AP Inter 2nd Year Botany Question Paper May 2018

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AP Inter 2nd Year Botany Question Paper May 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 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.
Differentiate osmosis from diffusion.
Answer:

Diffusion	Osmosis
1. The movement of gases or molecules from high concentrated place to low concentrated place is called diffusion.	1. The movement of water from low concentrated place to high concentrated place through a semi-permeable membrane is called osmosis.

Question 2.
Define the law of limiting factors proposed by Blackman.
Answer:
In a process participated by a number of separate factors the rate of the process is limited by the factor which is present in minimal value is called law of limiting factor.

Question 3.
What are pleomorphic bacteria? Give one example.
Answer:
Some bacteria keep on changing their shape depending upon the type of environment and nutrients available are called pleomorphic bacteria. Ex: Acetobacter.

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. It results in a diseased condition called sickle cell anemia.

Question 5.
What is the difference between exons and introns?
Answer:

Exons	Introns
1. Expressed sequences.	1. Intervening sequences.
2. They appear in nature or processed RNA.	2. They do not appear in nature or processed RNA.

Question 6.
Distinguish between Heterochromatin and Euchromatin. Which of the two is transcriptionally active?
Answer:

Hetero chromatin	Euchromatin
a. Coiled tightly.	a. Coiled loosely.
b. Takes more stain.	b. Takes less stain.
c. More coils.	c. Few coils.
d. More DNA is present Euchromatin is transcriptionally active.	d. Less DNA is present.

Question 7.
How does one visualize DNA on 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 tobacco plant. Name the strategy adopted to prevent this infestation.
Answer:
Meloidegyne incognita. A novel strategy adopted to prevent this infestation is RNA interference.

Question 9.
Which two species of sugarcane were crossed for better yield?
Answer:
Sacharum barben and Sacharum officinanim were crossed for better yield, high sugar etc.,

Question 10.
What are fermentors?
Answer:
Large vessels in which microbes are grown in large numbers on an individual scale are called fomenters.

Section – B
6 x 4 = 24

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

Question 11.
‘Transpiration is necessary evil”. Explain.
Answer:
Beneficial effects:

1. It helps in passive absorption of water.
2. It also helps in passive absorption of mineral salts by mass flow mechanism.
3. It is the main force for ascent of sap.
4. It regulates the temperature of plant today and provides cooling effect.
5. 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 obstructing gaseous exchange

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

1. Roots of legumes release sugars and 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, 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 lead to nodule formation.
• A mature nodule is complete with ascu1ar tissues continuous with those of the root.

Question 13.
Explain the different types of cofactors.
Answer:
cofactors are of two types.
They are:
1. Metal ion cofactors: Metallic Cations get tightly attached. to the apoenzyme are called metalloenzymes.
Ex: Cu²⁺ Cytochrome oxidase.
2. Organic cofactors: They are of two types.

• Co-enzymes: They are small organic molecules which are loosely associated with the apoenzyme.
• Prosthetic group: They are the organic cofactors which are tightly bound to the apoenzyme. Ex: Haem is the group for Peroxidase.

Question 14.
Write a note on agricultural/Horticultural applications of Auxins.
Answer:

1. IBA, NAA and LAA help to initiate rooting in stem cuttings. widely used for plant propagation in horticulture.
2. Auxins like 2, 4- D, 2, 4, 5 – T act as herbicides and kills broad-leaved dicot weeds to prepare weed-free lawns.
3. Auxins stimulate fruit growth. Ex: Tomato.
4. Auxins induce flowering in pineapple.
5. Auxins prevent premature leaf and fruit drop.

Question 15.
Explain the structure of T even Bacteriophages.
Answer:

1. The viruses which attack bacteria are called Bacteria phages. They were discovered by Twort (1915).
2. Felix d’Herelle (1917) coined the term Bactei1ophage.
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 capsomeres, 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.
Explain the law of dominance using a monohybrid cross.
Answer:
Characters are controlled by discrete units called factors. They occur in pairs. In a dissimilar pair of factors pertaining to a character, one number of the pair dominates the other. The law of dominance is used to explain the expression of only one of the parental characters in a monohybrid cross in the F1 and the expression of both in the F₂ generations. It also explains the proportion of 3: 1 obtained at the F₂ generation.

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 strandnds of nucleotides.
2. Deoxyribose sugar is present.	2. Ribose sugar is present.
3. Thymine, cytosin 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 genetic material.	6. RNA is non genetic material.
7. It does not involve 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 purrng is A=U,G = C.
10. It is present more in 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 a 1:1 ratio.

Question 18.
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 resistance. Ex : Pqaya is resistant to papayaring spot virus. But 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 Phytophthera.

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 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 ušefull 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 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 molecular of pyruvic acid is called Glycolysis. It was given by gustav Embden, Otto Mayerhof and J. Parnas 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 2 NADPH+H are the end products. In aerobic respiration, pyruvic acid, 2 NADPH+H⁺ are completely oxidised through TCA cycle, ETS pathway and produce 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 →F 1,6Bip + ADP

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

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

6. 2 molecules of G3P undergoes dehydrogenation in the presence of dehydrogenase to form 2 molecules of 1,3 DPGA.
2G3P+2NAD⁺ → 2-1,3DPGA+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 2 PGA in the presence of mutage.
2-3 PGA →12-2 PGA

9. 2- PGA losses water molecules to form 2- phosphoeno1- 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 + 2ADP → 2ATP + 2PA

Question 20.
Give a brief account of the tools 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 molecules at a particular pair by recognising a specific sequence of six base pairs, called recogniton sequence for Hind II. Today more than 900 restrict 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 (coli) the Iater 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 enzymes belong to a larger class of enzymes called nucleases.

They are of two types:

• a) Exonucleases which remove nucleotides from the ends of the DNA.
• b) 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 GAATT’3 sites on the DNA and cut it between G & A results in the formation of sticky ends or cohesive end pieces. This stiçkyness 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 a 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: (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 tetra cycline or kanamycin etc. are useful selectable markers for E. coil.

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 pH is adjusted to 5.6 to 6.0. Growth regulators like auxins, and 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) sterilization: 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 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, 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 shoot (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.