2nd PUC Biology Previous Year Question Paper June 2015

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Karnataka 2nd PUC Biology Previous Year Question Paper June 2015

Time: 3 hrs 15 min
Max. Marks: 70

General Instructions

  • This question paper consists of four parts A, B, C and D. Part – D consists of two sections. Section – I and Section – II.
  • All the parts are compulsory.
  • Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word or one sentence each. (10 × 1 = 10)

Question 1.
Define embryogenesis.
Answer:
The process of development of the embryo from the zygote is known as embryogenesis.

Question 2.
Mention the asexual reproductive structures of Penicillium.
Answer:
Conidia.

Question 3.
Name the fungus that produces cyclosporin A.
Answer:
Trichoderma polysporum.

2nd PUC Biology Previous Year Question Paper June 2015

Question 4.
Lactose is termed as an inducer in the lac operon. Give reason.
Answer:
Lactose is the substrate for the enzyme β-galactosidase and it regulates switching on and off of the operon. Hence, it is termed as inducer.

Question 5.
What are eurythermal organisms?
Answer:
The organisms which can tolerate and thrive in a wide range of temperatures are called eurythermal organisms.
e.g. toad, cyclops, lizards etc.

Question 6.
Define transcription.
Answer:
The process of transfer of a genetic message from DNA to mRNA in nucleotide language is called Transcription.

Question 7.
Retroviruses are disarmed before using to deliver desirable genes into animal cells. Why?
Answer:
Retroviruses 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.

Question 8.
What is saltation?
Answer:
A single step large mutation which can result in speciation is known as saltation.

2nd PUC Biology Previous Year Question Paper June 2015

Question 9.
Disposal of sewage into the water without proper treatment may cause an outbreak of serious diseases. Give reason.
Answer:
Sewage 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.

Question 10.
How does fragmentation of large habitats due to human activities lead to population decline?
Answer:
When 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.

Part – B

Answer any five of the following questions in 3 to 5 sentences each: Wherever applicable: (5 × 2 = 10)

Question 11.
Mention any two natural methods of. birth control.
Answer:

  1. Calendar method
  2. Coitus interuptus.

Question 12.
Mention the possible genotypes of A and B blood group individuals.
Answer:

  • IAIA or IAi for ‘A’ blood group.
  • IBIB or IBi for ‘B’ blood group.

Question 13.
Write two examples of loss of biodiversity due to alien species invasions.
Answer:

  1. 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.
  2. The recent illegal introduction of the African catfish Clarias gariepinus for aquaculture purposes is posing a threat to the indigenous catfishes.

2nd PUC Biology Previous Year Question Paper June 2015

Question 14.
The placenta is the structural and functional unit between the foetus and the maternal body. Substantiate the statement.
Answer:

  • 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.
  • 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.

Question 15.
What is gene therapy? Name the disorder to which clinical gene therapy was given first.
Answer:

  • Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in a child or embryo.
  • SCID (Severe Combined Immuno Deficiency) due to ADA deficiency.

Question 16.
The shape of RBCs of sickle cell anaemia patients changes to the elongated sickle-like structure. Give reason.
Answer:
In 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.

Question 17.
Classify restriction enzymes. Mention the function of each.
Answer:

  • Restriction enzymes are the nucleases. They are of two types namely exonucleases and endonucleases.
  • Exonucieases remove nucleotides from the ends of the DNA and endonucleases can cut at specific points within the DNA itself.

Question 18.
Name the pioneer species in primary succession and primary succession in water.
Answer:

  1. Crustose lichens on rocks.
  2. Phytoplanktons in water.

Part – C

Answer any five of the following questions in about 40 to 80 words each wherever applicable: (5 × 3 = 15)

Question 19.
What is a heterothallic condition? Differentiate between staminate and pistillate flowers.
Answer:

  • If plants possess male and female reproductive structures on different plants (unisexual) it is termed as a heterothallic condition.
  • 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.

2nd PUC Biology Previous Year Question Paper June 2015

Question 20.
What are cleistogamous flowers? Write their significance.
Answer:

  • 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.
  • Cleistogamous flowers produced assured seed-set even in the absence of pollinators.

Question 21.
How do the following assisted reproductive technologies assist the couples to have children?

  1. In-vitro Fertilization
  2. Zygote Intra Fallopian Transfer
  3. Gamete Intra Fallopian Transfer

Answer:

  1. 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.
  2. In ZIFT, zygote or early embryos (with upto 8 blastomeres) is transferred into the fallopian tube.
  3. 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.

