Telangana Inter 2nd Year 2024 : Botany Paper-I Important Question with Answers

Telangana Inter 2nd Year Class 12 exams have started and you have very little time left for the Botany - Paper 1 exam. Therefore, we are providing important questions in this article. You can study them well and score well in your exams.

Students often find themselves in need of effective study materials to prepare for challenging subjects like Botany - Paper 1. Recognizing the importance of comprehensive revision important short, and long question have emerged as a valuable resource for students.

Very Short Answer Type Questions

Question 1. What are porins? What role do they play in diffusion? [March 2018]

Answer:

1. Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria.
2. Molecules upto the size of small proteins to pass through the channels formed by porins due to facilitated diffusion.

Question 2. Define water potential. What is the value of water potential of pure water?

Answer:

1. Water potential is a relative term, which refers to the chemical potential of pure water to that of chemical potential of water in a solution.
2. Water potential of pure water not under pressure is zero (0) at standard temperature.

Question 3. Define hydroponics.

Answer:

1. The technique of growing plants in a specified nutrient solution is known as hydroponics.
2. Julius Von Sachs (1860) demonstrated this technique for the first time.

Question 4. How do you categorize a particular essential element as a macro or micronutrient?

Answer: 1. Macronutients : Elements needed in high quantities and are present is large amounts (in excess of lOmmole Kg^-1 of dry matter) in plant tissues.

2. Micronutrients : Trace elements that are needed in very small amounts and present is less than lOmmole Kg^-1 of dry matter in plants.

Question 5. How are prosthetic groups different from co-factors?

Answer: 1. Prosthetic group : An organic co-factor that is tightly bound to the apoenzyme. Eg. Hemee in peroxidase enzyme.

2. Cofactor : Non protein part of a holo enzyme. It may be a metal ion or orgain compound. Eg. Zinc in carboxy peptidase.

Question 6. What is meant by ‘feedback inhibition’?

Answer:

1. Feedback inhibition: The end product of a chain of enzyme catalysed reactions inhibits the enzyme of the first reaction as part of homeostatic control of metabolism.
2. Eg. Pyruvic acid in Glycolysis.

Question 7. Name the processes which take place in the grana and stroma regions of chloroplasts.

Answer: 1. Grana : Light reactions – trapping the light energy by membrane systems and synthesis of ATP and NADPH.

2. Stroma : Dark reactions – carbon fixation leading to synthesis of sugars.

Question 8. Can chloroplasts be passed on to progeny? How?

Answer:

1. Yes, nearly equal number of chloroplasts among 2 daughter cells formed from a mother cell during mitosis.
2. Maternal inheritance (cytoplasm of egg) in a sexually reproducing plant.

Question 9. Energy is released during the oxidation of compounds in respiration. How is this energy stored and released as and when it is needed?

Answer:

1. Energy contained in respiratory substrates is released in a series of slow step wise reactions controlled by enzymes and stored as ATP.
2. ATP is broken down whenever and wherever energy needs to be utilised.

Question 10. Explain the term ‘Energy currency’. Which substance acts as energy currency in plants and animals?

Answer:

1. ATP is broken down whenever and wherever energy needs to be utilised is cells of living organisms.
2. ATP acts as energy currency in plants and animals.

Question 11. Define plasticity. Give an example.

Answer:

1. Plasticity is the ability of plants to follow different pathways in response to the environment or phases of life to form different kinds of structures.
2. Heterophylly is an example of plasticity.

Question 12. What is the disease that formed the basis for the identification of gibberellins in plants? Name the causative fungus of the disease.

Answer:

1. “Bakane” (foolish seedling) disease of rice seedlings.
2. It is caused by a fungal pathogen Gibberella fujikuroi.

Question 13. Write briefly on the occurrence of micro-organisms.

Answer:

1. Microorganisms are ubiquitous. They are found in soil, water, air, and inside living beings.
2. They occur in a variety of foods and can withstand extreme cold, heat, and drought conditions.

Question 14. Define Microbiology.

Answer:

1. Microbiology is a branch of biological science that deals with the study of microorganisms like protozoa, microscopic algae, fungi (yeasts and molds) bacteria and viruses.
2. It is concerned with structure, function, classification, ways to control and using the activities of microorganisms.

Question 15. Mention the living and non-living characters of viruses.

Answer: Living characters:

1. They contain nucleic acid.
2. They maintain genetic continuity through multiplication and undergo mutations.
3. The live as obligate intracellular parasites.

Non-living characters:

1. They do not exhibit most of the life processes like growth, irritability.
2. They are inert life less molecules.
3. They are acellular.
4. They do not have metabolic system.

Question 16. What is the shape of T₄ phage? What is its genetic material?

Answer: T₄ phage is tadpole shaped with a large head and a tail.

1. Genetic material is double stranded DNA.

Question 17. What is the cross between the Fa progeny and the homozygous recessive parent called? How is it useful?

Answer:

1. The cross between the Ft progeny -having dominant phenotype and the homozygous recessive parent is called Test cross.
2. The progenies of such a cross can be easily analysed and the genotype of the test organism can be determined.

Question 18. Do you think Mendel’s laws of inheritance would have been different if the characters that he chose were located on the same chromosome?

Answer:

1. Yes, Mendel’s law of independent assortment would not be true for the genes that are located on the same chromosome.
2. Linkage refers to the physical association of genes on a chromosome and are called linked genes.

Question 19. What is the function of histones in DNA packaging?

Answer:

1. Histones are positively charged basic proteins. They are organized to form a unit of eight molecules called histone octamer.
2. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleOsome.

Question 20. Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active? [Mar. 2020]

Answer: 1. Euchromation : Regions of chromatin that are loosely packed and stained lightly. It is transcriptionally active.

2. Heterochromation : Regions of chromatin that are densely packed and stained dark. It is transcriptionally inactive.

Question 21. Define biotechnology.

Answer:

1. The European Federation of Biotechnology (EFB) defines Biotechnology as the intergration of natural science and organisms, cells, parts thereof, and molecular analogues for products and services.
2. Biotechnology is the science of utilising the properties and uses of microorganisms or to exploit cells and the cell constituents at industrial level for generating useful products essential to life and human welfare.

Question 22. What are molecular scissors? Where are they obtained from?

Answer:

1. Molecular scissors are the restrition endonucleases that recognize and cut specific nucleotide sequences of DNA.
2. They are usually obtained from Bacteria. For the first time, Herbert Boyer (1969) isolated two restriction enzymes from E.coli.

Question 23. Expand GMO. How is it different from a hybrid?

Answer:

1. GMO means Genetically Modified Organisms.
2. Hybrids are new crop varieties produced by crossing two genetically different parents whereas GMO are plants, bacteria, fungi or animals whose genes have been altered by manipulation.

Question 24. Give different types of cry genes and pests which are controlled by the proteins encoded by these genes.

Answer:

1. Genes Cry I Ac and Cry II Ab control the cotton bollworms.
2. Genes Cry I Ab controls corn borer.

Question 25. What is meant by ‘hidden hunger’?

Answer:

1. About 3 billion people over the globe are suffering from deficiencies of micro nutrients, protein and vitamins. This situation is said to be ‘Hidden hunger’.
2. More than 840 million people in the world do not have adequate food to meet their daily food and nutritional requirements.

Question 26. Name two semi-dwarf varieties of rice developed in India. [Mar. 2020]

Answer:

1. Jaya and Ratna, are two better yielding semi – dwarf varieties of rice developed in India.
2. They are derivatives of semi – dwarf varieties derived from crossing IR – 8 with TN – 1.

Question 27. Why does ‘Swiss cheese’ have big holes? Name the bacteria responsible for it. [Mar. 2020, 18; May 14]

Answer:

1. Large holes in ‘Swiss cheese’ are due to the production of a large amount of CO₂ by a Bacterium.
2. Propionibacterium sharmanii is responsible for it.

Question 28. What are fermentors? [May 17; Mar. 14]

Answer:

1. Very large vessels that are used to grow microbes for production of valuable products on an industrial scale.
2. Beverages like wine, beer, whisky, brandy and rum are produced through fermentation of malted careals and fruit juices by yeast.

Short Answer Type Questions

Question 1. Define and explain water potential. [Mar. 2019, ’18]

Answer: Water potential (Ψ_w) is a relative term to understand water movement. It represents the difference between chemical potential of water in a system and that of a pure water. Water molecules possess kinetic energy. The greater the concentration of water in a system, the greater is its kinetic energy or water potential. Hence, pure water will have greatest water potential. The water potential of pure water at standard temperatures which are not under any pressure, is taken to be zero. Solute potential (Ψ_s ) and pressure potential (Ψ_p) are the two main components that determine water potential.

