2nd PUC Biology Previous Year Question Paper March 2015

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Karnataka 2nd PUC Biology Previous Year Question Paper March 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.
What is Biomagnification?
Answer:
The process of accumulation of certain pollutants in tissues with increased concentration along the food chain is called Biomagnification.

Question 2.
Name a causative organism of pneumonia.
Answer:
Streptococcus pneumonia/Haemophilus influenza.

Question 3.
Give an example of adaptive radiation.
Answer:
Darwin’s finches.

2nd PUC Biology Previous Year Question Paper March 2015

Question 4.
Define parthenogenesis.
Answer:
The phenomenon where-female gametes develop into new organisms without fertilisation is known as parthenogenesis.

Question 5.
Papaya plants exhibit xenogamy only. Why?
Answer:
Papaya plants are dioecious in nature, thus they exhibit xenogamy.

Question 6.
What is detritus?
Answer:
Dead plant remains such as leaves, bark, flowers and dead remains of animals including faecal matter constitute detritus.

Question 7.
Mention the use of Cyclosporin A.
Answer:
It is used as an immunosuppressive agent in organ-transplant patients.

Question 8.
Name the plant from which cocaine is obtained.
Answer:
Erythroxylum coca [coca plant].

2nd PUC Biology Previous Year Question Paper March 2015

Question 9.
Mention the technique used to separate DNA fragments in rDNA technology.
Answer:
Agarose gel electrophoresis.

Question 10.
Why is oxytocin necessary for parturition?
Answer:
It causes stronger uterine contractions which leads to the expulsion of the baby out of the uterus.

Part – B

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

Question 11.
Differentiate between the menstrual cycle and the oestrus cycle.
Answer:
Non-primates like cows, sheep, deer etc exhibit cyclic changes in their reproductive phase called the oestrous cycle. It consists of a short period of oestrus or heat period followed by prolonged anoestrus period or passive or rest period. The female receives the male for copulation only in heat period (seasonal breeders).
Primates like monkeys apes and humans are continuous breeders. They breed throughout the year and their reproductive cycle is called the menstrual cycle.

Question 12.
Mention any four sexually transmitted diseases.
Answer:

  1. Syphilis
  2. Gonorrhoea
  3. Genital herpes
  4. AIDS.

Question 13.
tRNA is an adaptor molecule. Comment.
Answer:
tRNA can read the code on mRNA and on other hand would bind to specific amino acids, thus it is regarded as an adapter molecule.

Question 14.
Distinguish between Benign tumour and Malignant tumour.
Answer:
Benign tumours: Tumours result from abnormal and persistent cell division consists of cancer cells that remain localized at the spot of origin and do not spread to distinct sites. These are not a great threat to life (non – fatal), but some can be fatal as in brain tumour, Well-differentiated cells generally characterize benign tumours.

2nd PUC Biology Previous Year Question Paper March 2015

Malignant tumours: Tumours result from abnormal and persistent cell division consist of cancer cells that may be carried by the circulating fluid (blood or lymph) or by direct penetration to another part of the body and changing the neighbouring tissues where they induce secondary’ tumours. Malignant tumours usually contain undifferentiated cells. These are fatal to the organism. The phenomenon of formation of secondary tumours is called metastasis.

Question 15.
List any four hormones secreted by the placenta.
Answer:
Progesterone, human chorionic gonadotropin, human chorionic somatotropin and placental lactogen. .

Question 16.
What are the conclusions drawn by T.H. Morgan from the crossing experiment in Drosophila with respect to linkage?
Answer:

  1. When the two genes in a dihybrid cross are located on the same chromosome the proportion of parental gene combinations in the progeny was much higher than the non-parental or combination of genes.
  2. The proportion of recombinants varies, even if the two genes are present on the same chromosome.
  3. If the linkage is stronger between two genes the frequency of recombination is low and vice versa.

Question 17.
Discuss the role of fungi as biofertilizers.
Answer:
Fungi are also known to form symbiotic associations with plants (mycorrhiza). The fungal symbiont absorbs phosphorus from the soil and passes it to the plant. Also, it shows other benefits like resistance to root borne pathogens, tolerance to salinity and drought.

Question 18.
Write a note on single-cell protein.
Answer:
Single-cell protein (SCP):
Single-cell protein is one of the alternative sources of proteins for the nutrition of humans and animals.
Microbes are being grown on an industrial scale as a source of good protein.

