Section A

(10 × 2 = 20)
Note: Answer all questions. Each answer may be limited to 5 lines.

  1. Name two elements whose symptoms of deficiency first appear in younger leaves.
    Answer:
    • Nitrogen
    • Sulfur
      Explanation: Deficiency of nitrogen and sulfur in plants typically affects the younger leaves first, as these elements are essential for protein synthesis and chlorophyll formation.
  2. What is the primary acceptor of CO2 in C3 plants? What is the first stable compound formed in the Calvin cycle?
    Answer:
    • The primary acceptor of CO2 in C3 plants is Ribulose bisphosphate (RuBP).
    • The first stable compound formed in the Calvin cycle is 3-phosphoglycerate (3-PGA).
      Explanation: In C3 plants, CO2 is fixed to RuBP by the enzyme RuBisCO, resulting in the formation of two molecules of 3-PGA.
  3. What is lysozyme and what is its function?
    Answer:
    • Lysozyme is an enzyme that acts as an antimicrobial agent.
    • It functions by breaking down the cell wall of certain bacteria, specifically by cleaving the glycosidic bonds in peptidoglycan.
      Explanation: Lysozyme is part of the immune system and helps protect the body from bacterial infections.
  4. What will be the phenotypic ratio in the offsprings obtained from the following crosses? a) Aa x aa
    b) AA x aa
    c) Aa x Aa
    d) Aa x AAAnswer:
    a) 1:1
    b) All are dominant (100% AA)
    c) 3:1
    d) All are dominant (100% Aa)
    Explanation:
    • In cross Aa x aa, there is a 50% chance of getting dominant (A) and recessive (a) alleles, resulting in a 1:1 ratio.
    • In AA x aa, all offspring will be heterozygous (Aa), leading to 100% dominant phenotypes.
    • In Aa x Aa, the expected phenotypic ratio is 3:1 (3 dominant, 1 recessive).
    • In Aa x AA, all offspring will inherit the dominant allele from the AA parent, leading to 100% dominant phenotypes.
  5. Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?
    Answer:
    • Heterochromatin is tightly packed, dark-stained chromatin that is transcriptionally inactive.
    • Euchromatin is loosely packed, light-stained chromatin that is transcriptionally active.
      Explanation: Euchromatin is where active transcription occurs, whereas heterochromatin is generally associated with gene silencing.
  6. Given below is the sequence of the coding strand of DNA in a transcription unit.
    5’A A T G C A G C T A T T A G G-3′
    Write the sequence of:
    a) Its complementary strand.
    b) The mRNA.Answer:
    a) Complementary strand (template strand):
    3’T T A C G T C G A T A A T C C-5′
    (or)
    5’C C T A A T A G C T G C A T T-3′
    b) mRNA sequence:
    5’AAUG C AGCUAUUAG G-3′
    Explanation:
    • The complementary strand is the one that is used for transcription to synthesize mRNA.
    • The mRNA is synthesized in the 5′ to 3′ direction using the template strand of DNA.
  7. Give different types of cry genes and pests which are controlled by the proteins encoded by these genes.
    Answer:
    • Different types of cry genes include cryIAc, cryIIAb, and cryIAb.
    • These genes control pests such as cotton bollworms, corrupt cotton moths, and other insect pests like caterpillars.
      Explanation: Cry proteins are produced by Bacillus thuringiensis (Bt) and are used as biological control agents to control insect pests.
  8. Name the nematode that infects the roots of tobacco plants. Name the strategy adopted to prevent this infestation.
    Answer:
    • The nematode that infects tobacco plant roots is Meloidogyne incognita.
    • The strategy adopted to prevent this infestation is RNA interference (RNAi), which silences the gene responsible for nematode reproduction.
      Explanation: RNAi is used as a genetic method to suppress the expression of nematode-specific genes, thereby reducing the damage caused by the nematode.
  9. Why does ‘Swiss cheese’ have big holes? Name the bacteria responsible for it.
    Answer:
    • ‘Swiss cheese’ has big holes due to the production of carbon dioxide (CO2) by bacteria during fermentation.
    • The bacteria responsible for this are Propionibacterium shermanii.
      Explanation: These bacteria convert lactic acid into propionic acid and CO2, which forms bubbles and causes the characteristic holes in the cheese.
  10. Name an immunosuppressive agent. From where it is obtained?
    Answer:
  • Cyclosporine A is an immunosuppressive agent.
  • It is obtained from the fungus Trichoderma polysporum.
    Explanation: Cyclosporine A is used to prevent organ transplant rejection by suppressing the immune response.

