CBSE Class 12 Biology: 2024 board paper, set 57/1/1

1. Maize is monoecious: male (tassel) and female (cob) flowers occur on the same plant, so pollen can reach another flower of the same plant (geitonogamy) or a different plant (xenogamy).

   1 mark

2. Papaya, date palm and spinach are dioecious, so only xenogamy is possible in them.

Answer: (C) Maize

1. The placenta secretes hCG, which keeps the corpus luteum active so that it continues secreting progesterone. This maintains the endometrium and therefore the pregnancy.

   1 mark

Answer: (B) Human Chorionic Gonadotropin

1. Ovulation occurs around day 14. The fertile period, about day 10 to day 17, is when the chance of fertilisation is highest, so the couple should avoid coitus during these days.

   1 mark

Answer: (C) Day 10 to 17

1. 44 + XXY is Klinefelter's syndrome. The person is a sterile male showing gynaecomastia, not an overall female, so the match in (B) is incorrect.

   1 mark

2. The other options are correct matches: 45 + XX (trisomy 21) is Down's syndrome, 44 + XO is Turner's syndrome, and 44 + XY is a normal male.

Answer: (B)

1. With 12 different bases, a codon of n bases gives $12^n$ combinations. For n = 1 this is 12, which is less than 96. For n = 2 it is 144, which is at least 96.

   1 mark

2. Therefore the minimum codon length is 2 bases.

Answer: (C) 2

1. In a nucleotide, the nitrogenous base is attached to the 1' carbon of the pentose sugar by an N-glycosidic linkage. The dotted S–B line therefore represents this bond.

   1 mark

Answer: (C) N-glycosidic linkage

1. In allergy, IgE antibodies trigger mast cells to release histamine and serotonin, which cause the allergic symptoms.

   1 mark

Answer: (B) Allergy

1. Lactobacillus (LAB) obtains energy from organic matter such as the lactose in milk, so it is a heterotrophic bacterium.

   1 mark

Answer: (D) Heterotrophic bacteria

1. Transgenic mice are being developed for testing vaccine safety, including that of the polio vaccine, before it is used on humans.

   1 mark

Answer: (D) Mice

1. Hind II always cuts DNA at a specific recognition sequence of six base pairs.

   1 mark

Answer: (A) Six base pairs

1. Golden rice is engineered to make β-carotene, the precursor of vitamin A, which also gives the grains their golden colour.

   1 mark

Answer: (D) High vitamin-A content

1. The rate of formation of new organic matter by consumers is secondary productivity.

   ½ mark

2. The biomass available to herbivores and decomposers is net primary productivity (GPP minus respiration loss).

   ½ mark

Answer: (C)

1. The assertion is true: legal adoption is permitted in India and is one of the best options for childless couples.

   ½ mark

2. The reason is false: emotional, religious and social factors do act as deterrents to adoption.

   ½ mark

Answer: (C)

1. Linked genes lie on the same chromosome, so they do not assort independently. They are inherited together, giving mostly parental combinations.

   ½ mark

2. Because there is no independent assortment, the 9:3:3:1 ratio is not obtained. Both statements are true, and R correctly explains A.

   ½ mark

Answer: (A)

1. The assertion is false: Agrobacterium tumefaciens infects dicot plants, not monocots.

   ½ mark

2. The reason is true: it transfers its T-DNA into plant cells and causes tumours (crown gall).

   ½ mark

Answer: (D)

1. The assertion is true: GEAC was set up to decide the validity of GM research and the safety of introducing GM organisms for public use.

   ½ mark

2. The reason is false: GM organisms can have unpredictable effects when introduced into an ecosystem, which is exactly why regulation is needed.

   ½ mark

Answer: (C)

1. A – Pericarp is the wall of the fruit, which develops from the wall of the ripened ovary.

   ½ mark

2. B – The single cotyledon in the seed of the grass family (e.g. maize, wheat) is called the scutellum.

   ½ mark

3. C – The embryonal axis bears the radicle (root tip) and the plumule (shoot tip). The portion above the cotyledon is the epicotyl and the portion below it is the hypocotyl.

