Class 12 Biology Solved Paper 2024 PDF

BIOLOGY Solved Paper 2024 Time : 3 Hours Class-XII Max. Marks : 70 General Instructions : Read the following instructions very carefully and strictly follo...

BIOLOGY Solved Paper 2024 Time : 3 Hours Class-XII Max. Marks : 70 General Instructions : Read the following instructions very carefully and strictly follow them: (i) This question paper contains 33 questions. All questions are compulsory. (ii) Question paper is divided into FIVE sections. Section A, B, C, D and E. (iii) Section A – question numbers 1 to 16 are Multiple Choice type questions. Each question carries 1 mark. (iv) Section B – question numbers 17 to 21 are Very Short Answer type questions. Each question carries 2 marks. (v) Section C – question numbers 22 to 28 are Short Answer type questions. Each question carries 3 marks. (vi) Section D – question numbers 29 and 30 are Case-based questions. Each question carries 4 marks. Each question has subparts with internal choice in one of the subparts. (vii) Section E – question numbers 31 to 33 are Long Answer type questions. Each question carries 5 marks. (viii) There is no overall choice. However, an internal choice has been provided in section B, C and D of question paper. A candidate has to write answer for only one of the alternatives in such questions. (ix) Kindly note that there is a separate question paper for Visually Impaired candidates. (x) Wherever necessary, neat and properly labelled diagrams should be drawn. Delhi Set-I 57/5/1 SECTION - A mastia (iv) are not typically associated with Down syndrome.  1 Question No. 1 to 16 are Multiple Choice type questions, 3. Observe the schematic representation of assisted carrying 1 mark each. 16×1=16 reproductive technology given below: 1. A single gene that controls the expression of more than one trait is said to show (A) Multiple allelism (B) Polygenic inheritance (C) Incomplete dominance (D) Pleiotropism Ans. Option (D) is correct. Explanation: Pleiotropy is the phenomenon in which one gene controls many traits. For example, the gene in pea plants that controls the round and wrinkled texture of seeds also influences the Identify the most appropriate technique depicted phenotypic expression of starch grain size. in the above diagram. 2. A person with trisomy of 21st chromosome shows (A) IUT (B) IUI  (i) Furrowed tongue (C) ICSI (D) ZIFT (ii) Characteristic palm crease Ans. Option (C) is correct. (iii) Rudimentary ovaries Explanation: ICSI (intracytoplasmic sperm injection) (iv) Gynaecomastia is a laboratory procedure in which a single sperm Select the correct option from the choices given below: (from a male partner) is injected directly into an egg (A) (ii) and (iv) (B) (i), (ii) and (iv) (from a female partner). Then the fertilised egg is (C) (ii) and (iii) (D) (i) and (ii) implanted into the woman’s uterus. Ans. Option (D) is correct. 4. Interferons are proteins secreted by Explanation: Down syndrome is caused by trisomy (A) RBC of the 21st chromosome. Furrowed tongue (i) and (B) WBC characteristic palm crease (ii) are a common physi- (C) Bacteria infected cell cal feature of individuals with Down syndrome. However, rudimentary ovaries (iii) and gynaeco- (D) Virus infected cell SOLVED PAPER - 2024 (BIOLOGY) [ 669 Ans. Option (D) is correct. Explanation: Interferons are proteins secreted by virus–infected cells. They help alert neighbouring cells to the presence of a viral infection, stimulate immune responses and enhance the body's ability to fight off viruses. 5. During biological treatment of sewage, the masses of bacteria held together by fungal filaments to form mesh like structures are called (A) primary sludge (B) flocs (C) activated sludge (D) anaerobic sludge Ans. Option (B) is correct. Explanation: During the biological treatment of (A) A – zygote, B – degenerating synergids, C – sewage, flocs are masses of bacteria held together degenerating antipodals, D – PEN by fungal filaments to form mesh–like structures. This process typically occurs in the secondary (B) A – degenerating synergids, B – zygote, C – treatment stage of wastewater treatment, where PEN, D – degenerating antipodals microorganisms are employed to break down (C) A – degenerating antipodals, B – PEN, C – degenerating synergids, D – zygote organic matter present in the sewage. (D) A – degenerating synergids, B – zygote, C – 6. Which one of the following statements is correct degenerating antipodals, D – PEN in the context of observing DNA separation by Ans. Option (B) is correct. agarose gel electrophoresis? Explanation: In the given figure of a fertilised (A) DNA can be seen in visible light. embryo sac of an angiosperm, the part labelled as A is degenerating synergids, B is zygote, C is (B) DNA can be seen without staining in visible primary endosperm nucleus and D is degenerating light. antipodals. (C) Ethidium bromide stained DNA can be seen in visible light. (D) Ethidium bromide stained DNA can be seen under UV light. Ans. Option (D) is correct. Explanation: Ethidium bromide intercalates between the base pairs of DNA molecules and when exposed to UV light, it fluorescence, emitting visible light. This fluorescence allows the visualisation of DNA bands appears orange in the gel. 7. A phenomenon where a male insect mistakenly identified the patterns of a orchid flower as the female insect partner and tries to copulate and thereby pollinates the flower is said to be: 9. Study the pedigree chart of a family showing the (A) Pseudocopulation inheritance pattern of a certain disorder. Select the option that correctly identifies the nature of the (C) Pseudopollination trait depicted in the pedigree chart. (B) Pseudoparthenocarpy (D) Pseudofertilisation 7. Option (A) is correct. Explanation: Pseudocopulation is a phenomenon in which a male insect mistakenly identifies the patterns or characteristics of a flower, often an orchid, as those of a female insect partner. In this process the flower mimics the sexual pheromones. (A) Dominant X-linked 8. Identify the correct labellings in the figure of (B) Recessive X-linked a fertilised embryo sac of an angiosperm given (C) Autosomal dominant below: (D) Autosomal recessive 670 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th Ans. Option (B) is correct. Question numbers 13 to 16 consist of two statements Explanation: The trait depicted in the pedigree Assertion (A) and Reason (R). Answer these questions chart is recessive X–linked. selecting the appropriate option given below: (A) Both (A) and (R) are true and (R) is the correct X-linked recessive disorders are caused by gene explanation of (A). variants on the X chromosome, and are more (B) Both (A) and (R) are true, but (R) is not the correct common in males than females. explanation of (A). 10. Match the following genes of the lac operon listed (C) (A) is true, but (R) is false. in column 'A' with their respective products listed (D) (A) is false, but (R) is true. in column 'B': 13. Assertion (A): AIDS is a syndrome caused by HIV. A B  Gene Products Reason (R): HIV is a virus that damages the immune system with DNA as its genetic material. a. 'i' gene (i) β–galactosidase Ans. Option (A) is correct. b. 'z' gene (ii) lac permease Explanation: HIV is a virus that causes AIDS and c. 'a' gene (iii) repressor it damages the immune system using RNA as its d. 'y' gene (iv) transacetylase genetic material. 