TS Inter 2nd Year Botany Question Paper May 2015

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TS Inter 2nd Year Botany Question Paper May 2015

Time: 3 Hours
Max Marks: 60

Note: Read the following instructions carefully.

• Answer all questions of Section A. Answer any six questions out of eight in Section B and answer any two questions in Section C.
• In Section-A, questions from Sr. Nos. 1 to 10 are of ‘Very Short Answer Type”. Each question carries two marks. Every answer may be limited to five lines. Answer all these questions at one place in the same order.
• In Section B, questions from Sr. Nos. 11 to 18 are of Short Answer Type”. Each question carries four marks. Every answer may be limited to 20 lines.
• In Section-C, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 60 lines.
• Draw labelled diagrams wherever necessary for questions in Section-B and C.

Section-A
10 x 2 = 20

Note: Answer all questions. Each answer may be limited to 5 lines.

Question 1.
What are Porins? What role do they play in diffusion?
Answer:
Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria. They allows molecules up to the size of small proteins to pass through.

Question 2.
Which element is regarded as the 17th essential element? Name a disease caused by its deficiency.
Answer:
Nickel in 17th essential element. its deficiency causes Mouse ears in Pecan.

Question 3.
Mention the differences between Virulent phages and Temperate phages.
Answer:

Virulent phages	Temperate phages
1. Phages that attack the bacterium E-coli cause lysis of the cells.	1. “The phage DNA gets integrated into the Bacterial DNA and remains latent”.
2. They show lytic cycle. Ex: T-even Bacteriophages.	2. They show the Lysogenic cycle. Ex: Lambde phage.

Question 4.
What is the cross between the F₁ progeny and the homozygous recessive parent called? How is it useful?
Answer:
Test cross. It is used to test whether an Individual in Hornozygous (pure) or Heterozygous (hybrid).

Question 5.
Name any three viruses which have RNA as the genetic material.
Answer:

1. TMV
2. qB Bacteriophage
3. HIV

Question 6.
Given below is the sequence of coding strands of DNA in a transcription unit. 5’AAT GCA GCT ATT AGG —3’ Write the sequence of:
(a) its complementary strand,
(b) the mRNA.
Answer:
a) 3’ – TTA CGT CGA TAA TCC -5’
b) 3’-UUACGUCGAUAAUCC-5’

Question 7.
How does one visualize DNA on an agarose gel?
Answer:
The DNA fragments can be visualized only after staining the DNA with Ethidium Bromide followed by exposure to UV radiation.

Question 8.
Name the Nematode that infects the roots of a tobacco plant. Name the strategy adopted to prevent this infestation.
Answer:
Meloidegyne incognitia A novel strategy adopted to prevent this infestation is RNA interference.

Question 9.
Name two examples of wheat varieties introduced in India, which are high-yielding and disease resistant.
Answer:
Sonalika, Kalyan sona.

Question 10.
Name any two industrially important enzymes.
Answer:
Lipase, Streptokinase.

Section-B
6 x 4 = 24

Note: Answer any six questions. Each answer may be limited to 20 lines.

Question 11.
How does ascent of sap occur in tall trees?
Answer:
The upward movement of water through the xylem against gravitational force is called ascent of sap. The transpiration-driven ascent of xylem sap depends on

• cohesion – mutual attraction between water molecules
• Adhesion – attraction of water molecules to polar surfaces
• transpiration pull – driving force for upward movement of water.

These properties give water high tensile strength and high capillarity. In plants, capillarity is aided by the small diameter of the tracheary elements. As water evaporates through the stomata. Since the thin film of water over the cells is continuous, it results in pulling of water, molecule by molecule into the leaf from the xylem. Also, because of lower concentration of water vapour in the atmosphere, water diffuses into the surrounding air. This creates transpiration pull. The forces generated by the transpiration can create pressure sufficient to lift a xylem-sized column of water over 130 metres high.

