ISC Class 12 Biology Board Exam Question Paper 2018 with Solutions

ISC Class 12 Biology Board Exam Question Paper with Solutions

 

PART - I (20 Marks)

Answer all questions.

 

Question 1

(a) Answer the following questions briefly and to the point: [8 Marks]

 

(i) Give a significant point of difference between Oestrous and Menstrual cycle. [1 Mark]

Answer:
Oestrous cycle takes place in non-primate females during the breeding season, whereas the menstrual cycle takes place in primates and occurs monthly throughout the reproductive phase.

Teacher's Note:
a) Remember that oestrous cycles involve reabsorption of endometrium if conception fails, while menstrual cycles involve shedding of the endometrium (menstruation).
b) Students often confuse the mammalian groups; ensure you specify non-primates for oestrous and primates for menstrual cycles.

 

(ii) Give the biological name of the organism causing typhoid. [1 Mark]

Answer:
The organism causing typhoid is Salmonella typhi.

Teacher's Note:
a) Binomial nomenclature rules must be followed: genus name starts with a capital letter and species name with a small letter, and both must be italicized or underlined separately when handwritten.
b) Spelling errors in scientific names lead to a deduction of marks in board examinations.

 

(iii) If the haploid number of chromosomes in a plant species is 20, how many chromosomes will be present in the cells of the shoot tip? [1 Mark]

Answer:
The number of chromosomes in the cells of the shoot tip will be 40, as shoot tip cells are meristematic somatic cells which are diploid (\(2n\)).

Teacher's Note:
a) Haploid number (\(n\)) is given as 20. Somatic cells are diploid (\(2n\)), so \(2 \times 20 = 40\).
b) Clearly state that somatic tissues undergo mitotic division and maintain the diploid chromosome number.

 

(iv) Name a plant which flowers every twelve years. [1 Mark]

Answer:
Strobilanthes kunthianus (Neelakurinji).

Teacher's Note:
a) This is a classic example of masting and perennial flowering behavior documented in certain plant species.
b) Mentioning either the binomial name or the common local name is accepted, but writing both ensures full credit.

 

(v) Name the diagnostic test for AIDS. [1 Mark]

Answer:
The diagnostic test for AIDS is ELISA (Enzyme-Linked Immunosorbent Assay).

Teacher's Note:
a) ELISA detects the presence of antibodies produced against HIV antigens in the patient's serum.
b) Always write the full expanded form alongside the acronym for complete clarity.

 

(vi) Name the terminal stage of ageing in the life cycle of plants. [1 Mark]

Answer:
Senescence is the terminal stage of ageing in the life cycle of plants.

Teacher's Note:
a) Senescence marks the deterioration of functional characteristics in living organisms prior to death.
b) Do not confuse senescence with maturation; maturation is the attainment of developmental full growth.

 

(vii) Which organisms constitute the last trophic level? [1 Mark]

Answer:
Top carnivores or tertiary/quaternary consumers (and decomposers at the base/end of energy flow pathways) constitute the highest trophic levels.

Teacher's Note:
a) Trophic levels represent sequential feeding positions in a food chain from producers to apex consumers.
b) Ensure accurate terminology depending on whether the question refers to apex predators or decomposers.

 

(viii) What is emasculation? [1 Mark]

Answer:
The removal of anthers from a bisexual flower before the dehiscence of anthers to prevent self-pollination in plant breeding programs is called emasculation.

Teacher's Note:
a) Emasculation is a crucial step in artificial hybridization techniques in plants.
b) Mentioning that it is done on bisexual flowers is necessary for a complete definition.

 

(b) Each of the following questions has four choices. Choose the best option in each case: [4 Marks]

 

(i) Length of DNA with 23 base pairs is: [1 Mark]
(1) 78.4 Å
(2) 78.2 Å
(3) 78 Å
(4) 74.8 Å

Answer: (2) 78.2 Å

The distance between two successive base pairs in a B-DNA molecule is \(3.4\text{ \AA}\) (or \(0.34\text{ nm}\)). Total length = \(23 \times 3.4\text{ \AA} = 78.2\text{ \AA}\).

