Tamil Nadu 11th Biology Model Question Paper 2 English Medium

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TN State Board 11th Biology Model Question Paper 2 English Medium

General Instructions:

1. 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: 3:00 Hours
Maximum Marks: 70

Bio – Botany [Maximum Marks: 35]

PART – I

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
Red sea is red colour due to ……………………
(a) Dermacarpa sps
(b) Trichodesmium sps
(c) Scytonema sps
(d) Gloeocapsa sps.
Answer:
(b) Trichodesmium sps

Question 2.
Sago is obtained from …………………..
(a) Cycas revoluta
(b) Pinus roxburghil
(c) Pinus insularis
(d) Cedrus deodara
Answer:
(a) Cycas revoluta

Question 3.
Plants growing on the water surface are called as …………………… type of aquatic plants.
(a) Emergent
(b) Submerged
(c) Free floating
(d) Mangroves
Answer:
(c) Free floating

Question 4.
Synapsis occur between
(a) mRNA and ribosomes
(b) Spindle fibres and centromeres
(c) Two homologous chromosomes
(d) A male and a female gamete
Answer:
(c) Two homologous chromosomes

Question 5.
An example of feedback inhibition is ………………………
(a) Cyanide action on cytochrome
(b) Sulpha drug on folic acid synthesiser bacteria
(c) Allosteric inhibition of hexokinase by glucose-6-phosphate
(d) The inhibition of succinic dehydrogenase by malonate
Answer:
(c) Allosteric inhibition of hexokinase by glucose-6-phosphate

Question 6.
Rubber is a …………………..
(a) Latex
(b) Resin
(c) Alkaloid
(d) Drug
Answer:
(a) Latex

Question 7.
Identify the correct statement:
(I) Sulphur is essential for amino acids Cystine and Methionine
(II) Low level of N, K, S and Mo affect the cell division
(III) Non-leguminous plant Aims which contain bacterium Frankia
(IV) Denitrification carried out by Nitrosomonas and Nitrobacter
(a) (I), (II) are correct
(b) (I), (II), (III) are correct
(c) (I) only correct
(d) All are correct
Answer:
(b) (I), (II), (III) are correct

Question 8.
Which step is irrelevant with respect to aerobic respiration?
(a) Glycolysis
(b) Pyruvate oxidate
(c) Fermentation
(d) TCA cycle
Answer:
(c) Fermentation

PART – II

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

Question 9.
How milk is changed into curd, if a few drops of curd is added to it? What is the reason for its sourness?
Answer:
The change is brought by Lactobacillus lactis, a bacterium present in the curd. The sourness is due to the formation of lactic acid.

Question 10.
What are pyrenoids? Mention its role?
Answer:
Pyrenoids are proteinaceous bodies found in chromatophores of algae and assist in the synthesis and storage of starch.

Question 11.
From which type of flowers does the aggregate fruit develops?
Answer:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is develops into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit.

Question 12.
Define magnification. How will you calculate it?
Answer:
The optical increase in the size of an image is called magnification. It is calculated by the following formula;
Magnification =

Question 13.
Define Diffusion Pressure Deficit (DPD)?
Answer:
The difference between the diffusion pressure of the solution and its solvent at a particular . temperature and atmospheric pressure is called as Diffusion Pressure Deficit (DPD).

Question 14.
Name the two forms of phycobilins and also give an example?
Answer:
Phycobilins exists in two form. They are:

1. Phycocyanin found in Cyanobacteria.
2. Phycoerythrin found in red algae.

PART – III

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

Question 15.
Do you think shape of chloroplast is unique for algae? Justify your answer?
Answer:
Variation among the shape of the chloroplast is found in members of algae. It is Cup shaped (Chlartiydomonas), Discoid (Chara), Girdle shaped (Ulothrix), reticulate (Oedogonium), spiral (Spirogyra), stellate (Zygnema) and plate like (Mougeoutia).

Question 16.
Name few molecular markers used in molecular taxonomy?
Answer:
Allozymes, mitochondrial DNA, micro satellites, RFLP (Restriction Fragment Length Polymorphism), RAPD (Random amplified polymorphic DNA), AFLPs (Amplified Fragment Length Polymorphism), Single nucleotide Polymorphism – SNP, microchips or arrays.