14. Assertion (A): In molecular diagnosis, single Select the correct option: stranded DNA or RNA tagged with radioactive a b c d molecule is called a probe. (A) (i) (iii) (ii) (iv) Reason (R): A probe always searches and (B) (iii) (i) (ii) (iv) hybridises with its complementary DNA in a (C) (iii) (i) (iv) (ii) clone of cells. (D) (iii) (iv) (i) (ii) Ans. Option (C) is correct. Ans. Option (C) is correct. Explanation: In molecular diagnosis, single– Explanation: The correct matching pair is: stranded DNA or RNA tagged with a radioactive (a) 'i' gene – (iii) repressor molecule is indeed called a probe. Probes are used to detect specific nucleotide sequences, such as those (b) 'Z' gene – (i) ß–galactosidase associated with particular genes or pathogens, (c) 'a' gene – (iv) transacetylase in a sample. A probe does not always search and (d) 'y' gene – (ii) lac permease hybridise with its complementary DNA in a clone 11. If both the parents are carriers for thalassaemia, of cells. Probes are designed to hybridise with the chances of an afflicted child to be born to them complementary sequences wherever they may be is: present, not specifically within a clone of cells. (A) 25% (B) 50% 15. Assertion (A): In birds, the sex of the offspring is (C) 75% (D) 100% determined by males. Ans. Option (A) is correct. Reason (R): Males are homogametic while females are heterogametic. Explanation: If both parents are carriers for thalas- Ans. Option (D) is correct. semia, they each have one normal allele and one mutated allele for the gene associated with thalas- Explanation: In most birds, the females are semia. When they have children, there's a 25% heterogametic (designated as Z W) and males are homogametic (designated as Z Z). Therefore, it is chance that both parents will pass on their mutated the female that determines the sex of the offspring alleles, resulting in an affected child. based on whether she contributes a Z (resulting 12.If the sequence of nitrogen bases of the coding strand in a male offspring) or a W (resulting in a female in a transcription unit is 5' – ATGAATG – 3', the offspring) chromosome. sequence of bases in its RNA transcript would be  16. Assertion (A): Communities that comprise more species tend to be more stable. (A) 5' – AUGAAUG – 3' Reason (R): A higher number of species results in (B) 5' – UACUUAC – 3' less year to year variation in total biomass. (C) 5' – CAUUCAU – 3' Ans. Option (B) is correct. (D) 5' – GUAAGUA – 3' Explanation: Communities with more species Ans. Option (A) is correct. tend to be more stable than those with less Explanation: Given the coding strand sequence: species because they are able to resist occasional 5'–ATGAATG–3', the corresponding RNA transcript disturbance. David Tilman's long term experiments sequence would be 5'– AUGAAUG –3' as Thymine showed the plots with more species, experience less replaced by Uracil. year to year variation in total biomass. SOLVED PAPER - 2024 (BIOLOGY) [ 671 (b) 'K' is the carrying capacity of a habitat for SECTION - B members of a given sort of organism. 17. (a)  "Farmers prefer apomictic seeds to hybrid 20. Explain how are plants benefitted by their seeds." Justify giving two reasons. association with "Glomus species". OR Ans. Plants benefit from their association with Glomus (b) Mention one advantage and one disadvantage species through mycorrhizal symbiosis. Glomus of amniocentesis. fungi form mycorrhizal networks with plant Ans. (a) Advantages of apomictic seeds to a farmer are: roots, enhancing nutrient uptake, particularly (i) Reduces the cost of hybrid breeding phosphorus and nitrogen. This relationship programmes. improves plant growth, stress tolerance and overall health. Additionally, Glomus species contribute to (ii) Desired traits can be maintained without losing soil structure and fertility, promoting sustainable superiority of hybrids over parents. Farmers ecosystems and agricultural productivity. can replant these seeds year after year. OR 21. If the base adenine constitutes 31% of an isolated DNA fragment, then write what will be the (b) Advantage: Amniocentesis aids in detecting expected percentage of the base cytosine in it. genetic defects in the fetus. Explain how did you arrive at the answer given. Disadvantage: Amniocentesis is also employed  for determining the sex of the unborn child, contributing to an increase in cases of female Ans. Given: Cytosine= 19% foeticide. According to Chargaff’s rule in a DNA molecule 18. 5' – G↓AATTC – 3' A+T+G+C=100% 3 – CTTAA↑ G – 5' Percentage of A=Percentage of T (a) Name the restriction enzyme that recognises Percentage of G=Percentage of C the given specific sequence of bases. What are Therefore, If A=31% then, T=31% such sequence of bases referred to as? G+C=38% (b) What are the arrows in the given figure Therefore, the total percentage of adenine and indicating? Write the result obtained thereafter. thymine together is 31% + 31% = 62%.  Since the total percentage of all four bases (adenine, Ans. (a) Restriction enzyme: EcoRI. thymine, cytosine and guanine) in DNA is 100% Palindromic sequences. and adenine and thymine together constitute 62%, (b)The arrows indicate the restriction sites for the remaining percentage for cytosine and guanine EcoRI. together would be 100% – 62% = 38%. Result: Since guanine is equal to cytosine according 5'–G AATTC–3' to Chargaff's rules, we divide the remaining 3'–CTTAA G–5' percentage (38%) by 2 to get the percentage of 19. Observe the population growth curve and answer cytosine alone. the questions given below: 38% / 2 = 19% Therefore, the expected percentage of cytosine in the DNA fragment is 19%. SECTION - C 22. Identify a, b, c, d, e and f in the table given below:  Sl. No. Organism Bioactive Use Molecule 1. Monascus a b (a) State the conditions under which growth purpureus curve 'A' and growth curve 'B' plotted in the 2. c d Antibiotic graph are possible. 3. e Cyclosporin A f (b) Mention what does 'K' in the graph represent.  Ans. a– Statin Ans. (a)  Growth A is possible when the resources are b– blood cholesterol lowering agent. sufficient and population has enough space to c– Penicillium notatum grow under no competition. Such curves are generally unrealistic in nature. Growth B is possible d– Penicillin when the population have enough resources to e– Trichoderma polysporum grow which gradually deplete with time. f– Immunosuppressive agent. 672 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th 23. (a) Tropical regions harbour more species than the Example: Marine ecosystem: In marine ecosystems, temperate regions. How have biologists tried to phytoplanktons are the primary producers in explain this in their own ways? Explain. these ecosystems and support vast populations OR of zooplankton, which in turn support smaller (b) (i) What does an ecological pyramid represent? populations of small fish, followed by larger predators. In this case, the biomass or energy (ii)  The ecological pyramids may have an at the higher trophic levels (zooplankton and 'upright' or an 'inverted' shape. Justify fish) may exceed that of the lower trophic levels with the help of suitable examples. (phytoplankton), resulting in an inverted pyramid. Ans. (a) There are three different hypothesis proposed by scientists for explaining species richness in the tropics. (i) Tropical latitude receives more solar energy than temperate regions, which lead to high productivity and high species diversity. (ii) Tropical regions have less seasonal variations and have more or less constant environment. This promotes the niche specialisation and thus, high species richness. (iii) Temperate regions were subjected to glaciations during ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.1+1+1 Figure: Pyramid of biomass in sea OR 24. (a) What are transgenic animals? (b) (i) Ecological Pyramids: The representation of a (b) Name the transgenic animal having the largest food chain in the form of a pyramid is called number amongst all the existing transgenic ecological pyramids.1 animals. (ii) Depending on the ecosystem and the specific (c) State any three reasons for which these types trophic relationships within it, ecological of animals are being produced. 