   ½ mark

4. D – Remains of the nucellus that persist in some seeds (e.g. black pepper, beet) form the perisperm.

   ½ mark

Answer: A – Wall of fruit (ripened ovary wall); B – Scutellum; C – Radicle/Plumule (epicotyl, hypocotyl); D – Perisperm

1. The naturalist is Jean-Baptiste Lamarck.

   ½ mark

2. Lamarck held that evolution of life forms is driven by the use and disuse of organs: organs that are used more get strengthened or enlarged, and those not used weaken.

   ½ mark

3. In his example, the ancestors of giraffes stretched their necks repeatedly to reach and forage leaves on tall trees. As a result, their necks gradually elongated within their lifetime.

   ½ mark

4. This acquired character (a longer neck) was passed on to the next generation. Over many generations, the inheritance of acquired characters produced the present-day long-necked giraffes.

   ½ mark

Answer: Lamarck; use and disuse of organs, with acquired characters (elongated neck) inherited over generations

1. Step 1: Ova are collected from the wife or a female donor, and sperms from the husband or a male donor.

   ½ mark

2. Step 2: The ova and sperms are induced to fuse and form a zygote under simulated conditions in the laboratory (in vitro fertilisation).

   ½ mark

3. Step 3: The zygote or early embryo is transferred into the female's body (fallopian tube by ZIFT, or uterus by IUT) for further development.

   ½ mark

4. It is called a test tube baby programme because fertilisation and early embryo formation take place outside the body, in the laboratory (in vitro).

   ½ mark

Answer: Collect gametes, fertilise in vitro to form zygote, transfer zygote/embryo into female; called test tube baby because fertilisation occurs in the laboratory (in vitro).

1. (i) The doctor injected anti-venom (antitoxin), which contains readymade, preformed antibodies against the snake venom.

   ½ mark

2. (i) Snake venom acts very fast, so a quick immune response is needed. There is no time for the body to produce its own antibodies, so preformed antibodies that neutralise the toxin immediately are given.

   1 mark

3. (ii) The immunity provided is passive immunity.

   ½ mark

Answer: (i) Anti-venom/antitoxin with preformed antibodies, given because a quick response is needed; (ii) Passive immunity

1. Organic farmers believe that complete eradication of pests is not only impossible but also undesirable. Their aim is to keep pests at manageable levels through a balanced, holistic system.

   1 mark

2. Reason: Beneficial predatory and parasitic insects depend on the pests as food or hosts, so they cannot survive without them. Organic farmers also hold that biodiversity furthers health: the greater the biodiversity, the more sustainable the ecosystem.

   1 mark

Answer: Complete eradication is undesirable because beneficial predators and parasites depend on pests as food/hosts; biodiversity keeps the ecosystem healthy and sustainable.

1. Level A (primary consumers): zooplankton, or small fishes feeding on phytoplankton.

   ½ mark

2. Level B (primary producers): phytoplankton.

   ½ mark

3. Type of pyramid: inverted pyramid of biomass, because the small standing crop of phytoplankton supports a larger biomass of consumers.

   ½ mark

4. Ecosystem: sea (aquatic) ecosystem.

   ½ mark

Answer: A – Zooplankton; B – Phytoplankton; inverted pyramid of biomass; sea (aquatic) ecosystem

1. Oral contraceptive pills contain progestogens alone, or a progestogen–estrogen combination, in small doses.

   ½ mark

2. Mode of action 1: they inhibit ovulation.

   ½ mark

3. Mode of action 2: they inhibit implantation.

   ½ mark

4. Mode of action 3: they alter the quality of cervical mucus, which prevents or retards the entry of sperms.

   ½ mark

5. Schedule: the pills are taken daily for 21 days, starting preferably within the first five days of the menstrual cycle.

   ½ mark

6. After a gap of 7 days (during which menstruation occurs), the course is repeated in the same pattern for as long as the female wishes to prevent conception. (Saheli, a non-steroidal pill, is taken once a week.)

   ½ mark

Answer: Progestogen/progestogen–estrogen pills inhibit ovulation and implantation and alter cervical mucus; taken daily for 21 days from within the first 5 days of the cycle, then a 7-day gap, repeated.

