{"id":27568,"date":"2020-07-04T14:53:09","date_gmt":"2020-07-04T09:23:09","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27568"},"modified":"2021-07-02T15:07:07","modified_gmt":"2021-07-02T09:37:07","slug":"2nd-puc-biology-previous-year-question-paper-june-2014","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-biology-previous-year-question-paper-june-2014\/","title":{"rendered":"2nd PUC Biology Previous Year Question Paper June 2014"},"content":{"rendered":"<p>Students can Download 2nd PUC Biology Previous Year Question Paper June 2014, Karnataka <a href=\"https:\/\/kseebsolutions.guru\/2nd-puc-biology-model-question-papers-with-answers\/\">2nd PUC Biology Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n<h2>Karnataka 2nd PUC Biology Previous Year Question Paper June 2014<\/h2>\n<p>Time: 3 hrs 15 min<br \/>\nMax. Marks: 70<\/p>\n<p>General Instructions<\/p>\n<ul>\n<li>This question paper consists of four parts A, B, C and D. Part &#8211; D consists of two sections. Section &#8211; I and Section &#8211; II.<\/li>\n<li>All the parts are compulsory.<\/li>\n<li>Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.<\/li>\n<\/ul>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; A<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer the following questions in one word or one sentence each. (10 \u00d7 1 = 10)<\/span><\/p>\n<p>Question 1.<br \/>\nWhat is menopause?<br \/>\nAnswer:<br \/>\nIn human beings, the menstrual cycle ceases around 50 years of age and it is termed as menopause.<\/p>\n<p>Question 2.<br \/>\nName the physical and physiological connection between maternal body and embryo (foetus).<br \/>\nAnswer:<br \/>\nPlacenta.<\/p>\n<p>Question 3.<br \/>\nWhat is adaptive radiation?<br \/>\nAnswer:<br \/>\nAdaptive radiation is an evolutionary process in which an ancestral stock gives rise to new species in a given geographical area, starting from a point and literally radiating to other geographical areas or habitats.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 4.<br \/>\nMention the cells involved in cell-mediated immunity.<br \/>\nAnswer:<br \/>\nT-Lymphocytes [T-helper cells, T-killer cells]<\/p>\n<p>Question 5.<br \/>\nWhat is Benign tumour?<br \/>\nAnswer:<br \/>\nthe localised tumour which remains at the spot of origin and do not spread to distant sites is a benign tumour.<\/p>\n<p>Question 6.<br \/>\nWhat is totipotency?<br \/>\nAnswer:<br \/>\nThe ability of a plant cell to give rise to an entire plant on the medium is called totipotency.<\/p>\n<p>Question 7.<br \/>\nWhat are bio fertilisers?<br \/>\nAnswer:<br \/>\nBio fertilisers are the organisms that enrich the nutrient quality of the soil. The main source of biofertilizers are bacteria, fungi and cyanobacteria.<\/p>\n<p>Question 8.<br \/>\nWhat is a polymerase chain reaction?<br \/>\nAnswer:<br \/>\nIt is a technique in which a gene of choice is amplified into many copies using primers &amp; DNA polymerase enzyme.<\/p>\n<p>Question 9.<br \/>\nDefine natality.<br \/>\nAnswer:<br \/>\nIt refers to the number of births during a given period in the population that are added to the initial density.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 10.<br \/>\nGive an example for the gaseous cycle.<br \/>\nAnswer:<br \/>\nNitrogen cycle.<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; B<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any five of the following questions in 3 to 5 sentences each: Wherever applicable: (5 \u00d7 2 = 10)<\/span><\/p>\n<p>Question 11.<br \/>\nWhat are hermaphrodites? Give an example.<br \/>\nAnswer:<br \/>\nThe animals in which both the sex organs are present in the same body are called hermaphrodites.<br \/>\neg: Planaria.<\/p>\n<p>Question 12.<br \/>\nMention two symptoms of Turner\u2019s syndrome.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Broad chest.<\/li>\n<li>Low hairline at the back of the neck.<\/li>\n<li>Shortened fourth and fifth fingers.<\/li>\n<li>Short neck with a webbed appearance.<\/li>\n<\/ul>\n<p>Question 13.<br \/>\nWhat is a Homologous structure? Give an example.<br \/>\nAnswer:<br \/>\nHomology is the relationship among organs of different groups of organisms that show similarity in their basic structure and embryonic development but perform different functions.<br \/>\neg: Forelimbs of whales, birds, amphibians and mammals.<\/p>\n<p>Question 14.<br \/>\nDraw a neat labelled diagram of an antibody.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65126\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q14.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014 Q14\" width=\"369\" height=\"311\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q14.png 369w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q14-300x253.png 300w\" sizes=\"(max-width: 369px) 100vw, 369px\" \/><\/p>\n<p>Question 15.