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

1. Emasculation (removal of anthers) is needed only in bisexual flowers. Papaya is dioecious, so its female flowers are unisexual and carry no anthers; only bagging is needed. Rice, wheat and pea are bisexual.

   1 mark

Answer: (D) Papaya

1. The immune system includes lymphoid organs such as bone marrow, lymphocytes and plasma proteins such as antibodies. RBCs only carry oxygen and play no part in immunity.

   1 mark

Answer: (C) Red Blood Cells

1. (i) is true: the corpus luteum secretes progesterone. (iii) is true: the mid-cycle LH surge causes ovulation.

   ½ mark

2. (ii) is false because both LH and FSH peak at mid-cycle. (iv) is false because the luteal phase is the secretory phase, not the proliferative phase.

   ½ mark

Answer: (B) (i) and (iii) are correct

1. India has more than 50,000 genetically different strains of rice, so the pair 'less than 1000' is wrong. The other three pairs are correct: 12 mega-diversity countries, Robert May's estimate of about 7 million species, and zoos as ex-situ conservation.

   1 mark

Answer: (B)

1. Pomato is a somatic hybrid made by fusing protoplasts of tomato and potato.

   1 mark

Answer: (A) Tomato and Potato

1. The vertical distribution of different species occupying different levels in an ecosystem is called stratification. For example, trees, shrubs and herbs form separate layers in a forest.

   1 mark

Answer: (A) Stratification

1. The Khasi and Jaintia hills of Meghalaya have sacred groves. Bastar, Chanda and Sarguja are in Madhya Pradesh and Chhattisgarh.

   1 mark

Answer: (A) Jaintia hills

1. The two homologous pairs can line up in two equally likely ways at Metaphase I. This gives four allele combinations in the gametes (AB, ab, Ab, aB), which shows independent assortment.

   1 mark

Answer: (B) Independent assortment of genes

1. A biolistic (gene) gun is used to transform plant cells (a–v). Chitinase digests fungal cell walls (b–iv). The Ti plasmid is the tumour-inducing plasmid (c–ii). Ca${}^{2+}$ makes bacterial cells competent to take up DNA (d–i).

   1 mark

Answer: (C) a-v, b-iv, c-ii, d-i

1. Lamarck proposed that organs develop with use and degenerate with disuse, using the giraffe's long neck as his example.

   1 mark

Answer: (D) Lamarck

1. Plasmids replicate autonomously, independently of the chromosomal DNA. So the statement that their replication is controlled by chromosomal DNA is incorrect.

   1 mark

Answer: (B)

1. The ovaries are held to the pelvic wall and uterus by ligaments.

   1 mark

Answer: (B) Ligaments

1. Industrial melanism raised the numbers of dark moths, so A is true. Soot darkened the tree trunks and killed the lichens, which camouflaged dark moths from predators. So R is true and correctly explains A.

   1 mark

Answer: (A)

1. A is true: Rosie's milk was more balanced for human babies. R is false because the protein was human alpha-lactalbumin, not 'alpha lactaglobulin'.

   1 mark

Answer: (C)

1. A is true: flocs are masses of bacteria associated with fungal filaments in the aeration tank. R is false because flocs digest organic matter aerobically, reducing the BOD. Anaerobic digestion happens later in the sludge digesters.

   1 mark

Answer: (C)

1. A is false: in terrestrial ecosystems a much larger fraction of energy flows through the detritus food chain. R is true, because the grazing food chain is linked to the detritus food chain at some levels.

   1 mark

Answer: (D)

1. (a) Humoral immune response is the immune response carried out by antibodies. B-lymphocytes produce an army of these proteins (antibodies) and release them into the blood (body fluids, or humor) to fight the pathogens.

   1 mark

2. (b) Any two antibody types: IgG and IgA (others: IgM, IgE).

   1 mark

Answer: (a) Antibody-mediated immunity in which B-lymphocytes secrete antibodies into the blood against pathogens; (b) IgG, IgA (also IgM, IgE)

1. (i) Dormant seeds are dehydrated and metabolically inactive, so they can be safely stored for long periods. The farmer can therefore keep them as seed to raise the crop in the next season, and also use them as food or sell them commercially.

