CBSE Class 12 Biology Question Paper 2026 Solved Code 57-1-2

SECTION A

1. Match Column-I with Column-II and choose the correct option : [1 Mark]
Column-I | Column-II
a. Biolistic gun | i. Bacterial cell
b. Chitinase | ii. Tumour inducing
c. Ti | iii. Animal cell
d. Ca⁺⁺ | iv. Fungal cell
| v. Plant cell
Options :
a | b | c | d
(A) i | ii | iii | iv
(B) ii | v | i | iii
(C) v | iv | ii | i
(D) v | i | iv | ii

Answer: (C) v | iv | ii | i

Teacher's Note:
a) Biolistic (gene) gun is used for plant cells, chitinase dissolves the fungal cell wall, and Ti stands for tumour inducing.
b) Bacterial cells are made competent to take up DNA by treating them with Ca⁺⁺ ions.

2. Microbes commonly used in kitchen are [1 Mark]
(A) Lactobacillus and Yeast
(B) Penicillium and Yeast
(C) Microspora and E.coli
(D) Rhizopus and Lactobacillus

Answer: (A) Lactobacillus and Yeast

Teacher's Note:
a) Lactobacillus turns milk into curd, and yeast (Saccharomyces cerevisiae) is used to make bread dough rise.
b) Penicillium is used to make an antibiotic, not household food.

3. The features of some structures of human male reproductive system are given below. Choose the correct option that matches the features with the structures. [1 Mark]
(i) It opens into the Vasa efferentia through rete testis.
(ii) It carries semen as well as urine.
(iii) It maintains temperature (2 - 2.5°C) lower than body temperature for sperm formation.
(iv) It leads to vas deferens that ascends to the abdomen.
(i) | (ii) | (iii) | (iv)
(A) Seminal vesicle | Urethra | Scrotum | Seminiferous tubule
(B) Prostrate | Scrotum | Testes | Vas Deferens
(C) Seminiferous tubule | Urethra | Scrotum | Epididymis
(D) Seminiferous tubule | Vas Deferens | Urethra | Sertoli cells

Answer: (C) Seminiferous tubule | Urethra | Scrotum | Epididymis

Teacher's Note:
a) Remember the sperm pathway: seminiferous tubules, rete testis, vasa efferentia, epididymis, vas deferens, urethra.
b) The urethra is the common passage for both urine and semen in males.

4. Exploration of molecular, genetic and species level diversity for gaining products of economic importance is called [1 Mark]
(A) Exploitation
(B) Bio-prospecting
(C) Bio-patenting
(D) Bio-piracy

Answer: (B) Bio-prospecting

Teacher's Note:
a) Bio-prospecting is the legal search for useful products from biodiversity.
b) Do not confuse it with bio-piracy, which is the use of bio-resources without proper permission or payment.

5. Which connective tissue connects ovary to pelvic wall and uterus ? [1 Mark]
(A) Tendons
(B) Ligaments
(C) Cartilage
(D) Bone

Answer: (B) Ligaments

Teacher's Note:
a) Each ovary is held to the pelvic wall and uterus by ligaments.
b) Tendons join muscle to bone, so they are not the answer here.

6. Which of the following is not a component of detritus food chain ? [1 Mark]
(A) Dead Leaves
(B) Bacteria
(C) Fungi
(D) Zooplankton

Answer: (D) Zooplankton

Teacher's Note:
a) The detritus food chain begins with dead organic matter and involves decomposers like bacteria and fungi.
b) Zooplankton are primary consumers of the grazing food chain in aquatic ecosystems.

7. Identify the plant in which emasculation is not required for artificial hybridization. [1 Mark]
(A) Rice
(B) Wheat
(C) Pea
(D) Papaya

Answer: (D) Papaya

Teacher's Note:
a) Papaya is dioecious (male and female flowers on separate plants), so there are no anthers to remove from the female flower.
b) Emasculation is needed only in bisexual flowers such as rice, wheat and pea.

