TamilNadu Class 12th Botany Exam 2023 : 2 Marks Important Question with Solution

TamilNadu Class 12th Botany Exam 2023 : 2 Marks Important Question with Solution

Botany is one of the important subjects for those who want to pursue a career as a botanist. TN Board Class 12 Botany Important 2 Marks Questions are a valuable resource for revision purpose.

It boosts their confidence and can also self-analyze their preparation level so that they don't miss any topic from the syllabus. Students can solve these TN Board Class 12 Botany important questions for practice.

2 Mark Questions

Q.1. Most of the plants produce single type of flowers but Viola, Commelina and Oxalis produce two types of flowers. Explain

Ans. Chasmogamous flowers :- Flowers with exposed anther and stigma Cleistogamous flowers :- Closed flowers

Q.2. A microsporangium is surrounded by four layers. Name the first three layers and write their function

Ans. Epidermis, Endothecium & Middle layers

Function :- Protection & dehiscence

Q.3. Flowering plants have developed many devices to discourage self pollination and to encourage cross pollination. List out such features found in plants (any 2 point)

Ans. Self incompatibility (genetic mechanism to prevent pollen germination / pollen tube formation) / Pollen release & stigma receptivity not synchronized ( Anther & stigma mature at different time) / Anther & stigma are placed at different positions / Unisexual flowers. (any 2 points)

Q.4. Differentiate true fruits from false fruits. Give one example for each.

Ans. True fruit – develop from the ovary . e.g., mango ,coconut etc (any 1 example)

False fruit – develop from the part of the flower other than ovary. e .g., Apple , Strawberry , Cashew etc (any 1 example) 

Q.5. A typical angiosperm embryo sac is 7 celled and 8 nucleate.

a. Explain monosporic type of embryo sac development

Ans. Embryosac developed from single haploid megaspore.

b. Name haploid cells present in embryosac other than egg cell

Ans. Synergids and Antipodals 

Q.6. Observe the events given below (Embryogenesis , Gametogenesis , Syngamy) Choose a post fertilization event from the above and define it

Ans. Embryogenesis Development of embryo from zygote

Q.7. Zygote is dormant for some time in fertilized ovule. Why?

Ans. Endosperm provide food to developing embryo. So zygote divide only after certain amount of endosperm is formed. 

Q.8. Apomictic seeds are used in hybrid industry. Why?

Ans. Apomixis helps in the production of hybrid seeds with a combination of desirable characters. In apomictic hybrid seeds, there is no segregation of characters . Farmers do not have to buy hybrid seeds every year because apomixis preserve good characters over generations for crop plants. 

Q.9. Give reason

a. Hybrid seeds have to be produced year after year

Ans. The plants grown from the hybrid seeds show segregation of traits and do not maintain the hybrid characters.

b. Ground nut seeds and Castor seeds are dicot seeds. But ground nut seeds are exalbuminous and castor seeds are albuminous.

Ans. In ground nut, endosperm is not present in the seed. In castor, some amount of endosperm left in the seeds.

Q.10. Fusion of polar nuclei with male gamete in double fertilisaton result in the formation of endosperm.

a. Write down the function of endosperm

Ans. Endosperm provide nourishment (food) to developing embryo 

b. Write briefly about the endosperm development in coconut.

Ans. PEN undergoes free nuclear divisions to form many triploid nuclei . This is liquid endosperm / nuclear endosperm (Coconut water from tender coconut ). Then cell wall formed from periphery towards centre to form solid endosperm /multicellular endosperm. (White kernel)

Q.11. The early stages of embryo development are similar in both dicots and monocots. However, mature embryos have differences. Write the difference between dicot embryo and monocot embryo.

Ans. Dicot embryo – Two cotyledon, Monocot embryo – 1 cotyledon (scutellum) 

Q.12. (a) What is meant by Seed dormancy?

Ans. Temporary inactive state of a viable seed. 

(b)Write any two significance of seed dormancy

Ans. helps the storage of seeds / help to use as food throughout the year / helps to raise crops in next season / hard seed coat provide protection to young embryo (any 2)

Q.13. (a) Pollengrains are preserved as fossil. Why?

