Category: 1st PUC

Students can Download 1st PUC Biology Previous Year Question Paper March 2018 (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 2018 (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.
What is a herbarium?
Answer:
It is a storehouse of collected plant specimens that are dried, pressed, and preserved on sheets.

Question 2.
Give an example for palmately compound leaf,
Answer:
Silk cotton tree leaf.

Question 3.
Name the living mechanical tissue.
Answer:
Collenchyma.

Question 4.
Where is nucleolus found?
Answer:
The nucleolus is found within the nucleus.

Question 5.
What is cytokinesis?
Answer:
Division of the cytoplasm is known as cytokinesis.

Question 6.
What happens when fresh grapes are soaked in sugar solution?
Answer:
Grapes become plasmolyzed due to osmosis.

Question 7.
Define hydroponics.
Answer:
The technique of growing plants in a nutrient solution.

Question 8.
What are alveoli?
Answer:
Structural and functional units of lungs where exchange of gases takes place.

Question 9.
What is a cardiac cycle?
Answer:
The sequential cyclic events of systole and diastole together are known as one cardiac cycle.

Question 10.
Name the structural and functional unit of kidneys.
Answer:
Nephrons.

PART-B

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

Question 11.
Write any four characters of fungi.
Answer:
(a) They are heterotrophic eukaryotic organisms either parasites or saprophytes.
(b) They are either unicellular (yeast) or multicellular and filamentous.
(c) The body consists of long slender, thread-like structures called hyphae.
(d) The network of hyphae is called mycelium
(e) Some hyphae of some members are coenocytic.

Question 12.
Differentiate between Diploblastic and Triploblastic conditions.
Answer:
Animals in which cells are arranged in two embryonic layers namely the outer ectoderm and inner endoderm are called diploblastic animals and the animals in which cells are arranged in three embryonic layers namely outer ectoderm, inner endoderm, and middle mesoderm are called triploblastic animals.

Question 13.
Name the four types of tissues present in the animals.
Answer:
(a) Epithelial tissue
(b) connective tissue
(c) muscular tissue
(d) neural tissue

Question 14.
Draw a neat labeled diagram of mitochondria.
Answer:

Section through mitochondrion (diagrammatic)
The elementary particles have been shown only on one crista.

Question 15.
What are plant growth inhibitors? Give two examples.
Answer:
The hormones which inhibit (retard) growth rate are called growth inhibitors, e.g.: Abscisic acid (ABA) & ethylene.

Question 16.
Define (a) Tidal Volume (b) Expiratory Capacity.
Answer:
(a) Tidal volume: Volume of air inspired or expired during normal respiration.
(b) Expiratory capacity: Total volume of air a person can expire after a normal inspiration.

Question 17.
What are Urecotelic animals? Give an example.
Answer:
Animals that excrete nitrogenous wastes as uric acid are called Uriotelic animals.
e.g. reptiles, birds.

Question 18.
What is osteoporosis? Mention the common cause that leads to Osteoporosis.
Answer:
Osteoporosis is an age-related disorder characterized by decreased bone mass and increased chances of fracture. Decreased level of estrogen is the main cause.

PART-C

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

Question 19.
Write the taxonomic hierarchy of man.
Answer:
Kingdom – Animalia
Phy/Div – Chordata
Class – Mammalia
Order – Primates
Family – Hominidae
Genus – Homo
Species – sapiens

Question 20.
What is phyllotaxy? Explain any two types.
Answer:
The pattern of arrangement of leaves on the stem or branches is known as phyllotaxy.
Types:
(a) Alternate: Only one leaf at each node is arranged alternately.
(b) Opposite: The leaves are arranged in pairs at each node opposite to each other.
(c) Whorl: Three or more leaves arise at each node in whorls.

Question 21.
List any three differences between tracheids and trachea.
Answer:
Tracheids
(a) Primitive in origin
(b) cells are elongated with tapering ends.
(c) cells are longer in length
(d) possess narrow lumen
(e) less thickened secondary wall
(f) less efficient in conduction

Question 22.
Name the five sub-phases of prophase – I in meiosis.
Answer:
(a) leptotene
(b) zygotene
(c) pachytene
(d) diplotene
(e) diakinesis

Question 23.
List any three applications of auxins.
Answer:
Auxins:

• Auxiñs were the first hormones to be discovered in plants.
• F.W. Went confirmed and isolated the auxins from coleoptiles of Avena saliva.
• Chemically it ¡s 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 24.
Mention the three major types of cells present in gastric glands. Give their secretions.
Answer:
(a) Parietal /oxyntic cells – secrete HCl and intrinsic factor
(b) Peptic cells/chief cells – proenzyme, pepsinogen
(c) mucus neck cells – secrete mucus.

Question 25.
What is systemic circulation? Write the pathway diagram of systematic circulation.
Answer:
The circulation of blood between the heart and all body parts.

Question 26.
Write the names of bones or the forelimb of man.
Answer:
Humerus, radius, ulna, carpals, metacarpals, and phalanges.

Part-D ( Section -I ).

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

Question 29.
Draw a neat labeled diagram of the digestive system of cockroaches.
Answer:

Question 30.
What is centromere? Explain the types of chromosomes based on the position of the centromere.
Answer:
Primary constriction of the chromosome (or) The region where kinetochores are present is known as the centromere.

Types of Chromosomes:
I. Classification of chromosomes based on the number of centromeres
(a) Aceidric chromosome: It is a chromosome without centromere.
(b) Monocentric chromosome: Here there is one centromere to hold the chromatids together.
(c) Ilicentric chromosome: It is the presence of two centromeres in a chromosome.
(d) Polvcentric chromosome: It is more than three centromeres in a chromosome.

II. Classification of chromosomes based on the position of centromere:
(a) Metacentric type
Here the centromere is exactly at the center of two chromatids. It looks V-shaped during anaphase.

(b) Sub metacentric type
Here the ‘centromere is eccentric in position so that one of the chromatids is long and the other is shorter. It looks L – shaped during anaphase.

(c) Acrocentric type
Here the centromere is almost towards one end of the chromatid to form a very long arm and another very short arm. It looks hook-shaped during anaphase.

(d) Telocentric type
Here the centromere is towards one end of the chromatid such that one chromatid is only present. It looks rod-shaped during anaphase.

III. Classification of chromosomes based on their functions:
(a) Autosomes [AA]: They are also called somatic chromosomes that control body characteristics.
(b) Allosomes [X or VI]: They are also called sex chromosomes that determine the gender of an individual.

Functions of chromosomes:

1. Chromosomes are very important in the higher animals for the phenomenon of sex determination.
2. Chromosomes play an active role in the metabolic process of a cell.
3. They carry the heredity information from parents to offspring in the form of genes.

Question 31.
Explain the classification of enzymes.
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 32.
Describe the thistle funnel experiment to demonstrate osmosis with a labeled diagram.
Answer:
Thistle Funnel Experiment: Take a thistle funnel having a long stem. Close the mouth of the funnel with a membrane of animal bladder or parchment paper by means of a thread.

Pour 10% sugar solution in the thistle funnel till it stands at about 1/3 of the height of the stem. Now dip the covered end of the thistle funnel in a beaker containing water. Support it in its position by means of a stand. Mark the level of sugar solution as A by means of glass marking pencil or gummed paper. Note that after a few minutes the level of the sugar solution has risen in the stem of the thistle funnel to a point B.

Results: The rise of sugar solution in the thistle funnel can only be due to the entry of water into it through the animal bladder.

Section – II

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

Question 33.
Give the schematic representation of the nitrogen cycle.
Answer:
Nitrogen Cycle:

• Nitrogen ¡s 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.

Question 34.
Describe the structure of chloroplast with a neat labeled diagram.,
Answer:

Electron microscopic structure of the chloroplast

Chlorophyll-containing plastids are called Chloroplasts. 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. ¡t 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 interconnected 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 35.
Write the schematic representation of the steps 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 36.
Draw a neat labeled diagram of multipolar myelinated neurons.
Answer:

Question 37.
List the hormones secreted by the anterior lobe of the pituitary with one function of them.
Answer:
1. Somatotrophic hormone or Human growth hormone (STH or HGH): Promotes the growth of muscles. It stimulates the uptake of amino acids by tissues and their synthesis into proteins.

• Hypo secretion in childhood causes Pituitary dwarfism. Such an individual will be abnormally dwarf and is called a midget.
• Hypersecretion in childhood causes Pituitary gigantism. Such an individual will be abnormally tall.
• Hypersecretion in adolescence causes acromegaly which is characterized by the formation of disproportionately large hands, feet, cheekbones, jaws, etc.,

2. Thyroid-stimulating Hormone (TSH): Controls secretion of thyroid hormones by the thyroid gland and also regulates iodine intake by the thyroid gland.

• Hypo secretion leads to goiter, cretinism, and myxoedema.
• Hypo secretion leads to hyperthyroidism.

3. Adrenocorticotrophic hormone (ACTH): It controls the secretion of hormones (Cortisone) by the adrenal cortex.

• Hypersecretion of ACTH causes Cushing’s syndrome.
• Hypo secretion causes Addison’s disease.

4. Follicle Stimulating Hormone (FSH): In females the induces growth and maturation of Graffian follicle and stimulates follicular secretion of estrogen. In males, it stimulates the testis to produce sperm.

5. Luteinizing Hormone or Interstitial Cell Stimulating Hormone (EH or ICSH): In females, it stimulates ovulation and the formation of the corpus luteum. In males, it stimulates interstitial cells in the testis to secrete testosterone.

6. Prolactin or Lactogenic or Luteotropic Hormone (LTH): In females, it causes growth and development of breasts during pregnancy. It stimulates milk production and secretion after childbirth, it inhibits ovulation during pregnancy and breastfeeding.

7. Melanocyte Stimulating Hormone (MSH): Increases skin pigmentation by stimulating dispersion of melanin in amphibians but its exact role in humans is unknown.

Students can Download 1st PUC Biology Previous Year Question Paper March 2018 (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 2018 (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 a taxon?
Answer:
A taxon is a unit of classification.

Question 2.
Name the type of root modification in banyan.
Answer:
Prop roots.

Question 3.
Name the family in which the flowers show papilionaceous corolla.
Answer:
Family Fabaceae.

Question 4.
What is a closed vascular, bundle?
Answer:
Avascular cambium in which cambium is absent is known as a closed vascular bundle.

Question 5.
Name the type of plastids that store carbohydrates.
Answer:
Amyloplasts.

Question 6.
During which stage does synapsis occur?
Answer:
Zygotene.

Question 7.
What are porins?
Answer:
Porins are proteins that form huge pores in the outer membranes of plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through.

Question 8.
Name the enzyme which helps in nitrogen reduction in prokaryotes.
Answer:
Nitrogenase.

Question 9.
Which type of cells of gastric glands secrete HCl.
Answer:
Parietal cells or oxyntic cells of gastric glands secrete HCl.

Question 10.
What is serum?
Answer:
Plasma without cloning factors is known as serum.

PART-B

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

Question 11.
Mention any two major groups of protozoans with an example for each.
Answer:

1. Amoeboid protozoans e.g. Amoeba, Entamoeba,
2. Flagellated protozoans eg: Trypanosoma
3. Ciliated protozoans eg. Paramecium
4. sporozoans eg: plasmodium (Any two)

Question 12.
Distinguish between diploblastic and triploblastic animals. Give an example for each.
Answer:
Animals in which the cells are arranged in two embryonic layers namely ectoderm and endoderm are known as diploblastic animals, e.g. coelenterates.

Question 13.
What is phyllotaxy? Name any two types.
Answer:
Patterns of arrangement of leaves on the stem or branch are known as phyllotaxy.
Types of phyllotaxy include – alternate, opposite and whorled.