Question 17.
List out the criteria for being as essential minerals?
Answer:
Amon and Stout (1939) gave criteria required for essential minerals:

1. Elements necessary for growth and development.
2. They should have direct role in the metabolism of the plant.
3. It cannot be replaced by other elements.
4. Deficiency makes the plants impossible to complete their vegetative and reproductive phase.

Question 18.
Draw a simplified diagram showing the various regions of root?
Answer:

Question 19.
What do you mean by the term – Basipetal transport and Acropetal transport?
Answer:
Basipetal means transport through phloem from shoot to root and acropetal means transport through xylem from root to shoot.

PART – IV

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

Question 20.
Describe in detail about the lytic cycle of phages with diagram?
Answer:
Lytic cycle:
During lytic cycle of phage, disintegration of host bacterial cell occurs and the progeny virions are released. The steps involved in the lytic cycle are as follows:

(I) Adsorption:
Phage (T₄) particles interact with cell wall of host (E coli). The phage tail makes contact between the two, and tail fibres recognize the specific receptor sites present on bacterial cell surface. The lipopolysaccharides of tail fibres act as receptor in phages.

The process involving the recognition of phage to bacterium is called landing. Once the contact is established between tail fibres of phage and bacterial cell, tail fibres bend to anchor the pins and base plate to the cell surface. This step is called pinning.

(II) Penetration:
The penetration process involves mechanical and enzymatic digestion of the cell wall of the host. At the recognition site phage digests certain cell wall structure by viral enzyme (lysozyme). After pinning the tail sheath contracts (using ATP) and appears shorter and thicker.

After contraction of the base plate enlarges through which DNA is injected into the cell wall without using metabolic energy. The step involving injection of DNA particle alone into the bacterial cell is called Transfection. The empty protein coat leaving outside the cell is known as ‘ghost’.

(III) Synthesis:
This step involves the degradation of bacterial chromosome, protein synthesis and DNA replication. The phage nucleic acid takes over the host biosynthetic machinery. Host DNA gets inactivated and breaks down.

Phage DNA suppresses the synthesis of bacterial protein and directs the metabolism of the cell to synthesis the proteins of the phage particles and simultaneously replication of phage DNA also takes place.

(IV) Assembly and Maturation:
The DNA of the phage and protein coat are synthesized separately and are assembled to form phage particles. The process of assembling the phage particles is known as maturation. After 20 minutes of infection about 300 new phages are assembled.

(V) Release:
The phage particle gets accumulated inside the host cell and are released by the lysis of the host cell wall.

[OR]

Draw a flow chart depicting the Bentham and Hooker Classification?
Answer:

Question 21.
With the help of diagram explain the possible route of water across root cells?
Answer:
There are three possible routes of water.
They are:-

1. Apoplast
2. Symplast and
3. Transmembrane route.

1. Apoplast:
The apoplast (Greek: apo = away; plast = cell) consists of everything external to the plasma membrane of the living cell. The apoplast includes cell walls, extra cellular spaces and the interior of dead cells such as vessel elements and tracheids.

In the apoplast pathway, water moves exclusively through the cell wall or the non-living part of the plant without crossing any membrane. The apoplast is a continuous system.

2. Symplast:
The symplast (Greek: sym = within; plast = cell) consists of the entire mass of cytosol of all the living cells in a plant, as well as the plasmodesmata, the cytoplasmic channel that interconnects them.

In the symplastic route, water has to cross plasma membrane to enter the cytoplasm of outer root cell; then it will move within adjoining cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane, till it reaches xylem.

3. Transmembrane route:
In transmembrane pathway water sequentially enters a cell on one side and exits from the cell on the other side. In this pathway, water crosses at least two membranes for each cell. Transport across the tonoplast is also involved.

