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Students can Download 1st PUC Biology Previous Year Question Paper March 2016 (North), Karnataka 1st PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Previous Year Question Paper March 2016 (North)

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C, and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
Name the famous botanical garden of England.
Answer:
Kew Botanical garden.

Question 2.
What is gamopetalous corolla?
Answer:
If petals are united it is known as gamopetalous corolla.

Question 3.
Name the site of production of blood cells.
Answer:
Bone marrow.

Question 4.
Mitochondria are called powerhouses of a cell. Give reason.
Answer:
They produce cellular energy in the form of ATP, hence they are called ‘power houses’.

Question 5.
Name the most abundant animal protein in the world.
Answer:
Collagen.

Question 6.
What are Porins?
Answer:
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.

Question 7.
What is chlorosis?
Answer:
Chlorosis is the loss of chlorophyll leading to yellowing in leaves.

Question 8.
Mention the main cause for ‘Gout’ disease.
Answer:
Accumulation of uric acid crystals.

Question 9.
Which part of the ovary secretes progesterone.
Answer:
Corpus luteum.

Question 10.
What are metacentric chromosomes?
Answer:
The chromosome which has the middle centromere forming two equal arms of the chromosome is a metacentric chromosome.

PART-B

Answer any FIVE of the following questions in 3-5 sentences each, wherever applicable. (5 × 2 = 10)

Question 11.
What are Plasmogamy and Karyogamy?
Answer:

• The fusion of protoplasms between two motile or non-motile gametes is plasmogamy.
• The fusion of two nuclei is called karyogamy.

Question 12.
State any two universal rules of binomial nomenclature.
Answer:
(a) Biological names are generally in Latin and written in italics. They are Latinised or derived from Latin irrespective of their origin.
(b) The first word in a biological name represents the genus while the second component denotes the specific epithet.

Question 13.
Distinguish between microsporophyll and megasporophyll of gymnosperms.
Answer:
Microsporangia bearing sporophylls are known as microsporophylls and megasporangia bearing sporophylls are known as megasporophylls.

Question 14.
What are pseudocoelomates? Give an example.
Answer:
In some animals, the body cavity is not lined by mesoderm, instead, the mesoderm is present as scattered pouches in between ectoderm and endoderm.

The animals possessing them are called pseudocoelomates.
e.g. Aschelminthes.

Question 15.
Mention any four functions of the plasma membrane.
Answer:

1. Cell membrane s a selectively perméable membrane so that, it permits water and other selected materials to get in or out.
2. It helps in Pinocytosis.
3. It helps in Phagocytosis.
4. It helps in the absorption of ions and molecules. (Active transport).

Question 16.
Define glycolysis. Mention any two intermediate 3 carbon compounds produced during glycolysis.
Answer:
Partial oxidation of glucose to form two molecules of pyruvic acid is referred to as glycolysis. 3-phosphoglyceric acid and phosphoenolpyruvic acid.

Question 17.
Removal of shoot tips, make the hedge in tea plantations. Give reasons.
Answer:
In most higher plants, the growing apical bud inhibits the growth of the lateral (auxiliary) buds, a phenomenon called apical dominance. Removal of shoot tips (decapitation) usually results in the growth of lateral buds, thus it is widely applied in tea plantations for hedge making.

Question 18.
Mention four functional properties of muscle tissue.
Answer:
(a) Each muscle is made of many long, cylindrical fibres arranged in parallel arrays.
(b) Fibres are composed of numerous fine fibrils called myofibrils.
(c) Muscle fibres contract (shorten) in response to stimulation, then relax (lengthen) and return to their uncontracted state.
(d) Their action moves the body to adjust to the changes in the environment and to maintain the positions of the various parts of the body.

PART-C

Answer any FIVE of the following questions in 40-80 words each, wherever applicable: (5 × 3 = 15)

Question 19.
Describe the Diplontic life cycle pattern of the plant kingdom.
Answer:

Diplontic life cycle

• The dominant phase in the life cycle is the independent sporophytic plant.
• Meiosis occurs in the spore mother cells during spore formation and the gametophyte is short- and is, dependent on the sporophyte.
  e.g. Fucus (brown alga), gymnosperms and angiosperms.

Question 20.
Classify the flowers based on the position of the thalamus giving an example each.
Answer:
(a) Hypogynous flower – gynoecium occupies the highest position while the other parts are situated below it. Eg. mustard, brinjal.
(b) Perigynous flower – gynoecium is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level, e.g. plum rose.
(c) Epigynous flowers – margin of the thalamus grows upward enclosing the ovary completely and getting fused with it, the other parts of flower arise above the ovary. Eg. guava and cucumber.

Question 21.
Explain the steps involved in the catalytic cycle of enzyme action.
Answer:
(a) First the substrate binds to the active site of the enzyme-fitting into the active site.
(b) The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.
(c) The active site of the enzyme, now in close proximity of the substrate breaks the chemical bonds of the substrate and the new enzyme-product complex is formed.
(d) The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and run through the catalytic cycle again.

Question 22.
What is fermentation? Name any two enzymes involved in this process.
Answer:
k can be defined as a process of incomplete breaking down of food molecules into Ethyl alcohol, CO₂ and lactic acid by the activity of microorganisms or enzymes in the absence of molecular oxygen;

The equation for alcoholic fermentation is

Pyruvic acid decarboxylase and alcohol dehydrogenase.

Question 23.
List out any three physiological roles of Gibberellins in plant growth.
Answer:
Gibberellins:

• Discovered by Kurasawa in paddy seedlings, while investigating the cause for Bakane disease.
• Yabuta, and Suzuki isolated gibberellins from the fungi Gibberellafuzikor
• Gibberellins break genetic dwarfism in dwarf varieties of pea and beans.
• Can induce internode elongation and early flowering in rosette plants like cabbage and cauliflower. This is called bolting.
• Can break seed dormancy and bud dormancy.
• Can induce parthenocarpic fruits.
• Gibberellins can substitute cold treatment i.e., can replace vernalisation.
• Can be used ¡n inducing flowers even in off-seasons in long-day plants.

Question 24.
Mention the importance of the following with reference to the Human Digestive System,
(i) Epiglottis
(ii) Villi
(iii) Saliva
Answer:
(i) Epiglottis prevents the entry of food into the glottis.
(ii) Villi produce microvilli and increase the surface area for absorption of nutrients.
(iii) Saliva helps in lubricating and adhering to the masticated food particles
Salivary amylase helps in the breakdown of starch.

Question 25.
Describe the mechanism of gaseous exchange in alveolus and pulmonary capillary.
Answer:

External respiration:
The exchange of gases takes place between the alveoli of the lungs and the blood is known as ‘External respiration’. It takes place mainly by the process of diffusion. The air pressures are expressed as partial pressure. The partial pressure of O₂ is represented as P and the partial pressure of CO₂ is PO₂. The diffusion takes place from the region of their higher partial pressure to the region of their lower partial pressure.

The exchange of O₂ and CO₂ between the alveoli and blood can be expressed as follows. The alveoli of the lungs contain 107 mm of Hg of oxygen and the blood contains 45 mm of Hg of oxygen. Hence due to more partial pressure of O₂ in the alveoli, O₂ diffuses into the blood.

Meanwhile, CO₂ diffuses in the opposite direction. The P of the blood capillaries is around 46mm of Hg. White the oxygenated blood capillaries are around 46mm of Hg. While the oxygenated blood has a P of about 36 mm of Hg. Therefore CO₂ diffuses from blood capillaries to alveoli.

