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Karnataka 2nd PUC Biology Previous Year Question Paper March 2019

Time: 3.15 Hours
Max Marks: 70

General Instructions:

  1. This question paper consists of four parts A, B, C and D. Part D consists of two parts, Section – I and Section – II.
  2. All the parts are Compulsory.
  3. 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 x 1 = 10 )

Question 1.
Name the reproductive cycle that occurs in non-primates.
Answer:
Oestrous cycle.

Question 2.
Name the layer of the uterus that exhibits strong contractions during parturition.
Answer:
Myometrium

Question 3.
Which chromosome of man has the least number of genes?
Answer:
Y chromosome.

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Question 4.
Write the infectious forms of Plasmodium which enter human body through mosquito bite.
Answer:
Sporozoites.

Question 5.
Give an example for an inter-specific hybrid animal.
Answer:
Mule.

Question 6.
Write the scientific name of the source organism for citric acid.
Answer:
Aspergillus niger.

Question 7.
Mention a gene that codes for insecticidal protein in Bt cotton.
Answer:
cryIAc/cryIIAb Crygene

Question 8.
Name the plasmid present in Agrobacterium tumefaciens.
Answer:
TI plasmid / Tumor inducing plasmid.

Question 9.
What is the main aim of Montreal Protocol?
Answer:
To reduce the depletion or thinning of ozone layer.

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Question 10.
Name the type of food chain that is the major conduit for energy flow in an aquatic ecosystem.
Answer:
Grazing food chain / GFC.

Part – B

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

Question 11.
What is embryogenesis? Mention two important events that occur during embryogenesis.
Answer:
“ The process of the development of embryo from the zygote”.
The two important events in embryogenesis are cell division and cell differentiation.

Question 12.
Define point mutation. Give an example for point mutation.
Answer:
When mutation arises due to change in a single pair of DNA it is termed as point mutation Ex;- Sickle cell anaemia, Genetic disorders.

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Question 13.
What are the conclusions drawn by T.H. Morgan from the crossing experiments in Drosophila with respect to linkage.
Answer:
Physical association of genes on a chromosome is termed as linkage.
The more closely the genes are found on chromosomes greater is the linkage less closely the genes are found on chromosome lesser is the linkage.
Original needed (O.S of Human Health and dieases strategies in improvement of crop).

Question 14.
What is an allergy? Name the two chemicals released by mast cells in the body during allergy.
Answer:
The exaggerated response of the immune system to. certain antigen (foreign substance) percent in the environment.
Drugs: Antihistamine, Adrenalin and Steroids.

Question 15.
Mention the four traits for which plant breeding is done.
Answer:

  1. Increased yield.
  2. Improved quality.
  3. Tolerance to environmental stress
  4. Resistance to pathogen/diseases

Question 16.
With an example, explain the convention for naming restriction endonucleases scientifically.
Answer:
Naming REN, Principles:
Three letter code is given for each REN. First letter represents the genus name of Bacteria with capital letter and subsequent two letters with small letter will be the species name. Ex: Eco, Hin etc., Next letter represents the strain, and roman number indicates the order in which the enzymes were isolated from the same organism. Ex: Hind III Eco RI Bam HI.
Each REN cut DNA at specific 4 to 6 base pairs sequence known as ‘Palindrome Suquences’.

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Question 17.
A prey develops defense against a predator. Justify the statement with two examples in animals. Prey species have evolved to avoid Predators. Ex: Camouflage
Answer:
1. Some frogs and insects are cryptically coloured to avoid being detected easily by Predators.
2. Some species are poisonous to avoid Predators. Ex: Monarch butterfly is highly distasteful to predator bird. This distaste is due to chemical present in the body which is acquired during their caterpillar stage by feeding on a poisonous weed.

Question 18.
Mention any two mechanisms how human body compensates low oxygen availability at higher altitude.
Answer:
The human body compensate low oxygen availability at higher altitude by increasing red blood cell production, decreasing the binding affinity of hemoglobin and by increasing breathing rate.

Part – C

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

Question 19.
Write three advantages offered by the seeds to angiosperms.
Answer:

  1. Seed formation is a more dependable phenomenon since pollination and fertilization are independent of water.
  2. Reserve food in seeds nourishes the young seedlings until they are capable of photosynthesis on their own.
  3. Seeds have better adaptive strategies for dispersal to new habitats.
  4. The hard seed coats protect the young embryo till it develops completely.
  5. Seeds produce new genetic combinations leading to variations as they are products of sexual reproduction.

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Question 20.
Draw a neat labelled diagram of the sectional view of human mammary eland.
Answer:

2nd PUC Biology Previous Year Question Paper March 2019 1

Question 21.
DNA is the better genetic material than RNA. Justify the statement with three comparative reasons.
Answer:

  • DNA is more stable because even if the two complementary strands are separated by factors like heat, they can come together again. But RNA is more labile and easily degradable because of an additional – OH group in the 2 position of ribose in every nucleotide.
  • DNA does not acy as enzyme. But some RNA molecules act as enzymes and therefore, are more reactive.
  • DNA has Thymine which gives additional stability. RNA molecules do nit have thymine.
  • Both RNA and DNA can mutate. But RNA mutates at a faster rate as it is less stable.

Question 22.
Enumerate on convergent and. divergent evolution with suitable examples.
Answer:
Convergent Evolution: It creates analogous structures that evolve for same function and hence having similarities.
Eg: Sweet potato (root, modification) and Potato (Stem modification)

Divergent Evolution: The same structure developed along different direction due to adaptations to different need. This is divergent evolution and these structures are homologous.
Eg: Thorns and tendrils of Bougainvillea and Cucurbita.

Question 23.
What are carcinogens? Mention any two groups of carcinogens with one example for each.
Answer:
The agents which causes cancer are called carcinogens.

  1. Physical carcinogens- Ionising radiations, X-rays
  2. Chemical carcinogens- Tobacco Smoke
  3. Biological Carcinogens- Oncogenic viruses

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Question 24.
a) Mention four tools required for recombinant DNA technology.
Answer:
Tools required for recombinant DNA technology.
Restriction enzymes, cloning vector, competent host, polymerase enzymes, ligases.

b) With reference to gel electrophoresis, what is elution?
Answer:
The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece is called elution.

Question 25.
Ecological pyramids have limitations. Justify the statement with three reasons.
Answer:

  • It does not take into account the same species belonging to two or more trophic levels.
  • It assumes a simple food chain which is rare in nature.
  • It does not accommodate a food web.
  • Saprophytes (decomposers) are not given any place in ecological pyramids.

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Question 26.
Describe three factors which affect on decomposition. Factors which affect decomposition:
Answer:

  1. In a particular climatic condition, decomposition rate is slower if detritus is rich in lignin and chitin and quicker, if detritus is rich in nitrogen and water soluble substances like sugars.
  2. Temperature and soil moisture are the most important climatic factors that regulate decomposition through their effects on the. activities of soil microbes.
  3. Warm and moist environment favour decomposition whereas low temperature and anaerobiosis inhibit
    decomposition resulting in build up of organic materials

Part – D

Section – I

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

Question 27.
What is megasporogenesis? , Explain the development of eight nucleate embryosac in flowering plants.
Answer:
The process of formation of megaspore from Megaspore Mother Cell, (MMC), through meiosis is known as megasporogenesis. MMC is large nucleated with dense cytoplasm. It undergoes meiosis, which results in the production of 4 megaspores. Out of 4, 3 degenerates and one megaspore is functional which develops into female gametophyte. “Method of embryo sac formation from a single megaspore is known as Monosporic type of development”.

Developments of Female Gametophyte
Megaspore mother cell undergoes 3 successive mitotic division to form 8 nucleated embryo sac.

Development of Embryosac
First the functional megaspore nucleus divides mitotically to form 2 nuclei which move towards opppsite

2nd PUC Biology Previous Year Question Paper March 2019 2

pole forming 2- Nucleate embryo sac. The developing embryo sac enlarges in size due to formation of vacuole in the centre.

Each 2 nulei, divides and redivides, forming 4 neulei at micropylar end and 4 nuclei at chalazal end, here they undergo free nuclear division. At 8 celled stage cell wall formation.

Out of 4 nuclei at micropylar end, 3 differentiate to produce Egg apparatus = 1 Egg + 2 Synergid cells. 4 nuclei at chalazal, 3 remains at, polar region and form Antipodals.

One nucleus from micropylar end, one from chalazal end migrate towards the centre and fuse to form (2n) diploid secondary nucleus.
Thus a typical angiosperm embryosac at maturity though 8 nucleate, it is 7 celled.

