AP Inter 2nd Year Botany Question Paper May 2015

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AP 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 Sections B and C.

Section-A
10 x 2 = 20

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

Question 1.
Differentiate Osmosis from Diffusion.
Answer:

Osmosis	Diffusion
1) “The movement of water from low concentrated place to High concentrated place through a semi-permeable membrane”.	1) The movement of gases from high concentrated place to low concentrated place spontaneously.

Question 2.
Describe the terms quiescence and dormancy.
Answer:
The condition of a seed when it is unable to germinate because of external conditions normally required are not present is called quiescence. The condition of a seed when it fails to germinate because of internal conditions even though external conditions are suitable Is called dormancy.

Question 3.
What is transducer on? Who discovered it and in which organism?
Answer:
The transfer of genetic material from one bacterium to another through bacteria phase is called transduction. It was discovered by Lederberg and Zinder (1951) in Salmonella typhimurium.

Question 4.
Who proposed the chromosome theory of Inheritance?
Answer:
Walter Sutton and Theodore Boyen.

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

Question 6.
In a typical DNA molecule, the proportion of Thymine is 30% of the N bases. Find out the percentages of the other Nbases.
Answer:
A = T, G ≡ C
30 = 30 20 ≡ 20
Adenine = 30%, Guanine = 20%, Cytosine = 20%

Question 7.
What are cloning vectors? Give an example.
Answer:
Vectors used for multiplying the foreign DNA sequences are called cloning vectors. Ex: Plasmids, Bacteriophages, and Cosmids.

8. What is Green Revolution? Who is regarded as father of green revolution?
Answer:
A substantial and dramatic Increase in agricultural production is called green revolution. Norman Borlaug is regarded as the Father of Green Revolution.

Question 9.
Give two examples of fungi used In SCP production.
Answer:

1. Candida utilise
2. Saccharomyces Cerevisiae
3. Chaetomium Cellulolyticum.

Question 10.
What Nucleopolyhedro virus is being used for nowadays?
Answer:
The viruses are excellent organisms for species-specific, narrow-spectrum insecticidal properties. They have been shown no negative affect on plants, mammals, birds, fishes, etc.

Section-B
6 x 4 = 24

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

Question 11.
What is meant by plasmolysis? How ¡s it useful in our daily life?
Answer:
The phenomenon of shñnkage of protoplast due to the osmotic diffusion of water from the cells into the surrounding environment is called plasmolysis. When a cell is placed in a hypertonic solution, water moves out due to osmosis, resulting in shrinkage of protoplast away from the walls. Such cells is said to be plasmolysed. It is useful in our daily life, as the salting of pickles and preserving of fish and meat in salt.

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

1. Roots of legumes release sugars and amino acids which attached Rhizobium. They get attached to epidermal and root hair cells of the nos.
2. The root hair curls and the bacteria invade the root hair.
3. An infection thread produced carrying the bacteria into 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 13.
Explain different types of co-factors.
Answer:
Cofactors are two types.
1. Metal Ion cofactor: Metallic cations get tightly attached to the apoenzyme are called metalloenzymes. Eg: Cu⁺² cytochrome oxidase.
2. Organic cofactors: They are two types.

• Coenzyme: They are small organic molecules which are loosely associated with the apoenzyme. Eg: Thiamine pyrophosphate, Vitamin B.
• Prosthetic group: They are the organic cof actors which are tightly bounded to the apoenzyme. Eg: Haeme is the prosthetic group of enzyme Peroxidase.

Question 14.
Define R.Q. Write a short note on R.Q.
Answer:
The respiratory quotient is defined as the ratio between the volume of CO₂ released to volume of O₂ absorbed during respiration. It is measured by Ganong’s respirometer.
R.Q = \(\frac{\text { Number of } \mathrm{CO}_2 \text { molecules released }}{\text { Number of } \mathrm{O}_2 \text { molecules absorbed }}\)
The R.Q. value depends on the type of respiratory substrate used.