This can be schematically represented as follows.

Question 26.
Name any three excretory structures in invertebrates.
Answer:
Flame cells, Malpighian tubules, metanephridia.

Part-D (Section – I).

I. Answer any FOUR of the following questions in 200-250 words each. (4 × 5 = 20)

Question 27.
List out any five salient features of phylum Echinodermata.
Answer:

1. They are exclusively marine.
2. Echinoderms are rounded, cylindrical or star-shaped organisms.
3. The body is triploblastic, acoelomate, and unsegmented with radial symmetry. (But larvae are bilaterally symmetrical).
4. The body is generally covered by skin or exoskeleton supported by an endoskeleton made, up of calcareous plates/ossicles. From these plates, spines may arise which project through the epidermis (skin).
5. The Head is absent, and the body is distinguishable into oral, and aboral surfaces.
6. The coelom is extensive, enterococcus filled with coelomic fluid, and amoebocytes.

Question 28.
Draw a neat labelled diagram of T.S. of Dicot Root.
Answer:

Question 29.
Describe the structure of the nucleus with a neat labelled diagram.
Answer:

The nucleus is a dynamic celt organelle that actively and randomly controls the functioning of all other cell organelles either electrochemically or neurochemically. The shape of the nucleus varies in different cells. Normally it is spherical, but it may be oval, discoid, kidney-shaped or lobed. The size of the nucleus is also variable. Generally, there is a single nucleus ¡na cell, but some cells also contain two or more nuclei.

The nucleus is composed of the following components:
(a) Nuclear membrane: The nucleus is bounded by a doublç layered nuclear membrane with pores at intervals. The space between the two membranes is known as the per nuclear space.

(b) Nucleoplasm: Within the boundary of the nuclear membrane, a specialized cytoplasm is presently known as nucleoplasm or karyoplasm, which forms the matrix. Bathed in the nucleoplasm are present the nucleolus and chromosomes.

(c) Nucleolus: The nucleolus is a spherical dark body, seen in contact with a specific chromosome at a point called nucleolar organizer. The peripheral portion of the nucleolus is amorphous and contains RNA. The central portion is crystalline and contains DNA. It controls the activities of nucLeus and other cell organdies and is thus regarded as the ‘Nucleus of the nucleus’.

(d) Chromosomes: Chromosomes are the thread-like coloured bodies that are in the intranuclear position which act as the vehicles of heredity and variations. They are self-replicating and exhibit cyclic change ¡n size and shape. They are visible during cell division: Waldeyer described them in 1888.

Chromosomes are the self-reproducing components of the nucleus with DNA, which become visible during cell division. They exhibit a cyclic changes in size & shape.

Question 30.
Describe the events in phases of the cell cycle.
Answer:
Cell cycle:
The sequence of events that occur between the formation of a cell and its division into daughter cells, is called a cell cycle. It consists of two stages, non-living, growing interphase or interphase and a short dividing mitotic or M – phase.
1. Interphase (L Inter -between, Br. Phasis – aspect). The period between two mitotic divisions is called interphase.

• G₁ Phase: It is the postmitotic phase and takes place at the end of cell division. During this, the active synthesis of RNA and protein takes place.
• S phase: It is the synthesis phase. During this phase, duplication of DNA and centriole take place.
• G₂ Phase: It is the premitotic gap phase. The synthesis of RNA and protein continues in this phase.

2. Mitotic division: Mitotic division has two processes occurring simultaneously.
They are
(A) Karyokinesis and
(B) Cytokinesis.

A. Karyokinesis: (Or: Karyon-nucleus; kinesis-movement): It is a continuous process and for convenience divided into four stages. They are prophase, metaphase, anaphase and telophase.

B. Cytokinesis: (Gr:kytes-hollow; kinesis-movement): The division of the cytoplasm is called cytokinesis. A furrow develops in the middle of the cell. They deepen centripetally, till the cytoplasm is divided into two equal parts. It forms two daughter cells.

Question 31.
Explain the major events of cyclic photophosphorylation in photosynthesis.
Answer:
Cyclic Photophosphorylation:

1. It is a cyclic path of electrons expelled from chlorophyll through a series of substrates that are arranged in a suitable oxidation-reduction potential. The energy in the electrons is used for the phosphorylation of ADP to ATP.
2. In PS I the absorbed photons of light excite chlorophyll-a 700 to eject energised electrons on makes it positively charged and unstable. Electrons pass through the sequence → FRS → FD → Cyt b₆ → Cytf → PC and generate ATP at two places. Finally, an electron from PC returns to chi-a restoring its stability.

Question 32.
Name any five hormones secreted by the pituitary gland.
Answer:
(a)FSH
(b) Luteinizing hormone
(c) Prolactin
(d) Oxytocin
(e) Vasopressin

Section – II

II. Answer any THREE of the following questions in 200-250 words each, wherever applicable. (5 × 3 = 15)

Question 33.
Explain the structure of areolar connective tissue with a labelled diagram.
Answer:
Areolar tissue:
(a) it is present beneath the skin and in between the muscle fibres.
(b) Areolar tissue is made up of three major components.

1. Matrix.
2. Cells.
3. Fibres.

1. Matriz: It is a homogenous and transparent substance. It is also known as a ground substance. It contains five types of cells and two types of fibres.

2. Cell types :
(a) Fibroblasts or Fibrocytes: These are large and elongated cells with branched ends. They have an oval nucleus and granular cytoplasm. They secrete white and yellow fibres.

(b) Macrophages: These are large cells that move about. They swallow (ingest) foreign bodies like bacteria, thus provide on serving as a defence mechanism.

(c) Mast cells: These are spherical or oval-shaped cells with a spherical nucleus and cytoplasm. They secrete anticoagulatory substances like heparin. They also secrete histamine and serotonin.

(d) Adipose cells or Adipocytes: They are spherical cells and contain a large number of lipids.

(e) Plasma cells: They are oval-shaped cells with agranular cytoplasm and a small nucleus. They produce antibodies.

3. Fibres:
(a) White fibres: They are also called collagen fibres. as they are made up of collagen. They are tough and provide strength.
(b) Yellow fibres: They are also called elastin fibres as they are made up of elastin. They are branched and provide elasticity.

Question 34.
Comment on the following:
(i) Symplastic pathway
Answer:
The symplastic pathway is the movement of water from one cell to another through their cytoplasm and intercellular movement through the plasmodesmata.

(ii) Guttation
Answer:
The loss of water in liquid form through specialized structures called water stomata or hydathodes is called guttation.

(iii) Plasmolysis
Answer:
Plasmolysis is the phenomenon of cell shrinkage when the cell is placed in a hypertonic solution.

(iv) Transpiration
Answer:
It is the phenomenon by which excess water is eliminated, in the form of vapours through the aerial parts of the plant body.

(v) Osmosis
Answer:
It is a special type of diffusion in which only the solvent or water molecules move from a region of their higher concentration to a region of their lower concentration through a semi-permeable membrane.

Question 35.
Describe the process of nodule formation in soybean with diagrams.
Answer:

• When the root hair of a leguminous plant comes in contact with Rhizobium, it becomes curled or deformed due to the chemicals secreted by the bacterium.
• The rhizobia enter these deformed root hair and proliferate within the root hair.

Root Nodule Formation in a leguminous plant.

