TS Inter 2nd Year Botany Question Paper May 2017

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TS Inter 2nd Year Botany Question Paper May 2017

Time: 3 Hours
Max. Marks: 60

Note: Read the following instructions carefully:

1. Answer all questions of Section ‘A’. Answer any six questions out of eight in Section ‘B’ and answer any two questions out of three in Section ‘C’.
2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of ‘Very Short Answer Týpe”. Each question carries two marks. Every answer may be limited to 5 lines. Answer all the questions at one place in the same order.
3. 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.
4. 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.
5. 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.
Distinguish between Apoenzyme and Cofactor.
Answer:
The protein part of a holoenzyme is called Apoenzyme. The non-protein part of a holoenzyme is called a co-factor.

Question 2.
What is meant by bolting? Which hormone causes bolting?
Answer:
The sudden elongation of internodes prior to flowering is called bolting. It is caused by Gibberellins.

Question 3.
What is a plasmid? What is its function?
Answer:
Plasmid is a small, seif-replicating, circular, naked, double-stranded DNA molecule present in addition to a genophore in bacterial cell. They are used as agents in modern genetic engineering technique.

Question 4.
Explain the terms phenotype and genotype.
Answer:
The physical or external appearance of a character is called phenotype. The genetic makeup of an individual is called genotype.

Question 5.
What are the components of a transcription unit?
Answer:
A promoter, the structural gene, and A terminator.

Question 6.
What are the components of a nucleotide?
Answer:
A nitrogenous base, a pentose sugar, and a phosphate group.

Question 7.
What are molecular scissors? From where are they obtained?
Answer:
Molecular scissors are the restriction enzymes which cut the DNA at specific locations. They are obtäined from Bacteria.

Question 8.
What is GEAC and what are its objectives?
Answer:
GEAC stands for Genetic Engeneering Approval Committee. It make decisions regarding the validity of GM research and the safety of introducing GM organisms for public services.

Question 9.
Give two examples of fungi used in SCP production.
Answer:
Candida utilis, Saccharomyces cerevisiae, Chaetomium cellulolyticum.

Question 10.
Why does Swiss cheese have big holes? Name the bacteria responsible for it.
Answer:
The large holes in Swiss cheese are due to the production of large amount of CO₂ by propionibacterium propionic.

Section B
6 x4 = 24 M

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

Question 11.
Transpiration is a necessary evil, explain.
Answer:
Beneficial effects:

• It helps in passive absorption of water.
• It also helps in passive absorption of mineral salts by mass flow mechanism.
• It is the main force for ascent of sap.
• It regulates the temperature of plant body and provides cooling effect.
• Maintain shape and structure of the plants by keeping cell Turgid.

Harmful Effects:

1. Excessive transpiration makes the cells flaccid which retards growth.
2. Excessive transpiration leads to closure of stomata thus constructing gaseous exchange.

Question 12.
Tabulate any eight differences between C₃ and C₄ plants/cycles.
Answer:

C3 plants	C4 plants.
1. Kranz’s anatomy is not shown by the leaves.	1. Leaves show Kranz anatomy.
2. Chloroplast dimorphism is absent.	2. Chloroplast dimorphism is seen.
3. Only Calvin cycle occurs.	3. C4 cycle occurs in mesophyll cells and Calvin cycle occurs in bundle sheath cells.
4. The primary acceptor of CO2 is RUBP.	4. The primary acceptor of CO2 is PEPA.
5. The first formed product is PGA(3C).	5. The first formed product is OAA (4C).
6. Less efficient in utilizing atmospheric CO2.	6. More efficient in utilizing atmospheric CO2.
7. Photo respiration is high.	7. Photo respiration is not detectable.
8. The optimum temperature for this pathway is 15°C to 25°C.	8. The optimum temperature for this pathway is 30°C-45°C.
9. Photosynthetic yield is low.	9. Photosynthetic yield is more.
10. 18 ATP’ are required to synthesize one Glucose molecule.	10. 30 ATP are required to synthesize one Glucose molecule,
11. Water use efficiency is low.	11. Water use efficiency is high very high.
12. CO2 compensation point is very high.	12. CO2 compensation point is low.