Solute potential : All solutions have a lower water potential than pure water. The magnitude of this lowering due to dissolution of a solute is called solute potential denoted by Ψ_s. It is always negative. The more the solute molecules, the lower is the Ψ_s. For a solution at atmospheric pressure water potential Ψ_w = solute potential Ψ_s.

Pressure potential : The pressure which developed due to entry of water in a cell is called pressure potential. It is always positive, it is denoted as Ψ_p.

Water potential Ψ_w of a cell is affected by both solute potential and pressure potential.

Ψ_w = Ψ_s + Ψ_p

Question 2. Write short notes on facilitated diffusion.

Answer:

1. Membrane protein provides sites at which certain molecules can cross the membrane. This process is called facilitated diffusion.
2. 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.
3. Concentration gradient must already be present for molecules to diffuse even if facilitated by the proteins.
4. In’facilitated diffusion, special proteins help to move substance across the membrane without expenditure of ATP energy.
5. Facilitated diffusion cannot cause net transport of molecules from a low to a high concentration this would require input of energy.
6. Transport rate reaches a maximum when all of the protein transporters are being used.
7. Facilitated diffusion is very specific, it allows cell to select substance for uptake.
8. It is sensitive to inhibitors which react with protein side chain.

Question 3. “All elements that are present in a plant need not be essential for its survival.” Justify.

Answer: Most of the elements present in the soil enter plants through roots. All the elements that are present in a plant need not be essential for its survival. The criteria for essentiality of elements are given below :

1. The element must be absolutely necessary for supporting normal growth and reproduction. In the absence of the element the plant do not complete their life cycle or set the seeds.
2. The requirement of the element must be specific and not replaceable by another element. In other words, deficiency of any one element cannot be met by supplying some other element.
3. The element must be directly involved in the metabolism of the plant.

Question 4. Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.

Answer: Deficiency symptoms in plants are chlorosis, necrosis, inhibition of cell division, delay flowering.

Chlorosis : Chlorosis is the loss of chlorophyll leading to yellowing in leaves. This symptom is caused by the deficiency of elements N, K, Mg, S, Fe, Mn, Zn and Mo.

Necrosis : It is the death of tissue, particularly leaf tissue. It is due to the deficiency of Ca, Mg, Cu, K.

Inhibition of cell division : Cell division stops. It is due to the deficiency of lack or low level of N, K, S, Mo.

Delay flowering : Flowering is delayed due to low concentration of N, S, Mo.

Mottled leaf in citrus : It is due to deficiency of Zn,

Question 5. Explain how pH affects enzyme activity with the help of a graphical representation.

Answer:

1. Generally enzymes function in a narrow range of pH value.
2. Each enzyme shows its highest activity at a particular pH called optimum pH.
3. Activity declines both below and above the optimum value.

Question 6. Explain the importance of (ES) complex formation.

Answer: The formation of (ES) complex is essential for catalysis.

1. First, the substrate binds to the active site of the enzyme, fitting into the active site.
2. The binding of the substrate induces the enzyme to alter its shape; fitting more tightly around the substrate.
3. The active site of the enzyme, now in close proximity to the substrate, breaks the chemical bonds of the substrate and the new enzyme product complex is formed.
4. The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and runs through the catalytic cycle once again.

Question 7. Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic CO₂ requirements?

Answer:

1. Succulent plants have only one kind of photosynthetic cell in which CO₂ is fixed during night and used to make glucose during day.
2. In succulent plants there is an alternative pathway of CO₂ fixatibn called Crassulacean Acid Metabolism (CAM). E.g.: Cacti Crassulaceae is one family of succulent plants.
3. CAM pathway is similar to C₄ pathway in that CO₂ is trapped by highly efficient PEP carboxylase during night time.
4. During day time, the malic acid undergoes oxidative carboxylation to form Pyruvic acid & CO₂. CO₂ is used for photosynthesis.

Question 8. Chlorophyll ‘a’ is the primary pigment for light reaction. What are accessory pigments? What is their role in photosynthesis?

Answer:

1. Chlorophyll ‘a’ is the primary pigment for light reaction other thylakoid pigments like chlorophyll b, xanthophylls and carotenoids are accessory pigments.
2. Accessory pigments also absorb light and transferthe energy to chlorophyll a.
3. They utilise wider range of wavelength of incoming light for photosynthesis.
4. They also protect chlorophyll a from photo oxidation.

Question 9. What is meant by the statement ‘Aerobic respiration is more efficient’?

Answer:

1. Aerobic respiration is the process that leads to a complete oxidation of organic substance in the presence of oxygen.
2. It releases CO₂, water and a large amount of energy present in the substrate.
3. 686 k.cal of energy is released where as in anaerobic respiration only 56 k.Cal. is released.
4. In aerobic respiration 36 ATP molecules are formed. Where as in anaerobic respiration only 2 ATP molecules are formed.

Question 10. Pyruvic acid is the end product of glycolysis. What are the three metabolic fates of pyruvic acid under aerobic and anaerobic conditions?

Answer: The fate of pyruvate depends upon the availability of O₂ and the type of organism. The three metabolic fates of pyruvic acid are

1. lactic acid fermentation
2. alcoholic fermentation under anaerobic condition and
3. aerobic respiration under aerobic condition.

Question 11. Write a note on agricultural/horticultural applications of auxins. [Mar. ’17,’14; May ’14]

Answer:

1. Auxins help to initiate rooting in stem cuttings, an application widely used for plant propagation in horticulture.
2. Auxins promote flowering e.g. in pineapples.
3. Auxins help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.
4. Removal of shoot tips usually results in the growth of lateral buds. This is widely applied in tea plantations.
5. Auxins also induce parthenocarpy. e.g. in tomatoes.
6. Auxins (2, 4D) are widely used as herbicides.
7. Auxins control xylem differentiation and help in growth.

Question 12. Write the physiological responses of gibberellins in plants. [Mar. 2019]

Answer:

1. Gibberellins promote bolting (internode elongation just before flowering) in beet, cabbages and many plants with rosette habit.
2. The ability to cause an increase in the length of axis is used to increase the length of grapes stalks.
3. Gibberellins cause fruits like apple to elongate and improve their shape.
4. Gibberellins delay senescence. Thus, fruits can be left on the tree longer so as to extend the market period.
5. GA₃ is used to speed up the malting process in brewing industry.
6. Spraying Gibberellins on sugarcane crop increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre.
7. Spraying juvenile conifers with Gibberellins fastens the maturity period, thus leading to early seed production.

Question 13. Explain the importance of Microbiology.

Answer: Microbiology is a branch of biological science that deals with the study of micro-organisms. Micro-organisms are useful in several ways for the welfare of human society.

1. Soil fertility : Micro-organisms decompose dead plants and animals thereby enrich the soil with nutrients which are utilized by plants. Microbes also play vital role in recycling of elements like C, N, O, S, and P.

2. Antibiotics : Alexander Fleming isolated an antibiotic Penicillin from a fungus, Penicillium notatum. Waksman isolated an antibiotic Streptomycin from a bacterium, Streptomyces griseus.

3. Industrial products : Industrial products like enzymes, amino acids, vitamins, organic acids and alcohols are commercially produced using microbes.

4. Dairy products : Lactobacillus, commonly known as Lactic Acid Bacteria (LAB) grows in milk and convert it to curd, which also improves its natural quality by increasing vitamin B12, food stuff, like cheese, yogurt are the byproducts of microbial growth.

5. Mining : Metals like Uranium can be extracted with 50% reduced cost by using microbes.

6. Tools in Genetic Engineering : Microbes are used as tools in altering the genetic make up of organisms.

7. Biocontrol agents : Micro-organisms like Bacillus thuringiensis are used to control plant diseases and pests.,

8. Production of Biogas : Biogas, a mixture of gases (predominantly methane) produced by the microbial activity. It may be used as fuel.

9. Exomicrobiology : Microbes are used for the exploration of life in the outer space.

10. Sewage disposal : Bacteria and fungi are also used in sewage disposal.

Question 14. How are bacteria classified on the basis of number and distribution of flagella?

Answer: Based on the number and distribution of flagella bacteria are classified into 4 types.

1. Monotrichous : A single flagellum is present on one side of the cell.
2. Lophotrichous : A tuft of flagella are present on one side of the cell.
3. Amphitrichous : Tufts of flagella or a single flagellum on either end of the cell.
4. Peritrichous : Many flagella are distributed all over the cell surface.

Question 15. What is ICTV? How are viruses named?

Answer: ICTV means International Committee on Taxonomy of Viruses. It regulates the norms of classification and nomenclature of viruses.