Mechanism of Water Absorption Kramer (1949) recognized two distinct mechanisms which independently operate in the absorption of water in plants. They are,

1. Active absorption
2. Passive absorption.

[OR]

Explain the types of parasitic mode of nutrition in angiosperms?
Parasitic mode of nutrition in angiosperms:
Answer:
Organisms deriving their nutrient from another organism (host) and causing disease to the host are called parasites.

a. Obligate or Total parasite – Completely depends on host for their survival and produces haustoria.

(I) Total stem parasite:
The leafless stem twine around the host and produce haustoria. Example: Cuscuta (Dodder), a rootless plant growing on Zizyphns, Citrus and so on.

(II) Total root parasite:
They do not have stem axis and grow in the roots of host plants produce haustoria. Example: Rajflesia, Orobanche and Balanophora.

b. Partial parasite – Plants of this group contain chlorophyll and synthesize carbohydrates. Water and mineral requirements are dependent on host plant.

(I) Partial Stem Parasite:
Example: Loranthus and Viscum (Mistletoe) Loranthus grows on fig and mango trees and absorb water and minerals from xylem.

(II) Partial root parasite:
Example: Santalum album (Sandal wood tree) in its juvenile stage produces haustoria which grows on roots of many plants.

Bio – Zoology [Maximum marks: 35]

PART – I

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
Which is not a unit of taxonomic category?
(a) Series
(b) Glumaceae
(c) Class
(d) Phylum
Answer:
(b) Glumaceae

Question 2.
Study of ticks and mites is ……………………
(a) Acarology
(b) Entomology
(c) Malacology
(d) Carcinology
Answer:
(a) Acarology

Question 3.
Match the List – I and List – II.

List – I	List – II
1. Silver fish	(i) Book lung
2. Scorphion	(ii) Pharyngeal gills
3. Sea squirt	(iii) Lungs
4. Dolphin	(iv) Trachea

Answer:
(a) 1 – (iv), 2 – (i), 3 – (iii), 4 – (ii)
(b) 1 – (iv), 2 – (i), 3 – (ii), 4 – (iii)
(c) 1 – (i), 2 – (iii), 3 – (iv), 4 – (ii)
(d) 1 – (i), 2 – (iv), 3 – (iii), 4 – (ii)

Question 4.
MSH stands for ……………………
(a) Melanocyte stimulating hormone
(b) Malic stimulating hormone
(c) Myosin stimulating hormone
(d) Metabolic stimulating hormone
Answer:
(a) Melanocyte stimulating hormone

Question 5.
Which of the following is an correct statement?
(a) Ehler’s – Danlos syndrome – Affects collagen and results in facial abnormalities.
(b) Rhabdomyosarconca – Life threatening soft tissue tumour of head, neck and urinogenital tract.
(c) Rheumatoid arthritis – Progressive inability to secrete saliva and tears.
(d) Alzheiner’s disease – A degenerative disorder of the nervous system that affects movement often including tremors.
Answer:
(b) Rhabdomyosarconca – Life threatening soft tissue tumour of head, neck and urinogenital tract.

Question 6.
Which of the following is an incorrect statement?
(a) The functions of frontal region are Behaviour, intelligence, memory and movement.
(b) The functions of parietal region is intelligence and memory.
(c) The functions of temporal region are speech, hearing and memory.
(d) The functions of occipital region is visual processing.
Answer:
(b) The functions of parietal region is intelligence and memory.

Question 7.
Synovial fluid is present in …………………….
(a) Spinal cavity
(b) Cranial activity
(c) Freely movable joints
(d) Fixed joints
Answer:
(c) Freely movable joints

Question 8.
Glands responsible for secreting tears are ………………………….
(a) Glands of moll
(b) Lacrimal glands
(c) Meibomian glands
(d) Glands of Zeis
Answer:
(b) Lacrimal glands

PART – II

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

Question 9.
Why are spongin and spicules important to a sponge?
Answer:
Spongin and spicules provide support and supports the soft body parts of the sponges. The spicules give the sponges rigidity and form to the sponges.

Question 10.
What is water vascular system?
Answer:
The system which helps in nutrition and respiration in echinoderms is called water vascular system. Water enters into the body through special organs.