Ans. Due to the presence of Sporopollenin (Organic material resistant to high temperature, strong acids & alkali & cannot be degraded by enzymes) 

(b) How do pollengrains of many species affect human health ?

Ans. Cause allergies & lung disorders (asthma, bronchitis). 

Q.14. Explain pollen pistil interaction

Ans. All events from pollen deposition of stigma until the entry of pollen tube into ovule / Through chemical conversation between pollen & stigma Pistil recognize right pollen and wrong pollen, then pistil accept (Promote pollen germination) or reject (prevent pollen germination) pollengrain.

Q.15. If the female parent produces bisexual flowers, emasculation is necessary in artificial hybridization.

a. What is emasculation

Ans. Removal of anther from the flower buds of female flower before maturity

b. Write down the importance of emasculation

Ans. prevent self pollination.

Q.16. Give reasons.

(a) Pollen tablets are in use by people these days

Ans. Pollen grains are rich in nutrients . Pollen consumption increase the performance of athletes & race horses 

(b) Self incompatibility discourage self pollination.

Ans. Self incompatibility is the genetic mechanism to prevent pollen germination / pollen tube formation. 

Q.17. A typical angiosperm embryosac is 7 celled 8 nucleate stage.

(a) Name the cells that constitute egg apparatus?

Ans. Eggcell & 2 synergids

(b) Name the diploid cell present in embryosac

Ans. Polar nuclei / Secondary nucleus

Q.18. Different stages of development in a dicot embryo are given below. Arrange them in the correct sequential order (Heart shaped embryo , Globular embryo , Mature embryo, Proembryo)

Ans. Proembryo – Globular embryo – Heart shaped embryo – Mature embryo 

Q.19. Name the special cellular thickening present in the synergids at micropylar end. Write its function

Ans. Filiform apparatus It guide pollentube into synergid 

Q.20. Polyembryony is the formation of more than one embryo in the seed.

a. What are the reasons for polyembryony

Ans. Presence of more than one egg cells & all get fertilized / Presence of more than one embryosac / Many embryos develop from parts like synergids, antipodals, nucellus, integuments etc

b. Give one example

Ans. Seeds of orange, citrus, mango (any 1)

Q.21. How is it possible in Oxalis and Viola plants to produce assured seed set even in the absence of pollinators?

Ans. Cleistogamous flowers are closed flowers, anthers & stigma lie close to each other. When anther dehisce in the flower buds, pollen grains fall on the stigma and fertilization is effected .

Q.22. What are the features of tapetum?

Ans. Innermost wall layer of microsporangium / Nutritive tissue which nourishes the pollen grain / Tapetal cells possess dense cytoplasm / more than one nuclei (multinucleate) present.

Q.23. (a) Which enzymes are known as ‘ molecular scissors’?

(b) What is the use of these enzymes in Genetic engineering?

Ans. Restriction endonuclease Used to cut DNA into fragments at specific sites within molecules.

Q.24. How can we visualize DNA fragments in Gel electrophoresis?

Ans. Separated DNA fragments can be visualised after staining with Ethidium bomide followed by exposure to UV light. bright orange coloured bands obtained. 

Q.25. Distinguish Spooling from Elution

Ans. Spooling :- Process by which DNA threads wind on a reel. Last step of Isolation of DNA Elution - Separated bands of DNA are cut out from the agarose gel and extracted from gel piece. Last step of separation using Gel electrophoresis.

Q.26. Write notes on

(a) Microinjection

Ans. Microinjection - Recombinant DNA is directly injected into the nucleus of an animal cell using micropipette.

(b) Biolistics

Ans. Biolistics / Gene gun - High velocity microparticles of gold or tungsten coated with DNA and is bombarded into the host cell  (plant cell) 

Q.27. Observe the figure and answer the questions

a. Identify the instrument in the figure

Ans. Bioreactor

b. Write the role of this instrument in biotechnology.

