CBSE Class 12 Biology: 2026 board paper, set 57/5/1

1. The apple is a false fruit: the large fleshy outer part (P) develops from the thalamus, and the central five chambers hold the seeds (Q).

   ½ mark

2. Around the chambers lie the true fruit layers: the inner cartilaginous endocarp (R) and the mesocarp (S). So the order is Thalamus, Seed, Endocarp, Mesocarp, which is option (B).

   ½ mark

Answer: (B) P: Thalamus, Q: Seed, R: Endocarp, S: Mesocarp

1. Signals from the fully developed foetus and the placenta cause mild uterine contractions; this is the foetal ejection reflex.

   ½ mark

2. The reflex makes the maternal pituitary release oxytocin, which causes strong uterine contractions for parturition. The answer is option (B).

   ½ mark

Answer: (B) oxytocin, maternal pituitary

1. 160 base pairs contain $160\times 2=320$ bases. By Chargaff's rule G = C = 30%, so A + T = 40% and A = 20%.

   ½ mark

2. Number of adenine bases $=20\%\times 320=64$. The answer is option (B).

   ½ mark

Answer: 64, option (B)

1. A VV × vv cross gives only Vv offspring.

   ½ mark

2. Violet (V) is dominant, so every Vv plant is violet. Hence 100% of the offspring are violet, option (D).

   ½ mark

Answer: (D) 100%

1. Phase II (days 6–14) is the follicular phase, in which the primary follicle matures into a Graafian follicle. Statement (A) is correct.

   ½ mark

2. Phase III is the luteal phase. The corpus luteum secretes large amounts of progesterone, which also maintains pregnancy, so statement (B) is correct. Statement (C) is wrong because there is no corpus luteum in phase I. The answer is (D).

   ½ mark

Answer: (D) Both (A) and (B)

1. Invertebrates were formed and active about 500 million years ago, and jawless fish arose around 350 million years ago.

   ½ mark

2. Seaweeds and a few plants existed around 320 million years ago. The order is therefore (ii), (iii), (i), option (C).

   ½ mark

Answer: (C) (ii), (iii), (i)

1. Monascus purpureus produces statins.

   ½ mark

2. Statins lower blood cholesterol by competitively inhibiting the enzyme that synthesises cholesterol. The answer is option (D).

   ½ mark

Answer: (D) Statins

1. The top strand is cut between A and G, giving the fragments 5′ CTGCA and G 3′.

   ½ mark

2. The bottom strand is cut between G and A, giving 3′ G and ACGTC 5′. The result is sticky ends, which matches option (D).

   ½ mark

Answer: (D)

1. A +/0 interaction, where one species benefits and the other is unaffected, is called commensalism.

   ½ mark

2. The clownfish gains protection among the stinging tentacles of the sea anemone, and the anemone gets no benefit or harm. Mycorrhiza is mutualism, the tortoise–goat case is competition and Cuscuta is a parasite. The answer is (B).

   ½ mark

Answer: (B) Sea anemone and clownfish

1. A transgenic animal carries a foreign gene (DNA) that has been introduced and expressed in it.

   ½ mark

2. The foreign DNA is introduced at an early embryonic stage, so it is present in all of the animal's cells. The answer is (C).

   ½ mark

Answer: (C)

1. Options A, B and C are analogous structures or forms that arose by convergent evolution.

   ½ mark

2. The thorns of Bougainvillea and the tendrils of Cucurbita are homologous organs, so they show divergent evolution. This makes (D) the odd one out.

   ½ mark

Answer: (D)

1. The Nile perch was introduced into Lake Victoria. As an alien species it preyed on the native fish.

   ½ mark

2. This led to the extinction of more than 200 cichlid species. The answer is (C).

   ½ mark

Answer: (C) Nile Perch

1. The assertion is true: an embryo with 8–16 blastomeres is called a morula.

   ½ mark

2. The reason is false: the morula divides further into a blastocyst (trophoblast plus inner cell mass), not into trophoblast alone. The answer is (C).

   ½ mark

Answer: (C)

1. Both statements are true: repetitive sequences form a very large part of the human genome, and they have no direct coding function.

   ½ mark

2. Lacking a coding function does not explain why they are so abundant, so R is not the correct explanation of A. The answer is (B).

   ½ mark

Answer: (B)

1. Both statements are true: Trichoderma is a free-living fungus common in root ecosystems, and it is an effective biocontrol agent.