   1 mark

2. (ii) Pea seed is non-endospermic (non-albuminous): the endosperm is completely used up during embryo development, so none is left in the seed. Castor seed is endospermic (albuminous): part of the endosperm remains, because it is not fully consumed during embryo development.

   1 mark

Answer: (i) Dormant seeds can be stored and sown next season, or used as food or for trade; (ii) pea is non-albuminous (no residual endosperm), castor is albuminous (residual endosperm present)

1. The primary oocyte completes its first meiotic division in the tertiary follicle (as it matures towards the Graafian follicle).

   1 mark

2. This unequal division gives two haploid products: a large secondary oocyte, which keeps most of the cytoplasm, and a tiny first polar body.

   1 mark

Answer: Tertiary follicle; the products are the secondary oocyte and the first polar body

1. (a) These are palindromic nucleotide sequences: they read the same on both strands when each is read in the same direction (5'→3').

   ½ mark

2. They are recognised by a restriction endonuclease, here EcoRI.

   ½ mark

3. (b) The restriction endonuclease binds its specific recognition (palindromic) site. It cuts each strand a little away from the centre of the palindrome, between the same two bases on the opposite strands; EcoRI cuts between G and A in 5'-GAATTC-3'. This leaves single-stranded overhanging stretches called sticky ends on each fragment.

   1 mark

Answer: (a) Palindromic sequences; restriction endonuclease (EcoRI); (b) it cuts both strands between G and A, away from the centre of the palindrome, producing sticky ends

1. Case I (XX–XO): the male produces two kinds of gametes (with X and without X), so this is male heterogamety.

   ½ mark

2. Example: grasshopper (many insects).

   ½ mark

3. Case II (ZW–ZZ): the female produces two kinds of gametes (Z and W), so this is female heterogamety.

   ½ mark

4. Example: birds, e.g. hen.

   ½ mark

Answer: Case I: male heterogamety, e.g. grasshopper; Case II: female heterogamety, e.g. birds (hen)

1. (i) A DNA nucleotide contains deoxyribose sugar and one of the bases A, G, C or T. An RNA nucleotide contains ribose sugar and one of the bases A, G, C or U (uracil in place of thymine).

   1 mark

2. (ii) N-glycosidic linkage, between the nitrogen base and the pentose sugar (C-1').

   ½ mark

3. Phosphoester linkage, between the phosphate group and the 5'-OH of the sugar.

   ½ mark

Answer: (i) DNA nucleotide: deoxyribose with T; RNA nucleotide: ribose with U; (ii) N-glycosidic bond and phosphoester bond

1. (i) Because the bars show standing crop and the consumer bar is wider than the producer bar, this is an inverted pyramid of biomass.

   ½ mark

2. Such a pyramid occurs in aquatic ecosystems such as the sea. There, a small standing crop of phytoplankton reproduces very fast and supports a larger standing crop of zooplankton or fish.

   ½ mark

3. (ii) Any two limitations: (1) It does not account for the same species occupying two or more trophic levels. (2) It assumes a simple food chain and cannot accommodate a food web.

   1 mark

4. (Another accepted limitation: saprophytes or decomposers are given no place in the pyramid, though they play a vital role in the ecosystem.)

Answer: (i) Inverted pyramid of biomass, e.g. in the sea a small standing crop of phytoplankton supports a larger standing crop of zooplankton or fish; (ii) it ignores species at more than one trophic level, and it cannot show food webs (or it omits saprophytes)

1. (i) Loss of biodiversity in the area would lead to (1) a decline in plant production and (2) lowered resistance to environmental perturbations such as drought. (A third possible effect: increased variability in ecosystem processes such as water use and pest or disease cycles.)

   1 mark

2. (ii) Two major causes: (1) habitat loss and fragmentation, e.g. clearing forest for the project; (2) over-exploitation of natural resources. (Others: alien species invasion, co-extinction.)

   1 mark

Answer: (i) Decline in plant productivity, and lowered resistance to disturbances such as drought; (ii) habitat loss and fragmentation, and over-exploitation (also alien species invasion, co-extinction)

1. (a) No, pollen viability differs from species to species.

   ½ mark

2. Example 1: in rice and wheat (cereals), pollen grains lose viability within about 30 minutes of their release.

   1 mark

3. Example 2: in some members of Rosaceae, Leguminosae and Solanaceae, pollen grains stay viable for months.

   1 mark

4. (b) In pollen banks, pollen grains are stored by cryopreservation in liquid nitrogen at −196 °C, so they remain viable for years for crop-breeding programmes.

