2nd PUC Biology Previous Year Question Paper June 2014

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Karnataka 2nd PUC Biology Previous Year Question Paper June 2014

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 menopause?
Answer:
In human beings, the menstrual cycle ceases around 50 years of age and it is termed as menopause.

Question 2.
Name the physical and physiological connection between maternal body and embryo (foetus).
Answer:
Placenta.

Question 3.
What is adaptive radiation?
Answer:
Adaptive radiation is an evolutionary process in which an ancestral stock gives rise to new species in a given geographical area, starting from a point and literally radiating to other geographical areas or habitats.

2nd PUC Biology Previous Year Question Paper June 2014

Question 4.
Mention the cells involved in cell-mediated immunity.
Answer:
T-Lymphocytes [T-helper cells, T-killer cells]

Question 5.
What is Benign tumour?
Answer:
the localised tumour which remains at the spot of origin and do not spread to distant sites is a benign tumour.

Question 6.
What is totipotency?
Answer:
The ability of a plant cell to give rise to an entire plant on the medium is called totipotency.

Question 7.
What are bio fertilisers?
Answer:
Bio fertilisers are the organisms that enrich the nutrient quality of the soil. The main source of biofertilizers are bacteria, fungi and cyanobacteria.

Question 8.
What is a polymerase chain reaction?
Answer:
It is a technique in which a gene of choice is amplified into many copies using primers & DNA polymerase enzyme.

Question 9.
Define natality.
Answer:
It refers to the number of births during a given period in the population that are added to the initial density.

2nd PUC Biology Previous Year Question Paper June 2014

Question 10.
Give an example for the gaseous cycle.
Answer:
Nitrogen cycle.

Part – B

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

Question 11.
What are hermaphrodites? Give an example.
Answer:
The animals in which both the sex organs are present in the same body are called hermaphrodites.
eg: Planaria.

Question 12.
Mention two symptoms of Turner’s syndrome.
Answer:

  • Broad chest.
  • Low hairline at the back of the neck.
  • Shortened fourth and fifth fingers.
  • Short neck with a webbed appearance.

Question 13.
What is a Homologous structure? Give an example.
Answer:
Homology is the relationship among organs of different groups of organisms that show similarity in their basic structure and embryonic development but perform different functions.
eg: Forelimbs of whales, birds, amphibians and mammals.

Question 14.
Draw a neat labelled diagram of an antibody.
Answer:
2nd PUC Biology Previous Year Question Paper June 2014 Q14

Question 15.
Mention any two measures to improve the quality and quantity of milk production.
Answer:

  1. Selection of quality breeds.
  2. Feeding of the cattle show is done in a scientific manner, considering the quantity and quality of the folder.

Question 16.
What is commensalism? Give an example.
Answer:
It is an interaction in which one species benefits and the other is neither benefited nor harmed.
eg: An orchid growing as an epiphyte on a range branch.

2nd PUC Biology Previous Year Question Paper June 2014

Question 17.
List out any two effects of loss of biodiversity.
Answer:

  • Habitat loss and Fragmentation
  • Overexploitation.

Question 18.
What are sacred groves? Give an example.
Answer:
The tracts of forest conserved and protected based on the religious faith ground.
eg: Khasi and Jaintia Hills in Meghalaya.

Part – C

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

Question 19.
Mention the units of vegetative propagation in plants.
Answer:
Root tubers
Stem: Runners, suckers, stolon, rhizome, corms, bulbs & tubers.
Leaf: Vegetative buds, Bulbils.

Question 20.
Draw a neat labelled diagram of a longitudinal section of a typical flower.
Answer:
2nd PUC Biology Previous Year Question Paper June 2014 Q20

Question 21.
What is male heterogamety? Explain sex determination in humans.
Answer:
Heterogametic sex (gametic sex) refers to the sex of a species in which the sex chromosomes are not the same. For example, in humans, males, with an X and a Y sex chromosome, would be referred to as the heterogametic sex, and females having two X sex chromosomes would be referred to as the homogametic sex.

