{"id":27472,"date":"2020-07-01T15:57:36","date_gmt":"2020-07-01T10:27:36","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27472"},"modified":"2021-07-02T15:07:09","modified_gmt":"2021-07-02T09:37:09","slug":"2nd-puc-biology-previous-year-question-paper-march-2018","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-biology-previous-year-question-paper-march-2018\/","title":{"rendered":"2nd PUC Biology Previous Year Question Paper March 2018"},"content":{"rendered":"<p>Students can Download 2nd PUC Biology Previous Year Question Paper March 2018, 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 March 2018<\/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 implantation?<br \/>\nAnswer:<br \/>\nAttachment of blastocyst to the endometrium of uterus is called as implantation.<\/p>\n<p>Question 2.<br \/>\nWrite the restriction site of EcoRI enzyme.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65056\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q2.png\" alt=\"2nd PUC Biology Previous Year Question Paper March 2018 Q2\" width=\"177\" height=\"70\" \/><\/p>\n<p>Question 3.<br \/>\nDefine \u2018Totipotency\u2019.<br \/>\nAnswer:<br \/>\nThe developmental capacity of a cell to regenerate into an embryo or organism under suitable conditions is called totipotency.<\/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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 4.<br \/>\nMention the role of Methanobacterium in the rumen of cattle.<br \/>\nAnswer:<br \/>\nMethanobacterium help in the breakdown of cellulose present in the food of cattle and thus play an important role in the nutrition of cattle.<\/p>\n<p>Question 5.<br \/>\nWhat is Foetal Ejection Reflux?<br \/>\nAnswer:<br \/>\nThe signals for parturition originate from the fully developed foetus and the placenta which induce mild uterine contractions called foetal ejection reflex.<\/p>\n<p>Question 6.<br \/>\nDefine \u2018Saltation\u2019.<br \/>\nAnswer:<br \/>\nSingle-step large mutation leading to speciation is called saltation.<\/p>\n<p>Question 7.<br \/>\nName the type of antibodies produced during allergy.<br \/>\nAnswer:<br \/>\nImmunoglobulin E (IgE)<\/p>\n<p>Question 8.<br \/>\nWhat are Eurythermal organisms?<br \/>\nAnswer:<br \/>\nThe organisms which can tolerate and thrive in a wide range of temperature fluctuations are Eurythermal organisms.<\/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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 9.<br \/>\nDefine Biomagnification.<br \/>\nAnswer:<br \/>\nThe process of accumulation of certain pollutants in tissues of organisms with increased concentration along the food chain.<\/p>\n<p>Question 10.<br \/>\nWrite the name of the toxic substance responsible for fever and chill in Malaria.<br \/>\nAnswer:<br \/>\nDue to rupture of RBCs by the malarial parasite releases a toxin called hemozoin which causes fever and chilly in malaria.<\/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 homogametic and heterogametes?<br \/>\nAnswer:<br \/>\nGametes which are similar in their size, shape and structure (appearance) and which cannot be categorised into male and female gametes are called homogametic (isogametes), eg. chlodophora.<\/p>\n<p>Question 12.<br \/>\nMention any two examples of evolution by anthropogenic action.<br \/>\nAnswer:<br \/>\n(a) Industrial melanism: During post industrialisation period, the tree trunks become dark due to industrial smoke and soot. Under this condition, the white-winged moth did not survive due to predators and dark-winged or melanic moth survived. But before industrialisation thick growth of almost white coloured lichens covered the trees, in that background the white-winged moth survived but the dark coloured moth was picked out by predators.<\/p>\n<p>(b) Excess use of herbicides, pesticides etc. has only resulted in the selection of resistant varieties in a much lesser time scale.<\/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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 13.<br \/>\nDistinguish between homozygous and heterozygous plants.<br \/>\nAnswer:<br \/>\nHomozygous condition is a situation of a plant in which a particular character is controlled by a pair of similar genes.<\/p>\n<p>Heterozygous condition is a situation of an individual in which a particular character is controlled by alleles which express contrasting traits.<\/p>\n<p>Question 14.<br \/>\nWhat is innate immunity? Mention any two types of innate immunity barriers.<br \/>\nAnswer:<br \/>\nInnate immunity is a non-specific type of defence, that is present at the time of birth. It prevents entry of the foreign agents into our body through different types of barriers.