<br \/>\nMention any two measures to improve the quality and quantity of milk production.<br \/>\nAnswer:<\/p>\n<ol>\n<li>Selection of quality breeds.<\/li>\n<li>Feeding of the cattle show is done in a scientific manner, considering the quantity and quality of the folder.<\/li>\n<\/ol>\n<p>Question 16.<br \/>\nWhat is commensalism? Give an example.<br \/>\nAnswer:<br \/>\nIt is an interaction in which one species benefits and the other is neither benefited nor harmed.<br \/>\neg: An orchid growing as an epiphyte on a range branch.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 17.<br \/>\nList out any two effects of loss of biodiversity.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Habitat loss and Fragmentation<\/li>\n<li>Overexploitation.<\/li>\n<\/ul>\n<p>Question 18.<br \/>\nWhat are sacred groves? Give an example.<br \/>\nAnswer:<br \/>\nThe tracts of forest conserved and protected based on the religious faith ground.<br \/>\neg: Khasi and Jaintia Hills in Meghalaya.<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; C<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any five of the following questions in about 40 to 80 words each wherever applicable: (5 \u00d7 3 = 15)<\/span><\/p>\n<p>Question 19.<br \/>\nMention the units of vegetative propagation in plants.<br \/>\nAnswer:<br \/>\nRoot tubers<br \/>\nStem: Runners, suckers, stolon, rhizome, corms, bulbs &amp; tubers.<br \/>\nLeaf: Vegetative buds, Bulbils.<\/p>\n<p>Question 20.<br \/>\nDraw a neat labelled diagram of a longitudinal section of a typical flower.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65127\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q20.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014 Q20\" width=\"262\" height=\"274\" \/><\/p>\n<p>Question 21.<br \/>\nWhat is male heterogamety? Explain sex determination in humans.<br \/>\nAnswer:<br \/>\nHeterogametic sex (gametic sex) refers to the sex of a species in which the sex chromosomes are not the same. For example, in humans, males, with an X and a Y sex chromosome, would be referred to as the heterogametic sex, and females having two X sex chromosomes would be referred to as the homogametic sex.<\/p>\n<p>In the human being, the chromosomal mechanism of sex determination is of XX-XY type. In human the nucleus of each cell contains 46 chromosomes or 23 pairs of chromosomes of body characters) and 1 pair is of sex chromosomes (responsible for determination of sex).<br \/>\n1. In the female, two homomorphic sex chromosomes are XX.<br \/>\n2. In the male, two heteromorphic sex chromosomes are XY.<br \/>\n3. The genotypes of female and male are<br \/>\nFemale: 46 Chromosomes = 44 Autosomes + XX sex chromosomes<br \/>\nMale: 46 Chromosomes = 44 Autosomes + XY sex chromosomes<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 22.<br \/>\nDraw a labelled diagram of a diagrammatic representation of Miller\u2019s experiment.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65128\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q22.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014 Q22\" width=\"462\" height=\"361\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q22.png 462w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q22-300x234.png 300w\" sizes=\"(max-width: 462px) 100vw, 462px\" \/><\/p>\n<p>Question 23.<br \/>\nList any three types of Innate Immunity barriers and give an example for each.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Surface barriers. eg: Skin, mucus membrane<\/li>\n<li>Cellular barriers. eg: Phagocytosis, NK Cells<\/li>\n<li>Biochemical barriers. eg: Interferons.<\/li>\n<\/ul>\n<p>Question 24.<br \/>\nExplain any three types of major abiotic factors influencing the life of organisms.<br \/>\nAnswer:<br \/>\n1. The DNA from the desired segment to be amplified, an excess of the two primer molecules, the four deoxyribose triphosphates and DNA polymerase are mixed together in a reaction mixture in an Eppendorf tube with sufficient quantities of Mg. The Eppendorf tube is placed in the PCR unit and the following operations are performed sequentially.<\/p>\n<p>2. Denaturation: The reaction mixture is first subjected to a temperature between 90-98\u00b0C (commonly 94\u00b0C). It results in the separation of two strands of DNA due to the breakage of hydrogen bonds. This is called denaturation. Each strand of DNA acts as a template strand for DNA synthesis. The duration of this step in the first cycle of PCR is usually 2 minutes at 94\u00b0C.