In the human being, the chromosomal mechanism of sex determination is of XX-XY type. In human the nucleus of each cell contains 46 chromosomes or 23 pairs of chromosomes of body characters) and 1 pair is of sex chromosomes (responsible for determination of sex).
1. In the female, two homomorphic sex chromosomes are XX.
2. In the male, two heteromorphic sex chromosomes are XY.
3. The genotypes of female and male are
Female: 46 Chromosomes = 44 Autosomes + XX sex chromosomes
Male: 46 Chromosomes = 44 Autosomes + XY sex chromosomes

2nd PUC Biology Previous Year Question Paper June 2014

Question 22.
Draw a labelled diagram of a diagrammatic representation of Miller’s experiment.
Answer:
2nd PUC Biology Previous Year Question Paper June 2014 Q22

Question 23.
List any three types of Innate Immunity barriers and give an example for each.
Answer:

  • Surface barriers. eg: Skin, mucus membrane
  • Cellular barriers. eg: Phagocytosis, NK Cells
  • Biochemical barriers. eg: Interferons.

Question 24.
Explain any three types of major abiotic factors influencing the life of organisms.
Answer:
1. The DNA from the desired segment to be amplified, an excess of the two primer molecules, the four deoxyribose triphosphates and DNA polymerase are mixed together in a reaction mixture in an Eppendorf tube with sufficient quantities of Mg. The Eppendorf tube is placed in the PCR unit and the following operations are performed sequentially.

2. Denaturation: The reaction mixture is first subjected to a temperature between 90-98°C (commonly 94°C). It results in the separation of two strands of DNA due to the breakage of hydrogen bonds. This is called denaturation. Each strand of DNA acts as a template strand for DNA synthesis. The duration of this step in the first cycle of PCR is usually 2 minutes at 94°C.

3. Annealing (anneal join): The mixture ¡s now cooled to a low temperature (40-60°C). During this step, two oligonucleotide primers, complementary to a region of DNA, anneal (hybridize) one to each 3 ends of the DNA strand. The duration of the annealing step is usually one minute during the first as well as the subsequent cycles of PCR.

Question 25.
Comment: Compressed Natural Gas (CNG) is better than diesel.
Answer:
CNG Bums most efficiently unlike petrol or diesel in the automobiles and very little of it is left unburnt.
Moreover, it is cheaper than petrol and diesel. It cannot be siphoned off by thieves and adulterated like petrol or diesel.

2nd PUC Biology Previous Year Question Paper June 2014

Question 26.
What is biomagnification? Name two well studied toxic substances causing this phenomenon.
Answer:
Biomagnification is defined as the accumulation of a particular substance in the body of the organisms at different trophic levels of a food chain. One example of biomagnification is the accumulation of insecticide DDT which gets accumulated in zooplanktons. Small fishes consume these zooplanktons. Small fishes are consumed by large fish which are finally consumed by fish-eating bird present at the highest position in the tropic level. Hence, the organism at the highest trophic level has the maximum accumulation of toxic substances.

Human Health:
Humans become more susceptible to cancers, liver and kidney failure, respiratory disorders, birth defects in pregnant women, brain damage, and heart diseases are a result of mercury, cadmium, lead, cobalt, chromium and other chemical poisonings. For instance, diseases like hepatitis and cancer have been attributed to consuming seafood that has been poisoned by mercury and polycyclic aromatic hydrocarbons (PAPIs).

Marine Creatures:
The various toxic elements get accumulated in the vital organs of the various aquatic creatures, thus affecting their reproduction and development. Selenium and heavy metals such as mercury also affect the reproduction of aquatic creatures such as fish as it destroys their reproductive organs. Besides, PCBs (polychlorinated biphenyls) also biomagnify and impair reproduction and are considerably high in aquatic systems.

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.
What is double fertilization? Describe fertilized embryosac with a neat labelled diagram.
Answer:
(a) It is the process in which one of the two male gametes fuses with the egg cell to form a zygote and the other with the secondary nucleus to form P.E.N.
2nd PUC Biology Previous Year Question Paper June 2014 Q27

Question 28.
What is spermatogenesis? Explain the process of spermatogenesis.
Answer:
Formation of haploid sperms from the diploid spermatogonial cells of the testes is called spermatogenesis. Primordial germ cells give rise to spermatogonial cells and šertoli cells. It is the spermatogonial cells that develop into the spermatozoa. Spermatogenesis involves four phases. viz..
a. Multiplication Phase
b. Growth Phase
c. Maturation Phase
d. Spermiogenesis

2nd PUC Biology Previous Year Question Paper June 2014

(a) Multiplication Phase: The diploid spermatogonial cells of the seminiferous tubules divide repeatedly by mitosis to form more spermatogonial cells. Among them, only a few will enter into the growth phase and others are kept in reserve.