8. Pomato was produced by fusing protoplasts of [1 Mark]
(A) Tomato and Potato
(B) Pomegranate and Tomato
(C) Pomegranate and Potato
(D) Pomegranate, Potato and Tomato

Answer: (A) Tomato and Potato

Teacher's Note:
a) Pomato is a somatic hybrid made by fusing protoplasts of potato and tomato.
b) The name itself is a mix of "potato" and "tomato", which is an easy way to remember it.

9. Which of the following sacred groves is found in Meghalaya ? [1 Mark]
(A) Jaintia hills
(B) Bastar
(C) Chanda
(D) Sarguja

Answer: (A) Jaintia hills

Teacher's Note:
a) Khasi and Jaintia hills in Meghalaya have well-known sacred groves.
b) Bastar, Chanda and Sarguja are sacred grove areas of Madhya Pradesh and Chhattisgarh.

10. Messelson and Stahl's experiment carried out centrifugation in CsCl₂ density gradient to separate : [1 Mark]
(A) DNA from RNA
(B) DNA from protein
(C) RNA from protein
(D) Normal DNA from heavy DNA

Answer: (D) Normal DNA from heavy DNA

Teacher's Note:
a) Heavy DNA contained ¹⁵N and normal DNA contained ¹⁴N, so they settled at different densities.
b) This experiment proved that DNA replicates semi-conservatively.

11. Which one of the following is not a component required for Polymerase Chain Reaction (PCR) ? [1 Mark]
(A) Template DNA
(B) Primers
(C) DNA Ligase
(D) DNA Polymerase

Answer: (C) DNA Ligase

Teacher's Note:
a) PCR needs template DNA, two primers, deoxynucleotides and a thermostable DNA polymerase (Taq polymerase).
b) DNA ligase joins DNA fragments during cloning; it is not used in PCR.

12. In the following figure, two ways of pairing of two homologous pairs of chromosomes are shown. Which of the following phenomena is expressed ? [1 Mark]

[Figure: Two equally probable arrangements at Metaphase I (Possibility 1 and Possibility 2) of two homologous chromosome pairs carrying alleles A/a and B/b. Possibility 1 gives cells at Metaphase II with A-B and a-b, forming gametes AB (Combination 1) and ab (Combination 2). Possibility 2 gives cells with A-b and a-B, forming gametes Ab (Combination 3) and aB (Combination 4). The note in the figure reads: "Two equally probable arrangements at Metaphase I give rise to different chromosome combinations".]

(A) Linkage of genes
(B) Independent assortment of genes
(C) Multiple alleles
(D) Incomplete dominance

Answer: (B) Independent assortment of genes

Teacher's Note:
a) Random orientation of chromosome pairs at Metaphase I gives four types of gametes: AB, ab, Ab and aB.
b) This is the chromosomal basis of Mendel's law of independent assortment.

13. Assertion (A) : When we see stars, we are apparently peeping into the past.
Reason (R) : What we see today is an object whose emitted light started its journey millions of years back reaching our eyes now. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).

Teacher's Note:
a) Light from distant stars takes millions of years to reach us, so we see the star as it was long ago.
b) This idea is used in the chapter on evolution to explain how studying the universe is like looking into the past.

14. Assertion (A) : Flocs are masses of bacteria associated with fungal filaments in secondary treatment of sewage.
Reason (R) : Flocs help in digestion of solid waste by anaerobic respiration. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Flocs are formed in aeration tanks and consume organic matter aerobically, which reduces the BOD.
b) Anaerobic digestion happens later in anaerobic sludge digesters, not by flocs.

15. Assertion (A) : In terrestrial ecosystem much larger fraction of energy flows through grazing food chain.
Reason (R) : Grazing food chain may be connected to Detritus food chain at some levels. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (D) Assertion (A) is false, but Reason (R) is true.

Teacher's Note:
a) In terrestrial ecosystems, a much larger fraction of energy flows through the detritus food chain, not the grazing food chain.
b) The two food chains are linked because organisms of the detritus food chain can be prey for animals of the grazing food chain.