<\/p>\n<p>3. Annealing (anneal join): The mixture \u00a1s now cooled to a low temperature (40-60\u00b0C). During this step, two oligonucleotide primers, complementary to a region of DNA, anneal (hybridize) one to each 3 ends of the DNA strand. The duration of the annealing step is usually one minute during the first as well as the subsequent cycles of PCR.<\/p>\n<p>Question 25.<br \/>\nComment: Compressed Natural Gas (CNG) is better than diesel.<br \/>\nAnswer:<br \/>\nCNG Bums most efficiently unlike petrol or diesel in the automobiles and very little of it is left unburnt.<br \/>\nMoreover, it is cheaper than petrol and diesel. It cannot be siphoned off by thieves and adulterated like petrol or diesel.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 26.<br \/>\nWhat is biomagnification? Name two well studied toxic substances causing this phenomenon.<br \/>\nAnswer:<br \/>\nBiomagnification is defined as the accumulation of a particular substance in the body of the organisms at different trophic levels of a food chain. One example of biomagnification is the accumulation of insecticide DDT which gets accumulated in zooplanktons. Small fishes consume these zooplanktons. Small fishes are consumed by large fish which are finally consumed by fish-eating bird present at the highest position in the tropic level. Hence, the organism at the highest trophic level has the maximum accumulation of toxic substances.<\/p>\n<p>Human Health:<br \/>\nHumans become more susceptible to cancers, liver and kidney failure, respiratory disorders, birth defects in pregnant women, brain damage, and heart diseases are a result of mercury, cadmium, lead, cobalt, chromium and other chemical poisonings. For instance, diseases like hepatitis and cancer have been attributed to consuming seafood that has been poisoned by mercury and polycyclic aromatic hydrocarbons (PAPIs).<\/p>\n<p>Marine Creatures:<br \/>\nThe various toxic elements get accumulated in the vital organs of the various aquatic creatures, thus affecting their reproduction and development. Selenium and heavy metals such as mercury also affect the reproduction of aquatic creatures such as fish as it destroys their reproductive organs. Besides, PCBs (polychlorinated biphenyls) also biomagnify and impair reproduction and are considerably high in aquatic systems.<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; D<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Section &#8211; I<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any four of the following questions in about 200 to 250 words each, wherever applicable: (4 \u00d7 5 = 20)<\/span><\/p>\n<p>Question 27.<br \/>\nWhat is double fertilization? Describe fertilized embryosac with a neat labelled diagram.<br \/>\nAnswer:<br \/>\n(a) It is the process in which one of the two male gametes fuses with the egg cell to form a zygote and the other with the secondary nucleus to form P.E.N.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65129\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q27.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014 Q27\" width=\"223\" height=\"196\" \/><\/p>\n<p>Question 28.<br \/>\nWhat is spermatogenesis? Explain the process of spermatogenesis.<br \/>\nAnswer:<br \/>\nFormation of haploid sperms from the diploid spermatogonial cells of the testes is called spermatogenesis. Primordial germ cells give rise to spermatogonial cells and \u0161ertoli cells. It is the spermatogonial cells that develop into the spermatozoa. Spermatogenesis involves four phases. viz..<br \/>\na. Multiplication Phase<br \/>\nb. Growth Phase<br \/>\nc. Maturation Phase<br \/>\nd. Spermiogenesis<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>(a) Multiplication Phase: The diploid spermatogonial cells of the seminiferous tubules divide repeatedly by mitosis to form more spermatogonial cells. Among them, only a few will enter into the growth phase and others are kept in reserve.<\/p>\n<p>(b) Growth Phase: Spermatogonial cells obtained from multiplication phase grow in size, however, they still remain diploid. These cells are now called primary spermatocytes.<\/p>\n<p>(c) Maturatio\u00f1 Phase: This phase involves two successive divisions, viz., Meiosis I and Meiosis II. The meiosis I is reductional and two haploid cells are formed from each primary spermatocyte. The resultant cells of I meiotic division are called secondary spermatocytes. These secondary spermatocytes undergo meiosis II which is<br \/>\nequational. As a result, each secondary spermatocyte produces two haploid cells of equal size called spermatids. Hence, four haploid spermatids are formed during maturation phase from each diploid primary spermatocyte.