Microbes can be grown easily on materials like wastewater from potato processing plants (containing starch), straw, molasses, animal mañure and even sewage, to produce large quantities and can serve as food, rich in protein, minerals, fats, carbohydrate and vitamins.

Note: It has been calculated that a 250 kg cow produces 200 g of protein per day whereas 250 g of Methylophilus methylotrophs produce 25 tonnes of protein.

2nd PUC Biology Previous Year Question Paper March 2015

The advantages are that:

  1. SCPs are rich in proteins, minerals, vitamins and carbohydrates and low in fats.
  2. They can be easily grown with cheaper materials like wastewater from potato-processing plants, animal manure, molasses, etc.
  3. The use of waste materials (as culture medium) reduces pollution.
  4. They reduce the pressure on agriculture (for the supply of pesticides, fertilizers, etc.) e.g., 250 g of Methylophilus methylotrophic bacterium has been used to produce 25 tonnes of proteins.

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 aneuploidy? Write any four symptoms of Down’s syndrome.
Answer:
If there is excess or absence of one or more chromosome’s in the total complement is known as aneuploidy.

Symptoms:

  • Small penis
  • Diminished pubic, axillary, and facial hair.
  • Enlarged breast tissue (called gynecomastia).
  • Learning disabilities
  • Simian crease (a single crease in the palm)
  • Abnormal body proportions (long legs, short trunk)
  • Small firm testicles
  • Sexual dysfunction
  • Tall Stature
  • Personality impairment

Note: The severity of symptoms may vary.

Tests may include:

  • Karyotyping showing 47 chromosomes with XXY
  • Semen exam showing low sperm count
  • Decreased serum testosterone level
  • Increased serum luteinizing hormone and increased serum follicle-stimulating hormone.

Question 20.
RNA polymerases in eukaryotes show division of labour. Substantiate.
Answer:
In eukaryotes, there are at least three RNA polymerases in the nucleus with a clear cut division of labour

  • RNA polymerase I transcribe rRNAs
  • RNA polymerase II transcribes the precursor of mRNA (i.e., hnRNA)
  • RNA polymerase III catalyses the transcription of tRNA.

Question 21.
Suggest any three assisted reproductive technologies to overcome infertility.
Answer:

  1. IVF-ET
  2. ZIFT
  3. GIFT.

Question 22.
Name asexual reproductive structures in

  1. Hydra
  2. Chlamydomonas
  3. Penicillium.

Answer:

  1. Budding
  2. Zoospores
  3. Conidia.

2nd PUC Biology Previous Year Question Paper March 2015

Question 23.
Mention three types of Innate barriers of defence with an example each.
Answer:

  1. Physical barriers – skin, mucous membrane.
  2. Physiological barriers – saliva
  3. Cellular barriers – WBC

Question 24.
Draw a neat labelled diagram of a typical Biogas plant.
Answer:
2nd PUC Biology Previous Year Question Paper March 2015 Q24

Question 25.
(a) What is endemism?
Answer:
The species which are confined toa specific geographical region and not found elsewhere.

(b) Mention ‘The Evil Quartet’ of biodiversity loss.
Answer:

  • Habitat loss and fragmentation
  • Alien species invasions
  • Co-extinctions.

Question 26.
How does pollination is achieved in Vallisneria and seagrasses?
Answer:
Cross-pollination or allogamy: When the pollen grains of one flower pollinates the stigma of another flower located on the same plant or on the different plant of the same kind, it is called cross-pollination. Cross-pollination is of three kinds.

  1. Geitonogamy: Cross-pollination which takes place between two flowers borne on the same plant is called Geitonogamy. From the genetical point of view there is a little difference between geitonogamy and autogamy.
  2. Xenogamy: Cross-pollination which takes place between two flowers borne on two different plants of the same species is called xenogamy.
  3. Hybridization: Cross-pollination between the two flowers borne on two different species or genera is called hybridisation.

The following contrivances ensure cross-pollination:

1. Dicliny: It is the condition where one of the two sexes is absent in the flower and flower becomes unisexual male or female (diclinous). Such diclinous flowers may be borne either on the same plant or on the two different plants. in such cases, cross-pollination is the rule. e.g.. maize, Cucurbita.