• The plant responds by forming an infection thread, that grows inwards to deliver the bacteria to the tissues.
• It is believed that the cytokinin is produced by the bacteria, and the auxin is produced by the plant.
  Cells stimulate cell division, and enlargement to form a nodule.
• The nodule establishes contact with the vascular tissues of the host for absorption of nutrients.
• The formation of root nodules and nitrogen fixation occurs under the control of nod genes of legumes, and the nod, if, and fix genes of bacteria.

Question 36.
Describe the structure of the human heart.
Answer:

(a) Heart chambers: The human heart is typically four-chambered. The upper chambers are called uncles or atria The lower chambers are called ventricles. The walls of the auricles are thin, whereas the walls of the ventricles are thick. The auricles are separated by inter auricular septum. Ventricles are separated by the interventricular septum, whereas the auricles and ventricles are separated by the auricula ventricular septum.

(b) Blood vessels associated with the heart: Two large veins called superior vena cava and inferior vena cava come together and form sinus venous. This in turn opens into the right atrium. Two pairs of pulmonary veins (a pair from each lung) open separately into the left atrium from the lungs.

The pulmonary artery arises from the right ventricle and divides into two branches just above the heart. The largest artery called the aorta arises from the left ventricle and runs upward-turns to the left and runs posteriorly as the dorsal aorta.

(c) Valves of the heart: Various types of valves are present at various openings to check the backward flow of blood. The left atrioventricular opening is guarded by two flap-like structures called the bicuspid valves. The free edges (cusps) of these valves are attached to the walls of the ventricles by” Chordae tend Mae”. The opening of the pulmonary artery into the right ventricle and the opening of the aorta into the left ventricle are guarded by semilunar valves.

Question 37.
Draw a neat labelled diagram of a neuron.
Answer:

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Karnataka 1st PUC Biology Previous Year Question Paper March 2016 (South)

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C, and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
Define Metabolism.
Answer:
Sum total of all chemical reactions occurring in our body is known as metabolism.

Question 2.
What is placentation?
Answer:
The arrangement of ovules within the ovary is known as placentation.

Question 3.
What are Goblet Cells?
Answer:
Unicellular isolated glandular cells in the alimentary canal that secrete mucously.

Question 4.
Gram-negative bacteria do not take a stain – Give reason.
Answer:
Glycocalyx envelop do not retain the primary stain, thus it is gram-negative.

Question 5.
Name the monomers of the protein molecules.
Answer:
Amino acids.

Question 6.
Define imbibition.
Answer:
A special type of diffusion when water is absorbed by solids (colloids) causing them to enormously increase in volume

Question 7.
What is the ‘critical concentration’ of essential elements?
Answer:
“Concentration of essential element below which plant growth is retarded.

Question 8.
Name the flat bone on the ventral midline of the Thorax of the Human body.
Answer:
Sternum.

Question 9.
Define Hormone.
Answer:
Hormones are Non-nutrient chemicals that act as intercellular messengers and are produced in trace amounts.

Question 10.
What are Polyribosomes?
Answer:
Several ribosomes attached to a single mRNA forming a chain are Polyribosomes.

PART-B

Answer any FIVE of the following questions in 3-5 sentences each, wherever applicable. (5 × 2 = 10)

Question 11.
Define species. Write the biological names of any two plants.
Answer:
A group of individual organisms with fundamental similarities is known as a species. v Mangifera indica – Mango Solarium tuberosum – potato

Question 12.
Write any four characters of the virus.
Answer:
Living characters:
(a) Like organisms they contain protein and nucleic acid.
(b) Like organisms they unçergo mutation.

Non – living characters:

1. They are non-cellular.
2. They do not show metabolism.

Question 13.
Enumerate four salient features of Pteridophytes.
Answer:
(a) The main plant body is a sporophyte which is differentiated into true root, stem, and leaves.
(b) They possess well-differentiated vascular tissues.
(c) The leaves are small (microphylls) or large (macrophytes)
(d) The sporophytes bear sporangia that are subtended by leaf-like appendages called sporophylls.

Question 14.
Mention the function of the following organs.
(i) Malpighian tubules
(ii) Comb plates
Answer:
(i) Excretion
(ii) Locomotion

Question 15.
Classify the chromosomes based on the position of the centromere.
Answer:
(a) Metacentric
(b) Submetacentric
(c) aerocentric
(d) telocentric

Question 16.
What is a Respiratory Quotient? Mention the R.Q. value of proteins.
Answer:
It is the ratio of the volume of CO₂ evolved to the volume of O₂ consumed to break a molecule of respiratory substrate.
RQ = \(\frac{\text { volume of } \mathrm{CO}_{2} \text { evolved }}{\text { volume of } \mathrm{O}_{2} \text { consumed }}\)
RQ of protein = 0.9

Question 17.
Distinguish between Dedifferentiation and Redifferentiation
Answer:
Cells that have lost the capacity of division, regaining the capacity of division at certain conditions is called Dedifferentiation.

Dedifferentiated cells lose the capacity of division, mature, and perform specific functions in redifferentiation.

Question 18.
Name any four types of Synovial joints.
Answer:
Ball and socket, hinge joint, pivot joint, and saddle joint.

PART-C

Answer any FIVE of the following questions in 40-80 words each, wherever applicable: (5 × 3 = 15)

Question 19.
Pigments are the basis of the classification of Algae – Justify the statement.
Answer:
Pigments vary in different classes of algae, this is the basis for classification.
(a) Class: Chlorophyceae – chlorophyll a, b
(b) Class: Phaeophyceae – chlorophyll a, c, fucoxanthin.
(c) Class: Rhodophyceae – chlorophyll a, d, phycoerythrin.

Question 20.
Give the meaning of the following with an example each.
(a) Parallel venation
(b) Opposite Phyllotaxy
(c) Trimerous flower
Answer:
(a) Parallel venation: Veins running parallel to each other e.g. grass, coconut, maize.
(b) Opposite Phyllotaxy: Pair of leaves arise at each node that is opposite to each other, e.g. Calatropis, guava.
(c) Trimerous flower: Floral appendages are in multiples of three e.g.: Onion, coconut.

Question 21.
Mention any six secondary metabolites of plants.
Answer:
Carotenoids morphine, codeine, oils, rubber, resins.

Question 22.
Discuss the steps and the major pathways of Anaerobic respiration.
Answer:
The first step of anaerobic respiration is glycolysis, in which glucose is incompletely oxidized to 2 pyruvic acid molecules with 2 ATP and 2NADH₂ formed as by-products.

Pyruvic acid formed is used for alcoholic fermentation or lactic acid fermentation.

In alcoholic fermentation, the pyruvic acid is first decarboxylated to acetaldehyde liberating CO₂. Later acetaldehyde is reduced to ethyl alcohol by NADH₂ of glycolysis.

This process occurs in anaerobes like yeast and bacteria.

In lactic acid fermentation, the pyruvic acid is directly reduced to lactic acid by NADH₂ of glycolysis. NO CO₂ is liberated. It occurs in lactobacilli and skeletal musclçs.

Question 23.
Mention the three phases of growth.
Answer:
(a) Cell division or meristematic phase.
(b) Cell elongation
(c) Cell maturation.

Question 24.
Describe the process of carbohydrate digestion in the small intestine.
Answer:
Pancreatic juice contains pancreatic amylase which can hydrolyze starch into disaccharides.