(b) Growth Phase: Spermatogonial cells obtained from multiplication phase grow in size, however, they still remain diploid. These cells are now called primary spermatocytes.

(c) Maturatioñ Phase: This phase involves two successive divisions, viz., Meiosis I and Meiosis II. The meiosis I is reductional and two haploid cells are formed from each primary spermatocyte. The resultant cells of I meiotic division are called secondary spermatocytes. These secondary spermatocytes undergo meiosis II which is
equational. As a result, each secondary spermatocyte produces two haploid cells of equal size called spermatids. Hence, four haploid spermatids are formed during maturation phase from each diploid primary spermatocyte.

(d) Spermiogenesis: Spermatids are not gametes they are ordinary haploid cells. During spermiogenesis or spermateliosis each spermatid in association with Sertoli cells become tadpole-like, flagellated and highly motile gamete called spermatozoan or sperm.

Question 29.
(a) List three reasons for population explosion.
Answer:

  • The decline in the death rate
  • The decline in Maternal Mortality Rate (MMR)
  • The decline in Infant Mortality Rate (IMR) and
  • Increase in the number of people in the reproductive age.

(b) Name human male and female sterilization procedures.
Answer:
Vasectomy and tubectomy.

Question 30.
State the law of independent assortment. Write a schematic representation of the dihybrid cross.
Answer:
Based on the results of the dihybrid cross, he formulated the second law of heredity namely, the Law of Independent Assortment. It states that “When genes or alleles for two or more separate pairs of contrasting characters are brought together in a hybrid, they separate independently of each other.
2nd PUC Biology Previous Year Question Paper June 2014 Q30

Question 31.
Describe the Lac operon concept with a labelled sketch.
Answer:
2nd PUC Biology Previous Year Question Paper June 2014 Q31
Regulation of gene action is well studied in E.coli bacteria with regard to utilization of lactose by the bacteria.
The utilisation of lactose in E.coli needs three enzymes namely B-galactosidase, B-galactosidase permease and B-galactoside transacetylase. These are produced by Z, Y and genes respectively. Enzyme RNA polymerase enzyme initiates the synthesis of these 3 enzymes.

The mechanism of lac-operon can be studied under two steps namely,
(a) Switched OFF mechanism: When lactose is absent in the medium, the regulator gene produces repressor protein which binds with the operator gene. This prevents the movement of RNA polymerase on the structural genes is blocked o there is no synthesis of mRNA from the structural genes Z, Y and a, so there is no synthesis of enzymes. This is called a switched OFF mechanism

(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.

2nd PUC Biology Previous Year Question Paper June 2014

Question 32.
(a) What is poultry? Mention two important components of Poultry Farm Management.
Answer:
Poultry Farm Management:
Poultry is rearing of domesticated birds (fowl) for meat and eggs. Poultry typically includes chicken and ducks and sometimes turkey and geese.

The important components of poultry farm management include:

  • Selection of disease-free and suitable breeds.
  • Proper feed and water for the birds.
  • Proper and safe farm conditions.
  • Hygiene and health care of the birds.

(b) Define Inter-specific hybridization? Give an example.
Answer:
If hybridization is performed between two plants belonging to two different species then it is called as Inter-specific hybridization.
eg: Crossing wheat and rye to get Triticale.

Section – II

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

Question 33.
Summarize major events of DNA fingerprinting technique.
Answer:
2nd PUC Biology Previous Year Question Paper June 2014 Q33
This is a technique which helps in the establishment of the identity of a person by detecting and comparing spécifie nucleotide sequences in the DNA of individuals. This technique was developed by Alec Jeffreys and his associates.
Steps:
a. Collection of sample: Sample like blood, semen, hair, saliva or any other tissue even in the dried state are collected.

b. Extraction of DNA: The material collected is subjected to high-speed centrifugation which separates the DNA from the cell.

c. Fragmenting DNA: The DNA ¡n the sample is treated with restriction endonucleases to obtain smaller restriction fragments.

d. Separating DNA fragments: The smaller fragments of DNA are made to run on an agarose gel-plate by electrophoresis technique in which depending on molecular size and shape, the nucleotides produce distinct bands in the gel.

e. Extracting DNA from gel: The double-stranded DNA fragment in the gel it denatured into single strands by exposing the gel into an alkaline solution (NaOH). Then by Southern blotting technique, these single-stranded DNA fragments are transferred onto a nitrocellulose filter or a nylon membrane.

f. Tagging radioactive probes: A probe is radioactively labelled synthetic DNA. The nitrocellulose filter is now incubated with these radioactive probes. The probes act as specific nucleotide detectors and bind or hybridise with the specific nucleotides present in DNA fragments on the filter.

