{"id":27547,"date":"2020-07-03T16:35:51","date_gmt":"2020-07-03T11:05:51","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27547"},"modified":"2021-07-02T15:07:07","modified_gmt":"2021-07-02T09:37:07","slug":"2nd-puc-biology-previous-year-question-paper-june-2015","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-biology-previous-year-question-paper-june-2015\/","title":{"rendered":"2nd PUC Biology Previous Year Question Paper June 2015"},"content":{"rendered":"<p>Students can Download 2nd PUC Biology Previous Year Question Paper June 2015, 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 2015<\/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 \/>\nDefine embryogenesis.<br \/>\nAnswer:<br \/>\nThe process of development of the embryo from the zygote is known as embryogenesis.<\/p>\n<p>Question 2.<br \/>\nMention the asexual reproductive structures of Penicillium.<br \/>\nAnswer:<br \/>\nConidia.<\/p>\n<p>Question 3.<br \/>\nName the fungus that produces cyclosporin A.<br \/>\nAnswer:<br \/>\nTrichoderma polysporum.<\/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 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 4.<br \/>\nLactose is termed as an inducer in the lac operon. Give reason.<br \/>\nAnswer:<br \/>\nLactose is the substrate for the enzyme \u03b2-galactosidase and it regulates switching on and off of the operon. Hence, it is termed as inducer.<\/p>\n<p>Question 5.<br \/>\nWhat are eurythermal organisms?<br \/>\nAnswer:<br \/>\nThe organisms which can tolerate and thrive in a wide range of temperatures are called eurythermal organisms.<br \/>\ne.g. toad, cyclops, lizards etc.<\/p>\n<p>Question 6.<br \/>\nDefine transcription.<br \/>\nAnswer:<br \/>\nThe process of transfer of a genetic message from DNA to mRNA in nucleotide language is called Transcription.<\/p>\n<p>Question 7.<br \/>\nRetroviruses are disarmed before using to deliver desirable genes into animal cells. Why?<br \/>\nAnswer:<br \/>\nRetroviruses in animals have the ability to transform normal cells into cancerous cells. Thus they are disarmed before using to deliver desirable genes into animal cells.<\/p>\n<p>Question 8.<br \/>\nWhat is saltation?<br \/>\nAnswer:<br \/>\nA single step large mutation which can result in speciation is known as saltation.<\/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 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 9.<br \/>\nDisposal of sewage into the water without proper treatment may cause an outbreak of serious diseases. Give reason.<br \/>\nAnswer:<br \/>\nSewage contains large amounts of organic matter and microbes. Many of which are pathogenic. Thus their disposal into the water without proper treatment may cause an outbreak of serious diseases.<\/p>\n<p>Question 10.<br \/>\nHow does fragmentation of large habitats due to human activities lead to population decline?<br \/>\nAnswer:<br \/>\nWhen large habitats are broken up into small fragments due to various human activities, mammals and birds requiring large territories and certain animals with migrating habits are badly affected, leading to population declines.<\/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 \/>\nMention any two natural methods of. birth control.<br \/>\nAnswer:<\/p>\n<ol>\n<li>Calendar method<\/li>\n<li>Coitus interuptus.<\/li>\n<\/ol>\n<p>Question 12.<br \/>\nMention the possible genotypes of A and B blood group individuals.<br \/>\nAnswer:<\/p>\n<ul>\n<li>I<sup>A<\/sup>I<sup>A<\/sup> or I<sup>A<\/sup>i for \u2018A\u2019 blood group.<\/li>\n<li>I<sup>B<\/sup>I<sup>B<\/sup> or I<sup>B<\/sup>i for \u2018B\u2019 blood group.<\/li>\n<\/ul>\n<p>Question 13.<br \/>\nWrite two examples of loss of biodiversity due to alien species invasions.<br \/>\nAnswer:<\/p>\n<ol>\n<li>Introduction of Nile perch into Lake Victoria in East Africa led eventually to the extinction of an ecologically unique assemblage of more than 200 species of child fish in the lake.<\/li>\n<li>The recent illegal introduction of the African catfish Clarias gariepinus for aquaculture purposes is posing a threat to the indigenous catfishes.<\/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 June 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 14.<br \/>\nThe placenta is the structural and functional unit between the foetus and the maternal body. Substantiate the statement.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Placenta facilitates the supply of oxygen and nutrients to the embryo and also removal of carbon dioxide and excretory waste materials produced by the embryo.