Ans. Large scale production of recombinant proteins / extraction of biological products in large quantities 

Q.28. Observe the nucleotide sequence given below

a. Name this kind of nucleotide sequence

Ans. Palindromic sequence

b. Define this sequence

Ans. sequence of base pairs that read the same on the two strands when orientation of reading is kept the same 

Q.29. Multiple copies of gene of interest can be synthesized through PCR. Expand PCR and write its steps.

Ans. Polymerase Chain reaction.

Steps :- Denaturation , annealing, & Extension 

Q.30. EcoRΙ is a restriction endonuclease. What do E, co, R, Ι represent?

Ans. E - First letter of genus from which it is isolated

Co - first two letters of species name of the prokaryotic cell from which they were isolated

R - denotes the strain

I - romen number , denotes the order in which they are isolated from that strain of bacteria.

Q.31. Briefly describe Down stream processing and Bioreactor.

Ans. Downstream Processing :- All process after biosynthetic phase is collectively called Downstream processing / Downstream processing include Separation of Products , Purification of Products, Addition of Preservative , Clinical trials for drugs, Quality control test.

Bioreactor : – Large vessel in which raw materials are biologically converted into products. 

Q.32. Bioreactor is an apparatus used for large scale production of proteins.

a. Name two types of bioreactors.

Ans. Stirred tank bioreactors & Sparged stirred-tank bioreactor

b. Write any two feature of bioreactors

Ans. Possess an agitator system / Oxygen delivery system / A foam control system / A temperature control system/ pH control system / Sampling ports for periodic withdrawal of culture (any two feature)

Q.33. Isolation of DNA from plant cell involves many steps. Explain the different steps.

Ans. The cells are treated with Cellulase to digest cellwall. Treated with protease , ribonuclease etc to remove other impurities. The purified DNA is precipitated on the addition of chilled ethanol and are seen as threads in suspension. Spooling :- Process by which DNA threads wind on a reel. Last step of Isolation of DNA.

Q.34. (a) Agrobacterium tumifaciens is a natural genetic engineer of plants. Justify

Ans. Agrobacterium can deliver T- DNA to host

(b) What is the role of retrovirus in rDNA technology?

Ans. Retrovirus is used as a disarmed pathogen vector

Q.35. What is meant by sticky ends? Why are they called sticky ends?

Ans. sticky ends - When a fragment of DNA is cleaved by restriction enzyme, Two separate strands obtained. Both are overhanging  piece of DNA. They called sticky ends because they form hydrogen bonds with their complementary cut counter parts. 

Q.36. Explain

(a) ‘Ori’

Ans. Origin of replication (Ori) :- Specific sequence from where replication starts. 

(b) Selectable marker

Ans. Selectable markers :- A gene which helps to identify and eliminate recombinants/ transformants from nonrecombinants/non-transformants

Q.37. Name two disarmed pathogen vectors used in rDNA technology.

Ans. Ti plasmid of Agrobacterium tumefaciens & Retrovirus 

Q.38. (a) What is meant by Insertional inactivation?

Ans. Inactivation of an enzyme due to inactivation of a gene as a result of insertion of foreign DNA

(b) How insertional inactivation is used to identify recombinants?

Ans. A recombinant DNA is inserted within the coding sequence of an enzyme β galactosidase. So the enzyme is inactivated. Recombinants do not produce colour, but Non - recombinants gives bluish coloured colonies.

Q.39. Mention the key tools (any 4) in biotechnology?

Ans. Restriction enzymes / Polymerase enzymes / Ligase or Molecular glue / Vectors / Host (any 4) 

Q.40. Cloning vectors are used to transfer gene from one cell to another.

a. Name an artificially reconstructed plasmid vector.

Ans. pBR322

b. What are the features of cloning vector?

Ans. Origin of replication Single recognition sequence / cloning site , Selectable marker

Q.41. DNA fragments of size 500 bp , 1600 bp & 2000 bp are separated by Gel electrophoresis. Which fragment will migrate fast? Why?

Ans. 500 bp.

DNA fragments are separated according to their size through a sieving effect provided by agarose gel . Smaller fragments move farther

Q.42. What is the basic principle of Gel electrophoresis?

