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TN State Board 12th Biology Model Question Paper 3 English Medium

General Instructions:

    1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
    2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
    3. All questions of Part I, II, III and IV are to be attempted separately.
    4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
    5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
    6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
    7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identity the mismatched pair regarding the anther walls.
(a) Epidermal layer – Protective in function
(b) Endothecium layer – Helps in dehiscence of anther
(c) Middle layer – Persistent layer
(d) Tapetum – Nutritive in function
Answer:
(c) Middle layer – Persistent layer

Question 2.
Extra nuclear inheritance is a consequence of presence of genes in _______.
(a) Mitochondria and chloroplasts
(b) Endoplasmic reticulum and mitochondria
(c) Ribosomes and chloroplast
(d) Lysosomes and ribosomes
Answer:
(b) Endoplasmic reticulum and mitochondria

Question 3.
How many map units separate two alleles A and B, if the recombination frequency is 0.09?
(a) 900 cM
(b) 90 cM
(c) 9 cM
(d) 0.9 cM
Answer:
(c) 9 cM

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 4.
Plasmids are _______.
(a) circular protein molecules
(b) required by bacteria
(c) tiny bacteria
(d) confer resistance to antibiotics
Answer:
(d) confer resistance to antibiotics

Question 5.
Solar energy used by green plants for photosynthesis is only ________.
(a) 2 – 8%
(b) 2 – 10%
(c) 3 – 10%
(d) 2- 9%
Answer:
(b) 2 – 10%

Question 6.
One of the chief reasons among the following for the depletion in the number of species making endangered is _____.
(a) over hunting and poaching
(b) green house effect
(c) competition and predation
(d) habitat destruction
Answer:
(d) habitat destruction

Question 7.
Assertion (A): Genetic variation provides the raw material for selection.
Reason (R): Genetic variations are differences in genotypes of the individuals.
(a) (A) is right and (R) is wrong
(b) (A) is wrong and (R) is right
(c) Both (A) and (R) are right
(d) Both (A) and (R) are wrong
Answer:
(c) Both (A) and (R) are right

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 8.
Groundnut is native of ______.
(a) Philippines
(b) India
(c) North America
(d) Brazil
Answer:
(d) Brazil

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
Give the phenotypic ratio of
(a) Dihybrid cross
(b) Dihybrid test cross
Answer:
(a) Dihybrid cross ratio = 9 : 3 : 3 : 1 ,
(b) Dihybrid test cross ratio = 1 : 1 : 1 : 1

Question 10.
What are the materials used to grow microorganism like Spirulina.
Answer:
Spirulina can be grown easily on materials like waste water from potato processing plants (containing starch), straw, molasses, animal manure and even sewage, to produce large quantities.

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 11.
Compare Redifferentiation with Dedifferentiation.
Answer:
Redifferentiation:
A process by which an already differentiated cell undergo further differentiation to form another type of cell.

Dedifferentiation:
A process of reversion of cells (differentiated cells) to meristematic cells leading to formation of callus.

Question 12.
Loamy soil is ideal for crop cultivation – Justify.
Answer:
Loamy soil is ideal soil for cultivation, since it consists of 70% sand and 30% clay or silt or both. It ensures good retention and proper drainage of water. The porosity of soil provides adequate aeration and allows the penetration of roots.

Question 13.
Mention any four environmental benefits of Rain Water Harvesting.
Answer:

  • Promotes adequacy of underground water and water conservation.
  • Mitigates the effect of drought.
  • Improves groundwater quality and water table / decreases salinity.
  • Reduces soil erosion as surface run-off water is reduced.

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 14.
If a person drinks a cup of coffee daily it will help him for his health. Is this correct? If it is correct, list out the benefits.
Answer:
Yes, drinking coffee in moderation enhances the health of a person. Caffeine enhances release of acetylcholine in brain, which in turn enhances efficiency. It can lower the incidence of fatty liver diseases, cirrhosis and cancer. It may reduce the risk of type 2 diabetes.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
How the flowers of salvia are adopted for mellitophily?
Answer:
Pollination in Salvia (Lever mechanism): The flower of Salvia is adapted for Bee pollination. The flower is protandrous and the corolla is bilabiate with 2 stamens. A lever mechanism helps in pollination. Each anther has an upper fertile lobe and lower sterile lobe which is separated by a long connective which helps the anthers to swing freely. When a bee visits a flower, it sits on the lower lip which acts as a platform.

It enters the flower to suck the nectar by pushing its head into the corolla. During the entry of the bee into the flower the body strikes against the sterile end of the connective. This makes the fertile part of the stamen to descend and strike at the back of the bee. The pollen gets deposited on the back of the bee. When it visits another flower, the pollen gets rubbed against the stigma and completes the act of pollination in Salvia.