Teacher's Note:
a) Use the standard structural parameter of B-DNA where pitch is \(34\text{ \AA}\) with 10 base pairs per turn.
b) Multiply the number of base pairs directly by \(3.4\text{ \AA}\) to find the total length.

 

(ii) Opium is obtained from: [1 Mark]
(1) Papaver somniferum
(2) Cannabis sativa
(3) Erythroxylum coca
(4) Datura metel

Answer: (1) Papaver somniferum

Opium is dried latex obtained from the unripe seed capsules of the poppy plant, Papaver somniferum.

Teacher's Note:
a) Morphine, codeine, and heroin are all derived or processed from opium obtained from this plant.
b) Associate each drug source correctly: Cannabis sativa yields cannabinoids, and Erythroxylum coca yields cocaine.

 

(iii) According to Abiogenesis, life originated from: [1 Mark]
(1) Non-living matter
(2) Pre-existing life
(3) Oxygen
(4) Extra-terrestrial matter

Answer: (1) Non-living matter

Abiogenesis (spontaneous generation) theorizes that life arose naturally from non-living organic and inorganic compounds.

Teacher's Note:
a) Abiogenesis contrasts with biogenesis, which states that life originates from pre-existing life.
b) Keep chemical evolution and Oparin-Haldane's theory in mind while evaluating primitive earth conditions.

 

(iv) The largest unit in which gene flow is possible is: [1 Mark]
(1) Organism
(2) Population
(3) Species
(4) Genes

Answer: (3) Species

Gene flow occurs through interbreeding between individuals, which can happen across populations within the same biological species.

Teacher's Note:
a) A species is defined as a group of interbreeding natural populations that are reproductively isolated from other such groups.
b) While populations experience gene flow internally, the species boundary marks the absolute limit for natural gene flow.

 

(c) Give one significant contribution of each of the following scientists: [4 Marks]

 

(i) P. Maheshwari [1 Mark]

Answer:
P. Maheshwari made significant pioneering contributions to plant embryology, test-tube fertilization, and plant tissue culture.

Teacher's Note:
a) He established the Department of Botany at the University of Delhi as an internationally renowned center for plant embryology.
b) Mentioning plant embryology or test-tube fertilization secures full credit.

 

(ii) E. Wilson [1 Mark]

Answer:
Edward O. Wilson is widely recognized as the father of biodiversity and sociobiology.

Teacher's Note:
a) He popularized the term biodiversity and contributed extensively to island biogeography theory.
b) Ensure exact terminology is used when attributing titles to scientists.

 

(iii) M. S. Swaminathan [1 Mark]

Answer:
M. S. Swaminathan is known as the father of the Green Revolution in India for introducing high-yielding wheat and rice varieties.

Teacher's Note:
a) His collaborative work averted mass famines in India during the mid-20th century through agricultural science.
b) Mentioning agricultural advancement or high-yielding crop varieties is essential.

 

(iv) H. Boyer [1 Mark]

Answer:
Herbert Boyer discovered restriction endonucleases, which laid the foundation for recombinant DNA technology.

Teacher's Note:
a) Working alongside Stanley Cohen, he successfully constructed the first recombinant DNA organism.
b) Keywords like restriction enzymes or molecular scissors must be included.

 

(d) Define the following: [2 Marks]

 

(i) Biopatent [1 Mark]

Answer:
A biopatent is a patent granted by a government to an inventor for biological entities, genetically modified organisms, biotechnological processes, or products derived from them, preventing unauthorized commercial use.

Teacher's Note:
a) Biopatents protect intellectual property rights associated with biological innovations.
b) Emphasize that it grants legal rights to prevent commercial exploitation by others.

 

(ii) Parthenocarpy [1 Mark]

Answer:
Parthenocarpy is the natural or artificially induced production of fruit without fertilization of ovules, resulting in seedless fruits.

Teacher's Note:
a) Common examples include bananas, seedless watermelons, and grapes.
b) The absence of seeds due to lack of fertilization is the key diagnostic feature.