<\/p>\n<p>(d) Spermiogenesis: Spermatids are not gametes they are ordinary haploid cells. During spermiogenesis or spermateliosis each spermatid in association with Sertoli cells become tadpole-like, flagellated and highly motile gamete called spermatozoan or sperm.<\/p>\n<p>Question 29.<br \/>\n(a) List three reasons for population explosion.<br \/>\nAnswer:<\/p>\n<ul>\n<li>The decline in the death rate<\/li>\n<li>The decline in Maternal Mortality Rate (MMR)<\/li>\n<li>The decline in Infant Mortality Rate (IMR) and<\/li>\n<li>Increase in the number of people in the reproductive age.<\/li>\n<\/ul>\n<p>(b) Name human male and female sterilization procedures.<br \/>\nAnswer:<br \/>\nVasectomy and tubectomy.<\/p>\n<p>Question 30.<br \/>\nState the law of independent assortment. Write a schematic representation of the dihybrid cross.<br \/>\nAnswer:<br \/>\nBased on the results of the dihybrid cross, he formulated the second law of heredity namely, the Law of Independent Assortment. It states that \u201cWhen genes or alleles for two or more separate pairs of contrasting characters are brought together in a hybrid, they separate independently of each other.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65130\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q30.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014 Q30\" width=\"662\" height=\"367\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q30.png 662w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q30-300x166.png 300w\" sizes=\"(max-width: 662px) 100vw, 662px\" \/><\/p>\n<p>Question 31.<br \/>\nDescribe the Lac operon concept with a labelled sketch.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65131\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q31.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014 Q31\" width=\"445\" height=\"426\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q31.png 445w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q31-300x287.png 300w\" sizes=\"(max-width: 445px) 100vw, 445px\" \/><br \/>\nRegulation of gene action is well studied in E.coli bacteria with regard to utilization of lactose by the bacteria.<br \/>\nThe utilisation of lactose in E.coli needs three enzymes namely B-galactosidase, B-galactosidase permease and B-galactoside transacetylase. These are produced by Z, Y and genes respectively. Enzyme RNA polymerase enzyme initiates the synthesis of these 3 enzymes.<\/p>\n<p>The mechanism of lac-operon can be studied under two steps namely,<br \/>\n(a) Switched OFF mechanism: When lactose is absent in the medium, the regulator gene produces repressor protein which binds with the operator gene. This prevents the movement of RNA polymerase on the structural genes is blocked o there is no synthesis of mRNA from the structural genes Z, Y and a, so there is no synthesis of enzymes. This is called a switched OFF mechanism<\/p>\n<p>(b) Switched ON mechanism: When lactose is added to the culture medium, some of its molecules enter into the bacterial cell and one of them, binds itself with repressor. It induces the repressor protein to undergo structural change and makes it inactive. This inactive repressor becomes detached from the operator region. Now the RNA polymerase moves along the DNA and as a result the structural genes Z, Y and a produce mRNA. This mRNA synthesises 3 enzymes which are necessary for lactose metabolism. This is called switched ON mechanism.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 32.<br \/>\n(a) What is poultry? Mention two important components of Poultry Farm Management.<br \/>\nAnswer:<br \/>\nPoultry Farm Management:<br \/>\nPoultry is rearing of domesticated birds (fowl) for meat and eggs. Poultry typically includes chicken and ducks and sometimes turkey and geese.<\/p>\n<p>The important components of poultry farm management include:<\/p>\n<ul>\n<li>Selection of disease-free and suitable breeds.<\/li>\n<li>Proper feed and water for the birds.<\/li>\n<li>Proper and safe farm conditions.<\/li>\n<li>Hygiene and health care of the birds.<\/li>\n<\/ul>\n<p>(b) Define Inter-specific hybridization? Give an example.<br \/>\nAnswer:<br \/>\nIf hybridization is performed between two plants belonging to two different species then it is called as Inter-specific hybridization.<br \/>\neg: Crossing wheat and rye to get Triticale.<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Section &#8211; II<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any three of the following about 200 to 250 words each wherever applicable: (3 \u00d7 5 = 15)<\/span><\/p>\n<p>Question 33.<br \/>\nSummarize major events of DNA fingerprinting technique.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65132\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q33.