2. Herkogamy: (flakes = barrier) It is the condition where the style of the gynoecium extend far beyond the anthers or stamens may face outward or pollens may aggregate into pollinia. In such cases, self-pollination is impossible. e.g. gloriosa, calotropis.

3. Dichogamy: It is the condition where androecium and gynoecium in a flower mature at different times. In such a case, self-pollination is ineffective, however, it may take place at a later stage if cross-pollination fails. Dichogamy may be of two types:

  • Protandry: Anthers mature earlier than the carpels. e.g. sunflower, cotton.
  • Protogyny: Carpels mature earlier than anthers. e.g. Michela, ficus.

4. Self sterility: It is the condition where pollen grains fail to germinate on the stigma of the same flower. In such cases, self-pollination is ineffective and cross-pollination is a must. e.g. tobacco, potato.

2nd PUC Biology Previous Year Question Paper March 2015

5. Heterostyly: It is the condition where the flowers on the same plant have styles of different sizes, where one has short style and long stamens, while another has long style and short stamens. In such cases, self-pollination ¡s impossible and cross-pollination is a must. e.g. oxalis, primula.

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.
Draw a neat labelled diagram of a sectional view of the human female reproductive system.
Answer:
2nd PUC Biology Previous Year Question Paper March 2015 Q27

Question 28.
(a) Write the steps involved in multiple ovulation and embryo transfer.
Answer:
Multiple ovulation and embryo transfer.
The steps in the method are as follows:

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

This technology has 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 within a short time.

(b) Differentiate between outcrossing and cross-breeding.
Answer:
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.

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.

2nd PUC Biology Previous Year Question Paper March 2015

Question 29.
Describe the process of DNA replication with the help of a diagram.
Answer:
The process of formation of a new copy of DNA from the existing one is called replication of DNA. It takes place in the semi-conservative method. i.e, one strand of parent molecule is conserved in the daughter molecule of DNA. This mechanism was proposed by Watson and crick and experimentally proved by Meselson and Stahl.

Requirements: DNA templates. Unwind as, RNA primer, RNA polymerase, DNA polymerase I and III, DNA ligase and deoxyribonucleotides
2nd PUC Biology Previous Year Question Paper March 2015 Q29
Steps:
1. Unwinding of DNA: Due to the activity of REN nicking of DNA strand takes place at a specific point to form orisite. Unwinding bf DNA takes place from orisite, with the help of enzyme unwinds or helicase The two separated strands of DNA look like ‘y’-shape and is called the replication fork. The separated DNA strands act as templates for the synthesis of a new strand.

2. Synthesis of RNA primer: RNA primer is a short segment of RNA (8-10 nucleotides) synthesized on DNA template with the help of RNA polymerase or primase. The synthesis of RNA primer is required as the enzyme DNA polymerase III can not directly initiate the synthesis of new DNA strand on the template.

3. Formation of leading and lagging strands: Formation of new DNA strands by DNA polymerase enzyme can occur only in 5’ to 3’ direction. Synthesis of new complementary strain’s takes place on both the templates ¡n opposite directions. The new complementary strand that is formed towards the replication fork on Template strand is continuous and is called the leading strand or a continuous strand.

The synthesis of the other new complementary strand from the replication fork occurs discontinuously on the parental strand and is called lagging strand or a discontinuous strand. This lagging strand will have small segments of RNA primer to which DNA nucleotides are joined. These are called Okazaki fragments.

4. Formation and separation of the daughter DNA molecule: ln this step RNA primers are removed by the DNA polymerase I and the gaps are filled up by the DNA nucleotides by the same enzyme. Then short segments of DNA are joined by DNA lipase forming a continuous strand of DNA. Thus each daughter DNA molecule will have one old parental strand and one new strand and hence it follows a semi-conservative method of DNA replication.

Question 30.
Explain the transverse section of young anther with a labelled diagram.
Answer:
Structure of a mature anther:
A mature anther internally contains four microsporangia, two in each anther lobe. Each microsporangium consists of an outer anther wall and inner anther locule.
2nd PUC Biology Previous Year Question Paper March 2015 Q30
1. Anther wall: It is the protective covering of anther or microsporangium. It is composed of epidermis, endothelium, middle layer and tapetum.