16. Assertion (A) : The milk produced by transgenic cow 'Rosie' was nutritionally more balanced product for human babies than natural cow milk.
Reason (R) : It was human protein enriched milk containing human alpha lactaglobulin. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Rosie's milk contained human alpha-lactalbumin, not "alpha lactaglobulin", so the Reason is false.
b) Read protein names in the Reason carefully; one wrong word makes the whole statement false.

SECTION B

17. (a) What is the role of LAB (Lactic Acid Bacteria) in preparation of curd ?
(b) Mention two health benefits of consuming curd. [2 Marks]

Answer:
1. (a) LAB produce acids (lactic acid) that coagulate and partially digest the milk proteins, which converts milk into curd.
2. (b) Curd has more vitamin B₁₂, which improves its nutritional quality. Also, LAB in our stomach check the growth of disease-causing microbes.

Teacher's Note:
a) The key words for (a) are "acid", "coagulate" and "partially digest milk proteins".
b) Vitamin B₁₂ and checking disease-causing microbes are the two benefits examiners look for.

18. (a) Name the muscular and glandular layers of female uterus. Mention one function of each. [2 Marks]

Answer:
1. The muscular layer is the myometrium. It shows strong contractions during delivery of the baby.
2. The glandular layer is the endometrium. It undergoes cyclical changes during the menstrual cycle (it also helps in implantation of the blastocyst and in formation of the placenta).

Teacher's Note:
a) Do not mix up the names: "myo" means muscle and "endo" means inner lining.
b) The outer thin layer is the perimetrium; it is not asked here.

OR

(b) (i) Why are STI (Sexually Transmitted Infections) not detected in time ? (2 pts.)
(ii) Write two complications caused in later stages. [2 Marks]

Answer:
1. (i) STIs are not detected in time because there are no symptoms or only mild symptoms in the early stages of infection, and because of the social stigma attached to STIs.
2. (ii) In later stages, STIs can cause pelvic inflammatory disease (PID) and infertility.

Teacher's Note:
a) Other complications accepted are abortions, still births, ectopic pregnancy and cancer of the reproductive tract.
b) Give exactly two points in each part, as asked.

19. Write the importance of following in biotechnology :
(a) Ori (Origin of Replication)
(b) Enzyme Polymerase used in Polymerase Chain Reaction [2 Marks]

Answer:
1. (a) Ori is a specific DNA sequence where replication starts. It also controls the copy number of the linked DNA.
2. (b) The thermostable DNA polymerase (Taq polymerase) extends the primers and amplifies DNA. It stays active during the high temperature used to denature double-stranded DNA.

Teacher's Note:
a) For Ori, "initiates replication" and "controls copy number" are the key phrases.
b) Mention that the polymerase is heat-stable; this is why it can be used in repeated PCR cycles.

20. (a) Given below is a pyramid found in an ecosystem, where each bar represents the standing crop available in the trophic level.
(i) Identify the kind of pyramid and with the help of an example explain the conditions where this kind of pyramid is possible in nature.
(ii) Write any two limitations of ecological pyramids. [2 Marks]

[Figure: A pyramid of two horizontal bars. The lower bar, labelled PP, is short. The upper bar, labelled PC, is much wider than the lower bar.]

Answer:
1. (i) It is an inverted pyramid of biomass. It is found in aquatic ecosystems, where a small standing crop of phytoplankton supports a large standing crop of zooplankton or fish.
2. (ii) Ecological pyramids do not take into account the same species belonging to two or more trophic levels, and they do not accommodate a food web.

Teacher's Note:
a) PP means primary producers and PC means primary consumers; the wider top bar shows the pyramid is inverted.
b) Another accepted limitation: saprophytes are given no place in ecological pyramids.

OR

(b) In an ecosystem there was loss of biodiversity due to some project in that area.
(i) How will biodiversity be affected ? (2 points)
(ii) List two major causes of loss of biodiversity. [2 Marks]

Answer:
1. (i) There will be a decline in plant production, and lowered resistance of the ecosystem to environmental disturbances such as drought.
2. (ii) Two major causes are habitat loss and fragmentation, and alien species invasion.

Teacher's Note:
a) Increased variability in processes like water use and pest and disease cycles is another accepted effect.
b) Remember the four causes as "The Evil Quartet": habitat loss, over-exploitation, alien species invasion and co-extinction.