The ICTV scheme has only three hierarchial levels

– Family (including some sub – families)

– Genus

– Species

The family names end with the suffix ‘viridae’ while the genus names with ‘virus’ and the species names are common English expressions describing their nature.

Viruses are named after the disease they cause. Eg : Polio virus

Using the ICTV system, the virus that causes Acquired Immune Deficiency Syndrome (AIDS) in human beings is classified as :

Family : Retroviridae Genus : Lentivirus

Species : Human Immune Deficiency Virus (HIV).

Question 16. Explain the chemical structure of viruses. [Mar. 2020]

Answer:

1. All viruses consists of two basic components 1) core 2) capsid.
2. Core is the nucleic acid that forms the genome. Capsid is the surrounding protein coat.
3. Capsid gives shape and protection. It is made up of protein subunits called capsomeres. The no. of capsomeres is characteristic for each type of virus.
4. Virus contains its genetic information in either double stranded (ds) DNA or single stranded (ss) DNA.
5. Generally, viruses that infect plants have single stranded RNA and viruses that infect animals have double stranded DNA.
6. Bacteriophages are usually ds DNA.
7. Viral nucleic acid molecules are either circular or linear.
8. Most viruses have a single nucleic acid molecule, but a few have more than one (Eg : HIV which has two identical molecules of RNA).

Question 17. Differentiate between the following :

a) Dominant and Recessive [Mar. 2020]

b) Homozygous and Heterozygous [Mar. 2020]

c) Monohybrid and Dihybrid.

Answer: a) The character which is expressed in F₁ generation is called dominant and that which is unexpressed is called recessive.

b) Parents carrying similar genes such as TT or tt are called homozygous and the parents with unlike genes like Tt are called heterozygous.

c) Cross involving two parents differing in only one character is called Monohybrid cross. For example, cross between tall (TT) and dwarf (tt) parents. Cross involving two parents differing in two characters is called a Dihybrid cross. For example, cross between pea plant having yellow round seeds (YYRR) with homozygous pea plant having green wrinkled seeds (yyrr).

Question 18. In a Mendelian monohybrid cross, the F₂ generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.

Answer: Incomplete dominance : It is the phenomenon in which neither of the genes is completely dominant or completely recessive. As a result, the hybrid shows intermediate character. For example, the inheritance of flower colour in the dog flower. (Snapdragon or Antirrhinum sp). The cross between true breeding (homozygous) red flower (RR) and true breeding (homozygous) white flower plant (rr) F₁ (R r) was pink.

The phenotypic ratio deviates from Mendelian monohybrid ratio of 3 : 1 to 1 : 2 : 1 (Red flower – 1, Pink flowers – 2, White flower -1) since the heterozygous / hybrid shows a different phenotype genotype ratio remains the same as Mendelian ratio 1 : 2 : 1.

Question 19. Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as genetic material.

Answer: Transformation is defined as the uptake of a naked DNA molecule of the fragments of a bacterial cell and the incorporation of this DNA molecule into the recepient chromosome in a heritable form.

In 1928 Frederick Griffith performed the experiments on Bacterial transformation with Streptococcus pneumonia the bacterium causing pneumonia. During the course of his experiment he found that a living organism (bacteria) could change in physical form.

• He observed two strains of this bacterium, one forming smooth colonies with capsule (S-type) and the other forming rough colonies without capsule (R-type).
• The S-type cells are virulent while R-type cells are non-virulent.
• When live S-type cells are injected into the mice, they suffered from pneumonia and died.
• When live R-type cells are injected into the mice, the disease did not appear and the mice survived.
• When heat killed S-type cells were injected, the disease did not appear.
• When heat killed S-type were mixed with live R-type cells and injected into the mice, the mice died of pneumonia and live S-type cells were isolated from the body of mice.
• He concluded that the R-strain had some how been transformed by the heat killed S- strain bacteria, which must be due to the transfer of genetic material the transforming principle.

Question 20. Who revealed the biochemical nature of the transforming principle? How was it . done? Oswald Avery, Colin Mac Lead, and Madyn Me carty.

Answer: Avery Mac Lead and Me carty revealed the biochemical nature of the transforming principle in a Griffith experiment.

• They purified biochemicals like proteins, DNA and RNA from the heat killed S-cells to see which one could transform live R cells into S cells.
• When their fraction were added to the culture of live R-cells, DNA was able to cause transformation of R-cells into S-cells.
• They also found that protein digesting enzymes and RNA digesting enzymes did not affect transformation indicating that transforming substance is not a protein or RNA.
• Digestion with DNase did inhibit transformation, this suggests that the DNA cause transformation.

Question 21. Write short notes on restriction enzymes.

Answer: Restriction enzymes or molecular scissors belong to a class of enzymes called nucleases. It always cut DNA molecules at a particular point by recognizing a specific sequence of six base pairs known as recognition sequence.

They are of two types.

1. Exonucleases, which remove nucleotides from the ends of DNA.
2. Endonucleases, which cut the DNA at specific portions anywhere within its length.

Each restriction endonuclease recognizes a specific palindromic nucleotide sequence in the DNA Palindrome is a group of letters that forms the same words when read both forward and backward, eg: MALAYALAM. It is a sequence in DNA of base pairs that reads same on the two strands. When orientation of reading is kept same.

For example, the following sequence reads the same on the two strands in 5′ → 3′ direction as well as 3′ → 5′ direction

5′ – GAATTC – 3′

3′ – CTTAAG – 5′

The restriction enzymes are named as follows.

• The first letter of the name comes from the genus and the next two letters from the name of the species of the prokaryotic cell from which it is isolated.
• The next letter comes from the strain of the prokaryote.
• The Roman numbers following these four letters indicate the order in which the enzymes were isolated from that strain of the bacterium.

  eg : EcoRI from Escherichia coli RY13.

  Hind 11 from Haemophilus influenza.

  Bam H1 from Bacillus amyloliquefaciens.

Question 22. List out the beneficial aspects of transgenic plants. [March 2019, Mar. ’18; May ’17, ’14]

Answer: Beneficial aspects of transgenic plants are

I. Transgenic crop plants having resistance to pathogens and pests :

1. Transgenic papaya is resistant to papaya ring spot virus.
2. Bt. cotton is resistant to insects.
3. Transgenic tomato plants are resistant to the bacterial pathogen pseudomonas.
4. Transgenic potato plants are resistant to the fungus phytophthora.

II. Transgenic plants suitable for food processing technology :

Transgenic tomato ‘Flavr Savr’ is bruise resistant i.e., suitable for storage and transport due to delayed ripening and offers longer shelf life.

III. Transgenic plants with improved nutritional value :

Transgenic golden rice obtained from ‘Taipei’ is rich in vitamin A and prevents blindness.

IV. Transgenic plants useful for hybrid seed production :

Male sterile plants of Brassica napus are produced. This will eliminate the problem of manual emasculation and reduce the cost of hybrid seed production.

V. Transgenic plants tolerant to abiotic stresses caused by chemicals, cold, drought, salt, heat etc.

1. Basmati variety of rice was made resistant against biotic and abiotic stresses.
2. Round up ready soyabean is herbicide tolerant.

Question 23. What are some bio-safety issues concerned with genetically modified crops? [March 2017, 2014]

Answer: Biosafety issues concerned with genetically modified crop are :

1. There is fear of transferring allergins or toxins to humans and animals as side effects.
2. There is a rick of changing the fundamental nature of vegetables.
3. They may pose a harmful effect on biodiversity and have an adverse impact on environment.
4. There is a risk of gene pollution due to transfer of the new genes into related wild species through natural out-crossing. This may result in the development of super weeds which may be fast growing than the crops and may be resistant to weedicides.
5. They may bring about changes in natural evolutionary pattern.

Question 24. What is meant by germplasm collection? What are its benefits?

Answer: The entire collection of plants / seeds, having all the diverse alleles for all genes in a given crop is called germplasm collection.

1. Cell and tissue cultures of many plant species can be preserved maintained in a viable state for several years and used when required.
2. Plant materials from endangered species can be conserved using this method.
3. It is an ideal method for long term conservation of cell cultures producing secondary metabolites such as antibiotics.
4. Seeds which loose their viability or storage can be maintained for a long period of time.
5. Disease free plant material can be frozen and propagated whenever required.
6. Conservation of Somaclonal variations in cultures.
7. Rare germplasms developed by using Somatic hybridisation other genetic manipulation techniques can be stored.
8. Pollen conservation for enhancing longevity.
9. Germplasm banks to facilitate the exchange of information of international level.

Question 25. Why are the floes important in the biological treatment of waste water?