<\/li>\n<li>The placenta is connected to the embryo through umbilical cord which helps in the transport of substances to and from the embryo. Thus placenta is the structural and functional unit between the foetus and maternal body.<\/li>\n<\/ul>\n<p>Question 15.<br \/>\nWhat is gene therapy? Name the disorder to which clinical gene therapy was given first.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in a child or embryo.<\/li>\n<li>SCID (Severe Combined Immuno Deficiency) due to ADA deficiency.<\/li>\n<\/ul>\n<p>Question 16.<br \/>\nThe shape of RBCs of sickle cell anaemia patients changes to the elongated sickle-like structure. Give reason.<br \/>\nAnswer:<br \/>\nIn patients with sickle-celled anaemia substitution of glutamic acid by valine at the sixth position of the beta-globin chain of the haemoglobin molecule. The substitution of amino acid in the globin protein is due to the single base substitution at the sixth codon of the beta-globin gene from GAG to GUG. The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension causing the change in the shape of RBC from biconcave disc to elongated sickle-like structure.<\/p>\n<p>Question 17.<br \/>\nClassify restriction enzymes. Mention the function of each.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Restriction enzymes are the nucleases. They are of two types namely exonucleases and endonucleases.<\/li>\n<li>Exonucieases remove nucleotides from the ends of the DNA and endonucleases can cut at specific points within the DNA itself.<\/li>\n<\/ul>\n<p>Question 18.<br \/>\nName the pioneer species in primary succession and primary succession in water.<br \/>\nAnswer:<\/p>\n<ol>\n<li>Crustose lichens on rocks.<\/li>\n<li>Phytoplanktons in water.<\/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 \/>\nWhat is a heterothallic condition? Differentiate between staminate and pistillate flowers.<br \/>\nAnswer:<\/p>\n<ul>\n<li>If plants possess male and female reproductive structures on different plants (unisexual) it is termed as a heterothallic condition.<\/li>\n<li>If a flowering plant possesses unisexual male flowers with the only androecium it is called as staminate flower and if it possesses unisexual female flower with the only gynoecium it is called pistillate flower.<\/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 June 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 20.<br \/>\nWhat are cleistogamous flowers? Write their significance.<br \/>\nAnswer:<\/p>\n<ul>\n<li>The flowers which do not open at all are known as cleistogamous flowers. In such flowers, the anthers and stigma lie close to each other. When anthers dehisce in the flower buds, pollen grains come in contact with the stigma to effect pollination.<\/li>\n<li>Cleistogamous flowers produced assured seed-set even in the absence of pollinators.<\/li>\n<\/ul>\n<p>Question 21.<br \/>\nHow do the following assisted reproductive technologies assist the couples to have children?<\/p>\n<ol>\n<li>In-vitro Fertilization<\/li>\n<li>Zygote Intra Fallopian Transfer<\/li>\n<li>Gamete Intra Fallopian Transfer<\/li>\n<\/ol>\n<p>Answer:<\/p>\n<ol>\n<li>IVF is popularly known as a test-tube baby programme. Here, ova from the wife\/donor female and sperms from the husband\/donor male are collected and are induced to form zygote under simulated conditions in the laboratory. The further embryo is transferred into the uterus.<\/li>\n<li>In ZIFT, zygote or early embryos (with upto 8 blastomeres) is transferred into the fallopian tube.<\/li>\n<li>Transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one but can provide a suitable environment for fertilization and further development is a GIFT.<\/li>\n<\/ol>\n<p>Question 22.<br \/>\nName the six stages of human evolution in the correct sequence.<br \/>\nAnswer:<\/p>\n<ol>\n<li>Dryopithecus and Ramapithecus.<\/li>\n<li>Homo habilis (Handyman)<\/li>\n<li>Homo erectus<\/li>\n<li>Homo neanderthalensis<\/li>\n<li>Homo sapiens sapiens<\/li>\n<\/ol>\n<p>Question 23.<br \/>\nWhat is contact inhibition? Mention the types of tumours.<br \/>\nAnswer:<br \/>\nNormal cells show a property called contact inhibition by virtue of which contact with other cells inhibits their uncontrolled growth. Cancer cells appear to have lost this property. As a result of this, cancerous cells just continue to divide giving rise to masses of cells called tumours.