Ans. Negatively charged DNA molecules are forced to move towards anode under an electric field through agarose medium

Q.43. (a) Mention the technique of genetic engineering

Ans. Creation of recombinant DNA , Gene cloning (making identical copies) and Gene transfer.

(b) What is recombinant DNA / rDNA ?

Ans. Artificially made DNA that is composed of a combination of DNA sequences from two or more organisms. 

Q.44. Observe the sketch of stirred - tank bioreactor and label the parts A, B, C & D

Ans. A – Motor. B – Foam breaker. C- Flat bladed impeller. D – Acid / Base for pH control. 

Q.45. (a) What are the two kinds of nuclease enzyme?

Ans. Exonuclease & Endonuclease

(b) Write their role in rDNA technology

Ans. Exonucleases - Cut the DNA at the end Endonucleases - Cut at specific site within the DNA

Q.46. Bt- toxin does not kill bacillus, but it kill insects. Write the reason.

Ans. The Bt toxin exist as inactive protoxins in bacillus. When an insect ingest the inactive toxin, it is converted into an active toxin

Q.47. Expand ELISA. What is the principle of this process?

Ans. Enzyme Linked Immuno-Sorbent Assay Antigen – Antibody reaction

Q.48. Write four uses of transgenic animals.

Ans. To study how genes are regulated & how they affect normal functions of the body and its development (Study of complex factors involved in growth) / to study how genes contribute to the development of disease. ( Transgenic models exist for cancer, cystic fibrosis, rheumatoid arthritis, & Alzheimers) / To Produce biological products / To test the safety of vaccines / To test safety of chemicals (toxicity / safety testing) (any 4 response) 

Q.49. How gene therapy is practiced for a permanent cure of disease?

Ans. Gene therapy at embryonic stage /ADA gene from bonemarrow is introduced into cells at early embryonic stage

Q.50. How cancer due to mutation can be detected by molecular diagnostic method?

Ans. Clone having mutated gene will not appear on photographic film because probe will not have complimentarity with the mutated gene

Q.51. First clinical gene therapy was given in 1990 to a 4 year girl with Adenosine deaminase (ADA) deficiency.

a. What is gene therapy?

Ans. collection of methods that allows correction of a gene defect that has been diagnosed in a child / embryo. 

b. What are the two alternative methods to cure ADA deficiency?

Ans. Bone marrow transplantation . Enzyme replacement therapy( functional ADA is given by injection) . 

Q.52. Bt-cotton is an example of genetically engineered plant.

a. Name the gene responsible for Bt-toxin production.

Ans. cry gene

b. Cry protein is harmless to bacillus. Why?

Ans. The Bt toxin exist as inactive protoxins in bacillus.

Q.53. Explain how Bt-toxin kill the insect.

Ans. When an insect ingest the inactive toxin, it is converted into an active toxin due to the alkaline pH of the gut which solubilise the crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually leads to death of the insect.

Q.54. RNA can suppress the activity of a gene. Explain it with suitable examples.

Ans. RNA interference (RNAi) , a cellular defence mechanism suppress the activity of a gene / RNAi is mRNA silencing or Silencing of mRNA due to complementary double stranded RNA (dsRNA). We can prevent nematode infestation through RNAi in transgenic tobacco plant. 

Q.55. The first clinical gene therapy was given to a 4 year old girl child.

a. What was her disorder?

Ans. ADA deficiency

b. What is the cause of this disorder?

Ans. The gene which produce ADA enzyme is deficient .

Q.56. How does the inactive protoxin of Bacillus thuringiensis gets converted into active toxin when an insect ingest it?

Ans. When an insect ingest the inactive toxin, it is converted into an active toxin due to the alkaline pH of the gut which solubilise the crystals. 

Q.57. Early diagnosis is essential for the effective treatment of a disease. Write molecular diagnostic methods.

Ans. PCR based on amplification of DNA , ELISA based on antigen antibody reaction , Radioactive probe and rDNA technology.

Q.58. Write any two critical research areas of biotechnology

Ans. Applications of biotechnology in agriculture Applications of biotechnology in medicine

Q.59. What is the structural difference between proinsulin and mature insulin?