Succus entericus contain disaccharidases that act on disaccharides.

Question 25.
Summarise the mechanism of transport of oxygen in the blood.
Answer:
(a) Oxygen can bind with hemoglobin in a reversible manner to form oxyhemoglobin.
(b) Each hemoglobin molecule can carry a maximum of four molecules of O₂.
(c) Binding of oxygen with hemoglobin is primarily related to the partial pressure of O₂, partial pressure of CO₂, hydrogen ion concentration, and temperature.
(d) Every 100 ml of oxygenated blood can deliver around 5 ml of O₂ to the tissues under normal physiological conditions.

Question 26.
Mention any three disorders of the human excretory system.
Answer:
(a) Uremia
(b) Renal failure
(c) Renal calculi

PART-D (Section – I)

I. Answer any FOUR of the following questions in 200-250 words each, wherever applicable. (4 × 5 = 20)

Question 27.
Enumerate any four salient features of class Amphibia, giving any two examples.
Answer:
Class Amphibia:
The class amphibian (“Amphi-both” brain-life) includes cold-blooded vertebrates adapted to both land and water, as such they have made a “compromise between the t environments”.

1. Amphibians are co4id blooded (Poildiotberms) aquatic or terrestrial, vertebrates with limbs, and lungs.
2. The body is divided into head and trim (but in a few forms it is divided into head, neck, trunk, and tail).
3. The skin is smooth without scales and is always moist (due to the presence of glands). It is highly vascularized, thus cutaneous respiration is seen in an amphibian.
4. The paired fins of fishes are modified into limbs ¡n amphibia. Two pairs of limbs (forelimbs, and hind limbs) are present which are pentadactyl.
5. At the head region a pair of eyes, a pair of external nostrils, mouth, and a pair of eardrums are present.
6. The mouth is provided with teeth on one or both the jaws, and a movable tongue.
7. Eyes are often provided with a well-developed nictitating membrane.
8. Amphibians are devoid of the exoskeleton, but they possess a large bony endoskeleton.
9. The digestive system is well developed.
10. Respiration is by gills, book lungs, and skin or by the lining of the buccal cavity.
11. The circulatory system comprises a three-chambered heart and veins. The heart has double circulation (i.e. system and pulmonary circulation) RBC are oval, biconcave, and nucleated.
12. The nervous system has a brain and a pair of cranial nerves.
13. The excretory system and reproductive system are associative to form a urinògenital system. The excretory system consists of a pair of mesonephric kidneys and a large urinary bladder. Presence of cloacae into which digestive, excretory, and reproductive systems open out, through the cloacal aperture.
14. Sexes are separate. The reproductive system comprises paired gonads and products. The majority of them are oviparous and the eggs are non-Cleidoic. Fertilization is external.
15. Development is indirect. Larva ¡s called tadpole.

Note 1: During indirect development metamorphosis takes place.
Note 2: Pedogenesis ¡s seen in a few members like Ambystoma, where the development of ova into a young one takes place.

Limbless amphibians – Ichthyophis.
Tailed amphibians – Salamander, Newts, and Necturus.
Tailless amphibians – Frogs, Toads (Bufo, Rana, Hyla).

Question 28.
Draw a neat labeled diagram of T.S. of Dorsiventral leaf.
Answer:
Anatomy of a Dicot leaf or Dorsiventral leaf:

Question 29.
Explain the structure of chloroplast with a neat labeled diagram.
Answer:

Electron microscopic structure of the chloroplast

Chlorophyll-containing plastids are called Chloropbsts. They are present in the cells of all green plants and abundant in the leaf mesophyll cells. They are elliptical or oval in shape.

Chloroplast is bounded by two membranes with the intermembrane space called peri plastidial space, containing peri plastidial fluid. It is made up of H₂O, mineral ions, proteins, etc. It is a lubricating fluid that avoids friction between the two membranes. The inner chamber is filled with a colorless proteinaceous fluid matrix called the stroma. Besides abundant proteins, stroma contains 70s ribosomes, circular DNA, and all the enzymes of the Calvin cycle.

Embedded in the stroma, there are green-colored bodies called grana which are inter-connected by frets. Grana is the site of light reaction, each granum consists of a stack of lipoprotein membrane discs called Thylakoids. Each thylakoid contains several photosynthetic centers called Quantasomes. Each quantasome contains about 250 chlorophyll pigments and a few xanthophylls and carotenes.

Only chlorophyll- is capable of harvesting light energy into photosynthesis and hence it is called the primary photosynthetic pigment. All the other pigments, merely absorb light energy and pass it on to chlorophyll, and hence are called accessory photosynthetic pigments.

Note: As chloroplasts contain circular DNA 70s ribosomes and as they go for protein synthesis, they are regarded as semi-autonomous cell organelles.

Question 30.
Write any five differences between Mitosis and Meiosis.
Answer:

Mitosis	Meiosis
(a) Equational division	(a) Reductional division
(b) Somatic cell division	(b) Reproductive cell division
(c) Two cells are formed	(c) Four cells are formed
(d) One-step division	(d) Two consecutive divisions occur
(e) Chromosomes align on an equatorial plate and the centromeres divide and the chromatids start moving towards the two opposite poles	(e) Homologous chromosomes align on the equatorial plate. Half the number of chromosomes will move to opposite poles without division of centromeres.

Question 31.
Explain Joseph-Priestlyls experiments on photosynthesis using a candle and a plant.
Answer:

Priestley’s experiment

(a) Priestly performed an experiment to reveal the essential role of air in the growth of green plants.
(b) Priestly observed that a candle burning in a closed space – a bell jar soon gets extinguished.
(c) Similarly a mouse would soon suffocate in a closed space.
(d) He concluded that a burning candle or an animal that breathes the air somehow damages the air.
(e) But when he placed a mint plant in the same bell jar he found that the mouse stayed alive and the candle continued to bum. He concluded that plants restore the air whatever breathing animals and burning candles remove.

Question 32.
Mention any one Hormone secreted by the following glands.
(i) Adrenal gland1
(ii) Thymus gland
(iii) Pineal gland
(iv) Ovary
(v) Testes
Answer:
(i) Adrenal gland – Adrenaline/noradrenaline
(ii) Thymus gland – Thymosins
(iii) Pineal gland – Melatonin
(iv) Ovary – estrogen .
(v) Testis – Testosterone

Section – II

II. Answer any THREE of the following questions in 200-250 words each, wherever applicable. (3 × 5 = 15)

Question 33.
Mention the function of the following organs.
(i) Forewings of Cockroach
(ii) Crop in the alimentary canal of Cockroach
(iii) Lyphlosole in the alimentary canal of Earthworm
(iv) Setae in Earthworm
(v) Spiracles of Cockroach.
Answer:
(i) Protecting and covering hindwings
(ii) Storage of food
(iii) Increase the area of absorption
(iv) Locomotion
(v) Exchange of gases

Question 34.
Describe the structure of the stomatal apparatus in Dicot plants.
Answer:

Dicot stomatal apparatus includes,
(a) Stoma: a minute opening present between the guard cells.
(b) Guard cells: Specialised epidermal cells surrounding the stoma. These are two in number and are kidney-shaped. Their outer walls are thin and elastic but inner walls are thick and less elastic.