Question 14.
What are nephridia? Mention their types.
Answer:
Nephridia are excretory organs in earthworms. Types are septal, integumentary and pharyngeal

Question 15.
Draw a neat labelled diagram of a mitochondrion.
Answer:

Section through mitochondrion (diagrammatic)
The elementary particles have been shown only on one crista.

Question 16.
Write the physiological effects of cytokinins in plants.
Answer:
Physiological effects of cytokinins:

1. helps to produce new leaves, chloroplast in leaves, lateral shoot growth and adventitious shoot formation.
2. helps to overcome apical dominance.
3. promotes nutrient mobilization which helps in the delay of leaf senescence.
4. promotes shoot initiation.

Question 17.
What are joints? Name any two types of joints in man.
Answer:
Joints are points of contact between bones or between bones and cartilages.
Types:

1. Filariousjoints
2. Cartilaginous joints
3. Synovial joints (any two)

Question 18.
What is (a) tetany (b) arthritis?
Answer:
(a) Rapid spasms in muscles due to low Ca+2 in body fluids.
(b) Arthritis: inflammation of joints.

PART-C

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

Question 19.
Briefly explain any three types of taxonomical aids.
Answer:
All those collections of actual live specimens or preserved specimens, which help in the identification or verification of a species, are called taxonomic aids.
Some of them include,

1. Herbarium
2. Botanical Gardens
3. Museums
4. Zoological Parks
5. Keys
6. Monographs
7. Manuals
8. Floras.

1. Herbarium:
A herbarium is defined as a collection of plants, that have been dried, pressed and preserved on sheets.
These sheets are arranged in accordance with any accepted system of classification. They can be used for future references.

2. Botanical Gardens:
These are collections of living plants maintained for reference, i.e., they are meant for identification and classification. Both local and exotic plants are grown.

• These are considered natural and economical reference systems.
• Following are some famous botanical gardens in India.
  1. Indian Botanical Garden, Howrah.
  2. National Botanical Research Institute, Lucknow.

3. Museums:

• Museums are those places that have collections of preserved animals and plants for taxonomic studies.
• The plant and animal specimens are kept in chemical solutions of formalin and are preserved for a longer duration.
• Plant and animal specimens may also be preserved as dry specimens.
• Insects are preserved in insect boxes. The collected insects are dried and pinned in these boxes.
• Larger açimals like birds and mammals are usually preserved as stuffed specimens.
• Skeletons of animals are also collected in the museums.
• The specimens have been correctly identified, labelled and stored.
• A catalogue of specimens is prepared for future use.

4. Zoological Parks:

• These are the places where wild animals are protected in conditions very similar to their natural habitats. Food and shelter are provided. The health of the animal is monitored and a chance for reproduction (mating) is given.
• The scientific purpose of such parks is to breed animals under captivity to avoid extinction.

Question 20.
Draw a neat labelled diagram of phloem.
Answer:

Question 21.
Mention the significance of mitosis.
Answer:

• 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 behaviour during the different stages of mitosis maintains a constant chromosome number.
• 11 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 lower organisms eg: Amoeba, Euglena etc.
• Mitosis produces new cells for the healing of wounds and for regeneration.

Question 22.
What is photoperiodism? Classify the plants based on photoperiodism.
Answer:
Plants, in order to flower, requìre a particular day length or light period called photoperiod and the response of the plants to photoperiod in terms of flowering is called photoperiodism. Photoperiodism was first studied by W.W. Garner, and HA. Allard.

Based on their photoperiodic responses, plants are classified into the following groups:

1. Long Day Plants: These flower in photoperiod more than critical day length. e.g: Wheat, oats etc.
2. Short Day Plants: These flower in photoperiod less than critical day length. e.g: Tobacco, Chrysanthemums etc.
3. Day Neutral Plants: These are the plants that are not influenced by the dùration of light for their flowering.
   e.g: Tomato, cucumber, Cotton etc.

Question 23.
Briefly explain the role of enzymes in the digestion of disaccharides.
Answer:
(a) Lactose – helps in the breakdown of lactose into a molecule of glucose and a molecule of galactose.
(b) Sucrase – helps in the breakdown of sucrose into a molecule of glucose and a molecule of fructose.
(c) Maltase – helps in the breakdown of maltose into 2 molecules of glucose.
or
Sucrase:
It converts sucrose into glucose and fructose.

Maltase :
It converts maltose into glucose and glucose.

Lactase:
It converts Lactose into Glucose and Galactose.

Question 24.
Briefly describe the transport of CO₂ in the blood.
Answer:
CO₂ is produced within the body cell as a metabolic waste product of oxidative metabolism.
CO₂ is transported from cells to the alveoli through the following forms.
1. In the form of carbonic acid (7%).
2. In the form of carbamino haemoglobin (23%).
3. In the form of bicarbonates of Sodium and Potassium (70%).

Question 25.
Write a note on any three disorders of the circulatory system.
Answer:
(a) High blood pressure (Hypertension): B.P that is higher than the normal value of 120/80 mm Hg is known as hypertension.
(b) Coronary Artery Disease (CAD): Lumen of blood vessels become narrow due to deposition of calcium, fat, cholesterol etc. affecting the blood supply to the heart.
(c) Angina: symptoms of acute pain when enough O₂ is not reaching the heart muscle.

Question 26.
Briefly explain the role of lungs, liver and skin as excretory organs.
Answer:
Role of lungs: Lungs removes large amounts of CO₂ and also water.
Liver: removes bilirubin, biliverdin, cholesterol, degraded steroids, hormones, vitamins and
Skin-sweat glands help in the removal of sweat containing NaCl, small amounts of urea, lactic acid etc. Sebaceous glands help in the removal of sterols, hydrocarbons and waxes through

Part-D ( Section -I ).

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

Question 27.
List any four important characters of gymnosperms. Give two examples.
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 organised into fertile structures called strobili or cones.

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

• Green photosynthetic leaves (Foliage).
• Brown coloured scale leaves.

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

Question 28.
Write the general characters of class Aves.
Answer:

1. Birds are homeothermic (warm-blooded) animals.
2. The body is boat-shaped and is adapted in such a way as to offer the least resistance while flying
3. The body is differentiated into an ahead, long neck, a trunk and a short tail.
4. The body is covered by feathers, which provide insulation.
5. Forelimbs are modified into wings for flight. Wings are provided with feathers, which are bad conductors of heat.
6. The hind limbs are shifted forwards to balance the body while walking. They are variously modified for perching, walking, running, swimming etc.,
7. Hind limbs have 4 toes and are covered, with epidermal scales.
8. The endoskeleton is fully ossified (bony). Bones are tubular and hollow fond have air cavities or air sacs (pneumatic bones).
9. The bones of the skull are fused.
10. Jaws are devoid of teeth but covered by horny sheaths and are modified to form a beak.
11. The beak is variously modified to suit individual food habits.

Question 29.
Draw a neat labelled diagram of the male reproductive system of cockroaches.
Answer:

Question 30.
Describe the fluid mosaic model of the cell membrane.
Answer:
Fluid mosaic Model By Singer & Nicholson:
(a) According to this model, the plasma membrane consists of a double layer of lipid molecules and globular protein molecules and sterols distributed at random. In comparative terms, it can be said that the plasma membrane is formed of ‘protein icebergs in the sea of lipids’.
(b) The protein molecules are globular proteins and these proteins penetrate or lie at the periphery to form a mosaic pattern.
(c) Heads of the phospholipid molecules of the two layers are directed ¡n the opposite directions while tails of the two layers face each other.
(d) In animal cells glycolipids and cholesterol are also present along with proteins.

Question 31.
(a) Briefly describe the factors affecting enzyme activity. (3M)
(b) What is homopolymer? Give an example. (2M)
Answer:
(a) Factors affecting enzyme activity:

1. Temperature and pH: Enzymes generally function within a narrow range of temperature and pH. Each enzyme shows the highest activity at an optimum temperature and pH. At low or high temperatures and pH, enzymes become inactive.
2. The concentration of substrate: With the increase in substrate concentration, the velocity of the enzymatic activity rises at first, reaches a maximum velocity beyond which it does not increase further when the concentration of substrate is increased.

(b) Homopolymer: A polysaccharide consisting of only one type of monosaccharides. e.g. cellulose, insulin etc.

Question 32.
Transpiration and photosynthesis in plants is a compromise. Substantiate.
Answer:
(a) Transpiration creates transpiration pull for absorption and transport implants.
(b) Transpiration pull supplies water for photosynthesis.
(c) Transports minerals from the soil to all parts of the plants.
(d) Transpired water cools leaf surfaces.
(e) Entry of water due to transpiration maintains the shape and structure of the plants by keeping cells turgid.
(f) Plants need water for photosynthesis and it can be limited by the availability of water which can be swiftly depleted by transpiration. The humidity of rainforests is due to the cycling of water.
(g) The evolution of the CAM photosynthetic pathway helps in maximising the availability of C02 while minimising water loss.

Section – II

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

Question 33.
Describe the nodule formation in leguminous plants.
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.

Question 34.
Write the schematic representation of 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 utilised.
One molecule of ATP is formed.
Two molecules of NADPFI₂ are formed.
One molecule of O₂ is formed,

Question 35.
(a) What is respiratory quotient (RQ)
Mention the RQ values for carbohydrates, fats and proteins during aerobic respiration. (4M)
Answer:
Respiratory quotient (R-Q)
The respiratory quotient can be defined as the ratio of the volume of CO₂ released, to O₂ consumed to break one molecule of respiratory substrate completely.
RQ = \(\frac{\text { Volume of } \mathrm{CO}_{2} \text { released }}{\text { Volume of } \mathrm{O}_{2} \text { consumed }}\)
RQ of carbohydrate is 1.
RQ of proteins ¡s 0.7
RQ of fats ¡s 0.71
RQ of organic acid is more than 1(1.2 or 1.3).

(b) The respiratory pathway in plants is called the amphibolic pathway. Why? (1M)
Answer:
Since respiration involves the breakdown of organic compounds, it has been considered as a catabolic process and the respiratory pathway as a catabolic process. But as different substances enter the pathway as respiratory substrates, and certain intermediate of the respiratory pathway is also withdrawn for the synthesis of other compounds.

For e.g: Fatty acids would be broken down into acetyl COA for entering the pathway, acetyl C, A from the respiratory pathway will be taken out ¡f fatty acids have to be synthesised.

Similarly to the respiratory pathway, intermediates are involved in the synthesis of proteins, because the respiratory pathway involves both anabolism and catabolism it is called an amphibolic pathway.

Question 36.
Draw a neat labelled diagram of the sagittal section of the human brain.
Answer:

Question 37.
Name the hormones which are responsible for the following functions.
(a) Stimulates the resorption of water and electrolytes in kidneys.
(b) Maintains sleep-wake cycle.
(c) Regulates the blood calcium level.
(d) Provides immunity
(e) Supports pregnancy.
Answer:
(a) ADH or vasopressin.
(b) Melatonin
(c) Calcitonin
(d) Thymosins
(e) Progesterone.

Students can Download 1st PUC Biology Previous Year Question Paper March 2019 (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 2019 (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.
What is biodiversity?
Answer:
Biodiversity refers to the variety of organisms present on the Earth (species richness).,

Question 2.
Where do you find pneumatophores?
Answer:
In mangrove species.

Question 3.
Write the scientific name of Mango.
Answer:
Mangifera Indica

Question 4.
Where do you find motor cells?
Answer:
In monocot leaf

Question 5.
In which stage of the cell cycle does DNA replication occur?
Answer:
S-phase of interphase.