Ans. prohormone (Proinsulin) - contains an extra stretch called the C peptide. This C peptide is removed during maturation . 

or

C peptide present in proinsulin . C peptide absent in mature insulin

Q.60. Define

(a) Biopatent

Ans. Biopatent – Patent granted for biological products and processes. 

(b) Biopiracy

Ans. Biopiracy- Unauthorised use of bioresources by multinational companies & other organisations ,without compensatory payment. 

Q.61. The first clinical gene therapy was given to a 4 year old girl child.

a. Briefly describe the clinical procedure adopted in this case.

Ans. Lymphocytes from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA is then introduced into these lymphocytes using retroviral vector & are returned to the patient. Lymphocytes are not immortal, so periodic transfusion is necessary. 

b. What is the role of Adenosine deaminase enzyme ?

Ans. Activate immune system

Q62. Expand GEAC . Mention their aim

Ans. a. Genetic Engineering Approval Committee

b. Take decision regarding validity of GM research and safety of introducing GM organisms for public services.

Q.63. Observe the diagram and answer the questions.

a. Identify the growth model a & b

Ans. Exponential /Geometric growth model / J shaped curve Logistic growth model / verhulst pearl logistic growth / S shaped growth curve.

b. Which type of growth model is considered to be more realistic? Why?

Ans. Logistic growth model. Resources for growth for most animal populations are finite & become limiting sooner or later.

Q.64. Differentiate ectoparasites and endoparasites. Give an example for ectoparasites & endoparasites

Ans. Ectoparasites :- depend on the external surface of the host.

Endoparasites :- take shelter within the body of the host organism

Ectoparasite, example :- head lice on humans / ticks on dogs / Copepods on marine fish / Cuscuta / Loranthus 

Endoparasite, example – Liverfluke / Plasmodium.

Q.65. Write any two methods to measure population density?

Ans. Counting number of individuals of a population / measure % cover / biomass ( e.g., 200 parthenium plant and one huge banyan tree in an area. The role of banyan tree in that community is greater than parthenium) / measure relative population density. ( e .g., The number of fish caught /trap is used to measure total population density of fish in the lake) / Counting the colonies in a bacterial culture / Indirect method – In tiger reserves tiger census is done on pug marks (foot prints) and fecal matter (any 2 method)

Q.66. What is meant by interspecific competition? Give example.

Ans. Competition between unrelated species for the same resource.

eg., in some shallow south American lakes visiting flamingoes & native fishes compete for common food, zooplankton

Q.67. List out four adaptations of prey to protect from predation in plants.

Ans. Thorns in cactus, opuntia, Acacia etc / Poisonous chemical content kill insects / Poisonous cardiac glycosides seen in calotropis / Presence of alkaloids (nicotine, caffeine, quinine, strychnine , opium etc ) against grazers & browsers.

Q.68. Write any two physical factors that affect population density ?

Ans. Food availability /Predation pressure / Weather (any 2) 

Q.69. The density of population in a given habitat during a given period fluctuates due to changes in four basic process.

a. List out them

Ans. Natality, Mortality, Immigration & Emigration.

b. How do these process affect the population density.

Ans. Natality & Immigration leads to increase in population. Mortality & Emigration leads to decrease in population.

Q.70. Define the following.

(a) Natality

Ans. Natality – Number of births in a population during a given period

(b) Mortality

Ans. Mortality – Number of deaths in the population in a given period.

(c) Immigration

Ans. Immigration – Number of individuals of the same species that have come into the habitat from elsewhere during a given period. population Increase.

(d) Emigration

Ans. Emigration – Number of individuals of the population who left the habitat during a given period.

Q.71. Predation is an interaction which has great significance in nature. Write down two significance of predation.

Ans. Predation act as a channel for energy transfer across trophic levels / Predators keep prey population under control / Predators help in maintaining species diversity (by reducing competition among competing prey species) / If predator overexploits the prey, then the prey become extinct. When there is a shortage of prey the predator will also become extinct .(any 2)

Q.72. What is brood parasitism? Give one example for it.