<br \/>\nTypes:<\/p>\n<ul>\n<li>Physical barriers<\/li>\n<li>Physiological barriers<\/li>\n<\/ul>\n<p>Question 15.<br \/>\nWhich are the important components of Poultry Farm Management?<br \/>\nAnswer:<br \/>\nPoultry Farm Management:<\/p>\n<ol>\n<li>Poultry is rearing of domesticated birds (fowl) for meat and eggs.<\/li>\n<li>Poultry typically includes chicken and ducks and sometimes turkey and geese.<\/li>\n<li>The important components of poultry farm management include:\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<\/li>\n<\/ol>\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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 16.<br \/>\nWrite the methods to introduce alien DNA into Host cells.<br \/>\nAnswer:<br \/>\n(a) Electroporation (chemical method): The bacterial cell is placed in a solution with cold CaCl<sub>2<\/sub> solution followed by placing them at 42\u00b0C and then putting them on ice. This results in the development of pores in the cell membrane.<br \/>\n(b) Microinjection: It is the direct injection of the desired gene into the nucleus of an animal cell by microsyringe.<br \/>\n(c) Biolistics: Here a suitable plant cell is bombarded with high-velocity microparticles of gold or tungsten coated with DNA in order to introduce DNA into the cell.<\/p>\n<p>Question 17.<br \/>\nDefine Endemism. Name any two regions of accelerated habitat loss in India.<br \/>\nAnswer:<br \/>\nThe species which are confined to a specific geographical region and not found elsewhere is known as endemism.<br \/>\nTwo regions of accelerated habitat loss in India are:<\/p>\n<ul>\n<li>Western ghats and Srilanka<\/li>\n<li>Indo Burma and Himalaya.<\/li>\n<\/ul>\n<p>Question 18.<br \/>\nInvasion of alien animal species eliminates the native animal species. Give two examples.<br \/>\nAnswer:<br \/>\nTwo examples are:<\/p>\n<ol>\n<li>The Nile perch introduced into Lake Victoria in East Africa led eventually to the extinction of an ecologically unique assemblage of more than 200 species of cichlid fish in the lake.<\/li>\n<li>The illegal introduction of the African catfish Clarias gariepinus for aquaculture purposes is posing a threat to the indigenous catfishes in our rivers.<\/li>\n<\/ol>\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 vegetative propagules of the following plants.<\/p>\n<ol>\n<li>Water Hyacinth<\/li>\n<li>Agave<\/li>\n<li>Banana<\/li>\n<\/ol>\n<p>Answer:<\/p>\n<ol>\n<li>Offset<\/li>\n<li>Bulbil<\/li>\n<li>Rhizome<\/li>\n<\/ol>\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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 20.<br \/>\nDraw a neat labelled diagram of T.S. of young Anther.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65057\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q20.png\" alt=\"2nd PUC Biology Previous Year Question Paper March 2018 Q20\" width=\"615\" height=\"286\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q20.png 615w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q20-300x140.png 300w\" sizes=\"(max-width: 615px) 100vw, 615px\" \/><\/p>\n<p>Question 21.<br \/>\nDescribe Haplodiploid sex-determination system in Honey Bees.<br \/>\nAnswer:<br \/>\nHaplodiploid sex-determination<br \/>\nThis type of sex-determination is seen in honeybees; \u00a1t is based on the number of sets of chromosomes, an individual receives. When the ovum is fertilised by a male gamete, the diploid the zygote (21\/32) develops into a female, i.e., a queen or worker.\u00a0 When the ovum develops .by parthenogenesis, i.e., without fertilisation, a male individual, called drone is formed. The male honeybee (drone) is haploid (with n = 16 chromosomes) and forms sperms by mitosis.<\/p>\n<p>The sex-determination is as follows:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65058\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q21.png\" alt=\"2nd PUC Biology Previous Year Question Paper March 2018 Q21\" width=\"445\" height=\"266\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q21.png 445w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q21-300x179.png 300w\" sizes=\"(max-width: 445px) 100vw, 445px\" \/><\/p>\n<ul>\n<li>The special characteristic features of this system of sex-determination are:<\/li>\n<li>The male honeybees don&#8217;t have a father but have a grandfather.<\/li>\n<li>They do not have sons but can have grandsons.<\/li>\n<\/ul>\n<p>Question 22.<br \/>\nState Hardy-Weinberg principle of genetic equilibrium. Write any four factors affecting the equilibrium.<br \/>\nAnswer:<br \/>\nHardy Weinberg Law or Hardy Weinberg Equilibrium: This law was proposed by British Mathematician Hardy and German Physicist E. Weinberg. The Law states that \u201cGene frequency of a Mendelian population remains constant through generations unless there are chromosomal aberrations, mutation etc to alter the genetic equilibrium\u201d.