1. (a) DNA fingerprinting is based on repetitive DNA / satellite DNA, particularly VNTRs (Variable Number of Tandem Repeats).

   1 mark

2. (a) Characteristics (any two):

   • It does not code for any protein.
   • It forms a large portion of the human genome.
   • It shows a high degree of polymorphism, which makes it unique to each individual.
   1 mark

3. (b) Applications (any two):

   • Forensic investigations, i.e. identifying criminals.
   • Paternity (parentage) testing.
   • Determining population and genetic diversity, and studying evolution.
   1 mark

Answer: (a) Satellite (repetitive) DNA/VNTR – non-coding, large part of genome, highly polymorphic; (b) forensics, paternity testing (also population/genetic diversity studies)

1. (a) In 1953, S.L. Miller created early-earth conditions in a closed flask containing CH${}_4$, H${}_2$, NH${}_3$ and water vapour at 800°C, and passed electric discharges through it. Amino acids were formed.

   ½ mark

2. (a) Significance: the experiment showed that organic molecules can form from inorganic molecules under primitive earth conditions. It thus supported the theory of chemical evolution of life.

   1 mark

3. (a) The hypothesis was proposed by Oparin and Haldane.

   1 mark

4. (b) Analysis of meteorites revealed similar organic compounds. This indicates that similar chemical processes occur elsewhere in space, which supports the hypothesis.

   ½ mark

Answer: (a) Showed formation of amino acids from inorganic gases, supporting chemical evolution; Oparin and Haldane; (b) meteorites contain similar compounds, so similar processes occur in space

1. A (pneumonia symptoms): fever, chills, cough, headache; in severe cases the lips and fingernails turn grey to bluish.

   ½ mark

2. B: Salmonella typhi.

   ½ mark

3. C (common cold symptoms): nasal congestion and discharge, sore throat, hoarseness, cough, headache, tiredness.

   ½ mark

4. D: fungi of the genera Microsporum, Trichophyton or Epidermophyton.

   ½ mark

5. E (ascariasis symptoms): internal bleeding, muscular pain, fever, anaemia and blockage of the intestinal passage.

   ½ mark

6. F: amoebiasis (amoebic dysentery).

   ½ mark

Answer: A – fever, chills, cough, headache, bluish lips/nails; B – Salmonella typhi; C – nasal congestion, sore throat, cough; D – Microsporum/Trichophyton/Epidermophyton; E – internal bleeding, anaemia, intestinal blockage; F – Amoebiasis

1. No. The son has inherited colour blindness from his mother, while the daughter has inherited it from both mother and father.

   ½ mark

2. Colour blindness is an X-linked recessive trait. A son receives his X chromosome only from his mother and his Y from his father, so his defective X ($X^c$) must come from the mother. A colour-blind daughter must be $X^c{X}^c$: one $X^c$ from the father and one from the mother. A normal-visioned mother with a colour-blind child must therefore be a carrier.

   ½ mark

3. Parents: Mother $X^{c}X$ (normal, carrier) × Father $X^{c}Y$ (colour blind).

   1 mark

4. Gametes: mother gives $X^c$ or $X$; father gives $X^c$ or $Y$.

   	$X^c$	$Y$
   $X^c$	$X^c{X}^c$ colour-blind daughter	$X^{c}Y$ colour-blind son
   $X$	$X^{c}X$ carrier daughter	$XY$ normal son

   The colour-blind son ($X^{c}Y$) received $X^c$ from the mother, and the colour-blind daughter ($X^c{X}^c$) received $X^c$ from each parent.

   1 mark

Answer: No – the son inherited it from the mother (carrier $X^{c}X$); the daughter inherited it from both parents. Cross: $X^{c}X\times X^{c}Y$.

1. (i) Transgenic animals are animals whose DNA has been manipulated so that they possess and express an extra (foreign) gene.

   1 mark

2. (ii) The first transgenic cow was Rosie (1997).

   1 mark

3. (ii) Importance: Rosie produced milk enriched with the human protein alpha-lactalbumin (2.4 g per litre). This milk is nutritionally more balanced for human babies than natural cow milk.