Question 35.
Schematically represent the Nitrogen cycle and explain.
Answer:
Nitrogen Cycle:

• Nitrogen is a limiting nutrient for both natural, and agricultural ecosystems.
• It exists as two nitrogen atoms held together by strong triple covalent bonds (N = N).
• The nitrogen cycle involves the following steps:
  1. Nitrogen fixation
  2. Ammonification
  3. Nitrification and
  4. Denitrification.

1. Nitrogen Fixation:

• The process of conversion of nitrogen into ammonia and) or other nitrogen compounds, is known as nitrogen fixation.
• In nature, lighting and ultraviolet radiation provide energy, to convert nitrogen into nitrogen compounds like N₂O, NO, and NO₂.
• The atmosphere also gets some amount of nitrogen oxides from forest fires, automobile exhausts, industrial combustion, and power generating stations.

2. Ammonification:

• The process by which organic nitrogenous compounds arc decomposed to produce ammonia is known as ammonification.
• Some ammonia volatilizes and re-enters the atmosphere.
• Most of the ammonia is converted into nitrites and nitrates.

3. Nitrification:

• Nitrification is the process of converting ammonia first into nitrite, and then into nitrate.
  2NH₃ + 3O₂ → 2NO₂ + 2H + 2H₂O
  2NO₂ + O₂ → 2NO₃
• These processes are carried out by soil bacteria that are chemoautotrophs.
• Ammonia is oxidized into nitrite by Nitrosomonas and N enterococcus.
• Nitrite is oxidized to nitrate by Nitrobacter,
• The nitrates are absorbed by the plants. and reduced to nitrites.
• The nitrites are transported to the leaves and reduced to ammonia, which forms the amino group of amino acids.

4. Denitrification :

• It is the process of conversion or reduction of nitrates into free nitrogen.
• It ¡s carried out by bacteria like Pseudomonas and Thiobacillus.

Question 36.
(a) Name the formed elements of Blood.
Answer:
Erythrocytes, Leucocytes, Thrombocytes

(b) Write a note on Erythroblastosis fetal.
Answer:
This disease is due to the incompatibility of Rh+ve, and Rh-ve blood which may result in the death of the child before or soon after birth. This problem arises when an Rh-ve woman marries a Rh+ve man and bears a Rh+ve fetus. Sometimes due to some placental defect, some of the fetal RBC carrying Rh antigens may pass into the mother’s bloodstream, and cause the production of anti-Rh antibodies. The concentration of anti-Rh antibodies is gradually built up in the mother, and she becomes sensitized for anti-Rh antibodies. There is no risk in the first pregnancy because enough antibodies are not found to destroy the fetus.

In the second or subsequent pregnancies involving a Rh+ve child, these anti-Rh antibodies may return to the fetus through the placenta and destroy the Rh antigen carrying RBC of the fetus and is referred to as Haemolysis. This condition is called the hemolytic disease of newborns (HDN), and the condition is identified as erythroblastosis fetal. The disease tends to become worse in each subsequent pregnancy.

Question 37.
Draw a neat labeled diagram of the sagittal section of the Human Brain.
Answer:

Students can Download 1st PUC Biology Previous Year Question Paper March 2017 (North), Karnataka 1st PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Previous Year Question Paper March 2017 (North)

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C, and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
Name the modification of leaves for climbing.
Answer:
LeafTendrils.

Question 2.
Give an example of a medicinal plant of the family Solanaceae.
Answer:
Belladonna/ashwagandha.

Question 3.
Name the dense regular connective tissue which attaches the skeletal muscles to bone.
Answer:
Tendons.

Question 4.
What is a telocentric chromosome?
Answer:
A chromosome with a terminal centromere is called a telocentric chromosome.

Question 5.
Why mitosis is called ‘equational division’?
Answer:
Since the number of chromosomes in the parent and progeny cells is the same, mitosis is called equational division.

Question 6.
Name the phase of Prophase-1 of meiosis during which synapsis takes place.
Answer:
Zygotene.

Question 7.
What are porins?
Answer:
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.

Question 8.
The deficiency symptoms of Nitrogen in plants, tend to appear first in the older ‘ tissues, why?
Answer:
Nitrogen is readily mobilized from older dying leaves to younger leaves, thus deficiency symptoms tend to appear first in the older tissues.

Question 9.
What happens when a plant cell is placed in a hypertonic solution to the protoplasm.
Answer:
The protoplasm will shrink away from the walls due to plasmolysis.

Question 10.
Define transamination.
Answer:
It involves the transfer of an amino group from one amino acid to the keto group of a ketoacid.

PART-B

Answer any FIVE of the following questions in 3-5 sentences each, wherever applicable. (5 × 2 = 10)

Question 11.
Name the phases of the life cycle in plants, which alternate with each other.
Answer:
Gametophyte and sporophyte.

Question 12.
Differentiate chordates from Non-chordates by giving two differences.
Answer:

Chordata	Non-Chordata
(a) Presence of a solid notochord or a vertebral column.	(a) Absence of Chordadorsalis.
(b) Presence of a dorsal and tubules nerve chord.	(b) Presence of ventral and solid nervechord.
(c) Presence of pharyngeal gillsiits.	(c) Absesnce of gilislits.
(d) Presence of ventral heart.	(d) Presence of dorsal heart.

Question 13.
What is Venation? Mention the types.
Answer:
The arrangement of veins and veinlets in the lamina of the leaf is termed as venation, types :
(a) Reticulate
(b) Parallel.

Question 14.
List the key features of the Metaphase of mitosis.
Answer:

1. Spindle fibers attach to kinetochores of chromosomes.
2. Chromosomes are moved to the spindle equator and get aligned along with the metaphase plate through spindle fibers to both poles.

Question 15.
What are micronutrients? Name any two of them.
Answer:
Mineral elements required in very small amounts ie, less than 10 mmole kg^-1 of dry matter are known as micronutrients, eg: Zinc, copper.

Question 16.
What is biological nitrogen fixation? Give an example each for free-living and symbiotic nitrogen-fixing bacteria.
Answer:
The reduction of nitrogen to ammonia by living organisms is called biological nitrogen fixation.

• Free-living nitrogen-fixing bacteria: A acetobacter.
• Symbiotic nitrogen-fixing bacteria: Rhizobium.

Question 17.
All the minerals cannot be passively absorbed by the roots give two reasons.
Answer:
All minerals cannot be passively absorbed by the roots as

1. Minerals are present in the soil as charged particles (ions) that cannot move across cell membranes.
2. The Concentration of minerals in the soil is usually lower than the concentration of minerals in the root.

Question 18.
Differentiate guttation from transpiration.
Answer:

Transpiration	Guttation
1. Transpiration is a universal phenomenon that occurs in all plants.	1. Guttation occurs only in herbs, especially in the members of the Milygraminae.
2. It occurs through stomata, cuticles,s, or lenticels.	2. it occurs through hydathodes or water Stomata.
3. Amount of water lost is enormous.	3. Water loss is negligible.
4. Transpired water is pure, and is in the form of vapor.	4. Guttated water contains nitrogenous wastes, amino acids, etc. it is impure and is in the form of droplets.

PART-C

Answer any FIVE of the following questions in 40-80 words each, wherever applicable: (5 × 3 = 15)

Question 19.
List any six taxonomic categories.
Answer:
Species, Genus, Family, order, class, phylum, kingdom.

Question 20.
Classify the plastids on the basis of pigments.
Answer:
(a) Chloroplasts: Contain more chlorophylls and few carotenoids.
(b) Chromoplasts: Contain carotenoids.
(c) Leucoplasts: Colourless plastids with stored nutrients.