21. (a) Analyse the following two cases of sex determination in different organisms :
Case I : Males have (XO) sex chromosomes and females have two copies of same sex chromosome (XX).
Case II : Females have two different sex chromosomes (ZW) and males have two copies of same sex chromosome (ZZ).
Identify the type of heterogamety in each case, giving one example of each. [2 Marks]

Answer:
1. Case I shows male heterogamety. Example: grasshopper (many insects).
2. Case II shows female heterogamety. Example: birds (hen).

Teacher's Note:
a) The sex that makes two different kinds of gametes is called heterogametic.
b) Give one correct example with each case to get the full 2 marks.

OR

(b) (i) Differentiate between a DNA and RNA nucleotide.
(ii) Name the two types of bonds present within a single nucleotide. [2 Marks]

Answer:
1. (i) A DNA nucleotide has deoxyribose sugar, while an RNA nucleotide has ribose sugar.
2. (ii) The two bonds are the N-glycosidic linkage (between the sugar and the nitrogenous base) and the phosphoester linkage (between the sugar and the phosphate group).

Teacher's Note:
a) Another accepted difference: DNA has the bases A, G, C, T while RNA has A, G, C, U.
b) Do not write the phosphodiester bond; it joins two nucleotides, not the parts of one nucleotide.

SECTION C

22. A farmer grew two varieties of corn crop in field A and B. He grew normal corn crops in field A and GM corn crops in field B.
He observed that corn borers attacked only in field A. To control it, spores of Bt were sprayed on field A.
(a) Name the gene in the spores responsible for control of pest.
(b) What effect will the spores of Bt have on insect pest ?
(c) How has corn plants of field B developed resistance against this pest ? Explain. [3 Marks]

Answer:
1. (a) The gene is cry IAb.
2. (b) The Bt spores are mixed with water and sprayed on field A. The insect larvae eat them, the toxin is released in the gut of the insect, and the insect gets killed.
3. (c) The GM corn plants of field B carry and express the Bt (cry) gene, so they make the Bt toxin themselves. When the pest eats the plant, the toxin kills it.

Teacher's Note:
a) The gene cry IAb controls corn borer; cry IAc and cry IIAb control cotton bollworms.
b) In (b), write all three steps: eaten by larvae, toxin released in the gut, insect killed.

23. Observe the given picture carefully. A mixture of DNA with fragments ranging from 100 base pairs to 1800 base pairs were separated by electrophoresis on agarose gel with the following arrangement :
(a) What result will be obtained in staining with ethidium bromide ? Explain with reasons.
(b) The above setup was modified as shown below and a band with 100 base pairs was obtained at X.
What changes were made to the previous design to get a band at 'X'. Why did the band appear at 'X' ? [3 Marks]

[Figure: (a) An agarose gel set-up with the positive terminal (+) at the top, next to a row of five wells, and the negative terminal (-) at the bottom. A downward arrow runs from + to -. A thick arrow points to the third well. No bands are shown. (b) The modified set-up with the negative terminal (-) at the top next to the five wells and the positive terminal (+) at the bottom. Bands are seen below the wells: two bands in lane 1, two bands in lane 3 and two bands in lane 5. One band in lane 3, marked X, lies farthest from the wells.]

Answer:
1. (a) No bands will be seen in the agarose gel. DNA fragments are negatively charged, so they will not move towards the negative end (cathode) at the bottom; they stay near the positive end (anode) at the wells.
2. (b) The positions of the positive terminal (anode) and the negative terminal (cathode) were interchanged, so the wells are now at the cathode end.
3. The band at X is the smallest DNA fragment (100 bp), so it moved farthest towards the anode through the gel.

Teacher's Note:
a) DNA always moves towards the anode (positive end); the wells must be kept at the cathode end.
b) Smaller fragments move faster and farther through the gel pores, which is the sieving effect of agarose.