2nd PUC Biology Previous Year Question Paper June 2014

g. Auto radiography: An x-ray film is placed on the filter to obtain autoradiographic prints. The film is then processed to observe visible hybridised pattern of bands. This is the genetic fingerprint or DNA fingerprint.

h. Comparison: Comparison of two genetic fingerprints with great care can provide information about the degree of similarities or differences.

Question 34.
Explain the different stages involved in sewage treatment.
Answer:
Sewage 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, ¡t has to be treated in Sewage Treatment Plants (STPs). Treatment of wastewater is done by the heterotrophic microbes naturally present in the sewage. This treatment is carried out in two stages: Primary treatment and secondary treatment. The wastewater can be passed into rivers after secondary treatment. In some cases, tertiary treatment is also carried out which removes nutrients by a chemical process.

Primary Treatment: This step involves the physical removal of floating and suspended solids from sewage through filtratión 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.

Secondary treatment or Biological treatment: The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of useful aerobic microbes into floes (masses of bacteria associated with fungal filaments to form mesh-like structures). While growing, these microbes consume the major part of the organic matter in the effluent. This significantly reduces the BOD (biochemical oxygen demand) of the effluent. BOO refers to the amount of the oxygen that would be consumed if all the organic matter in one litre of water were oxidised by bacteria.

The sewage water ¡s 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.

2nd PUC Biology Previous Year Question Paper June 2014

Once the BOD of sewage or wastewater is reduced significantly, the effluent is then passed into a settling tank where the bacterial ‘floes’ are allowed to sediment. This sediment is called activated sludge. A small part of the activated sludge ¡s pumped back into the aeration tank to serve as the incolulum. The remaining major part of the sludge is pumped into large tanks called anaerobic sludge digesters. Here, other kinds of bacteria, which grow anaerobically, digest the bacteria and the fungi in the sludge.

During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide. These gases form biogas and can be used as a source of energy as it is inflammable. The effluent from the secondary treatment plant is generally released into natural water bodies like rivers and streams.

Question 35.
(a) Write a diagrammatic representation of Recombinant DNA technology.
Answer:
2nd PUC Biology Previous Year Question Paper June 2014 Q35(a)

(b) Write a note on downstream processing.
Answer:
Downstream Processing:
The products formed in a bioreactor have to be subjected through a series of processes before they are ready for marketing as finished products. The various processes used for the recovery of useful products are collectively called downstream processing. The processes include separation and purification of the product, the addition of suitable preservatives and a stringent quality control testing etc. Such formulation has to undergo strict clinical trials as in the case of drugs. These quality control testing and clinical trials vary from product to product.

Question 36.
(a) What is Biopiracy? Explain it with reference to Basmati rice.
Answer:
Biopiracy: It is a term used to refer to the use of bio-resources by multinational companies and other organisations without proper authorisation from the countries and people concerned without compensatory payment.
e.g.: Basmati rice grown in India is distinct for its unique flavour and aroma, but an American company got patent rights on Basmati through the US patent and trademark office. The new variety of Basmati has been developed by this company by crossing an Indian variety with the semi-dwarf varieties.

(b) What are genetically modified organisms? Name two Bt-toxin producing plants.
Answer:
An organism developed by introducing a desired foreign gene is called Genetically modified ‘organism’.
eg: BT Cotton, Bt Soybean.

2nd PUC Biology Previous Year Question Paper June 2014

Question 37.
Define ecological pyramid? Describe the pyramid of energy flow with an example.
Answer:
Ecological Pyramid:
It is the graphic representation of trophic structure and function of a food chain. It is so-called due to its superficial resemblance to the Egyptian pyramid. The Pyramid concept in ecology was proposed by Charles Elton and hence they are also called Eltonian pyramids. It is usually broad below and tapering to on apex above. Producers occupy the broad base. while animals are consumers occupy successive steps tapering into the apex of the pyramid.
2nd PUC Biology Previous Year Question Paper June 2014 Q37
Pyramid of Energy: it is a graphic representation of available energy at successive trophic levels in a food chain. The energy level is expressed as kilocalorie. It is the typical upright pyramid. It illustrates decreasing energy level at successive trophic levels from producers at the base to the consumers. The energy level is maximum in producers, less in consumers. e.g.: Pond food chain.

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