AP Inter 2nd Year Botany Question Paper March 2019

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AP Inter 2nd Year Botany Question Paper March 2019

Time: 3 Hours
Max. Marks: 60

Note: Read the following instructions carefully:

1. Answer all the 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 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 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 Sections 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 apoplast and symplast?
Answer:
The path of water within the plant system that moves without crossing membranes is called apoplast. it is a non-living path. The path of water movement in the plant systems that crosses the membrane is called symplast. It is a living system.

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.

Question 3.
What is a genophore?
Answer:
The bacterial chromosome which is the main genetic material is called genophore.

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

Question 5.
What are the components of a transcription unit?
Answer:
Components of transcription unit are

• A promoter
• The structural gene and
• A terminator.

Question 6.
Define stop codon. Write the codons.
Answer:
The genetic codes which do not code for amino acids is called stop codons. Stop codons are UGA, UAG, UAA.

Question 7.
What is downstream processing?
Answer:
After completion of the biosynthetic stage, the product has to be subjected through a series of processes before it is ready for marketing as a finished product. The processes include separation and purification, which are collectively referred to as downstream processing.

Question 8.
Can a disease be detected before its symptoms appear? Explain the principle involved.
Answer:
Yes. Amplification of nucleic acid through PCR even though microorganisms are in low concentration.

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

Question 10.
Name any two industrially important enzymes.
Answer:
Pectinase, Protease, Streptokinase, Upase.

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 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 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:
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 body 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 14.
Write the physiological responses of gibberellins in plants.
Answer:

1. Gibberellins delay senescence.
2. Spraying of gibberellins on sugarcane crops increases the length of the stem thus increasing the yield as much as 20 tonnes per acre.
3. GA hastens the maturity period of conifers thus leading to early seed production.
4. GA also promotes bolting in cabbage, beet, etc.
5. They also promote parthenocarpic fruits in grapes and tomatoes.
6. Gibberellins favour the formation of male flowers in cucurbit.
7. Gibberellins cause fruits like apples to elongate and improve their shape.
8. Gibberellins cause to increase in length of grape stalks.

Question 15.
Explain the lytic cycle with reference to certain viruses.
Answer:
T-even bacteriophages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle, which is a five-step process.

They are:
a) Adsorption: The tail fibers of phages help in attachment to the complementary receptor sites on the bacterial cell wall.

b) Penetration: The tail sheath of phage contracts and the tail core is driven in through the bacterial cell wall. When the tip of the core reaches the plasma membrane, the DNA of the phage passes through the tail core and enters into the bacterial cell. The capsid remains outside the bacterial cell and is called ghost.

c) Biosynthesis: When the phage DNA reaches the cytoplasm of the host cell, many copies of phage DNA, enzymes and capsid proteins are synthesized, using the cellular machinery of the hot cell.

d) Maturation: Bacteriophage DNA and capsids are assembled into complete virions. The period of time between the infection and the appearance of the mature virus is called the eclipse period.

e) Release: The plasma membrane of the host cells gets dissolved or lysed due to the viral enzyme, Lysozyme. The bacterial cell wall breaks releasing the newly, produced phage particles or virions.

Question 16.
Explain the law of Dominance using a monohybrid cross.
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 the 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.
Write briefly on nucleosomes.
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 genetic 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 purring 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 nucleus.
11. Purines and pyrimidines exist in 1: 1 ratio.	11. Purines and pyrimidines does not exist in 1: 1 ratio.

Question 18.
List out the beneficial aspects of transgenic plants.
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 inaçtive 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 creates 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 IAb and cry hAb control the cotton bollworms and cry lAb controls corn borer.

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 Krebs cycle.
Answer:
Glucose is broken down into 2 molecules 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 organisms. 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 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 isomerized 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-t-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. DHAP does not undergo oxidation in further reactions, so gets isomerised to another G3P in the presence of Isomerase.
1 DRAP → 1G₃P

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

7. 2 mol. of 1, 3 DPGA undergoes dephosphorylation in the presence of phosphor 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.
3PGA → 2-2 PGA

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

10. 2 PEPA mols. are phosphorylates 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 is endonuclease. Hind il which cut DNA molecules at a particular pair by recognising a specific sequence of six base pairs, is 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 (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:

• 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 helx at different locations such a cleavage is known as staggered cut. E.CORI recognizes 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 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 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, 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 plantš. Similarly, retroviruses have also been disarmed and are now used to deliver desirable genes into animal cells.

Question 21.
Describe the tissue culture technique. 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 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) 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) Acciamatization 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 the 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.