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 16.
What is the difference between mis-sense mutation and non-sense mutation?
Answer:
Mis-sense Mutation:
The mutation where the codon for one amino acid is changed into a codon for another amino acid is called Missense or non-synonymous mutations.

Non-sense Mutation:
The mutations where codon for one amino acid is changed into a termination or stop codon is called Nonsense mutation.

Question 17.
Give an account on germplasm conservation. .
Answer:
Germplasm conservation refers to the conservation of living genetic resources like pollen, seeds or tissue of plant material maintained for the purpose of selective plant breeding, preservation in live condition and used for many research works.

Germplasm conservation resources is a part of collection of seeds and pollen that are stored in seed or pollen banks, so as to maintain their viability and fertility for any later use such as hybridization and crop improvement. Germplasm conservation may also involve a gene bank and DNA bank of elite breeding lines of plant resources for the maintenance of biological diversity and also for food security.

Question 18.
How are microbial innoculants used to increase the soil fertility?
Answer:
Biofertilizers or microbial innoculants are defined as preparations containing living cells or latent cells of efficient strains of microorganisms that help crop plants uptake of nutrients by their interactions in the rhizosphere when applied through seed or soil. They are efficient in fixing nitrogen, solubilising phosphate and decomposing cellulose.

They are designed to improve the soil fertility, plant growth, and also the number and biological activity of beneficial microorganisms in the soil. They are ecofriendly organic agro inputs and are more efficient and cost effective than chemical fertilizers. .

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 19.
Shape of pyramid in a particular ecosystem is always different in shape. Explain with example.
Answer:
In a forest ecosystem the pyramid of number is spindle in shape, it is because the base (T1) of the pyramid occupies large sized trees (Producer) which are lesser in number. Herbivores (T2) (Fruit eating birds, elephant and deer) occupying second trophic level, are more in number than the producers. In final trophic level (T4), tertiary consumers (lion) are lesser in number than the secondary consumer (T3) (fox and snake).

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Find out the molecular explanation for the wrinkled pea seeds used by Mendel.
Answer:
The protein called starch branching enzyme (SBEI) is encoded by the wild-type allele of the gene (RR) which is dominant. When the seed matures, this enzyme SBEI catalyzes the formation of highly branched starch molecules. Normal gene (R) has become interrupted by the insertion of extra piece of DNA (0.8 kb) into the gene, resulting in r allele. In the homozygous mutant form of the gene (rr) which is recessive, the activity of the enzyme SBEI is lost resulting in wrinkled peas. The wrinkled seed accumulates more sucrose and high water content.

Hence the osmotic pressure inside the seed rises. As a result, the seed absorbs more water and when it matures it loses water as it dries. So it becomes wrinkled at maturation. When the seed has at least one copy of normal dominant gene heterozygous, the dominant allele helps to synthesize starch, amylopectin an insoluble carbohydrate, with the osmotic balance which minimises the loss of water resulting in smooth structured round seed.
Tamil Nadu 12th Biology Model Question Paper 3 English Medium 1

[OR]

(b) Write in detail about Remote sensing and its uses.
Answer:
Remote Sensing is the process of detecting and monitoring the physical characteristics of an area by measuring its reflected and emitted radiation at a distance from the targeted area. It is an tool used in conservation practices by giving exact picture and data on identification of even a single tree to large area of vegetation and wild life for classification of land use patterns and studies, identification of biodiversity rich or less areas for futuristic works on conservation and maintenance of various species including commercial crop, medicinal plants and threatened plants. .

Specific uses:

  • Helps predicting favourable climate, for the study of spreading of disease and controlling it.
  • Mapping of forest fire and species distribution.
  • Tracking the patterns of urban area development and the changes in Farmland or forests over several years.
  • Mapping ocean bottom and its resources.

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 21.
(a) Explain various edaphic factors that affect vegetation.
Answer:
The important edaphic factors which affect vegetation are as follows:

  1. Soil moisture: Plants absorbs rain water and moisture directly from the air.
  2. Soil water: Soil water is more important than any other ecological factors affecting the distribution of plants. Rain is the main source of soil water. Capillary water held between pore spaces of soil particles and angles between them is the most important form of water available to the plants.
  3. Soil reactions: Soil may be acidic or alkaline or neutral in their reaction. pH value of the soil solution determines the availability of plant nutrients. The best pH range of the soil for cultivation of crop plants is 5.5 to 6.8.
  4. Soil nutrients: Soil fertility and productivity is the ability of soil to provide all essential plant nutrients such as minerals and organic nutrients in the form of ions.
  5. Soil temperature: Soil temperature of an area plays an important role in determining the geographical distribution of plants. Low temperature reduces use of water and solute absorption by roots.
  6. Soil atmosphere: The spaces left between soil particles are called pore spaces which contains oxygen and carbon-di-oxide.
  7. Soil organisms: Many organisms existing in the soil like bacteria, fungi, algae, protozoans, nematodes, insects and earthworms, etc., are called soil organisms.