   ½ mark

Answer: (a) No: rice/wheat pollen stays viable about 30 min, while pollen of Rosaceae, Leguminosae and Solanaceae stays viable for months; (b) cryopreservation in liquid nitrogen at −196 °C

1. (a) Pedigree analysis is the study of how a trait is inherited over several generations of a family, using a family tree drawn with standard symbols.

   1 mark

2. Importance: controlled crosses are not possible in humans, so pedigree analysis is used to trace the inheritance of a trait, abnormality or genetic disease. It shows whether the trait is dominant or recessive, autosomal or sex-linked, and helps in genetic counselling.

   1 mark

3. (b)(i) Normal parents have affected children, so the trait is recessive and both parents are carriers. An affected daughter is born to a normal father; this rules out X-linked recessive inheritance, because the father would then have to be affected. The pattern is therefore autosomal recessive.

   ½ mark

4. (ii) Example: sickle-cell anaemia (also phenylketonuria or thalassemia).

   ½ mark

Answer: (a) Study of inheritance of a trait over several generations of a family; used to trace the inheritance of traits and diseases; (b)(i) autosomal recessive; (ii) sickle-cell anaemia

1. (a) The Hardy–Weinberg principle states that allele frequencies in a population are stable and stay constant from generation to generation, and that the sum of all allele frequencies is 1.

   ½ mark

2. Let one gene have two alleles, A and a, with frequencies $p$ and $q$. Then $p+q=1$.

   ½ mark

3. In a diploid individual, the chance of A appearing on both homologous chromosomes is the product of the probabilities, $p\times p=p^2$, which is the frequency of AA. Likewise, the frequency of aa is $q^2$.

   ½ mark

4. Heterozygotes Aa can arise in two ways (A from either parent), so their frequency is $2pq$. Adding all genotype frequencies gives $(p+q{)}^2=p^2+2pq+q^2=1$.

   ½ mark

5. (b) Any two factors that disturb genetic equilibrium: gene migration (gene flow) and genetic drift. (Others: mutation, genetic recombination, natural selection.)

   1 mark

Answer: (a) With $p+q=1$, AA = $p^2$, aa = $q^2$ and Aa = $2pq$, so $(p+q{)}^2=p^2+2pq+q^2=1$; (b) gene flow and genetic drift (also mutation, recombination, natural selection)

1. Draw an anatropous ovule in L.S.: an oval body curved back so that the micropyle lies close to the stalk. Label any six of the parts below.

2. Funicle: the stalk attaching the ovule to the placenta. Hilum: the junction where the body of the ovule fuses with the funicle.

   1 mark

3. Outer and inner integuments: two protective layers around the nucellus. Micropyle: the small opening at the tip of the integuments, next to the funicle.

   1 mark

4. Nucellus: the mass of cells enclosed by the integuments. Embryo sac (female gametophyte): inside the nucellus, showing the egg apparatus (egg cell plus 2 synergids) at the micropylar end, 2 polar nuclei in the centre and 3 antipodals at the chalazal end. Chalaza: the basal part of the ovule, opposite the micropyle.

   1 mark

Answer: Labelled L.S. of an anatropous ovule showing funicle, hilum, micropyle, outer and inner integuments, nucellus, embryo sac and chalaza (any six)

1. (a) Source plant: poppy, Papaver somniferum.

   ½ mark

2. Part used: the latex obtained from the plant (its unripe capsules).

   ½ mark

3. (b) Heroin (smack) is chemically diacetylmorphine. It is obtained by acetylation of morphine.

   1 mark

4. Effect of morphine: it is a very effective sedative and painkiller, useful for patients who have undergone surgery.

   ½ mark

5. Effect of heroin: it is a depressant and slows down body functions.

   ½ mark

Answer: (a) Papaver somniferum, latex; (b) heroin is acetylated morphine (diacetylmorphine); morphine is a sedative and painkiller, heroin a depressant that slows body functions

1. (a) The corn borer is controlled by the protein encoded by the gene *cry*IAb.