3 pyramids can exhibit either an upright or an Ans. (a) Transgenic Animals: These are the animals inverted shape. whose genome has been altered by the (a) Upright Ecological Pyramid: In most ecosystems, introduction of an extra (foreign) gene by ecological pyramids have an upright shape, manipulation. 1 meaning that energy or biomass decreases as we move up the trophic levels. (b) Over 95% of all existing transgenic animals are mice. ½ Example: Terrestrial ecosystems, such as forests (c) (i) Normal physiology and development. or grasslands, typically have upright ecological (ii) Study of disease. pyramids. For instance, in a forest ecosystem, (iii) Biological products. producers (plants) form the broad base of the pyramid, followed by primary consumers (iv) Vaccine safety. (herbivores), secondary consumers (carnivores) (v) Chemical safety testing. (Any three) and so on. Each successive trophic level supports 25. If the cells in the leaves of a maize plant contain fewer organisms and less biomass, reflecting the 10 chromosomes each, write the number of loss of energy as it moves up the food chain. chromosomes in its endosperm and zygote. Name and explain the process by which an endosperm and a zygote are formed in maize. Ans. Endosperm (3n): If the cells in the leaves of the maize plant contain 10 chromosomes each, the endosperm would contain 10 chromosomes x 3 = 30 chromosomes.1 Zygote (2n): If the cells in the leaves of the maize plant contain 10 chromosomes each, the zygote would contain 10 chromosomes x 2 = 20 chromosomes.1 The process by which an endosperm and zygote are formed in maize is called double fertilisation. It involves the fusion of two sperm cells from the pollen grain with two different nuclei in the (b) Inverted Ecological Pyramid: In some ecosystems, embryo sac. One sperm cell fertilises the egg cell particularly aquatic ecosystems, ecological pyramids to form the diploid zygote, while the other sperm may exhibit an inverted shape, where biomass or cell fuses with two polar nuclei to form the triploid energy increases as we move up the trophic levels. endosperm nucleus. SOLVED PAPER - 2024 (BIOLOGY) [ 673 26. (a) Why does DNA replication occur within a Now, this transcribed hnRNA undergoes additional replication fork and not in its entire length processing called capping and tailing. In capping, simultaneously? methyl guanosine triose phosphate is added to the (b) "DNA replication is continuous and 5' end and in tailing, 200–300 adenylate residues are discontinuous on the two strands within the added at 3’ end of spliced RNA. replication fork." Explain with the help of a The processing of hnRNA into mature mRNA occurs schematic representation. predominantly within the nucleus of eukaryotic Ans. (a) DNA being very long, requires high energy for cells. Once the mRNA is fully processed, it is opening along its entire length. exported from the nucleus to the cytoplasm, where it can be translated into protein by ribosomes. (b) Schematic representation of DNA replication: 28. The world is facing accelerated rates of species extinction largely due to human activities. Explain any three human activities responsible for accelerated rates of species extinction. Ans. (i)  Habitat loss and fragmentation: Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agricultural, mining and urbanisation. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange 2 between populations leading to a declination of Explanation: (i) The process of DNA replication species. For example: begins at a point called the origin of replication Tropical rain forests once covering 14% Earth (ori), to form a replication fork. surface, now reduced to 6%. Thousands hectares (ii)  The separated strands act as templates for the of rain forests are being lost within hours. synthesis of new strands. The Amazon rain forest is being cut for cultivating (iii) DNA replicates in the 5’ → 3’ direction. soya beans or for conversion of grass lands for (iv) dNTPs (Deoxyribonucleotide triphosphate) act cattle.  (Any one e.g.) 1 as substrate and also provide energy for the (ii) Over–exploitation: Due to over–hunting polymerisation of nucleotides. and over–exploitation of various plants and (v)  DNA polymerase is an enzyme that assembles animals by humans, many species have a new DNA strand that is complementary to become endangered or extinct. Many species the template strand. like Stellar’s sea cow, Passenger pigeon, etc. extinct due to over exploitation. 1 (vi) DNA polymerase continues to move along the template strand and add new nucleotides to (iii) Alien species invasions: Accidental or the growing or complementary strand until intentional introduction of non–native species the entire genome is replicated. into a habitat has led to the declination or extinction of indigenous species. Alien species (vii) The DNA polymerase forms one new strand cause decline or extinction of indigenous (leading strand) in a continuous stretch in the species. For example: 5’ → 3’ direction (continuous synthesis). The Nile Perch introduced in Lake Victoria (viii) T  he other new strand is formed in small (East Africa) caused extinction of more than 200 stretches (Okazaki fragments) in the 5’ → 3’ species of native fish, cichlid fish in the lake. direction (discontinuous synthesis). Invasive weed species like carrot grass (ix)  The Okazaki fragments are then joined (Parthenium), Lantana and water hyacinth together to form a new strand by an enzyme, (Eicchornia) caused damage to our native DNA ligase. This new strand is called the species. lagging strand. The illegal introduction of the African Catfish 27. Explain the processing of heterogeneous nuclear (Clarias gariepinus) for aquaculture is posing a RNA (hnRNA) into a fully functional mRNA in threat to the indigenous catfishes in our rivers. eukaryotes. Where does this processing occur in (any one e.g.) the cell? Ans. The hnRNA in eukaryotes needs to change for SECTION - D converting it into functional mRNA. The hnRNA Q. No. 29 and 30 are case based questions. Each question contains both exons and introns. The exons are has three sub-questions with internal choice in one sub- functional coding segments while introns are non– question. functional and non–coding sequences. This hnRNA undergoes processing where the introns are 29. In a human female, the reproductive phase removed and exons are joined by a process called starts on the onset of puberty and ceases around splicing. middle age of the female. Study the graph given 674 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th