 

(e) Give a reason for each of the following: [2 Marks]

 

(i) Pollen grains of wind-pollinated flowers are produced in large quantities. [1 Mark]

Answer:
Pollen grains in wind-pollinated flowers are produced in massive quantities to compensate for the very high wastage and random dispersal in the open air during transit to receptive stigmas.

Teacher's Note:
a) Wind pollination (anemophily) is an undirected process with low pollination efficiency.
b) Mentioning pollen wastage and random transport ensures complete credit.

 

(ii) Equilibrium of a forest ecosystem can be disturbed by uncontrolled hunting of big predators. [1 Mark]

Answer:
Uncontrolled hunting of apex predators leads to an unchecked surge in herbivore populations, resulting in overgrazing, depletion of vegetation, and trophic cascades that destabilize the ecosystem.

Teacher's Note:
a) Predators maintain top-down control over prey population densities in food webs.
b) Explain the ecological consequence of population explosion among herbivores.

 

PART II

Section A (14 Marks)

Answer all questions.

 

Question 2

(a) A woman with blood group O married a man with blood group AB shows the possible blood groups of the progeny. List the alleles involved in this inheritance. [2 Marks]

Answer:
The alleles involved in ABO blood group inheritance are \(I^A\), \(I^B\), and \(i\).
Mother with blood group O has genotype \(ii\).
Father with blood group AB has genotype \(I^A I^B\).
Punnet Square:

	\(I^A\)	\(I^B\)
\(i\)	\(I^A i\) (Blood Group A)	\(I^B i\) (Blood Group B)
\(i\)	\(I^A i\) (Blood Group A)	\(I^B i\) (Blood Group B)

Possible phenotypes of progeny: Blood Group A and Blood Group B in a 1:1 ratio.

Teacher's Note:
a) This problem demonstrates multiple allelomorphism and codominance combined with recessive alleles.
b) Always write parental genotypes and clearly construct the Punnet square.

OR

(b) If the mother is a carrier of colour blindness and the father is normal, the possible genotype and phenotype of the offspring of the next generation, with the help of a punnet square. [2 Marks]

Answer:
Colour blindness is an X-linked recessive disorder.
Genotype of carrier mother: \(X^C X\)
Genotype of normal father: \(XY\)
Punnet Square:

	\(X^C\)	\(X\)
\(X\)	\(X^C X\) (Carrier daughter)	\(XX\) (Normal daughter)
\(Y\)	\(X^C Y\) (Affected son)	\(XY\) (Normal son)

Possible Genotypes: \(X^C X\), \(XX\), \(X^C Y\), \(XY\).
Possible Phenotypes: Carrier daughter, Normal daughter, Affected son, Normal son.

Teacher's Note:
a) X-linked recessive disorders show criss-cross inheritance from mother to son.
b) Clearly superscript the recessive allele on the X chromosome in all crosses.

 

Question 3

Define life span. Give the life span of an elephant. [2 Marks]

Answer:
Life Span definition: The period from birth to the natural death of an organism represents its life span.
Life span of an elephant:
Asian elephant - 48 years.
African elephant - 60 to 70 years.

Teacher's Note:
a) Life span is species-specific and does not necessarily correlate with organism size.
b) Mentioning both Asian and African elephant lifespans guarantees complete credit.

 

Question 4

Give two characteristic features of each of the following: [2 Marks]
(a) Ramapithecus
(b) Cro-Magnon man

Answer:
(a) Ramapithecus:
1. It was an ape-like primate that walked more erect on its hind limbs.
2. Its fossils (teeth and jaw bones) were discovered in the Siwalik hills.
(b) Cro-Magnon man:
1. It was an extinct modern man (Homo sapiens fossilis) with a cranial capacity of about 1600 cc.
2. They lived in caves, made advanced stone and bone tools, and practiced art and burial customs.

Teacher's Note:
a) Human evolution stages must be accompanied by accurate anatomical and cultural descriptions.
b) Point out that Ramapithecus showed more hominid-like dental features than modern apes.