Question 6.
Define imbibition.
Answer:
Imbibition is a special kind of diffusion that involves the movement of water molecules along a diffusion gradient from a region of higher concentration onto a suitable solid matrix (adsorption).
or
It is the adsorption of water molecules by hydrophilic colloids.

Question 7.
What is nitrogen fixation?
Answer:
The process of conversion of nitrogen into ammonia and/or other nitrogen compounds is known as nitrogen fixation.

Question 8.
Name any one muscle protein.
Answer:
Actin

Question 9.
Mention the type of movement that macrophage exhibit.
Answer:
Amoeboid movement.

Question 10.
Define reflex action.
Answer:
Reflex action is an automatic response to any kind of stimulus without conscious involvement of the brain.

PART-B

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

Question 11.
What is binomial nomenclature? Who proposed it?
Answer:
Naming of living organisms by scientific name universally by two names is binominal nomenclature. The two names include generic name and species name. It was Proposed by Carolus Linnaeus.

Question 12.
Name the fungal and algal components of lichens.
Answer:
Chemosynthetic bacteria are autotrophic as they can manufacture organic food from inorganic raw materials, with the help of chemical energy obtained by them by oxidising inorganic chemicals present in their environment.

Question 13.
Mention two types of body forms exhibit by Coelenterata.
Answer:

1. Polyp form (asexual reproduction).
2. Medusoid form (sexual reproduction).

Question 14.
Earthworms are called “Farmer’s friends.” Justify it.
Answer:
Earthworms are called friends of farmers because they make burrows in the soil and make it porous which helps in respiration and penetration of developing plant roots. The fertility of soil is also increased by their vermicompost.

Question 15.
Differentiate tendons from Ligaments.
Answer:
1. Tendons: They are exclusively composed of white collagenous fibres arranged in regular bundles. In between these fibres, a few fibroblasts may occur. Usually, they connect muscle to bone.

2. Ligaments: They may contain both white and yellow fibres and these fibres are arranged in less orderly patterns. Usually, they connect bone to bone.

Question 16.
Name the two types of lymphocytes responsible for immunity.
Answer:
T and ‘B’ lymphocytes.

Question 17.
Write a note on coronary artery diseases.
Answer:
Atherosclerosis: It is a disease of the arterial wall in which the inner layer becomes thick due to the cholesterol deposition in the form of plaques on the inner lining of the arteries. It is called coronary artery disease (CAD) or ischemic heart disease (IHD).

Question 18.
What is osteoporosis? What is the cause for it?
Answer:

1. Amoeboid Movement: Macrophages and Neutrophils.
2. Ciliary Movement: Epithelium of nasal tract, urinary tubule and fallopian tube.

PART-C

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

Question 19.
Mention any three salient features of class mammalian.
Answer:

1. Class Mammalia includes warm-blooded and highly organized animals possessing mammary (mi1k) 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 ¡Lis covered by hairs.

Question 20.
What is placentation? Mention any two types.
Answer:
The arrangement of ovules within the ovary on a fertile tissue.
Marginal, Axile, Parietal, Basal, Central, Free central.

Question 21.
Write any three differences between prokaryotes and eukaryotes.
Answer:

Procaryotes	Eucaryotes
(a) Primitive and indefinite nucleus	(a) True and definite nucleus
(b) Membrane bound organelles are absent	(b) Present
(c) 70S types of Ribosomes	(c) 80S type of Ribosomes

Question 22.
Mention the significance of meiosis.
Answer:
Significance of Meiosis:

1. Meiosis brings genetic crossing over and random distribution of paternal and maternal chromosomes to daughter cells.
2. Recombination produces variations and varia’Iîons are the sources of organic evolution.

Question 23.
Explain the carbohydrate digestion in the small intestine.
Answer:
a. Digestion in the mouth: The salivary amylase or ptyalin converts starch and glycogen into maltose units.

b. Digestion in the intestine:
Pancreatic juice: Carbohydrase secreted by the pancreas is called pancreatic amylase. It converts starch and glycogen into maltose units.

Intestinal juice: It consists of three types of carbohydrates.
Maltase: Maltase catalyses the hydrolysis of maltose (disaccharide) into glucose units (monosaccharides).

Sucrase: Sucrase catalyses the splitting of sucrose (disaccharide) into glucose, and fructose units (monosaccharides).

Lactase: Lactase acts on milk sugar lactose (disaccharide), and splits lactose into galactose, and glucose units (monosaccharides).

Question 24.
Define the following:
(a) Tidal volume
(b) Residual volume
(c) Vital capacity
Answer:
(a) Tidal volume: It is the volume of air inhaled or exhaled with each inhalation and exhalation without any extra effort. Its volume is 500 ml.
(b) Residual volume: It is the volume of air that remains inside the lungs at the end of forceful expiration. Its volume is 1100 – 1200 ml.
(c) Vital capacity: It is. the total volume of air that can be expelled from the lungs during maximum exhalation.
VC = TV+ IRV + ERV
= 500 + 2500 + 1100
= 4100 ml approx.

Question 25.
What are uricotelic animals? Give two examples.
Answer:
Ureotelic animals: Animals that excrete urea are called ureotelic animals. Urea is produced by the liver during the ornithine cycle. Urea is stored in the urinary bladder in a dissolved state called urine.
e.g: Cartilaginous fish (Shark), Mammals and Amphibians.

Question 26.
What is meristem? Mention the types based on position.
Answer:
Meristems are young immature cells capable of cell division.
Based on position, there are mainly three types:
(a) Apical Meristems: They are found at the tip of the stem and root. The activity of this, meristem leads to an increase in length.
(b) Intercalary Meristems: These are found between permanent tissues, at the base of the internode and leaf. The activity of this meristem leads to an increase in the length of the internode.
(c) Lateral Meristems: These are meristem is-situated parallel to the sides of the organs and permanent tissues. Their activity leads to secondary growth.

PART-D (Section – I)

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

Question 27.
Write the salient features of kingdom Mycota with two examples.
Answer:
Kingdom -Fungi (Mycota):
Fungi are eukaryotic, achlorophyllous organisms that are but, generally multicellular a few are unicellular (yeast).

1. The fungal body consists of long, slender, thread-like structures called hyphae, which form a network called mycelium.
2. The hyphae have a cell wall made up of chitin.
3. They may be aseptate and multi-nucleate i.e., coenocytic or septate.
4. All fungi are heterotrophs -they are saprotrophs or parasites or live as symbionts in the roots of higher plants (mycorrhizae) and in lichens (mycobiont).
5. They reproduce vegetatively by fragmentation, fission or budding.
6. Asexual reproduction is by the formation of zoospores as in Saprolegnia and Pythium by Aplanospores/sporangiospores formed endogenously (within the sporangium) as in Rhizopus and Mucor or by conidia, which are produced exogenously on the swollen tip of the conidiophore as in Penicillium and Aspergillus.

Question 28.
Give the functions of the following :
(a) Adipocytes
(b) Macrophages
(c) Fibroblasts
(d) Plasma cells
(e) Mast cells.
Answer:
(a) Stores fat
(b) Phagocytosis
(c) Synthesis collagen
(d) Single kind of antibody production
(e) Provide immune tolerance.

Question 29.
Describe the fluid mosaic model of the plasma membrane.
Answer:
Plasma Membrane:

(A) Schematic representation of a section through a eukaryote fluid mosaic membrane.
(B) The fluid mosaic model of Singer and Nicholson (1972)

All living cells are enclosed in a thin, living and selectively permeable membrane called the plasma membrane. Its thickness is about 75 – 100 A Plasma membrane consists of 30 – 40% lipids and 60 – 70% proteins. In animal cells, it surrounds the protoplasm and in plant cells, it is next to the cell wall.

Fluid mosaic Model By Singer & Nicholson:
(a) According to this model, the plasma membrane consists of a double layer of lipid molecules and globular protein molecules and sterols distributed at random. In comparative terms, it can be said that the plasma membrane is formed of ‘protein icebergs in the sea of lipids’.
(b) The protein molecules are globular proteins and these proteins penetrate or lie at the periphery to form a mosaic pattern.
(c) Heads of the phospholipid molecules of the two layers are directed in the opposite directions while tails of the two layers face each other.
(d) In animal cells glycolipids and cholesterol are also present along with proteins.

Question 30.
Describe any five classes of enzymes and the type of reaction they catalyse.
Answer:
Classification of enzymes:

• Oxidoreductases: Act on many chemical groupings to add or remove hydrogen atoms.
• Transferases: Transfer functional groups between donor and acceptor molecules. Kinases are specialised 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.
• Isomerases: 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 the ATP diagram of stomata and describe the structure.

Question 31.
Draw a labelled diagram of stomata and describe the structure.
Dicot Stomatal Apparatus:

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 32.
Explain the process involved in root nodule formation.
Answer:
1.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.

2.The rhizobia enter these deformed root hair and proliferate within the root hair.

Section – II

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

Question 33.
Explain the steps involved in 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 34.
(a) Write any two differences between a long day and short-day plant,
Answer:

1. Long Day Plants: These flower in photoperiod more than critical day length, eg: Wheat, oats etc.
2. Short Day Plants: These flower in photoperiod less than critical day length.
   e. g: Tobacco, Chrysanthemums etc.

(b) Mention the role of Auxins in plants,
Answer:
Auxins

• 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 35.
Draw a labelled diagram of the sagittal section of the human brain.
Answer:

Question 36.
Write the schematic representation of hormonal action.
Answer:
Mechanism of hormone action: Hormones exert their effects on target cells by interacting with specific receptors of these cells. These receptors are proteins or lipoproteins. The location of the hormone receptor depends on the chemical nature of the hormone. The water-soluble hormones (polypeptides, glycoproteins and catecholamines) cannot pass through the plasma membrane, hence their receptors are located on the outer surface of the membrane.

Water-soluble hormone action: The lipophilic hormones (steroids and thyroxine) can pass through the plasma membrane and enter into their target cells, hence the receptors for these hormones are located within the nucleus and cytoplasm.

The water-soluble hormones have their action mediated through second messengers. When these hormones bind to their specific receptors of the plasma membrane specific membrane proteins become activated to generate second messengers

Cyclic AMP, phospholipase C, inositol triphosphate, diacylglycerol and Ca+ act as second messengers.
Cyclic AMP second messenger system has the following sequence of events to affect the action of the hormone.

1. The hormone binds to the specific membrane receptor of the target cell.
2. G proteins linked to specific receptors activate the enzyme adenylate cyclase.
3. Activated adenylate cyclase, converts ATP to cyclic AMP.
4. Cyclic AMP activates protein kinase enzymes of the cytoplasm.
5. Protein Kinase phosphorylates (adding phosphate groups too) are the other enzymes of the cytoplasm.
6. The phosphorylation of specific enzymes may enhance or inhibit their activity.
7. This alters the metabolism of the target cell in the required direction as per the specific hormone action.

Lipophilic hormone action:

Mechanism of action of a steroid hormone The lipophilic steroid hormones and thyroid hormones pass through the plasma membrane and get bound to a specific intracellular receptor. These receptors are commonly located within the nucleus. Once these nuclear hormone receptors are activated by their binding to the specific hormones, they alter the gene expression. They may switch on or off specific genes on the nuclear DNA. The transcription of DNA occurs and the newly formed mRNA directs the synthesis of specific enzyme proteins. These enzymes alter the metabolism of the cell in the required direction as per the intended action of the hormone.