below regarding menstrual cycle and answer the Implanting the early embryo, up to 3 questions that follow: blastomeres, during this phase ensures that Hormone levels the uterus is in a receptive state, maximising the chances of successful implantation and Pituitary FSH LH ultimately leading to a successful pregnancy. Therefore, in ART procedures, embryo transfer typically takes place during the luteal phase of Ovarian events Developing Regressing Developing follicle Mature follicle corpus luteum corpus luteum the menstrual cycle. (c) The hormone responsible for the events Ovulation occurring during the proliferative phase of Hormone levels the menstrual cycle is estrogen. Estrogen is Ovarian Estrogen Progesterone primarily produced by the developing ovarian follicles in the ovaries Follicular (Proliferative) phase: In this phase, Uterne events Menses the primary follicle matures into Graafian follicles. This causes the regeneration of Days 1 3 5 Menstruation 7 9 11 Follicular 13 15 17 19 21 23 25 27 29/1 Luteal Next cycle endometrium. (proliferative phase) (secretory phase) begins Estrogens stimulate Proliferation of ruptured uterine endometrium (a) Name the hormones and their source organ, and mucous lining of oviduct and vagina. which are responsible for menstrual cycle at puberty. Development of secondary sexual characters. (b) For successful pregnancy, at what phase of Estrogen inhibits the secretion of FSH and the menstrual cycle an early embryo (upto stimulates the secretion of LH. It also causes 3 blastomeres) should be Implanted in the the thickening of the uterine endometrium. 2 Uterus (IUT) of a human female who has OR opted for Assisted Reproductive Technology (c) After ovulation, the remains of the Graafian follicle (ART)? Support your answer with a reason. 1 get converted into the yellow endocrine mass (c) Name the hormone and its source organ called corpus luteum. It secretes progesterone. responsible for the events occurring during Progesterone hormone helps in maintenance and proliferative phase of menstrual cycle. preparation of endometrium for the implantation Explain the event. of the embryo. During pregnancy all events of OR menstrual cycle stop and there is no menstruation. (c) In a normal human female, why does In the absence of fertilisation, corpus luteum menstruation only occurs if the released ovum degenerates and becomes a whitish mass called is not fertilised? Explain. corpus albicans. Since, it has no secretory ability, Ans. (a) Follicle–stimulating hormone (FSH): Source: progesterone level in blood decreases. It leads to Pituitary gland menstruation indicating a new cycle. Luteinising hormone (LH): Source: Pituitary 30. Read the following passage and answer the gland questions that follow: Estrogen: Ovaries (specifically, the developing "Mosquitoes are drastically affecting the human ovarian follicles) health in almost all the developing tropical Progesterone: Source: Ovaries (specifically, the countries. Different species of mosquitoes cause corpus luteum) very fatal diseases so much so that many humans (b) For successful pregnancy through Assisted loose their life and if they survive, are unable to Reproductive Technology (ART), such as in put in productive hours to sustain their life. With vitro fertilisation (IVF), the early embryo (up the result, the health index of the country goes to 3 blastomeres) should be implanted in the down." uterus (UT) during the luteal phase of the (a) Name the form in which Plasmodium gains menstrual cycle. entry into (i) human body (ii) the female (i) During the luteal phase of the menstrual Anopheles body. cycle, when progesterone levels are elevated, (b) Why do the symptoms of malaria not appear in the endometrium (uterine lining) undergoes a person immediately after being bitten by an changes that prepare it for embryo implantation infected female Anopheles? Give one reason. during assisted reproductive technology (ART) Explain when and how do the symptoms of procedures such as in vitro fertilisation (IVF). the disease would appear. (ii) The thickened and enriched endometrium, along with the presence of optimal blood vessel OR development, creates an ideal environment (b) Explain the events which occur within a for successful embryo implantation and female Anopheles mosquito after it has sucked subsequent pregnancy. blood from a malaria patient. SOLVED PAPER - 2024 (BIOLOGY) [ 675 (c) Name the species of mosquito other than sporozoites into the person's bloodstream. This female Anopheles and the disease, for which it transmission of sporozoites along with saliva carries the pathogen. initiates the infection of the new human host with Ans. (a) (i) Entry into the human body: Plasmodium malaria. gains entry into the human body in the form of (c) Aedes aegypti: It carries the pathogens responsible sporozoites. for dengue fever. (ii) Entry into the female Anopheles body: Plasmodium gains entry into the female Anopheles body in the SECTION - E form of gametocytes. 31. (a) (i)  Draw a schematic diagram of the cloning (b) The symptoms of malaria do not appear immediately vector pBR 322 and label (1) Bam HI site (2) after being bitten by an infected female Anopheles gene for ampicillin resistance (3) 'ori' (4) 'rop' mosquito due to the incubation period of the gene. malaria parasite within the human body which is 7-30 days. (ii) State the role of 'rop' gene. One reason for the delay in symptom onset is that (iii) A cloning vector does not have a selectable after being bitten by an infected mosquito, the marker. How will it affect the process of Plasmodium parasites initially enter the bloodstream cloning? in the form of sporozoites. These sporozoites travel (iv) Why is insertional inactivation preferred to the liver, where they undergo replication and over the use of selectable markers in cloning maturation within hepatocytes (liver cells). During vectors? this pre–erythrocytic phase, the parasites multiply OR without causing noticeable symptoms. This phase (b) (i)  Name the nematode (scientific name) that of the infection can last from several days to a few infects the roots of tobacco plant and reduces its weeks, depending on the species of Plasmodium. yield. After the pre–erythrocytic phase, the matured (ii) Name the vector that is used to introduce parasites are released from the liver into the nematode-specific genes into the host plant bloodstream, where they invade red blood cells (tobacco). (erythrocytes) and undergo further replication. It (iii) How do sense and anti-sense RNAs function? is during this erythrocytic phase of the infection that symptoms of malaria typically appear. The (iv) Why could parasite not survive in a transgenic destruction of red blood cells by multiplying tobacco plant? parasites leads to the release of toxins and triggers Ans. (a) (i) Schematic diagram of pBR322: an immune response, resulting in symptoms such as fever, chills and headache, fatigue and muscle aches. OR (b) (i) When a female Anopheles mosquito feeds on the blood of an infected human host, it ingests red blood cells (RBCs) containing gametocytes, the sexual stage of the malaria parasite. (ii) After ingestion, further development of the gametocytes occurs in the mosquito's stomach wall, where male and female gametes fuse to form a zygote. The zygote undergoes further development and differentiation