2nd PUC Biology Previous Year Question Paper June 2017

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Karnataka 2nd PUC Biology Previous Year Question Paper June 2017

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 emasculation?
Answer:
Removal of anthers from the flower bud before the anther dehisces using a pair of sterile forceps is referred to as emasculation.

Question 2.
Define genetic drift.
Answer:
Random changes in the allelic frequencies of a population, occurring only by chance constitute genetic drift.

Question 3.
Name the causative agent of Elephantiasis.
Answer:
Wuchereria bancrofti and W. malayi.

Question 4.
From which plant Morphine is extracted?
Answer:
Poppy plant or Papaver somniferum.

Question 5.
Name a new breed of sheep developed in Punjab by cross-breeding technique.
Answer:
Hisardale by crossing Bikaneri ewes and Marino rams.

Question 6.
What are floes?
Answer:
Masses of bacteria associated with fungal filaments to form mesh-like structures.

Question 7.
What is downstream processing?
Answer:
Collection of product from the bioreactor, its purification and its processing to form a finished product is known as Downstream processing.

Question 8.
State Allen’s rule.
Answer:
Mammals from colder climates generally have shorter ears and limbs to minimise heat loss. This is called Aliens’ rule.

Question 9.
Mention a biodiversity hot-spot in India.
Answer:
Western Ghats/Himalayan region/Indo Burma region.

Question 10.
Name the chemical compound responsible for the depletion of the ozone layer.
Answer:
Chloro flouro carbons (CFCs) or Freons.

Part – B

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

Question 11.
What are vegetative propagates? Give two examples.
Answer:
Vegetative parts/fragments of the plant which can detach & under suitable conditions develop into an independent plant are known as vegetative propagates.
eg: Rhizome & tuber.

Question 12.
Differentiate between Geitonogamy and Xenogamy.
Answer:
Geitonogamy is the transfer of pollen grains from the anther of a flower to the stigma of another flower of the same plant. (Genetically it is similar to autogamy).
Xenogamy is the transfer of pollen grains from the anther of a flower on one plant to the stigma of a flower on a different plant. (Brings genetically different pollen grains to the stigma).

Question 13.
Mention the criteria? for a molecule that can act as genetic material.
Answer:
A molecule that can act as a genetic material must fulfil the following criteria.

• It should be able to generate its replica (replication)
• It should chemically and structurally be stable.
• It should provide the scope for slow changes (mutation) that are required for evolution.
• It should be able to express itself in the form of “Mendelian characters”. (any two points).

Question 14.
What are Mendelian disorders? Give two examples.
Answer:
The disorders which arise mainly by alteration or mutation in a single gene referred to as Mendelian Disorders, eg: Haemophilia, Cystic fibrosis.

Question 15.
List any two methods of HIV transmission.
Answer:

1. By use of multiple sexual partners.
2. During blood transfusion.

Question 16.
Mention the accessory ducts of the male reproductive system.
Answer:

1. Vasa efferentia
2. Vas deferens
3. Ejaculatory duct
4. Urethra.

Question 17.
Define Apiculture. List any two economic importance of it.
Answer:
Beekeeping or apiculture is the maintenance of hives of honeybees for the production of honey. Economic Importance.
(a) Honey is a food of high nutritive value and also finds use in the indigenous systems of medicine
(b) Bees was is used in the preparation of cosmetics and polishes of various kinds.

Question 18.
Define the polymerase chain reaction. Name the enzyme used in it.
Answer:
The reaction in which multiple copies of the gene (or DNA) of interest is synthesised in-vitro using two sets of primers & enzyme DNA polymerase is known as polymerase chain Reaction. The enzyme used is thermostable DNA polymerase isolated from the bacterium, Thermus aquaticus.

Part – C

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

Question 19.
(a) Differentiate between external fertilization and internal fertilization.
(b) Name the plant which produces flowers once in 12 years.
Answer:
(a)

External fertilisation	Internal fertilisation
When the fusion of gametes (syngamy) occurs outside the body of the organism, it is called external fertilisation.	When fusion of gametes (syngamy) occurs inside the body of the organism, it is called internal fertilisation.
A large number of gametes are released in the surrounding medium by such animals e.g., bony fishes, amphibians, etc.	The number of ova produced is less, but a large number of male gametes are formed, as many of them fail to reach the ova. e.g., birds, mañimais, earthworm, etc.