Section B

(6 × 4 = 24)
Note: Answer any six questions. Answer may be limited to 20 lines.

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

Answer:
Transpiration, the process of water loss from plants through evaporation primarily from the stomata, is often referred to as a “necessary evil” because it has both beneficial and harmful effects on plants.

  • Necessary: Transpiration plays a vital role in the plant’s water and nutrient uptake. It creates a suction force that helps draw water and minerals from the roots to the leaves (a process known as the “transpiration stream”). This also helps in cooling the plant and maintaining its temperature, preventing overheating.
  • Evil: Excessive transpiration can lead to water loss, especially under hot or dry conditions. This can cause dehydration and hinder the plant’s growth and development. In extreme cases, it may lead to wilting.

Therefore, while transpiration is essential for nutrient transport and temperature regulation, it can be detrimental if water loss exceeds the plant’s ability to absorb water.

12. Explain the mechanism of opening and closing of stomata.

Answer:
Stomata are small pores found on the surface of leaves and stems that allow gas exchange (CO₂ in, O₂ out) and transpiration. The opening and closing of stomata are regulated by the guard cells surrounding each stoma.

  • Opening of Stomata: When the guard cells absorb water and become turgid, they swell and bend apart, causing the stoma to open. This process is driven by the active transport of potassium ions (K⁺) into the guard cells, which increases their osmotic pressure, drawing water from surrounding cells.
  • Closing of Stomata: When the guard cells lose water and become flaccid, they collapse and close the stoma. This happens when potassium ions are actively transported out of the guard cells, leading to a decrease in their osmotic pressure, causing water to leave by osmosis.

The opening and closing of stomata are influenced by environmental factors such as light (which causes stomata to open during the day for photosynthesis) and humidity (which affects water loss).

13. Write briefly about enzyme inhibitors.

Answer:
Enzyme inhibitors are molecules that bind to enzymes and decrease their activity. There are two main types of enzyme inhibitors:

  • Competitive Inhibitors: These inhibitors resemble the enzyme’s substrate and compete for binding to the active site of the enzyme. When they bind, they prevent the substrate from binding, thereby inhibiting enzyme activity. The inhibition can be overcome by increasing the substrate concentration.
  • Non-Competitive Inhibitors: These inhibitors bind to a site other than the enzyme’s active site, known as the allosteric site. Binding causes a conformational change in the enzyme, reducing its activity. This type of inhibition cannot be overcome by increasing the substrate concentration.

Enzyme inhibitors play important roles in regulating metabolic pathways and are also used as drugs to treat diseases, such as antibiotics inhibiting bacterial enzymes.

14. What are the physiological processes that are regulated by ethylene in plants?

Answer:
Ethylene is a plant hormone that regulates several key physiological processes in plants, including:

  • Fruit Ripening: Ethylene is widely known for its role in promoting fruit ripening. It activates the expression of genes responsible for the conversion of starches into sugars, the breakdown of cell walls, and the production of pigments.
  • Leaf Abscission: Ethylene triggers the process of leaf shedding (abscission), particularly in deciduous plants, by weakening the cell walls in the abscission zone.
  • Flowering: Ethylene influences the flowering of certain plants. For example, it can promote flowering in pineapples and some other plants.
  • Stress Response: Ethylene helps plants respond to stress factors such as drought, flooding, and mechanical wounding by regulating the expression of stress-related genes.

Overall, ethylene acts as a key regulatory hormone in plant development and stress responses.

15. How are bacteria classified on the basis of number and distribution of flagella?

Answer:
Bacteria are classified based on the number and distribution of flagella, which are tail-like structures that help bacteria move. The main types of flagella arrangement are:

  • Monotrichous: A single flagellum at one pole of the bacterium.
  • Lophotrichous: A group of flagella at one or both poles of the bacterium.
  • Amphitrichous: One flagellum at each pole of the bacterium.
  • Peritrichous: Flagella are distributed over the entire surface of the bacterium.

Flagella play an essential role in bacterial motility, enabling bacteria to move toward or away from environmental stimuli (a process known as taxis).