24. (a) Mention the scientific name of the source plant and the part from which opioids are extracted.
(b) How are morphine and heroin related ? Mention the effect each one of them has on human body. [3 Marks]

Answer:
1. (a) The source plant is Papaver somniferum (poppy plant), and opioids are extracted from its latex.
2. (b) Heroin is formed by acetylation of morphine (heroin is diacetylmorphine).
3. Morphine is an effective sedative and pain killer, useful for patients who have undergone surgery. Heroin is a depressant and slows down body functions.

Teacher's Note:
a) Write the scientific name correctly: genus with a capital letter, species in small letters.
b) The word "acetylation" carries 1 full mark in part (b).

25. Give the schematic representation of oogenesis in human female. Indicate chromosome number also. [3 Marks]

Answer: The schematic diagram should show these stages with the chromosome number of each:
1. Fetal life: oogonia (2N, 46 chromosomes) multiply by mitosis and differentiate into primary oocytes (2N, 46 chromosomes), which begin the first meiotic division and stay arrested.
2. At puberty: the primary oocyte completes the first meiotic division (before ovulation) to form a large secondary oocyte (N, 23 chromosomes) and a small first polar body.
3. Adult reproductive life: the secondary oocyte undergoes the second meiotic division to form the ovum (egg cell or ootid, N, 23 chromosomes) and a second polar body.

Teacher's Note:
a) Marks are given for oogonia, primary oocyte, secondary oocyte and ovum, and for writing 46 and 23 at the correct stages.
b) Show the life phases (fetal life, puberty, adult life) on one side of the diagram to make it complete.

26. (a) Do all pollen grains remain viable for the same length of time ? Support your answer with two suitable examples.
(b) How are pollen grains stored in pollen banks ? [3 Marks]

Answer:
1. (a) No, all pollen grains do not remain viable for the same length of time.
2. In rice and wheat (cereals), pollen grains lose viability within 30 minutes of their release, while in some members of Rosaceae, Leguminosae and Solanaceae they remain viable for months.
3. (b) Pollen grains are stored by cryopreservation in liquid nitrogen at \( -196^{\circ}C \).

Teacher's Note:
a) Start with a clear "No" as it carries marks on its own.
b) Remember the temperature of liquid nitrogen: \( -196^{\circ}C \).

27. (a) What is pedigree analysis ? Mention its importance in human genetics. (2 pts.)
(b) Analyse the following pedigree and write the
(i) Pattern of inheritance.
(ii) Give one example of disease showing such an inheritance pattern. [3 Marks]

[Figure: A pedigree chart. Generation I: an unaffected male and an unaffected female. Generation II: their six children in order are an unaffected son, an affected son (shaded square), an unaffected daughter, an affected daughter (shaded circle) and an unaffected son; the unaffected daughter is married to an unaffected male. Generation III: this couple has an affected son (shaded square) and two unaffected daughters.]

Answer:
1. (a) Pedigree analysis is the study of the inheritance of a trait over several generations of a family.
2. Its importance is that it helps to trace the inheritance of a trait, abnormality or disease in a family.
3. (b) (i) The pattern is autosomal recessive. (ii) Example: sickle-cell anaemia.

Teacher's Note:
a) Unaffected parents having affected children, both sons and daughters, shows an autosomal recessive trait.
b) Thalassemia and phenylketonuria are also accepted examples.

28. (a) How is Hardy-Weinberg expression (p² + 2pq + q²) = 1 derived ?
(b) List any two factors that disturb the genetic equilibrium. [3 Marks]

Answer:
1. (a) Let two alleles A and a have frequencies p and q. The sum of all allelic frequencies in a population is 1, so \( p + q = 1 \).
2. The chance of allele A appearing on both chromosomes of a diploid individual is the product of the probabilities, so the frequency of AA is \( p^2 \). Similarly, the frequency of aa is \( q^2 \) and that of Aa is \( 2pq \). Hence \( (p + q)^2 = p^2 + 2pq + q^2 = 1 \).
3. (b) Two factors are gene migration (gene flow) and genetic drift.

Teacher's Note:
a) Each step (p + q = 1, \( p^2 \), \( q^2 \), \( 2pq \)) carries half a mark, so write all of them.
b) Mutation, genetic recombination and natural selection are also accepted factors.

