus cycle. [1 Mark]

Answer:
(i) (a) Progesterone: Plotline I (peaks during the luteal phase).
(b) Luteinizing hormone (LH): Plotline II (shows a sharp surge right before ovulation around Day 14).
(ii) Differences:

Teacher's Note:
a) LH surge triggers ovulation, after which the ruptured follicle develops into the corpus luteum which secretes progesterone.
b) Students must clearly distinguish reproductive cycles of primates versus non-primates.

Question 10 [3 Marks]
As a part of an experiment, Kavya, David and Abdul had to cut a plasmid which is 650 bp long to insert a gene of interest. All of them worked on this task individually. As per the protocol, the restriction enzyme was incubated for two hours with the plasmid. After some time, they loaded their sample onto the same agarose gel for electrophoresis.
Given below is an image of the agarose gel where sample A belongs to Kavya, sample B to David and sample C to Abdul.
[Figure 9: Shows agarose gel electrophoresis with marker bands and lanes for Sample A (two bands at 450 bp and 200 bp), Sample B (one band at 700 bp), and Sample C (two bands at 450 bp and 200 bp)]
(a) If Abdul has performed the experiment correctly, what is the reason for the difference in the band pattern in Kavya's and David's samples? [1.5 Marks]
(b) According to Abdul, only a gene of the size of 200 bp can be inserted into this plasmid as this is the size of DNA that has been cut out from the plasmid. Is he correct? Justify by giving one reason. [1.5 Marks]

Answer:
(a) Kavya's plasmid was successfully cut by the restriction enzyme into two fragments (450 bp and 200 bp), whereas David's sample shows a band at 700 bp indicating that his plasmid was uncut (undigested) or the restriction enzyme failed to cut his plasmid.
(b) No, Abdul is incorrect. Any gene fragment equal to or smaller than the excised fragment size (up to 200 bp or even larger genes if using appropriate cloning vectors with multiple restriction sites) can be inserted, provided sticky ends match, and foreign inserts do not necessarily have to match the exact excised size.

Teacher's Note:
a) Agarose gel electrophoresis separates DNA fragments according to their size (molecular weight) through an electric field.
b) Uncut circular plasmids migrate differently on agarose gels compared to linear DNA fragments.

Question 11 [3 Marks]
The figure given below shows a mushroom that is connected to the roots of another tree growing near it.
[Figure 10: Shows mycorrhizal association between a mushroom/fungus and tree roots with bidirectional arrows]
(i) Identify the type of interaction between the mushroom and the roots of the tree. [1 Mark]
(ii) Mention any two advantages of this type of interaction. [2 Marks]

Answer:
(i) Mycorrhiza (Mutualism / Symbiosis).
(ii) Advantages:

1. The fungal symbiont absorbs essential nutrients (especially phosphorus) and water from the soil and makes them available to the plant root.
2. The plant provides the fungus with food (carbohydrates/photosynthates) and shelter, while the fungus also protects the host plant against root-borne pathogens and drought.

Teacher's Note:
a) Mycorrhizal association is a classic example of mutualism where both partners benefit (+, +).
b) Genera such as Glomus form vesicular-arbuscular mycorrhizae (VAM) with plant roots.

Question 12 [3 Marks]
The homozygous Andalusian chicken exhibits black feathers (BB) and white feathers (WW), while the heterozygous bird has bluish feathers.
(i) If two heterozygous Andalusian chickens are crossed, what are the possible genotypic and phenotypic ratios of their offspring? [2 Marks]
(ii) What kind of dominance does the gene for feather colour exhibit? Give a reason to support your answer. [1 Mark]

Answer:
(i) Cross: Heterozygous blue (\( BW \)) × Heterozygous blue (\( BW \))
Progeny: 1 BB (Black) : 2 BW (Blue) : 1 WW (White)
- Genotypic ratio: 1 (BB) : 2 (BW) : 1 (WW)
- Phenotypic ratio: 1 Black : 2 Blue : 1 White
(ii) Type of dominance: Incomplete dominance.
Reason: The heterozygous condition (\( BW \)) results in an intermediate phenotype (bluish feathers) rather than expressing either the dominant black or recessive white trait fully.

Teacher's Note:
a) Incomplete dominance is a deviation from Mendel's principles where neither allele is completely dominant over the other.
b) The phenotypic and genotypic ratios in incomplete dominance are identical (1:2:1).

Question 13 [3 Marks]
Name the category of microorganisms that aid in biogas production. Provide an example of this category.
Mention any four benefits of biogas compared to other energy sources.

Answer:
Category of microorganisms: Methanogens.
Example: Methanobacterium.
Four benefits of biogas:

1. It is a clean, eco-friendly, and renewable source of energy that produces minimum pollution.
2. It burns without smoke, leaving no residue, thereby reducing indoor air pollution compared to wood or dung cakes.
3. The spent slurry can be directly used as a high-quality organic fertilizer rich in nitrogen and phosphorus.
4. It utilizes animal dung and organic wastes efficiently, helping in rural waste management and sanitation.

