2nd PUC Biology Previous Year Question Paper June 2016

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Karnataka 2nd PUC Biology Previous Year Question Paper June 2016

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.
Mention the function of Sertoli cells of testes.
Answer:
Sertoli cells provide nutrition to the germ cells during spermatogenesis.

Question 2.
What are plasmids?
Answer:
Plasmids are extrachromosomal circular double-stranded DNA capable of self-replication, and self-regulation.

Question 3.
Name the hormone secreted by Corpus Luteum.
Answer:
Progesterone.

Question 4.
Which bioactive molecule is used as an immunosuppressive agent?
Answer:
Cyclosporin A.

Question 5.
What are interferons?
Answer:
Interferons are the antiviral proteins/drugs secreted by virally infected cells. They prevent multiplication of the virus.

Question 6.
Name one microbe used in the production of single-cell protein.
Answer:
Spirulina.

Question 7.
Name the theory which attempts to explain ‘origin of the universe’.
Answer:
Big Bang theory.

Question 8.
Define Biochemical Oxygen Demand (BOD).
Answer:
It is the amount of oxygen required for oxidising all organic matter present in one litre of water by the aerobic microbes.

Question 9.
Name the diagnostic test for AIDS.
Answer:
ELISA.

Question 10.
State Gause’s ‘Competitive Exclusion Principle’.
Answer:
Gause’s Competitive Exclusion Principle states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior will be eliminated eventually.

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 two examples.
Answer:
The animals which possess both male and female reproductive organs-in them are called as hermaphrodites.
Eg. Tapeworm, earthworm.

Question 12.
Write four symptoms of Down’s Syndrome.
Answer:

1. Short saturated with a small round head.
2. Furrowed tongue and partially open mouth.
3. Palm is broad with characteristic palms crease.
4. Physical psychomotor and mental development is retarded.

Question 13.
What is Adaptive Radiation? Give an example.
Answer:
Evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats) is called adaptive radiation.eg. Darwin’s finches.

Question 14.
Draw a labelled diagram of Biogas plant.
Answer:

Question 15.
‘Alien species invasion leads to the extinction of indigenous species’ – Justify this statement by considering two animals as examples.
Answer:

1. Introduction of Nile Perch into Lake Victoria in East Africa led eventually to the extinction of more than 200 species of orchid fishes in the lake.
2. Introduction of African catfish Claris gariepinus for aquaculture purposes is posing a threat to the indigenous catfishes in our rivers.

Question 16.
What is Ex-Situ conservation? Mention two examples.
Answer:
Protecting threatened animals and plants out of their natural habitat and placing them in special settings where they can be protected and given special care is called ex-situ conservation.
e.g.: zoological parks and botanical gardens.

Question 17.
What is cross-breeding? Name a new breed developed by crossing Bikaneri ewes and Marino rams.
Answer:
It is a method of outbreeding in which superior males of one breed are mated with 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. Hisardale is a new breed of sheep developed by crossing bikaneri ewes and merino rams.

Question 18.
Mention two techniques to make Host competent in rDNA technology.
Answer:

• Microinjection
• Biolistics or gene gun.

Part – C

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

Question 19.
Write the Homozygous Genotype of ‘A’, ‘B’ and ‘O’ blood groups.
Answer:

Blood Group	Homozygous Genotype
A	IAIA
B	IBIB
O	i i

Question 20.
Explain the structure of matured pollen grain.
Answer:

1. When the pollen grain is matured it contains two cells, the vegetative cell and generative cell.
2. The vegetative cell is bigger, has abundant food reserve and a large irregularly shaped nucleus.
3. The generative cell is small and floats in the cytoplasm of the vegetative cell. It is spindle-shaped with dense cytoplasm and a nucleus.

Question 21.
Name the vegetative propagules of the following plants.

1. Agave
2. Ginger
3. Bryophyllum

Answer:

1. Bulbil
2. Rhizome
3. leaf buds

Question 22.
Differentiate between homologous and analogous organs by giving plant example.
Answer:
Homologous organs are those whose structure and origin are the same but functionally they differ. They exhibit divergent evolution.
eg. In plants the thorns of bougainvillaea and tendrils of Cucurbita exhibit homology. Analogous organs are those whose structure and origin are not similar but functionally, they are same. They exhibit convergent evolution.

Question 23.
Draw a neat labelled diagram of simply stirred tank bioreactor.
Answer:

Question 24.
What are biogeochemical cycles? Mention two types by giving an example for each.
Answer:
Cyclic movements of essential chemical elements between the environment and living organisms, establishing relationship among living organisms, soil and chemical elements are called biogeochemical cycles.

The two types of biogeochemical cycles are:

• Gaseous cycle eg. carbon cycle
• Sedimentary cycle eg. Phosphorus cycle.

Question 25.
Name the pathogen, vector and symptom of malaria.
Answer:
Plamodium vivax, P. malariae and P. falciparum.