 

Question 5

(a) List any four effects of global warming. [2 Marks]

Answer:
1. Melting of polar ice caps and glaciers, leading to a rise in sea levels and coastal flooding.
2. Increased carbon dioxide concentration enhancing the rate of photosynthesis in certain plants (CO2 fertilization effect).
3. Accelerated climatic changes leading to extreme weather events and erratic rainfall patterns.
4. Disruption of ecosystems and loss of biodiversity due to changing temperature thresholds.

Teacher's Note:
a) List direct climatic and ecological consequences.
b) Avoid writing vague points; use standard geographical and biological impacts.

OR

(b) State any four measures to control noise pollution. [2 Marks]

Answer:
1. Use of sound absorbents and acoustic materials in industrial units and public auditoriums.
2. Planting dense rows of trees (green belts) along highways and residential sectors to absorb noise.
3. Enforcement of strict legal regulations banning horns in silent zones near hospitals and schools.
4. Imposing permissible decibel limits and time restrictions on the use of loudspeakers and heavy vehicles.

Teacher's Note:
a) Noise control involves both technological modifications and legislative measures.
b) Mentioning green belt creation and horn-free zones fetches full marks.

 

Question 6

Define BOD. What is its significance in an aquatic ecosystem? [2 Marks]

Answer:
Biochemical Oxygen Demand (BOD): It is the amount of dissolved oxygen consumed by aerobic bacteria in decomposing organic matter present in a given volume of water sample over a set period at 20°C.
Significance in an Aquatic Ecosystem: BOD is a direct measure of organic water pollution. Higher BOD indicates severe organic pollution and rapid depletion of dissolved oxygen, leading to the suffocation and death of aquatic flora and fauna.

Teacher's Note:
a) Emphasize that higher BOD values correlate inversely with water quality.
b) Explain that microbial decomposition consumes oxygen dissolved in water.

 

Question 7

Give one significant difference between each of the following pairs: [2 Marks]
(a) Humoral immunity and cell mediated immunity.

Answer:

Humoral Immunity (Antibody-mediated)	Cell-Mediated Immunity (CMI)
Mediated by B-lymphocytes that differentiate into plasma cells to secrete antibodies into body fluids.	Mediated directly by T-lymphocytes (such as cytotoxic T cells) to destroy infected or foreign cells.

Teacher's Note:
a) Humoral immunity targets pathogens in body fluids, whereas CMI targets intracellular pathogens and graft rejections.
b) Always format difference questions as a clear comparative table.

 

(b) Benign tumour and malignant tumour. [2 Marks]

Answer:

Benign Tumour	Malignant Tumour
Non-cancerous growths that remain confined to their original site, are usually encapsulated, and do not metastasize.	Cancerous growths that invade surrounding normal tissues and spread to distant body parts via metastasis.

Teacher's Note:
a) Metastasis is the hallmark property distinguishing malignant tumors from benign growths.
b) Highlight encapsulation and invasiveness as the key distinguishing features.

 

Question 8

Give four causes of infertility in males. [2 Marks]

Answer:
1. Cryptorchidism: Failure of testes to descend into the scrotum, impairing spermatogenesis due to high body temperature.
2. Oligospermia: Abnormally low sperm count in semen.
3. Asthenospermia: Reduced sperm motility preventing successful migration to the ovum.
4. Hormonal Imbalance: Deficiencies in testosterone or gonadotropins required for normal sperm production.

Teacher's Note:
a) Mention specific medical terminology such as oligospermia and cryptorchidism for full credit.
b) Ensure causes relate strictly to male reproductive dysfunctions.

 

Section B (21 Marks)

Answer all questions.

 

Question 9

(a) Draw a labelled diagram of L.S. of human testis. [3 Marks]

Answer:
[Figure: Longitudinal section of human testis showing testicular lobules, seminiferous tubules, rete testis, vasa efferentia, epididymis, vas deferens, tunica albuginea, and tunica vaginalis.]
1. The testis is covered by a dense fibrous capsule called tunica albuginea.
2. Each testis contains 250 compartments called testicular lobules, each housing 1 to 3 seminiferous tubules.
3. Tubules converge to form the rete testis, leading into vasa efferentia and the epididymis.