<br \/>\nTumours are of two types<\/p>\n<ul>\n<li>Benign<\/li>\n<li>Malignant tumours<\/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 June 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 24.<br \/>\nDraw a labelled diagram of plasmid pBR-322.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65114\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q24.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2015 Q24\" width=\"345\" height=\"349\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q24.png 345w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q24-297x300.png 297w\" sizes=\"(max-width: 345px) 100vw, 345px\" \/><\/p>\n<p>Question 25.<br \/>\nIn which ways genetically modified plants are useful?<br \/>\nAnswer:<br \/>\nApplication of biotechnology in Agriculture:<br \/>\nAgricultural practice is of three types:<\/p>\n<ul>\n<li>Agrochemical-based agriculture.<\/li>\n<li>Organic agriculture and<\/li>\n<li>Genetically engineered crop-based agriculture.<\/li>\n<\/ul>\n<p>The use of genetically modified plants has been useful in the following ways:<\/p>\n<ul>\n<li>Genetic modification has made the crops more tolerant to abiotic stresses like cold, heat, drought, salinity etc.<\/li>\n<li>Post-harvest losses are much reduced.<\/li>\n<li>Food produced from GM (Genetically modified) crops have enhanced nutritional value and taste.<\/li>\n<li>As the plants have increased efficiency of mineral usage by plants, the early exhaustion of fertility of the soil is prevented.<\/li>\n<\/ul>\n<p>Question 26.<br \/>\n(a) Single-cell proteins can be used as an alternate source of proteins for human nutrition. Illustrate with two examples.<br \/>\n(b) How can tissue culture be applied to recover healthy plants from virus-infected plants?<br \/>\nAnswer:<br \/>\n(a) (i) Microbes like spirulina can be grown easily on materials like wastewater from potato processing plants, straw, molasses etc. to produce large quantities of food rich in proteins<br \/>\n(ii) 250 g of microorganism, Methylophilus methylotrophs can produce about 200 g of protein per day. Thus SCP can be used as an alternate source of proteins for human nutrition.<\/p>\n<p>(b) Although the plant is infested with a virus, the meristem is free of virus. Hence, the meristem can be grown in in-vitro conditions to obtain virus-free plants.<\/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 \/>\nWrite the events of development of female gametophyte incorrect order.<br \/>\nAnswer:<br \/>\nDevelopment of female gametophyte:<\/p>\n<ul>\n<li>The female gametophyte develops from a single megaspore and hence is described as monosporic development.<\/li>\n<li>The functional megaspore is the first cell of the female gametophyte of angiosperms.<\/li>\n<li>It enlarges to form the female gametophyte, also called embryo sac.<\/li>\n<li>Its nucleus undergoes a mitotic division and the two nuclei move to the opposite poles, forming the 2-nucleate embryo sac.<\/li>\n<li>Two successive mitotic divisions in each of these two nuclei result in the formation of an 8-nucleate embryo sac.<\/li>\n<li>Cell wall formation starts at the eight-nucleus stage, resulting in the formation of a typical female gametophyte.<\/li>\n<li>Three nuclei are grouped together at the micropylar end to form the egg apparatus, consisting of two synergids and a female gamete\/egg cell by wall formation.<\/li>\n<li>The remai\u00f1ing two nuclei are called polar nuclei, they move to the centre of the embryo sac and fuse to form a diploid secondary nucleus. The remaining three nuclei aggregate at the chalazal end to form three antipodal cells.<\/li>\n<li>Thus a typical angiosperm embryo sac is 8-nucleate and celled.<\/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 June 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 28.<br \/>\nWhat is incomplete dominance? Explain the inheritance of flower colour in snapdragon.<br \/>\nAnswer:<br \/>\nIt is a process where the dominant gene is incompletely dominant over the recessive gene and produces a phenotype which is intermediate to the parental type.<\/p>\n<p>In Mirabilis or 4\u00b0 clock plant, incomplete dominance has been reported with respect to the colour of the flower. In these plants, some produce red flowers and some others produce white flowers. When a plant producing red flowers is crossed with a plant producing white flowers, in the F<sub>1<\/sub> generation all plants produced pink flowers. It is a heterozygous condition and the pink character is an intermediate or a blend between red and white This colour is produced because the red is being only partially dominant over white.