Ans. Brood parasitism – seen in parasitic bird e.g., between cuckoo & crow . Cuckoo lays its eggs in the nest of its host, Crow for incubation, hatching &rearing of youngones. The eggs of cuckoo resembles the eggs of crow in size, shape & colour. Crow cannot detect the foreign egg.

Q.73. Nt+1 = Nt + [(B+I) - (D + E)] is the equation to find the population density at a given time. What does B, I, D & E stands for?

Ans. B – Natality.

I – Immigration

D – Mortality

E – Emigration 

Q.74. What are the effects of parasites on their host?

Ans. Reduce the survival of host / Growth and reproductive rate of host reduced / Render the host vulnerable to its predators by making them weak / Reduce the population density of host.

Q.75. (a) How does monarch butterfly protect itself from predation?

Ans. Monarch butterfly is distasteful to its predator (bird) due to poisonous chemical content in the body. 

(b) How do some species of insects and frogs avoid being easily detected by the predators?

Ans. Camouflaged (cryptically coloured)

Q.76. Observe the equation     dN/dt =rN [K-N/K]

a. What do 'N', 'r' and 'K' represent?

Ans. N – population density , r- intrinsic rate of natural increase , K - Carrying capacity.

b. Define K

Ans. Maximum number of individuals of a particular species that can live in a particular area.

Q.77. Population growth may be exponential or logistic. Differentiate between them.

Ans. Exponential /Geometric growth model :- Growth curve is J shaped. Initial slow growth or lag phase, followed by rapid growth . Growth stops suddenly due to death of individuals. There is an unlimited supply of resources .There is no environmental resistance/ check Logistic growth model / verhulst pearl logistic growth :- S shaped growth curve. Initial slow growth (lag phase) followed by rapid growth (exponential /log phase) . When environmental resistance come into play / when carrying capacity reaches, growth slows down (Stationary phase).

Q.78. Fill in the blanks

Ans. Competition ( - -)

Mutualism (+ +)

Commensalism ( + 0)

Ammensalism (- 0) 

Q.79. Population has certain attributes that an individual organism does not. What are they?

Ans. Birth rate, death rate, sex ratio and age group.

Q.80. Parasites evolved special adaptations to live on host. What are they?

Ans. Loss of unnecessary sense organs / Hooks and sucker / Loss of digestive system / High Reproductive capacity.

Q.81. Explain life history variations with examples.

Ans. Certain organism breed only once in their life time( eg., bamboo, pacific salmon fish). Others breed many times during their life time. eg., Most birds & mammals Some produce large number of small- sized off spring ( eg., Oysters, pelagic fishes). Others produce small number of largesized offspring ( eg., birds, mammals).

Q.82. What are the limitations of ecological pyramids ?(4 points)

Ans. Ecological pyramid does not accommodate food web / Do not take into account same species belonging to two or more trophic levels. / Assumes simple food chain, which never exist in nature, / Saprophytes are not included.

Q.83. What are the different trophic levels?

Ans. Producers – first trophic level

Primary consumers – second trophic level

Secondary consumer – third trophic level

Teritiary consumer – fourth trophic level

Q.84. Decomposition takes place through different steps. Mineralisation is the last step. Write the other four steps.

Ans. Fragmentation , Leaching , Catabolis, Humification. 

Q.85. Observe the equation given below NPP = GPP- Respiration.

a. What does NPP and GPP stand for?

Ans. NPP – Net primary productivity. GPP – Gross primary productivity.

b. GPP depends on various factors. Write any two factors.

Ans. Sunlight, temperature, moisture, plants in that area, photosynthetic capacity, availability of nutrients etc.

Q.86. What are the factors which affect decomposition?

Ans. Warm and moist environment favour decomposition Decomposition rate becomes high if detritus, rich in nitrogen and water soluble substances. Decomposition rate is slow in detritus, rich in lignin & chitin. Low temperature and lack of Oxygen inhibit decomposition.

Q.87. The figure depicts pyramid of energy.

a. Pyramid of energy is always upright, can never be inverted. Justify.

Ans.  Flow of energy is always unidirectional. Only 10% of energy is transferred to next trophic level, rest is lost as heat.

b. Which are the other two ecological pyramids?