Question 22.
Name the six stages of human evolution in the correct sequence.
Answer:

  1. Dryopithecus and Ramapithecus.
  2. Homo habilis (Handyman)
  3. Homo erectus
  4. Homo neanderthalensis
  5. Homo sapiens sapiens

Question 23.
What is contact inhibition? Mention the types of tumours.
Answer:
Normal 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.
Tumours are of two types

  • Benign
  • Malignant tumours

2nd PUC Biology Previous Year Question Paper June 2015

Question 24.
Draw a labelled diagram of plasmid pBR-322.
Answer:
2nd PUC Biology Previous Year Question Paper June 2015 Q24

Question 25.
In which ways genetically modified plants are useful?
Answer:
Application of biotechnology in Agriculture:
Agricultural practice is of three types:

  • Agrochemical-based agriculture.
  • Organic agriculture and
  • Genetically engineered crop-based agriculture.

The use of genetically modified plants has been useful in the following ways:

  • Genetic modification has made the crops more tolerant to abiotic stresses like cold, heat, drought, salinity etc.
  • Post-harvest losses are much reduced.
  • Food produced from GM (Genetically modified) crops have enhanced nutritional value and taste.
  • As the plants have increased efficiency of mineral usage by plants, the early exhaustion of fertility of the soil is prevented.

Question 26.
(a) Single-cell proteins can be used as an alternate source of proteins for human nutrition. Illustrate with two examples.
(b) How can tissue culture be applied to recover healthy plants from virus-infected plants?
Answer:
(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
(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.

(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.

Part – D

Section – I

Answer any four of the following questions in about 200 to 250 words each, wherever applicable: (4 × 5 = 20)

Question 27.
Write the events of development of female gametophyte incorrect order.
Answer:
Development of female gametophyte:

  • The female gametophyte develops from a single megaspore and hence is described as monosporic development.
  • The functional megaspore is the first cell of the female gametophyte of angiosperms.
  • It enlarges to form the female gametophyte, also called embryo sac.
  • Its nucleus undergoes a mitotic division and the two nuclei move to the opposite poles, forming the 2-nucleate embryo sac.
  • Two successive mitotic divisions in each of these two nuclei result in the formation of an 8-nucleate embryo sac.
  • Cell wall formation starts at the eight-nucleus stage, resulting in the formation of a typical female gametophyte.
  • 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.
  • The remaiñing 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.
  • Thus a typical angiosperm embryo sac is 8-nucleate and celled.

2nd PUC Biology Previous Year Question Paper June 2015

Question 28.
What is incomplete dominance? Explain the inheritance of flower colour in snapdragon.
Answer:
It 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.

In Mirabilis or 4° 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 F1 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.

P1 Red flowers × White flowers
F1 Pink flower
When F1 pink flower plants are self cross the raise F2 progeny 3 types of plants are produced i.e, Red, Pink and White, in the ratio of 1 : 2 : 1 respectively.
By using appropriate symbols, the incomplete dominance can be represented as shown below.
This cross is illustrated as indicated below by using appropriate letters.

Thus, in this case, parental characters are expressed in homozygous conditions and intermediate character in heterozygous condition.

Question 29.
Write the diagrammatic representation of the life cycle of HIV.
Answer:
2nd PUC Biology Previous Year Question Paper June 2015 Q29

Question 30.
Describe the parts of a biogas plant with a labelled diagram.
Answer:
2nd PUC Biology Previous Year Question Paper June 2015 Q30

Question 31.
Describe Frederick Griffth’s experiment to show transformation in bacteria.
Answer:
In 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.
2nd PUC Biology Previous Year Question Paper June 2015 Q31
In 1944, Avery, MacLeod and Mc Carty extended Griffith’s experiment to identify’ the transforming principle. They mixed ‘R’ strain with DNA extracted from ‘S’ 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.

2nd PUC Biology Previous Year Question Paper June 2015

Question 32.
(a) Define the following terms:
(i) Commensalism.
(ii) Brood parasitism.
(iii) Mutualism.
Answer:
(i) Commensalism:
It is a relationship between two living individuals of different species in which one is benefitted. while the other is neither harmed nor benefitted.

(ii) Brood parasitism:
Brood 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’s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest.
e.g. cuckoo lays eggs in the nest of crow.

(iii) Mutualism:
It 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.

(b) Mention four basic processes that fluctuate the density of a population in a given habitat.
Answer:
Natality, Mortality, Immigration and Emigration.