<\/p>\n<p>This law describes that when the population is in equilibrium there is no evolution. Evolution occurs only when the equilibrium is altered. The mendelian population is a closely interbreeding group of organisms sharing a common gene pool.<\/p>\n<p>Evolutionary factors like mutation, migration, genetic drift, gene flow, and sexual reproduction are factors affecting the equilibrium of a population.<\/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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 23.<br \/>\nWrite a short note on ecosystem services.<br \/>\nAnswer:<br \/>\nEcosystem services are<\/p>\n<ol>\n<li>Healthy forest ecosystems purify air and water.<\/li>\n<li>Mitigate droughts and floods.<\/li>\n<li>Cycle help in nutrients.<\/li>\n<li>Generatefertilesoils.<\/li>\n<li>Provide wildlife habitat.<\/li>\n<li>Maintain biodiversity.<\/li>\n<li>Help to Pollinate crops.<\/li>\n<li>Provide storage site for carbon and also provide aesthetic, cultural and spiritual values.<\/li>\n<\/ol>\n<p>Question 24.<br \/>\nSketch the diagrammatic representation of the Replication of Retrovirus inside an animal cell.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65059\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q24.png\" alt=\"2nd PUC Biology Previous Year Question Paper March 2018 Q24\" width=\"423\" height=\"514\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q24.png 423w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q24-247x300.png 247w\" sizes=\"(max-width: 423px) 100vw, 423px\" \/><\/p>\n<p>Question 25.<br \/>\nWhat is Ecological succession? How Hydrarch succession is different from Eerarch succession?<br \/>\nAnswer:<br \/>\nThe gradual and fairly predictable change in the species composition of a given area is called ecological succession.<br \/>\nHydrarch succession takes place in wetter areas and the successional series progress from hydric to the mesic conditions.<\/p>\n<p>It involves the pioneer stage, submerged stage, floating stage, reed swamp stage, marsh meadow stage, woodland stage and forest stage. Xerarch succession takes place in dry areas and the series progress from xeric to mesic conditions. The stages include crustose lichen stage, foliose lichen stage, moss stage, herb stage, shrub stage and tree stage.<\/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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 26.<br \/>\nDraw a neat labelled diagram of plasmid pBR322.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65060\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q26.png\" alt=\"2nd PUC Biology Previous Year Question Paper March 2018 Q26\" width=\"342\" height=\"351\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q26.png 342w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q26-292x300.png 292w\" sizes=\"(max-width: 342px) 100vw, 342px\" \/><\/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 \/>\nDraw and describe the structure of a matured embryosac of angiosperms.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65061\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q27.png\" alt=\"2nd PUC Biology Previous Year Question Paper March 2018 Q27\" width=\"635\" height=\"263\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q27.png 635w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q27-300x124.png 300w\" sizes=\"(max-width: 635px) 100vw, 635px\" \/><br \/>\nEmbryosac: It is the female gametophyte present in the nucleus towards the micropylar end. It consists of three antipodal cells towards the chalazal end, secondary nucleus at the centre and egg apparatus towards the micropylar end. Egg apparatus consists of a central egg cell and two lateral synergid cells.<\/p>\n<p>Question 28.<br \/>\nDraw a neat labelled diagrammatic sectional view of the female reproductive system.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65062\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q28.png\" alt=\"2nd PUC Biology Previous Year Question Paper March 2018 Q28\" width=\"425\" height=\"327\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q28.png 425w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q28-300x231.png 300w\" sizes=\"(max-width: 425px) 100vw, 425px\" \/><\/p>\n<p>Question 29.<br \/>\nExplain the inheritance of one gene with reference to stem height of Garden Pea plant.<\/p>\n<p>Question 30.<br \/>\n(a) What are the features of an ideal contraceptive?<br \/>\nAnswer:<br \/>\nFeatures of an ideal contraceptive are:<\/p>\n<ul>\n<li>should be user-friendly, easily available, effective and reversible with no or least side effects.