AP Inter 2nd Year Botany Model Paper Set 3 with Solutions

Thoroughly analyzing AP Inter 2nd Year Botany Model Papers Set 3 helps students identify their strengths and weaknesses.

AP Inter 2nd Year Botany Model Paper Set 3 with Solutions

Time : 3 Hours
Max Marks:60

Section – A
10 x 2 = 20 Marks

Answer all questions.

Question 1.
What are potins? 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, allowing molecules up to the size of small proteins to pass through.

Question 2.
What is apical dominance? Name the growth hormone that causes it.
Answer:
Growing Apical bud inhibits the growth of Axillary buds is called Apical dominance. It is caused by Auxins.

Question 3.
What is conjugation? Who discovered it and in which organism?
Answer:
Transfer of DNA from one Bacterium (Donor) to the recipient Bacterium through direct contact is called conjugation. It was first observed in Escherichia cou by Iedeeberg and Tatum (1946).

Question 4.
What is point Mutation? Give an example.
Answer:
Mutation which occurs in a single base pair of DNA is called point mutation. E.g: Sickle cell anemia.

Question 5.
Who proved that DNA is genetic material? What is the organism they worked on?
Answer:
Alfred Hershey and Martha Chase (1952). They worked with viruses that infect Bacteria, bacteriophages.

Question 6.
What are the components of a nucleotide?
Answer:
A nucleotide has three components-a nitrogenous base, a pentose sugar, and a phosphate group.

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, Cosrnids.

Question 8.
What is downstream processing?
Answer:
Separation and purification of products before they are ready for marketing is called downstream processing.

Question 9.
Give two examples of fungi used in SCP production.
Answer:
Candida citilis (Tolula yeast), saccharomyces cerevisiae (Baker’s yeast), chaetomium cellulolyticum.

Question 10.
How do Biofertilizers enrich the fertilities of the soil’?
Answer:
Biofertilizers are organisms that enrich the nutrient quality of the soil. E.g.:

1. Rhizobium bacteria present in the root nodules of legumes, helps in nitrogen fixation.
2. Mycorrhiza facilitates phosphorous absorption by the plant from the soil.
3. Blue-green algae add organic matter to the soil and increase its fertility.

Section-B
6 x 4 = 24 Marks

Answer any six questions.

Question 11.
Explain pressure flow hypothesis of translocation of sugars in plants.
Answer:
Glucose is prepared at the source as converted to sucrose, then moved into the companion cells and then into the living phloem sieve tube cells by active transport. Water in the adjacent xylem moves into the phloem by osmosis. As osmotic pressure builds up, the phloem sap moves to the cells which will use the sugar converting it into energy starch or cellulose. As sugars are removed, the osmotic pressure decreases and water moves out of the phloem.

To explain this, Munch conducted one experiment. In this, he took two osmometres A and B. ¡n ‘B’ he took concentrated sugar solution and in A, he took pure water. These two bulbs A and B are connected by a tube ‘C, He then placed A and B bulbs in X and Y water tubes which are connected by the ‘Z’. Due to osmosis. Water moves from A to B, B to A through ‘C’ and then A-X and finally X-Y through Z occurs until the solution becomes isotonic. In this, he compared ‘B’ bulb as source ‘A’ bulb as sink. ‘C’ tube as phloem X and Y tubes are xylem.

Question 12.
Explain in brief how plants absorb essential elements.
Answer:
The process of absorption can be demarcated into two main phases. In the I phase, there is an initial uptake of ions into the free space or outer space of cells – the apoplast. It is a passive process. In the second phase of uptake, the ions are taken in slowly into the inner space – the symplczst. The passive movement of ions into the apoplast from the cell along the concentration gradient usually occurs through ion channels. The entry or exit of ions to and from the symplast against the concentration gradient requires metabolic energy which is an active process.

Question 13.
Compare and contrast the absorption spectrum of chlorophylls and Carotenoids.
Answer:
Pigments are substances that have an ability to absorb light at specific wavelengths. Chlorophyll absorbs lights of different wavelengths and is the major pigment. Other pigments, chlorophyll b and carotenoids also absorbs light energy and transmit it to chlorophyll a more over they protect the chlorophyll a form photo oxidation.