Question 11.
Difference between chordates and non-chordates?
Answer:

S.No	Chordates	Non-Chordates
1.	Notochord is present.	Absence of notochord.
2.	Pharynx perforated by gill slits.	Gill slits absent.
3.	A post anal tail is present.	Post anal tail is absent.

Question 12.
Draw the diagram of head region of Periplaneta americana?
Answer:

Question 13.
Name the different types of movement?
Answer:

1. Amoeboid movement.
2. Ciliary movement.
3. Flagellar movement.
4. Muscular movement.

Question 14.
What is endomysium?
Answer:
The connective tissue surrounding the muscle fibre is called the endomysium.

PART – III

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

Question 15.
What are the classical taxonomical tools?
Answer:
Taxonomical tools are the tools for the study of classification of organisms.
They include:-

Taxonomical keys:
Keys are based on comparative analysis of the similarities and dissimilarities of organisms. There are separate keys for different taxonomic categories.

Museum:
Biological Museums have collection of preserved plants and animals for study and ready reference. Specimens of both extinct and living organisms can be studied.

Zoological parks:
These are places where wild animals are kept in protected environments under human care. It enables us to study their food habits and behavior.

Marine parks:
Marine organisms are maintained in protected environments.

Printed taxonomical tools:
It consist of identification cards, description, field guides and manuals.

Question 16.
Differentiate Ehler’s – Danlos syndrome and stickler syndrome?
Answer:
Ehler’s – Danlos syndrome is the defect in the synthesis of collagen in the joints, heart valves, organ walls and arterial walls:
Stickler syndrome is a group of hereditary conditions affecting collagen and results in facial abnormalities.

Question 17.
Define Purkinje fibres?
Answer:
Two special cardiac muscle fibres originate from the auriculo ventricular node and are called the bundle of his which runs down into the interventricular septum and the fibres spread into the ventricles. These fibres are called the Purkinje fibres.

Question 18.
Draw the diagram of pelvic girdle with lower limb?
Answer:

Question 19.
What are the symptoms of acromegaly?
Answer:
Acromegaly is caused due to excessive secretion of growth hormone in adults. The symptoms of acromegaly are:

1. Overgrowth of hand bones, feet bones, jaw bones.
2. Malfunctioning of gonads.
3. Enlargement of viscera, tongue, lungs, heart, liver, splean and endocrine glands like thyroid, or adrenal glands.

PART – IV

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

Question 20.
Describe the digestive system of Lamipto mauritii?
Answer:
The digestive system of the earthworm consists of the alimentary canal and the digestive glands. The alimentary canal runs as a straight tube throughout the length of the body from the mouth to anus.

The mouth opens into the buccal cavity which occupies the 1s^stand 2^2nd segments. The buccal cavity leads into a thick muscular pharynx,which occupies the 3^rd and 4^th segments and is surrounded by the pharyngeal glands.

A small narrow tube, oesophagus lies in the 5^th segment and continues into a muscular gizzard in the 6^th segment. The gizzard helps in the grinding of soil particles and decaying leaves. Intestine starts from the 7^th segment and continues till the last segment.

The dorsal wall of the intestine is folded into the cavity as the typhlosole. This fold contains blood vessels and increases the absorptive area of the intestine. The imier epithelium consists of columnar cells and glandular cells. The alimentary canal opens to the exterior through the anus.

The ingested organic rich soil passes through the digestive tract where digestive enzymes breakdown complex food into smaller absorbable units. The simpler molecules are absorbed through the intestinal membrane and are utilized.

The undigested particles along with earth are passed out through the anus, as worm castings or vermicasts. The pharyngeal or salivary gland cells and the glandular cells of the intestine are supposed to be the digestive glands which secrete digestive enzymes for digestion of food.

[OR]

What are the effects of smoking?
Answer:
Today due to curiosity, excitement or adventure youngsters start to smoke and later get addicted to smoking. Research says about 80% of the lung cancer is due to cigarette smoking. Smoking is inhaling the smoke from burning tobacco. There are thousands of known chemicals which includes nicotine, tar, carbon monoxide, ammonia, sulphur-dioxide and even small quantities of arsenic.