1. Carbohydrates: When carbohydrates are used as substrates, the volume of CO₂ molecules released are equal to the volume of O₂ molecules consumed. So the R.Q value is one
C₆H₁₂O₆ + 6O₂ + 6H₂O → 6CO₂ + 12H₂O + energy
RQ = \(\frac{6 \mathrm{CO}_2}{6 \mathrm{O}_2}=\frac{6}{6}\) = 1

2. Fats: When fats are used in Respiration, the R.Q value is less than one because the volume of CO2 molecules released are less than the amount of 02 molecules consumed.
2 (C₅₁H₉₈O₆) + 145 O₂ → 102 CO₂ + 98 H₂O + energy
RQ = \(\frac{102}{145}\) = 7

3. Proteins: When proteins are used as respiratory substrates, R.Q value would be about 0.9.

Question 15.
Explain the structure of T-even bacteriophage.
Answer:

1. The viruses which attack bacteria are called Bacteriophages. They were discovered by Twort (1915).
2. Felix ‘d’Herelle (1917) coined the term Bacteriophage.
3. Bacteriophages are tadpole-shaped with a large head and a tail.
4. The head is hexagonal and is capped by hexagonal pyramid, measures about 65 x 95 nm.
5. The head Is formed with several cap-someres, each of which is a single protein.
6. The head protein forms a semipermeable membrane enclosing the folded double-stranded DNA which is 1000 times longer than the phage.
7. The tail is composed of several parts present around central core.

Question 16.
Mention the advantages of selecting pea plant for experiment by Mendel.
Answer:

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 male and female parts.
4. It can be self-fertilized conveniently.
5. It has a short life cycle and produces large number of offsprings.

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 or 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 codes for methionine and also acts as initiator codon. (Dual role).

Question 18.
List out the beneficial aspects of transgenic plants.
Answer:
Plants with desirable characteristics created through gene transfer methods are called Transgenic plants. Beneficial aspects are:
a) Transgenic crop plants are efficient because they have many beneficial traits like virus resistance, insect resistance, and herbicide resistance.

1. Papaya in resistant to papaya ring spot virus
2. Bt. cotton is resistant to insects.
3. Transgenic tomato plants are resistant to bacterial pathogen pseudomonas.
4. Transgenic potato plants are resistant to fungus Phytophthora.

b) Transgenic plants which are suitable for food processing are produced with improved nutritional quality.
Eg:

• Transgenic tomato “Flavr Savr” are bruise resistant i.e., suitable for storage and transport due to delayed ripening and offers longer shelf life.
• Transgenic Golden Rice ‘Taipei’ is rich in vitamin A and prevents blindness.

c) Transgenic plants are used for hybrid seed production.
Eg.: Male sterile plants of Brassica napus are produced. This will eliminate the problem of manual emasculation and reduce the cost of hybrid seed production.

d) Transgenic plants have been shown to express the genes of insulin, interferon, human growth hormones, antibiotics, antibodies etc.

e) Transgenic plants are used as bio-reactors for obtaining commercially useful products, specialised medicines, and antibodies on large scale is called molecular farming.

f) Transgenic plants tolerant to abiotic stresses caused by chemicals, cold, drought, salt, heat, etc.,

• Basmati variety of rice was made resistant against biotic and abiotic stresses.
• Round-up-ready soybeans is herbicide tolerant.

Section-C
2 x 8 = 16

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

Question 19.
Explain the reactions of Kelvin cycle.
Answer:
Melvin Calvin, Andrew Benson, and James Basham discovered this pathway in chiorella and hence named as Calvin cycle. It is also called Photosynthetic Carbon Reduction cycle [PCR cycle] or reductive pentose phosphate pathway.