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Question 28.
What are contraceptives? Explain four different non- surgical contraceptive methods.
Answer:
1. Contraceptive are methods or devices which prevent conception or pregnancy.
Non – surgical contraceptive methods:

2. Natural method : It works on the principle of avoiding the chances of sperms and ovum meeting together. There are two types – periodic abstinence (physiological or rhythm method),Abstinence from intercourse during this ‘fertile period’ (danger period), i.e., between 10th and 17th day of menstrual cycle) and withdrawal method (coitus interruptus – Male partner withdrawing his penis from vagina just before ejaculation).

3. Barrier Method : Barriers are contraceptives which act as mechanical obstructers and prevent the entry of semen into the fallopian tube.
Ex: Condom, diaphragm, cervical caps, vaults, etc.

4. Intra – uterine devices (IUDs) :
Intra uterine device is a small flexible plastic or rubber or copper object which is inserted into the uterus to prevent conception. They may be non-medicated IUDs (e.g., Lippes loop), copper releasing IUDs (CuT, CU7, Multiload 375) or hormone releasing IUDs (progestasert, LNG – 20). They bring about contraception by different ways.

5. oral Contraception (Hormonal method) : This is contraceptive method in which ‘pills’ are taken orally by females. These pills contain progestogens or progestogen – estrogen combinations that inhibit the secretion of FSH and LH and thus prevent the maturation of graafian follicle and ovulation.

6. Implants and inheritable : These are progestrogens or progestogen – estrogen combinations that are uses under the skin by females,. Their function similar to that of oral’ contraceptives (prevent maturation of graafian follicle and ovulation)

Question 29.
What is incomplete dominance? Explain it with reference to flower colour in snapdragon.
Answer:
The expression of intermediate trait in the hybrid obtained by crossing two pure breeding varieties with contrasting characters is called in complete dominance.
Eg: Parent – Red flower x White flower
Parent – Red flower x White flower
2nd PUC Biology Previous Year Question Paper March 2019 3
Phenoty pic ratio -1 : 2 : 1
Genoty pic ratio – 1 RR : 2 Rr : 1 rr

Question 30.
(a) Explain the role of three , organisms as biocontrol agents.
(b) What is the significance of BOD?
(c) Given an example for fungus found in mycorrhia.
Answer:
(a) The ladybird beetle is used to control aphids. Dragonflies are used to control mosquitoes.
Dried spores of these to bacteria Bacillus thuringiensis are mixed with water and sprayed onto vulnerable plants such as brassicas and fruit trees. Their toxin is released in the gut of insect (butterfly) larvae which feed on the spores. This kills the caterpillars.

Trichoderma species, a free-living fungus in the roots of many plants is used to control several plant pathogens.
Baculoviruses like nucleo- polyhedrovirus are species specific and are used as biocontrol agents against insects and other arthropods.
(b) It is a measure of organic matter present in water / sewage Or it indicates whether the water sample is polluted or not
(c) Glomus.

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Question 31.
Explain five benefits of creating transgenic animals.
Answer:
Benefits of creating transgenic animals
1. Study of normal physiology and development:

  • Useful to study gene regulation, their effect on the normal functions of the body and its development
  • For example, study of complex growth factors like insulin-like growth factors

2. Study of disease:

  • Study of genes which are responsible for diseases in human and their treatment.
  • Transgenic models have been developed for many human diseases like cancer, cystic fibrosis, rheumatoid arthritis and Alzheimer’s disease.

3. Biological products:

  • Useful biological products can be produced by introducing into transgenic animals the portion of DNA (or genes) which codes for a particular product.
  • For example, human protein(a- 1-antitrypsin) is used to treat emphysema.
  • In 1997, the first transgenic cow , Rosie produced human protein- enriched milk (2.4g/L).
  • The milk contained the human alpha-lactalbumin and was more nutritionally balanced for human babies than natural cow milk.

4. Vaccine safety:

  • Transgenic mice are developed to test safety of vaccines before being used on humans. Eg: polio vaccine

5. Chemical safety testing:

  • Transgenic animals are made to carry genes which make them more sensitive to the toxic substances than non-transgenic animals.

Question 32.
(a) Tropical region has greater biodiversity than temperate region. Justify the statement with three reasons.
(b) The use of CNG is better than petrol or diesel. Give four reasons.
Answer:
(a) Temperate regions have been subjected to frequent glaciations in the past whereas, tropical latitudes have remained relatively undisturbed for millions of years. Therefore, tropical regions had a long evolutionary time for species diversification.

Tropical environments are less seasonal, relatively more constant and predictable.
Such constant environments promote niche specialisation and lead to a greater species diversity.
There is more solar energy available in the tropics, which contributes to higher productivity.
This in turn might contribute indirectly to greater diversity.