  • Epidermis: It is the outermost prot&tive Layer of anther composed of a single layer of cells. The cells are often stretched and flattened in a mature anther.
  • Endothecium: It is the uniseriate layer composed of radially elongated cells with callose thickening on their inner tangential walls. Along the line of dehiscence of anther lobe, endothecium is composed of thin-walled cells called stomium.
  • Middle layers: These are 2-3 concentric layers of cells present in between the endothecium and tapetum. In mature anther, the cells of the middle layers are flattened and collapsed due to the pressure exerted in the anther locule.
  • Tapetum: This is the innermost layer of the anther wall composed of secretory cells. These cells secrete nutrients for developing microspore mother cells! pollen grains in the anther locule.

2nd PUC Biology Previous Year Question Paper March 2015

2. Anther locule: It is the sporangial cavity containing a number of microspore tetrads/pollen grains formed from microspore mother cells by meiosis. (Microspores after the separation from tetrad are called pollen grains)

Question 31.
Explain Brood parasitism and ‘sexual deceit’ as an interaction of species with an example each.
Answer:
Brood parasitism in birds is a fascinating example of parasitism in which the parasitic bird 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 aird ejecting them from the nest. Eg: Cuckoo lays eggs in the nest of crow.

The Mediterranean orchid ophrys employs sexual deceit to get pollination done by a species of bee. One petal of its flower bears a un carry resemblance to the female of the bee in size, colour and markings. The male bee is attracted to what it perceives as a female, ‘pseudo copulates’ with the flower and during that process is dusted with pollen from the flower. When this same bee pseudocopulation with another flower, it transfers pollen to it and thus pollinates the flower.

Question 32.
With reference to flower colour in snapdragon, explain incomplete dominance.
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 40 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 crossed to 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.
2nd PUC Biology Previous Year Question Paper March 2015 Q32
Thus, in this case, parental characters are expressed in homozygous conditions and intermediate character in heterozygous condition.

Section – II

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

Question 33.
Give the diagrammatic representation of Miller’s experiment.
Answer:
In 1953, Stanley Miller under the guidance of his professor Harold Urey conducted an experiment to investigate the conditions under which the simplest organic compounds were formed from the gases of the primitive earth. Miller designed a glass apparatus called “Spark-discharge apparatus” to conduct the experiment: Spark discharge apparatus is roughly rectangular in shape with a Spark discharge chamber, water boiler, U-trap condensers all of which were connected by a rectangular side tube as shown in the diagram.
2nd PUC Biology Previous Year Question Paper March 2015 Q33
The side tube is connected to the vacuum pump or stop cock, water was taken in the round bottom flask and the entire apparatus was evacuated to remove the free molecular oxygen. A mixture of gases like methane, hydrogen, and ammonia in the ratio of 2 : 1 : 2 by volume were passed into the apparatus. The water in the round bottom flask was boiled to produce steam which was circulated clockwise in the apparatus.

The spark discharge chamber contained two tungsten electrodes which were connected to a sparking coil that produced a continuous discharge of sparks which served as sources of energy similar to the lighting of the primitive Earth. The convectional current created by the circulation of the steam carried the gases across the electrodes. The cooling jacket or condenser present below the sparking chamber condensed the steam and the contents were collected in the U-trap.

2nd PUC Biology Previous Year Question Paper March 2015

When the products were analyzed it was found that there were a number of simpler organic compounds such as amino acids, acetic acid, propionic acid etc. The acids formed were alanine, glycine, aspartic acid, glutamic acids etc. Formation of amino acid is an important step in the origin of life. This Miller’s experiment provided the vital proof that organic compound clean be formed from simple inorganic molecules under the conditions that existed on the primitive earth. This proves the origin of life from chemical substances.

Question 34.
(a) Development of Bt cotton has reduced the use of insecticides. Justify.
Answer:

  1. Bacillus thuringiensis produces crystal proteins called Cry proteins which are toxic to larvae of insects like tobacco budworm, armyworm, beetles and mosquitoes.
  2. Cry proteins exist as inactive protoxins and get converted into active toxin when ingested by the insect, as the alkaline pH of the gut solubilises the crystals.
  3. The activated toxin binds to the surface of epithelial cells of midgut and creates pores.
  4. This causes swelling and lysis of cells leading to the death of the insect (larva).
  5. 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.
  6. The proteins encoded by the following cry genes control the corresponding pests.
    • Cry I AC and Cry II Ab control cotton bollworms.
    • Cry I Ab controls corn borer.
    • Cry III Ab controls Colorado potato beetle.
    • Cry III Bb controls corn rootworm.