<\/p>\n<p>P<sub>1<\/sub> Red flowers \u00d7 White flowers<br \/>\nF<sub>1<\/sub> Pink flower<br \/>\nWhen F<sub>1<\/sub> pink flower plants are self cross the raise F<sub>2<\/sub> progeny 3 types of plants are produced i.e, Red, Pink and White, in the ratio of 1 : 2 : 1 respectively.<br \/>\nBy using appropriate symbols, the incomplete dominance can be represented as shown below.<br \/>\nThis cross is illustrated as indicated below by using appropriate letters.<\/p>\n<p>Thus, in this case, parental characters are expressed in homozygous conditions and intermediate character in heterozygous condition.<\/p>\n<p>Question 29.<br \/>\nWrite the diagrammatic representation of the life cycle of HIV.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65115\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q29.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2015 Q29\" width=\"354\" height=\"451\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q29.png 354w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q29-235x300.png 235w\" sizes=\"(max-width: 354px) 100vw, 354px\" \/><\/p>\n<p>Question 30.<br \/>\nDescribe the parts of a biogas plant with a labelled diagram.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65116\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q30.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2015 Q30\" width=\"391\" height=\"406\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q30.png 391w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q30-289x300.png 289w\" sizes=\"(max-width: 391px) 100vw, 391px\" \/><\/p>\n<p>Question 31.<br \/>\nDescribe Frederick Griffth\u2019s experiment to show transformation in bacteria.<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 smooth strain (S) with a gelatinous coat and a rough 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. 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-65117\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q31.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2015 Q31\" width=\"196\" height=\"330\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q31.png 196w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q31-178x300.png 178w\" sizes=\"(max-width: 196px) 100vw, 196px\" \/><br \/>\nIn 1944, Avery, MacLeod and Mc Carty extended Griffith\u2019s experiment to identify\u2019 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><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 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 32.<br \/>\n(a) Define the following terms:<br \/>\n(i) Commensalism.<br \/>\n(ii) Brood parasitism.<br \/>\n(iii) Mutualism.<br \/>\nAnswer:<br \/>\n(i) Commensalism:<br \/>\nIt is a relationship between two living individuals of different species in which one is benefitted. while the other is neither harmed nor benefitted.<\/p>\n<p>(ii) Brood parasitism:<br \/>\nBrood parasitism in birds is a fascinating example of parasitism in which the parasitic birds lays its eggs in the nest of its host and lets the host incubate them. During the course of evolution, the eggs of the parasitic bird have evolved to resemble the host\u2019s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest.<br \/>\ne.g. cuckoo lays eggs in the nest of crow.<\/p>\n<p>(iii) Mutualism:<br \/>\nIt is an interaction or relationship between two organisms of different species where both the partners are benefitted with none of the two capable of living separately.<\/p>\n<p>(b) Mention four basic processes that fluctuate the density of a population in a given habitat.<br \/>\nAnswer:<br \/>\nNatality, Mortality, Immigration and Emigration.<\/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 \/>\nDraw a neat labelled diagram of the human male reproductive system.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65118\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q33.png\" alt=\"2nd PUC Biology Previous Year Question Paper June 2015 Q33\" width=\"508\" height=\"361\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q33.png 508w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/07\/2nd-PUC-Biology-Previous-Year-Question-Paper-June-2015-Q33-300x213.png 300w\" sizes=\"(max-width: 508px) 100vw, 508px\" \/><\/p>\n<p>Question 34.<br \/>\nDescribe the process of translation of mRNA.<br \/>\nAnswer:<br \/>\nTranslation:<br \/>\nThe process of translation of the nucleotide language of mRNA into the aminoacid language by tRNA is called translation. This process occurs in the cytoplasm of the cell both in prokaryotes and eukaryotes.