   ½ mark

2. Its biocontrol role does not explain why it is free-living in root ecosystems, so R is not the correct explanation of A. The answer is (B).

   ½ mark

Answer: (B)

1. Yeast is a fungus, and its cell wall is made of chitin, so chitinase is needed to break it. Spirogyra is an alga with a cellulose wall, which needs cellulase instead.

   ½ mark

2. Both statements are true and R explains A. The answer is (A).

   ½ mark

Answer: (A)

1. (i) The tassels (silk) of a corn cob are the long, feathery styles and stigmas of the female flowers. They hang out and wave in the wind, which helps them trap wind-borne pollen grains (anemophily).

   1 mark

2. (ii) The scutellum is the single, shield-shaped cotyledon of the monocot (maize/grass) embryo. It absorbs food from the endosperm and passes it to the growing embryo, so it nourishes the embryo during germination.

   1 mark

Answer: (i) Tassels (styles and stigmas) trap wind-borne pollen; (ii) scutellum absorbs food from the endosperm and nourishes the embryo.

1. Reason 1: Hybrid seeds must be bought every year, because sowing seeds collected from a hybrid crop gives offspring in which the characters segregate, so the hybrid traits are lost. Apomictic seeds form without meiosis and fertilisation, so the progeny keeps the hybrid characters unchanged. Farmers can then keep using their own seed year after year.

   1 mark

2. Reason 2: Hybrid seed is costly to produce and to buy. Apomictic hybrid seed would remove this repeated expense, so it is far more cost-effective for farmers.

   1 mark

Answer: No segregation of hybrid characters, so seed need not be bought every year; apomictic seed is cheaper and cost-effective.

1. Amniocentesis is a prenatal diagnostic technique. A small amount of amniotic fluid containing foetal cells is withdrawn and analysed (chromosomes, enzymes) to detect genetic disorders such as Down's syndrome in the developing foetus.

   1 mark

2. The same test also reveals the sex of the foetus. It was being misused for sex determination, followed by abortion of female foetuses (female foeticide). A statutory ban on its use for sex determination was imposed to check this menace and the falling sex ratio.

   1 mark

Answer: Amniocentesis is analysis of foetal cells in amniotic fluid to detect genetic disorders. It is banned for sex determination to prevent female foeticide.

1. The organ is the spleen, a large bean-shaped secondary lymphoid organ.

   1 mark

2. As a lymphoid organ, the spleen contains mainly lymphocytes and phagocytes. It acts as a filter of the blood by trapping blood-borne microorganisms, which the immune cells then destroy.

   1 mark

Answer: Spleen; it filters blood by trapping blood-borne microorganisms.

1. Remove the meristem (apical or axillary) from the diseased sugarcane plant. The meristem stays free of virus even when the rest of the plant is infected.

   1 mark

2. Grow this meristem in vitro by tissue culture on a suitable nutrient medium. The plantlets that regenerate are virus-free healthy plants.

   1 mark

Answer: Meristem culture: excise the virus-free apical/axillary meristem and grow it by tissue culture to get virus-free plants.

1. Treat the bacterial cells with a specific concentration of a **divalent cation such as calcium ($C{a}^{2+}$)**. This increases the efficiency with which DNA enters through pores in the cell wall.

   ½ mark

2. Incubate the treated cells with the recombinant DNA on ice.

   ½ mark

3. Give a brief heat shock at 42 °C.

   ½ mark

4. Put them back on ice. This enables the bacteria to take up the recombinant DNA.

   ½ mark

Answer: Ca²⁺ treatment, incubation with rDNA on ice, heat shock at 42 °C, then back on ice.

1. Grazing food chain (GFC)	Detritus food chain (DFC)
   Starts with producers (green plants).	Starts with dead organic matter (detritus) and its decomposers.

   1 mark

2. Grazing food chain (GFC)	Detritus food chain (DFC)
   Major conduit of energy flow in aquatic ecosystems.	Major conduit of energy flow in terrestrial ecosystems.

   1 mark

Answer: GFC starts from producers and is the major energy route in aquatic ecosystems. DFC starts from detritus and is the major energy route in terrestrial ecosystems.

1. Ecological pyramids do not take into account that the same species may belong to two or more trophic levels.

   1 mark

2. They assume a simple food chain, which almost never exists in nature, and so do not accommodate a food web. Also, saprophytes/decomposers are given no place in the pyramids even though they play a vital role in the ecosystem.