Question 13.
Explain the steps involved in the formation of root nodule.
Answer:

1. Roots of legumes release sugars,Amino acids which attached Rhizobium. They get attached to epidermal and root hair cells of the host.
2. The roots hair curl and the bacteria invade the roots hair.
3. An infection thread produced carrying the bacteria into the cortex of the root.
4. Bacteria initiate nodule formation in the cortex of the root. Then the bacteria released from the thread into the cortical cells of the host and stimulate the host cells to divide. Thus leads to the differentiation of specialized nitrogen-fixing cells.
5. The nodule thus formed establishes a direct vascular connection with the host for exchange of nutrients.

Development of root nodules in soyabean:

• Rhizobium bacteria contact a susceptible root hair, and divide near it.
• Successful infection of the root hair causes it to curl.
• Infected thread carries the bacteria to the inner cortex. The bacteria get modified into rod-shaped bacteroids and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation.
• A mature nodule is complete with vascular tissues continuous with those of the root.

Question 14.
What are the physiological processes that are regulated by ethylene in plants?
Answer:

1. Ethylene promotes the ripening of fruits.
2. Ethylene promotes the senescence and abscission of leaves and flowers.
3. Ethylene breaks bud and seed dormancy initiates germination in peanut seeds and sprouting of potato tubers.
4. Ethylene promotes rapid internode/petiole elongation in deep-water rice plants.
5. It also promotes root growth and root hair formation, thus helping plants to increase their absorption surface. :
6. Ethylene is used to initiate flowering (Mango) and for synchronising fruit set in pineapples.
7. It promotes female flowers in cucumbers, thereby increasing the yield.

Question 15.
Explain the structure of TMV.
Answer:
a) Franklin et. Al have described the structure of TMV Capsomeres
b) It is rod shaped virus. It is about 300 nm long and 18 or 19 nm in diameter with a Protein molecular weight of 39 x 10⁶ daltons.
c) The capsid is made up of 2,130 protein subunits of identical Helical Symmetry size. They are called capsomeres.
d) The protein subunits are arranged in a helical manner around a central core of 4 nm. Each protein subunit is made up of a single polypeptide chain with 158 amino acids.

e) Inside the protein capsid there is a single-stranded RNA molecule which is also spirally coiled to form helix. RNA of TMV consists of 6500 nucleotides.

Question 16.
Explain the incomplete dominance with example.
Answer:
Incomplete dominance: It is the condition where one allele of a gene is not completely dominant over the other allele and results in the heterozygotes having phenotypes different from the dominant and recessive homozygotes. Ex: In a cross between a true-breeding read flowered plant (RR) and true breeding white-flowered plant (rr), the F₁ was pink (Rr). When the F₁ was self-pollinated, the F₂ resulted in the ratio of
RR:Rr: rr
1:2 :1
Red Pink White

Here the genotype ratios were as in monohybrid cross of Mendel but the phenotypic ratio had changed from 3: 1 because ‘R’ was not completely dominant over ‘r’ and is possible to distinguish ‘Rr’ as pink from ‘RR and ‘rr’.

Question 17.
What are the differences between DNA and RNA?
Answer:

DNA	RNA
1. DNA consists of 2 strands of nucleotides.	1. It consists of 1 strand of nucleotides.
2. Deoxyribose sugar is present.	2. Ribose sugar is present.
3. Thymine, cytosine are pyrimidines.	3. Uracil, cytosine are pyrimidines.
4. DNA is made of 4 millions nucleotides.	4. RNA is made of 75-2000 nucleotides.
5. It undergoes self-replication.	5. It do not undergo self replication.
6. DNA is generic material.	6. RNA is non genetic material.
7. It does not involved in protein synthesis directly.	7. It involves in protein synthesis.
8. Metabolically DNA is one type.	8. Metabolically RNA is 3 types.
9. Base puring is A = T,G ≡ C.	9. Base puring is A = U, G ≡ C.
10. It is present more in the nucleus and little in chloroplasts and mitochondria.	10. It is present more in cytoplasm and little in the nucleus.
11. Purines and pyrimidines exist in 1: 1 ratio.	11. Purines and pyrimidines does not exist in 1: 1 ratio.

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 (flies, 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 Boliworm, 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 ‘cry’. The proteins encoded by the genes cry lAc and cry IIAb control the cotton bollworms and cry lAb controls corn borer.

Section – C
(2 × 8 = 16 M)

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?
Answer:
Glucose is broken down into 2 molecular of pyruvic acid is called Glycolysis. It was given by gustav Embden, Otto Mayerhof and J.Parnas so called EMP Parthway. It occurs in the cytoplasm of the cell and takes place in all living organism. In this, 4 ATP are formed of which two are utilized and 2 NADPH+H⁺ are formed.