• Vector: Female anopheles mosquito.
• Symptoms: Recurrent high fever and chills for 3 to 4 days.

Question 26.
Write the diagrammatic representation of the pyramid of energy.
Answer:
Diagrammatic representation of the pyramid of energy as shown in the figure.

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.
Describe the structure of anatropous ovule with a labelled diagram.
Answer:

Question 28.
What is contraception? Briefly explain two barrier methods and two surgical methods, to prevent conception.
Answer:
Contraception: Prevention of conception is called contraception. There are three major categories of contraception. They are natural, mechanical and chemical methods.

Mechanical contraception or Barrier methods.
Prevention of conception by using mechanical devices is called mechanical contraception. Mechanical methods of conception include the use of condom, diaphragm and intrauterine devices (IUD).

(a) Condom (Sheath): This is used by men. A condom is a nonporous, elastic, tight-fitting sheath of rubber which is unrolled over the penis before intercourse. The ejected semen is trapped in the condom and thus the entry of sperm into the vagina is prevented.

(b) Diaphragm (cap): This is used by women: Diaphragm is a flexible dome-shaped device made up of thin rubber. It is inserted into the.vagina and positioned over the cervix before intercourse. It prevents the entry of sperms into the uterus.
Cervical caps and vaults are similar to the diaphragm in function but are structurally modified. They are known as female condoms.

(c) Intrauterine device (IUD): Intrauterine devices are used by women. The device is in the form of a small object made up of plastic, copper or stainless steel. It is in the form of a loop, coil or T. The IUD is inserted to the uterus with the help of an insertion tube by an experienced gynaecologist. The IUD is provided with strings which are useful to check and confirm its position. They are also useful in removing the IUD. An IUD could remain in the uterus for up to three years. It should be replaced after 3 years. It prevents implantation. It is a reliable method of birth control. Loop and copper T are very widely used by women in our country.

Sterilization: Surgical sterilization is the mõst effective and reliable method of birth control. Both men and women can undergo sterilization. Surgical sterilization in men is called vasectomy and in women tubectomy and laparoscopic sterilization. Sterilization normally does not affect sexual performance.

Question 29.
Explain incomplete dominance with reference to flower colour in Snapdragon.
Answer:
It is a process where the dominant gene is incompletely dominant over the recessive gene and produces a phenotype which is intermediate to the parental type.
In Mirabilis or 4° clock plant, incomplete dominance has been reported with respect to the colour of the flower. In these plants, some produce red flowers and some others produce white flowers. When a plant producing red flowers is crossed with a plant producing white flowers, in the F₁ generation all plants produced pink flowers. It is a heterozygous condition and the pink character is an intermediate or a blend between red and white This colour is produced because the red is being only partially dominant over white.
P₁ Red flowers × White flowers
F₁ Pink flower
When F₁ pink flower plants are self crossed to raise F₂ progeny 3 types of plants are produced
Le, 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.

Phenotypic and Genotypic ratio 1 : 2 : 1
Thus, in this case, parental characters are expressed in homozygous conditions and intermediate character in heterozygous condition.

Question 30.
Explain the working of Lac-operon with diagrammatic representation.
Answer:
RgtiIation 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 – ga1actosde 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 also 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.

Question 31.
List out the salient features of the double-helix model of DNA.
Answer:
Watson and Crick in 1953, proposed a model to explain the structure of DNA called the double helix model.
According to this model,
(a) DNA is composed of two polynucleotide chains coiled spirally around a central axis to form a double helix.

(b) The two polynucleotide chains of a double helix arc opposite in direction. One chain runs in 5¹ → 3¹ direction and the other is 3¹ → 5¹ direction.

(c) Hydrogen bonds are formed between nitrogenous bases. The base-pairing in DNA consists of one Purine and one pyrimidine. Further Adenine (A) always pairs with Thymine (T) and guanine (G) always pairs with cytosine. This base pairing is said to be complementary base pairing. Two hydrogen bonds are formed between A and T and between G and C three hydrogen bonds are formed.

(d) The total amount of A is equal to that ofT and the total amount of G is equal to that of C. in DNA molecules. This relationship between base pairs is known as Chargoffs’s rule. It is represented as A = T, C = G.

(e) The backbone of the DNA helix is made up of sugar and phosphate.

(f) One double-helical turn is ‘S’ shaped and contains one major groove and a minor groove. One complete turn of the helix is 34A° in length and it consists of 10 base pairs. Thus the distance between two adjacent bases is 3.4A°. The constant distance between the two strands is 20 A°.

Question 32.
Draw a neat labelled diagram of a sectional view of the human female reproductive system.
Answer:

Section – II

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

Question 33.
Give the meaning of the following terms.