Question 21.
Name the three major types of cells and their secretions of gastric glands.
Answer:
(a) Mucus neck cells: secrete mucus.
(b) Peptic or chief cells: secrete proenzyme pepsinogen.
(c) Parietal or oxyntic cell: secrete HCl and intrinsic factor essential for the absorption of vitamin B12.

Question 22.
Name any three respiratory disorders.
Answer:
Asthma, Emphysema, and occupational respiratory disorders.

Question 23.
Briefly explain the structure of myosin.
Answer:
Myosin (Thick) Filament: length is about 1.6 nm and breadth the 15nm.

1. Myosin is a polymeric protein, whose monomeric proteins are called meromyosins.
2. Each meromyosin has two important parts, a globular head with a short arm and a tail.
3. The head with the short arm is called the heavy meromyosin (HMM) and the tail is called light meromyosin (LMM).
4. The HMM component projects outwards at a regular distance at an angle from the surface of the polymeric meromyosin; it is known as a cross arm.
5. The globular head functions as an ATPase enzyme and has the binding sites for ATP and active sites of actin.

Question 24.
List the hormones secreted by the Parsdistalis region of the pituitary.
Answer:
Growth hormone [GH], prolactin [PRL]
Thyroid-stimulating hormone [TSH]
Adreno corticotrophic hormone [ACTH]
Luteinizing hormone [LH]
Follicle-stimulating hormone [FSH],

Question 25.
Which are the organs involved in the regulation of kidney functioning.
Answer:
Hypothalamus, Juxta Glomerulus (diurnal) rhythm of our body.

Question 26.
Mention any three functions of melatonin.
Answer:

1. Regulation of a 24 hour (diurnal) rhythm of our body.
2. Maintaining the normal rhythms of sleep-wake cycle body temperature.
3. Influences metabolism, pigmentation, and menstrual cycle.

Part-D (Section – I).

I. Answer any FOUR of the following questions in 200-250 words each. (4 × 5 = 20)

Question 27.
List any five general characteristics of Pteridophytes.
Answer:
Pteridophytes are characterized by the following features:
1. Their life cycle, comprises a distinct diploid phase represented by a sporophyte and a haploid phase by a gametophyte exhibiting heteromorphic alternation of generations.

2. Sporophyte represents the dominant phase of the life cycle, which is an entirely independent plant body. It bears roots, stems, and leaves. It reproduces asexually by sporangia.

3. The sporangia may be arranged in specific groups called sorus. In a few cases, sporangial structures aggregate on a specialized reproductive structure called strobilus or cone.

4. In sporophyte the vegetative structure shows highly specialized conducting tissues namely xylem and phloem. These tissues are organized into the vascular system called stele.

5. The gametophyte through independence is short-lived. It bears both male and female reproductive structures namely, antheridia and archegonia respectively.

6. Gametophytes thus show Oogamous types of sexual reproduction like Bryophytes. Water is essential for fertilization. The zygote develops into a multicellúlar embryo. Hence, the group is treated under embryophytes Siphonogama (Siphonogama – as it possesses vasculature).

Question 28.
Write any four salient features of the phylum. Arthropoda and name any two economically important insects.
Answer:
(a) General characters:
1. Arthropods are bilaterally symmetrical, triploblastic, metamerically segmented, and coelomates.

2. Segmentation of the body is less apparent and the number of segments is fewer in number.

3. They show the highest degree of cephalization, that is the formation of a well-developed head. The Head is usually formed by the fusion of six body segments. Following the head, the region is the trunk which is usually divisible into an anterior thorax and a posterior abdomen.

4. Their body segment usually bears paired lateral and jointed appendages. A segment of the appendage is known as podomere. The appendages are variously modified in different groups and even in different parts of the same animal.

• A few appendages close to the mouth are modified into jaws and are helpful in feeding.
• A few appendages are sensory in function

Some appendages help, in locomotion. Due to the presence of jointed appendages or legs for locomotion, the phylum gets the name Arthropoda, which means jointed feet.

(b) Apis (honey bee) and bombyx (silkworm)

Question 29.
Mention any five important characteristics of Prokaryotic cells.
Answer:

1. Cell envelop consists of a tightly bound three-layered structure i.e., the outermost glycocalyx followed by the cell wall and then the plasma membrane.
2. A special membranous structure is a mesosome that is formed by the extensions of the plasma membrane into the cell.
3. The fluid matrix falling from the cell is the cytoplasm.
4. There is no well-defined nucleus. The genetic material is basically naked. Many bacteria have small circular DNA outside the genomic DNA called plasmids.
5. No organelles except for ribosomes.
6. Reserve food material is present in the form of inclusion bodies.

Question 30.
What are ribozymes? Describe the catalytic cycle of enzyme action.
Answer:
Some nucleic acids (RNA) behave like enzymes. These are called ribozymes.
The catalytic cycle of enzyme action can be described in the following steps:

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 of 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 bond to another molecule of the substrate is ready to bond to another molecule of the substrate and run through the catalytic cycle once again.

Question 31.
Draw a neat labeled diagram of the anatomy of the Dicot leaf.
Answer:

Helianthus lear

Question 32.
Brief explain Calvin cycle with schematic representation.
Answer:
It occurs in the grana of the chloroplast.
It utilizes the assimilatory powers i.e., ATP and NADPH₂ produced in 11gm reaction.

Steps:

1. Phosphorylation
2. Carbon dioxide fixation
3. Phosphorylation and reduction.
4. Regeneration and RUMP
5. Glucose formation.

Note: To complete a Calvin cycle, 18 ATPs and 12 NADPH₂ molecules are required.

Section – II

Answer any THREE of the following questions in 200-250 words each, wherever applicable. (5 × 3 = 15)

Question 33.
List any five physiological effects of auxins in plants.
Answer:
Auxins:

• Auxins were the first hormones to be discovered in plants.
• F.W. Went confirmed and isolated the auxins from coleoptiles of Avena saliva.
• Chemically it is called indole-3-Acetic Acid and is derived from amino acid tryptophan.
• Auxins can stimulate cell division, cell elongation, and cell maturation.
• It promotes Apical dominance.
• It initiates Root formation.
• Can promote parthenocarpy.
• Prevent premature fall of flowers, buds, and leaves.
• 2, 4D, and 2, 4, 5T are used as selective weedicides.
• Promote phototrophic and hydrotropic movements.
• Promote xylem differentiation.

Question 34.
Explain the structure of the alimentary canal of Cockroach.
Answer:

Question 35.
Explain the process of double circulation in humans.
Answer:
Double circulation:
The human heart acts as a double pump.

The right side of the heart receives, and pumps deoxygenated blood and the left side of the heart receives and pumps oxygenated blood and hence both the routes are kept completely separate and there is no mixing of oxygenated and deoxygenated blood. This type of circulation is said to be complete double circulation.

Double circulation involves shorter pulmonary circulation and longer systematic circulation.