(b) Neelakuranji (Strobilanthus kunthiana)

Question 20.
Briefly explain sex-determination in human being.
Answer:
Human beings have XX_XY sex-determination system. In this, the males have XY sex chromosomes. They produce two types of gametes having X and Y chromosome. The females have an XX chromosome. They produce only one type of gamete having an X chromosome. The sex of a child is determined by the type of chromosome he/she receiving from the father. If sperm is having an X chromosome, then the child will be a girl. If sperm is having a Y chromosome, then the child will be a boy. The Y chromosome carries factors responsible for triggering male development. The X chromosome carries factors responsible for triggering female development.

Question 21.
Mention any three applications of DNA fingerprinting technique.
Answer:

1. It is a very useful identification tool in forensic applications.
2. It is the basis of paternity testing in case of disputes.
3. It helps to reunite the lost children to their respective parents.

Question 22.
(a) Differentiate between active immunity and passive immunity.
(b) Define allergy.
Answer:
(a) Active Immunity: Immunity obtained by production of antibodies upon exposure to living or dead microbes or other proteins.
Passive Immunity: Immunity obtained when ready-made antibodies are directly given to protect the body against foreign agents.

(b) The exaggerated response of the immune system to certain antigens present in the environment is called allergy.

Question 23.
Draw a neat labelled diagram of PBR – 322 plasmids.
Answer:

Question 24.
(a) Mention the functional components of the ecosystem.
(b) What is a trophic level?
Answer:
(a) (i) Productivity
(ii) Decomposition
(iii) Energy flow
(iv) Nutrient cycling.

(b) Based on the source of their nutrition or food, organisms occupy a specific place in the food chain that is known as their trophic level.

Question 25.
(a) Mention any two causes of biodiversity losses
(b) What are endemic species?
Answer:
(a) (i) Habitat destruction and fragmentation
(ii) Co-extinctions

(b) The species which are confined toa specific geographical region and not found elsewhere.

Question 26.
Write the schematic representation of the phosphorus cycle.
Answer:

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.
(a) Explain any four outbreeding devices that prevent autogamy.
(b) What is apomixis?
Answer:
(a) Contrivances for cross-pollination: [Out breeding devices]
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 (didhnous). 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: (Herkos = 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. Dlchogamy: 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: Carpets 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.

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

(b) Apomixis is the phenomenon of production of seeds without fertilization.

Question 28.
Explain the inheritance of ‘two genes’ in pea plants with two suitable examples.
Answer:
It is a cross between two plants differing in 2 pairs of contrasting characters. Mendel selected shape of the seed and colour of the cotyledons as the two pairs of contrasting characters. The round seed coat is dominant over wrinkled seed coat and yellow colour of cotyledon is dominant over the green colour.

Mendel crossed a pure breeding garden pea plant producing round and yellow seeds with a pure breeding garden pea plant producing wrinkled and green seeds.

Round seed coat and yellow cotyledons = 9
Wrinkled seed coat and yellow cotyledons = 3
Round seed coat and green cotyledons = 3
Wrinkled seed coat and green cotyledons = 1
Dihybrid phenotypic ratio = 9 : 3 : 3 : 1.
Genotypicratio = 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1.
Thus in the F₂ generation, he got Yellow Round, Yellow wrinkled, Green round and Green wrinkled in the ration of 9 : 3 : 3 : 1 respectively.

Question 29.
Draw a heat labelled diagram of human sperm.
Answer:

Question 30.
What is infertility? Explain IVF-ET and ZIFT methods to assist the infertile couple to have children.
Answer:
Infertility: An woman who is unable to conceive is called infertile. This infertility is caused due to defects in men or women or both.

IVF and ET:
These can be adopted when the ovaries are defective or fallopian tube is defective. This technique involves the collection of sperms. In the former case, the ovary is stimulated to produce oocytes by the administration of proper doses of FSH and LH, then many oocytes must be collected by laparoscope and they are graded, selected and preserved in an incubator. Collection of many oocytes from the ovary by administrating artificial FSH and LH is known as superovulation.

In the latter case, sperms are also collected from a donor or from a sperm bank. These sperms are centrifuged
and then incubated. The fast-moving sperms are collected in a pipette. The sperms and ova are relèased into the Petri dish or test tube containing a nutrient medium. Then both male and female gametes are fused in an artificial medium. Such a fusion process is called in-vitro fertilization. This results in the many zygotes in invitro medium undergo cleavage resulting ubblastocysts. The embryos can also be cultured in an artificial medium.