1. Explant
2. Totipotency
3. Some clones
4. Somatic hybrids
5. Micropropagation.

Answer:

1. Any part of a plant taken out and grown in a test tube under sterile conditions in special nutrient media is called explant.
2. The capacity to generate a whole plant from any cell/explant is called totipotency.
3. Plants which are genetically identical to the original plant from which they are grown are called some clones.
4. Isolated protoplasts from two different varieties of plants each having a desirable character can be fused to get hybrid protoplasts, which can be further grown to form a new plant. These hybrids are called somatic hybrids.
5. Producing thousands of plants through tissue culture is called micropropagation.

Question 34.
Explain the role of microbes in household food products.
Answer:
Microbes in Household Products: Microbes have been used in different ways by mankind from time immemorial in the preparation of household products.
Some of the household products obtained from the microbial activity are:

• Dairy Products: A number of bacteria like Lactobacillus and some others commonly called Lactic Acid Bacteria (LAB) are used in the preparation of several dairy products.
• Lactic acid fermentation: Lactic acid causes coagulation of milk protein casein. Milk is converted into products like curd, yoghurt and cheese.

1. Curd: For making curd, milk is boiled and then the temperature of the milk is brought to about 40°C by keeping it in a cool place. Now a small amount of curd (the inoculum or starter containing Lactobacillus acidophilus) is added to the milk. The fermentation process begins which converts milk sugar lactose into lactic acid. The acid coagulates milk protein by partially digesting it so that a thick concentrated curd is prepared. Curd is more nutritious than milk as it contains an increased quantity of vitamin B₁₂.

2. Yoghurt (Yogurt): It is prepared by curdling milk with the help of Streptococcus Thermophilus and Lactobacillus Bulgarians. The temperature is maintained at about 45°C for four hours.

3. Butter Milk: The acidulated liquid left after churning out butter from curd is buttermilk.

4. Cheese: It is a protein-rich nutritive preparation obtained after fermentation and curdling of milk. It consists of milk curd that has separated from the liquid part or whey. Depending upon the water content present in milk/curd, cheese is of three types.

• Raw cheese
• Unripened cheese (cottage cheese), and
• Ripened cheese.

Bread: A small amount of baker’s yeast (Saccharomyces cerevisiae) is added to wheat flour at the time of its kneading. The kneaded flour or dough is kept for a few hours in a warm place. It results in swelling of the dough called leavening. The leavening is caused by the actions of three types of enzymes secreted by the yeast. These are amylase, maltase and zymase.

Question 35.
What is population growth? Explain the factors responsible for fluctuating population density.
Answer:
Population growth: The size of the population of any species in not a static parameter. It keeps on changing with time, depending on various factors such as availability of food, predation pressure and prevailing weather.

Such changes give some idea of what is happening to the population, whether it is increasing or decreasing. Though a number of factors affect population size, the density of a population in a given habitat during a given period fluctuates due to four basic processes-natality, mortality, immigration and emigration.

The population density is the number of individuals of a species per unit area/space at a given time.
Mathematically population density is expressed as:
D = \(\frac{N}{S}\)
where D stands for the population density, N denotes the number of individuals of a species at a specific time and S represents the number of units of the space.

• Natality: It refers to the number of births during a given period in the population that are added to the initial density.
• Mortality: It is the number of deaths in the population during a given period.
• Immigration: It is the number of individuals of the same species that have come into the habitat from elsewhere during the time period under consideration.
• Emigration: It is the number of individuals of the same species who have left the habitat and gone elsewhere during the time period under consideration. Nation and immigration contribute to an increase in population density, while mortality and emigration to a decrease.

Therefore, population growth or change in the size of the population in a given time is determined by the above factors.
Change in population size = (Births + Immigration ) – (Deaths + Emigration)
If N is the population density at time t, then its density at time t + 1 is N_{t + 1} = N_t + [(B + I) – (D + E)]

It is clear from the above equation. that population density increases if the number of births plus the number of immigrants (B + I) is more than the number of deaths plus the number of emigrants (D + E), otherwise, it will decrease. Under normal conditions, births and deaths are the most important factors influencing population density. If a new habitat is just being colonized, then immigration may contribute more significantly to the population growth than birth rates.

Question 36.
“One of the applications of biotechnology is the production of insect-resistant crop plants.” Justify the statement with reference to Bt-cotton.
Answer:
Bacillus thuringiensis produces crystal proteins called Cry proteins which are toxic to larvae of insects like tobacco budworm, armyworm, beetles and mosquitoes.

Cry proteins exist as inactive protoxins and get converted intó active toxin when ingested by the insect, as the alkaline pH of the gut solubilises the crystals.

The activated toxin binds to the surface of epithelial cells of midgut and creates pores.
This causes swelling and lysis of cells leading to the death of the insect (larva).

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.

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.

Question 37.
Discuss the consequences of global warming and ozone-layer depletion.
Answer:
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 0-6° 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 15cm 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 0-88 m 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 the 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.

Effects of ozone depletion:
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: UV radiation damages nucleic 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.