<\/li>\n<li>It should in no way interfere with the sexual drive, desire and\/or the sexual act of the user.<\/li>\n<\/ul>\n<p>(b) Mention the natural methods of contraception.<br \/>\nAnswer:<br \/>\nNatural methods of contraception include:<\/p>\n<ul>\n<li>Periodic abstinence<\/li>\n<li>withdrawal method or coitus interruptus<\/li>\n<li>lactational amenorrhea.<\/li>\n<\/ul>\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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 31.<br \/>\nExplain the following terms:<\/p>\n<ol>\n<li>Inbreeding depression (1)<\/li>\n<li>Interspecific hybridisation (1)<\/li>\n<li>Biofortification (1)<\/li>\n<li>Micropropagation (1)<\/li>\n<li>Some clones (1)<\/li>\n<\/ol>\n<p>Answer:<\/p>\n<ol>\n<li>Inbreeding depression: Continuous breeding reduces fertility and even productivity. It is called inbreeding depression.<\/li>\n<li>Interspecific hybridisation: Male and female animals of two different related species are mated, progeny have desirable features of both the parents.<\/li>\n<li>Biofortification: It is breeding of crops with higher levels of vitamins, minerals, proteins and healthy fats in order to improve public health.<\/li>\n<li>Micropropagation: The method of producing thousands of plants through tissue culture is called micropropagation.<\/li>\n<li>Somoclones: Plants produced by the method of micropropagation are morphologically and genetically identical to the original plants. Such plants are called some clones.<\/li>\n<\/ol>\n<p>Question 32.<br \/>\nExplain the regulation of Lac operon in the absence and presence of Lactose as an inducer.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65063\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q32.png\" alt=\"2nd PUC Biology Previous Year Question Paper March 2018 Q32\" width=\"435\" height=\"424\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q32.png 435w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q32-300x292.png 300w\" sizes=\"(max-width: 435px) 100vw, 435px\" \/><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 &#8211; 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 \u00a1s 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><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 March 2018\" width=\"172\" height=\"16\" \/><\/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 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 questions in about 200 to 250 words each wherever applicable: (3 \u00d7 5 = 15)<\/span><\/p>\n<p>Question 33.<br \/>\nOswald Avery and others have continued Griffth\u2019s transforming principle to prove DNA as genetic material. Substantiate.<br \/>\nAnswer:<br \/>\nIn 1928 Griffith working on the pathogenicity of streptococcus pneumonia discovered a process called transformation. This bacterium is responsible for a form of pneumonia. killing mice. There are two strains of bacteria &#8211; smooth strain (S) with a gelatinous coat and a \u2018rough strain (R) without the gelatinous coat. Smooth strain bacteria are virulent type and kill the mice while rough strain bacteria are avirulent type and do not kill the mice. If live rough strain bacteria (R) are injected the mouse does not die.<\/p>\n<p>When live smooth bacteria are injected the mouse dies. But if heat-killed smooth bacteria (S) are injected, the mouse does not die. If a mixture of heat-killed smooth bacteria (S) and live rough bacteria (R) is injected into the mice it dies. Some property of heat-killed smooth bacteria had transformed R bacterium to become virulent types. This activity was called the transforming principle by Griffith.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65064\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-March-2018-Q33.png\" alt=\"2nd PUC Biology Previous Year Question Paper March 2018 Q33\" width=\"161\" height=\"275\" \/><br \/>\nIn 1944, Avery, MacLeod and Mc Carty extended Griffith\u2019s experiment to identify the transforming principle. They mixed \u2018R\u2019 strain with DNA extracted from \u2018S\u2019 strain of bacteria. When the enzyme (DNAase) which digests or breaks down DNA was added to this mixtures, there was no transformation of R to S type. But when the enzyme proteases that digest proteins were added to the medium the transformation of R to S type was not prevented. This confirmed that the transforming principle was DNA.<\/p>\n<p>Question 34.<br \/>\nDescribe the role of Microbes in the sewage treatment plant.<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. 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, it has to be treated in Sewage Treatment Plants (STPs).