A⁺ the end of glycolysis, 2 PA, 2 ATP and 2 NADPH+H are formed as end products. The ATP and NADPH+H are utilised for fixation of CO₂.

Glyclosis occur in cytoplasm, pyruvic acid, 2ATP 2 NADPH + H are the end products. In aerobic respiration, pyruvic acid, 2 NADPH+H⁺ are completely oxiclised through TCA cycle, ETS pathway and produces 36 ATP molecules. In anaerobic respiration, pyruvic acid is partially oxidised results in the formation of ethyl alcohol and CO₂.

Reactions:
1. Glucose is phosphorylated in the presence of Kinase to form glucose-6-phosphate..
Glucose + ATP → Glucose-6-phosphate + ADP

2. G-6P is isomerised to Fructose-6-phosphate in the presence of Isomerase
G-6P → F6P

3. Fructose 6 phosphate is phosphorylated in the presence of hexokinase to form Fructose 1, 6 Biphosphate.
G6P+ATP → F1,6BiP+ADP

4. F 1,6 BiP undergoes cleavage in the presence of Aldolase to form 1 Dihydroxy acetone phosphate and 1 Glyceraldehyde-3 phosphate
F1,6BiP → DHAP+1G₃P

5. DRAP does not undergo oxidation in further reactions, so gets isomerised to another G₃P in the presence of Isomerase.
DHAP →1G₃P

6. 2 molecules of G₃P undergoes dehydrogenation in the presence of dehydrogenase to form 2 molecules of 1, 3 DPGA.
2G₃P + 2NAD⁺ → 2- 1,3 DPGA + 2NADH + H⁺

7. 2 mol. of 1, 3 DPGA undergoes dephosphorylation in the presence of phospho Glycerokinase to form 2 mol. of 3 PGA
2-1,3DPGA+2ADP → 2-3PGA+2ATP

8. 2 mol. of 3PGA are converted into 2 mol. of 2PGA in the presence of mutage.
2-3PGA → 2-2 PGA

9. 2-PGA looses water molecules to form 2-phosphoenol pyruvic acid in the presente of Enolase.
2-2PGA → 2-PEPA + H₂O

10. 2 PEPA mols. are phosphorylated in the presence of pyruvic kinase to form 2 pyruvic acid molecules.
2 PEPA + 2ADP → 2ATP + 2PA

Question 20.
Explain briefly the various processes of recombinant DNA technology.
Answer:
Key tools are:
1. Restriction enzymes: Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coli 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 endonuclease. Hind II which cut DNA molecule at a particular pair by recognising a specific sequence of six base pairs, called 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 letter comes from the genus (escherichia) and the second two letters from the species of the prokaryotic cell (cou) the letter ‘R’ is derived from the name of strain. Roman numbers indicate the order in which the enzymes were isolated from the strain of Bacteria. Restriction enzyme 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 locations within the DNA.

Most restriction enzymes cut the two stands of DNA double helix at different locations sucli a cleavage 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 facilities the action of enzyme DNA ligase.

Cloning vectors: The DNA used as a carrier for transferring 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, bacterio-phages, 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 ‘ori the vector requires a selectable maker, which help 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, Chloramphenicol, tetracycline, 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 and what are the advantages of tissue culture over conventional method of plant breeding in crop improvement programmes.
Answer:
Tissue culture Technique: 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 cytokinins are added to the medium. The nutrient medium is poured 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 attracts the growth of microorganisms. The culture medium is autoclaved for 15 mins, at 120°c or 15 pounds of pressure to make aseptic.

c) Preparation of explant: Any living part of plant can be used as explant. The explants must be cleaned with liquid detergent and in running water and surface sterilized 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, the cells of the explant divide and redivide, producing a mass of tissue called callus. The callus is transferred to another medium containing growth regulators to initiate the formation of roots and leafy shoots (organogenesis). 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. The production of exact copies of plants that produce particularly good flowers, fruits or have other desirable traits.
2. To quickly produce mature plants.
3. The production of multiples of plants in the absence of seeds or necessary pollinators to produce seeds.
4. The regeneration of whole plants from plant cells that have been genetically modified.
5. The production of plants from seeds that otherwise have very low chances of germinating and growing i.e., orchids and nepenthes.
6. To clean particular plants of viral and other infections and to quickly multiply these plants as cleaned stock for Horticulture and Agriculture.