   1 mark

Answer: (i) Animals with manipulated DNA carrying and expressing a foreign gene; (ii) Rosie – produced human alpha-lactalbumin-enriched milk, more balanced for human babies

1. (i) EcoRI is isolated from Escherichia coli strain RY13. E stands for the genus Escherichia.

   ½ mark

2. (i) co stands for the species coli.

   ½ mark

3. (i) R stands for the strain RY13.

   ½ mark

4. (i) I (Roman numeral) shows the order in which the enzyme was isolated from that strain; EcoRI was the first.

   ½ mark

5. (ii) EcoRI recognises the palindrome
   $5^{\prime }$–G A A T T C–$3^{\prime }$
   $3^{\prime }$–C T T A A G–$5^{\prime }$
   and cuts each strand between G and A.

   ½ mark

6. (ii) Products:
   $5^{\prime }$–G AATTC–$3^{\prime }$
   $3^{\prime }$–CTTAA G–$5^{\prime }$
   This leaves single-stranded overhanging sticky ends (AATT).

   ½ mark

Answer: (i) E – Escherichia (genus), co – coli (species), R – strain RY13, I – first enzyme isolated from that strain; (ii) EcoRI cuts GAATTC between G and A, giving sticky ends

1. (a) The Mediterranean orchid Ophrys uses sexual deceit to get pollinated by one specific species of bee.

   ½ mark

2. (a) One petal of its flower closely resembles the female bee of that species in size, colour and markings.

   ½ mark

3. (a) The male bee is attracted to what it takes to be a female and pseudocopulates with the flower. In doing so, it gets dusted with pollen.

   ½ mark

4. (a) When the same bee pseudocopulates with another flower, it transfers the pollen to it and pollinates it.

   ½ mark

5. (b) Co-evolution: the flower and its pollinator evolve together. If the colour pattern of the female bee changes even slightly during evolution, pollination success falls unless the orchid's petal also evolves to keep resembling the female bee.

   1 mark

Answer: (a) Ophrys petal mimics female bee, so male bee pseudocopulates, picks up pollen and transfers it to another flower; (b) if female bee's pattern changes, the flower must co-evolve to keep resembling her

1. (a) [First alternative] Region III represents tropical latitudes, so species richness is maximum there. These regions have remained relatively undisturbed for millions of years and had a long evolutionary time for species diversification. They are also less seasonal, with a more constant and predictable environment, and receive more solar energy, which gives higher productivity.

   1 mark

2. (a) [Second alternative] Region I represents temperate (higher-latitude) regions, so species richness is minimum there. These regions were subjected to frequent glaciations in the past and got less evolutionary time for diversification. They are more seasonal and less predictable, and receive less solar energy, so productivity and diversity are lower.

   1 mark

3. (b) This uneven distribution is called the latitudinal gradient in diversity: species diversity decreases from the equator towards the poles.

   1 mark

4. (c) Conventional taxonomic methods are based on morphology. They are not suitable for identifying microbial (prokaryotic) species, which look alike.

   1 mark

5. (c) Many prokaryotic species cannot be cultured under laboratory conditions, so they cannot be studied and counted. Their diversity is therefore not estimated, though biochemical and molecular methods suggest it may run into millions.

   1 mark

Answer: (a) Region III – tropics: long undisturbed evolutionary time, constant environment, more solar energy; Region I – temperate: glaciations, seasonal, less solar energy; (b) Latitudinal gradient; (c) conventional taxonomy unsuitable for microbes and many are not culturable

1. (a) Colony 4 is the transformed colony with recombinant DNA. It grows on ampicillin (plate M) but fails to grow on the tetracycline-containing plate N.

   ½ mark

2. (a) Justification: the foreign DNA was inserted within the tetracycline resistance gene. This inactivated the gene (insertional inactivation), so recombinants lose tetracycline resistance but keep ampicillin resistance.

   ½ mark

3. (b) The ampicillin and tetracycline resistance genes are selectable markers. The sites within them where foreign DNA is inserted are restriction (recognition) sites, e.g. BamHI in tet${}^R$ and PstI in amp${}^R$ of pBR322.