<\/p>\n<p>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><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 March 2018\" width=\"172\" height=\"16\" \/><\/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 the 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.<\/p>\n<p>BOD 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. The sewage water is 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>Once the BOD of sewage or wastewater is reduced significantly, the effluent is then passed into a settling tank where the bacterial \u2018flocs\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 inoculum. 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 \u00a1s generally released into natural water bodies like rivers and streams.<\/p>\n<p>Question 35.<br \/>\nOne of the applications of biotechnology is to get pest-resistant plants. Justify the statement with reference to Bt. cotton.<br \/>\nAnswer:<br \/>\nBt cotton:<\/p>\n<ol>\n<li>Bacillus thuringiensis produces crystal proteins called Cry proteins which are toxic to larvae of insects like tobacco budworm, armyworm, beetles and mosquitoes.<\/li>\n<li>Cry proteins exist as inactive protoxins and get converted int\u00f3 active toxin when ingested by the insect, as the alkaline pH of the gut solubilises the crystals.<\/li>\n<li>The activated toxin binds to the surface of epithelial cells of midgut and creates pores.<\/li>\n<li>This causes swelling and lysis of cells leading to the death of the insect (larva).<\/li>\n<li>The genes (cry genes) encoding this protein are isolated from the bacterium and incorporated into several crop plants like cotton, tomato, corn, rice, soybean, etc.<\/li>\n<li>The proteins encoded by the following cry genes control the corresponding pests.\n<ul>\n<li>Cry I AC and Cry II Ab control cotton bollworms. .<\/li>\n<li>Cry I Ab controls corn borer.<\/li>\n<li>Cry III Ab controls Colorado potato beetle.<\/li>\n<li>Cry III Bb controls corn rootworm.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 36.<br \/>\n(a) What are Ectoparasites and Endoparasites? (2)<br \/>\nAnswer:<br \/>\nEctoparasites are the parasites that live on the outer surface of the host and generally attach themselves during feeding, e.g. leech, ticks, bed bugs etc.<\/p>\n<p>Endoparasites are parasites that live inside the body of the host. e.g. malarial parasite in RBC, Ascaris and Taenia in the gut of man etc.<\/p>\n<p>(b) List any three parasitic adaptations in animals. (3)<br \/>\nAnswer:<br \/>\nParasites develop special arrangements such as the presence of adhesive organs or suckers to cling to the bost, loss of digestive systems and high reproductive capacity and loss of unnecessary sense organs.<\/p>\n<p>Question 37.<br \/>\nWrite a note on the following:<br \/>\n(a) Radioactive wastes (3)<br \/>\n(b) Joitt forest management. (2)<br \/>\nAnswer:<br \/>\n(a) Radioactive wastes:<br \/>\nRadiation, which is given off by nuclear waste is extremely damaging to organisms because it causes mutations at a very high rate. At high doses nuclear radiation is lethal but at lower doses, it creates various disorders, the most frequent of all being cancer. Therefore, nuclear waste is in extremely potent pollutant and has to be dealt with utmost caution. It has been recommended that storage of nuclear waste, after sufficient pre-treatment, should be done in suitably shielded containers, buried within the rocks, about 500 m deep below the earth\u2019s surface.<\/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 March 2018\" width=\"172\" height=\"16\" \/><\/p>\n<p>(b) You may have heard of the Chipko Movement of Garhwal Himalayas in 1974, local women showed enormous bravery in protecting trees from the axe of contractors by hugging them. People all over the world have acclaimed the Chipko movement. Realising the significance of participation by local communities, the Government of India in the 1980s has introduced the concept of Joint Forest Management (JFM) so as to work closely with the local communities for protecting and managing forests. In return for their services to the forest, the communities get the benefit of various forest products. (e.g., fruits, gum, rubber, medicine, etc). and thus the forest can be conserved in a sustainable manner.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Students can Download 2nd PUC Biology Previous Year Question Paper March 2018, 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 March 2018 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\/27472"}],"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=27472"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27472\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}