BOTANY, Paper – II

(English Version)

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.

   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.  

Section: A

Marks: 10 x 2 = 20

SECTION-A

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

1. Define water potential. What is the value of water potential of pure water?

• Water Potential: It is the measure of the free energy of water molecules in a system. It determines the direction of water movement.
• Value for Pure Water: The water potential of pure water at standard temperature and pressure is zero (Ψw = 0).

2. Distinguish between apoenzyme and cofactor.

• Apoenzyme: The protein part of an enzyme. It is inactive on its own.
• Cofactor: The non-protein part of an enzyme. They can be organic (coenzymes) or inorganic (metal ions). Cofactors are essential for the activity of the apoenzyme.

3. What are pleomorphic bacteria? Give an example.

• Pleomorphic Bacteria: These are bacteria that exhibit a variety of shapes and sizes.
• Example: Mycoplasma are pleomorphic bacteria as they lack a rigid cell wall.

4. Who proposed the Chromosome Theory of Inheritance?

• Walter Sutton and Theodor Boveri are credited with independently proposing the Chromosome Theory of Inheritance.

5. What is the function of the codon-AUG?

• AUG is the start codon. It signals the beginning of protein synthesis and codes for the amino acid methionine.

6. What are the components of a nucleotide?

• A nucleotide consists of three components:
  • Pentose Sugar: A five-carbon sugar (either ribose in RNA or deoxyribose in DNA).
  • Nitrogenous Base: A purine (adenine, guanine) or pyrimidine (cytosine, thymine/uracil).
  • Phosphate Group: One or more phosphate groups attached to the sugar.

7. Name the nematode that infects the roots of tobacco plants. Name the strategy adopted to prevent this infestation.

• Nematode: Meloidogyne incognita (Root-knot nematode) is a common pest that infects tobacco roots.
• Prevention Strategy: Crop rotation, using resistant varieties, and soil fumigation are some strategies to prevent root-knot nematode infestations.

8. What is green revolution? Who is regarded as Father of Green Revolution?

• Green Revolution: It refers to a period of significant increases in agricultural production worldwide, particularly in the developing world, from the 1940s to the 1970s.
• Father of Green Revolution: Dr. Norman Borlaug is regarded as the Father of the Green Revolution.

9. Why does ‘Swiss Cheese’ have big holes? Name the bacteria responsible for it.

• Reason for Holes: The holes in Swiss cheese are formed by the production of carbon dioxide gas by the bacteria Propionibacterium freudenreichii.

10. What is Nucleopolyhedro virus being used for nowadays?

• Nucleopolyhedrovirus (NPV): These are a group of viruses that infect insects. They are being used as biological control agents for pest management in agriculture. They are specific to certain insect pests and have minimal impact on other organisms.

Section: B

Marks: 6 x 4 = 24

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

11. “Transpiration is a necessary evil”. Explain.

Answer:

The statement “Transpiration is a necessary evil” highlights the dual nature of this process in plants.

Necessary:

• Water and Mineral Transport: Transpiration is the driving force behind the upward movement of water and minerals from roots to leaves through the xylem. This is crucial for plant growth and survival.
• Cooling Effect: Transpiration helps to cool the plant by evaporating water from the leaves, preventing overheating, especially in hot and sunny conditions.
• Turgor Pressure Maintenance: Transpiration helps maintain turgor pressure in plant cells, which is essential for various physiological processes, including cell expansion and growth.

Evil:

• Water Loss: Transpiration leads to significant water loss from plants, which can be detrimental in arid or drought-prone environments.
• Wilting: Excessive transpiration can lead to wilting, especially in plants with poorly developed root systems or under water stress.
• Reduced Photosynthesis: Severe water loss due to transpiration can limit the availability of water for photosynthesis, thereby reducing plant productivity.

Therefore, while transpiration is essential for plant survival and growth, it also comes with certain drawbacks. Hence, the statement “Transpiration is a necessary evil.”

12. Describe in brief photorespiration.

Answer:

Photorespiration is a metabolic pathway that occurs in plants, particularly in C3 plants. It is a wasteful process that competes with photosynthesis.