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014 Q33\" width=\"320\" height=\"579\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q33.png 320w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q33-166x300.png 166w\" sizes=\"(max-width: 320px) 100vw, 320px\" \/><br \/>\nThis is a technique which helps in the establishment of the identity of a person by detecting and comparing sp\u00e9cifie nucleotide sequences in the DNA of individuals. This technique was developed by Alec Jeffreys and his associates.<br \/>\nSteps:<br \/>\na. Collection of sample: Sample like blood, semen, hair, saliva or any other tissue even in the dried state are collected.<\/p>\n<p>b. Extraction of DNA: The material collected is subjected to high-speed centrifugation which separates the DNA from the cell.<\/p>\n<p>c. Fragmenting DNA: The DNA \u00a1n the sample is treated with restriction endonucleases to obtain smaller restriction fragments.<\/p>\n<p>d. Separating DNA fragments: The smaller fragments of DNA are made to run on an agarose gel-plate by electrophoresis technique in which depending on molecular size and shape, the nucleotides produce distinct bands in the gel.<\/p>\n<p>e. Extracting DNA from gel: The double-stranded DNA fragment in the gel it denatured into single strands by exposing the gel into an alkaline solution (NaOH). Then by Southern blotting technique, these single-stranded DNA fragments are transferred onto a nitrocellulose filter or a nylon membrane.<\/p>\n<p>f. Tagging radioactive probes: A probe is radioactively labelled synthetic DNA. The nitrocellulose filter is now incubated with these radioactive probes. The probes act as specific nucleotide detectors and bind or hybridise with the specific nucleotides present in DNA fragments on the filter.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>g. Auto radiography: An x-ray film is placed on the filter to obtain autoradiographic prints. The film is then processed to observe visible hybridised pattern of bands. This is the genetic fingerprint or DNA fingerprint.<\/p>\n<p>h. Comparison: Comparison of two genetic fingerprints with great care can provide information about the degree of similarities or differences.<\/p>\n<p>Question 34.<br \/>\nExplain the different stages involved in sewage treatment.<br \/>\nAnswer:<br \/>\nSewage refers to the municipal wastewater generated in cities and towns that contains human and animal excreta and other domestic wastes. Large quantities of wastewater are generated every day in cities and towns. Sewage contains a large amount of organic matter and microbes.<\/p>\n<p>Many of the microbes present in sewages are pathogenic. Therefore, the sewage cannot be discharged into natural water bodies like rivers and streams directly. To make the sewage less polluting, \u00a1t has to be treated in Sewage Treatment Plants (STPs). Treatment of wastewater is done by the heterotrophic microbes naturally present in the sewage. This treatment is carried out in two stages: Primary treatment and secondary treatment. The wastewater can be passed into rivers after secondary treatment. In some cases, tertiary treatment is also carried out which removes nutrients by a chemical process.<\/p>\n<p>Primary Treatment: This step involves the physical removal of floating and suspended solids from sewage through filtrati\u00f3n and sedimentation. Initially, floating debris is removed through sequential filtration. The filtrate is kept in large open settling tanks where grit (sand, silt and small pebbles) are removed by sedimentation. Sometimes, alum or iron sulphate is added for flocculation and settling down of solids. The sediment is called primary sludge, while the supernatant is called effluent. The primary sludge is subjected to composting or landfill. The effluent from the primary settling tank is taken for secondary treatment.<\/p>\n<p>Secondary treatment or Biological treatment: The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of useful aerobic microbes into floes (masses of bacteria associated with fungal filaments to form mesh-like structures). While growing, these microbes consume the major part of the organic matter in the effluent. This significantly reduces the BOD (biochemical oxygen demand) of the effluent. BOO refers to the amount of the oxygen that would be consumed if all the organic matter in one litre of water were oxidised by bacteria.<\/p>\n<p>The sewage water \u00a1s treated until the BOD is reduced. The BOD test measures the rate of uptake of oxygen by micro-organisms in a sample of water and thus, indirectly. BOD is a measure of the organic matter present in the water. The greater the BOD of wastewater more is its polluting potential.