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

Answer:
Mendel chose the pea plant (Pisum sativum) for his experiments on inheritance because it had several advantages:

  1. Distinctive Traits: Pea plants exhibit clear, easily distinguishable traits (e.g., seed color, pod shape, flower position) that could be tracked through generations.
  2. Self-pollination: Peas can self-pollinate, ensuring purebred lines. They can also cross-pollinate, allowing for controlled hybridization experiments.
  3. Short Life Cycle: Pea plants have a relatively short life cycle, allowing Mendel to observe multiple generations in a short period.
  4. Large Number of Offspring: Pea plants produce a large number of seeds, providing a statistically significant sample size for analysis.
  5. Simple to Grow and Maintain: Peas are easy to cultivate, making them ideal for repeated experiments.

These advantages made the pea plant an excellent model organism for studying inheritance patterns.

17. How many types of RNA polymerases exist in cells? Write their names and functions.

Answer:
In eukaryotic cells, there are three main types of RNA polymerases:

  1. RNA Polymerase I:
    • Function: It is responsible for synthesizing ribosomal RNA (rRNA), which forms the major structural component of ribosomes.
  2. RNA Polymerase II:
    • Function: It synthesizes messenger RNA (mRNA), which carries genetic information from DNA to the ribosome for protein synthesis. RNA polymerase II also synthesizes small nuclear RNAs (snRNAs) involved in splicing.
  3. RNA Polymerase III:
    • Function: It synthesizes transfer RNA (tRNA) and other small RNAs, such as 5S rRNA and some other non-coding RNAs.

In prokaryotes, there is a single type of RNA polymerase that synthesizes all RNA molecules.

18. What is a bio-reactor? Describe briefly the stirring type of bio-reactor.

Answer:
A bio-reactor is a vessel or container used to carry out biological reactions, where microorganisms, cells, or enzymes are used to produce desired products such as drugs, biofuels, or chemicals. Bio-reactors are essential in industrial microbiology, biotechnology, and fermentation processes.

Stirring Type Bio-reactor:
In a stirring bio-reactor, the contents are continuously stirred to ensure proper mixing and aeration, which is crucial for the growth of microorganisms and the efficient production of the desired product. These bio-reactors are equipped with a motorized stirrer or impeller that maintains uniform distribution of nutrients, gases, and temperature. This type of bio-reactor is commonly used in fermentation processes where aerobic conditions are required for the microorganisms to thrive.

Section C

(2 × 8 = 16)
Note: Answer any two questions. Answer may be limited to 60 lines.

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:
Glycolysis is the first step in cellular respiration, breaking down glucose (a six-carbon sugar) into two molecules of pyruvate (three carbons each). It occurs in the cytoplasm of the cell and is an anaerobic process (does not require oxygen). Glycolysis involves a series of 10 enzyme-catalyzed reactions and can be divided into two phases:

  • Energy Investment Phase: Two ATP molecules are consumed to phosphorylate glucose and its derivatives.
  • Energy Payoff Phase: Four ATP molecules are produced, along with two NADH molecules, resulting in a net gain of two ATP and two NADH.

The steps of glycolysis:

  1. Glucose is phosphorylated to glucose-6-phosphate by hexokinase.
  2. This molecule is converted into fructose-1,6-bisphosphate.
  3. Fructose-1,6-bisphosphate is split into two molecules of glyceraldehyde-3-phosphate (G3P).
  4. Each G3P is processed to form pyruvate, producing NADH and ATP in the process.

End products of glycolysis:

  • 2 pyruvate molecules
  • 2 ATP molecules (net gain)
  • 2 NADH molecules

Fate of the products:

  • In aerobic respiration (with oxygen):
    • Pyruvate is transported into the mitochondria and converted into acetyl-CoA by the enzyme pyruvate dehydrogenase.
    • Acetyl-CoA enters the citric acid cycle (Krebs cycle), where it is further oxidized to produce more ATP, NADH, and FADH₂.
    • NADH and FADH₂ are used in the electron transport chain to generate a large amount of ATP via oxidative phosphorylation.
  • In anaerobic respiration (without oxygen):
    • Pyruvate undergoes fermentation.
      • In lactic acid fermentation, pyruvate is converted to lactic acid (or lactate), regenerating NAD⁺ for glycolysis.
      • In alcoholic fermentation, pyruvate is converted to ethanol and CO₂, also regenerating NAD⁺.

Thus, glycolysis is the same in both aerobic and anaerobic conditions, but the fate of pyruvate and the amount of ATP produced differ depending on the presence or absence of oxygen.

20. Give a brief account of the tools of recombinant DNA technology.

Answer:
Recombinant DNA (rDNA) technology involves the manipulation and combination of DNA from different sources to create new genetic combinations. Various tools are used in this process to isolate, manipulate, and introduce genes into organisms.