Question 37.
Explain any five factors that affect the rate of photosynthesis.
Answer:
Factors affecting photosynthesis:
The following factors influence photosynthesis:
1. Light: Sunlight is used as the primary source of energy. Its three aspects affecting th; process are:
(a) Light Intensity: Optimum light intensity for photosynthesis is 2000-2500 foot candles Higher light intensity bleaches chlorophyll and is called solarisation.
(b) Light Quality: Photosynthetic rate is maximum in red light, next maximum in blue light, and least in green light.
(c) Duration: Photosynthetic rate is more efficient in intermittent light than in continuous light supply.

2. CO₂: Used as a raw material in photosynthesis. Its concentration in air is 0.03%. Its increase up to 0.5% increases the rate of the process but above 0.5% inhibits photosynthesis.

3. O₂: Liberated as a by-product in photosynthesis. Its increase above the normal 21% in air, decreases the rate of the process. It is called the Warburg effect.

4. Temperature: It affects photosynthesis through its influence on the enzyme controlled dark reaction. The optimum temperature is 18 – 40°C for photosynthesis.

5. Water: Used as a raw material in photosynthesis. In the dehydrated state of cells, photosynthesis is inhibited.

Students can Download 1st PUC Biology Previous Year Question Paper March 2019 (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 2019 (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 species.
Answer:
Species is the lowest level of organization and organisms belonging to a particular species have similar morphological, anatomical & physiological characters and can interbreed to produce fertile offsprings.

Question 2.
What is the hypogynous flower?
Answer:

1. In a hypogynous flower, the thalamus is convex of flat and gynoecium occupies topmost position and other floral parts originate below.
2. In epigynous flowers, the thalamus is cup-shaped enclosing the ovary and the floral parts will be arranged above it.
3. In hypogynous flowers, the ovary is superior in epigynous flowers, the ovary is inferior.

Question 3.
Who proposed the ‘fluid mosaic model’ of the plasma membrane?
Answer:
S.J. Singer and G Nicolson.

Question 4.
What is mycorrhiza?
Answer:
All fungi are heterotrophs -they are saprotrophs or parasites or live as symbionts in the roots of higher plants (mycorrhizae) and in lichens (mycobiont).

Question 5.
Mention any two nitrifying bacteria.
Answer:
Double-stranded DNA.

Question 6.
What is Asthma?
Answer:
It is the parietal blocking of bronchi and bronchioles, resulting in breathlessness, chest tightness, wheezing and dry cough.

Question 7.
Which part of the brain contains corpora Quadrigemina?
Answer:
Midbrain.

Question 8.
How trichomes are helpful to plants.
Answer:
In a dicot stem, some of the cells of the epidermis give out multicellular epidermal hairs called stem hairs or trichomes.

Question 9.
At the end of meiosis how many haploid cells are formed?
Answer:
Four cells

Question 10.
What is resting potential?
Answer:
At the resting stage i.e., when the neuron is not involved in the conduction of impulses, the membrane of the neuron is more permeable to K+ ions than to Na+ ions. The membrane is impermeable to the negatively charged proteins of the axoplasm and also Cl” ions.

PART-B

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

Question 11.
What are lichens? Mention its components.
Answer:
Lichens are composite organisms, composed of algae and certain fungi that live in a symbiotic biotic association.
The algal partner is called phycobiont and the fungal partner is called mycobiont.

Question 12.
List any four physiological effects of Auxins.
Answer:
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 13.
Write the role of pepsin and rennin in the stomach.
Answer:
1. Pepsin: It is an enzyme produced in an inactive form called pepsinogen. Pepsinogen is activated by HCI. Pepsin is an endopeptidase and it breaks inner peptide bonds of the protein molecule to form proteoses, peptones, and polypeptides.

2. Renin: Renin is also secreted in an inactive form called prorenin. It is activated by HCI. Renin acts on the milk protein called casein and convert it into paracasein. Paracasein in turn combines with calcium salts and forms calcium Paracaseinate (curdy precipitate).

This process of conversion into calcium Paracaseinate is called the curdling of milk.

Question 14.
Mention the proteins found in thick and thin filaments of myofibrils.
Answer:
Acting myosin.

Question 15.
Differentiate between exocrine and endocrine glands.
Answer:
Exocrine gland:
Duct bearing glands, which release their secretions to the target organs directly through the ducts are called exocrine glands e.g: Salivary glands.

Endocrine glands:
Ductless glands which release their secretions to the blood circulation are called endocrine glands. e.g: Pituitary gland.

Question 16.
Write any four salient features of phylum Ctenophora.
Answer:
They are commonly known as sea walnuts or comb jellies.

• They are exclusively marine.
• They are of tissue level of body organization.
• They have the diploblastic conditions and are radially symmetrical.
• The body bears eight external rows of ciliated comb plates which helping in the locomotion.
• Digestion is both extracellular and intracellular.
• Reproduction takes place only by sexual means.
• Fertilization is external and development is indirect.
• Special property in ctenophores is Bioluminescence.

Question 17.
Mention the factors which help in the binding of O₂ with hemoglobin.
Answer:
PO₂PCO₂, Temp, H+ ion cone

Question 18.
Classify chromosomes based on the position of the centromere.
Answer:
Classification of chromosomes based on the position of centromere:
(a) Metacentric type: Here the centromere is exactly at the center of two chromatids. It looks V-shaped during anaphase.
(b) Sub metacentric type: Here the centromere is eccentric in position so that one of the chromatids is long and the other is shorter. It looks L – shaped during anaphase.
(c) Acrocentric type: Here the centromere is almost towards one end of the chromatid to form a very long arm and another very short arm. It looks hook-shaped during anaphase.
(d) Telocentric type: Here the centromere is towards one end of the chromatid such that one chromatid is only present. It looks rod-shaped during anaphase.

PART-C

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

Question 19.
Assign the following plants to their respective families.
(a) Groundnut
(b) Tomato
(c) Onion
Answer:
(a) Papillionaceae
(b) Solanaceae
(c) Liliaceae.

Question 20.
Write the events of the Telophase of mitosis.
Answer:
Telophase (Gr:telos-end; phases-stage):
It is characterized by the following features.

1. Here, reversal of the prophasic events occur. The daughter chromosomes move and reach the opposite poles where they become thin and thread-like again. These threads overlap one another to form a fine chromatin network.
2. The spindle fibers disintegrate and disappear.
3. Reconstitution of a new nuclear envelope occurs. Nucleolus reappear.

Question 21.
Write any three uses of ethylene.
Answer:

1. Only gaseous plant hormone.
2. Product of metabolism of amino acid methionine.
3. Ethylene is a fruit ripening agent.
4. Promotes senescence of flowers, leaves, and fruits.
5. Initiates flowering in pineapple.

Question 22.
Briefly explain the process of coagulation of blood.
Answer:
The mechanism of blood clotting is explained by various theories out of these ‘Best and Taylor’s theory is one.

Chemical events leading to the formation of blood clots as suggested by ‘Best and Taylor’s theory can be represented as follows.

The reaction shown above indicates that the inactive Prothrombin is converted into active thrombin by the catalytic activity of enzyme Thromboplastin in the presence of Ca²⁺. The activated thrombin reacts with soluble fibrinogen, resulting in the formation of insoluble fibrin. The fibrin threads form a mesh-like network in which the blood components get entangled, and results in blood clots and thus prevents bleeding.

Question 23.
Mention and explain any three taxonomical aids.
Answer:
The collection of specimens that help in the identification of a species are called taxonomic aids. Some of the major taxonomic aids are:

1. Herbarium: A herbarium is defined as a collection of plants that have been dried, pressed, and preserved on sheets. These sheets are arranged in accordance with any accepted system of classification and they form a repository for future use. The date and place at collection are marked on sheets.
2. Botanical Gardens: These specialized gardens have collections of living plants for reference. Plant species in these gardens are grown for identification purposes and each plant is labeled indicating its scientific name and its family.
3. Museum: Museums are those places that have collections of preserved animals and plants for taxonomic studies.
4. Zoological parks: These are places that have wild animals kept in protected environments under human care and which enable us to learn about their food habits and behavior.
5. Keys: A key is an analytical scheme for the identification of plants and animals based on similarities and differences.

Question 24.
Dicot root does not possess cambium, yet it shows secondary growth. Comment on.
Answer:
In the dicot root, the vascular cambium is completely secondary in origin. It originates from the tissue located just below the phloem bundles, a portion of pericycle tissue, above the protoxylem forming a complete and continuous wavy ring, which later becomes circular, further events are similar to dicotyledon stem.

Question 25.
Bile juice, enterokinase, and HCl are essential for digestion. Give reasons.
Answer:
Bile salts emulsify the fats by lowering their surface tension and convert them into numerous tiny droplets. Bile is rich in sodium bicarbonate; hence it neutralizes the acidic chyme. It also helps in the absorption of fat-soluble vitamin K.

Enterokinase: It is a nondigestive intestinal enzyme and acts as an activator. It activates the inactive form trypsinogen of pancreatic juice into trypsin.

HCl present in the gastric juice gives acidic PH for the action of enzymes and also kills harmful bacteria.
HCl activates and converts pepsinogen into pepsin, and prorenin into rennin.

Question 26.
Why transmission of impulse across an electrical synapse is always faster?
Answer:
Electrical Synapses:

1. The membranes of the pre-synaptic and post-synaptic neurons are in close proximity and there is no synaptic cleft.
2. Electrical current can flow directly from one neuron to the other.
3. Impulse conduction is faster.
4. Electrical synapses are rare in the human system.

Part-D (Section -I)

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

Question 27.
Give an account of sporophytic generation in Pteridophytes.
Answer:
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 multicellular embryo.

Question 28.
Explain the structure of the chloroplast with a diagram.
Answer:

Electron microscopic structure of the chloroplast Chlorophyll containing plastids is called Chloroplasts. 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 H20, 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 interconnected 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- a 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.

Question 29.
Describe the functions of any five essential nutrient elements in the plant cells.
Answer:
(A) Nitrogen :
It is absorbed as NO^3-, NO^2- and NH⁴⁺ ions.

Functions:
It is a major constituent of amino acids, proteins, nucleic acids, vitamins, etc.
(B) Sulphur:
It is obtained as sulfates (50/- ions).

Functions:

1. It is the constituent of amino acids, methionine, and cystine.
2. It also forms part- of ferredoxin, and vitamins like thiamine, biotin, and coenzyme-A.

(C) Phosphorus:
It is absorbed by plants in the form of phosphate ions, either as H₂PO₄^– or HPO₄^3-

Functions:

1. It is a constituent of nucleotides and nucleic acids.
2. It is present in cell membranes as phospholipids.
3. It is involved in phosphorylation reactions and energy metabolism ATP.

(D) Potassium: Potassium is taken in as potassium ions (K⁺). It is more abundant in meristematic tissues like buds, root-tips, and youn£ leaves.

Functions:

1. It determines the anion-cation balance in cells.
2. It controls the opening and closing of stomata.
3. It maintains turgidity (osmotic balance) in cells.
4. It activates a number of enzymes.
5. It is also involved in protein synthesis.

(E) Magnesium: It is absorbed as Mg⁺ ions by the plants.

Functions:

1. It is a constituent of chlorophyll (forms the central atom of the porphyrin ring).
2. It maintains the ribosome structure for protein synthesis.
3. It activates the enzymes of phosphate metabolism in respiration and photosynthesis.
4. It is also involved in the synthesis of DNA, RNA, and the activation of enzymes.

Question 30.
Sketch and label the L.S. of the human kidney.
Answer:

Question 31.
What are proteins? Describe the four levels of protein structure. (3M)
Answer:
Definition: Proteins are the building block is of the living body, and make up the structure and functional component of the living system. They are polymers made up of monomers called amino acids, linked by peptide bonds.