SECTION D

29. A student performed some crosses in plants and represented the result in the form of bar graphs as shown below. Each graph displays the phenotypic proportion of the progeny. Study the graphs and answer the questions :

[Figure: A bar graph with "Crosses" (I, II, III) on the x-axis and "Number of Plants" (0 to 600) on the y-axis. Black bars show Tall and white bars show Dwarf. Cross I: 500 Tall, no Dwarf. Cross II: 375 Tall and 125 Dwarf. Cross III: 250 Tall and 250 Dwarf.]

(a) What can you infer about the genotype of parents in crosses I and II ? [1 Mark]

Answer: In cross I, the parents are TT and tt (or TT and Tt, or TT and TT), since all progeny are tall. In cross II, both parents are Tt, since tall and dwarf appear in a 3 : 1 ratio.

Teacher's Note:
a) A 3 : 1 ratio (375 : 125) always points to two heterozygous parents.
b) All-tall progeny means at least one parent must be homozygous tall (TT).

OR

(a) Which genetic cross is represented by these crosses ? [1 Mark]

Answer: These are monohybrid crosses, as only one character (plant height) is studied.

Teacher's Note:
a) Only one pair of contrasting traits (tall and dwarf) is shown, so it is a monohybrid cross.
b) A dihybrid cross would show four phenotypes, not two.

(b) Looking at bar graph of cross III, identify the type of cross performed and its importance in genetics. [2 Marks]

Answer:
1. Cross III is a test cross (Tt crossed with tt gives tall and dwarf in a 1 : 1 ratio).
2. A test cross is done to find the genotype of an organism that shows the dominant phenotype, by crossing it with the recessive parent.

Teacher's Note:
a) A 1 : 1 ratio (250 : 250) is the signature of a test cross with a heterozygous plant.
b) If the tall plant were TT, all progeny of the test cross would be tall.

(c) What conclusion can you draw from the results of bar graphs of crosses I and II ? Name the genetic principle illustrated. [1 Mark]

Answer: The results show that the tall trait is dominant over the dwarf trait. The principle illustrated is the Law of Dominance.

Teacher's Note:
a) In cross I, the dwarf trait is completely hidden, which shows dominance.
b) Write the name of the law exactly: "Law of Dominance".

30. The diagram below shows the distribution of two barnacle species, Chthamalus and Balanus on a rocky sea shore.
When Balanus is experimentally removed, Chthamalus expands its range in lower intertidal zone.

[Figure: A rocky sea shore showing barnacles Balanus balanoides and Chthamalus stellatus, with three zones between the highest tides and the lowest tides. Upper intertidal zone: "Balanus dessicates in this zone, allowing Chthamalus to thrive". Middle intertidal zone: "Where it can survive, Balanus outcompetes Chthamalus for space". Lower intertidal zone: "Balanus is subject to predation in the lower tidal zone".]

(a) Identify and define the ecological phenomenon demonstrated by this observation. [1 Mark]

Answer: The phenomenon is competitive release. A species whose distribution is restricted to a small area because of a competitively superior species expands its range when the superior competitor is removed.

Teacher's Note:
a) Naming "competitive release" and giving its definition carry half a mark each.
b) Here Balanus is the superior competitor and Chthamalus is the species released.

OR

(a) State the principle that explains the elimination of one species. [1 Mark]

Answer: Gause's competitive exclusion principle: two closely related species competing for the same resources cannot co-exist indefinitely, and the competitively inferior one will be eliminated eventually.

Teacher's Note:
a) Write the scientist's name (Gause) along with the name of the principle.
b) The key phrase is "cannot co-exist indefinitely".

(b) Two different species can compete for the same resource. Give another example of it. [1 Mark]

Answer: In some shallow South American lakes, visiting flamingos and resident fishes compete for their common food, the zooplankton.

Teacher's Note:
a) This example shows that even totally unrelated species can compete for the same resource.
b) Any other correct example of interspecific competition is also accepted.