   1 mark

4. (b) Role: selectable markers help identify and select transformants from non-transformants. Insertion of foreign DNA into one marker inactivates it, which helps distinguish recombinants from non-recombinants.

   1 mark

5. (c) [First alternative] Two enzymes: restriction endonuclease (cuts DNA at specific sites) and DNA ligase (joins DNA fragments). Taq DNA polymerase, used in PCR, is also important.

   1 mark

6. (c) [Second alternative] The recombinant DNA is inserted within the coding sequence of β-galactosidase, which inactivates it. With a chromogenic substrate, non-recombinant colonies with active enzyme appear blue, while recombinant colonies appear colourless (white). This identifies recombinants.

   1 mark

Answer: (a) Colony 4 – grows on ampicillin but not tetracycline due to insertional inactivation of tet^R; (b) selectable markers (restriction sites within them) – help select transformants/recombinants; (c) restriction endonuclease and DNA ligase / inactivated β-galactosidase gives colourless recombinant colonies

1. (i) In a young anther, the sporogenous tissue occupies the centre of each microsporangium. Each cell of this tissue is a potential microspore mother cell (pollen mother cell).

   ½ mark

2. (i) Each microspore mother cell undergoes meiosis to form four haploid microspores. This process is called microsporogenesis.

   ½ mark

3. (i) The microspores stay arranged in a cluster of four cells, the microspore tetrad.

   ½ mark

4. (i) As the anther matures and dehydrates, the microspores dissociate and develop into pollen grains, the male gametophytes.

   ½ mark

5. (i) Each pollen grain undergoes an asymmetric mitotic division to form a large vegetative cell and a small generative cell. In about 40% of angiosperms, pollen is shed at this 2-celled stage. In the rest, the generative cell divides mitotically to give two male gametes, so the pollen is shed at the 3-celled stage.

   ½ mark

6. (ii) Diagram (described): a rounded pollen grain with a two-layered wall. Label 1: exine, the thick outer layer of sporopollenin.

   ½ mark

7. (ii) Label 2: germ pore, a gap in the exine.

   ½ mark

8. (ii) Label 3: intine, the thin inner layer of cellulose and pectin.

   ½ mark

9. (ii) Label 4: vegetative cell with its large, irregular vegetative nucleus.

   ½ mark

10. (ii) Label 5: two male gametes (sperm cells), small spindle-shaped cells with dense cytoplasm, lying in the cytoplasm of the vegetative cell.

    ½ mark

Answer: (i) Microspore mother cell → meiosis → microspore tetrad → pollen grains → asymmetric mitosis forms vegetative and generative cells; generative cell → 2 male gametes; (ii) labelled 3-celled pollen: exine, intine, germ pore, vegetative nucleus/cell, two male gametes

1. (i) Diagram (described): an oval section of the ovary with stroma, showing stages of follicle development around the periphery. Label 1: blood vessels, entering at the hilum and running through the medulla.

   ½ mark

2. (i) Label 2: primary follicle, a primary oocyte surrounded by a single layer of granulosa cells, near the surface.

   ½ mark

3. (i) Label 3: tertiary follicle, a larger follicle with a fluid-filled cavity (antrum) and theca layers.

   ½ mark

4. (i) Label 4: ovum, the secondary oocyte released from the ruptured Graafian follicle.

   ½ mark

5. (ii) Primary follicles are formed at the embryonic (foetal) stage. Oogonia formed in the foetal ovary enter meiosis I and are arrested as primary oocytes surrounded by granulosa cells.

   ½ mark

6. (iii) Both LH and FSH reach their peak in the middle of the cycle, around day 14.

   ½ mark

7. (iii) The rapid secretion of LH to its maximum level, the LH surge, induces rupture of the Graafian follicle and release of the ovum. This is ovulation.

   ½ mark

8. (iii) Ovulation is followed by the luteal phase, in which the remaining parts of the Graafian follicle transform into the corpus luteum.

   ½ mark

9. (iii) The corpus luteum secretes large amounts of progesterone, which is essential for maintaining the endometrium for implantation.