<br \/>\nThe process of translation involves the following steps.<\/p>\n<ol>\n<li>Activation of amino acids.<\/li>\n<li>Initiation of the polypeptide chain.<\/li>\n<li>Elongation of the polypeptide chain.<\/li>\n<li>Termination of the polypeptide chain.<\/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 June 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>1. Activation of amino acids: Amino acids are present in the cytoplasm and are in an inactive form. These are combined with ATP in presence of an enzyme aminoacyl synthetase and Mg ions forming activat\u00e7d aminoacid.<br \/>\nAmino acid + ATP \u2192 Amino acid-AMP &#8211; enzyme complex<br \/>\nThis activated amino acid then combines with 3 ends of tRNA forming aminoacyl-tRNA complex or charged RNA and AMP is released.<\/p>\n<p>2. Initiation of polypeptide chain: In this process, mRNA gets attached to 30s ribosomal subunit. Now the tRNA carrying activated amino acid, methionine gets attached to the first codon of mRNA. It takes place in presence of GTP and initiation factors [IF1 IF2 IF3]. This is the initiation complex. Now, 50s ribosomal sub-unit combines with 30s ribosomal unit forming 70s ribosome. The first tRNA-aminoacid complex is now at \u2018P\u2019 site of the ribosome.<\/p>\n<p>3. Elongation of polypeptide chain: The \u2018A\u2019 site of 50s ribosome is now free and so the second tRNA with second activated amino acid is get attached at \u2018A\u2019 site. A peptide bond is formed between 1st and 2nd amino acid in presence of peptidyl transferase at \u2018P\u2019 site. Now the ribosome moves a distance of one codon so that the \u2018A\u2019 site become vacant. Meanwhile, the 3rd tRNA with 3rd amino acid gets attached to \u2018A\u2019 site. Again the ribosome moves on mRNA by<br \/>\none codon and bonding takes place between 2\u201d and 3td aminoacid at \u2018P\u2019 site. 1st tRNA is released through \u2018E\u2019 site. This process continues until all the codons on mRNA are completely read. Thus polypeptide chain elongates along the mRNA.<\/p>\n<p>4. Termination and release of polypeptide chain: The process of protein synthesis continues till the arrival of terminator codon on mRNA at \u2018A\u2019 Site. The terminator codon may be UAA, UAG and UGA. These codons do not code for any amino acids. During this process, the enzyme peptidyl synthetase catalyses the cleavage of polypeptide chain from the last tRNA. Here releasing factors RF<sub>1<\/sub>, RF<sub>2<\/sub>, and RF<sub>3<\/sub> are involved. Thus the polypeptide chain is finally released.<\/p>\n<ul>\n<li>The speed of transcription is 30 nucleotides per second in bacteria.<\/li>\n<li>The speed of translation 20 amino acids per second in bacteria.<\/li>\n<\/ul>\n<p>Question 35.<br \/>\nExplain the important steps involved in the process of decomposition.<br \/>\nAnswer:<br \/>\nDetritus is the raw material for decomposition.<br \/>\nFollowing are the steps in the process of decomposition:<\/p>\n<ol>\n<li>Fragmentation: It is the process of breaking the detritus into smaller particles by detritivores.<\/li>\n<li>Leaching: It is the process in which water-soluble inorganic substances run down into soil profile and get precipitated as unavailable salts.<\/li>\n<li>Catabolism: The enzymatic conversion of the detritus into simple organic compounds and then into inorganic compounds, is called catabolism. The enzymes are secreted by the decomposers like bacteria and fungi.<\/li>\n<li>Humification: Humification during decomposition leads to the accumulation of a dark coloured amorphous substance called humus.<\/li>\n<li>Mineralisation: It is the process in which the humus is degraded by certain microbes and the inorganic nutrients are released.<\/li>\n<\/ol>\n<p>Decomposition is largely are aerobic process, i.e., it requires oxygen.<br \/>\nThe factors affecting decomposition are:<\/p>\n<ul>\n<li>The chemical composition of detritus and<\/li>\n<li>The climatic factors.<\/li>\n<\/ul>\n<p>If detritus is rich in lignin and chitin, decomposition is slow and it is faster if detritus is rich in nitrogen and water-soluble substances.<br \/>\nTemperature and soil moisture are the important climatic factors that regulate decomposition through their effects on the activities of soil microbes.<br \/>\nWarm and moist environment favours decomposition, while low temperature and anaerobic conditions inhibit decomposition.