Carbon monoxide and nicotine damage the cardiovascular system and tar damages the gaseous exchange system. Nicotine is the chemical that causes addiction and is a stimulant which makes the heart beat faster and the narrowing of blood vessels results in raised blood pressure and coronary heart diseases.

Presence of carbon monoxide reduces oxygen supply. Lung cancer, mouth cancer and larynx is more common in smokers than non – smokers. Smoking also causes cancer of the stomach, pancreas and bladder and lowers sperm count in men.

Smoking can cause lunb diseases by damaging the airways and alveoli and results in emphysema and chronic bronchitis. These two diseases along with asthma are often referred as Chronic Obstructive Pulmonary Disease (COPD).

When a person smokes, nearly 85% of the smoke released is inhaled by the smoker himself and others in the vicinity, called passive smokers, are also affected. Guidance or counselling should be done in such users to withdraw this habit.

Question 21.
Explain the structure of spinal cord?
Answer:
The spinal cord is a long, slender, cylindrical nervous tissue. It is protected by the vertebral column and surrounded by the three membranes as in the brain. The spinal cord that extends from the brain stem into the vertebral canal of the vertebral column up to the level of 1^st or 2^2nd lumbar vertebra.

So the nerve roots of the remaining nerves are greatly elongated to exit the vertebral column at their appropriate space. The thick bundle of elongated nerve . roots within the lower vertebral canal is called the cauda equina (horse’s tail) because of its appearance.

In the cross section of spinal cord, there are two indentations: the posterior median sulcus and the anterior median fissure. Although there might be slight variations, the cross section of spinal cord is generally the same throughout its length. In contrast to the brain, the grey matter in the spinal cord forms an inner butterfly shaped region surrounded by the outer white matter.

The grey matter consists of neuronal cell bodies and their dendrites, intemeurons and glial cells. White matter consists of bundles of nerve fibres. In the center of the grey matter there is a central canal which is filled with CSF. Each half of the grey matter is divided into a dorsal horn, a ventral hom and a lateral horn.

The dorsal horn contains cell bodies of interneurons on which afferent neurons terminate. The ventral hom contains cell bodies of the efferent motor neurons supplying the skeletal muscle. Autonomic nerve fibres, supplying cardiac and smooth muscles and exocrine glands, originate from the cell bodies found in the lateral horn.

In the white matter, the bundles of nerve fibres form two types of tracts namely ascending tracts which carry’ sensory impulses to the brain and descending tracts which carry motor impulses from the brain to the spinal nerves at various levels of the spinal cord. The spinal cord shows two enlargements,one in the cervical region and another one in the lumbosacral region. The cervical enlargement serves the upper limb and lumbar enlargement serves the lower limbs.

[OR]

What are the stages involved in rearing of chicken?
Answer:
Stages involved in rearing:
There are some steps involved in rearing of chicken.

1. Selection of the best layer:
An active intelligent looking bird, with a bright comb, not obese should be selected.

2. Selection of eggs for hatching:
Eggs should be selected very carefully. Eggs should be fertile, medium sized, dark brown shelled and freshly laid eggs are preferred for rearing. Eggs should be washed, cleaned and dried.

3. Incubation and hatching:
The maintenance of newly laid eggs in optimum condition till hatching is called incubation. The fully developed chick emerges out of egg after an incubation period of 21 – 22 days.

There are two types of incubation namely natural incubation and artificial incubation. In the natural incubation method, only a limited number of eggs can be incubated by a mother hen. In artificial incubation, more number of eggs can be incubated in a chamber (Incubator).

4. Brooding:
Caring and management of young chicks for 4 – 6 weeks immediately after hatching is called brooding. It can also be categorized into two types namely natural and artificial brooding.

5. Housing of Poultry:
To protect the poultry from sun, rain and predators it is necessary to provide housing to poultry. Poultry house should be moisture proof, rat proof and it should be easily cleanable and durable.

6. Poultry feeding:
The diet of chicks should contain adequate amount of water,carbohydrates, proteins, fats, vitamins and minerals.