Key points:

• Rubisco’s Dual Function: Rubisco, the enzyme responsible for carbon fixation in photosynthesis, can also bind to oxygen under certain conditions.
• Oxygenation Reaction: When Rubisco binds to oxygen instead of carbon dioxide, it catalyzes the oxygenation of RuBP (ribulose-1,5-bisphosphate), resulting in the formation of phosphoglycolate.
• Energy Expenditure: Photorespiration involves a series of reactions that consume ATP and NADPH, reducing the efficiency of photosynthesis.
• Carbon Loss: Photorespiration results in the release of carbon dioxide, further reducing photosynthetic output.

Factors Favoring Photorespiration:

• High temperatures
• High oxygen concentrations
• Low carbon dioxide concentrations

Significance:

While photorespiration is generally considered wasteful, it may have some protective functions under certain conditions. For example, it may help to protect plants from photooxidative damage under stress conditions.

13. Explain the steps involved in the formation of root nodule.

Answer:

Root nodules are specialized structures formed on the roots of legumes (e.g., beans, peas, lentils) in symbiosis with nitrogen-fixing bacteria (primarily Rhizobium). Here are the steps involved in root nodule formation:

1. Recognition and Attachment: The legume root secretes flavonoids that attract Rhizobium bacteria. The bacteria attach to the root hairs.
2. Root Hair Curling and Infection Thread Formation: The bacteria cause the root hair to curl, forming an infection thread that grows into the cortex of the root.
3. Bacterial Release and Nodule Development: The bacteria are released from the infection thread and differentiate into bacteroids within the plant cells. These infected cells divide and enlarge, forming the root nodule.
4. Nitrogen Fixation: The bacteroids convert atmospheric nitrogen (N2) into ammonia (NH3) in a process called nitrogen fixation. The plant provides the bacteria with carbohydrates and other nutrients, while the bacteria provide the plant with fixed nitrogen.

Key points:

• The interaction between the legume and Rhizobium is a mutualistic symbiosis, benefiting both organisms.
• Root nodules are essential for the nitrogen cycle, as they contribute significantly to the fixation of atmospheric nitrogen into a form usable by plants.

14. Write a note on agricultural/horticultural applications of auxins.

Answer:

Auxins are plant hormones that play a crucial role in various aspects of plant growth and development. They have numerous applications in agriculture and horticulture:

• Root Initiation: Auxins stimulate root formation from cuttings, enhancing the success of vegetative propagation techniques like grafting and layering.
• Fruit Development: Auxins promote fruit set and development. They can be used to prevent fruit drop in some crops.
• Weed Control: Synthetic auxins like 2,4-D are used as herbicides to selectively kill broadleaf weeds in cereal crops.
• Apical Dominance: Auxins produced by the apical bud inhibit the growth of lateral buds, maintaining the apical dominance of the plant.
• Parthenocarpy: Auxins can induce parthenocarpy, the development of fruit without fertilization, in some plants.

15. Explain the structure of T-even bacteriophages.

Answer:

T-even bacteriophages, such as T4, are complex viruses that infect bacteria (specifically E. coli). They have a characteristic structure:

• Head: An icosahedral head that contains the viral DNA genome.
• Neck: A short connecting region between the head and the tail.
• Tail: A long, contractile tail that injects the viral DNA into the host cell.
• Tail Fibers: These fibers attach the phage to the bacterial cell surface.
• Base Plate: A structure at the end of the tail that helps in attaching to the host cell.
• Tail Pins: Short, spike-like structures that penetrate the bacterial cell wall.

This complex structure allows the T-even bacteriophages to efficiently attach to, inject their DNA into, and replicate within their bacterial host.

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

Answer:

Mendel chose pea plants (Pisum sativum) for his experiments due to several advantages:

• Easy to Grow and Cultivate: Peas are relatively easy to grow and maintain in experimental gardens.
• Short Life Cycle: Peas have a short life cycle, allowing Mendel to conduct multiple generations of experiments within a short period.
• Self-Pollinating: Pea plants are naturally self-pollinating, which allowed Mendel to control the mating of plants and obtain pure-breeding lines.
• Distinct Traits: Peas exhibit several easily observable and contrasting traits, such as flower color, seed shape, and pod shape.
• Large Number of Offspring: Pea plants produce a large number of offspring, providing sufficient data for statistical analysis.

These characteristics made pea plants an ideal model system for Mendel’s experiments on inheritance.