Answer:

1. The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it.
2. This allows vigorous growth of useful aerobic microbes into floes.
3. Floes are masses of bacteria associated with fungal filaments to form mesh like structures.
4. While growing, these microbes consume the major part of the organic matter in the effluent.
5. This significantly reduces the BOD (Biochemical Oxygen Demand) of the effluent.
6. When BOD of sewage is reduced, the effluent is passed into a settling tank, where the floes are allowed to form the activated sludge.

Long Answer Type Questions 

Question 1. Explain how plants absorb water.

Answer: Water is absorbed along with mineral solutes by the root hairs purely by diffusion. Once water is absorbed by the root hair, it moves by two distinct pathways.

a) Apoplast pathway

b) Symplast pathway

a) Apoplast pathway :

it is “the system of adjacent cell walls that is continuous throughout plant, except at the casparian strips of endodermis in roots.” Apoplastic movement of water takes place through intercellular spaces and cell walls. Movement through apoplast does not include crossing cell membrane. It is dependent on the gradient. It does not provide any barrier to water movement. The water movement is through mass flow. When the water evaporates into intercellular spaces or atmosphere, a tension develops in continuous stream of water in apoplast. Therefore mass flow of water takes place as a result of adhesive and cohesive properties of water.

b) Symplast system :

It is “the system of interconnected protoplasts.” The adjacent cells are connected through cytoplasmic strands that extend through plasmodesmata. In it, water travels through the cells – their cytoplasm; intercellular movement is through plasmodesmata. Water enters the cells only through the cell membrane, so the movement is relatively slower. The movement is again down a potential gradient. It may cytoplasmic streaming.

Most of the water flow in roots takes place by apoplast as cortical cells are loosely packed. They offer no resistance to water movement. Endodermis is impervious to water due to casparian strip. Water moves through symplast and again crosses a membrane to reach the cells of the xylem. In the endodermis, the movement of water is symplastic. This is the only way water and other solutes can enter the vascular cylinder.

Question 2. Define transpiration. Explain the structure and mechanism of opening and closing of stomata.

Answer: Transpiration is defined as the loss of water in the form of vapour from the living tissues of aerial parts of the plants.

Structure of Stomata :

Each stoma is composed of two bean shaped cells known as guard cells. In grasses, the guard cells are dumb-bell shaped. The outer walls of guard cells are thin and inner walls are thickened. The guard cells possess chloroplasts and regulate the opening and closing of stomata. Sometimes, a few epidermal cells, in the vicinity of the guard cells become specialised in their shape and size and are known as subsidiary cells. The stomatal aperture, guard cells and the surrounding subsidiary cells are together called stomatal apparatus.

Opening and closing of stomata :

The stoma is the turgor operated valve in the epidermis.

1. The immediate cause of opening or closing of the stomata is a change in the turgidity of the guard cells.
2. The inner wall of each guard cell towards the pore or stomatal aperture is thick and elastic.
3. When turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape,
4. The opening of the stoma is also aided by the orientation of the microfibrils in the cell walls of the guard cells.
5. Cellulose microfibrils are oriented radially rather than longitudinally, making it easier for the stoma to open.
6. When the guard cells lose turgor due to water loss the elastic inner walls regain their original shape, the guard cells becomes flaccid and the stoma closes.

Levitt proposed K⁺ pump theory to explain the mechanism of opening and closing of photoactive stomata.

Opening of stomata :

1. According to this theory accumulation of K⁺ ions into the guard cells from the subsidiary cells occurs in the presence of light.
2. This coupled with efflux of protons leads to increase in pH of the guard cells.
3. Accumulation of K⁺ ions into the guard cells is associated with passive influx of Cl^– ions thereby decreasing the water potential of the guard cells.
4. Water thereby enters the guard cells making them turgid.
5. As the outer walls are thin and elastic, the guard cells expand outwardly, leaving a minute pore at the centre open.

Closing of stomata :

1. At night, in the absence of light, the K⁺ and Cl^– ions move out of the guard cells due to which the water potential of guard cells increases and water starts moving out of them leading to closure of stomata.
2. Under water stress conditions, abscisic acid (ABA) a natural anti-transpirant drives the K⁺ ions out of guard cells making them close.
3. In Succulent plants, the water potential gradient established due to accumulation of organic acids at night makes the guard cells become turgid, hence stomata opens at night.

Question 3. Explain the nitrogen cycle, giving relevant examples. [Mar. 2020]

Answer: The cyclic movement of nitrogen from the atmosphere to soil and from soil back into the atmosphere through plants, animals and micro-organisms is termed as nitrogen cycle. Nitrogen cycle involves five steps :

1. Nitrogen fixation
2. Nitrogen assimilation
3. Ammonification
4. Nitrification
5. Denitrification

1) Nitrogen fixation : The process of conversion of molecular nitrogen (N₂) to ammonia or nitrogen oxides, nitrites and nitrates is termed as nitrogen – fixation. It occurs both by biological and physical method.

Biological method : Conversion of molecular nitrogen into ammonia by prokaryotes is called biological methed.

Eg : Free – living nitrogen – fixing aerobic microbes – Azotobacter – Beijernickia.

Free living nitrogen – fixing anaerobic microbes – Rhodospirillum

Cyanobacteria (blue green algae) – Nostoc & Anabaena

Symbiotic bacteria – Rhizobium (roots of leguminous plant)

Symbiotic bacteria – Frankia (roots of non – leguminous plant)

Physical or abiological method :

In nature lightning and ultraviolet radiation provide enough energy to convert nitrogen to nitrogen oxides (NO, NO₂, NO₃). Industrial combustions, forest fires, automobile exhausts and power – generating stations are also sources of atmospheric nitrogen oxides.

N₂ + O₂ → 2NO

2NO + O₂ → 2NO₂

2NO₂ + H₂O → HNO₂ + HNO₃

HNO₃ + Ca/K salts → Ca or K nitrates

2) Nitrogen assimilation :

1. The process of absorbing nitrates, ammonia to produce organic nitrogen constitutes is called nitrogen assimilation.
2. Nitrates and ammonia formed in 1st step are absorbed by plants and converted into organic nitrogen constitute like proteins, enzymes, nucleic acid etc.
3. When plants are eaten by animats, this organic nitrogen is passed into animal body.

3) Ammonification :

1. Decomposition of organic nitrogen of dead plants and animals into ammonia is called ammonification.
2. Bacteria responsible for this are called ammonifying bacteria.

4) Nitrification :

The conversion of ammonia into nitrites and nitrates by bacteria is called nitrification. Such bacteria are called nitrifying bacteria (chemo auto trophs). It occurs in two steps.

1. Ammonia is first oxidised to nitrite by the bacteria Nitrosomonas and Nitrococcus.

   2NH₃ + 3O₂ → 2NO₂^– + 2H⁺ + 2H₂O

2. The nitrite is further oxidised to nitrate with the help of the bacterium Nitrobacter.

   2NO₂^– + O₂ → 2NO₃^–

The nitrate thus formed is absorbed by plants and is transported to the leaves. In leaves, it is reduced to form ammonia that finally forms the amine group of amino acids.

5) Denitrification :

Conversion of nitrates from soil into molecular nitrogen is called denitrification. Denitrification is done by bacteria like Pseudomonas and Thiobacillus.

Question 4. Trace the events starting from the coming in contact of Rhizobiurri with a leguminous root till nodule formation. Add a note on the importance of leg haemoglobin.

Answer: Various stages of nodule formation :

1. Roots of legume plant secrete sugars, amino acids etc.
2. Attracted by this, Rhizobium bacteria move to the root. It multiply and colonise the surroundings of roots and get attached to epidermal and root hair cells.
3. The root – hairs curl and the bacteria invade the root – hair.
4. An infection thread is produced, carrying the bacteria into the cortex of the root.
5. Bacteria initiate nodule formation in the cortex of the root.
6. Then the bacteria released from the thread into the cortical cells of the host stimulate the host cells to divide. Thus leads to the differentiation of specialised nitrogen fixing cells.
7. The nodule thus formed establishes a direct vascular connection with the host for exchange of nutrients.

Development of root nodules in soyabean :

(a) Rhizobium bacteria contact a susceptible root hair, divide near it. (b) Successful infection of the root hair causes it to curl, (c) 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, (d) A mature nodule is complete with vascular tissues continuous with those of the root.

The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leg-haemoglobin. The enzyme nitrogenase is a Mo-Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia, the first stable product of nitrogen fixation.

N₂ + 8H⁺ + 8e^– + 16 ATP → 2NH₃ + H₂ + 16 ADP + 16 Pi

The enzyme nitrogenase is highly sensitive to the molecular oxygen, it requires anaerobic conditions. To protect these enzymes, the nodule contains an oxygen scavenger called leg -haemoglobin. These microbes live as aerobes under free living conditions but during nitrogen – fixing events, they adapt to anaerobic conditions, thus protecting the nitrogenase enzyme.