Protein structure:
The protein structure is referred to as a three-dimensional structure or conformation, which determines the biological function. Four basic structural levels are assigned to proteins.

They are primary structure, secondary structure, tertiary structure, and quaternary structure.
(a) Primary structure of protein: The primary structure of a protein refers to the linear sequence of amino acids in the polypeptide chain. The amino acids are linked by polypeptide bonds only.

(b) Secondary structure of protein: The folding of the polypeptide chain into a specific coiled three-dimensional structure like α helix or β sheet is called the secondary structure. The coiled configuration is produced by hydrogen bonding between carboxyl and amino groups of individual amino acids.

(c) Tertiary structure of protein: The- helix of protein is further folded into specific loops and bends to produce specific three-dimensional configurations to carry out specialized functions.
This is called the tertiary structure of the protein.

The tertiary structure is stabilized by four kinds of bonds.

1. Hydrogen bonds.
2. Hydrophobic bonds.
3. Ionic or electrostatic bonds.
4. Disulfide bonds.

(d) Quaternary structure of protein: When a protein is made up of 2 or more polypeptide chains as in hemoglobin, it is known to possess a quaternary structure. The polypeptide chains are held together by hydrogen bonds and electrostatic or salt bonds, formed between amino acids.

Question 32.
Differentiate between Chondrichthyes and Osteichthyes.
Answer:

Chondrichthyes	Osteichthyes
1. Skeleton is made up of cartilage.	1. Skeleton is made up of bones.
2. Exclusively marine.	2. Both marine and freshwater forms are seen.
3. They possess placoid scales.	3. Possess cycloid or ctenoid scales.
4. Mouth and nostrils are situated on the ventral side of the body.	4. Mouth and nostril are terminal in their Position.
5. Passes 5 pairs of gill slits.	5. Possess 4 pairs of. gill slits which are covered by the operculum.
6. Tail is heterocercal.	6. Tail is homocercal.
7. Devoid of air bladders.	7. Possess air bladders.
8. Males usually possess a pair of claspers. eg: Scilodon (Shark) Narcine (Electric ray) Trygon (Stingray).	8. Claspers are absent. eg: Anguilla (Freshwater Eel) Clarias (Catfish) Hippocampus (Sea horse).

Section-II

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

Question 33.
Explain the apoplast movement of water.
Answer:
The apoplast is the system of adjacent cell walls that is continuous throughout the plant, except at the Casparian strips of the endodermis in the rods. This movement exclusively through the intercellular spaces and cell walls. It does not involve crossing the cell membrane. This movement is dependent on the gradient water movement is through the mass flow of water occurs due to the adhesive cohesive properties of water.

Question 34.
State Blackman’s ‘Law of Limiting Factors’ and explain the four factors affecting photosynthesis.
Answer:
Blackman’s Law of limiting Factors: The law states that “When a process is conditioned to its rapidity by the number of separate factors the role of the process is limited by the pace of the slowest factor”.

Factors affecting photosynthesis: The following factors influence photosynthesis:
1. Light: Sunlight is used as a primary source of energy. Its three aspects affecting the process are:
(a) Light Intensity: Optimum light intensity for photosynthesis is 2000-2500 foot candles/ Higher light intensity bleaches chlorophyll and is called solarisation.
(b) Light Quality: Photosynthetic rate is maximum in red light, next maximum in blue light, and least in green light.
(c) Duration: Photosynthetic rate is more efficient in intermittent light than in continuous light supply.

2. CO₂: Used as a raw material in photosynthesis. Its concentration in air is 0.03%. Its increase up to 0.5% increases the rate of the process but above 0.5% inhibits photosynthesis.

3. O₂: Liberated as a by-product in photosynthesis. Its increase above the normal 21% in air, decreases the rate of the process. It is called the Warburg effect.

4. Temperature: It affects photosynthesis through its influence on the enzyme-controlled dark reaction. The optimum temperature is 18 – 40°C for photosynthesis.

5. Water: Used as a raw material in photosynthesis. In the dehydrated state of cells, photosynthesis is inhibited.

Question 35.
Draw a neat labeled diagram of the alimentary canal of the Cockroach.
Answer:

Question 36.
Give the schematic representation of the citric acid cycle.
Answer:

Note: To account for two molecules of acetyl CoA produced from 1 molecule of glucose the entire reaction has to be multiplied by two.

Bioenergetics:
1. Number of AlPs produced = 2
2. Number of NADH₂ produced = 6 (6 × 3 = 18ATPs)
3. Number of FADH₂ produced = 2 (2 × 2 = 4 ATPs)
The total yield of ATPs = 24.
Note: Efficiency of Krebs cycle along with its predatory reaction.is 30 ATPs.

Question 37.
Write one function for each of the following hormones :
(a) TSH
(b) ACTH
(c) ADH
(d) PTH
(e) Progesterone.
Answer:
(a) TSH: Stimulate thyroid hormones
(b) ACTH: Stimulate steroid hormones
(c) ADH: Reabsorption of water and electrolytes for a kidney.
(d) PTH: Increases calcium ions in the blood.
(e) Progesterone: Support pregnancy, acts on mammary glands

Students can Download 1st PUC Biology Previous Year Question Paper March 2020 (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 2020 (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.
Define species.
Answer:
Species is the lowest level of organization and organisms belonging to a particular species have similar morphological, anatomical & physiological characters and can interbreed to produce fertile offsprings.

Question 2.
What is a parthenocarpic fruit?
Answer:
Seedless fruit.

Question 3.
Name the living mechanical tissue in plants.
Answer:
Collenchyma.

Question 4.
Name the Plastids which store proteins.
Answer:
Aleuroplast.

Question 5.
What are bio-macromolecules?
Answer:
Biomolecules whose molecular weight is less than a thousand Daltons are referred to as macromolecules while those with a molecular weight of more than a thousand, Daltons are called biomacmmokcùles.
Micromolecules include one

1. amino acids
2. sugars
3. nucleotides and
4. lipids

Biomacromolecules include

1. polysaccharides
2. nucleic acids and
3. proteins.

Question 6.
Define plasmolysis.
Answer:
Plasmolysis is the phenomenon of cell shrinkage when the cell is placed in a hypertonic solution.

Question 7.
Name the oxygen scavenger present inside the root nodules of Leguminous plants.
Answer:
Leghaemoglobin

Question 8.
Define tidal volume.
Answer:

Question 9.
Name the red coloured, iron continuing complex protein present in RBC’s.
Answer:
Haemoglobin.

Question 10.
Mention the hormone secreted by the pineal gland.
Answer:
Melatonin.

PART-B

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

Question 11.
Write any two-living characteristics of viruses.
Answer:
Living characters:
(a) Like organisms they contain protein and nucleic acid.
(b) Like organisms they undergo mutation.

Non – living characters:

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

Question 12.
Assign the following animals to their respective phyla.
(a) Jellyfish
(b) Starfish
Answer:
(a) Colecnterate
(b) Echinodermata.

Question 13.
Distinguish between tendons and ligaments.
Answer:
1. Tendons: They are exclusively composed of white collagenous fibres arranged in regular bundles. In between these fibres, a few fibroblasts may occur. Usually, they connect muscle to bone.

2. Ligaments: They may contain both white and yellow fibres and these fibres are arranged in a less orderly patterns. Usually, they connect bone to bone.

Question 14.
Write any two functions of the Endoplasmic Reticulum.
Answer:
1. Smooth Endoplasmic Reticulum:

• Without ribosomal particles.
• Present mainly near the cell membrane.
• Mainly consists of tubules.
• Formed from rough ER by the loss of ribosomes.

2. Rough ER (Granular ER):

• Ribosomes are present on the surface of the ER.
• Mainly present near the nucleus.
• Consists of cisternae.
• Formed from nuclear membrane.

Question 15.
Mention any two physiological effects of Gibberellins.
Answer:
(a) Gibberellins are helpful in the elongation of internodes in genetically dwarf plants.
(b) Gibberellins are used to break down the dormancy of seeds and buds.

Question 16.
Briefly explain the transportation of O₂ by the blood during respiration.
Answer:
Haemoglobin is called an oxygen carrier because about 97% of the oxygen is transported by RBCs. RBCs contain haemoglobin. Haemoglobin has an affinity towards oxygen. Hence haemoglobin binds with oxygen molecules to form oxyhaemoglobin.

Oxyhaeinoglobm is a very unstable compound. Hence it dissociates into haemoglobin, and molecular oxygen, when it reaches tissues.

Question 17.
Mention any two disorders of the skeletal system.
Answer:
Osteoporosis: It is all age-dependent disorder of the bones. characterised by low bone mass, deterioration of the microarchitecture of the bone and increased fragility.

Myasthenia gravis:
It is an auto-immune disorder, affecting the neuromuscular junction leading to progressive weakening and paralysis of skeletal muscles. It is caused due to non-transmission of nerve impulses due to the non-functioning of acetylcholine.

Question 18.
Name the two contractile proteins of muscle.
Answer:
Actin and Myosin.

PART-C

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

Question 19.
Write any three universal rules of binomial nomenclature.
Answer:

1. Scientific names are, generally, written in Latin or derived from Latin, irrespective of their origin.
2. The scientific names are written ¡n Italics or underlined. The first word denotes the name of the genus, and the second word denotes the species.
3. The generic name starts with a capital letter, while the specific name starts with a small letter.
4. The name of the author is written in an abbreviated form after the specific name e.g. Mangifera indica Linn.-indicates that this species was first described by Linnaeus.
5. The name should be short, precise and easy to pronounce.

Question 20.
What is aestivation? Explain any two types.
Answer:
The mode of arrangement of sepals, petals or even tepals in a flower bud is called aestivation. The different kinds of aestivation are as follows:

1. Valvate aestivation – When sepals, petals or tepals are not overlapping.
2. Imbricate aestivation – When out of the total number of sepals, petals or .tepals, one is completely out, one is completely in and the rest are in and out.

Question 21.
Differentiate between heartwood and sapwood.
Answer:

Heartwood	Sapwood
(a) Dark in colour	(a)Light in colour
(b) No conduction of water	(b) Involved in conduction of water
(c) Nucleic acids	(c) Peripheral layer.

Question 22.
Name the chemical bonds present in
(a) Polysaccharides
(b) Proteins
(c) Nucleic acids
Answer:
(a) Polysaccharide: Glycosidic bond.
(b) Protein: Peptide bond.
(c) Fat: Ester bond.

Question 23.
Mention the three phases of growth in plants.
Answer:

1. Cell division
2. Cell elongation
3. Cell differentiation

Question 24.
Explain the role of gastric juice indigestion.
Answer:
Gastric juice: Pepsin, and Renin are the proteases present in gastric juice
1. Pepsin: It is an enzyme produced in an inactive form called pepsinogen. Pepsinogen is activated by HCI. Pepsin is an endopeptidase and it breaks inner peptide bonds of the protein molecule to form proteoses, peptones, and polypeptides.

2. Renin: Renin is also secreted in an inactive form called prorenin. It is activated by HCI. Renin acts on the milk protein called casein and convert it into paracasein. Paracasein in turn combines with calcium salts and forms calcium Paracaseinate (curdy precipitate).

This process of conversion into calcium Paracaseinate is called the curdling of milk.

Question 25.
Briefly explain the mechanism of blood clotting.
Answer:
It is a clot formed of threads called fibrous where dead blood and damages formed elements are trapped.
(1) Prothrombin→ Thrombokinase Thrombin ______________

Question 26.
Classify the animals on the basis of excretory material and give one example for each.
Answer:
Ammonoteilic Ex: Fish;
Ureotelic Ex: Human
Uricotelic Ex: Birds.