The Calvin cycle can be described in three stages:

1. Carboxylation
2. Reduction and
3. Regeneration.

1. Carboxylation: In this, six CO₂ molecules reacts with six molecules of RUBP to form 12 molecules of PGA in the
presence of water and RUBISCO, an enzyme
6CO₂ + 6RUBP + 6H₂O → 12 PGA

2) Reduction:
1. 12 molecules of PGA are phosphorylated to 12 molecules of DPGA in the presence of phosphoglycerokinase.
12PGA + 12ATP →12 DPGA + 12 ADP

2. 12 DPGA molecules are reduce to 12 G₃P molecules in the presence of G₃P dehydrogenase.
12 DPGA + 12 NADPH + H⁺ 12 G₃P + 12NADP + 12H₂PO_4^–

3. Regeneration: of 12 G₃P molecules, 2G₃P molecules are transported from chloroplast to cytosol. Here, they are utilised In the synthesis of hexose and then sucrose. Remaining 10 G₃P are utilised to regenerate RUBP molecules.

4. Of 10 G₃P molecules, 4 are isomerised to 4 DHAP (Dihydroxy acetone phosphate) in the presence of Isomerase.
4G₃P → 4DHAP

5. 2 molecules of G3P condense with 2 DHAP molecules to form fructose 1, 6 Biphosphate in the presence of aldolase.
2G₃P + 2DHAP → 2F₁, 6BiP

6. 2 molecules of F1, 6Bip undergoes dephosphorylation to form F6P in the presence of phosphotase.
2F₁6Bip → 2F6P + 2iP

7. 2 molecules of F6P combines with 2 molecules of G₃P to form two molecules of Erythrose 4 phosphate and 2 molecules of xylulose-5-phosphate in the presence of transketolase.
2F6P + 2G₃P → 2XMP + 2EMP

8. Two molecules of EMP combines with 2DHAP to form 2 molecules of SHDP (Sedoheptulose -1, 7 biphosphate) in the presence of aldolase.
2EMP+2DHAP → 2SHDP

9. Two SHDP molecules undergoes dephosphorylation to 2SHMP in the presence of phosphatase inorganic phosphate is released:
2SHDP → 2SHMP + 2iP

10. 2SHMP molecules combines with 2G₃P to form 2 molecules of ribose monophosphate (RMP) and 2 molecules of XMP in the presence of transketolase.
2SHMP+2G₃P → 2RMP+2XMP

11. 4 molecules of XMP (7, 10) are epimerised to form 4 molecules of RUMP in the presence of epimerase.
4XMP → 4RUMP

12. 2 RMP molecules are isomerised to 2 RUMP molecules in the presence of Isomerase
2RMP → 2RUMP

13. 6 RUMP molecules are phosphorylated to form 6 RUBP molecules in the presence of kinase.
6RUMP+6ATP → 6RUBP+6ADP

The overall equation is
6CO₂ + 6RUBP + 6H₂O + 18ATP + 12 NADPH + H’ → Glucose + 6 RUBP + 18 ADP + 12 NADP
Hence for every CO₂ molecules entering Into the Calvin cycle, 3 ATP and 2 NADPH are required.

Enzymes are :

1. RUBISCO
2. Phosphoglycerokinase
3. G₃P dehydrogenase
4. Triosephosphate Isomerase
5. Aldolase
6. Fructose 1, 6 Biphosphotase
7. Transketolase
8. Aldolase
9. Sedoheptulose – 1,7 biphosphotase
10. Transketolase
11. Epimerase
12. Ribose 5 Phosphate Isomerase
13. Ribulose -5 – phosphokinase

Question 20.
Explain briefly the various processes of recombinant DNA technology.
Answer:
The important methods in recombinant DNA technology are performed through genetic engineering.

They are:

• Isolation of a desired gene
• Insertion of Isolated gene into a suitable vector
• Introduction of recombinant vector into a host and
• Selection of the transformed host cells.