Question 5. Write an account of the classification of enzyme.

Answer: Enzymes have been classified into different groups based on the type of reactions they catalyse. Enzymes are divided into 6 classes. Each class is again divided into sub-class and sub-subclasses. They are

1) Oxidoreductases / dehydrogenases :

Enzymes which catalyse oxidoreduction between two substrates S and S’.

E.g.: S.reduced + S’ oxidised → S oxidised + S’ reduced

2) Transferases :

Enzymes catalysing a transfer of a group G (other than hydrogen) between a pair of substrate S and S’.

E.g.: S — G + S’ → S + S’ — G

3) Hydrolases :

Enzymes catalysing hydrolysis of ester, ether, peptide, glycosidic, C – C, C – halide or P – N bonds.

4) Lyases :

Enzymes that catalyse removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds.

5) Isomerases :

Includes all enzymes catalysing inter – conversion of optical, geometric or positional isomers.

6) Ligases :

Enzymes catalysing the linking together of 2 compounds. E.g.: enzymes which catalyse joining of C – O, C – S, C – N, P – O etc., bonds.

The above classification provides for a four digit code to identify individual enzymes. For E.g.: Glucose – 6 – Phosphotransferase has the enzyme code (E.C.) 2,7.1.2

The first digit of code indicates the major class.

The second digit of code indicates the subclass.

The third digit of code indicates the sub-subclass.

The fourth digit indicates the serial number of the enzyme in a particular sub-subclass.

Question 6. Explain the mechanism of enzyme action. [Mar. 2020]

Answer:

1. During the enzyme action the enzyme (E) combines with its specific substrates (S) to form a enzyme – substrate complex (E – S) which is short – lived.
2. Energy that is required for a substrate to react inorder to get converted into end product is called “activation energy”.
3. This activation energy is available in different forms like heat, ATP etc.
4. The activation energy of the formation of this E – S complex is low, hence many molecules can react and participate in the reaction, leading to the formation of products (P).
5. The E – S complex dissociates into its products P and the unchanged enzyme E with an intermediate formation of the enzyme – product complex (EP).
6. The formation of the ES complex is essential for catalysis. E + S → (ES) → (EP) → E + P
7. Formation of (ES) complex has been explained with ‘Lock and Key’ hypothesis by Emil Fisher and later with “Induced Fit” hypothesis by Daniel E. Koshland (1973).
8. According to this theory every enzyme possess “Active sites”.
9. The substrate (S) gets attached to the active site of the enzyme (E) and forms an enzyme substrate (ES) complex.
10. The enzyme remains unchanged while the substrate is broken into products (P).

Question 7. The entire process of photosynthesis consists of a number of reactions. Where in the cell do each of these take place?

a) Synthesis of ATP and NADPH

b) Photolysis of water

c) Fixation of CO₂

d) Synthesis of sugar molecule

e) Synthesis of starch

Answer:

a) This occur in Grana thylakoid.

b) PS II – Oxygen Evolving Complex (OEC) is associated with the PS II, which itself is physically located on the inner side of the membrane of the thylakoid.

c) Stroma of chloroplast.

d) Cytoplasm

e) Cytoplasm

Question 8. Which property of pigments is responsible for its ability to initiate the process of photosynthesis? Why is the rate of photosynthesis higher in red and blue regions of the spectrum of light?

Answer:

1. Ability to absorb light at specific wavelength initiates the process of photosynthesis.
2. Absorption spectrum of chlorophyll a show maximum absorption of light at blue and red regions.
3. Action spectrum shows the rate of photosynthesis it is maximum at blue and red region.
4. Absorption spectrum and action spectrum clearly indicate that chlorophyll a is the chief pigment associated with photosynthesis. The rate of photosynthesis is higher in red and blue regions of the spectrum due to it absorption of light wavelengths.

Question 9. Explain Mitchell’s chemiosmosis in relation to oxidative phosphorylation.

Answer: The synthesis of ATP in respiration is associated with the consumption of oxygen, hence it is referred as oxidative phosphorylation. The mechanism of mitochondrial ATP synthesis can be explained by Mitchell’s chemiosmosis. The transfer of electrons from NADH or FADH to oxygen through electron transport system results in proton transfer from matrix to inner membrane space of mitochondria. Due to this, proton concentration gradient is established across the inner mitochondrial membrane (more number of H+ on inner membrane space side and less on the matrix side).

The inner membrane Of mitochondria is virtually impermeable to protons and thus prevents the return of proton into the matrix.

However the ATP synthase (complex V) consists of the major components F₁ and F₀ .

F₀ is the integral membrane protein complex that forms the channel through which protons cross the inner membrane.

F₁ head piece is a peripheral membrane protein complex and contains the site for the synthesis of ATP from ADP and inorganic phosphate.

When H⁺ move down the gradient, energy is released some amount of energy helps in combining ADP and iP leading to the form of ATP. The energy of 3H⁺ moving down the potential gradient is sufficient to form one ATP molecule.

Diagramatic presentation of ATP synthesis in mitochondria

Question 9. Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.

Answer: The energy stored in NADH + H⁺ and FADH₂ are oxidised through electron transport system. The electrons are passed on to O₂ resulting in the formation of H₂O.

The metabolic pathway through which an electron passes from one carrier to another is called the Electron Transport System (ETS). It is present in the inner mitochondrial membrane.

1. Electrons from NADH produced in the mitochondrial matrix during the citric acid cycle are oxidised by an NADH dehydrogenase (complex I) and the electrons are transferred to ubiquinone located within the inner membrane.
2. Ubiquinone also receives reducing equivalents via FADH₂ (complex II) that is generated during oxidation of succinate in the citric acid cycle.
3. The reduced ubiquinone is then oxidised with the transfer of electrons to cytochrome c via cytochrome bcx complex (Complex III).
4. Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for the transfer of electrons between complex III and IV.
5. Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3 and two copper centres.

When the electrons pass from one carrier to another via complex I to IV in the electron transport chain, they are coupled to ATP synthase (complex V) for the production of ATP from ADP and inorganic phosphate.

The number of ATP molecules synthesised depends upon the nature of electron donor. Oxidation of one molecule of NADH gives rise to 3 molecules of ATP, while that of one molecule of FADH₂ produces 2 molecules of ATP. Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. The presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system.

Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilised for the production of proton gradient required for phosphorylation.

In respiration it is the energy of oxidation – reduction utilised for the same process. It is for this reason that the process is called oxidative phosphorylation.

Question 10. Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.

Answer: 1) Growth :

Growth is defined as a permanent or irreversible increase in dry weight, size, mass or volume of a cell, organ or organism.

2) Differentiation :

The cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions. This act leading to maturation is termed as differentiation.

3) Dedifferentiation :

The living differentiated cells, that have lost the capacity to divide, can regain the capacity of division under certain conditions. This phenomenon is called dedifferentiation, e.g.: Formation of meristems – interfascicular cambium and cork cambium from parenchyma cells.

4) Redifferentiation :

In the process of redifferentiation meristems/tissues are able to divide and produce cells that once again lose the capacity to divide but mature to perform specific functions. That means they get redifferentiation.

5) Determinate growth :

Leaves are specialised organs characterised by defined developmental destiny and determinate growth.

6) Meristem :

It is located at root and shoot tips. The meristem refers to the cells that remain dividing.

7) Growth rate :

Increased growth per unit time is termed as growth rate.

Question 11. List five natural plant growth regulators. Write a note on discovery, physiological functions an agricultural/horticultural applications of any one of them.

Answer:

Question 12. Explain different methods of sexual reproduction in Bacteria.

Answer: Sexual reproduction :

True sexual reproduction is absent in bacteria. However the exchange of genetic material is reported through other methods.

Three types of genetic recombinations are reported in different species of bacteria. They are

1. Conjugation
2. Transformation and
3. Transduction.

1. Conjugation :

1. The transfer of genetic material (DNA) through direct cell to cell contact is known as conjugation.
2. It was first reported by Lederberg and Tatum in 1946 in Escherichia coli.
3. In E.coli, in addition to the bacterial chromosome or genophore which is the main genetic material, bacteria contain small circular, double stranded DNA molecules called plasmids or ‘F’ factor.
4. E.coli having F factor are called F⁺ cells or donor cells and the cells without F factor are called F^– cells or acceptor cells.
5. F⁺ cells have pilus or sex pilus. During conjugation the F⁺ and F strains come close together. Once contact is established, the pilus shortens to bring the two bacteria close together.
6. A conjugation tube is established, The plasmid in F⁺ replicates and forms a copy of it, which moves to the acceptor cell (F). Thus donor bacterium generally retain a copy of genetic material that is being transferred.