Ans. Pyramid of number and Pyramid of energy.

Q.88. (a) Define productivity

Ans. Productivity :- Rate of biomass /Organic matter produced per unit area during a given period of time.

(b) Distinguish primary productivity from secondary productivity

Ans. Primary productivity - Rate of biomass produced per unit area during a given period of time by plants through photosynthesis /Rate of biomass production at producer level.

Secondary productivity :- Rate of biomass production at consumer level

Q.89. Ecological pyranids are usually upright. Meanwhile some pyramid of biomass is inverted. Give reason with example

Ans. In aquatic ecosystem, pyramid of biomass is inverted because biomass of phytoplankton is less as compared with that of small herbivorous fish, that feed on these producers. 

Q.90. What is an ecological pyramid? Name the different types of ecological pyranids.

Ans. Ecological pyramids :- Representation of food chain in the form of pyramid Pyramid of number , Pyramid of biomass and Pyramid of energy.

Q.91. Ecological pyranids are usually upright. Meanwhile some pyramid of number is inverted. Give reason with example.

Ans. In tree ecosystem, pyramid of number is inverted because tree is single in number, primary consumers are more in number.

Q.92. What is 10% law?

Ans. 10% law of energy transfer :- 10% of the energy in the food is fixed into their flesh & is available to next trophic level. 90% of energy is utilized for life activities & released as heat energy.

Q.93. Flow of energy in plant is based on the law of thermodynamics. Explain.

Ans. i. Solar energy captured by plants and flows through different organisms in an ecosystem as food energy. (first law of thermodynamics- - Energy can neither be created nor destroyed it can only be transformed from
one form to another.

ii. Only 10% of energy is transferred to next trophic level, rest is lost as heat (Second law of thermodynamics states that Whenever energy is transferred fron one form to another, there is a decrease in the amount of useful energy).

Q.94. A list of different organisms in an ecosystem are given below. Arrange them in 1st, 2nd, 3rd, and 4th trphic levels. (Phytoplanktom, Man, Fish, Zooplankton).

Ans. Phytoplankton – first trophic level

Zooplankton – second trophic level

Fish – third trophic level

Man – fourth trophic level

Q.95. Given number of individuals in a grassland ecosystem. Grasshopper - 1500, Grass-5842000, Wolf-28, Birds - 215

a. Draw a pyramid of number showing various trophic levels.

Ans.

b. Explain trophic level.

Ans. Trophic levels :- Based on the source of food, organisms occupy a specific place in the food chain.

Q.96. In a marine ecosystem, a population of phytoplankton (150000) supports a standing crop of fishes (4000)

a. Draw a pyramid of biomass

Ans.

b. Draw a pyramid of number

Ans.

Q.97. Humification leads to accumulation of a dark coloured amorphous substance.

a. Identify the substance

Ans. Humus 

b. Write its peculiarities

Ans. partially decomposed dark coloured amorphous substance. Humus is resistant to microbial action. colloidal in nature so it undergoes slow decomposition. Humus is Reservoir of nutrients. 

Q.98. Match the following

Ans. Fragmentation - Breakdown of detritus into smaller particles Leaching - Water soluble inorganic nutrients go down into the soil Catabolism - Bacterial and fungal enzymes degrade detritus into simpler inorganic substances Mineralisation – Formation of inorganic nutrients from humus 

Q.99. Distinguish Grazing food chain from Detritus food chain.

Ans. i. Grazing food chain (GFC)– Begins from plants. Takes place in major aquatic ecosystems. Less fraction of energy flow. eg., Grass →Goat→Lion→Hawk.

ii. Detritus food chain (DFC) – Energy transfer begins from detritus. It include Saprophytes which take food from detritus. Takes place in major terrestrial ecosystem. Large fraction of energy flow takes place.

Q.100. (a) Construct a grazing food chain using the following organisms. ( Frog, Grass, Grasshopper, Snake)

Ans. Grass →Grass hopper→Frog→Snake 

(b) Write the trophic level of Grass hopper and Snake.

Ans. Grass hopper – second trophic level Snake – Fourth trophic level

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