   ½ mark

10. (iii) If fertilisation does not occur, the corpus luteum degenerates and progesterone falls. The endometrium then disintegrates, leading to menstruation and the start of a new cycle.

    ½ mark

Answer: (i) Labelled ovary section showing blood vessels, primary follicle, tertiary follicle, ovum; (ii) embryonic (foetal) stage; (iii) LH and FSH peak; LH surge ruptures Graafian follicle (ovulation); corpus luteum forms and secretes progesterone; without fertilisation it degenerates, endometrium breaks down and menstruation begins

1. The scientist was Frederick Griffith (1928).

   ½ mark

2. He used two strains of Streptococcus pneumoniae and injected them into mice.

   ½ mark

3. S strain: smooth, shiny colonies because of a mucous (polysaccharide) coat; virulent, causing pneumonia.

   ½ mark

4. R strain: rough colonies without the mucous coat; non-virulent.

   ½ mark

5. Experiment:

   • Live S strain injected → mice die.
   • Live R strain injected → mice live.
   ½ mark
6. • Heat-killed S strain injected → mice live.
   ½ mark
7. • Heat-killed S strain + live R strain injected → mice die, and live S-strain bacteria are recovered from the dead mice.
   ½ mark

8. Conclusion 1: the R-strain bacteria had been transformed by the heat-killed S-strain bacteria.

   ½ mark

9. Conclusion 2: some 'transforming principle' passed from the heat-killed S strain to the R strain and enabled it to make a smooth polysaccharide coat and become virulent. This must be due to the transfer of the genetic material.

   ½ mark

10. Conclusion 3 (stability): the heat that killed the S bacteria did not destroy the properties of their genetic material. The genetic material is therefore stable and kept its ability to transform the R cells.

    ½ mark

Answer: Frederick Griffith – transformation experiment with S (virulent) and R (non-virulent) Streptococcus pneumoniae in mice; heat-killed S + live R killed mice; conclusion: transforming principle (genetic material) passed from heat-killed S to R and survived heat, showing its stability

1. Possible genotypes. Tall (T) is dominant over dwarf (t), and violet (V) is dominant over white (v). A tall, violet plant can therefore have any one of four genotypes: TTVV, TTVv, TtVV, TtVv. Selfing the plant and studying its progeny tells us which one it is.

   2 marks

2. Case I: TTVV × TTVV (selfing)

   Gametes: TV only, so all progeny are TTVV.

   • Genotypic ratio: all TTVV (1)
   • Phenotypic ratio: all tall violet (1)

   If no segregation occurs, the plant is homozygous for both traits (TTVV).

3. Case II: TtVV × TtVV

   Gametes: TV, tV

   ♀ \ ♂	TV	tV
   TV	TTVV (tall violet)	TtVV (tall violet)
   tV	TtVV (tall violet)	ttVV (dwarf violet)

   • Genotypic ratio: TTVV : TtVV : ttVV = 1 : 2 : 1
   • Phenotypic ratio: tall violet : dwarf violet = 3 : 1

   Dwarf plants appear but no white-flowered ones, so the plant is TtVV.

   1 mark

4. Case III: TtVv × TtVv

   Gametes: TV, Tv, tV, tv

   ♀ \ ♂	TV	Tv	tV	tv
   TV	TTVV	TTVv	TtVV	TtVv
   Tv	TTVv	TTvv	TtVv	Ttvv
   tV	TtVV	TtVv	ttVV	ttVv
   tv	TtVv	Ttvv	ttVv	ttvv

   • Phenotypic ratio: tall violet : tall white : dwarf violet : dwarf white = 9 : 3 : 3 : 1
   • Genotypic ratio: TTVV : TTVv : TtVV : TtVv : TTvv : Ttvv : ttVV : ttVv : ttvv = 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1

   All four phenotypes appear, so the plant is heterozygous for both traits (TtVv).

   1 mark

5. Case IV: TTVv × TTVv

   Gametes: TV, Tv

   ♀ \ ♂	TV	Tv
   TV	TTVV (tall violet)	TTVv (tall violet)
   Tv	TTVv (tall violet)	TTvv (tall white)

   • Genotypic ratio: TTVV : TTVv : TTvv = 1 : 2 : 1
   • Phenotypic ratio: tall violet : tall white = 3 : 1

   White-flowered plants appear but no dwarf ones, so the plant is TTVv.