2. Transformation :

Transformation is uptake of naked DNA fragments from the surrounding environment and the expression of that genetic information in the

recipient cell. That is, the recipient cell has now acquired a characteristic that is previously lacked. This mode of bacterial genetic recombination was discovered by Frederick Griffith in streptococcus pneumoniae.

3. Transduction :

The transfer of genetic material from one bacterium to another through bacteriophage is known as transduction.

Question 13. “Bacteria are friends and foes of man” – discuss.

Answer: Bacteria are known to be the casual agents of plant, animal and human diseases. At the same time there are many bacteria which are directly or indirectly beneficial to man. Thus, these organisms can be considered both as ‘friends and foes of man’.

Beneficial activities:

1. Microbes are now used in extracting valuable metals like uranium from rocks. The process is known as Bio-mining. The use of microbes in mining reduces the cost of production by more than 50%.
2. DNA components from bacteria are used as Biosensors that can detect biologically active toxic pollutants.
3. Microbes also find application in medical diagnostics, food and fermentation operations.
4. The most important development in Biotechnology depends on the possibility of altering the genetic makeup of bacteria through genetic engineering.
5. Microbes in household products :

   a) A common example is the production of curd from milk micro-organisms such as Lactobacillus and others, commonly called lactic acid bacteria (LAB), grow in milk and convert into curd. It also improves its nutritional quality by increasing vitamin B12.

   b) Some of our food stuffs like cheese, yogurt are also actually the by- products of microbial growth.

6. Microbes are used as biocontrol agents. Biocontrol refers to the use of biological methods for controlling plant diseases and pests.
7. Many industrial products like enzymes, amino acids, vitamins, organic acid and alcohols are commercially produced by micro organisms.
8. Microorganism decompose dead plants and animals and enrich the soil nutrients which can be used by plants. They play an important role in recycling of elements.
9. Microbes cause diseases in plants and human beings. On the other hand, they help in creating disease free world by producing antibiotics and vaccinations. Penicillin was the antibiotic discovered by Alexander Fleming from a fungus, penicillin notatum. The antibiotic obtained from bacteria streptomyces griseus is known as streptomycin.
10. Biogas is a mixture of gases (containing predominantly methane) produced by the microbial activity and which may be used as fuel.

Harmful activities:

Some bacteria that cause human diseases are

Bacteria causes plant diseases :

Bacteria also cause animal diseases :

Question 14. Write about the discovery and structural organization of viruses.

Answer:

1. Viruses have been causing diseases in humans, animals and plants from the ancient time.
2. ‘Germ theory of disease’ was put forth by Louis Pasteur but the agent responsible for the diseases are not known.
3. In 1892 for the first time, the Russian pathologist Dmitri Iwanowski, while studying tobacco mosaic disease, filtered the “sap of diseased tobacco leaf” through filter which was designed to retain bacteria. However the infectious agent passed through the pores of the filter. After injecting the filtered sap into the healthy plant, he found the development of symptoms of mosaic disease in it. Unable to see any microorganism in a sap, he reported that the filterable agent was responsible for the disease.
4. Martinus Beijerinck repeated Iwanowski’s experiments and concluded that the disease causing agent was a contagious living fluid (contagium vivum fluidum).
5. W.M. Stanley (1935) purified the sap and announced that the virus causing mosaic disease in tobacco could be crystallized. It was named as Tobacco Mosaic Virus (TMV).
6. Fraenkel Conrat (1956) confirmed that the genetic material of the TMV is RNA.

Question 15. Describe the process of multiplication of viruses.

Answer: The process of multiplication of viruses is done by two alternative mechanisms.

a) Lytic cycle b) Lysogenic cycle

a) Lytic cycle :

T – even phages that attack the bacterium E.coli cause lysis of the cells and are called virulent phages. They show lytic cycle. It involves 5 step process. They are 1. attachment 2. penetration 3. biosynthesis 4. maturation and 5. release.

1. Attachment:

1. Contact of the virion to the surface of host bacterium is called attachment or adsorption.
2. The phages use tail fibres for attachment to the complementary receptor sites on the bacterial cell wall.

2. Penetration :

1. The injection of phage nucleic acid into the host cell is called penetration.
2. The phage DNA is injected into the bacterium through the tail core like a hypodermal syringe.
3. The capsid remains outside the bacterial cell and is referred to as ghost.

3. Biosynthesis :

1. Once 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 host cell.
2. Host cell do not contain any complete infective viruses.

4. Maturation :

1. In this process bacteriophage DNA and capsids are assembled into complete virions.
2. This period of time between the infection by a virus and the appearance of the mature virus within the cell is called the eclipse period.

5. Release :

1. The final stage of viral multiplication is the lysis phase of the host cell and the release of virions from the host cell.
2. The plasma membrane of the host cell gets dissolved or lysed due to the viral enzyme called lysozyme.
3. The bacterial cell wall breaks releasing the newly produced phage particles / virions.

1. In lysogenic cycle, some bacteriophages such as X (Lambda) phages do not cause lysis during multiplication.
2. Instead the phage DNA upon penetration into the E.coli gets integrated in to the circular bacterial DNA, becomes part of it and remains latent (inactive). Such phages are called temperate phages. This inserted phage DNA is now called prophage.
3. Every time the bacterial genetic material replicates, the prophase also gets replication. The prophage remains latent in the progeny cells.
4. However, rarely prophage gets disintegrated when they are exposed to UV light or some chemicals and enter into lytic cycle.
5. Thus temperate phase plays a role of transduction by the transfer of genetic material from one bacterium to another through bacteriophage.

Question 16. In a plant, tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents workout a dihybrid cross. What is standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?

Answer: In a plant tallness is dominant over dwarfness and red flower is dominant over the white cross between two parents with one parent, Tall with red flowers (TT RR) and other parent, Dwarf with white flowers (tt rr). Such a cross involving two parental plants differing in two character is called dihybrid cross.

When the F₁ hybrids were allowed to undergo self pollination, the F₂ generation progeny was showing four kinds of phenotypes in a definite pattern.

The phenotypic ratio is 9 : 3 : 3 : 1 and the genotypic ratio is 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1.

The F₂ progeny showed not only the parental combination i.e., Tall with red flowers and Dwarf with white flowers but also the new combination i.e., Tall white and Dwarf red.

These new combination are produced due to genetic recombination.

The F₂ progeny are Tall purple 9 ; Tall white 3 ; Dwarf purple : 3; Dwarf white 1.

The dihybrid ratio is 9 : 3 : 3 : 1.

Yes, the values would deviate if the two genes interact with each other.

Question 17. Replication was allowed to take place in the presence of radioactive Deoxy-nucleotide precursors in E.coli that was a mutant for DNA ligase. Explain how the newly synthesised radioactive DNA will be when purified.

Answer:

1. In the long DNA molecule, the replication occurs within a small opening of DNA Helix, called Replication fork.
2. On one strand, the template with polarity (3′ – 5′) the replication is continuous.
3. On the other strand, the template with polarity (5′ – 3′), it would be discontinuous.
4. The discontinuously synthesized fragments are joined by the DNA ligase.
5. When Replication was allowed to take place in the presence of radioactive Deoxy – nucleotide precusors in E.Coli that was a mutant for DNA ligase, their Okazaki fragments will not be joined.

Question 18. Write briefly about Griffith’s experiments on S. pneumoniae bacteria. What was his conclusion?

Answer: Frederick Griffith (1928) performed the experiments on Bacterial transformation with Streptococcus pneumoniae, the bacterium causing pneumonia.

• He observed two strains of this bacterium, one forming smooth colonies with capsule (S-type) and the other forming rough colonies without capsule (R-type).
• The S-type cells are virulent while the R-type cells are non-virulent.
• When live S-type cells were injected into the mice, they suffered from pneumonia and died.
• When live R-type cells were injected into the mice, the disease did not appear and the mice survived.
• When heat killed S-type cells were injected, the disease did not appear.
• When heat killed S-type cells were mixed with live R-type cells and injected into the mice, the mice died of pneumonia and live S-type cells were isolated from the body of the mice.
• He concluded that the R-strain had somehow been transformed by the heat killed S-strain bacteria, which must be due to transfer of genetic material, the transforming principle.