PART-D (Section-I)

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

Question 27.
Mention any five general characters of Bryophytes.
Answer:

1. They are non-vascular plants, exhibiting amphibious habitats.
2. The gametophytic phase of the life cycle is represented by a multicellular, haploid thallus, which is called gametophyte.
3. Gametophytic plant body bearing male and female sex organs, represents the dominant phase of life cycle unlike in higher plants.
4. The gametophyte is autotrophic and independent in nature.
5. The gametophyte produces sex organs, the antheridia and the archegonia which are multicellular.
   Antheridia may be embedded within the thallus or may be specifically located at the tips of the gametophyte. It produces biciliate curved antherozoids.
   Archegonia is a flask like structure. It bears a basal bulbous vendor and a narrow, hallow structure at it is summit called the neck. Within the ventçr, the female gamete or egg is situated along with a venter canal cell. The neck shows neck cells, neck canal cells and cover cells.

Question 28.
Write any four salient features of the class Aves. Mention on a flightless bird.
Answer:

1. Presence of feathers.
2. Presence of pneumatic bones: Pneumatic bones are long hollow bones with air cavities.
3. Forelimbs are modified into wings to help in flight.
4. Hind limbs generally have scales helping them to clasp the prey in-flight or a tree branch.
5. Excretion of uric acid and faeces is through a single opening, and excretion uses a very little amount of water, to reduce body weight.
6. An aerodynamically built body helps in flying (by reducing friction).

Question 29.
Draw a neat labelled diagram of a plant cell.
Answer:

Plant Cell – Ultrastructure

Question 30.
Draw a neat labelled diagram of the digestive system of cockroaches.
Answer:

Question 31.
List any five differences between Mitosis and Meiosis.
Answer:

Mitosis	Meiosis
1. Occurs in Somatic cells	1. In reproductive cells
2. Equational division	2. Reduction division
3. Helps in growth and repair	3. Formation of gametes
4. two daughter cell	4. four daughter cell
5. Diploid daughter cells	5. Haploid daughter cells

Question 32.
Explain the Mass flow hypothesis of translocation of sugar in plants.
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 ¡s 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.

Section-II

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

Question 33.
Mention any one role of the following macronutrients
(a) Nitrogen
(b) Phosphorus
(c) Potassium
(d) Calcium
(e) Magnesium
Answer:
(a) Component of proteins and Nucleic acid
(b) Component of cell membranes
(c) Opening and closing of stomata
(d) Component of the middle lamella of cell wall
(e) Component of chlorophyll.

Question 34.
Write the schematic representation of the Calvin cycle.
Answer:
It occurs in the grana of the chloroplast.
It utilises the theassimùatory powers i.e., ATP and NADPH₂ produced in light 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.

Question 35.
(a) List any three differences between aerobic and anaerobic respiration.
Answer:

Aerobic C.R	Anaerobic C.R
1. Occurs in the presence of molecular oxygen.	1. Occurs in the absence of oxygen.
2. Centered in the cytoplasm.	2. Centered in mitochondria and cytoplasm
3. Glucose is incompletely oxidised into C2H50H and CO2.	3. Glucose is completely oxidised into CO2 and H2O.
4. Generates 38 ATPs per glucose.	4. Generates only 2 ATPs per glucose.
5. Occurs in a great majority of organisms called anaerobes.	5. Occurs in only a few organisms called aerobes.

(b) Define RQ. Write the RQ values of fats.
Answer:
The respiratory quotient can be defined as the ratio of the volume of CO₂ released, to 0 consumed to break one molecule of respiratory substrate completely.
RQ = \(\frac{\text { Volume of } \mathrm{CO}_{2} \text { released }}{\text { Volume of } \mathrm{O}_{2} \text { consumed }}\)
RQ of carbohydrate is 1.
RQ of proteins is 0.7
RQ of fats ¡s 0.71
RQ of organic acid is more than 1(1.2 or 1.3).

Question 36.
Write the schematic representation of hormonal action.
Answer:
Mechanism of hormone action:
Hormones exert their effects on target cells by interacting with specific receptors of these cells. These receptors are proteins or lipoproteins. The location of the hormone receptor depends on the chemical nature of the hormone. The water-soluble hormones (polypeptides, glycoproteins and catecholamines) cannot pass through the plasma membrane, hence their receptors are located on the outer surface of the membrane.

Water-soluble hormone action:
The lipophilic hormones (steroids and thyroxine) can pass through the plasma membrane and enter into their target cells, hence the receptors for these hormones are located within the nucleus and cytoplasm.

The water-soluble hormones have their action mediated through second messengers. When these hormones bind to their specific receptors of the plasma membrane specific membrane proteins become activated to generate second messengers

Cyclic AMP, phospholipase C, inositol triphosphate, diacylglycerol and Ca⁺⁺ act as second messengers.
The cyclic AMP second messenger system has the following sequence of events to affect the action of hormones.

1. The hormone binds to the specific membrane receptor of the target cell.
2. G proteins linked to specific receptors activate the enzyme adenylate cyclase.
3. Activated adenylate cyclase, converts ATP to cyclic AMP.
4. Cyclic AMP activates protein kinase enzymes of the cytoplasm.
5. Protein Kinase phosphorylates (adding phosphate groups too) the other enzymes of the cytoplasm.
6. The phosphorylation of specific enzymes may enhance or inhibit their activity.
7. This alters the metabolism of the target cell in the required direction as per the specific hormone action.

Lipophilic hormone action:

The lipophilic steroid hormones and thyroid hormones pass through the plasma membrane and get bound to a specific intracellular receptor. These receptors are commonly located within the nucleus. Once these nuclear hormone receptors are activated by their binding to the specific hormones, they alter the gene expression. They may switch on or off specific genes on the nuclear DNA. The transcription of DNA occurs and the newly formed mRNA directs the synthesis of specific enzyme proteins. These enzymes alter the metabolism of the cell in the required direction as per the intended action of the hormone.

Question 37.
Name any five hormones for the pituitary gland and mention and one function of each.
Answer:
1. The pituitary gland weighs about 0.5 gm and is a pea-sized endocrine gland that lies on the ventral surface of the brain attached to the hypothalamus by a nervous stalk called the infundibulum.

2. It is called the master gland or conductor of the endocrine orchestra as several of its hormones control other endocrine glands directly. The hormones of the pituitary that influence other endocrine glands are called tropins or trophic hormones. However, the pituitary gland itself works under the influence of the hypothalamus through releasing factors. The pituitary gland is called the master gland because it controls the functions of other endocrine glands by secreting hormones.

3. Based on origin, it is divided into two parts namely Adenohypophysis and Neurohypophysis.

Students can Download 1st PUC Biology Previous Year Question Paper March 2020 (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 2020 (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:
It is the sum total of all molecular interactions that occur inside living organisms.

Question 2.
Which cell can fix atmospheric nitrogen in some of the cyanobacteria?
Answer:
Heterocyst.

Question 3.
What is venation?
Answer:
Arrangement of veins and veinlets on the lamina of a leaf.

Question 4.
Mitochondria are called Tower houses* of the cell. Why?
Answer:
They produce ATP.

Question 5.
Name the phase of the cell cycle which involves DNA replication.
Answer:
S-phase of interphase.

Question 6.
What is guttation?
Answer:
The loss of water ¡n liquid form through specialized structures called water stomata or hydathodes is called guttation.

Question 7.
Mention the technique of growing plants in a nutrient solution.
Answer:
Hydroponics.

Question 8.
Name the antibacterial agent present in saliva.
Answer:
Lysozyme.

Question 9.
What is tetany?
Answer:
It refers to the continued state of contraction or wild contractions of muscles due to low Ca in body fluids.

Question 10.
Define hormones.
Answer:
Hormone: Hormones are biochemical messengers secreted in minute quantities at a place and are transported to a target cell to induce a specific change in the target cell. Hormones have many functions.

PART-B

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

Question 11.
List any two differences between chordates and non-chordates.
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 gillsilts.	(c) Absesnce of gilislits.
(d) Presence of ventral heart.	(d) Presence of dorsal heart.

Question 12.
What are tendons and ligaments?
Answer:
Tendons: They are exclusively composed of white collagenous fibres arranged in regular bundles. In between these fibres, a few fibroblasts may occur. Usually, they connect muscle to bone.

Ligaments: They may contain both white and yellow fibres and these fibres are arranged in less orderly patterns. Usually, they connect bone to bone.

Question 13.
Why the Golgi apparatus remains in close association with the endoplasmic reticulum?
Answer:

1. At end of the prophase of mitosis in animal cells, centriole duplicate and new daughter centrioles are formed.
2. Give rise to basal bodies from which cilia and flagella arise.
3. Distal centriole of sperm gives rise to axonemes or axial filament of the sperm tail.

Question 14.
Differentiate hypotonic and hypertonic solutions.
Answer:
(a) Hypotonic solution: It is a solution whose solute concentration is less than the solute concentration of the cell sap.
(b) Hypertonic solution: It is a solution whose solute conceñtration is more than the solute concentration of the cell sap.

Question 15.
Abscisic acid (ABA) is called the ‘Stress hormone’. Justify.
Answer:
It helps to overcome adverse conditions.

Question 16.
Mention two types of circulatory systems.
Answer:
Pulmonary circulation:
It includes the collection of deoxygenated blood into the right chambers of the heart and then pumping the same to the lungs for oxygenation, and bringing back the oxygenated blood from the lungs into the left atrium.

The superior vena cava and inferior vena cava collect the deoxygenated blood into the right atrium from anterior and posterior regions of the body respectively. Then it is pumped into the right ventricle. This, in turn, pumps it to the lungs, through the pulmonary artery for purification.

The oxygenated blood leaves the lungs through pulmonary veins which open into the left atrium.

Systemic circulation:
It includes the collection of oxygenated blood into the left chambers of the heart, and the distribution of the same to various tissues through the aorta, and its branches.

A set of four pulmonary veins bring the oxygenated blood from the lungs into the left atrium. The left atrium pumps the oxygenated blood into the left ventricle which shows powerful contractions. So the blood is pumped into the aorta under definite pressure.

Aorta gives off various branches in different directions for distribution. The blood exchanges oxygen, and carbon dioxide in the tissues of the body. Thus the blood is again deoxygenated. This deoxygenated blood is collected by superior and inferior vena cava and sent into the right atrium.

Question 17.
Write antigens on RBC’s and antibodies in the plasma of the universal donor blood group and universal recipient blood group.
Answer:

Question 18.
Bile juice contains no digestive enzymes. Then how does it helps for digestion?
Answer:
Bile juice is secreted by the hepatic cells of the liver and stored in the gall bladder. It is released into the duodenum through the bile duct. Bile juice doesn’t contain any digestive enzymes. But it contains the digestive components ¡n the form of bile salts Sodium Glycocholate, and Sodium taurocholate, Bile salts emulsify the fats by lowering their surface tension and convert them into numerous tiny droplets. Bile is rich ¡n sodium bicarbonate; hence it neutralizes the acidic chyme. It also helps in the absorption of fat-soluble vitamin K.

PART-C

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

Question 19.
What is Binomial nomenclature? Write any two universal rules of nomenclature.
Answer:
Binomial Nomenclature:
This method was introduced by Carolus Linnaeus.
In this method, every organism is given a scientific name, which has two parts, the first is the name of the genus (generic name) and the second is the name of the species (specific epithet).
e.g. Mangifera indica for mangoes and Homo sapiens for human beings. In the above,

Man gera and Homo are generic names, while India and sapiens are the names of the species belonging to Man gilera and Homo respectively.

The classification consists of a hierarchy of steps where each step represents a range or category. The various ranges or categories used in classification are called Taxonomic categories.