to form sporozoites. (iv) These sporozoites migrate from the stomach wall to different organs in the mosquito's body cavity. (ii) The rop gene encodes a small regulatory protein, Many of them penetrate the salivary glands of the Rop (repressor of primer), in pBR322 plasmids. The mosquito, where they become concentrated and primary role of this protein is to control the copy await transmission to a new host. number of the plasmid within the bacterial host.  1 (v) When the infected female Anopheles mosquito (iii) The absence of a selectable marker in a cloning bites a healthy person, it injects saliva containing vector complicates the process of cloning by 676 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th making it difficult to distinguish between cells that OR have incorporated the vector with the desired DNA (b) (i) With the help of labelled diagram only, show insert and those that haven't. Selectable markers, the different stages of embryo development in like antibiotic resistance genes, help in the growth a dicot plant. and selection of transformed cells, simplifying the process by allowing the isolation of only those cells (ii) Endosperm development precedes embryo that have successfully taken up the vector.  1 development. Justify. (iv) Insertional inactivation is favoured over the use Ans. (a) (i) Structure of human sperm: of selectable markers in cloning vectors because it facilitates the straightforward identification of recombinant plasmids without requiring extra selection procedures. With insertional inactivation, the cloning vector harbours a reporter gene interrupted by the insertion site for DNA fragments. Successful insertion of a DNA fragment disrupts the reporter gene, causing a loss of function that can be visually observed. This approach streamlines the cloning process and eliminates the necessity for antibiotic selection, making it more convenient and economical. OR (b) (i) Meloidogyne incognita. (ii) Vector used is Agrobacterium tumefaciens. (iii) Sense RNA: Sense RNA refers to the RNA sequence that is complementary to the coding strand of DNA. It has the same sequence as the mRNA that will be translated into a protein. Sense RNA is transcribed from the template strand of DNA  (Any four parts) during transcription. (b) Functions of parts: Antisense RNA: Antisense RNA is the RNA a. Head: The head of the sperm contains the nucleus, sequence that is complementary to the coding which houses the genetic material (DNA) necessary (sense) strand of DNA and to the sense RNA. for fertilisation. The head is covered by a cap–like Antisense RNA molecules are often involved in structure called the acrosome, which contains the regulation of gene expression by binding to enzymes that aid in penetrating and fertilising the complementary sequences in mRNA or sense RNA egg. through base pairing. This binding can inhibit b. Midpiece: The midpiece of the sperm contains translation or induce degradation of the target numerous mitochondria, which provide the mRNA, leading to reduced protein expression.  1 energy (in the form of ATP) needed for the sperm's (iv) Parasites might not survive in transgenic tobacco movement (motility). Mitochondria are essential for plants due to the expression of nematode–specific powering the beating motion of the sperm's tail. genes introduced into the plant. These genes could c. Tail (Flagellum): The tail, also known as the encode proteins or RNAs that disrupt the parasite's flagellum, is responsible for the sperm's locomotion. life cycle, development, or its ability to infect the It propels the sperm forward through the female plant. Moreover, transgenic tobacco plants may reproductive tract towards the egg, allowing it to demonstrate heightened resistance mechanisms, reach and penetrate the egg for fertilisation. including modifications in cell wall composition or d. Neck: The neck of the sperm connects the head heightened production of defensive compounds, to the midpiece and plays a role in facilitating the which can impede parasite infection and survival. movement of genetic material from the head to the 32. (a) (i) Draw a diagram of a human sperm. Label any tail during sperm maturation and motility. four parts and write their functions. (ii) During fertilisation, the sperm induces changes in (ii) In a human female, probability of an ovum the zona pellucida and blocks the entry of other to get fertilised by more than one sperm is sperms. This ensures monospermy that only one impossible. Give reason. sperm fertilises the ovum and prevents polyspermy. SOLVED PAPER - 2024 (BIOLOGY) [ 677 OR (i) Stages in embryo development in a dicot plant: (ii) Endosperm development precedes embryo devel- (iii) Write the conclusion they arrived at the end of opment in flowering plants to provide essential their experiment. nutrients and support for the growing embryo. The Ans. (a) (i) Directional selection1 formation of endosperm through the fusion of male (ii) Before Industrialisation (1850s): There were more and female gametes ensures the establishment of a white–winged moths (Biston betularia) on trees nutritive tissue that supplies the developing em- than dark–winged or melanised moths (Biston bryo with carbohydrates, proteins and other essen- carbonaria). tial nutrients. Once the endosperm is established The reason for this was the presence of white and functional, embryo development commences, coloured lichen covering the trees, providing utilising the nutrients provided by the endosperm camouflage for the white–winged moths against for its growth and differentiation. This sequential predators. In contrast, the dark–winged moths process ensures proper nourishment and support stood out against the light background and were for the developing embryo, ultimately contributing more easily spotted and preyed upon. to the successful formation of viable seeds. After Industrialisation (1920): The situation reversed, with more dark–winged moths and fewer white–winged moths observed. 33. This change occurred due to the darkening of tree trunks caused by pollution from industrial smoke and soot. The presence of pollutants inhibited the growth of lichens on tree trunks. As a result, the light–coloured lichens disappeared and the tree trunks became darker. In this altered environment, the white–winged moths were now easily spotted against the dark background by predators, while the dark–winged moths were better camouflaged and had a survival advantage. This example of industrial melanism illustrates how environmental changes resulting from industrialisation can drive evolutionary shifts in populations, favouring individuals with traits that (a) Natural selection operates in different ways in provide better camouflage and survival advantages nature. in the altered environment. (i) Identify the type of natural selection depicted (iii) Anthropogenic actions, which are activities in the graph above. produced or caused by humans, can indeed (ii) In England after industrialisation, the enhance the rate of evolution and have significant population of dark winged moths were more impacts on natural developments. For example favoured than white winged moth. Explain.  (a) Industrial Melanism: The example of industrial (iii) Anthropogenic action can enhance the rate of melanism illustrates how anthropogenic activities, evolution. Explain with the help of an example. such as industrial pollution emitting smoke and  soot, can lead to changes in the environment that OR drive evolutionary shifts in populations. In this case, the darkening of tree trunks due to pollution (b) (i) Why did Hershey and Chase use '35S' and '32P' in favoured the survival of dark–winged moths over their experiment? Explain. white–winged moths, demonstrating how human– (ii) State the importance of (1) blending and (2) induced environmental changes can influence the centrifugation in their experiment. evolution of species. 678 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th (b) Use of Herbicides and Pesticides: Anthropogenic into the DNA component of the bacteriophage, activities such as the widespread use of herbicides Hershey and Chase could track the location of DNA and pesticides in agriculture have led to the during viral infection. selection of resistant varieties of plants and the Using these radioactive isotopes, Hershey and development of antibiotic–resistant bacteria. This Chase were able to determine that only the DNA rapid evolution of resistance in response to human of the bacteriophage, not the protein, enters the interventions demonstrates how anthropogenic bacterial cell during infection. actions can accelerate the rate of evolutionary change in populations, often within short time (ii) Blending: Blending facilitated the physical scales. separation of bacteriophage protein coats from OR bacterial cells after viral infection. (b) (i) Hershey and Chase used radioactive isotopes of Centrifugation: Centrifugation enabled the sulphur–35 (35S) and phosphorus–32 (32P) in their separation of bacteriophage protein coats from experiment to distinguish between protein and bacterial cells and their contents. By subjecting DNA in bacteriophages. the bacteriophage infected bacterial suspension to Sulphur–35 (35S) labels proteins, as proteins contain centrifugation, the heavier bacterial cells and their sulphur but no phosphorus. By incorporating 35S contents, including any viral DNA or RNA, were into the protein component of the bacteriophage, pelleted to the bottom of the tube, while the lighter Hershey and Chase could track the location of protein coats remained in the supernatant. proteins during viral infection. (iii) Based on the experiment, Hershey and Chase Phosphorus–32 (32P) labels DNA, as DNA contains concluded that DNA, not protein, is the genetic phosphorus but no sulphur. By incorporating 32P material responsible for heredity and inheritance. Delhi Set-2 57/5/2 Note: Except these, all other questions are available in Delhi - Set 1. (C) substance resistant to microbial action SECTION - A (D) colloidal substance 2. How many base pairs will be there in 20 Ans. Option (A) is correct. nucleosomes in a DNA double helix? Explanation: Humus is dark coloured colloidal (A) 4000 (B) 40 substance called humus. Humus is resistant to (C) 20 (D) 2000 microbial action and so decomposes very slowly. Ans. Option (A) is correct. Being colloidal in nature it serves as a reservoir of nutrients. Explanation: In general, a nucleosome typically Question number 13 to 16 consist of two statements – encompasses approximately 200 base pairs (bp) of Assertion (A) and Reason (R). Answer these questions the DNA helix. However, in the context of human selecting the appropriate option given below: nucleosomes, specifically, they consist of around 146 base pairs of the DNA helix. Therefore, for 20 (A) Both (A) and (R) are true and (R) is the correct nucleosomes: explanation of (A). Number of base pairs = 200 base pairs/nucleosome (B) Both (A) and (R) are true, but (R) is not the correct × 20 nucleosomes explanation of (A). = 200 × 20 = 4,000 base pairs (C) (A) is true, but (R) is false. 7. Which one of the following enzymes a fungal cell (D) (A) is false, but (R) is true. should be treated with to get the DNA along with 14. Assertion (A): The sugar–phosphate backbone other macro molecules released from it? of two chains in DNA double helix show anti- (A) Isozymes (B) Cellulose parallel polarity. (C) Ribonuclease (D) Chitinase Reason (R): The phosphor–diester bonds in one strand go from a 3' carbon of one nucleotide to a Ans. Option (D) is correct. 5' carbon of adjacent nucleotide, whereas those in Explanation: Chitinase is the enzyme that degrades complementary strand go vice versa. chitin, a major component of the fungal cell wall. Ans. Option (A) is correct. Treating fungal cells with chitinase helps to break down the cell wall, releasing DNA along with other Explanation: The DNA double helix features macro molecules from the cell. an antiparallel arrangement, where the sugar– phosphate backbones of the two strands run in 9. Which one of the following is not a characteristic opposite directions. This means that one strand feature of "humus" that is formed during runs from 5’ to 3’ while the complementary strand decomposition of detritus? runs from 3’ to 5’. The phosphodiester bonds within (A) amorphous, colloidal, dark coloured substance each strand connect the sugar of one nucleotide to (B) amorphous, colloidal, light coloured the phosphate of the next, forming a continuous substance backbone. SOLVED PAPER - 2024 (BIOLOGY) [ 679 SECTION - B Grazing food chain Detritus food chain Energy is derived Energy comes from organic 17.Answer the questions based on the typical biogas from the sun through matter or detritus generated plant diagram given below: photosynthesis by in trophic levels of the primary producers grazing food chain. Gas (plants). ‘Y’ It begins with primary It begins with detritus, such producers. as dead bodies of animals or fallen leaves. ‘X’ Grass → Rabbit → Fox Leaf litter → Bacteria → ‘Z’ (CH4+CO2 +H2S) Earthworm → Bird (b) The detritus food chain may be connected to the grazing food chain at some levels in an ecosystem. For example: Leaf litter → Bacteria → Earthworm → Bird  In this example, the detritus food chain Digester (involving earthworms as detritivores) is connected to the grazing food chain through the consumption of detritus by organisms that are part of the grazing food chain. 24. (a) Why must a cell be made 'competent' in biotechnology experiments? How does (a) Identify 'X', 'Y' and 'Z'. calcium ion help in doing so? (b) Why is dung preferred for the production of (b) State the role of "biolistic gun" in biotechnology biogas? experiments. Ans. (a)  In the given diagram of a biogas plant, X is Ans. (a) In biotechnology experiments, the cells must be sludge, Y represents biogas outlet pipe and Z made competent so that they can take up the represents dung + water. hydrophilic DNA molecule inside them from the external medium. Treatment of bacterial cells with divalent calcium cations makes them competent and helps them to take up the DNA through the pores in the cell wall. (b) Biolistic gene or gene gun is a method of introducing alien DNA into the plants cells. In this method, the host cells are bombarded with high–velocity micro–particles of gold or tungsten coated with DNA molecules, facilitating genetic modification without the need for complex tissue culture. 