Question 12.
Write briefly about enzyme inhibitors.
Answer:
The chemicals that can shut off enzyme activity are called inhibitors.
They are of 3 types.
1. CompetitIve Inhibitors: Substance which closely resembles the substrate molecules and inhibits the activity of the enzymes are called competitive inhibitors. E.g.: Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate.

2. Noncompetitive inhibitors: The inhibitors which have no structural similarity with the substrate and bind to an enzyme at locations other than the active site, so that the globular structure of the enzyme is changed are called non-competitive inhibitors. Eg: Metal ions of Copper, Mercury.

3. Feedback inhibitors: The end product of a chain of enzyme-catalysed reactions inhibit the enzyme of the first reaction as a part of homoeostatic control of metabolism are called feedback inhibitors. Eg: During respiration (Glycolysis) accumulation of Glucose-6 Phosphate occurs, It inhibits the Hexokinase.

Question 13.
Tabulate any eight differences between C₃ and C₄ plants/cycles.
Answer:

C3 plants	C4 plants
1. Leaves do not show Kranz’s anatomy.	1. Leaves show Krenz’s anatomy.
2. Chloroplast dimorphism is seen.	2. Chloroplasts are similar and do not show dimorphism.
3. The primary acceptor of CO2 = RUBP.	3. The primary acceptor of CO2 is PEPA
4. The first stable product is = PGA(3c).	4. The first stable product is = OAA(4c).
5. They are less efficient in utilising atmospheric CO2.	5. They are more efficient in utilizing atmospheric CO2.
6. Photo respiration is very high.	6. Photo respiration is not detectable.
7. The optimum temperature is 15-25°C	7. The optimum temperature is 30-45°C
8. They utilise 18 ATP molecules to synthesise 1 glucose.	8. Photosynthetic yield is very high.
9. Photosynthetic yield is low.	9. They utilise 30 ATP molecules to synthesize glucose.
10. They utilise water less efficiently.	10. They utilise water more efficiently.
11. CO2 compensation point is very High.	11. CO2 compensation point is low.

Question 14.
Write the physiological responses of Gibberellins in plants.
Answer:

1. Gibberellins delay senescence. Thus fruits can be left on the tree longer so as to extend the market period.
2. Spraying of Gibberellins on sugarcane crops increases the length of the stem, thus increasing the yield as much as 20 tonnes per acre.
3. GA hastens the maturity period of conifers thus leading to early seed production.
4. GA also promotes bolting ¡n cabbages, beet etc.
5. They also produce parthenocarpic fruits in grapes and tomato.
6. Gibberellins favour the formation of male flowers in cucurbita.
7. Gibberellins cause fruits like apples to elongate and Improve their shape.
8. Gibberellins cause to Increase in length of Grapes stalks.

Question 15.
What are the nutritional groups of bacteria based On their source of energy and carbon?
Answer:
According to the source of energy & carbon, the bacteria are classified into four types.
They are :
1. Photo autotrophs: These bacteria like green plants posses the chlorophyll and obtain energy from sunlight and carbon from atmospheric CO₂.
Ex: Purple sulphur Bacteria = chromium
Green sulphur Bacteria = chromium

2. Photo heterotrophs: These bacteria obtain energy from sunlight and carbon Is derived from organic sources.
Ex: Purple non-sulphur bacteria = Rhodospirillum, Rhodo microbium.

3. Chemo autotrophs: These bacteria derive the carbon from CO₂ and energy from oxidation of Inorganic substances.
They are

• Sulphur bacteria: It oxidises H₂S and releases sulphur. Ex: Beggiotoa
• Nitrifying bacteria: It oxidises ammonia into Nitrates. Ex: Nitrosomonas, Nitrobactor.

4. Chemoheterotrophs: These bacteria derive both carbon and energy from organic substances like glucose and amino acids.
They are of 2 types :

• a) Saprophytes: The bacteria which grown on dead organic matter are called saprophytes. Ex: Bacillus.
• b) Parasites: The bacteria which derive nutrients from a living host and cause diseases. Ex: Xanthomonas.