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>Once the BOD of sewage or wastewater is reduced significantly, the effluent is then passed into a settling tank where the bacterial \u2018floes\u2019 are allowed to sediment. This sediment is called activated sludge. A small part of the activated sludge \u00a1s pumped back into the aeration tank to serve as the incolulum. The remaining major part of the sludge is pumped into large tanks called anaerobic sludge digesters. Here, other kinds of bacteria, which grow anaerobically, digest the bacteria and the fungi in the sludge.<\/p>\n<p>During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide. These gases form biogas and can be used as a source of energy as it is inflammable. The effluent from the secondary treatment plant is generally released into natural water bodies like rivers and streams.<\/p>\n<p>Question 35.<br \/>\n(a) Write a diagrammatic representation of Recombinant DNA technology.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65133\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q35a.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014 Q35(a)\" width=\"301\" height=\"639\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q35a.png 301w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q35a-141x300.png 141w\" sizes=\"(max-width: 301px) 100vw, 301px\" \/><\/p>\n<p>(b) Write a note on downstream processing.<br \/>\nAnswer:<br \/>\nDownstream Processing:<br \/>\nThe products formed in a bioreactor have to be subjected through a series of processes before they are ready for marketing as finished products. The various processes used for the recovery of useful products are collectively called downstream processing. The processes include separation and purification of the product, the addition of suitable preservatives and a stringent quality control testing etc. Such formulation has to undergo strict clinical trials as in the case of drugs. These quality control testing and clinical trials vary from product to product.<\/p>\n<p>Question 36.<br \/>\n(a) What is Biopiracy? Explain it with reference to Basmati rice.<br \/>\nAnswer:<br \/>\nBiopiracy: It is a term used to refer to the use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment.<br \/>\ne.g.: Basmati rice grown in India is distinct for its unique flavour and aroma, but an American company got patent rights on Basmati through the US patent and trademark office. The new variety of Basmati has been developed by this company by crossing an Indian variety with the semi-dwarf varieties.<\/p>\n<p>(b) What are genetically modified organisms? Name two Bt-toxin producing plants.<br \/>\nAnswer:<br \/>\nAn organism developed by introducing a desired foreign gene is called Genetically modified &#8216;organism&#8217;.<br \/>\neg: BT Cotton, Bt Soybean.<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 37.<br \/>\nDefine ecological pyramid? Describe the pyramid of energy flow with an example.<br \/>\nAnswer:<br \/>\nEcological Pyramid:<br \/>\nIt is the graphic representation of trophic structure and function of a food chain. It is so-called due to its superficial resemblance to the Egyptian pyramid. The Pyramid concept in ecology was proposed by Charles Elton and hence they are also called Eltonian pyramids. It is usually broad below and tapering to on apex above. Producers occupy the broad base. while animals are consumers occupy successive steps tapering into the apex of the pyramid.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65134\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2014-Q37.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2014 Q37\" width=\"227\" height=\"264\" \/><br \/>\nPyramid of Energy: it is a graphic representation of available energy at successive trophic levels in a food chain. The energy level is expressed as kilocalorie. It is the typical upright pyramid. It illustrates decreasing energy level at successive trophic levels from producers at the base to the consumers. The energy level is maximum in producers, less in consumers. e.g.: Pond food chain.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Students can Download 2nd PUC Biology Previous Year Question Paper June 2014, Karnataka 2nd PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Biology Previous Year Question Paper June 2014 Time: 3 hrs 15 min Max. Marks: &hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27568"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=27568"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27568\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27568"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27568"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27568"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}