26. Expression of different genes for different traits may show dominance, incomplete dominance or co-dominance. Write about expression of such genes with the help of one example each. Ans. The expression of different genes for different traits can exhibit various patterns, including dominance, incomplete dominance, or co–dominance. (i) Dominance: Dominance occurs when one allele completely masks the expression of the (b) Cattle dung is preferred for the production other allele in a heterozygous individual. In of biogas due to its high content of organic this case, the dominant allele is expressed matter and methane–producing bacteria called phenotypically, while the recessive allele Methanobacterium. remains hidden. For example Mendel’s pea plant experiments with flower color. In pea SECTION - C plants, the allele for purple flowers (P) is dominant over the allele for white flowers (p). 22. (a) How is the grazing food chain different from When a plant inherits one allele for purple the detritus food chain? flowers (PP) and one allele for white flowers (b) "The detritus food chain may be connected (pp), it will have purple flowers because the to the grazing food chain at some levels in an dominant purple allele masks the expression of the recessive white allele. ecosystem." Give an example in support of the (ii) Incomplete Dominance: It is an inheritance statement. in which heterozygous offspring shows an Ans. (a)  Differences between grazing food chain and intermediate character between two parental detritus food chain: characteristics. For example Flower colour in 680 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th snapdragon (dog flower or Antirrhinum sp.). Ans. (a) (i) Schematic representation of spermatogenesis: In snapdragons, the allele for red flowers (R) is incompletely dominant over the allele for white flowers (r). When a plant inherits one allele for red flowers (RR) and one allele for white flowers (rr), it will have pink flowers, exhibiting a blend of the red and white colours. (iii) Co–dominance: It is the inheritance in which both alleles of a gene are expressed equally and independently in a hybrid, i.e., both the alleles are dominant. For example ABO blood grouping in humans. ABO blood groups are controlled by the gene. The gene (I) has three alleles IA, IB and i. However, a person can have any two of these three alleles. IA and IB both are dominant alleles while i is a recessive allele. The alleles IA and IB produce antigen A and antigen B respectively on the RBC surface while allele i doesn’t produce any antigen. When IA and IB are present together they both express their types of surface antigen A and B. This is due to co–dominance. SECTION - E (ii) GnRH acts on the anterior pituitary to 32. (a)(i) Describe the events of spermatogenesis with secrete luteinising hormone (LH) and follicle the help of a schematic diagram. stimulating hormone (FSH). (ii) Explain the role of hormones in spermatogenesis. LH acts at the Leydig cells and stimulates OR the synthesis and secretion of androgens. (b) (i) Show the development of megaspore mother Androgens, in turn, stimulate the process of cell upto the formation of mature embryo sac spermatogenesis. in flowering plants with the help of labelled diagrams only. FSH acts on the Sertoli cells and stimulates the (ii) How does geitonogamy differ from xenogamy? secretion of some factors which help in the (iii) Name the type of flowers that are invariably process of spermiogenesis. autogamous. OR (b) (i) SOLVED PAPER - 2024 (BIOLOGY) [ 681 (ii) Parts of the ovule showing a large megaspore the stigma of a flower on a different individual plant mother cell, (a) a dyad and a tetrad of megaspores; (b) of the same species. 2, 4 and 8–nucleate stages of embryo sac and a mature embryo sac; (c) A diagrammatic representation of (iii) The type of flowers that are invariably the mature embryo sac.Geitonogamy occurs when autogamous are called cleistogamous flowers. pollen from the anther of one flower is transferred to Cleistogamous flowers are a type of closed, the stigma of another flower on the same individual self–pollinating flower where the anthers and plant. Xenogamy, on the other hand, occurs when stigma are in close proximity, often within the pollen from the anther of one flower is transferred to same flower. Delhi Set-3 57/5/3 Note: Except these, all other questions are available in Delhi - Set 1 + 2. 1, which contains approximately 2,968 genes. On SECTION - A the other hand, the Y chromosome has the least number of genes among the autosomes, with 2. Which one of the following chromosomal event around 56 protein–coding genes. will not result in genetic variation amongst the Question numbers 13 to 16 consist of two statements offsprings? Assertion (A) and Reason (R). Answer these questions (A) Independent assortment selecting the appropriate option given below: (B) Crossing over (A) Both (A) and (R) are true and (R) is the correct (C) Linkage explanation of (A). (D) Mutation (B) Both (A) and (R) are true, but (R) is not the correct Ans. Option (C) is correct. explanation of (A). Explanation: When genes are tightly linked, (C) (A) is true, but (R) is false. they tend to be inherited as a unit, reducing the (D) (A) is false, but (R) is true. likelihood of genetic variation among the offspring. 14. Assertion (A): 'Biodiversity hotspots' are the 8. The source of 'Smack' is: regions which possess high levels of species (A) Leaves of Cannabis sativa richness, high degree of endemism. (B) Flowers of Datura Reason (R): Total number of biodiversity hotspots (C) Fruits of Erythroxylum coca in the world is 22 with two of these hotspots found (D) Latex of Papaver somniferum in India. Ans. Option (D) is correct. Ans. Option (A) is correct. Explanation: Heroin, commonly called smack Explanation: Biodiversity hotspots are the regions (chemical name diacetylmorphine) is a white, characterised by high levels of species richness and odourless, bitter crystalline compound. This is a high degree of endemism. Globally, there are 22 obtained by acetylation of morphine, which is recognised biodiversity hotspots. Notably, India is extracted from the latex of the poppy plant (Papaver home to four of these hotspots: the Western Ghats, somniferum). Indo–Burma region, the himalayas ans sundaland. 9. The first antibiotic was discovered accidentally by A while working on B. 'A' and 'B' are SECTION - B (A) A-Waksman; B-Streptococcus 20. List the events that reduce the Biochemical (B) A-Fleming; B-Penicillium notatum Oxygen Demand (BOD) of a primary effluent (C) A-Waksman; B-Bacillus brevis during sewage treatment. (D) A-Fleming; B-Staphylococci Ans. Events that lead to biogas production from waste Ans. Option (B) is correct. water with reduced BOD are: Explanation: The first antibiotic, penicillin, was (i) Once the BOD of wastewater is significantly discovered accidentally by Alexander Fleming reduced, the effluent is passed into a settling while he was working with the mold Penicillium tank for sedimentation. notatum. (ii) From the settling tank, the major part of 12. The human chromosome with the highest and sedimented material called activated sludge least number of genes in them are respectively: (bacterial flocs) is pumped into large tanks (A) Chromosome 21 and Y. called anaerobic sludge digester and a small part is pumped back into the aeration tank to (B) Chromosome 1 and X. serve as inoculum. (C) Chromosome 1 and Y. (iii) In these tanks, the sludge is anaerobically (D) Chromosome X and Y. digested by bacteria and fungi, biogas is Ans. Option (C) is correct. produced which is a mixture of methane, Explanation: The human chromosome hydrogen sulphide and CO2. The biogas can be with the highest number of genes is chromosome used as a source of energy as it is inflammable. 