[OR]

(b) Describe the procedure involved in Blue-White colony selection methods.
Answer:
Blue- White Colony Selection Method is a powerful method used for screening of recombinant plasmid. In this method, a reporter gene lacZ is inserted in the vector. The lacZ encodes the enzyme β-galactosidase and contains several recognition sites for restriction enzyme.

β-galactosidase breaks a synthetic substrates called X-gal (5-bromo-4-chloroindolyl- β-D- galacto-pyranoside) into an insoluble blue coloured product. If a foreign gene is inserted into lacZ, this gene will be inactivated. Therefore, no-blue colour will develop (white) because p-galactosidase is not synthesized due to inactivation of lacZ.

Therefore, the host cell containing r-DNA form white coloured colonies on the medium contain X-gal, whereas the other cells containing non-recombinant DNA will develop the blue coloured colonies. On the basis of colony colour, the recombinants can be selected.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identify the proper sequence.
(a) juvenile phase, senescent phase, vegetative phase
(b) juvenile phase, maturity phase, senescent phase
(c) vegetative phase, maturity phase, juvenile phase
(d) senescent phase, juvenile phase, vegetative phase
Answer:
(b) juvenile phase, maturity phase, senescent phase

Question 2.
Which of the following symbol is used in pedigree analysis to represent unspecified sex?
Tamil Nadu 12th Biology Model Question Paper 3 English Medium 2
Answer:
Tamil Nadu 12th Biology Model Question Paper 3 English Medium 3

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 3.
Darwin’s finches are an excellent example of _______.
(a) connecting links
(b) seasonal migration
(c) adaptive radiation
(d) geographical isolation
Answer:
(c) adaptive radiation

Question 4.
A 30 year old woman has bleedy diarrhoea for the past 14 hours, which one of the following organisms is likely to cause this illness?
(a) Streptococcus pyogenes
(b) Clostridium difficile
(c) Shigella dysenteriae
(d) Salmonella enteritidis
Answer:
(c) Shigella dysenteriae

Question 5.
Which of the following microorganism is used for the production of citric acid in industries?
(a) Lactobacillus bulgaris
(b) Penicillium citrinum
(c) Aspergillus niger
(d) Rhizopus nigricans
Answer:
(c) Aspergillus niger

Question 6.
The interaction in nature, where one gets benefit on the expense of other is __________.
(a) Predation
(b) Mutualism
(c) Amensalism
(d) Commensalism
Answer:
(d) Commensalism

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 7.
Which of the following region has maximum bio-diversity?
(a) Taiga
(b) Tropical forest
(c) Temperate rain forest
(d) Mangroves
Answer:
(b) Tropical forest

Question 8.
The thickness of stratospheric ozone layer is measured in ____________.
(a) Sieverts units
(b) Melson units
(c) Dobson units
(d) Beaufort Scale
Answer:
(c) Dobson units

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
What is parthenogenesis? Give two examples from animals.
Answer:
Development of an egg into a complete individual without fertilization is known as parthenogenesis. It was first discovered by Charles Bonnet in 1745.
E.g. Honey bees, Aphis.

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 10.
Mention the production site and action site of following hormones.
(a) GnRH (b) Relaxin
Answer:

Hormone Production Site Action Site
GnRH Hypothalamus Pituitary gland
Relaxin Placenta Pelvic joints and cervix

Question 11.
Differentiate foeticide and infanticide.
Answer:
Female foeticide refers to ‘aborting the female in the mother’s womb’.
Female infanticide is ‘killing the female child after her birth’.

Question 12.
State Van’t Hoff’s rule.
Answer:
Van’t Hoff’s rule states that with the increase of every 10°C, the rate of metabolic activity is doubled or the reaction rate is halved with the decrease of 10°C.

Question 13.
Name the active chemical found in the medicinal plant Rauwolfia vomitoria. What type of diversity does it belongs to?
Answer:
Rauwolfia vomitoria can be cited as an example for genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria.

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 14.
List any four adverse effect of noise.
Answer:

  1. High blood pressure
  2. Stress related ailments
  3. Sleep disruption
  4. Hearing impairment

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosome in females so that both males and females have only one functional X chromosome per cell.

Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Ban- body.

Question 16.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional activity of the structural gene.

  • The structural gene codes for proteins, rRNA and tRNA required by the cell.
  • Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
  • The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 17.
Explain the principles of Lamarckian theory.
Answer:

  • The theory of use and disuse – Organs that are used often will increase in size and those that are not used will degenerate. Neck in giraffe is an example of use and absence of limbs in snakes is an example for disuse theory.
  • The theory of inheritance of acquired characters – Characters that are developed during the life time of an organism are called acquired characters and these are then inherited.