Finally, the embryos are transferred directly into the uterus by using an embryo transfer catheter. Normally more than one selected embryos are transferred so as to increase the chances of implantation. This phenomenon is
known as embryo transfer or tubal embryo transfer. This leads to the pregnancy and finally, delivery is occurred by natural or by artificial method i.e, Caesarian. The baby born out of NF and ET is known as a test-tube baby. ft was first carried out by doctors Steptoe and Edward. The baby, Louis Joy Brown was born in 1978.

ZIFF:
Zygote intrafallopian transfer: In this case, after in-vitro fertilization, the zygote is cultured and transferred into the uterus. This leads to pregnancy. Afterwards, baby can be delivered either by Caesarian or by natural method.

Question 31.
Enumerate any five salient features of genetic code.
Answer:
The properties of genetic code are:

• Triplet code: The genetic code is a triplet code. It means that three bases of DNA code for one amino acid.
• Genetic code is universal: It means that a particular mRNA codon, codes for the same amino acid in all living organisms. For e.g.: AUG codes for amino acid methionine in all organisms.
• Genetic code is degenerate of redundancy: Some of the amino acids are coded by two or more codons. These redundant codons for the same amino acids are called degenerate codons and this property is called degeneracy.
• Genetic code is non-overlapping: In this property, the base of one codon is not shared by the neighbouring codons.
• Genetic code is comma less: Genetic code has no punctuation mark inside the message.
  Once the reading of message begins with the initiator codon AUG and it is continued till the terminator codon is reached.

Question 32.
What is tissue culture? Mention any four application.
Answer:
Separation of cells or tissue from plants and growing them on a synthetic nutrient medium under aseptic conditions is called tissue culture.

It involves the following steps:
1. Selection of plant material: The part of the plant used for culture is known as explant. It is isolated from leaves, seeds, roots, buds, nodes etc.

2. Preparation of the culture medium: A suitable culture medium ¡s essential for the growth of explants. A medium is prepared using several macros and micronutrients like sucrose, vitamins, minerals etc. Agar-agar is added as a solidifying agent. Antibiotics are added to prevent the growth of microbes. Hormones like Auxins and cytokinins are added to promote growth. This medium is sterilised ¡n culture tubes.
There are various nutrient media viz,
(a) M.S. medium (b) Nitsch’s medium etc

3. Inoculation: It is the process of transfer of explants onto the sterilised culture medium. It is done in a special chamber called Lamina airflow table.

4. Callus formation: It is an undifferentiated mass of cells produced by the explant. It is formed within a short period after inoculation.

5. Organogenesis: It is the process of differentiation of cells in the callus into different tissues and organs. Root initiation occurs auxin to cytokinin ratio is high and shoot can be initiated by increasing kinetin to auxin ratio. Embryoids thus obtained are called plantlets.

6. Hardening: Immediate exposure of plantlets obtained from tissue culture to direct sunlight, wind etc may kill them. Therefore they are grown in greenhouses for a brief period and then gradually transferred to their natural habitats. This is called hardening.

Advantages:
1. Micropropagation: The method of rapid vegetative propagation of desired plants is called micropropagation. A large number of medicinal, ornamental and forest plants can be produced through this technique.

2. Production of virus-free plants: Virus-free or disease-free plants can be produced through meristem culture, as the apical meristems are devoid of viruses.

3. Androgenic haploid plants: Haploid plants can be produced through anther or pollen culture. Diploid homozygous plants can be produced from the baploid plants by doubling the chromosome number.

4. Induction and selection of mutants: Mutations are induced in the cell cultures, and the mutants are then subjected to herbicides, toxins etc. The mutant cells that show resistance are selected and grown by tissue culture to raise resistant varieties.

Section – II

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

Question 33.
With the help of neat labelled diagram explain Stanley Miller’s experiment in support of chemical evolution.
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. 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.

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 compounded can 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.
Describe the role of microbes as biofertilizers.
Answer:
Biofertilizers are the organisms that enrich the nutrient quality of the soil. The main source of biofertilizers are bacteria, fungi and cyanobacteria. Bacteria and cyanobacteria have the property of nitrogen fixation, while mycorrhizal fungi help in the mineral uptake by the plant. Rhizobium forms a symbiotic association with the roots of leguminous plants. They develop the ability to fix nitrogen when they are present inside that root nodules.

Note: Some non-legume plants also show symbiotic association with some other nitrogen-fixing bacteria.
Example: Frankia is associated symbiotically with the root nodules of several non-legume plants like Casuarina, Alnus.