Guidelines/Principles for nomenclature:

1. Scientific names are, generally, written in Latin or derived from Latin, irrespective of their origin.
2. The scientific names are written in italics or underlined. The first word denotes the name of the genus, and the second woixi denotes the species.
3. The generic name starts with a capital letter, while the specific name starts with a small letter.
4. The name of the author is written in an abbreviated form after the specific name e.g. Mangifera indica Linn.-indicates that this species was first described by Linnaeus.
5. The name should be short, precise and easy to pronounce.

Question 20.
‘Flower is modified shoot’. Substantiate your answer.
Answer:
The flower is a modified shoot: When the stem takes on the role of general reproduction it is modified into a flower. That is why flower ¡s often called a modified shoot.

Question 21.
Write about the significances of meiosis.
Answer:
Significance of Meiosis:

1. Meiosis brings genetic crossing over and random distribution of paternal and maternal chromosomes to daughter cells.
2. Recombination produces variations and variations are the sources of organic evolution.

Question 22.
How is mycorrhizal association helpful in the absorption of water and minerals in plants?
Answer:
The hyphae have a large surface area from an Ammonotebic large volume of soil.

Question 23.
List any three physiological roles of cytokinins.
Answer:

• Discovered by Skoog and Miller
• Chemically these are 6-fur furyl amino purlins.
• Synthesised in young fruits, leaves, buds, root tips and translocated through the xylem.
• Cytokinins stimulate cell division.
• Cytokinins prevent early ageing or senescence by stabilizing proteins and chlorophyll. This phenomenon is called Richmond – Lang effect.
• Cytokinins break seed dormancy and induce germination.

Question 24.
Briefly explain disorders of the Respiratory system.
Answer:

1. It is the inflammation of bronchi, leading to breathlessness, chest tightness, light sneezing, and cough with the expulsion of yellow or green sputum.
2. It is the parietal blocking of bronchi and bronchioles, resulting in breathlessness, chest tightness, wheezing and dry cough.
3. Emphysema is the condition, where the walls of the alveoli lose their elasticity and remain rigid due to a permanent air-filled condition.

Question 25.
Mention three types of body movements.
Answer:

1. Amoeboid Movement: Macrophages and Neutrophils.
2. Ciliary Movement: Epithelium of nasal tract, urinary tubule and fallopian tube.

Question 26.
Classify the animals based on nitrogenous wastes excreted by them.
Answer:
Ureotetic (Urecotelic).

Part-D (Section -I)

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

Question 27.
Write five salient features of Bryophytes.
Answer:
(1) They are non-vascular plants, exhibiting amphibious habitats.

(2) Gametophytic phase of the life cycle is represented by a multicellular, haploid thallus, which is called gametophyte.

(3) Gametophytic plant body bearing male and female sex organs, represents the dominant phase of life cycle unlike in higher plants.

(4) Gametophyte is autotrophic and independent in nature.

(5) The gametophyte produces sex organs, the antheridia and the archegonia which are multicellular Antheridia may be embedded within the thallus or may be specifically located at the tips of the gametophyte. It produces biciliate curved antherozoids.

Archegonia is a flask like structure. It bears a basal bulbous vendor and a narrow, hallow structure at it is summit called the neck. Within the venter, the female gamete or egg ¡s situated along with a venter canal cell. The neck shows neck cells, neck canal cells and cover cells.

(6) Fertilisation takes place in the presence of water, leading to the formation of a zygote which later develops into a multicellular embryo. Hence, the group is treated under embryophytes.

(7) Sexual reproduction is always of the Ougamous type.

(8) Embryo develops into a diploid sporophyte consisting of the foot, seta and capsule, which is dependent on the gametophyte.

(9) Diploid sporophyte reproduced by the formation of haploid spores (meiosis occur during spore formation). The spore germinates and produces gametophytes.

(10) In their life cycle, such distinct heteromorphic alternation of generations is shown by the members.

Question 28.
(a) Describe three types of simple epithelial tissues based on structural modification of the cells.
Answer:
Epithelial tissue ¡s a tissue, which lies on the upper and outer layer covering the surface of the organ or lining the cavity.

Squamous epithelium:
a. It occurs in the skin of vertebrates. Alveoli of lungs, the mouth cavity, etc.
b. The cells are nearly hexagonal flat and plate-like.
c. A spherical or oval nucleus is found to be present at the cell centre.
d. The cells are closely fitted with minimum intercellular space. Hence they look like mosaic pavement. Because of this, the squamous epithelium is also called pavement epithelium.
e. It forms a continuous cover and protects the inner tissues.

Columnar epithelium:
a. it occurs in the mucosa membrane of the stomach and the intestine.
b. The cells are elongated and pillar-like. The lower ends are narrow and the upper ends are broad.
c. The nucleus is oval and the basal cytoplasm is granular.
d. Cells are lodged over a basement membrane. Intercellular spaces are present at the basal region of the cells.
e. The columnar cells are absorptive and protective in function.

(b) Write any two functions of bones.
Answer:
Structure support.

Question 29.
Draw a neat labelled diagram of T.S. of Dicot Root.
Answer:

Question 30.
Explain any five classes of enzymes based on the type of reactions they catalyse.
Answer:

• Oxidoreductases: Act on many chemical groupings to add or remove hydrogen atoms.
• Transferases: Transfer functional groups between donor and acceptor molecules. Kinases are specialised 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 31.
Mention any one importance of the following macronutrients.
(a) Nitrogen (b) Phosphorus (c) Potassium
Answer:
Macronutrients: They are essential elements that occur in a concentration of 1-10 mg/g of dry matter, e.g., C, H, O, N, P, K.

Micronutrients: They are essential elements that occur in plants in a concentration of 0.1 mg or below per gin of dry matter, e,g,, Mn, Zn, Cu, Mo, Bo.

Beneficial Nutrients: They are non-essential elements that have some specific role in some plants, e.g., Na4 in Ç pLants, Si in grasses.

Toxic Elements: They are non-essential elements which if absorbed reduce plant growth, and yield. Many elements become toxic only after their concentration reaches a particular level, e.g., Lead, Aluminium, Mn.

Essential Elements: They are those elements that are involved in some vital structural, and functional role in plants so that their deficiency will cause disorders while their absence results in non-completion of the life cycle, e.g., C, l-I, O, N, P, K Ca, Mg, S, etc.

Question 32.
Draw a neat labelled diagram of multipolar Neuron.
Answer:

Section-II

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

Question 33.
List any five characters of class Aves.
Answer:

1. Birds are homeothermic (warm-blooded) animals.
2. The body is boat-shaped and is adapted in such a way as to offer the least resistance while flying
3. The body is differeñtiated into ahead, long neck, a trunk and a short tail.
4. The body is covered by feathers, which provide insulation.
5. Forelimbs are modified into wings for flight. Wings are provided with feathers, which are bad conductors of heat.

Question 34.
Draw a neat labelled diagram of the Animal cells.
Answer:

Animal Cell – Ultrastructure

Question 35.
(a) State Blackmail’s law of limiting factors.
Answer:
The law of limiting factor states that “when a process is governed by a number of separate factors, the rate of the process is limited by the pace of the slowest factor”. Photosynthesis is influenced by several factors such as CO₂ concentration, availability of water, sunlight etc. Any of these factors may become a limiting factor at any one moment and may limit the rate of photosynthesis.

(b) Mention external factors affecting photosynthesis.
Answer:
Light, temp H₂O, CO₂

Question 36.
Give the schematic representation of Glycolysis.
Answer:

1. It occurs in the cýtoplasm of the cell.
2. It is an enzymatic reaction, thus temperature sensitive.
3. It is a common reaction for both aerobic and anaerobic respiration.

Question 37.
Name any five hormones secreted by the pituitary gland and mention anyone function of each.
Answer:
1. The pituitary gland weighs about 0.5 gm and is a pea-sized endocrine gland that lies on the ventral surface of the brain attached to the hypothalamus by a nervous stalk called the infundibulum.

2. It is called the master gland or conductor of the endocrine orchestra as several of its hormones control other endocrine glands directly. The hormones of the pituitary that influence other endocrine glands are called tropins or trophic hormones. However, the pituitary gland itself works under the influence of the hypothalamus through releasing factors. The pituitary gland is called the master gland because it controls the functions of other endocrine glands by secreting hormones.

3. Based on origin, it is divided into two parts namely Adenohypophysis and Neurohypophysis.

Expert Teachers at KSEEBSolutions.com has created New Syllabus Karnataka 1st PUC Kannada Model Question Papers with Answers 2019-20 Pdf Free Download of 1st PUC Kannada Previous Year Board Model Question Papers with Answers are part of 1st PUC Model Question Papers with Answers. Here We have given the Department of Pre University Education (PUE) Karnataka State Board Syllabus Second Year Model Question Papers for 1st PUC Kannada Model Question Papers with Answers 2019-2020 Pdf. Students can also read 1st PUC Kannada Question Bank with Answers hope will definitely help for your board exams.

Karnataka 1st PUC Kannada Model Question Papers with Answers 2019-2020

• 1st PUC Kannada Model Question Paper 1 with Answers
• 1st PUC Kannada Model Question Paper 2 with Answers
• 1st PUC Kannada Model Question Paper 3 with Answers
• 1st PUC Kannada Model Question Paper 4 with Answers
• 1st PUC Kannada Model Question Paper 5 for Practice
• 1st PUC Kannada Previous Year Question Paper March 2019 (South)
• 1st PUC Kannada Previous Year Question Paper March 2019 (North)

Karnataka 1st PUC Kannada Blue Print of Model Question Paper

We hope the given New Syllabus Karnataka 1st PUC Class 11 Kannada Model Question Papers with Answers 2019-20 Pdf Free Download of 1st PUC Kannada Previous Year Board Model Question Papers with Answers will help you. If you have any queries regarding Karnataka State Board Syllabus First PUC Class 11 Model Question Papers for 1st PUC Kannada Model Question Papers with Answers 2019-2020 Pdf, drop a comment below and we will get back to you at the earliest.

Students can Download 1st PUC Biology Model Question Paper 9 with Answers, 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 Model Question Paper 9 with Answers

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 a herbarium?
Answer:
A herbarium is defined as a collection of plants, that have been dried, pressed and preserved on sheets.

Question 2.
Define venation.
Answer:
The mode of arrangement of veins and veinlets is called venation.

Question 3.
“Earthworms are called friends of farmers”. Justify the statement.
Answer:
As they increase soil fertility by their worm castings and by making the soil porous by burrowing.

Question 4.
What are chromoplasts?
Answer:
Plastids contain coloured pigments except for green pigment chlorophyll.

Question 5.
Name the organic compounds (non-protein constituents) present in certain enzymes that are tightly bound to the apoenzyme.
Answer:
Prosthetic group.

Question 6.
Why are mitochondria called “powerhouses of the cell”?
Answer:
Produce cellular energy in the form of ATP.

Question 7.
Define 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 8.
Name the essential element that is present in chlorophyll.
Answer:
Magnesium.

Question 9.
Name the red coloured oxygen storing pigment present in muscles.
Answer:
Myoglobin.

Question 10.
Mention the hormone that regulates the 24-hour (diurnal) rhythm of the human body.
Answer:
Melatonin.

PART-B

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

Question 11.
Mention two rules of binomial nomenclature.
Answer:

1. Scientific names are, generally, written in Latin or derived from Latin, irrespective of their origin.
2. The scientific names are written in italics or underlined. The first word denotes the name of the genus, and the second word denotes the species.
3. The generic name starts with a capital letter, while the specific name starts with a small letter.