Teacher's Note:
a) Methanogens are anaerobic bacteria that produce methane gas along with \( \text{CO}_{2} \) and hydrogen.
b) Biogas plants are widely promoted in rural areas under initiatives like Gobar-Gas plants.

Question 14 [3 Marks]
(i) Draw a well labelled diagram of L.S of anatropous ovule. [3 Marks]

Answer:
[Figure: A well-labeled diagram of L.S. of anatropous ovule showing funicle, hilum, micropyle, micropylar end, chalazal end, nucellus, embryo sac, synergids, egg cell, polar nuclei, and antipodals.]
1. The anatropous ovule is completely inverted during development such that the micropyle lies close to the funicle.
2. Key labels must include: Chalaza, Nucellus, Embryo sac (Female gametophyte), Egg apparatus, Antipodal cells, Polar nuclei, Integuments, Micropyle, Hilum, and Funicle.

Teacher's Note:
a) Anatropous ovule is the most common type of ovule found in about 82% of angiosperm families.
b) Correct and neat labeling of at least 4 to 6 key parts is required to secure full marks in diagram questions.

OR

(ii) Draw a well labelled diagram of mammalian testis. [3 Marks]

Answer:
[Figure: A well-labeled diagram of L.S. of mammalian testis showing tunica vaginalis, tunica albuginea, testicular lobules, seminiferous tubules, epididymis, vas deferens, and rete testis.]
1. The mammalian testis is enclosed in a capsule of fibrous connective tissue called tunica albuginea.
2. Key labels must include: Testicular lobules, Seminiferous tubules, Rete testis, Vasa efferentia, Epididymis, Vas deferens, and Blood vessels.

Teacher's Note:
a) Each testis contains 250 compartments called testicular lobules, each housing 1-3 highly coiled seminiferous tubules.
b) Ensure clear structural distinction between seminiferous tubules, rete testis, and epididymis in the diagram.

Question 15 [3 Marks]
Explain three different types of parasitism with one example each.

Answer:
1. Ectoparasitism: Parasites that feed on the external surface of the host organism. Example: Ticks on dogs or lice on humans.
2. Endoparasitism: Parasites that live inside the host's body at various sites (such as liver, gut, lungs). Example: Tapeworm (\( \text{Taenia solium} \)) or Ascaris in the human intestine.
3. Brood Parasitism: Parasitism in which the parasitic bird lays its eggs in the nest of its host and lets the host incubate them. Example: The koel laying eggs in the crow's nest.

Teacher's Note:
a) Parasitism is an interaction where one species (parasite) is benefitted while the other (host) is harmed (+, -).
b) Brood parasitism is a fascinating evolutionary adaptation where the cuckoo/koel eggs resemble the host's eggs in size and color to avoid detection.

SECTION D - 15 MARKS

Question 16 [5 Marks]
(i) Vallisneria is a submerged dioecious hydrophyte. The female flowers of Vallisneria reach the surface of water by their long stalk while the male flowers are released on to the surface of water due to bursting of the inflorescence.
(a) Give any two characteristic features of the pollen grains of hydrophilous flowers. [1 Mark]
(b) State whether Vallisneria is adapted for autogamy or xenogamy. Give one reason to justify your answer. [1 Mark]
(c) Give any three disadvantages of self-pollination. [3 Marks]

Answer:
(a) 1. Pollen grains are light, un-wettable, and usually covered with a mucilaginous covering to prevent wetting by water.
2. They are often long and ribbon-like or produced in large quantities.
(b) Xenogamy (Cross-pollination).
Reason: Vallisneria is dioecious (separate male and female plants), which completely prevents autogamy (self-pollination) and geitonogamy, ensuring cross-pollination.
(c) Disadvantages of self-pollination:

1. It leads to inbreeding depression, causing a gradual reduction in vigor, vitality, and productivity of offspring over generations.
2. It reduces genetic diversity and adaptability of offspring to changing environmental conditions since no new gene combinations are introduced.
3. Harmful recessive traits and deleterious mutations get accumulated and expressed easily in homozygous offspring.

Teacher's Note:
a) Hydrophily is a relatively rare mode of pollination found in about 30 genera of mostly monocotyledons.
b) Dioeciousness is a foolproof outbreeding device evolved by plants to prevent self-pollination.

