AP Inter 2nd Year Botany Question Paper March 2018

Thoroughly analyzing AP Inter 2nd Year Botany Model Papers and AP Inter 2nd Year Botany Question Paper March 2018 helps students identify their strengths and weaknesses.

AP Inter 2nd Year Botany Question Paper March 2018

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 Type”. Each question carries 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 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 the 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.
What is the primary acceptor of CO₂ in C₄ plants? What is the first compound formed as a result of primary carboxylation in the C₄ pathway?
Answer:
Primary acceptor of CO₂ in C₄ plants is phosphoenol pyruric acid (PEPA). First compound formed in C₄ pathway is = Oxalo Acetic Acid (OAA).

Question 3.
What is transduction? Who discovered it and in which organism?
Answer:
The transfer of genetic material from one bacterium to another through bacteriophage is known as transduction. It was discovered by Zinder and Hederberg in Salmonella typhi.

Question 4.
What is point mutation? Give an example.
Answer:

• Change of single base pair in the gene for beta globulin chain that results in the change of amino acid residue
• glutamate to valine is called point mutation. It results in diseased condition called sickle cell anaemia.

Question 5.
What are the components of a nucleotide?
Answer:
A nitrogenous base, a pentose sugar, and a phosphate group.

Question 6.
The proportion 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:
Ribonucleic acid (RNA). It consists of one strand.

Question 7.
What is the full form of PCR? How is it useful in Biotechnology?
Answer:
PCR stands for polymerase chain reaction. It can be used for the diagnosis of diseases like Aids, middle ear infection and Tuberculosis.

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

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

Question 10.
Why does ‘Swiss Cheese’ have big holes? Name the bacteria responsible for it.
Answer:
The Big holes in ‘Swiss cheese’ are due to the production of large amounts of CO₂. Propionibacterium sharmanii is responsible for it.

Section – B
6 x 4 = 24

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

Question 11.
Define and explain water potential.
Answer:
The measure of the relative tendency of water to move from one area to another is called water potential. It is denoted by the Greek symbol Psi or ψ and is expressed in pascals (Pa). It has two components namely solute potential and pressure potential.

a) Solute Potential: If some solute is added to pure water, the solution has fewer free water molecules and the concentration of water decreases reducing its water potential. The magnitude of this lowering due to dissolution of a solute is called solute potential. It is denoted as ‘ψ_s‘. It is always negative.

b) Pressure Potential: When water enters a plant cell due to diffusion, causing a pressure to build up against the cell wall. This makes the cell turgid. The magnitude of increase in water potential in such turgid cell is called pressure potential. It is usually positive and is denoted as ‘ψ _p‘.
‘ψ = ψ_s + ‘ψ _p‘

Question 12.
Explain the steps involved in the formation of root nodule.
Answer:

1. Roots of legumes release sugars. Amino acids which at tached Rhizobium. They get attached to epidermal and root hair cells of the host.
2. The root’s hair curls and the bacteria invade the root’s 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, 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 13.
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 – 6 phosphate occurs, inhibits the Hexokinase.

Question 14.
Write any four physiological effects of cytokinins 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:

1. Franklin et. All have described the structure of TMV.
2. It is rod-shaped virus. It is about 300 nm long and Tobacco Mosaic Virus 18 or 19 nm in diameter Helical Symmetry with a molecular weight of 39 x 10⁶ daltons. :
3. The capsid is made up of 2,130 protein subunits of identical size. They are called capsomeres.
4. 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.
5. 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.
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:

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

Question 17.
Write the important features of Genetic Code.
Answer:

1. The codon is triplet. of 64 codons, 61 codons code for 20 amino acids and 3 codons do not code for any amino acids (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 codon, 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.
   For Ex: UUU code for phenylalanine (phe) in bacteria and humans.
6. AUG code for methionine and also acts as initiator codon. (Dual role).

Question 18.
List out the beneficial aspects of transgenic plants.
Answer:
Plants with desirable characters created through gene transfer methods are called transgenic plants.
Beneficial aspects:
A) Transge is crop plants are efficient because they many, beneficial traits like virus resistance, instance arid ‘herbicide resistance. Ex : Papaya is resistant 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 “Flavr Savr” 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 prevent blindness.

D) Transgenic plants are used for hybrid seed production. Ex: Mate 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.

Section – C
2 x 8 = 16

Note: Answer any TWO questions. Each answer may be Limited to 60 tines.

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 molecular of pyruvic acid is called Glycolysis. It was given by Gustav Embden. Otto Mayerhof and J. Parnas’s so-called EMP Pathway. 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.

Glycolysis occur in cytoplasm, pyruvic acid, 2ATP, 2NADPH⁺, H are the end products. In aerobic respiration, pyruvic acid, 2 NADPH + H⁺ are completely oxidised through TCA cycle, ETS. 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 presene of hexokinase to form Fructose 1, 6 Biphosphate.
G6P+ATP → F 1,6 BiP+ADP

4. F 1,6 Bip undergoes cleavage in the presence of Aldolase to form I Dihydroxy acetone phosphate and 1 G1yceral aldehyde-3 phosphate.
F 1, 6 Bip → 1 DHAP + 1G₃P

5. DHAP does not undergo oxidation in further reactions, so gets isomerised to another G₃P in the presence of
Isomerase.
1DHAP → 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.
2-1,3 DPGA + 2ADP → 2 – 3 PGA+ 2 ATP

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

9. 2-PGA looses water molecules of form 2-phosphoenol pyruvic acid in the presence 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,+ 2 ADP → 2 ATP + 2 PA

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 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 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 letters from the species of the prokaryotic cell (cou) the latter R’ is derived from the name of strain. Rçman 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 location within the DNA.

Most restriction enzymes cut the two stands of DNA double helix at different locations such 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 facilitates 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, bacteriophages, cosmids, plasmids are extrachromosome 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 facilitates cloning into a vector:
a) Origin of replication: (on) 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, 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 tumefacient 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 characters. 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 and 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 the superior to both the parents are obtained. These are self-pollinated for several generations till they reach a homozygosity.

5. Testing, release and commercialisation of new characters:
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.