   ½ mark

2. (b) The Bt spores, sprayed with water on field A, contain inactive protoxin crystals. These crystals are eaten by the insect larvae along with the leaves.

   ½ mark

3. In the alkaline pH of the insect gut, the crystals dissolve and the inactive protoxin is converted into the active toxin.

   ½ mark

4. The active toxin binds to the midgut epithelial cells and creates pores. This causes swelling and lysis of the cells, and the insect dies.

   ½ mark

5. (c) The corn in field B is genetically modified: the cry gene from Bacillus thuringiensis has been introduced into it. These Bt corn plants produce the Bt toxin themselves, so a corn borer feeding on them takes in the toxin and is killed.

   1 mark

Answer: (a) cry IAb; (b) larvae eat the spores, the alkaline gut activates the toxin, it makes pores in the gut cells and the insect dies; (c) Bt corn carries the cry gene and produces its own Bt toxin, which kills the borer

1. (a) No bands will be seen in the gel after ethidium bromide staining.

   ½ mark

2. Reason: DNA fragments are negatively charged, so they move only towards the anode (+). Here the wells are at the positive end, so the DNA does not move through the gel towards the cathode (−). No fragments are separated within the gel.

   ½ mark

3. (b) Change made: the positions of the electrodes were interchanged, so the cathode (−) is now near the wells and the anode (+) is at the far end.

   1 mark

4. Why the band appears at X: the negatively charged DNA now moves through the gel towards the anode. Agarose acts as a sieve, so smaller fragments move faster and farther. The smallest fragment (100 bp) therefore travels farthest and appears at X.

   1 mark

Answer: (a) No bands, because negatively charged DNA stays at or moves towards the anode near the wells and does not migrate through the gel; (b) anode and cathode were interchanged; the 100 bp fragment, being smallest, moved farthest towards the anode to X

1. (a) The phenomenon is competitive release.

   ½ mark

2. Definition: a species whose distribution is restricted to a small area, because a competitively superior species is present, expands its range when the superior competitor is removed. Here Chthamalus spreads into the lower zones once Balanus is removed.

   ½ mark

3. (a) OR: Gause's competitive exclusion principle: two closely related species competing for the same resources cannot coexist indefinitely, and the competitively inferior one is eventually eliminated.

   1 mark

4. (b) In some shallow South American lakes, visiting flamingos and resident fishes compete for the same food, zooplankton.

   1 mark

5. (c) Species avoid competition by resource partitioning: they choose different times for feeding or different foraging patterns.

   1 mark

6. Example: MacArthur showed that five closely related species of warblers living on the same tree avoid competition and coexist. They do this by behavioural differences in foraging, each feeding in a different part of the tree.

   1 mark

Answer: (a) Competitive release (or Gause's competitive exclusion principle); (b) flamingos and fishes competing for zooplankton in South American lakes; (c) resource partitioning, e.g. MacArthur's warblers foraging in different parts of the same tree

1. (a) Cross I gives all tall progeny (500 : 0), so at least one parent is homozygous tall: TT × tt (also possible: TT × TT or TT × Tt).

   ½ mark

2. Cross II gives tall : dwarf = 375 : 125 = 3 : 1, so both parents are heterozygous: Tt × Tt.

   ½ mark

3. (a) OR: these crosses involve a single character (height), so they are monohybrid crosses.

   1 mark

4. (b) Cross III gives 250 tall : 250 dwarf = 1 : 1, the result of a test cross (Tt × tt).

   1 mark

5. Importance: a test cross crosses an organism showing the dominant phenotype with the homozygous recessive parent. It is used to find out whether the dominant organism is homozygous (TT) or heterozygous (Tt).

   1 mark

6. (c) In Cross I only tall plants appear, and in Cross II tall plants outnumber dwarf plants 3 : 1. So the tall trait is dominant over the dwarf trait, and dwarfness is expressed only in the homozygous (tt) condition.