OR

(ii) A couple is unable to have children naturally, despite the woman's ability to ovulate and the man having a normal sperm.
(a) Mention the two different types of abnormalities in sperm, which cause infertility in males. [2 Marks]
(b) Suggest any three methods of assisted reproductive technology except GIFT, available to the couple to consider. [3 Marks]

Answer:
(a) 1. Oligozoospermia: Low sperm count in the ejaculate.
2. Asthenozoospermia (or Teratozoospermia): Poor sperm motility or abnormal sperm morphology (defective acrosome or flagellum).
(b) Assisted Reproductive Technologies (ART) excluding GIFT:

1. In Vitro Fertilization and Embryo Transfer (IVF-ET): Fertilization is carried out outside the body in laboratory conditions, followed by embryo transfer (such as ZIFT or intrauterine transfer).
2. Intracytoplasmic Sperm Injection (ICSI): A single sperm is injected directly into the cytoplasm of the oocyte under a microscope to achieve fertilization.
3. Artificial Insemination (AI): Semen collected from the husband or a healthy donor is artificially introduced into the vagina or uterus of the female.

Teacher's Note:
a) Male infertility factors account for nearly 40-50% of total infertility cases.
b) ICSI is especially useful when sperm count or motility is severely compromised.

Question 17 [5 Marks]
What does the term biodiversity mean? Explain four major causes of loss of biodiversity due to human activities.

Answer:
Biodiversity definition: The combined diversity at all levels of biological organization, ranging from genetic variability within species and species diversity within communities to ecosystem-level variety across the biosphere (often referred to as genetic, species, and ecological diversity).
Four major causes of biodiversity loss ("The Evil Quartet"):

1. Habitat Loss and Fragmentation: Destruction of natural habitats (e.g., deforestation, conversion of tropical rainforests for agriculture or urban development) is the primary cause driving animals and plants to extinction.
2. Over-exploitation: When 'need' turns to 'greed', over-hunting, over-fishing, and excessive harvesting of natural resources lead to extinction of species (e.g., Steller's sea cow, passenger pigeon).
3. Alien Species Invasions: Introduction of non-native (exotic) species intentionally or unintentionally can become invasive and drive native species to extinction due to the absence of natural predators (e.g., Nile perch introduced in Lake Victoria, water hyacinth).
4. Co-extinctions: When a species becomes extinct, the plant and animal species associated with it in an obligatory relationship also become extinct (e.g., extinction of a host fish and its specific parasite).

Teacher's Note:
a) Biodiversity conservation is critical for ecosystem stability and resilience.
b) Students should remember the collective term "The Evil Quartet" to structure the four causes accurately.

Question 18 [5 Marks]
A research group studied various aspects of a particular region in their gene of interest and its mutated versions. Shown below is a part of the DNA sequence they were studying:
5' ATG TTG ACA TCA TCC AGC TGT 3'
[Figure 11: Standard Genetic Code table showing mRNA codons and corresponding amino acids]
(i) What is the DNA sequence complementary to the above sequence? [1 Mark]
(ii) Write the sequence of mRNA transcribed by this segment of DNA. [1 Mark]
(iii) Write the amino acid sequence coded by the mRNA sequence, using the genetic code shown in the figure given above. [1 Mark]
(iv) A mutant of this gene contains A instead of C in the 11th position in the given DNA sequence. What will be the amino acid sequence coded by this mutant gene? [1 Mark]
(v) What type of mutation would the change in nucleotide referred to in subpart (iv) lead to? Justify by giving one reason. [1 Mark]

Answer:
(i) Complementary DNA strand (written 3' to 5'):
3' TAC AAC TGT AGT AGG TCG ACA 5'
(Or written 5' to 3': 5' ACG TGA CTA CTA CAT TCAA 3' - standard convention is 3' to 5': 3' TAC AAC TGT AGT AGG TCG ACA 5')
(ii) mRNA sequence transcribed (assuming coding strand 5' to 3' template, template is 3' to 5' TAC AAC TGT AGT AGG TCG ACA 5'):
5' AUG UUG ACA UCA UCC AGC UGU 3'
(iii) Amino acid sequence using genetic code:
Met - Leu - Thr - Ser - Ser - Ser - Cys
(Codons: AUG = Met, UUG = Leu, ACA = Thr, UCA = Ser, UCC = Ser, AGC = Ser, UGU = Cys)
(iv) Original DNA (coding strand): 5' ATG TTG ACA TCC AGC TGT 3'
Mutant DNA (C at 11th position changed to A): 5' ATG TTG ACA TCA AGC TGT 3'
Corresponding mRNA: 5' AUG UUG ACA UCA UCC AGC UGU 3'
Amino acid sequence: Met - Leu - Thr - Ser - Ser - Ser - Cys
(v) Type of mutation: Silent mutation (or substitution mutation / point mutation).
Reason: Although the nucleotide changes from C to A, the resulting codon changes from UCC to UCA, both of which code for the exact same amino acid (Serine), resulting in no change in the protein product.

Teacher's Note:
a) Degeneracy of the genetic code allows multiple codons to specify the same amino acid, often preventing phenotypic effects from point mutations.
b) Always double-check codon tables carefully to map nucleotide triplets to correct amino acids.