   1 mark

Answer: Same species at more than one trophic level is ignored; they assume simple food chains (no food webs) and leave out saprophytes.

1. Asynchronous maturity: Pollen release and stigma receptivity are not synchronised. Either the pollen is released before the stigma becomes receptive, or the stigma becomes receptive much before pollen release. This prevents self-pollination, so pollen must come from another flower.

   1 mark

2. Different positions of anther and stigma: The anther and stigma are placed at different positions in the flower. Pollen therefore cannot come in contact with the stigma of the same flower, which favours cross-pollination.

   1 mark

3. Self-incompatibility: This genetic mechanism prevents pollen from the same flower, or from other flowers of the same plant, from germinating on the stigma or growing a pollen tube in the pistil. Only pollen from another plant can fertilise the ovules. (Production of unisexual flowers, or of male and female flowers on different plants as in papaya, also prevents self-pollination.)

   1 mark

Answer: Asynchronous maturity of pollen and stigma, different positions of anther and stigma, self-incompatibility (also unisexual flowers / dioecy).

1. Diagram (described): Draw a Y-shaped replication fork. The parental double helix (one strand 5'→3', the other 3'→5') opens at the fork point (helicase site).

   • On the template with 3'→5' polarity, the new leading strand is drawn as one continuous arrow, synthesised 5'→3' towards the fork.
   • On the template with 5'→3' polarity, the new lagging strand is drawn as short separate pieces (Okazaki fragments), each synthesised 5'→3' away from the fork and joined by DNA ligase.
   • Label the 5' and 3' ends of all four strands, mark DNA polymerase, and show the direction of replication with arrows.
   2 marks

2. DNA is a very long molecule. Separating its two strands along the entire length at once would need a very high amount of energy. Replication therefore occurs within a small opening of the helix, the replication fork, where the strands unwind only a little at a time.

   1 mark

Answer: Labelled fork showing 3'→5' and 5'→3' templates, continuous leading strand, discontinuous lagging strand (Okazaki fragments). A fork is used because separating the full length of the long DNA molecule at once needs very high energy.

1. (a) Psilophyton is the immediate ancestor of both ferns and conifers.

   ½ mark

2. (b) Seed ferns are the nearest ancestors of flowering plants.

   ½ mark

3. (c) Chlorophyte ancestors are the most primitive group, at the base of the chart.

   ½ mark

4. (d) Psilophyton gives common ancestry to ferns, conifers and seed ferns.

   ½ mark

5. (e) Tracheophyte ancestors are the common ancestor of Psilophyton and seed ferns.

   ½ mark

6. (f) Chlorophyte ancestors are the common ancestor of mosses and tracheophytes.

   ½ mark

Answer: (a) Psilophyton; (b) Seed ferns; (c) Chlorophyte ancestors; (d) Ferns, conifers, seed ferns; (e) Tracheophyte ancestors; (f) Chlorophyte ancestors

1. (a) Saccharomyces cerevisiae (brewer's/baker's yeast): bread/bakery products, or ethanol and alcoholic beverages.

   ½ mark

2. (b) Propionibacterium shermanni: Swiss cheese (its CO₂ produces the large holes).

   ½ mark

3. (c) Aspergillus niger: citric acid.

   ½ mark

4. (d) Trichoderma polysporum: cyclosporin A, an immunosuppressive agent used in organ-transplant patients.

   ½ mark

5. (e) Acetobacter aceti: acetic acid (vinegar).

   ½ mark

6. (f) Streptococcus: streptokinase, a 'clot buster' enzyme used to dissolve clots in blood vessels.

   ½ mark

Answer: (a) Bread/ethanol; (b) Swiss cheese; (c) Citric acid; (d) Cyclosporin A; (e) Acetic acid; (f) Streptokinase

1. Study of normal physiology and development: Transgenic animals can be designed to study how genes are regulated and how they affect normal body functions and development. An example is the study of complex growth factors such as insulin-like growth factor.

   1 mark

2. Study of disease: Many transgenic animals serve as models for human diseases such as cancer, cystic fibrosis, rheumatoid arthritis and Alzheimer's. These models help us understand how genes contribute to disease and help in investigating new treatments.