Question 36.
(a) Compare aerobic respiration and fermentation.
(b) What is the respiratory quotient? Write the RQ value of carbohydrates.
Answer:
(a) Aerobic respiration: It is a completely oxygen-dependent process, and occurs in the presence of molecular oxygen. It is centered in mitochondria. Here glucose is completely oxidized into CO₂ and water. It generates 38 ATP per glucose, and hence a more efficient energy-releasing process. It occurs in the majority of living organisms called aerobes.
C₆ H₁₂ O₆ + 6O₂ → 6CO₂ + 6H₂O + 686kcal + 38ATP mol

Anaerobic respiration: It is an oxygen-independent procéss and occurs in the absence of oxygen. It is centered in general cytoplasm. Here glucose is incompletely oxidized int0 ethyl alcohol. and CO₂ or lactic acid generates only 2 ATPs per glucose molecule and hence less efficient energy-releasing process. It occurs only, in a few orgánisms like bacteria, yeast, etc.

(b) It is defined as the ratio of the volume of CO₂ evolved to the volume of O₂ utilized.
RQ = \(\frac{\text { Volume of } \mathrm{CO}_{2} \text { released }}{\text { Volume of } \mathrm{O}_{2} \text { utilised }}\)
RQ value for carbohydrates is one.

Question 37.
Draw a neat labeled diagram of a neuron.
Answer:

Students can Download 1st PUC Biology Previous Year Question Paper March 2017 (South), Karnataka 1st PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Previous Year Question Paper March 2017 (South)

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C, and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
What is parthenocarpic fruit?
Answer:
If the fruit is formed without fertilization of the ovary.

Question 2.
Define Staminode.
Answer:
A sterile stamen is known as a staminode.

Question 3.
Earthworms are known as friends of farmers. Why?
Answer:
Because they make burrows in the soil and make it porous which helps in respiration and penetration of the developing plant roots.

Question 4.
Name the powerhouses of the cell.
Answer:
Mitochondria.

Question 5.
Define Osmosis.
Answer:
Osmosis can be defined as the diffusion of water across a semi-permeable membrane.

Question 6.
What is Leg-haemoglobin?
Answer:
Leg-haemoglobin is a pink-colored pigment present in the root nodule and acts as an oxygen scavenger.

Question 7.
Name the gaseous plant growth regulator.
Answer:
Ethylene.

Question 8.
What is plasticity?
Answer:
Plants follow different pathways in response to the environment or phases of life to form different kinds of structures. This is known as plasticity.

Question 9.
Expand ECG
Answer:
Electrocardiograph.

Question 10.
Name the region of the vertebrate eye where the optic nerve passes out of the retina.
Answer:
Blindspot.

PART-B

Answer any FIVE of the following questions in 3-5 sentences each, wherever applicable. (5 × 2 = 10)

Question 11.
Name the algal and’ fungal compnent of Lichens.
Answer:
The algal component is phycobiont and the fungal component is mycobiont.

Question 12.
Mention four major groups of protozoans.
Answer:
Four groups of protozoans are:
(a) Flagellated protozoans.
(b) Ciliated protozoans.
(c) Amoeboid protozoans.
(d) Sporozoan protozoans.

Question 13.
Write any two general characters of phylum Ctenophora.
Answer:
(a) Ctenophores are exclusively marine, radially symmetrical, diploblastic organisms with tissue level of organization.
(b) Bioluminescence (emit light) is well marked in ctenophores.

Question 14.
What are macronutrients? Mention any two of them.
Answer:
Nutrients required in large amounts (in excess of 10 m mole kg^-1 dry weight of the plant body are known as macronutrients, eg: carbon and nitrogen

Question 15.
Define
(a) Ammonification
(b) Hydroponics.
Answer:
(a) Decomposition of organic nitrogen of dead plants and animals into ammonia.
(b) Technique of growing plants in a nutrient solution.

Question 16.
List any two physiological effects of Auxins.
Answer:
(a) Auxins help to initiate rooting in stem cuttings.
(b) Auxins promote flowering eg: in pineapples.

Question 17.
Write the differences between Systole and Diastole.
Answer:
Systole is the contracted state of the chambers of the heart.
Diastole is the relaxed state of the chambers of the heart.

Question 18.
Name the antigens and antibodies present in the A and B blood groups.
Answer:

PART-C

Answer any FIVE of the following questions in 40 to 80 words each, wherever applicable. (5 × 3 = 15)

Question 19.
List and explain any two taxonomical aids.
Answer:
(a) Herbarium: It is a storehouse of collected plant specimens that are dried, pressed, and preserved on sheets. Further, these sheets are arranged according to a universally accepted system of classification.
(b) Botanical gardens: These specialized gardens have collections of living plants for reference.

Question 20.
Explain three types of Phyllotaxy.
Answer:
(a) Alternate phyllotaxy: a single leaf arises at each node, eg: china rose.
(b) Opposite Phyllotaxy: a pair of leaves arise at each node and lie opposite to each other eg: Calotropis.
(c) Whorled phyllotaxy: More than two leaves arise at a node and form a whorl, eg: Alstonia.

Question 21.
List any three types of chromosomes based on the position of the centromere.
Answer:
(a) Metacentric chromosome: chromosome with middle centromere forming two equal arms.
(b) Sub-metacentric: chromosome has centromere slightly away from the middle of the chromosome resulting in one shorter arm and one long arm.
(c) Acrocentric: centromere is situated close to its end forming one extremely short and one very long arm.
(d) Telocentric: chromosome has a terminal centromere.

Question 22.
Classify the plastids ba&d on the type of pigments.
Answer:

• Chloroplasts: contain more chlorophylls and few carotenoids.
• Chromoplasts: contain carotenoids.
• Leucoplast: colorless plastids with stored nutrients.

Question 23.
Name any three significances of Mitosis.
Answer:
Significance of Mitosis:

• The somatic chromosome number (2n) in all the cells of an organism is effectively maintained by mitosis. The duplication of chromosomes occurs during interphase and their orderly behavior during the different stages of mitosis maintains a constant chromosome number.
• The daughter cells produced as a result of mitosis are quantitatively and qualitatively identical.
• Mitosis of somatic cells helps in the growth and development of multicellular organisms.
• It forms a means of multiplication in some of the over organisms eg: Amoeba, Eugene, etc.
• Mitosis produces new cells for the healing of wounds and for regeneration.

Question 24.
Mention three steps in Urine formation.
Answer:
(a) Ultrafiltration
(b) reabsorption
(c) tubular secretion.

Question 25.
Write a note on Actin.
Answer:
Actin (Thin) filament: length is about 1.0 nm and breadth 9 nm globules

• Actin filament contains three proteins namely actin (F-actins and G-acts), tropomyosin, and troponin.
• Each actin filament consists of two F (filamentous)-actins, that are helically wound to each other.
• Each F-actin is a polymer of G-globular actins.
• Two filaments of tropomyosin also run close to the F-actins throughout their length.
• Troponin is a complex protein found at regular intervals on the tropomyosin.
• In the resting state, a subunit of troponin masks the actin-binding site of myosin.

Question 26.
Name three types of joints.
Answer:
(a) Fibcousjoint
(b) Cartilagenous joint
(c) Synovial joint

PART-D (Section – I)

I. Answer any FOUR of the following questions in 200-250 words each, wherever applicable. (4 × 5 = 20)

Question 27.
List any five general characteristics of Gymnosperms.
Answer:
1. The life cycle has a distinct, dominant, diploid, asexual phase represented by the well-differentiated evergreen woody plant, which is known as the sporophyte.

2. The sporophyte is heterosporous bearing microspores and megaspores within microsporangia and megasporangia respectively. These structures occur on leaf-like microsporophylls and megasporophylls. These are further organized into fertile structures called strobila or cones.