Question 19. Explain briefly the various processes of recombinant DNA technology. [Mar. 18 17, 14; May 14]

Answer: Recombinant DNA technology involves several steps in specific sequence such as

a) Isolation of DNA

b) Fragmentation of DNA by restriction endonucleases

c) Isolation of a desired DNA fragment

d) Ligation of the DNA fragment into a vector

e) Transferring the recombinant DNA into the host

f) Culturing the host cells in a medium at large scale

g) Extraction of the desired product.

a) Isolation of DNA :

1. DNA is enclosed within the membranes. To release DNA along with other macromolecules such as RNA, proteins, polysaccharides and lipids, bacterial cells / – plants or animal tissue are treated with enzymes such as lysozyme (bacteria) cellulose (plant cells), chitinase (fungus) to digest cell wall.
2. RNA can be removed by treatment with ribonuclease.
3. Proteins can be removed by treatment with protease.
4. Other molecules can be removed by appropriate treatments and purified DNA ultimately precipitates out after the addition of chilled ethanol.

b) Fragmentation of DNA by restriction endonucleases :

Cutting of DNA at specific locations can be done by using restriction enzymes. The purified DNA is incubated with the specific restriction-enzymes at conditions optimum for the enzyme to act.

c) Isolation of a desired DNA fragment :

Is carried out using agarose gel electrophoresis the DNA is negatively charged, it moves towards the positive electrode or anode and in the process, DNA separates out. The desired DNA fragment is eluted out. Amplification of the gene of interest: Using Polymerase Chain Reaction (PCR) is a reaction in which amplification of specific DNA sequences is carried out in vitro.

i) PCR technique requires :

1. A DNA template which is a double-stranded DNA that needs to be amplified.
2. Primers are small chemically made oligonucleotides of about 10-18 nucleotides that are complementary to a region of template DNA.
3. Enzymes used are Taq polymerase (from Thermus aquaticus) and the vent polymerase (from Thermococcus litoralis).

ii) Steps in PCR :

1. Denaturation of double-stranded DNA is carried out by applying high temperature of 95°C for 15 seconds. Each separated single-stranded strand acts as a template for DNA synthesis.
2. Annaling is carried out in two sets of primers. Which are added and anneal to the 3′ end of each separated strand. Primers act as initiator of replication.
3. Extension is done by DNA polymerase of primers by adding nucleotides complementary to the template provided in the reaction.
4. A thermostable DNA polymerase (taq polymerase) is used in the reaction which can tolerate the high temperature of the reaction.
5. These steps are repeated many times to obtain several copies of desired DNA.

d) Ligation of the DNA fragment into a vector requires a vector DNA and source DNA.

1. These are cut with the same endonuclease to obtain sticky ends.
2. Both are then ligated by mixing vector DNA, gene of interest and enzyme DNA ligase to form recombinant DNA.

e) Insertion of recombinant DNA into the host cell :

Organism occurs by several methods, before which the recipient cells are made competent to receive the DNA.

1. If recombinant DNA carrying antibiotic resistance (eg., ampjcillin) is transferred into E.coli cells, the host cell is transformed into ampicillin resistant cells
2. The ampicillin resistant gene can be called as selectable marker.
3. When transformed cells are grown on agar plates containing ampicillin, only transformants will grow and other will die.

f) Culturing the host cells in a medium at a large scale :

It is carried out in appropriate medium at optimal conditions. The DNA gets multiplied and express itself to form desired products.

g) Extraction of desired gene products :

It is carried out by following steps.

1. A protein encoded gene expressed in a heterologous host is called recombinant protein.
2. Cells having genes of interest can be grown on a small scale or on a large scale.
3. In small scale cells are grown on cultures and then expressed protein is extracted and purified by various separation methods.
4. In large scale cells are grown in a continuous culture system in which fresh medium is added from one side to maintain cells growth phase and the desired protein is collected from the other side.

Question 20. Give a brief account of the tools of recombinant DNA technology. [March 2020, March 2019, May 2017]

Answer: The tools of recombinant DNA technology

i) Restriction enzymes

ii) Polymerase enzymes

iii) Ligases

iv) Vectors

v) Host organism

i) Restriction enzymes :

Belong to a larger class or enzymes called Nucleases. These are two kinds.

a) Exonucleases :

Exonucleases remove nucleotides from the ends of the DNA.

b) Endonucleases :

Endonucleases make cuts at specific positions with in the DNA. Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands.

Eg : EcoRI recognises 5′ GAATTC 3′ sites on the DNA and cuts in between G and A.

Naming of Restriction Enzymes :

• The first letter of the name comes from the genus and the next two letters from the name of the species of the prokaryotic cell from which it is isolated.
• The next letter comes from the strain of the prokaryote.
• The Roman numbers following these four letters indicate the order in which the enzymes were isolated from that strain of the bacterium.

  Eg: EcoRI from Escherichia coli RY13

  Hind II is from Haemophilus influenzae

  Bam HI is from Bacillus amyloliquefaciens

ii) Polymerase enzymes :

Thermas aquaticus a bacterium yields DNA polymerase used in biotechnology.

1. This enzyme remains active during the high temperature applied during denaturation of double-stranded DNA.
2. It extends the primers using the nucleotides provided in the reaction and the genomic DNA as template.
3. Repeated amplification is achieved by this enzyme. The amplified fragments, if desired can be used to ligate with a vector for further cloning.

iii) Ligases :

The enzyme DNA ligase joins the complementary ends of the plasmid DNA with that of desired gene by covalent bonding to regenerate a circular hybrid called Recombinant (r) DNA or chimeric DNA.

iv) Vectors :

The DNA used as a carrier for transferring a fragment of foreign DNA into a suitable host called vector.

Vectors used for multiplying the foreign DNA sequence are called cloning vectors. Commonly used vectors are plasmids, bacteriophages, cosmids.

Properties of cloning vector

1. It must have low molecular weight.
2. It must have unique cleavage site for the activity of restriction enzymes at single point.
3. It must be able to replicate inside the host cell after its introduction (through ori gene – origin of replication).
4. The vectors requires a selectable marker, which helps in identifying and eliminating non transformants and selectively permitting the growth of transformants.

v) Host organisms :

Competent host for transformation with Recombinant DNA is required as tool.

Question 21. Give an account of bio-technological applications in agriculture and other fields.

Answer: Bio-technology application in agriculture : Involves following three options

1. Agro-chemical based agriculture
2. Organic agriculture
3. Genetically engineered crop based agriculture.

Genetically modified organisms (GMO) are plants, animals, bacteria and fungi whose genes have been altered by manipulation.

Genetic modification in organisms lead to following results :

1. Crops become more tolerant to abiotic stresses, such as cold, drought, salt, heat, etc.
2. Dependence on chemical pesticides reduced i.e., pest resistant crop.
3. Post harvest losses reduced.
4. Efficiency of mineral usage increased in plants (preventing loss of soil fertility).
5. Nutritional value of food enhanced eg: Vitamin A enriched rice.
6. Tailor-made plants are created to supply alternative resources to industries, in the form of starches, fuels and pharmaceuticals.

Some of the application of biotechnology in agriculture are the production of pest resistant plants, eg : Bt Cotton, Bt Corn, etc.

Bt. Cotton :

It is created by using some strains of a bacterium, Bacillus thuringiensis (Bt is short form).

1. This bacterium produces protein that kill certain insects such as lepidopterans (tobacco budworm, armyworm), coleopterans (beetles) and dipterans (flies, mosquitoes).
2. Bt toxin proteins exists as inactive protoxins but once insect ingest the inactive toxin it becomes active and leads to death of an insect.
3. Most Bt toxins are insect group specific, hence the toxin is coded by a gene named cry. For example, the proteins encoded by the genes Cry II Ac and Cry I Ab control the cotton bollworms and Cry I Ab controls corn borer.

Pest resistant Plants are developed by using biotechnology processes.

1. A nematode Meloidegyne incognitia infects the roots of tobacco plants which reduces the production of tobacco.
2. Agro bacterium vectors are used to introduce nematode-specific genes into the host plant. It produces both sense and anti – sense RNA in the host cells.
3. These two RNA are complementary to each other and forms a double stranded RNA (ds RNA) that initiate RNAi and hence silenced the specific mRNA of the nematode.
4. The parasite cannot survive in transgenic host, so prevents the plants from pest. The transgenic plant thus gets itself protected from the parasite.

Biotechnology and Environment:

Bio remediation :

In the process of using microbes and plants to break down or recycle environmental pollutants.

Utilization of sewage and agrowastes to produce biogas and vermicompost.

Biotechnological application in medicine :

Have made immense impact in the area of health care by enabling the mass production of safe and more effective therapeutic drugs.

Vitamins (A, B₁₂ etc.) and antibiotics (pencillin) are produced at low cost using microorganisms.

Genetically engineered insulin leads to sufficient availability of insulin for the management of adult onset diabetes.