   1 mark

3. Biological products: Transgenic animals can make useful products. For example, human protein α-1-antitrypsin is produced for treating emphysema. The transgenic cow Rosie (1997) produced milk rich in human alpha-lactalbumin, which is nutritionally more balanced for human babies. (Also: vaccine safety – transgenic mice are used to test the safety of polio vaccine; chemical safety testing – animals made more sensitive to toxic substances are used to study their effects.)

   1 mark

Answer: Any three: study of normal physiology/development, study of disease, biological products (α-1-antitrypsin, Rosie's milk), vaccine safety testing, chemical safety testing.

1. Statement: Gause's competitive exclusion principle states that two closely related species competing for the same limited resources cannot co-exist indefinitely.

   1 mark

2. The competitively inferior species is eventually eliminated, because the superior competitor uses the limited resource more efficiently and dominates.

   1 mark

3. Example: The Abingdon tortoise of the Galapagos Islands became extinct within about a decade after goats were introduced on the island. This was probably because the goats browsed more efficiently. (Another example: in some shallow South American lakes, visiting flamingoes and resident fishes compete for the same food, zooplankton.)

   1 mark

Answer: Closely related species competing for the same limited resource cannot co-exist indefinitely; the inferior one is eliminated. Example: Abingdon tortoise eliminated by introduced goats.

1. Genetic diversity: A single species can show high diversity at the genetic level over its range of distribution. Example: India has more than 50,000 genetically different strains of rice and about 1,000 varieties of mango. Another example is Rauvolfia vomitoria in different Himalayan ranges, which varies in the potency and concentration of reserpine.

   1 mark

2. Species diversity: This is the diversity in the number and variety of species in a region. Example: the Western Ghats have greater amphibian species diversity than the Eastern Ghats.

   1 mark

3. Ecological diversity: This is the diversity at the ecosystem level, meaning the variety of ecosystems in an area. Example: India, with deserts, rain forests, mangroves, coral reefs, wetlands, estuaries and alpine meadows, has greater ecosystem diversity than a Scandinavian country like Norway.

   1 mark

Answer: Genetic (e.g. 50,000 rice strains), species (Western Ghats amphibians vs Eastern Ghats), ecological (India's deserts, rain forests, coral reefs).

1. (a) RNA polymerase can polymerise nucleotides only in the 5'→3' direction. The RNA can be built 5'→3' only on the template strand with 3'→5' polarity, so this strand is transcribed.

   1 mark

2. (b)(i) hnRNA is the primary transcript. It contains both coding sequences (exons) and non-coding sequences (introns). Splicing removes the introns and joins the exons in a defined order, so that only coding sequence remains for translation.

   1 mark

3. (b)(ii) Capping: an unusual nucleotide, methyl guanosine triphosphate (mGppp), is added to the 5' end of hnRNA. Tailing: about 200–300 adenylate residues (poly-A tail) are added at the 3' end in a template-independent manner.

   1 mark

4. (c) Reason 1: If both strands were transcribed, one DNA segment would code for two different RNAs and hence two different polypeptides. This would complicate the genetic information transfer machinery, and one product might be useless to the cell.

   1 mark

5. (c) Reason 2: The two RNA molecules produced at the same time would be complementary to each other. They would pair to form double-stranded RNA, which cannot be translated into protein, so transcription would be futile.

   1 mark

Answer: (a) RNA polymerase works only 5'→3'; (b)(i) to remove introns and join exons; (b)(ii) capping and tailing; (c) two different proteins would complicate the machinery; complementary RNAs would form non-translatable dsRNA.

1. (a) Plasmids and bacteriophages can replicate within bacterial cells independently of the chromosomal DNA, and an alien piece of DNA can be ligated to them. The linked foreign DNA is therefore multiplied along with them, up to their copy number.

   1 mark

2. (b)(i) (I) The gene controlling copy number is ori (origin of replication). (II) The restriction site C in the rop gene is Pvu II.

   1 mark

3. (b)(ii) G is ampR (ampicillin-resistance gene). H is tetR (tetracycline-resistance gene). Both are selectable markers.

   1 mark

4. (c) Restriction sites in ampR: Pst I and Pvu I. Restriction sites in tetR: BamH I and Sal I.

   2 marks

Answer: (a) They replicate independently in bacteria and can carry linked foreign DNA; (b)(i) ori, Pvu II; (b)(ii) G = ampR, H = tetR; (c) ampR: Pst I, Pvu I; tetR: BamH I, Sal I

1. Fertilisation: It takes place in the ampullary (ampullary-isthmic junction) region of the fallopian tube, where the ovum and the sperm reach together.

   ½ mark

2. The secretions of the acrosome help the sperm dissolve the zona pellucida and plasma membrane and enter the ovum.

   ½ mark

3. When a sperm contacts the zona pellucida, it induces changes in the membrane that block the entry of additional sperms. This ensures that only one sperm fertilises the ovum (monospermy).

   ½ mark

4. Entry of the sperm induces the secondary oocyte to complete meiosis II, forming the ootid (ovum) and a second polar body.

   ½ mark

5. The haploid nucleus of the sperm fuses with the haploid nucleus of the ovum.

   ½ mark

6. This fusion forms a diploid zygote (2n). The sex of the baby is decided at this stage by the X or Y chromosome of the sperm.

   ½ mark

7. Implantation: The zygote divides mitotically (cleavage) to form a morula and then a blastocyst, which moves to the uterus. In the blastocyst, the outer trophoblast layer gets attached to the endometrium of the uterus.

   1 mark

8. The uterine cells divide rapidly and cover the blastocyst. As a result, the blastocyst becomes embedded in the endometrium. This embedding is implantation, and it leads to pregnancy.

   1 mark

Answer: Fertilisation in the ampulla: acrosomal entry, block to polyspermy, completion of meiosis II, fusion of nuclei to form a diploid zygote. Implantation: trophoblast attaches to the endometrium and uterine cells cover the blastocyst, embedding it.

1. (i) Sequence of secretion: FSH → LH → hCG → Relaxin.

   1 mark

2. (ii) FSH – Source: anterior pituitary. Function: stimulates the growth and development of ovarian follicles during the follicular phase.

   1 mark

3. LH – Source: anterior pituitary. Function: its mid-cycle surge causes rupture of the Graafian follicle and release of the ovum (ovulation). It also helps form the corpus luteum.

   1 mark

4. hCG (human chorionic gonadotropin) – Source: placenta. Function: supports foetal growth, brings about metabolic changes in the mother and maintains pregnancy.

   1 mark

5. Relaxin – Source: ovary, in the later phase of pregnancy. Function: facilitates parturition by relaxing the pelvic ligaments and softening the cervix.

   1 mark

Answer: (i) FSH, LH, hCG, Relaxin; (ii) FSH – anterior pituitary, follicle development; LH – anterior pituitary, ovulation; hCG – placenta, maintains pregnancy and supports foetal growth; Relaxin – ovary, facilitates parturition.

1. Cross 1 – Pea (flower colour): Parents: violet $VV$ × white $vv$. Gametes: $V$ and $v$.

   ½ mark

2. $F_1$: all $Vv$, violet. The violet allele masks white. $F_1$ is selfed: $Vv\times Vv$, giving gametes $V,v$ from each parent.

   ½ mark

3. $F_2$ Punnett square:

   ♀ \ ♂	V	v
   V	VV (violet)	Vv (violet)
   v	Vv (violet)	vv (white)

   ½ mark

4. Phenotypic ratio violet : white = 3 : 1. Genotypic ratio $VV:Vv:vv$ = 1 : 2 : 1.

   ½ mark

5. Comment: The $F_1$ resembles one parent and the $F_2$ phenotypic ratio is 3 : 1. This is complete dominance (violet is dominant over white).

   ½ mark

6. Cross 2 – Snapdragon (Antirrhinum): Parents: red $RR$ × white $rr$. Gametes: $R$ and $r$.

   ½ mark

7. $F_1$: all $Rr$, pink. The $F_1$ is intermediate and resembles neither parent. $F_1$ is selfed: $Rr\times Rr$.

   ½ mark

8. $F_2$ Punnett square:

   ♀ \ ♂	R	r
   R	RR (red)	Rr (pink)
   r	Rr (pink)	rr (white)

   ½ mark

9. Phenotypic ratio red : pink : white = 1 : 2 : 1, which is the same as the genotypic ratio.

   ½ mark

10. Comment: Neither allele is completely dominant, so the heterozygote shows an intermediate (pink) phenotype. This is incomplete dominance. Here the phenotypic and genotypic ratios are identical, whereas in the pea cross they differ.

    ½ mark

Answer: Pea: F₂ violet : white = 3 : 1 (genotypic 1 : 2 : 1) – dominance. Antirrhinum: F₂ red : pink : white = 1 : 2 : 1 – incomplete dominance.