3. Sporophyte shows the presence of a taproot system which is well developed. The stem possesses branches that are dichotomies. Leaves are well developed and are dimorphic (two types of leaves): viz,

• Green photosynthetic leaves (Foliage).
• Brown-colored scale leaves.

4. Microspore develops into male gametophyte and megaspore produces female gametophyte. The gametophytes represent the haploid phase and are highly inconspicuous in comparison with sporophytic generation.

5. Female gametophyte is enclosed within a megasporangium that in turn is covered by an integument. Such an integuments megasporangium possessing the female gametophyte is called the ovule.

Question 28.
Write any four general characteristics of class Mammalia. Mention an oviparous Mammal.
Answer:
(a) Class Mammalia:
1. Class Mammalia includes warm-blooded and highly organized animals possessing mammary (milk) glands for suckling their young ones. They live almost in all the environments of the world and undergo secondary adaptation to suit their habitats. They are also known as hair quadrupeds.

2. Mammals are all homoeothermic (warm-blooded) animals.

3. The body shape varies, and it is covered by hairs.

4. Body is divisible into the head, neck, trunk, and tail.

5. Skin is glandular and includes sebaceous, and sweat glands in both the sexes and mammary glands in the females.

6. Head region possesses a pair of eyes protected by eyelids and eyelashes, a pair of external ears or pinnae, and a pair of external nares.

7. They possess two pairs of pentadactyl legs, which are variously modified to perform different functions [like walking, running, climbing, swimming, burrowing, and lying. Digits bear horny claws, nails, or hoofs.

8. Teeth are placed in sockets [ThecodontJ. Teeth are heterodont [different types of teeth
i.e, all are not of the same morphology], and diphyodont [i.e. 2 sets of teeth milk teeth
and permanent teeth].

9. There is well-developed voice apparatus, the larynx.

10. m is a muscular diaphragm separating the thorax and abdomen.

11. Respiration is by lungs.

12. Circulatory syštem consists of a four-chambered head ‘which maintains complete double circulation. RBC‘s are non-nucleated.

13. Nervous system is well developed. The brain consists of a large cerebrum, cerebellum, and cerebral cortex. There are 12 pairs of cranial nerves.

14. Excretory organs comprise a pair of metanephric bean-shaped kidneys and a coronary bladder. Nitrogenous water products are excreted in the form of urea (present in urine fluid).

15. Sexes are separate, and sexual dimorphism is seen.

16. Fertilization is internal.

17. MajorIty are viviparous. Viviparity is common except in the egg-laying mammal called platypus.
eg: Lion, Tiger, Cat, Dog.

(b) Platypus.

Question 29.
(a) List any two differences between springwood and autumn wood,
Answer:

Springwood [Earlywood]	Autumn wood [late wood]
(a) Secondary xylem formed during the spring season is called springwood	(a) Secondary xylem formed during the autumn season is called autumn wood.
(b) Xylem vessels are thin-walled with a wide lumen	(b) Xylem vessels are thick-walled with narrow lumen

(b) Differentiate Heartwood from the sapwood.
Answer:

Heartwood (Duramen)	Sapwood (Alburnum)
1.  It is a central dark-colored nonfunctional part of the secondary xylem of an old stem	1. It is the outer, light-colored functional part of the secondary xylem of an old stem.
2. Tracheary elements are plugged by tyloses or get filled with resin gum or tannins.	2. Tracheary elements are not plugged.
3. It is heavier and more durable	3. It is lighter less durable.
4. It is less susceptible to the attack of pathogens and insects	4.It is more susceptible to the attack of pathogens and insects.

Question 30.
Draw a neat labeled diagram of an Animal cell.
Answer:

Animal Cell – Ultrastructure

Question 31.
Write the schematic representation of glycolysis.
Answer:

• It occurs in the cytoplasm of the cell.
• It is an enzymatic reaction, thus temperature sensitive.
• It is a common reaction for both aerobic and anaerobic respiration.

Question 32.
List the hormones secreted by the following glands.
1. Thyroid
2. Adrenal
3. Ovary
4. Testis
5. Pancreas.
Answer:
(a) Thyroid – Thyroxine, Tri-iodo thyronine
(b) Adrenal-Adrenaline, epinephrine
(c) Ovary – estrogen, progesterone
(d) Testis – Testosterone
(e) Pancreas – Insulin, Glucagons

Section – II

II. Answer any THREE of the following questions in 200-250 words each, wherever applicable. (3 × 5 = 15)

Question 33.
(a) Name three types of cell junctions.
(b) Mention any two specialized connective tissues.
Answer:
(a) Tight junction, adhering junction, and gap junction.
(b) Cartilage, Blood, Bones (any two).

Question 34.
Classify the enzymes based on the type of reaction they catalyze.
Answer:

• Oxidoreductases: Act on many chemical groupings to add or remove hydrogen atoms.
• Transferases: Transfer functional groups between donor and acceptor molecules. Kinases are specialized transferases that regulate metabolism by transferring phosphate from ATP to other molecules.
• Hydrolases: Add water across a bond, hydrolyzing it.
• Lyases: Add water, ammonia, or carbon dioxide across double bonds, or remove these elements to produce double bonds.
• Isorne rases: Carry out many kinds of isomerization: L to D isomerizations. Mutase reactions (shifts of chemical groups) and others.
• Ligases: Catalyze reactions in which two chemical groups are joined (or ligated) with the use of energy from ATP.

Question 35.
Describe pressure flow or mass flow Hypothesis.
Answer:
According to Munch, food materials are translocated from the source (leaves) to sink (roots) enmass, through a turgor pressure gradient that occurs between leaves and roots.

The mechanism of translocation can be explained as follows:
1. Sugars are formed in the mesophyll cells during photosynthesis and are loaded into the phloem sieve elements of veins. This is called vein loading.

2. The osmotic potential in the phloem becomes more negative, and as a result, water is drawn into phloem elements from xylem cells.

3. Turgor pressure increases in the sieve tubes of leaves, and at the same time the turgor. the pressure of roots becomes less. So a turgor pressures gradient occurs between mesophyll cells and root cells.

4. As a result, food molecules in solution form, are translocated from leaves to roots.

5. In the roots the food is consumed, and in the fruits, it is converted into insoluble starch. So osmotic potential and turgor pressure in the roots and storage organs decreases.

6. Thus turgor pressure gradient continues between leaves, and roots and translocation of food also continue.

Munch conducted a physical experiment to demonstrate the mass flow. He took two osmometers A and B, with concentrated sugar solution in A and B filled with a dilute solution. These are connected by a ‘U’ tube C. A B is immersed in a trough containing water.

Now water enters into ‘A’, due to high osmotic pressure and this creates turgor pressure. This results in mass flow of sugar solution from A to B through ‘C’ until equilibrium is attained and maintained. He compared osmometer A to leaves, B to roots, C to phloem, and water in vessels to xylem vessels. This clearly explains the mass flow hypothesis.

Question 36.
Explain the Z-scheme of light reaction.
Answer:
Non-cyclic photophosphorylation is also called ‘Z’ scheme electron transport.
In this cycle, the electrons lost by chlorophyll pigment do not return to the same molecule.
Both PS I and PS II are utilized.
One molecule of ATP is formed.
Two molecules of NADPH₂ are formed.
One molecule of O₂ is formed.

Question 37.
Draw a neat labeled diagram showing a sagittal section of the Human brain.
Answer: