TS Inter 2nd Year Botany Question Paper March 2017

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TS Inter 2nd Year Botany Question Paper March 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 SI. Nos. 1 to 10 are of “Very Short Answer Type”. Each question cames two marks. Every answer may be limited to 5 lines. Answer all these questions at one place in the same order.
3. In Section – B, questions from SI. 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 SI. 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 M)

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

Question 1.
What are the components of a transcription unit?
Answer:
A promoter, the structural gene, and a terminator.

Question 2.
What is Green revolution? Who is regarded as ‘Father of green revolution?
Answer:
A substantial and dramatic increase in agriculture production is called ‘green revolution. Norman Borlaug is regarded as the ‘Father of green revolution’.

Question 3.
What is downstream processing?
Answer:
Separation and purification of products before they are ready for marketing is called downstream processing.

Question 4.
Name any two industrially important enzymes.
Answer:
Lipases and streptokinase.

Question 5.
Write briefly on the occurrence of microorganisms.
Answer:
Microorganisms are omnipresent in vast numbers and are described as ubiquitous. They are found in soil, water, air, plants,
animals, food materials, clothes, leather, etc., and also can grow on extreme conditions like hot water springs, Ice Mountains, volcanic ash, and in sulphur springs.

Question 6.
What are Apoplast and Symplast?
Answer:
Apoplast is the path of movement of water within the plant that moves without crossing the membranes. It is a faster process. Symplast is the path of water movement in the plant that crosses the membranes. It is a slower process.

Question 7.
What are the physical properties of water responsible for the ascent of sap through xylem in plants?
Answer:

• Cohesion: Mutual attraction between water molecules.
• Adhesion: Attraction between water molecules and inner walls of tracheary elements.
• Transpiration pull: Driving force for upward movement of water.

Question 8.
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 is homozygous (pure) or heterozygous (hybrid).

Question 9.
The proportions of nucleotides in a given nucleic acid are:
Adenine 18%,
Guanine 30%,
Cytosine 42%
and Uracil 10%.
Name the nucleic acid and mention the number of strands in it.
Answer:
Nucleic acid is RNA. Only one strand is present.

Question 10.
How nucleopolyhedro viruses are used in these days?
Answer:
The viruses are excellent candidates for species-specific, narrow-spectrum insecticidal applications. They have been shown to have no negative impact on plants. mammals, birds, and fish, and even on-target insects.

Section – B
(6 x 4 = 24 M)

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

Question 11.
Write briefly about enzyme inhibitors.
Answer:
The chemicals that can shuts off enzyme activity are called enzyme inhibitors. They are of three types.
A. Competitive inhibitors: Substances which are closely resemble the substrate molecules and inhibits the activity of the enzyme are called Competitive inhibitors. Ex: Inhibition of Succinic dehydrogenase by Malonate which closely resembles the substrate Succinate.

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

C. Feedback inhibitors: The end product of a chain of enzyme catalysed reactions inhibits the enzyme of the first reaction as part of homeostatic control of metabolism is called feedback inhibitors. Ex: During respiration, accumulation of Glucose-6phosphate occurs, inhibits the Hexokinase.

Question 12.
Tabulate any four differences between C₃ and C₄ plants.
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 the 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. Photo synthetic 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.
12. CO2 compensation point is very high.	12. CO2 compensation point is very high.

Question 13.
Explain the structure of TMV
Answer:
TMV is the most extensively studied plant virus. It is elongated and rod-like. TMV is about 300 nm long and 18 nm in diameter with a molecular weight of 39 x 10⁶ daltons. The capsid is made up of 2,130 protein subunits called capsomeres, arranged helically around a central hollow space of 4 nm. Each capsomere is made up of a single polypeptide chain with 158 amino acids. The core nucleic acid is RNA (SS) contains 6500 nucleotides.

Question 14.
Write a short note on seed dormancy.
Answer:
The inability of seed to germinate or grow is called seed dormancy It may be due to either external or internal factors.
Internal factors:
A. Immature embryo: The embryo has not reached morphological maturity capable of germination. Ex: Ranunculus.
B. Hard seed coat: In Fabaceae members, seeds have hard seed coats which prevent uptake of oxygen or water. It can be broken by scarification.
C. Chemicals: Seeds of some plants contain chemicals which inhibit germination. Ex: Tomato.

External factors:
A. Low-temperature treatment: Many seeds will not germinate until they have been exposed to low temperatures. The practice of layering seeds in sand and peat is called stratification. Ex: Polygonum.

Question 15.
Mention the advantages of selecting pea plant for experiment by Mendel.
Answer:
The garden pea (Pisum sativum) is suitable for hybridization experiments due to the following advantages:

• It is an annual plant that has well-defined Characteristics.
• It can be grown and crossed easily.
• It has bisexual flowers containing both female and male parts.
• It can be self-fertilized conveniently.
• It has a short life cycle and produces large number of offspring.

Question 16.
How many types of RNA polymerases exist in cells? Write their names and functions.
Answer:
Three types of RNA polymerases exist in cells. They are:

1. RNA polymerase-I: It transcribes rRNA(28S,18S and 5.8S)
2. RNA polymerase-II: It transcribes the precursor of mRNA, the heterogenous nuclear RNA (hnRNA).
3. RNA polymerase-III: It is responsible for transcription of tRNA, 5s rRNA, and small nuclear RNAs.

Question 17.
List the beneficial aspects of transgenic plants.
Answer:
Plants with desirable characteristics created through gene transfer methods are called transgenic plants.
Beneficial aspects:
A) Transgenic crop plants are efficient because they have many beneficial traits like virus resistance, insect resistance and herbicide resistançe. Ex: Papaya is resistant to papaya ring spot virus. Bt cotton is resistant to insects.

B) Transgenic plants are resistant to bacterial and fungal pathogens. Ex: Transgenic Tomato plants are resistant to Pseudomonas bacterium. Transgenic potato is resistant to Fungus Phytophthora.

C) Transgenic plants which are suitable for food processing are produced with improved nutritional quality. Ex: Transgenic Tomato “Flaw SavrTM are bruise resistant. suitable for storage and transport due to delayed ripening and offers higher shelf life. Transgenic Golden Rice ‘Taipei” is rich in Vitamin-A, and prevents blindness.

D) Transgenic plants are used for hybrid seed production. Ex: Male sterile plants of Brassica napus are produced. This will be eliminate the problem of manual emasculation and reduces the cost of hybrid seed production.

E) Transgenic plants have been shown to express the genes of insulin, interferons, human growth hormones, antibiotics, and antibodies.

F) Transgenic plants are used as bioreactors for obtaining commercially useful products, specialized medicines, and antibodies on large scale is called molecular farming.

Question 18.
Write in brief how plants synthesize amino acids?
Answer:
A. Reductive amination reaction: In this reaction, ammonia combines with an alpha ketoglutanc acid, to produce a Glutamic acid. Alpha ketoglutaric acaid + NH₄⁺NADPH ……………… Glutamic acid +H₂O+NADP

B. Transamination reaction: The reaction involves transfer of amino group, from amino acid, to the keto acid.
(Keto acid) (Amino acid) (Amino acid) (Keto acid)
α-Ketoglutaric acid+ Aspartic acid … Glutamic acid + Oxaloacetic acid

In the above reaction, aspartic acid has transferred its amino group (NH₃) to the α – ketoglutaric acid to synthesize glutamic acid and release keto acid. The reaction is catalyzed by enzymes called transaminases.

Section – C
(2 x 8 = 16 M)

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

Question 19.
Explain the reactions of Krebs cycle.
Answer:

The acetyl CoA eaters into the [mitochondrial matrix] a cyclic pathway tricarboxylic acid cycle, more commonly called krebs cycle after the scientist Hans Krebs who first elucidated it.

1. Condensation: In this acetyl CoA condenses with oxaloacetic acid and water to yield citric acid in the presence of citrate synthetase and CoA is released.

2. Dehydration: Citric acid looses water molecule to yield cis aconitic acid in the presence of aconitase.

3. Hydration: A water molecule is added to cisaconic acid to yield isocitric acid in the presence of a conitase.

4. Oxidation I: Isocitric acid undergoes oxidation in the presence of dehydrogenase to yield succinic acid

5. Decarboxylation: Oxalosuccinic acid undergoes decarboxylation in the presence of decarboxylase to form α-keto glutaric acid.

6. Oxidation II, decarboxylation: α – keto glutaric acid undergoes oxidation and decarboxylation in the presence of dehydrogenase and condenses with co.A to form succinyl Co. A.

7. Cleavage: Succinyl Co.A splits into succinic acid and Co.A in the presence of thiokinase to form succinic acid. The energy released is utilised to from ATP from ADP and PI.

8. Oxidation – III: Succinic acid undergoes oxidation and forms Fumarc acid in the presence of succinic dehydrogenase.

9. Hydration: A water molecule is alcohol to Fumaric acid in the presence of Fumarase to form Malic acid.

10. Oxidation IV: Malic acid undergoes oxidation in the presence of malic dehydrogenase to form oxaloacetic acid.

In TCA cycle, for every 2 molecules of Acetyl Co.A undergoing oxidation, 2 ATP 8 NADPH+ H⁺, 2FADH₂ molecules are formed.

Question 20.
Give a brief account of the tools of recombinant DNA (rDNA) technology.
Answer:
Key tools are:
1) Restriction enzymes :
Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia cou 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 latter comes from the genus (escherichia) and the second two latters from the species of the prokaryotic cell (coil) the latter 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:

1. Exonucleases which remove nucleotides from the ends of the DNA.
2. 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 camera 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 ‘on’ 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, 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.
You are a Botanist, working in the area of plant breeding. Describe the various steps that you will undertake to release a new variety.
Answer:
The main steps in breeding a new genetic variety of a crop are:
1. Collection of Variability: Genetic variability is the root of any breeding programme. Collection and preservation of all the different wild varieties, species, and relatives of cultivated species is a prerequisite for effective exploitation of natural genes available in the populations. The entire collection having all the diverse alleles for all genes in, a given crop is called Germplasm collection.

2. Evaluation and selection of parents: The germplasm is evaluated so as to identify plants with desirable characteristics. The selected plants are multiplied and are used. Purelines are created wherever desirable.

3) Cross Hybridisation among the selected parents: After emasculation (Removal of Anthers from bisexual flower of a female parent) the female flowers are enclosed in a polythene bag to prevent undesired cross-pollination. Pollen grains are collected from the male parent with the help of a brush and are transferred to the surface of the stigma thus cross-pollination is affected artificially.

4) Selection and Testing of superior recombinants: It involves selecting among the progeny of hybrids, those plants that
have the desired character combination. The selection process requires careful scientific evaluation of the progeny. Due to this, plants that are superior to both the parents are obtained. These are self-pollinated for several generations till they reach homozygosity.

5) Testing, release, and commercialisation of new characters: The newly selected lines are evaluated for their yield and other traits of quality, disease resistance etc. It is done by growing these in research fields and recording their performance under ideal fertilizer application, irrigation, and other crop management practices. It is followed by testing the materials in farmers’ fields for at least 3 growing seasons at several places in the country, in all agroclimatic zones. Finally, they are distributed to farmers as a new variety.