I) Isolation of a desired gene:

• The desired gene is isolated from the donor cell. Normally bacteria are the source of desired genes.
• The cell walls of bacteria are degraded with the help of enzymes.
• The cell membranes are lysed with the help of detergents.
• By treating the cellular constituents with phenols and suitable nucleases and by subjecting to gradient centrifugation, pure DNA is isolated.
• The purified DNA is cut into a number of fragments by restriction endonucleases.
• The restriction enzymes cleave DNA molecules in two ways.

i) In one way they cut both strands of DNA at exactly opposite points to each other. This results in DNA fragments with blunt ends or flush ends, where two strands end at the same point. Such cut is generally termed as even cut.

ii) But commonly, most enzymes cut the two strands of DNA double helix at different locations. Such a cleavage is generally termed as staggered cut This generates protruding ends i.e., one strand of DNA double helix extends some bases beyond the other. Since the target site is palindromic in nature, the protruding ends generated by such a cleavage have complementary base sequence. As a result, they readily pair with each other and such ends are called cohesive or sticky ends. This stickyness of the ends facilitates the action of the enzyme DNA ligase. When cut by the same restriction enzyme, the resultant DNA fragments have the same kind of ‘sticky ends’ and these can be ‘joined together readily by using DNA ligases.

E.g.: The restriction enzyme E coRI.
E – The first letter, represents the name of genus Escherichia.
Co – The next two letters, represent the species Escherichia Coil.
The letter R is derived from the name of the strain.
Roman numbers following the names indicate the order in which the enzymes were Isolated from the strain of bacteria. This enzyme specifically recognises GAA sites on the DNA and cuts it between GandA(G↓A).

The resultant fragments are separated from each other by gel electrophoresis.
The desired fragments are selected by Southern blotting technique.

II) Insertion of Isolated gene into a suitable vector:

• The selected fragments of DNA are inserted into a suitable vector to produce a large number of copies of genes.
• This is called gene cloning.
• There are two major types of vectors, namely plasmids and bacteriophages.

• Among the two types, plasmids are the ideal cloning vectors.
• To isolate a plasmid, the Bacterial cell is treated with EDTA (Ethylene diamine tetra acetic acid) along with lysozyme
  enzyme to digest the cell wall.
• Then the bacterial cell is subjected to centrifugation in sodium lauryl sulphate to separate the plasmid.
• The plasmid DNA is cut with the help of restriction endonuclease.
• The circular plasmid is converted into a linear molecule having sticky ends.
• The two sticky ends of linear plasmid are joined to the ends of desired gene by DNA ligase.
• The plasmid containing foreign DNA segments is called recombinant DNA (r DNA) or Chimeric DNA.

III) Introduction of recombinant vector Into a suitable host:

• The rDNA molecule is introduced into suitable bacterial host cell by transformation.
• The cell containing r DNA is called transformed cell.
• Bacterial cell walls are not permeable to recombinant vectors, but keeping in dil. Calcium chloride renders the bacterial cell wall permeable to recombinant vectors.
• The rDNA replicates within the host cell.
• The transformed cell grows on the culture medium. Each daughter cell contains r DNA.

IV) Selection of transformed host cells:
1) The selection of transformed cells depends on the nature of gene which is cloned.
2) It can be done in two ways. They are:

• Without using probes
• By using probes.

a) Without using probes:

• If the gene is cloned for antibiotic resistance, the cells are first incubated on a medium without antibiotic for one hour, to allow the antibiotic resistance gene to be expressed.
• Then the cells are placed on a medium with an antibiotic for selection of colonies containing rDNA.
• The cells which have expressed the gene will survive and the others die.

b) By using probes: When transformed cells are cultured on the nutrient medium, several cells are produced. To select the cells containing the desired genecolony hybridization method is used. In this gene specific probes are used. A probe is a small fragment of single-stranded RNA or DNA which is tagged with radioactive, molecules. It can search out complimentary DNA sequences from an organism.

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 Hybndisatlon 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 a 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.