(c) How do species avoid competition in nature ? Explain with an example. [2 Marks]

Answer:
1. Species avoid competition by resource partitioning, that is, by choosing different times for feeding or different foraging patterns.
2. Example: MacArthur showed that five closely related species of warblers living on the same tree avoid competition and co-exist because of behavioural differences in their foraging activities.

Teacher's Note:
a) The term "resource partitioning" carries 1 mark on its own.
b) The warbler example is the standard NCERT example; learn it well.

SECTION E

31. (a) In angiosperms, the female gametophyte develops from a single cell of megasporangium.
(i) Name that cell of megasporangium and describe the sequential changes it undergoes to form a mature female gametophyte.
(ii) Draw a well labelled diagram of mature female gametophyte. Label any four parts. [5 Marks]

Answer:
1. (i) The cell is the megaspore mother cell (MMC). It undergoes meiosis to form four megaspores (a tetrad).
2. Three megaspores degenerate and one remains functional. The functional megaspore undergoes three successive mitotic divisions without cell wall formation, forming eight nuclei.
3. Three nuclei go to the micropylar end, three to the chalazal end and two stay in the centre. Cell walls are then laid down, forming the 7-celled, 8-nucleate female gametophyte (embryo sac).
4. (ii) The diagram shows an oval embryo sac with three antipodal cells at the chalazal end, two polar nuclei in the large central cell, and the egg apparatus at the micropylar end.
5. The egg apparatus has one egg cell and two synergids, and the synergids have a filiform apparatus at their tips. Labels: antipodals, polar nuclei, central cell, egg, synergids, filiform apparatus (any four).

Teacher's Note:
a) Stress "three mitotic divisions without cell wall formation"; this is free nuclear division.
b) Remember "7-celled, 8-nucleate", because the central cell has two polar nuclei.
c) In the diagram, show the egg apparatus at the micropylar end and the antipodals at the opposite (chalazal) end.

OR

(b) Given below are certain situations related to birth control. Analyse each situation carefully and suggest appropriate contraceptive method that could be used. Also explain its mode of action.
Situation 1 - To prevent entry of sperm into cervix.
Situation 2 - Devices that are inserted by doctors or expert nurses in the vagina.
Situation 3 - Effective emergency contraceptive.
Situation 4 - Permanent method for male partner.
Situation 5 - Surgical method in female to prevent pregnancy. [5 Marks]

Answer:
1. Situation 1: Barrier methods such as condoms (male) or diaphragms, cervical caps and vaults (female). They form a barrier so that sperm and ovum cannot meet.
2. Situation 2: Intra Uterine Devices (IUDs), such as Copper-T, inserted by doctors or trained nurses into the uterus through the vagina. Copper ions released reduce sperm motility and their fertilising capacity.
3. Situation 3: Pills having progestogen or a progestogen-estrogen combination, or IUDs, used within 72 hours of coitus. Pills inhibit ovulation and implantation and change the cervical mucus to stop the entry of sperms.
4. Situation 4: Vasectomy. A small part of the vas deferens is removed or tied up, which blocks the transport of sperms.
5. Situation 5: Tubectomy. A small part of the fallopian tube is removed or tied up, which blocks gamete transport and prevents conception.

Teacher's Note:
a) For each situation, name the method and give its mode of action; each carries half a mark.
b) Vasectomy is for males and tubectomy is for females; do not mix them up.
c) Emergency contraceptives must be used within 72 hours of coitus.

32. (a) (i) Name the protozoan species that is responsible for causing the most serious and even fatal malarial disease.
(ii) Name the host in which the parasite completes its sexual stages and explain the changes taking place.
(iii) How does the parasite damage the human body after entering the blood stream ?
(iv) Suggest two effective preventive measures to control the spread of this disease in endemic regions. [5 Marks]

Answer:
1. (i) Plasmodium falciparum.
2. (ii) The parasite completes its sexual stages in the female Anopheles mosquito. The mosquito picks up gametocytes while biting an infected person.
3. Fertilisation and development take place in the gut of the mosquito, and the sporozoites formed migrate from the gut to the salivary glands of the mosquito.
4. (iii) After entering the blood, the parasites first multiply in liver cells and then attack red blood cells, where they reproduce asexually. The RBCs rupture and release a toxic substance, haemozoin, which causes chills and high fever.
5. (iv) Avoid stagnation of water in and around homes, and use mosquito nets.