682 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th 27. (a) Construct a pyramid of biomass in sea with SECTION - C phytoplankton and fishes. Explain giving 23. (a) "Mother's milk is considered very essential for reasons about the characteristic of the the new born infant." Justify. constructed pyramid. (b) What is a 'vaccine'? Explain the principle on (b) In which condition will the pyramid remain which it works. always upright? Ans. (a)  The milk produced during the initial few Ans. (a) Pyramid of biomass in sea days of lactation is called colostrum which contains several antibodies essential to develop resistance for the new–born babies. Breast–feeding during the initial period of infant growth is recommended by doctors for bringing up a healthy baby. (b) Vaccine is an antigen protein of pathogens or inactivated or weakened pathogens or their toxin. When introduced into the body, vaccines stimulate the immune system to produce an immune response, including the production of antibodies and memory cells. Principle: During the primary response, the immune system generates antibodies specific to the antigen present in the vaccine. A pyramid of biomass is a graphical Additionally, memory cells are produced, representation of the total amount of living which are specialised immune cells that matter present at each trophic level of an “remember” the pathogen’s antigen. If the individual is exposed to the same pathogen ecosystem. Pyramid of biomass in sea is in the future, these memory cells quickly generally inverted because the biomass of recognise the antigen and initiate a rapid fishes far exceeds that of phytoplankton. and robust immune response, known as the (b) The pyramid of energy is always upright secondary response. This secondary response because energy decreases as it moves through is more rapid and effective compared to the trophic levels in an ecosystem due to the primary response, providing long–lasting second law of thermodynamics, which states immunity against the pathogen. that energy is lost as heat during each transfer. 25. State why plant breeders are interested in artificial hybridisation programme. How do they carry out SECTION - E this process? Ans. Plant breeders are interested in artificial 33. (a) Work out a dihybrid cross upto F2 generation hybridisation programmes because it allows them between pea plants bearing violet coloured to introduce desirable traits from different plant axial flowers and white coloured terminal varieties or species into a single plant, creating flowers using Punnet's square. Give their new cultivates with improved characteristics such F2 phenotypic ratio. State the Mendel's law as higher yield, disease resistance, pest resistance, of inheritance that was derived from such a better taste or adaptability to specific environmental cross. conditions. In this method, desired pollen grains OR are used for pollination. This is achieved by emasculation and bagging techniques. (b) Explain the process of transcription in Emasculation: Emasculation is the process prokaryotes. How is it different from of removal of anthers (using forceps) from transcription in eukaryotes? the bisexual flower bud without affecting the Ans. (a) Cross between pea plants bearing violet female reproductive part, i.e., pistil. coloured axial flowers and white coloured Bagging: Emasculated flowers are then terminal flowers: covered with a suitable bag (made up of butter paper) to prevent contamination of its stigma with unwanted pollen. This is called bagging.  When the stigma of the bagged flower attains receptivity, mature pollen grains collected from anthers of the male parent are dusted on the stigma. Then the flowers are rebagged and allowed to develop the fruits. If the female parent is unisexual, then there is no need for emasculation. In this case, the female flower buds are directly bagged before the flowers open. SOLVED PAPER - 2024 (BIOLOGY) [ 683 VA vA Va va VA VVAA VvAA VVAa VvAa Violet axial Violet axial Violet axial Violet axial vA VvAA vvAA VvAa vvAA Violet axial White Axial Violet axial White Axial Va VVAa VvAa VVaa Vvaa Violet axial Violet axial Violet terminal Violet terminal va VvAa vvAa Vvaa vvaa White Axial White Axial Violet terminal White terminal Phenotypes –  violet  :  violet  :  violet  :  White         axial     axial     terminal   terminal       Phenotype ratio– 9   :   3   :    3   :   1   =1 Law of Independent Assortment: When two pairs template for mRNA synthesis. of traits are combined in a hybrid segregation of The enzyme, RNA polymerase, utilises one pair of characters is independent of the other ribonucleoside triphosphates (ATP, GTP, pair of characters. UTP and CTP) as substrate and polymerises (b) The process by which the DNA message is copied them to form mRNA following the rule of into a strand of mRNA is called transcription. The complementarity. process of transcription is completed in three steps: This process of opening of helix and elongation (i) Initiation: Here, the enzyme RNA polymerase of polynucleotide chain continues until the binds at the promoter site of DNA and initiates enzyme reaches the terminator gene. the process of transcription. It causes the (iii) Termination: local unwinding of the DNA double helix. An RNA polymerase recognises the terminator initiation factor (σ) present in RNA polymerase gene by a termination–factor called rho (ρ) initiates the RNA synthesis. factor. (ii) Elongation: The RNA chain is synthesised in After RNA polymerase reaches the terminator the 5’–3’ direction. region, the newly synthesised mRNA transcript RNA polymerase unzips the DNA double helix along with enzyme is released. and forms an open loop. The proceeded mRNA leaves the nucleus and One of the strands, called sense strand, acts as enters the cytoplasm.             In eukaryotes, there are two additional (ii) Post transcriptional processing (occurs complexities: inside the nucleus): Eukaryotic pre–mRNA (i) There are three RNA polymerases: RNA undergoes post transcriptional processing, polymerase I, which transcribes rRNAs (28S, 18S which includes capping, splicing and and 5.8S), RNA polymerase II, which transcribes polyadenylation, before it is transported out the heterogeneous nuclear RNA (hnRNA) and of the nucleus. These modifications stabilise RNA polymerase III, which transcribes tRNA, the mRNA molecule, facilitate its export to the 5S rRNA and snRNAs (small nuclear RNAs). cytoplasm and regulate gene expression. 684 ] Oswaal CBSE 10 Previous years’ Solved Papers, Class–12th Outside Delhi Set-1 57/4/1 (A) Ovum – B, Morula – D, Blastocyst – F SECTION - A (B) Ovum – A, Morula – B, Blastocyst – G Question No. 1 to 16 are Multiple Choice type questions, (C) Ovum – A, Morula – E, Blastocyst – G carrying 1 mark each. 16×1=16 (D) Ovum – B, Morula – D, Blastocyst – G 1. In a fertilised ovule of an angiosperm, the cells in 2. Option (C) is correct. which n, 2n and 3n conditions respectively occur are: (A) antipodal, zygote and endosperm (B) zygote, nucellus and endosperm (C) endosperm, nucellus and zygote (D) antipodals, synergids and integuments Ans. Option (A) is correct. Explanation: Antipodal cells are indeed typically haploid (n). The zygote is diploid (2n) re