  1. Restriction Enzymes (or Restriction Endonucleases):
    • These enzymes act as molecular scissors that cut DNA at specific sequences. They are crucial for isolating the gene of interest from a larger DNA molecule and also for inserting this gene into a vector.
    • Example: EcoRI, BamHI.
  2. DNA Ligase:
    • This enzyme is used to join two strands of DNA by forming phosphodiester bonds between the sugar-phosphate backbone. It is essential for sealing the inserted gene into the vector (such as a plasmid).
  3. Vectors:
    • Vectors are DNA molecules used to carry the foreign DNA into the host cell. Common vectors include plasmids (circular DNA in bacteria), bacteriophages, and artificial chromosomes (BACs and YACs).
    • These vectors typically contain sequences that allow for the insertion of foreign DNA and selection markers for identifying recombinant DNA.
  4. Plasmids:
    • Plasmids are small, circular DNA molecules found in bacteria. They replicate independently of the bacterial chromosomal DNA. They are often used as vectors for gene cloning.
  5. Polymerase Chain Reaction (PCR):
    • PCR is a technique used to amplify specific DNA sequences. It uses DNA primers, a heat-stable DNA polymerase (like Taq polymerase), and cycles of temperature changes to create millions of copies of a targeted gene or DNA segment.
  6. Gene Cloning:
    • This technique involves the insertion of a gene of interest into a vector, followed by introducing the recombinant vector into a host cell (like bacteria). The host cell replicates the vector, producing copies of the gene.
  7. Electroporation and Heat Shock:
    • These methods are used to introduce recombinant DNA into host cells. Electroporation uses electrical pulses to increase cell membrane permeability, while heat shock is a brief exposure to high temperature that facilitates DNA uptake.
  8. Southern Blotting:
    • This technique is used for detecting specific DNA sequences in a DNA sample. The DNA is transferred from a gel to a membrane and hybridized with a labeled probe.
  9. Gene Gun and Microinjection:
    • These methods are used to introduce DNA into plant or animal cells. The gene gun uses high-pressure helium to shoot DNA-coated gold particles into cells, while microinjection involves directly injecting DNA into the nucleus of a cell using a fine needle.

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:
As a botanist working in plant breeding, the process of developing and releasing a new plant variety involves several key steps:

  1. Selection of Parental Plants:
    • The first step is to select parent plants with desirable traits, such as high yield, disease resistance, or drought tolerance. The plants may be of the same species or different species (hybridization).
    • Pure-line Selection or Cross-breeding can be used, depending on the breeding goals.
  2. Hybridization or Cross-Breeding:
    • The selected parent plants are cross-bred to combine their favorable traits. In cross-breeding, pollen from one plant is used to fertilize another plant’s flower, resulting in a hybrid offspring.
    • This step may involve controlled pollination to ensure the right combinations of genes are inherited.
  3. Selection of Offspring:
    • The hybrid plants are grown, and the offspring are evaluated for their traits. Plants showing superior characteristics such as increased yield, resistance to pests, and better adaptability to the environment are selected.
    • This process is repeated over several generations to stabilize the desirable traits.
  4. Back-Crossing:
    • If the hybrid offspring do not fully exhibit the desired traits, back-crossing is used to cross the hybrid with one of the parent plants. This helps in transferring specific traits from the parent plant back into the hybrid.
  5. Testing and Evaluation:
    • The new plant variety is tested in various environmental conditions to assess its performance under different climatic and soil conditions.
    • Field trials are conducted to ensure that the new variety is superior to existing varieties in terms of productivity, quality, and resistance to diseases or pests.
  6. Seed Production:
    • Once the new variety has been successfully developed and tested, large-scale seed production is undertaken. This ensures a consistent supply of seeds for commercial distribution.
  7. Registration and Approval:
    • The new variety must be registered with agricultural authorities and undergo evaluation for distinctness, uniformity, and stability (DUS test). The variety is also tested for its agronomic performance and compliance with regulatory standards.
  8. Release to Farmers:
    • After approval, the new variety is officially released to the public. Information on its cultivation and benefits is provided to farmers. Extension services may be provided to assist with the adoption of the new variety.
  9. Post-Release Monitoring:
    • After the release, the performance of the new variety is continuously monitored. This helps to address any potential issues and ensures that it continues to meet the needs of farmers.

By following these steps, a new variety with enhanced traits can be developed and made available to improve agricultural productivity and sustainability.