Question 14.
Write any four physiological effects of cytokinins in plants.
Answer:
Cytokinins induces cell division

1. They help to produce new leaves, chloroplasts in leaves, lateral shoot growth, and adventitious shoot formation.
2. Cytokinins help to overcome apical dominance.
3. They promote nutrient mobilisation which help in the delay of senescence.
4. Cytokinins help in the opening of stomata by increasing the concentration of K⁺ ions in guard cells.

Question 15.
Mention the differences between lytic and lysogenic cycles.
Answer:

Lytic cycle	Lysogenic cycle
1. At the end of lytic cycle, bacterial cell undergoes lysis.	1. Bacterial cell does not undergo immediate lysis.
2. The entry of viral DNA brings about the degradation of bacterial DNA.	2. Bacterial DNA is not destroyed and viral DNA get incorporated.
3. Prophages are not formed and the virulent phages do not allow bacteria to survive.	3. Prophages persist in close relationships for long period even when bacterial cell undergoes many division cycles.
4. The viruses are called virulent phages. Ex: T-even phages.	4. The viruses are called temperate phages. Ex: Coliphage- λ

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

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

Question 17.
Write briefly on nucleosomes.
Answer:
Nucleosome is a bead-like structure of chromosomes. It consists of eight histone molecules and a DNA segment of about 150 base pairs Each Nucleosome is separated from one another by a linker DNA sequence of about 50 base pairs. Nucleosome helps to fold DNA into a compact form in the interphase nucleus. Otherwise, the length of a chromosome, when linear is many order of magnitude greater than the diameter of the nucleus.

Question 18.
Write short notes on restriction enzymes.
Answer:
Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coil were isolated in the year 1963. One of these added methyl group of DNA and the other cut DNA. The latter was called restriction endonuclease. The first restriction endonuclease – Hind II which cut DNA molecules at a particular point 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 letter comes from the Genus (Escherichia), and the second two letters from the species of the Prokaryotic cell [coil], the letter ‘R is derived from the name of strain.

Roman number indicate the order in which the enzyme were isolated from that strain of Bacteria. Restriction enzymes belong to a larger class of enzymes called nucleases. They are of two types.

• Endonucleases which make cuts at specific location within the DNA.
• Exonucleases which remove nucleotides from the ends of the DNA.

Section – C
2 x 8 = 16 Marks

Answer any two of the following.

Question 19.
Enumerate the assumptions that we undertake in making the respiratory balance sheet. Are these assumptions valid for a living system? Compare fermentation and aerobic respiration in this context?
Answer:
It is possible to make calculations of the net gain of ATP for every glucose is oxidised. These calculations can be made only on certain assumptions that:

1. There is a sequential, orderly pathway functioning with one substrate forming the next and with glycolysis, TCA cycle, ETS pathway following one after another.
2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
3. None of the intermediates in the pathway are utilised to synthesize any other compound.
4. Only glucose is being transferred – no other alternative substrates are entering in the pathway at any of the intermediary stages. But these kind of assumptions are not really valid in a Living system all pathways work simultaneously and do not take place one after another.

Differences:

Fermentation	Aerobic respiration
1. Partial breakdown of glucose.	1. Complete oxidation of glucose.
2. End products are CO2 and Ethyl alcohol.	2. End products are CO2, and H2O.
3. 2 ATP molecules are formed.	3. 36 ATP are formed.

Question 20.
Give an account of biotechnological applications in agriculture and other fields.
Answer:
Green Revolution: Substantial and dramatic increases in agricultural production which was termed as Green Revolution by William Gaud, the director of United States Agency for International Development (USAID) Norman Börlaug is regarded as father of Green revolution Dr. M. Swaminathan and his team is the success of green revolution in our country. It was possible due to use of improved varieties, chemical fertilizers and pesticides, improved irrigation as facilities, adoption of better agricultural management strategies, reforms etc.

Gene revolution: It is the application of biotechnology in food production. It is of great potential to farmers as it provides them with disease-free planting material and develops plants that resist pests and diseases, reducing use of chemicals that harm free environment and human health.

It can provide diagnostic tools and vaccines that help control devastating animal diseases. It can also improve the nutritional quality of staple foods such as rich and create new products for health and industrial uses.