<\/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 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 36.<br \/>\n(a) Explain the types of outbreeding processes employed to improve desirable qualities of animals.<br \/>\nAnswer:<br \/>\nOutbreeding:<br \/>\nOutbreeding refers to the breeding of unrelated animals either of the same breed having no common ancestors for 4-6 generations (outcrossing) or of different breeds (cross-breeding) or even different species (interspecific).<\/p>\n<p>Outbreeding is of the following types:<\/p>\n<p>(i) Outcrossing:<\/p>\n<ul>\n<li>Outcrossing is the practice of mating of animals of the same breed, but that have no common ancestors on either side of their pedigree upto 4-6 generations.<\/li>\n<li>The offspring of outcrossing, is called an outcross.<\/li>\n<li>A single outcross helps to overcome inbreeding depression.<\/li>\n<li>It is the best breeding method for animals that are below average in productivity and growth rate.<\/li>\n<\/ul>\n<p>(ii) Cross-breeding:<\/p>\n<ul>\n<li>It is a method of outbreeding in which superior males of one breed are mated with the superior females of another breed of the same species. This helps in combining the desirable qualities of the two different breeds into the hybrid progeny.<\/li>\n<li>The hybrid progeny may be directly used for commercial production or they may be subjected to some form of inbreeding and selection, to develop new stable breeds.<\/li>\n<li>One example of cross-breeding is Hisardale, a new breed of sheep developed by crossing Bikaneri ewes and Marino rams.<\/li>\n<\/ul>\n<p>(iii) Interspecific hybridisation:<\/p>\n<ul>\n<li>It is a method of outbreeding in which male and female animals of two different species are crossed to combine the desirable features of both the parents into one.<br \/>\ne.g.: Mule is produced by a cross between a male donkey and a female horse.<\/li>\n<\/ul>\n<p>(b) Write any two processes deal with to increase the yield and quality of milk in dairy farm management.<br \/>\nAnswer:<br \/>\n1. It is a method to improve herds.<br \/>\nThe steps in the method are as follows:<\/p>\n<ul>\n<li>A cow is administered hormones (like FSI-I) to induce follicular maturation and superovulation, i.e., production of 6-8 ova in one cycle.<\/li>\n<li>The cow is mated with the selected bull or artificially inseminated.<\/li>\n<li>The fertilised eggs at 8-32-celled stages are recovered and transferred to surrogate mothers.<\/li>\n<\/ul>\n<p>This technology has have been used for cattle, rabbits, mares, etc.<br \/>\nHigh milk-yielding breeds of females and high-quality meat-yielding bulls have been bred successfully to increase the herd size in a short time.<\/p>\n<p>2. Artificial Insemination: The semen of a selected male is deposited in the reproductive tract of the female at the proper time in her reproductive cycle in order to effect fertilization and also to improve the quality of animals.<\/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 2015\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 37.<br \/>\nWhat Is Eutrophication? Discuss the phenomenon of accelerated eutrophication.<br \/>\nAnswer:<br \/>\nEutrophication is a natural process which means \u2018well-nourished or enriched\u2019. It is a natural process of ageing of water bodies due to the rich supply of nutrients. In young water bodies, water is cold, clear with moderate life forms. However, when abnormally high amounts of nutrients are released from sewage it results in excessive algal growth or bloom of microorganisms. Over the period of time because of dumping of industrial and house effluents as well as piling up of silt, the water body gradually becomes shallower, warmer, giving way to marshy plants and eventually converting into the land. This phenomenon of accelerated ageing of the water body is called Eutrophication.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Students can Download 2nd PUC Biology Previous Year Question Paper June 2015, 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 2015 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\/27547"}],"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=27547"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27547\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27547"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27547"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27547"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}