Section – II

Answer any three of the following about 200 to 250 words each wherever applicable: (3 × 5 = 15)

Question 33.
Draw a neat labelled diagram of the human male reproductive system.
Answer:
2nd PUC Biology Previous Year Question Paper June 2015 Q33

Question 34.
Describe the process of translation of mRNA.
Answer:
Translation:
The 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.
The process of translation involves the following steps.

  1. Activation of amino acids.
  2. Initiation of the polypeptide chain.
  3. Elongation of the polypeptide chain.
  4. Termination of the polypeptide chain.

2nd PUC Biology Previous Year Question Paper June 2015

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 activatçd aminoacid.
Amino acid + ATP → Amino acid-AMP – enzyme complex
This activated amino acid then combines with 3 ends of tRNA forming aminoacyl-tRNA complex or charged RNA and AMP is released.

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 ‘P’ site of the ribosome.

3. Elongation of polypeptide chain: The ‘A’ site of 50s ribosome is now free and so the second tRNA with second activated amino acid is get attached at ‘A’ site. A peptide bond is formed between 1st and 2nd amino acid in presence of peptidyl transferase at ‘P’ site. Now the ribosome moves a distance of one codon so that the ‘A’ site become vacant. Meanwhile, the 3rd tRNA with 3rd amino acid gets attached to ‘A’ site. Again the ribosome moves on mRNA by
one codon and bonding takes place between 2” and 3td aminoacid at ‘P’ site. 1st tRNA is released through ‘E’ site. This process continues until all the codons on mRNA are completely read. Thus polypeptide chain elongates along the mRNA.

4. Termination and release of polypeptide chain: The process of protein synthesis continues till the arrival of terminator codon on mRNA at ‘A’ 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 RF1, RF2, and RF3 are involved. Thus the polypeptide chain is finally released.

  • The speed of transcription is 30 nucleotides per second in bacteria.
  • The speed of translation 20 amino acids per second in bacteria.

Question 35.
Explain the important steps involved in the process of decomposition.
Answer:
Detritus is the raw material for decomposition.
Following are the steps in the process of decomposition:

  1. Fragmentation: It is the process of breaking the detritus into smaller particles by detritivores.
  2. Leaching: It is the process in which water-soluble inorganic substances run down into soil profile and get precipitated as unavailable salts.
  3. 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.
  4. Humification: Humification during decomposition leads to the accumulation of a dark coloured amorphous substance called humus.
  5. Mineralisation: It is the process in which the humus is degraded by certain microbes and the inorganic nutrients are released.

Decomposition is largely are aerobic process, i.e., it requires oxygen.
The factors affecting decomposition are:

  • The chemical composition of detritus and
  • The climatic factors.

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.
Temperature and soil moisture are the important climatic factors that regulate decomposition through their effects on the activities of soil microbes.
Warm and moist environment favours decomposition, while low temperature and anaerobic conditions inhibit decomposition.

2nd PUC Biology Previous Year Question Paper June 2015

Question 36.
(a) Explain the types of outbreeding processes employed to improve desirable qualities of animals.
Answer:
Outbreeding:
Outbreeding 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).

Outbreeding is of the following types:

(i) Outcrossing:

  • 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.
  • The offspring of outcrossing, is called an outcross.
  • A single outcross helps to overcome inbreeding depression.
  • It is the best breeding method for animals that are below average in productivity and growth rate.

(ii) Cross-breeding:

  • 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.
  • 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.
  • One example of cross-breeding is Hisardale, a new breed of sheep developed by crossing Bikaneri ewes and Marino rams.

(iii) Interspecific hybridisation:

  • 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.
    e.g.: Mule is produced by a cross between a male donkey and a female horse.

(b) Write any two processes deal with to increase the yield and quality of milk in dairy farm management.
Answer:
1. It is a method to improve herds.
The steps in the method are as follows:

  • A cow is administered hormones (like FSI-I) to induce follicular maturation and superovulation, i.e., production of 6-8 ova in one cycle.
  • The cow is mated with the selected bull or artificially inseminated.
  • The fertilised eggs at 8-32-celled stages are recovered and transferred to surrogate mothers.

This technology has have been used for cattle, rabbits, mares, etc.
High milk-yielding breeds of females and high-quality meat-yielding bulls have been bred successfully to increase the herd size in a short time.

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.

2nd PUC Biology Previous Year Question Paper June 2015

Question 37.
What Is Eutrophication? Discuss the phenomenon of accelerated eutrophication.
Answer:
Eutrophication is a natural process which means ‘well-nourished or enriched’. 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.

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