1. (i) Difference: A Mendelian disorder is caused by a mutation or alteration in a single gene and is passed on in the Mendelian pattern, so it can be traced through a pedigree. A chromosomal disorder is caused by the absence, excess or abnormal arrangement of one or more whole chromosomes (for example, through failure of chromatids to separate during cell division). It does not follow Mendelian inheritance.

   1 mark

2. Example of a Mendelian disorder: haemophilia, sickle-cell anaemia, colour blindness, phenylketonuria, thalassemia or cystic fibrosis (any one).

   ½ mark

3. Example of a chromosomal disorder: Down's syndrome (trisomy 21), Klinefelter's syndrome (XXY) or Turner's syndrome (XO) (any one).

   ½ mark

4. (ii) Phenomenon: Aneuploidy, specifically monosomy. It arises from non-disjunction during gamete formation, which gives a gamete with no sex chromosome.

   ½ mark

5. Disorder: Turner's syndrome.

   ½ mark

6. How affected individuals are affected (symptoms): They are females who are sterile because their ovaries are rudimentary. They show lack of (underdeveloped) secondary sexual characters and short stature.

   1½ marks

7. Karyotype: 44 + XO, a total of 45 chromosomes (one X chromosome is missing).

   ½ mark

Answer: (i) Mendelian disorder: single-gene mutation, follows Mendelian inheritance (e.g. haemophilia). Chromosomal disorder: gain or loss of whole chromosome(s), not Mendelian (e.g. Down's syndrome). (ii) Aneuploidy (monosomy); Turner's syndrome; sterile females with rudimentary ovaries, no secondary sexual characters and short stature; karyotype 44 + XO (45).

1. (i) Diagram (described): Draw a Y-shaped molecule made of four polypeptide chains. There are two long heavy (H) chains forming the stem and both arms. There are two short light (L) chains, one along the outer side of each arm. The chains are held together by disulphide (S–S) bonds. The tips of the two arms form the antigen-binding sites. Label any four of: (1) heavy chain, (2) light chain, (3) antigen-binding site, (4) disulphide bonds; also variable region / constant region (H₂L₂ structure).

   2 marks

2. (ii) Difference 1 (origin): In active immunity, the body's own B-lymphocytes produce antibodies in response to an antigen, through infection or vaccination. In passive immunity, ready-made antibodies are given to the body, e.g. anti-tetanus serum, or IgA in colostrum and antibodies crossing the placenta.

   1 mark

3. Difference 2 (speed): Active immunity is slow and takes time to give its full response. Passive immunity gives immediate/quick protection.

   1 mark

4. Difference 3 (duration and memory): Active immunity is long-lasting because memory cells are formed. Passive immunity is short-lived and no memory cells are formed.

   1 mark

Answer: (i) Y-shaped H₂L₂ antibody: two heavy chains, two light chains, disulphide bonds, antigen-binding sites. (ii) Active: own antibodies, slow, long-lasting, memory cells formed. Passive: ready-made antibodies, quick, short-lived, no memory cells.

1. (i)(I) The two filarial worms are Wuchereria bancrofti and Wuchereria malayi.

   1 mark

2. (II) The worms live in the lymphatic vessels for many years. They cause slowly developing chronic inflammation of the organs in which they live, usually those of the lower limbs. This leads to swelling of the legs (elephantiasis), and the genital organs may also become grossly deformed.

   1 mark

3. (III) The disease spreads from an infected person to a healthy person through the bite of a female mosquito vector (e.g. Culex).

   1 mark

4. (ii)(I) Histamine: Source: released by mast cells (and basophils). Role: it dilates blood vessels and increases their permeability. This produces the inflammatory response and brings defence cells to the site. When released excessively against harmless allergens, it causes the symptoms of allergy (sneezing, watery eyes, running nose).

   1 mark

5. (II) Interferons: Source: secreted by virus-infected cells. Role: they protect the neighbouring non-infected cells from further viral infection. This is part of the cytokine barrier of innate immunity.

   1 mark

Answer: (i)(I) Wuchereria bancrofti and W. malayi. (II) Chronic inflammation of lymphatic organs, leading to elephantiasis of the lower limbs and deformity of the genitals. (III) Bite of a female mosquito vector. (ii)(I) Histamine: from mast cells; causes inflammation/allergic response. (II) Interferons: from virus-infected cells; protect non-infected cells from viral infection.