   1 mark

Answer:

The plant can be one of four genotypes, each identified by the progeny of selfing:

• TTVV: all tall violet (1 genotype).
• TtVV: genotypes TTVV : TtVV : ttVV = 1 : 2 : 1; phenotypes tall violet : dwarf violet = 3 : 1.
• TTVv: genotypes TTVV : TTVv : TTvv = 1 : 2 : 1; phenotypes tall violet : tall white = 3 : 1.
• TtVv: genotypes 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1; phenotypes 9 : 3 : 3 : 1.

1. (i) The property is contact inhibition.

   ½ mark

2. (i) In normal cells, contact with neighbouring cells inhibits further uncontrolled growth and division. Cancer cells have lost this property, so they keep dividing and form masses of cells called tumours.

   1 mark

3. (ii) These genes are cellular oncogenes (c-onc), also called proto-oncogenes.

   ½ mark

4. (ii) When they are activated under certain conditions, for example by mutation caused by carcinogens or by oncogenic viruses, they cause oncogenic transformation of the cells, which then become cancerous.

   ½ mark

5. (iii)(1) Biopsy and histopathology: a piece of the suspected tissue is cut into thin sections, stained, and examined under a microscope by a pathologist. Abnormal, increased cell numbers or growth confirm cancer.

   1 mark

6. (iii)(2) MRI: it detects cancers of internal organs. It uses strong magnetic fields and non-ionising radiations to accurately detect pathological and physiological changes in living tissue.

   1½ marks

Answer: (i) Contact inhibition, which is lost in cancer cells, so they divide uncontrollably and form tumours.
(ii) Proto-oncogenes (cellular oncogenes); when activated (for example by mutation or carcinogens) they cause oncogenic transformation.
(iii)(1) Biopsy and histopathology: stained thin sections of the suspected tissue are examined microscopically for abnormal cell growth.
(iii)(2) MRI: uses strong magnetic fields and non-ionising radiation to detect changes in internal organs and tissues.

1. (i)(1) The primary effluent is constantly agitated and aerated in the aeration tanks. This allows vigorous growth of useful aerobic microbes into flocs (bacteria associated with fungal filaments). The flocs consume the major part of the organic matter in the effluent, which significantly reduces its BOD.

   1 mark

2. (i)(2) The sediment is called activated sludge.

   ½ mark

3. (i)(2) Significance: a small part of the activated sludge is pumped back into the aeration tank to serve as the inoculum of microbes.

   ½ mark

4. (i)(3) Once the BOD has been reduced sufficiently, the effluent is held in the settling tank so that the bacterial flocs sediment as activated sludge. The clear supernatant is then released into water bodies. The major part of the sludge is pumped into anaerobic sludge digesters, where anaerobic bacteria digest it and produce biogas (CH₄, H₂S, CO₂).

   1 mark

5. (ii) Rhizobium (bacterium, kingdom Monera): it lives symbiotically in the root nodules of leguminous plants. It fixes atmospheric nitrogen into organic forms that the plant uses as a nutrient.

   1 mark

6. (ii) Glomus (fungus, kingdom Fungi): it forms a mycorrhizal symbiotic association with the roots of higher plants. It absorbs phosphorus from the soil and passes it to the plant.

   (Alternative: cyanobacteria such as Anabaena or Nostoc fix atmospheric nitrogen and add organic matter to soil, for example in paddy fields. Note that cyanobacteria are also Monera, so they cannot be paired with Rhizobium here.)

   1 mark

Answer: (i)(1) For growth of aerobic microbial flocs that consume organic matter and reduce BOD.
(i)(2) Activated sludge; a small part is used as inoculum in the aeration tank.
(i)(3) Bacterial flocs settle as sludge, and the clear effluent is released into water bodies; most of the sludge goes to anaerobic digesters.
(ii) Rhizobium (bacterium) fixes atmospheric nitrogen in legume root nodules; Glomus (fungus) absorbs phosphorus for the plant as mycorrhiza.