Symbiotic nitrogen-fixing cyanobacteria: A number of cyanobacteria (blue-green algae) form a symbiotic association with several plants. e.g., Lichens, hornworts, Azolla (fern). Cycas roots. Azolla is a small fast-growing fern that occurs floating on water. Anabaena azollae. (a cyanobacterium) lives in the cavities of Azolla leaves. Azolla -Anabaena symbiotic association is of great importance to agriculture. The fern can grow ¡n rice fields because it does not interfere with the growth of crop plants.

Mycorrhiza: Mycorrhiza (pl. mycorrhizae) is a symbiotic association of certain fungi with the roots of certain seed-bearing plants. Many members of the genus Glomus form mycorrhiza. The fungal symbiont in these associations absorbs phosphorus from soil and passes it to the plant. Plants having such associations show other benefits also, like

• resistance to root borne pathogens.
• tolerance to salinity and drought, and
• an overall increase in plant growth and development.

Mycorrhizae can be broadly classified into two types: ectomycorrhizae and endomycorrhizal.

• Ectomycorrhiza: The mycelium of the fungus forms a mantle on the surface of the root. Ectomycorrhizae commonly occurs on the roots of trees such as pine, oak, peach and eucalyptus.
• Endomycorrhiza: The fungus lives between and within the cells of the cortex of the roots.

Question 35.
What is RNA interference? How does RNA interference help to develop resistance in tobacco plaint against Nematoda infection?
Answer:
Protection Against Nematodes:

• A nematode called Meloidogyne incognita infects plants and reduces their yield.
• The specific genes (in the form of cDNA) from the parasite are introduced into the plant using Agrobacterium as the vector.
• The genes are introduced in such a way that both sense/coding RNA and antisense RNA (complementary to the sense/coding RNA) are produced.
• Since these two RNAs are complementary, they form a double-stranded RNA (ds RNA).
• This silences the specific RNA of the nematode, by a process called RNA-interference. It prevents the translation of a specific mRNA (silencing).
• As a result, the parasite cannot live in the transgenic host and the transgenic plant is protected from the pest.

Question 36.
(a) List the responses of living organisms to abiotic factors
(b) What is commensalism? Give an example.
(c) Name the type of interaction between algae and fungi in lichens.
Answer:
(a) (i) Regulation, (ii) Conformation, (iii) Migration, (iv) Suspension.

(b) Commensalism: This is the interaction in which one species benefits and the other is neither harmed nor benefited.
Example: An orchid growing as an epiphyte on a mango branch, and barnacles growing on the back of a whale benefit, while neither the mango tree nor the whale derives any apparent benefit.

(c) Mutualism.

Question 37.
What is global warming? Mention the causes and effects of global warming.
Answer:
Increase in the level of greenhouse gases has led to considerable heating of Earth known as global warming.
Causes of global warming are increasing concentration of greenhouse gases (le., CO₂, CH₄, CFCs and N₂O) in the atmosphere.

Effect of global warming: Global warming, a consequence of the higher concentration of greenhouse gases, has the potential to affect weather and climate, stratosphere and thermosphere, rise in sea level, species distribution and food production.

1. Weather and climate: Global mean temperature rose by 0 – 6o°C during the 20th century. It is going to increase further by 1-4° – 5-8°C between 1990 – 2100 AD.

2. Stratosphere and thermosphere: Warming of troposphere will cause cooling of stratosphere and thermosphere. The whole atmosphere will shrink. Cooling of stratosphere will increase the size of ozone holes and thinning of ozone shield at other places. Cooling of thermosphere will disrupt communications and the shielding effect of the ionosphere.

3. Sea level change: It is believed that sea level has risen by 15 cm during the 20th century, a rise of 1-2 mm every year. By the year 2100 AD, the global mean sea level is going to rise by 0-88m over that of 1990 sea level. The rise in sea level is due to thermal expansion of oceans as the temperature rises, melting of glaciers and Greenland ice
sheets.

4. Range of species distribution: Each species has a particular range of temperature. Global warming will push tropics into temperate areas and temperate areas towards poles and higher altitudes in mountains A rise of 2° – 5° C can cause pushing of temperate range vegetation some 250 – 600 km towards the pole.

5. Food production: Rise in global temperature is going to have a negative impact on food production despite the beneficial effect of CO2 fertilization. The various reasons are an increase in basal rate of respiration by plants

• Lesser storage of food and hence less productivity
• The explosive growth of weeds
• Higher incidence of pathogens and pests
• Increased evaporation of soil water and higher transpiration from plants resulting in water deficiency.