Teacher's Note:
a) Diagrams must be neat, clean, and correctly labeled with standard anatomical pointers.
b) Essential labels include seminiferous tubules, testicular lobules, rete testis, and vas deferens.

OR

(b) Draw a labelled diagram of the mature embryo sac of angiosperms. [3 Marks]

Answer:
[Figure: 7-celled, 8-nucleate mature female gametophyte (embryo sac) showing egg cell, synergids with filiform apparatus, central cell with secondary polar nuclei, and antipodals at the chalazal end.]
1. The mature embryo sac is typically 7-celled and 8-nucleate at fertilization.
2. It contains an egg apparatus at the micropylar end (two synergids and one egg cell) along with a filiform apparatus.
3. Three antipodal cells are situated at the chalazal end, and two polar nuclei occupy the central cell.

Teacher's Note:
a) Ensure the distinction between cell count (7) and nuclear count (8) is clearly depicted.
b) The filiform apparatus in synergids is an important labeling point.

 

Question 10

Explain gene therapy, with reference to treatment of SCID. [3 Marks]

Answer:
1. Gene therapy is a corrective genetic engineering technique used to insert normal, functional genes into cells to correct genetic disorders.
2. Severe Combined Immunodeficiency (SCID) is caused by a deletion in the gene coding for the enzyme adenosine deaminase (ADA), which is crucial for immune system functioning.
3. In clinical treatment, patient lymphocytes are extracted, a functional ADA cDNA is introduced using a retroviral vector, and the genetically engineered cells are infused back into the patient.

Teacher's Note:
a) Mention the exact genetic defect (ADA deficiency) and the vector used.
b) Note that because lymphocytes are not immortal, periodic re-infusions are required unless bone marrow stem cells are treated at an embryonic stage.

 

Question 11

Study the table given below. Do not copy the table, but write the answers in the correct order. [3 Marks]

Scientific Name	Commercial Product	Use
(a) ____________	Streptokinase	(b) ____________
Monascus purpureus	(c) ____________	(d) ____________
(e) ____________	Lactic acid	(f) ____________

Answer:
(a) Streptococcus (or Streptococcus pyogenes / Haemolytic streptococci)
(b) Clot buster (dissolving blood clots / thrombolytic agent)
(c) Statins
(d) Blood-cholesterol lowering agent
(e) Lactobacillus (or Lactobacillus bulgaricus)
(f) Curdling of milk (or production of curd)

Teacher's Note:
a) This question tests knowledge of microbes in human welfare and their industrial applications.
b) Students should list answers corresponding to labels (a) through (f) clearly without rewriting the entire table.

 

Question 12

Explain industrial melanism. [3 Marks]

Answer:
1. Industrial melanism is a classic evolutionary phenomenon demonstrating natural selection in peppered moths (Biston betularia) in Great Britain.
2. Before the industrial revolution, light-coloured peppered moths predominated because they camouflaged well against lichen-covered tree trunks, escaping predatory birds.
3. Industrial pollution killed the lichens and blackened tree trunks with soot, giving a selective advantage to dark-coloured (melanic) mutant moths, whose populations subsequently surged while light forms declined.

Teacher's Note:
a) Emphasize that pollution changed environmental selective pressures favoring camouflage.
b) Clearly link the shift in moth frequencies to directional natural selection.

 

Question 13

Describe the tissue culture technique in plants. [3 Marks]

Answer:
[Figure: Plant tissue culture stages showing explant selection, callus formation, organogenesis, and micropropagation of multiple plantlets.]
1. Explant Selection and Sterilization: Any plant part (explant) is excised, sterilized, and inoculated onto a nutrient agar medium enriched with carbon sources and vitamins.
2. Callus Formation: Explant cells undergo active mitotic division to form an unorganized mass of parenchyma cells called a callus.
3. Organogenesis and Regeneration: The callus is treated with auxins and cytokinins in specific hormone ratios to induce shoot and root differentiation, generating plantlets.