Question 16.
Differentiate between the following:
(a) Dominant and Recessive
(b) Homozygous and Heterozygous
Answer:
a)

Dominant	Recessive
A character is expressed phenotypically in both homozygotes and heterozygotes.	A character which is not expressed phenotypically in Heterozygous condition.

b)

Homozygous	Heterozygous
Two similar or Identical alleles for a single character.	Two different alleles for a single character.

Question 17.
Write the Important features of the Genetic code.
Answer:

1. The code is a triplet. 61 codons code for 20 amino acids and 3 codons do not code for any amino acids called stop codons. [UAA, UAG, UGA].
2. One codon codes for only one amino acid hence it Is unambiguous and specific.
3. Some amino acids are coded by more than one codoñ hence the code is degenerate.
4. The codon is read In mRNA in a contiguous fashion. There are no punctuations.
5. The code is nearly universal. E.g.: UUU code for phenylalanine in bacteria and humans.
6. AUG has dual functions, It codes for Methionine (Met) and also acts as the initiator codon.

Question 18.
Give a brief account of Bt cotton.
Answer:
Some strains of Bacillus thuringiensis produce proteins that kill certain insects such as lepidopterans (tobacco budworm, armyworm), coleopterans (beetles) and dipterans (files, mosquitoes). Bacillus thuringiensis forms protein Crystals which contain a toxic insecticidal protein. The gene responsible for the production of this toxic protein is introduced genetically into the cotton seeds protects the plants from Bollworm, a Major pest of cotton. Once an insect Ingests the inactive toxin is converted into an active form of toxin due to the alkaline pH of the gut. The activated toxin binds to the midgut epithelial cells and create pores that cause cell swelling and lysis and cause the death of the Insect.

Use of Bt. Cotton has led to 3-27% Increase in cotton yield in countries where it Is grown. The toxin is coded by a gene named cryA. The proteins encoded by the genes cry lAc and cry hAb control the cotton bollworms and cry lAb controls corn borer.

Section-C
2 x 8 = 16

Note: Answer any two questions. Each answer may be limited to 60 lines.

Question 19.
Give an account of Glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration.
Answer:
Glucose is broken down into 2 molecules of pyruvic acid Is called Glycolysis. It was given by Gustav Embden, Mayerhof and Parnas so-called EMP Pathway. It occurs in the cytoplasm of the cell and takes place in all living organisms. At the end of Glycolysis, 2 PA, 2 ATP and 2 NADPH+H⁺ are formed as End products. In Aerobic Respiration, pyruvic acid, NADPH + H⁺’ are completely oxidised through TCA cycle, ETS and produces 36 ATP molecules. In Anaerobic respiration, pyruvic acid undergoes partial oxidation and releases CO₂ and Ethyl alcohol along with 56 K.cals of energy.

Question 20.
Give a brief account of the tools of recombinant DNA technology.
Answer:
Ans. Key tools are:
1) Restriction enzymes: Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coil were isolated in the year 1963. One of these added methyl group to DNA and the other DNA. The latter was called restriction endonuclease. The first restriction is endonuclease. Hind Ii which cut DNA molecules as a particular pair by recognising a specific sequence of six base pairs, is called the recognition sequence for Hind II. Today more than 900 restriction enzymes were isolated from over 200 strains of Bacteria each of which recognises a different recognition sequence.

E. CORI is a restriction enzyme in which, the first latter comes from the genus (escherichia) and the second two tatters from the species of the prokaryotic cell (coil) the latter ‘R is derived from the name of the strain. Roman numbers indicate the order in which the enzymes were isolated from the strain of Bactena. Restriction enzymes belong to a larger class of enzymes called nucleases. They are of two types:

• Exonucleases which remove nucleotides from the ends of the DNA.
• Endonucleases which makes cuts at specific location within the DNA.