Planst, animals, bacteria, and fungi whose genes have been altered by manipulation are called genetically modified organisms. GM plants made crops more tolerant to abiotic stresses (cold, drought, suit, heat).

They reduce reliance on chemical pesticides. E.g.: Bt cotton. They helped to reduce post-harvest losses. They increased efficiency of mineral usage by plants. They enhanced nutritional value of food. E.g. : Vitamin A-enriched rice. Pest resistant plants produced by biotechnology, which could decrease the amount of pesticides used. Bt Toxin gene has been cloned from Bacillus thuringiensis and has ‘been expressed in plants to provides resistance to insects. Without the need of insecticides.

Bt. Cotton : Some strains of Bacillus thuringiensis produce proteins that kill certain insects such as lepidopterans (tobacco budworm army worm), 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 Boll worm, a major pest of cotton. The worm feeding on the leaves of Bt. Cotton plants becomes lethargic and sleepy thereby causing less damage to the plant. 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 lAc and Cry II Ab control the cotton boliworms and cry lAb controls corn bores.

b) Pest-resistant plants: Several nematodes parasitize a wide variety of plants and animals. A nematode, Meloidegyne incognitia infects the roots of tobacco plants and causes a great reduction in yield. To prevents this infestation. Process of RNA interference was adopted. Using Agrobacterium vectors, nematode-specific genes were introduced into the host plant. The introduction of DNA was such that it produced both sense and anti-sense RNAS in the host cells. These two RNAS formed a double-stranded RNA that initiated RNAI and thus silenced the specific mRNA of the nematode. The parasite could not survive in a transgenic host expressing specific interfering RNA. The transgenic plant therefore got itself protected from the parasite.

Other applications:
Cloned DNAs are cultilized in the commercial synthesis of hormones like insulin, interferons, and vaccines. Microorganisms are commercially exploited for production of vitamins antibiotics and other commercial chemicals at lower costs.

Insulin used for diabetes was earlier extracted from pancreases of slaughtered cattle and pigs. Human insulin is now made in bacteria. Yet its structure is identical to natural molecule. Gene therapy is a corrective therapy to be taken for hereditary diseases.

Here genes are inserted into a person’s cells and tissues to treat a disease. rDNA technology, PCR and enzyme-linked immunosorbent assay are some of the techniques that serve the purpose of early diagnosis. ELISA is based on the principle of Antigen. Antibody interaction. Infection by a pathogen can be detected by the presence of Antigens. DNA fingerprinting successfully helped forensic science in the search of criminals and also solveing parentage disputes.

Question 21.
What are the three options to increase food production? Discuss each giving the sailent features, merits, and demerits.
Answer:
Three options to increase food production are

1. Development of disease resistance plants.
2. Development of insect pests resistance plants.
3. Development of plants with improved food quality.

Modern methods of Breeding plants:
1) Plant breeding for disease resistance: Several fungal, bacterial and viral pathogens affect the yield of cultivated crop species. In this situation, breeding and development of cultivars resistant to disease enhances food production. It also helps reduce the dependence on the use of fungicides and bacteriocides. Breeding is carried out by conventional breeding techniques or by mutation breeding. The method of breeding for disease resistance is that of hybridization and selection. By this some crop varieties are produced which were disease resistance to bacteria and fungi. They are

Crop	variety	Resistance to disease
1. Wheat	Himgiri	Leaf and stripe rust, hill bunt
2. Cauliflower	Pusa swarnim,	White rust
3. Cowpea	Pusa Komal	Dacterial blight Chilly Mosaic virus
4. Chilli	Pusa Sadabahar	Tobacco mosaic virus and leaf curl

Merits:

• It provides superior variety directly after selection.
• It protects the variability from genetic erosion by collecting germplasm.
• Very quick and economical method of crop improvement.
• It provides protection to the crop by introducing into new disease-free areas.
  Eg: Coffee and rubber.

Demerits:

• Introduction of noxious weeds.
• Introduction of disease.
  Eg: Late blight of potatoes from Europe in 1883.
• ornamentals turned weeds.
  Eg : Water hyacinth and lantana camera.
• Threat to ecological balance.