Question 18.
List the causative agent, mode of transmission and symptoms for Diphtheria and Typhoid.
Answer:
Tamil Nadu 12th Biology Model Question Paper 3 English Medium 4

Question 19.
ELISA is a technique based on the principles of antigen-antibody reactions. Can this technique be used in the molecular diagnosis of a genetic disorder such as Phenylketonuria?
Answer:
Yes, ELISA test can be done to diagnose phenylketonuria. The affected person does not produce the enzyme phenylalanine hydroxylase. If specific antibodies are developed against the enzyme and ELISA is performed, the unaffected person will show positive result due to antigen and antibody reaction, whereas the affected individual produces negative result.

[Note: phenylketonuria is an inherited metabolic disorder that causes the accumulation of Phenylalanine (an amino acid) in body cells due to defect in the synthesizing of an enzyme phenylalanine hydroxylase]

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) The following is the illustration of the sequence of ovarian events (a-i) in a human female.
Tamil Nadu 12th Biology Model Question Paper 3 English Medium 5
(a) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents.
(b) Name the ovarian hormone and the pituitary hormone that have caused the above- mentioned events.
(c) Explain the changes that occurs in the uterus simultaneously in anticipation.
(d) Write the difference between C and H.
Answer:
(a) A- Primordial follicle; B- Primary follicle; C- Secondary follicle; D-Tertiary follicle; E- Mature graafian follicle; F- Ovulation (release of egg); G- Empty Graafian follicle; H- Corpus luteum; I – Corpus albicans.

(b) Pituitary hormones: Follicle Stimulating Hormones (FSH) and Lutenizing Hormone (LH). Ovarian hormones: Estrogen and Progesterone.

(c) At the start of menstrual cycle, the endometrium of uterus starts regenerating through proliferation of cells induced by FSH and CH. After ovulation, the progesterone secreted by corpus luteum prepares the endometrium (uterine wall) to receive the egg if it is fertilized.

(d) C- Secondary follicle
H – Corpus luteum
During development of ovum, the primary follicle gets surrounded by many layers of granular cells and forms a new layer called secondary follicle.

Corpus luteum is the empty graafian follicle that remains after ovulation. It acts as a transitory endocrine gland secreting progesterone to maintain pregnancy.

[OR]

(b) Write short notes on the following.
(i) Brewer’s yeast
(ii) Ideonella sakaiensis
(iii) Microbial fuel cells
Answer:
(i) Brewer’s yeast – Saccharomyces cerevisiae is a widely used fungal species in preparation and softening of bakery products like dough.

(ii) Ideonella sakaiensis is a bacterium is used to recycle PET plastics. The enzyme PETase and MHETase in the bacterium breakdown the PET plastics into terephthalic acid and ethylene glycol.

(iii) A microbial fuel cell is a bio-electrochemical system that drives an electric current by using bacteria and mimicking bacterial interaction found in nature. Microbial fuel cells work by allowing bacteria to oxidize and reduce organic molecules. Bacterial respiration is basically one big redox reaction in which electrons are being moved around.

A MFC consists of an anode and a cathode separated by a proton exchange membrane. Microbes at the anode oxidize the organic fuel generating protons which pass through the membrane to the cathode and the electrons pass through the anode to the external circuit to generate current.
Tamil Nadu 12th Biology Model Question Paper 3 English Medium 6

Tamil Nadu 12th Biology Model Question Paper 3 English Medium

Question 21.
(a) Write the salient features of Human Genome Project.
Answer:

  • Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
  • An average gene consists of3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
  • The function of 50% of the genome is derived from transposable elements such as LINE and ALU sequence.
  • Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosome 13 and Y chromosome have lowest gene densities.
  • The chromosomal organization of human genes shows diversity. .
  • There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
  • Functions for over 50 percent of the discovered genes are unknown.
  • Less than 2 percent of the genome codes for proteins.
  • Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
  • Chromosome 1 has 2968 genes, whereas chromosome ‘Y’ has 231 genes.
  • Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotide polymorphism – pronounce as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

[OR]

(b) Tropical regions are rich in biodiversity. Why?
Answer:
The reasons for the richness of biodiversity in the Tropics are:

  • Warm tropical regions between the tropic of Cancer and Capricorn on either side of equator possess congenial habitats for living organisms.
  • Environmental conditions of the tropics are favourable not only for speciation but also for supporting both variety and number of organisms.
  • The temperatures vary between 25°C to 35°C, a range in which most metabolic activities of living organisms occur with ease and efficiency.
  • The average rainfall is often more than 200 mm per year.
  • Climate, seasons, temperature, humidity, photo periods are more or less stable and
    encourage both variety and numbers. .
  • Rich resource and nutrient availability.

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