Most restriction enzymes cut the two stands of DNA double helx at different locations such a clevage is known as staggered cut. E.CORI recognises 5 GAATT3’ sites on the DNA and cut it between G & A results in the formation of sticky ends or cohesive end pieces. This stickyness of the ends facilitates the action of enzyme DNA ligase.

Cloning vectors: The DNA used as a carrier for transfering a fragment of foreign DNA into a suitable host is called vector. Vectors used for multiplying the foreign DNA sequences are called cloning vectors. Commonly used cloning vectors are plasmids, bacteriophages, cosmids, plasmids are extra chromosome circular DNA molecules present in almost all bacteria species. They are inheritable and carry few genes are easy to isolate and reintroduce into the bacterium (host).

Features required to facilitate cloning into a vector:
a) Origin of replication : (ori) This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within host cells. It is also responsible for controlling the copy number of the linked DNA.

b) Selectable marker: In addition to ‘on the vector requires a selectable maker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the any transformants normally, the genes encoding resistance to Antibiotics such as Ampicillin and chloramphenicol. tetra cycline or Kanamycin etc. are useful selectable makers for E.coli.

c) Cloning sites: In order to link the alien DNA, the vector needs to have very few preferably single recognition sites for the restriction enzyme.

d) Molecular weight: The cloning vector should have low molecular weight.

e) Vectors for cloning genes in plants and animals: The tumour-inducing (Ti) plasmid of Agrobacterium tunifaciens has now been modified into a cloning vector such that It is no more pathogenic to plants. Similarly, retroviruses have also been disarmed and are now used to deliver desirable genes into animal cells.

Question 21.
Describe the tissue culture technique. What are the advantages of tissue culture over conventional methods of plant breeding in crop improvement programmes?
Answer:
Tissue culture techniques: It involves
a) Preparation of Nutrient medium: The nutrient medium is a mixture of various essential nutrients, Amino acids, vitamins and carbohydrates. These are mixed in distilled water and P’ is adjusted to 5.6 to 6.0 growth regulators like auxins, and cytokinins are added to the medium. The nutrient medium is powered in glass vessels and closed tightly with cotton plugs before sterilizing them in an Autoclave.

b) Sterilisation: The nutrient medium is rich in nutrients and therefore attacks the growth of microorganisms. The culture medium is auto-clad for 15 mins at 121°C or 15 pounds of pressure to make aseptic.

c) Preparation of explant: Any living of plant can be used as an explant. The explants must be deaned with liquid detergent and in running water and surface sterilised with sodium hypochlorite and rinsed with distilled water.

d) Inoculation of explants: The transfer of explants onto the sterilized nutrient medium is called inoculation. It is carried out under sterilized conditions.

e) Incubation: The culture vessels with inoculated explants are incubated in a culture room under controlled temperature, optimum light and humidity. The cultures are incubated for 3-4 weeks, and the cells of the explants divide and redivide, producing a mass of tissue called a callus. The callus is transferred to another medium containing growth regulators to initiate the formation of roots and leafy shoots. Sometimes embryo-like structures develop directly from the callus which are referred as somatic embryos. These can be encapsulated with sodium alginate to form synthetic or artificial seeds.

f) Acclimatization and transfer to pots: The plants produced through tissue culture are washed gently and are planted in pots kept in glass house for 1-2 weeks. Finally, they are transferred to field.

Advantages:

1. More number of plants are produced within a short period and in a little space. (Micropropagation).
2. Virus-free plants are produced through meristem culture.
3. Somaclonal variations in plants and used in plant breeding.
4. Synthetic or artificial seeds are produced by encapsulation of embryoids with sodium alginate.
5. The production of exact copies of plants that produce particularly good flowers, and fruits.
6. To quickly produce mature plants.
7. The production of multiples of plants in the absence of seeds or necessary pollinators to produce seeds.
8. The regeneration of whole plants from plant cell that have been genetically modified.
9. Virus-free plants are used as cleaned stock for horticulture.