Teacher's Note:
a) Remember: sexual stages occur in the mosquito, asexual stages occur in humans.
b) The word "haemozoin" is a key value point for part (iii).
c) Other accepted measures: regular cleaning of coolers, introducing Gambusia fish in ponds, spraying insecticides and using wire mesh.

OR

(b) (i) What are bio-fertilizers ?
(ii) Name the different types of microorganisms used as bio-fertilizers in organic farming and explain how each contributes to soil fertility.
(iii) Write two advantages of using bio-fertilizers over chemical fertilizers. [5 Marks]

Answer:
1. (i) Bio-fertilizers are organisms that enrich the nutrient quality of the soil, for example bacteria, fungi and cyanobacteria.
2. (ii) Bacteria: Rhizobium forms a symbiotic association with the roots of leguminous plants and fixes atmospheric nitrogen into organic forms; Azotobacter and Azospirillum fix nitrogen while free-living in the soil.
3. Fungi: Glomus forms mycorrhiza with plant roots and helps absorb phosphorus from the soil and pass it to the plant.
4. Cyanobacteria: Anabaena, Nostoc and Oscillatoria fix atmospheric nitrogen and are important bio-fertilizers in paddy fields.
5. (iii) Bio-fertilizers do not pollute the soil and ground water, and they conserve beneficial soil microbes.

Teacher's Note:
a) Name all three groups (bacteria, fungi, cyanobacteria) with one example and one role each.
b) Mycorrhizal plants also show resistance to root-borne pathogens and tolerance to salinity and drought.
c) Another accepted advantage: they do not harm human health.

33. (a) (i) Watson and Crick's discovery of double helical structure was based on which two findings. Also mention the name of the scientists associated with these findings.
(ii) Write down the salient features of double helix structure of DNA (any three points). [5 Marks]

Answer:
1. (i) X-ray diffraction studies of DNA by Maurice Wilkins and Rosalind Franklin.
2. Erwin Chargaff's finding that in double-stranded DNA, the ratios of Adenine to Thymine and of Guanine to Cytosine are constant and equal to one.
3. (ii) DNA is made of two polynucleotide chains. The backbone is made of sugar and phosphate, and the bases project inside.
4. The two chains have antiparallel polarity.
5. The bases of the two strands are paired by hydrogen bonds: Adenine forms two hydrogen bonds with Thymine, and Guanine forms three hydrogen bonds with Cytosine.

Teacher's Note:
a) In (i), each finding and the scientists' names carry half a mark each.
b) Other accepted features: right-handed coiling, pitch of 3.4 nm with about 10 bp per turn, 0.34 nm between base pairs, and stacking of base pairs.

OR

(b) Given below is a stretch of DNA showing the coding strand of structural gene of transcription unit.
5' - ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TTA 3'
(i) Write the corresponding template strand and m-RNA strand that will be transcribed along with its polarity.
(ii) If GUA of transcribed mRNA is an intron, then depict the sequence involved in formation of mRNA / mature / processed hnRNA strand :
(1) In a bacterium
(2) In humans
(iii) How many amino acids the resulting polypeptide will have after the process of translation in humans ? [5 Marks]

Answer:
1. (i) Template strand: 3' - TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT AAT - 5'
2. mRNA: 5' - AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA - 3'
3. (ii) (1) In a bacterium, there is no splicing, so the mRNA stays the same: 5' - AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA - 3'
4. (2) In humans, all GUA introns are removed by splicing, a cap is added at the 5' end and a poly A tail at the 3' end: 5' - mGppp AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA - poly A tail 3'
5. (iii) The polypeptide will have 10 amino acids.

Teacher's Note:
a) The mRNA has the same sequence as the coding strand, with U in place of T.
b) Always write the polarity (5' and 3') on each strand; it carries marks.
c) Bacteria have no introns (split genes) and no splicing, so their mRNA is not processed.

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