(b) CNG bums more efficiently than petrol or diesel.
Very little of it is left un burnt.
CNG is cheaper than petrol or diesel.
CNG cannot be siphoned off and adulterated like petrol.

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Section – II

Answer any three of the following questions in about 200 to 250 words each wherever applicable: ( 3 x 5 = 15 )
Question 33.
(a)Write the schematic representation of spermatogenesis. ( 3 )
(b) Write the two events that occur in the ovary and uterus during the follicular phase of menstrual cycle.  ( 2 )
Answer:

(a)
2nd PUC Biology Previous Year Question Paper March 2019 4
(b) Secretion of FSH and LH increases that induces the development of primary follicle into mature graafian follicle The endometrium of uterus regenerates by proliferation.The graafian follicle starts producing estrogens.

Question 34.
‘DNA replication is said to be semiconservative’. Why? Describe the experimental proof of Messelson and Stahl to show DNA replication is semiconservative.
Answer:
2nd PUC Biology Previous Year Question Paper March 2019 5

Meselson and Stahl conducted experiments on E.coli to prove that DNA replication is semi conservative.

  • They grew coli cells in a culture medium containing radioactive isotopes N15 (NH4cl). The 15N was incorporated into the newly synthesized DNA and other nitrogen containing compounds.
  • Bacteria labeled with 15N are transferred into a medium containing normal 14NH4
  • The analysis was done after every generation at regular intervals.
  • The DNA samples were separated by centrifugation on CSCl2
  • The hybrid DNA separated from fresh generations possesses one heavier strand contributed by parental with 15N isotope which is radioactive and lighter strand is newly synthesized with 14N isotope, is non-radioactive.
  • The DNA of the second generation showed that in one DNA molecule one strand was radioactive (1SN) and other was non radioactive (14N). Whereas both the strands of other DNA molecule were non­ radioactive (14N).
  • It is clear that the DNA of the first generation is intermediate it contains both 15N and 14N
  • In the second generation there were two types of DNA one heavier DNA with 15N and 14N isotopes while the lighter DNA with 14N and 14N, the ratios of two types of DNA was 50 : 50.
  • In the third generation the ratio of intermediate (15N and 14N) was reduced from 50% to 25% while the lighter fraction (14N and 14N) increased from (50% – 75%).
  • In the 4th generation the percentage of intermediate still reduced to 12.5% while that of lighter fraction increased from 75% to 87.5%.
  • These experiments prove that DNA shows semi conservative replication.

Question 35.
(a) Draw a neat labelled diagram of Miller’s experiment. ( 3 )
Answer:
2nd PUC Biology Previous Year Question Paper March 2019 6
(b) Mention two assumptions of Oparin and Haldane with reference to ‘Origin of Life’. 2
Answer:
The first form of life could have come from pre-existing non-living organic molecules.
Formation of life was preceded by chemical evolution (formation of diverse organic molecules from inorganic constituents).

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Question 36.
Name the technology that can successfully increase the herd size of cattle in a short time and explain the steps involved in this technology.
Answer:
MOET / Multiple ovulation Embryo Transfer technology.
Steps involved in the technique :

  • Cow is injected with hormones having FSH – like activity to induce follicular development and production of 6-8 eggs per cycle (Super ovulation).
  • The cow is artificially inseminated or mated with a pedigree bull.
  • Fertilised eggs are non-surgically or mated with a pedigree bull.
  • These embryos are transferred to surrogate cows.

Question 37.
(a) Mention four ‘Evil Quartet’ which cause depletion of biodiversity. 2
Answer:
Four ‘Evil Quartet’ which causes depletion of biodiversity are

  1. Habitat loss and fragmentation
  2. Over Exploitation
  3. Alien species invasion
  4. Co-extinction

(b) Write two suspended activities in animals against abiotic stresses with suitable examples. 2
Answer:
1. Hibernation (winter sleep):
It is a supressed metabolic state of an organism to avoid winter season
Eg: Bears going into hibernation in winter.

2. Aestivation (summer sleep):
Hibernation during the summer season is called aestivation
Eg: Fishes avoid summer related problems like heat and dessication.

3. Diapause:
It is the delay in the delopment during adverse enviornmental condition.
Eg: Many zooplanktons undergoes a stage of suspended development in unfavourable conditions

(c) Among vertebrates, which group of animals has the highest number in global biodiversity. 1
Answer:
Fish.