Teacher's Note:
a) Mention the key steps: explant isolation, callus induction, and organogenesis.
b) Highlight the role of plant growth regulators (auxins and cytokinins) in organ differentiation.

 

Question 14

Define the following: [3 Marks]
(a) Spermiogenesis
(b) Reproductive health
(c) Amenorrhea

Answer:
(a) Spermiogenesis: The process of transformation of non-motile, immature round spermatids into mature, motile spermatozoa (sperm).
(b) Reproductive Health: According to the WHO, it refers to a total well-being in all aspects of reproduction, encompassing physical, emotional, behavioural, and social dimensions.
(c) Amenorrhea: The abnormal absence or cessation of menstrual cycles in a woman of reproductive age.

Teacher's Note:
a) Ensure precise definitions for all reproductive physiology terms.
b) Do not confuse spermiogenesis with spermatogenesis; spermiogenesis is specifically the final metamorphosis of spermatids into sperm.

 

Question 15

(a) Define the following: [3 Marks]
(i) Hotspots
(ii) Ramsar Sites
(iii) Red data book

Answer:
(i) Biodiversity Hotspots: Regions with very high levels of species richness and endemic species that are under constant threat of habitat destruction.
(ii) Ramsar Sites: Wetlands designated as being of international ecological importance under the Ramsar Convention established in 1971.
(iii) Red Data Book: A documented record maintained by the IUCN listing rare, threatened, and endangered species of plants and animals.

Teacher's Note:
a) These ecological terms are important for conservation biology questions.
b) Mentioning endemic species for hotspots and IUCN for Red Data Book adds precision.

OR

(b) Define the following: [3 Marks]
(i) Biodiversity
(ii) Eutrophication
(iii) PAR

Answer:
(i) Biodiversity: The variability among living organisms from all sources, encompassing genetic, species, and ecosystem diversity.
(ii) Eutrophication: The natural or artificial nutrient enrichment of a water body leading to excessive algal blooms and severe depletion of dissolved oxygen.
(iii) PAR (Photosynthetically Active Radiation): The spectral range of solar radiation from 400 nm to 700 nm that photosynthetic organisms can use for photosynthesis.

Teacher's Note:
a) Give complete scientific definitions specifying wavelength ranges for PAR.
b) Highlight the ecological consequences of eutrophication in water bodies.

 

Section C (15 Marks)

Answer all questions.

 

Question 16

(a) Describe post transcriptional processing of RNA in eukaryotes. [5 Marks]

Answer:
1. In eukaryotic cells, primary transcripts (hnRNA) undergo extensive post-transcriptional modifications before functioning as mature mRNA.
2. Capping: A modified nucleotide (methylguanosine triphosphate) is added to the 5' end of the primary transcript.
3. Polyadenylation: A tail of 200-300 adenine residues (poly-A tail) is added at the 3' end in a template-independent manner.
4. Splicing: Non-coding intervening sequences (introns) are excised, and coding sequences (exons) are joined together in a precise sequence by spliceosomes.
5. Once fully processed, the mature mRNA is transported out of the nucleus into the cytoplasm for translation.

Teacher's Note:
a) Detail the three major processing events: capping, tailing, and splicing.
b) Explain why splicing is necessary for eukaryotic transcripts containing split genes (exons and introns).

OR

(b) Describe Avery, McLeod and McCarty’s experiment. State its significance. [5 Marks]

Answer:
1. Oswald Avery, Colin MacLeod, and Maclyn McCarty (1944) worked to determine the biochemical nature of the transforming principle in Griffith's pneumococcus experiment.
2. They purified biochemicals (proteins, DNA, RNA) from heat-killed virulent (S) bacteria to see which one could transform live non-virulent (R) bacteria into S strain.
3. They discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not inhibit transformation.
4. However, when DNA-digesting enzymes (DNases) were added, transformation was abolished, proving that DNA caused the transformation.
5. Significance: It conclusively established that DNA, and not protein, is the genetic material.