17. What are the differences between DNA and RNA?

Answer:

DNA and RNA are both nucleic acids, but they differ in several key aspects:

• Sugar: DNA contains deoxyribose sugar, while RNA contains ribose sugar.
• Nitrogenous Bases: DNA contains adenine (A), guanine (G), cytosine (C), and thymine (T), while RNA contains adenine (A), guanine (G), cytosine (C), and uracil (U).
• Structure: DNA is typically double-stranded, forming a double helix, while RNA is usually single-stranded.
• Function: DNA primarily stores genetic information, while RNA plays various roles in gene expression, including mRNA (messenger RNA) for protein synthesis, tRNA (transfer RNA) for amino acid transport, and rRNA (ribosomal RNA) for protein synthesis.

18. Give a brief account of Bt cotton.

Answer:

Bt cotton is a genetically modified crop that has been engineered to produce its own insecticide. It contains a gene from the bacterium Bacillus thuringiensis (Bt), which produces a protein toxic to certain insect pests like bollworms. This reduces the need for chemical pesticides, making it a more environmentally friendly and sustainable option for cotton farmers.

Section: C

Marks: 2 x 8 = 16

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

19. Give an account of glycolysis. Where does it occur? What are the end products?

Answer:

Glycolysis is the first stage of cellular respiration, a metabolic pathway that breaks down glucose into pyruvate. It is an anaerobic process, meaning it does not require oxygen.

Where does it occur?

Glycolysis occurs in the cytoplasm of all living cells.

End products:

• Pyruvate: The primary end product of glycolysis. It is a three-carbon molecule that can enter the Krebs cycle (aerobic respiration) or undergo fermentation (anaerobic respiration) under anaerobic conditions.
• ATP: A net gain of two ATP molecules is produced during glycolysis through substrate-level phosphorylation.
• NADH: Two molecules of NADH are also produced, which can be used to generate more ATP in the electron transport chain.

Key steps of glycolysis:

1. Energy Investment Phase: Glucose is phosphorylated twice, using two ATP molecules, to form fructose-1,6-bisphosphate.
2. Cleavage Phase: Fructose-1,6-bisphosphate is split into two three-carbon molecules of glyceraldehyde-3-phosphate (G3P).
3. Energy-Yielding Phase: Each G3P molecule is oxidized and phosphorylated, generating ATP and NADH. This phase results in the net production of four ATP molecules and two NADH molecules.

20. Give an account of the tools of recombinant DNA technology.

Answer:

Recombinant DNA technology involves manipulating DNA sequences to create new combinations of genetic material. Several key tools are essential for this technology:

• Restriction Enzymes: These enzymes cut DNA molecules at specific recognition sites, generating fragments with defined ends.
• DNA Ligase: This enzyme joins DNA fragments together, creating recombinant DNA molecules.
• Vectors: These are DNA molecules that can carry foreign DNA into a host cell. Examples include plasmids, bacteriophages, and cosmids.
• Host Cells: These are living cells that can replicate recombinant DNA molecules. Commonly used host cells include bacteria (e.g., E. coli), yeast, and mammalian cells.
• Polymerase Chain Reaction (PCR): This technique allows for the amplification of specific DNA sequences, making it possible to obtain large quantities of DNA from small samples.

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, also known as micropropagation, is a technique for growing plant cells, tissues, or organs in a sterile environment on a nutrient medium. It involves the following steps:  

 

1. Explant Selection: A small piece of plant tissue (explant) is taken from the parent plant.
2. Sterilization: The explant is sterilized to prevent contamination by microorganisms.
3. Culture Establishment: The explant is placed on a nutrient medium containing hormones and other growth factors.
4. Cell Division and Growth: The plant cells divide and grow to form a callus, a mass of undifferentiated cells.
5. Organogenesis or Somatic Embryogenesis: The callus can be induced to differentiate into shoots and roots, or directly into somatic embryos.
6. Plantlet Regeneration: The plantlets are transferred to soil and grown into mature plants.

Advantages of Tissue Culture Over Conventional Plant Breeding:

• Rapid Propagation: Tissue culture allows for the rapid multiplication of plants in a short period.
• Clonal Propagation: It produces genetically identical plants (clones), preserving desirable traits.
• Disease-Free Plants: Tissue culture can be used to produce disease-free plants by eliminating pathogens.
• Production of Novel Plants: It enables the production of genetically modified plants and the generation of novel varieties through genetic engineering.
• Conservation of Endangered Species: Tissue culture can be used to conserve rare and endangered plant species.