Gene therapy is a collection of methods that allows correction of gene defects diagnosed in a child or embryo. Genes are inserted into a person’s cells or tissue to treat a disease. Molecular diagnosis helps to solve the problem of early diagnosis and treatment of diseases.

i) Using conventional methods of diagnosis (serum and urine analysis) early detection of diseases is not possible.

ii) To overcome this problem, some molecular diagnosis techniques provide early detection of diseases. These are

a) Recombinant DNA technology

b) Polymerase Chain Reaction

c) Enzyme Linked Immuno – Sorbent Assay (ELISA)

Question 22. You are a Botanist working in the area of plant breeding. Describe the various steps that you will undertake to release a new variety. [Mar. ’18; May ’17, ’14]

Answer: The steps in breeding a new genetic variety of a crop are

1. Collection of variability.
2. Evaluation and Selection of parents.
3. Cross hybridisation among selected parents.
4. Selection and testing of superior recombinants.
5. Testing, release and commercialization of new cultivars.

i) Collection of variability :

1. Genetic variability is important for any breeding programme.
2. Pre-existing genetic variability available in wild varieties, species and relatives of crop species is collected and preserved.
3. Evaluation of their characteristics is a pre-requisite for the effective exploitation of natural genes available in the population.
4. The entire collection of plants / seeds having all diverse alleles for all genes in a given crop is called germplasm collection.

ii) Evaluation and selection of parents :

a) It is carried out by evaluating germplasm to identify plants with desirable combination of characters.

b) The selected plant is multiplied and hybridized.

c) By self-pollination, purelines are created wherever desired.

iii) Cross hybridization among selected parents :

1. Cross hybridization is carried out to combine desired genetic characters from two different plants (parents).
2. Cross hybridization is a time consuming and tedious process as it involves collection of pollen grains from the desired plants and other pollination techniques to incorporate desired traits.
3. It is also not certain the hybrids combine desired characters. The chances of desirable combination is usually only one in few hundred to a thousand crosses carried out.

iv) Selection and testing of superior combinants :

1. This step consists of selection of plants among the progeny of the hybrids with desired combination of characters.
2. It yields plants that are superior to both of the parents. This is known as hybrid vigour / heterosis.
3. They are self pollinated for several generations till they reach a state of uniformity or homozygosity so that characters will not segregate in the progeny.

v. Testing, release and commercialization of new cultivars :

1. Evaluation is done for newly selected lines for their yield and other agronomic traits of quality, disease resistance etc.
2. Selected plants are grown in research fields and their performance is recorded under ideal fertilizer application irrigation and other crop management practices.
3. Testing of hybrid line is done in farmer’s field after evaluation.
4. After testing, the crop is grown at different locations in the country with different agroclimatic zones for at least three growing seasons.
5. The tested material is evaluated in comparison to the best available local crop cultivar used as reference cultivar.
6. Release of tested material is done in bulk after selection and certification.

Question 23. Describe the tissue culture technique and what are the advantages of tissue culture over conventional method of plant breeding in crop improvement programmes? [Mar. 2020, 2019, 17, 14]

Answer: The technique of growing, culturing and maintaining plant cells, tissues and organs in vitro is called tissue culture. Tissue culture is done by following methods :

1) Preparation of nutrient culture medium :

The nutrient medium must provide a carbon source such as sucrose and also inorganic salts, vitamins, amino acids and growth regulators like auxins, cytokinins etc.

2) Sterilization of the culture medium :

1. The medium is rich in nutrients and therefore, attracts the growth of microorganisms. The medium should be sterilized.
2. Sterilization is carried out in a steam sterilizer called autoclave.
3. The culture medium is autoclaved for 15 min, at 121°C and 15 lbs pounds of pressure.

3) Preparation of explant:

Any part of the plant body which is used as inoculum is called explant.

4) Inoculation of explants :

The transfer of explants on to the sterilized nutrient culture medium is called inoculation and is carried out in an aseptic condition.

5) Incubation of growth :

1. The cultures are incubated for 3 to 4 weeks during which period the cells of the explant absorb the nutrients, grow and undergo repeated divisions to produce undifferentiated mass of cells known as Callus.
2. Sometimes sheets or roots may be produced directly. ,
3. The explants or callus cultured on different combinations of auxins and cytokinins will produce shoots or roots and this process is called organogenesis.
4. Alternately, embryo like structures develop from the callus and this phenomenon is known as somatic embryogenesis, the embryo-like structure which develop from the callus are called embryoids. Since these embryoids develop from somatic tissue they are also called Somatic embryos.

6) Acclimatization of plantlets and transfer to pots :

The plants generated through organogenesis (or) somatogenesis need to be acclimatized before they are transferred to pots.

Advantages of tissue culture :

1. More number of plants can be produced in a short time.
2. Disease free plants can be developed from diseased plants.
3. Seedless plants can be multiplied.
4. Female plants are selectively produced through tissue culture.
5. Somatic hybrids can be raised by tissue culture, where sexual hybridization is not possible.

Flow chart showing Plant Tissue culture Technique

Question 24. Modern methods of breeding plants can alleviate the global food ‘shortage’. Comment on the statement and give suitable examples.

Answer: The development of several high yielding varieties of wheat and rice in the mid 1960’s as a result of various plant breeding techniques led to dramatic increase in food production in our country. This phase is often referred as green revolution.

High yielding and disease resistant varieties were introduced in India e.g. Sonalika and Kalyan Sona.

• Semi dwarf varities of rice e.g.: IR-8 Taichung Nalive -1 were introduced in 1966.
• Better yielding semi-dwarf varieties i.e., Jaya and Ratna were developed in India.

Successful developed high yield varieties of maizy jowar and bajra were obtained by hybrid breeding.

Plant breeding for disease resistance had enhance food production breeding is carried out by convention breeding techniques or by mutation breeding, e.g. In mung bean, resistance to yellow mosaic virus and powdery mildew were induced by mutations.

Plant breeding for developing resistance to insect pests :

Major cause for large scale destruction of crop plant and crop produce is insect and pest infestation. Insect resistance in lost crop plants may be due to morphological, biochemical or physiological characteristics.

Examples:

1. Wheat – Hairy leaves – resistance to cereal leaf beetle
2. Maize – High aspartic acid and low nitrogen and sugar contents – resistance to stem borer
3. Wheat – Solid stem – resistance to sawfly
4. Cotton – Smooth leaves and nectar-less condition – resistance to bollworm
5. Cotton – Hairy leaves – resistance to. sawfly

Plant breeding for improved food quality :

Breeding crops with higher levels of vitamins and minerals or higher protein and healthier fats is called Biofortification. It is the practical means to improve public health.

Examples:

1. Lysine and tryptophan rich maize.
2. High protein rich wheat.
3. Iron fortified rice.
4. Vitamin enriched bitter gourd, tomato, mustard, bathua.
5. Iron and Calcium enriched spinach and bathua.
6. Vitamin A rich carrots, spinach and pumpkin.

Single Cell Protein :

Microbes are grown on an industrial scale and used as nutrient rich food. E.g.: Spirulina.

Tissue Culture:

Technique of regeneration of whole plant from any part of the plant by growing it on suitable culture medium under aseptic conditions in vitro.

Advantages:

1. 1) A number of plants can be grown in a short period of time, e.g.: Tomato, banana, apple, teak, eucalyptus etc.
2. Healthy disease free plant can be grown by meristem culture, e.g.: Banana, sugarcane, potato etc.
3. Somatic hybrid can be raised by tissue culture where sexual hybridisation is not possible.

   e.g.: Pomato

Question 25. a) What would happen if a large volume of untreated sewage is discharged into a river?

b) In what way is anaerobic sludge digestion important in sewage treatments?

Answer: a) The untreated sewage discharged directly into rivers, leading to their pollution and an increase in water born diseases.

b) Treatment of sewage involves two steps. 1) Primary treatment 2) Secondary treatment

Primary treatment :

It is a physical process of removal of small and large particles through filtration and sedimentation.

• The sewage is allowed to go into the primary setting tank, where the suspended material settle down to form primary sludge.
• The effluent is taken for secondary treatment. The anaerobic sludge digestion is important in secondary sewage treatment.

Secondary sewage treatment:

• It is a biological process that employs the heterotrophic bacteria naturally present in the sewage.

  The effluent from the primary treatment is passed into large aeration tanks, where it is constantly agitated and air is pumped in it.

• This allows the rapid growth of aerobic bacteria into floes, which consume the organic matter of sewage and reduce the BOD.
• The effluent is passed into a settling tank, where the floes are allowed to sediment forming the activated sludge.
• A small part of the activated sludge is pumped back into aeration tank as inoculum.
• The remaining major part of the sludge is pumped into sludge digestors, where the anaerobic bacteria digest the organic matter and produce a mixture of gases such as methane, hydrogen sulphide and CO₂. These gases form biogas which can be used as a source of energy as it is. inflammable.

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