   ½ mark

7. Principle illustrated: Law of Dominance.

   ½ mark

Answer: (a) Cross I: TT × tt (or TT × TT/Tt); Cross II: Tt × Tt, or the crosses are monohybrid crosses; (b) test cross, used to find the genotype of an organism with the dominant phenotype; (c) tall is dominant over dwarf, Law of Dominance

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

   ½ mark

2. The haploid nucleus of the sperm fuses with the haploid nucleus of the ovum, forming a diploid zygote.

   ½ mark

3. The zygote moves through the isthmus of the oviduct towards the uterus.

   ½ mark

4. On the way it undergoes mitotic divisions (cleavage), forming 2, 4, 8 and 16 daughter cells called blastomeres.

   ½ mark

5. The embryo with 8–16 blastomeres is called a morula.

   ½ mark

6. The morula keeps dividing and transforms into a blastocyst as it moves into the uterus.

   ½ mark

7. The blastomeres of the blastocyst are arranged into an outer layer, the trophoblast, and an inner cell mass attached to it.

   ½ mark

8. The trophoblast attaches to the endometrium. The uterine cells then divide rapidly and cover the blastocyst, so it becomes embedded in the endometrium. This is implantation, which leads to pregnancy.

   ½ mark

9. Site of fertilization: the ampullary region of the fallopian tube (ampullary–isthmic junction).

   1 mark

Answer: Sperm entry, completion of meiosis II (ootid), fusion of nuclei (zygote), cleavage while moving through the isthmus (blastomeres), morula (8–16 cells), blastocyst (trophoblast + inner cell mass), attachment of trophoblast to endometrium and embedding = implantation. Fertilization occurs in the ampullary region of the fallopian tube.

1. (i) Case I: the sperms are normal but cannot reach the ejaculate because the epididymis–vas deferens link is missing. Sperms can be retrieved and used in in vitro fertilization followed by ZIFT (embryo up to 8 blastomeres transferred into the fallopian tube). Alternatively, the collected semen can be used for artificial insemination / IUI. The female's reports are normal, so either is suitable.

   1 mark

2. Case II: the eggs and sperms are normal but the fallopian tube is blocked, so fertilization cannot occur in the body. Suggested method: IVF followed by IUT (intra-uterine transfer of an embryo with more than 8 blastomeres into the uterus), which bypasses the tube.

   1 mark

3. Case III: the male has a low sperm count with poor motility. Suggested method: ICSI (intra-cytoplasmic sperm injection, i.e. a sperm injected directly into the ovum in the laboratory) or artificial insemination / IUI with the husband's semen.

   1 mark

4. (ii) Copper-releasing IUDs: CuT and Multiload 375 (also Cu7).

   1 mark

5. They are effective because (1) they increase phagocytosis of sperms within the uterus, and (2) the Cu ions released suppress sperm motility and the fertilising capacity of sperms.

   1 mark

Answer: (i) Case I: IVF + ZIFT (or IUI/AI); Case II: IVF + IUT; Case III: ICSI (or AI/IUI); (ii) CuT, Multiload 375 (Cu7); they increase phagocytosis of sperms, and Cu ions suppress sperm motility and fertilising capacity

1. (i) Finding 1: X-ray diffraction data of DNA, produced by Maurice Wilkins and Rosalind Franklin.

   1 mark

2. Finding 2: Chargaff's rules, by Erwin Chargaff: in double-stranded DNA, the ratios A/T and G/C are constant and equal to 1.

   1 mark

3. (ii) Feature 1: DNA is made of two polynucleotide chains. The sugar–phosphate backbone lies outside and the nitrogen bases project inwards.

   1 mark

4. Feature 2: the two chains have antiparallel polarity (5'→3' and 3'→5').

   1 mark

5. Feature 3: bases on the two strands pair by hydrogen bonds: A pairs with T by two H-bonds, and G pairs with C by three H-bonds. Purines always pair with pyrimidines, keeping the distance between the strands uniform. (Other accepted features: the chains are coiled in a right-handed helix with a pitch of 3.4 nm, about 10 bp per turn and 0.34 nm between base pairs; base-pair planes stack over one another.)