Teacher's Note:
a) Describe the step-by-step enzymatic degradation process used in their in vitro assay.
b) Clearly state the final conclusion regarding DNA as the hereditary material.

 

Question 17

(a) Write a short note on Chipko Movement. [5 Marks]

Answer:
1. The Chipko Movement was a grassroots forest conservation movement that originated in the Garhwal Himalayas of Uttarakhand in 1974.
2. Local villagers, notably women led by activists like Sunderlal Bahuguna and Chandi Prasad Bhatt, protected trees from commercial logging by hugging them.
3. It was a non-violent, ecological movement protesting deforestation and asserting local community rights over forest resources.
4. The movement drew inspiration from historical events such as the Bishnoi community sacrifice in Khejarli, Rajasthan, in 1730.
5. Its success forced the government to issue a 15-year ban on commercial tree felling in the Himalayan forests in 1980.

Teacher's Note:
a) Mention key leaders (Sunderlal Bahuguna) and the geographic origin.
b) Highlight the ecological significance and the unique method of protest (hugging trees).

OR

(b) Write a short note on Joint Forest Management. [5 Marks]

Answer:
1. Joint Forest Management (JFM) is a strategy introduced by the Government of India in the 1980s for the protection and management of forests through local community involvement.
2. It operates on a cooperative framework where village communities work alongside state forest departments to regenerate degraded forest lands.
3. Participating villagers are given benefits of non-timber forest products and a share in timber revenues in exchange for guarding forests against grazing and illegal felling.
4. JFM projects have been successfully implemented across numerous states, including Odisha, West Bengal, Gujarat, and Karnataka.
5. It emphasizes sustainable resource management by integrating ecological conservation with rural livelihood needs.

Teacher's Note:
a) Explain the cooperative partnership between local villagers and forest departments.
b) State the incentives provided to communities to ensure sustainable participation.

 

Question 18

(a) What does PCR stand for? Describe the different steps of PCR. [5 Marks]

Answer:
PCR stands for Polymerase Chain Reaction.
[Figure: Three-step PCR cycle showing denaturation at 94°C, primer annealing at 50-65°C, and extension by Taq polymerase at 72°C.]
1. Denaturation: The double-stranded DNA template is heated to approximately 94°C to break hydrogen bonds, separating it into single strands.
2. Annealing: The reaction temperature is lowered (50-65°C) to allow short synthetic oligonucleotide primers to bind to complementary sequences on the single-stranded DNA templates.
3. Extension (Polymerization): The temperature is raised to 72°C, and a thermostable DNA polymerase (Taq polymerase) synthesizes new DNA strands by adding dNTPs complementary to the template.
4. Repeating these thermal cycles exponentially amplifies the target DNA segment millions of times.

Teacher's Note:
a) Detail the three temperature-controlled steps: denaturation, annealing, and extension.
b) Mention the role and source of thermostable Taq polymerase (Thermus aquaticus).

OR

(b) Give an account of the Blue-White Method of selection of recombinants. [5 Marks]

Answer:
1. The blue-white screening method is a rapid technique used in recombinant DNA technology to detect recombinant bacterial colonies.
2. The plasmid vector contains the coding sequence for the enzyme beta-galactosidase (\(lacZ\) gene), which metabolizes a chromogenic substrate called X-gal into a blue-coloured product.
3. When foreign DNA is inserted into the cloning site within the \(lacZ\) gene, the gene becomes inactivated (insertional inactivation), and functional beta-galactosidase is not produced.
4. Non-recombinant plasmids produce functional enzyme and grow into blue colonies in the presence of X-gal.
5. Recombinant plasmids with disrupted \(lacZ\) genes fail to produce the enzyme and appear as white colonies, allowing easy identification and isolation.

Teacher's Note:
a) Explain the principle of insertional inactivation of the \(lacZ\) gene clearly.
b) Differentiate clearly between blue colonies (non-recombinants) and white colonies (recombinants).