(b) Write a note on gene therapy.
Answer:
Gene therapy:
The technique of curing genetic diseases by replacing the defective genes by normal genes is called gene therapy W.F. Anderson is regarded as the father of gene therapy.

Types of gene therapy:
There are two main types of gene therapy, namely somatic cell gene therapy and germline gene therapy.
(a) Somatic cell gene therapy: It is the replacement of the defective gene by a normal gene in somatic cells of the body. This therapy helps for treating the defective organs or tissues parts such as blood cells, liver cells, skin cells, lung cells etc. The genetic defects can be rectified only in that individual. The replaced gene is not transmitted to the next generation; The diseases like SCID [Severe combined immuno deficiency], Cystic fibrosis, Haemophilia, Cancer etc can be cured by somatic cell gene therapy.

(b) Germline gene therapy: It is the replacement of the defective genes by normal genes in germ cells to rectify the genetic disorders. This includes gene therapy in sperms, eggs. zygotes or early embryos. Any change made in the germ cells by gene therapy will be transmitted to the offspring.

Question 35.
Explain the features of cloning vectors.
Answer:
Vectors: These are molecular vehicles which carry the desired gene from one organism to the other organism. There are different types of vectors namely plasmids, bacteriophages, cosmids, plasmids etc [cosmid means plasmid + cohesive ends of phage DNA, Plasmid means phage DNA + a small part of plasmid]
Vectors for cloning genes in plants and animals are Ti plasmid isolated from Agrobacterium tumefaciens in plants and retrovirus are now made non-pathogenic and are used to deliver a gene into animal cells.

2nd PUC Biology Previous Year Question Paper March 2015

Question 36.
(a) What are the effects of ozone depletion?
Answer:
Thinning of the ozone layer and development of ozone holes increases the amount of UV-B radiations reaching the earth’s surface. A 5% ozone depletion increases UV-B radiations by 10%. Increased incidence of UV-B radiations on earth will have the following adverse effects.

Skin Cancers: There is an increase in the incidence of skin cancers. 1% fall in ozone concentration increase UV load of the earth by 2% that causes the addition of 50,000 cases of skin cancer. in Australia which lies near the area of the ozone hole, every second middle-aged person suffers from skin cancer while in old persons the incidence is nearly 100%.

Blinding: Many land animals would lose their eye šight and become blind. In human beings, the cases of photo-burning, cataract and dimming of eyesight are on the increase. 1% fall of ozone concentration in the stratosphere will blind another 1 lakh persons.

Immune System: It is partially suppressed. Incidence of herpes and other immune system-related diseases are likely to increase.

Larval Stages: More larvae and young ones of aquatic animals will die.

Photosynthesis: Photosynthetic machinery is impaired. Photosynthesis decreases by 10 – 25%. There is a corresponding fall in the yield of crops.

Nucleic Acids: VV radiation damages nuçleic acids by forming dimers. Incidence of harmful mutations increases.

Phytoplankton: Both photosynthetic activities as well as the function of phytoplankton are disturbed by UV-radiations. 6-22% fall in productivity will occur.

Global Warming: Decreased primary productivity over land and inside oceans will increase carbon dioxide concentration resulting in global warming, despite a reduction in CO2 emissions from industries and automobiles.

(b) BOD is an index of water pollution. Comment.
Answer:
The BOD test measures the rate of uptake of oxygen by microorganisms 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.

2nd PUC Biology Previous Year Question Paper March 2015

Question 37.
(a) What is ecological succession?
Answer:
The occurrence of a relatively definite sequence of communities over a long period of time in the same area resulting in the establishment of stable or climax community is known as ecological or biotic succession.

(b) Represent an ideal pyramid of number in a grassland ecosystem.
Answer:
2nd PUC Biology Previous Year Question Paper March 2015 Q37(b)

(c) Name the two types of nutrient cycles with an example each.
Answer:

  1. Gaseous cycle
    eg: carbon cycle
  2. Sedimentary cycle
    eg: Phosphorous cycle.

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