   1 mark

Answer: (i) X-ray diffraction data (Wilkins and Franklin) and the base-equivalence rule A/T = G/C = 1 (Chargaff); (ii) two chains with an outer sugar–phosphate backbone; antiparallel polarity; complementary base pairing A=T, G≡C by H-bonds (also right-handed coil, 3.4 nm pitch, 10 bp per turn)

1. (i) Template strand. It is complementary to the coding strand and has the opposite polarity:

   3' – TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT AAT – 5'

   1 mark

2. (i) mRNA. The mRNA is copied from the template strand, so it has the same sequence and polarity as the coding strand, with U in place of T:

   5' – AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA – 3'

   1 mark

3. (ii)(1) In a bacterium. Prokaryotic genes are not split, so there is no splicing. The transcript itself works as mRNA:

   5' – AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA – 3'

   1 mark

4. (ii)(2) In humans. The hnRNA is processed in three ways:

   • Splicing: all three GUA introns are removed and the exons are joined.
   • Capping: methyl guanosine triphosphate (m7Gppp) is added at the 5' end.
   • Tailing: a poly-A tail is added at the 3' end.

   Mature mRNA: 5' – m7Gppp – AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA – poly A tail – 3'

   1 mark

5. (iii) The mature human mRNA has 10 codons after the start codon AUG: AUG, ACC, UUU, UCU, GUG, CCC, CUU, CAG, GCA, UUA. Each codon codes for one amino acid, so the polypeptide will have 10 amino acids.

   1 mark

Answer: (i) Template 3'-TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT AAT-5'; mRNA 5'-AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA-3'. (ii)(1) Bacterium: same mRNA, unchanged; (2) Humans: 5'-m7Gppp-AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA-poly A-3'. (iii) 10 amino acids

1. (i) Plasmodium falciparum causes malignant malaria, the most serious and often fatal form.

   ½ mark

2. (ii) Host. The sexual stages are completed in the female Anopheles mosquito.

   ½ mark

3. (ii) Changes in the mosquito.

   • When the mosquito bites an infected person, it takes up gametocytes along with the blood.
   • In the mosquito's gut the gametes form and fuse (fertilisation), and the parasite develops further, producing sporozoites.
   • The sporozoites migrate to the mosquito's salivary glands. From there they are injected into a new human host at the next bite.
   1½ marks

4. (iii) Sporozoites entering the blood first reach the liver cells and multiply there. They then attack red blood cells and multiply asexually inside them. The RBCs rupture and release a toxic substance, haemozoin. Haemozoin causes chills and high fever that recur every 3–4 days.

   1½ marks

5. (iv) Any two preventive measures:

   • Avoid stagnant water in and around homes, and clean water coolers regularly, so that mosquitoes cannot breed.
   • Use mosquito nets and wire mesh on doors and windows.
   • Introduce larvivorous fish such as Gambusia into ponds.
   • Spray insecticides in ditches and drains.
   1 mark

Answer: (i) Plasmodium falciparum; (ii) female Anopheles mosquito – gametocytes taken up, fertilisation and development in the gut, sporozoites move to the salivary glands; (iii) multiply in liver cells, then in RBCs; RBCs rupture releasing haemozoin, causing chills and recurring high fever; (iv) e.g. remove stagnant water, use mosquito nets / Gambusia fish

1. (i) Bio-fertilisers are living organisms that enrich the nutrient quality of the soil. The main sources are bacteria, fungi and cyanobacteria.

   1 mark

2. (ii) Bacteria.

   • Rhizobium forms a symbiotic association with the root nodules of leguminous plants and fixes atmospheric nitrogen into organic forms the plant can use.
   • Azotobacter and Azospirillum fix nitrogen while living freely in the soil.
   1 mark

3. (ii) Fungi. The genus Glomus forms mycorrhiza with plant roots. The fungus absorbs phosphorus from the soil and passes it to the plant. Such plants also resist root-borne pathogens, tolerate salinity and drought, and show better growth.

   1 mark

4. (ii) Cyanobacteria. Anabaena, Nostoc and Oscillatoria are autotrophic microbes that fix atmospheric nitrogen. They are important bio-fertilisers in paddy fields and also add organic matter to the soil.

   1 mark

5. (iii) Any two advantages over chemical fertilisers:

   • They do not pollute the soil or groundwater.
   • They conserve beneficial soil microbes.
   • They do not contaminate food or harm human health.
   1 mark

Answer: (i) Organisms that enrich soil nutrient quality; (ii) bacteria (Rhizobium symbiotic, Azotobacter free-living N-fixation), fungi (Glomus mycorrhiza – phosphorus absorption), cyanobacteria (Anabaena/Nostoc – N-fixation in paddy fields); (iii) eco-friendly/no pollution; conserve soil microbes and are safe for health