Question 12.
Write any two classes of Kingdom Fungi with one example for each class.
Answer:
Class: Zygomycetes eg., Rhizopus
Class: Ascomycetes eg., yeast.

Question 13.
(a) Why bryophytes are called amphibians of the plant kingdom?
(b) Cycas is considered a naked seeded plant. Give a scientific reason.
Answer:
(a) Bryophytes can live in soil but require water to complete sexual reproduction.
(b) Ovules are not enclosed by my ovary wall and remain exposed.

Question 14.
Write two differences between cartilaginous and bony fishes.
Answer:

Chondrichthyes	Osteichthyes
1. Skeleton is made up of cartilage.	1. Skeleton is made up of bones.
2. Exclusively marine.	2. Both marine and freshwater forms are seen.
3. They possess placoid scales.	3. Possess cycloid or ctenoid scales.
4. Mouth and nostrils are situated on the ventral side of the body.	4. Mouth and nostril are terminal in their position.

Question 15.
List two functions of Golgi bodies.
Answer:

1. It secretes various enzymes and hormones.
2. It forms secretory vesicles which store enzymes, proteins, carbohydrates etc.
3. It takes part in the formation of the cell plates and cell walls along with ER at the time of telophase in plant cells.
4. It helps in the formation of primary Lysosomes.

Question 16.
Define respiratory quotient, Write the RQ value for glucose.
Answer:
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 17.
Differentiate long-day plants and short-day plants.
Answer:

1. Long Day Plants: These flower in photoperiod more than critical day length, eg: Wheat, oats etc.
2. Short Day Plants: These flower in photoperiod less than critical day length. e.g: Tobacco, Chrysanthemums etc.

Question 18.
Mention two types of movements exhibited by the cells of the human body with an example for each.
Answer:
1. Amoeboid Movement/Pseudopodial movement :
Macrophages and leucocytes exhibit amoeboid movement. It is also seen in the amoeba. Amoeboid movement is affected by pseudopodia formation by cytoplasmic streaming. It also involves cytoskeletal elements like microfilaments.

2. Ciliary Movement:

• These movements are caused by fine versatile hair called cilia.
• Ciliary movement occurs in those hollow tubular visceral organs, that is internally lined by ciliated epithelium.
• Ciliary movement in the trachea helps in removing foreign substances or particles.
• Ciliary movement in the fallopian tube causes the ova to move to the uterus.

PART-C

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

Question 19.
Schematically represent the Haplontic life cycle in plants.
Answer:
Haplontic life cycle

1. The dominant phase in the life cycle is a free-living gametophyte (haploid).
2. The diploid phase is represented only by the single-celled zygote.
3. The zygote undergoes meiosis (zygotic meiosis) and forms haploid spores which germinate to form the gametophyte. e.g Spimgpra Chiamydomonas etc.

Question 20.
Explain any three types of aestivation in lowering plants.
Answer:
Aestivation of calyx and corolla:
The mode of arrangement of sepals, petals or even tepals in a flower bud is called aestivation. The different kinds of aestivation are as follows:

1. Valvate aestivation: When sepals, petals or tepals are not overlapping.
2. Imbricate aestivation: When out of the total number of sepals, petals or tepals, one is completely out, one is completely in and the rest are in and out.
3. Descending imbricate aestivation: When the standard petals are large and overlap the two wing petals, which in turn overlap the keel petals. It ¡s technically known as vexillary aestivation.
   Note: It is characteristic of the members of the subfamily Papilionoideae (Papilionaceae).
   e.g: Pea, Bean, Indigofera, Tephrosia etc.
4. Ascending imbricate aestivation – When the small standard petal is completely ¡n and is overlapped by the lateral wing petals which in turn are overlapped, by the keel petals.
   Note: it is characteristic of the sub-family Caesalpinioideae (Caesalpinae).
   e.g: Caesalpinia puicherhima, Delonix regia, etc.
5. Quincuncial aestivation – When out of the total number of sepals or petals, two are completely out, two are completely in and the rest are in and out.
6. Contorted or twisted aestivation – When all sepals or petals are both in and out.

Question 21.
Differentiate homopolymers and heteropolymers (polysaccharides) with an example each.
Answer:
Homopolymers are made up of only one type of monosaccharide, eg., cellulose, starch. Heteropolymers are made up of different types of monosaccharides, eg., chitin.

Question 22.
Write the schematic representation of steps involved in the formation of ethyl alcohol during fermentation of glucose.
Answer:
The first step of anaerobic respiration is glycolysis, in which glucose is incompletely oxidised 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 muscles.

Question 23.
Write about any three disorders of the digestive system.
Answer:
JAUNDICE IICTERUSI:
It is the most common digestive disorder. Jaundice relates to the yellow appearance of skin, urine, nails and cornea of the eye due to an increased Bilirubin pigment (Bile pigment).

Jaundice is of three types:
1. Haemolytic Jaundice or Neonatal Jaundice or Pre hepatic Jaundice: It is commonly found in newly born babies up to one week of age. It is due to excessive haemolysis (Destruction of RBC’s).

2. Infective or Hepatocellular Jaundice: It is due to the infection of the hepatitis virus of the liver. So, the liver becomes unable to process Bilirubin. It results in a heavy accumulation of Bilirubin which leads to Hepatocellular Jaundice.

3. Obstructive Jaundice or Extra Hepatic jaundice: It occurs due to the obstruction in the Bile duct. The obstruction of the Bile duct is due to gallstone formation. The bile duct is thus blocked, and bile juice cannot enter into the duodenum.

Causes:

1. Liver damage due to viral infection, and consumption of alcohol.
2. Increased Bilirubin pigment in the body.
3. Excessive haemolysis.
4. Blockage of bilk duct due to gallstone.
5. Presence of tumours in the bile duct.

Effects:

1. Yellow colouration of the skin, eyes and urine.
2. Fever, loss of appetite, nausea, vomiting and headache.
3. Infective jaundice leads to bleeding, and shrinkage of the liver [Cirrhosis]
4. Unconsciousness and finally it may lead to death.

Prevention:
It can be prevented by avoiding the consumption of alcohol, and also by avoiding the consumption of contaminated food and water.

Treatment :

1. Vaccines and anti-viral drugs are used against infective jaundice.
2. Obstructive jaundice can be treated by surgery.

Question 24.
What is: (a) Tidal volume, (b) Residual volume, (c) Vital capacity?
Answer:
(a) Tidal volume (VT): It is the volume of air inhaled or exhaled with each inhalation and exhalation without any extra effort. Its volume ¡s 500 ml.

(b) Residual volume (RV): It is the volume of air that remains inside the lungs at the end of forceful expiration. Its volume is 1100 – 1200 ml.

(c) Vital Capacity (VC): It is the total volume of air that can be expelled from the Lungs during maximum exhalation.
VC = TV + IRV + ERV
= 500 + 2500 + 1100
= 4100 ml approx.

Question 25.
How are differentiation, dedifferentiation and redifferentiation different from each other?
Answer:
Differentiation: It is the conversion (maturation) of meristematic cells, to regain the power to divide under certain conditions.

Dedifferentiation: Permanent cells like parenchyma can revert back to meristematic activities (cell division) to form secondary persistent. This conversion is called Dedifferentiation.

Redifferentiation: If the secondary meristem divides to produce new cells, which further differentiate to form micelle cells, and once again lose the capacity to divide, then it is called Redifferentiation.

Question 26.
Classify the animals with an example each based on the chief excretory product produced in them.
Answer:
Based on the type of excretory product produced, animals are divided into 3 groups:
1. Anunonoteic animals: Animals that excrete ammonia as nitrogenous waste products are called ammonotelic animals.
e.g: Aquatic invertebrates and bony fish, larvae of frog (Tadpole).

2. Ureotelic animals: Animals that excrete urea are called ureotelic animals. Urea is produced by the liver during the ornithine cycle. Urea is stored in the urinary bladder in a dissolved state called urine.
e.g: Cartilaginous fish (Shark), Mammals and Amphibians.

3. Uricotelic animals: Animals that excrete uric acid as nitrogenous waste products are called uricotelic animals.
e.g: Reptiles, birds.

PART-D (Section-I)

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

Question 27.
List five characteristic features of birds.
Answer:

1. Birds are homeothermic (warm-blooded) animals.
2. The body is boat-shaped and is adapted in such a way as to offer the least resistance while flying.
3. The body is differentiated into an ahead, long neck, a trunk and a short tail.
4. The body is covered by feathers, which provide insulation.
5. Forelimbs are modified into wings for flight. Wings are provided with feathers, which are bad conductors of heat.
6. The hind limbs are shifted forwards to balance the body while walking. They are variously modified for perching, walking, running, swimming etc.,
7. Hind limbs have 4 toes and are covered, with epidermal scales.
8. The endoskeleton is fully ossified (bony). Bones are tubular and hollow fond have air cavities or air sacs (pneumatic bones).
9. The bones of the skull are fused.
10. Jaws are devoid of teeth but covered by horny sheaths and are modified to form a beak.
11. The beak is variously modified to suit individual food habits.

Question 28.
Describe the important events that occur during anaphase and telophase of mitosis.
Answer:
Anaphase (Gr:ana-up; phases-stage):
Anaphase is characterized by the following features.

• The two chromatids of each chromosome separate comp letebç to become daughter chromosomes. The two daughter chromosomes move away from the equator towards the opposite poles.
• During the anaphasic movement of chromosomes, the centromeres lead the path and the arms trail behind. As a result, the anaphasic chromosomes appear V, L, J and L shaped.

Telophase (Gr:telos-end; phases-stage):
It is characterized by the following features.

• Here, reversal of the prophasic events occur. The daughter chromosomes move and reach the opposite poles where they become thin and thread-like again. These threads overlap one another to form a fine chromatin network.
• The spindle fibres disintegrate and disappear.
• Reconstitution of a new nuclear envelope occurs. Nucleolus reappear.

Question 29.
Draw a labelled diagram showing the anatomy of the leaf of a typical monocot plant.
Answer:

Question 30.
(a) Carbon, nitrogen, phosphorous, sulphur, calcium, etc, are considered essential elements. What are the criteria for considering them as essential? (3M)
(b) What is the role of leg-haemoglobin in leguminous plants? (1M)
(c) Name a soil bacterium that helps in nitrification. (1M)
Answer:
(a) The element must be absolutely necessary for supporting normal growth and reproduction, and in the absence of the element, the plants do not complete their life cycle.

• The requirement of the element must be specific and not replaceable by another element, i.e. deficiency of any one element cannot be met by supplying some other element.
• The element must be directly involved in the metabolism of the plant.

(b) Leg haemoglobin protects nitrogenase enzyme from oxygen and acts as an oxygen scavenger.
(c) Nitrosomonas.

Question 31.
Write the schematic representation of the Calvin cycle.
Answer:
It occurs ¡n grana of the chloroplast.
It utilises the assimilatory powers i.e., ATP and NADPH2 produced ¡n light 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.

Question 32.
Mention one function of each of the following hormones:
(a) ADH, (b) Thymosin, (c) Glucagon, (d) Atrial natruiuretic fact, (e) Erythropoietin.
Answer:
(a) ADH: Stimulates reabsorption of water and electrolytes in the kidney tubules.
(b) Thymosin: Promotes the differentiation of T – lymphocytes to provide cell-mediated immunity and promotes the production of antibodies to provide humoral immunity.
(c) Glucagon: Promotes glycogenolysis and increases blood sugar level.
(d) ANF: Causes dilation of the blood vessels and reduces blood pressures.
(e) Erythropoietin : stimulates erythropoiesis / formation of RBCS.

Section-II

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

Question 33.
Draw a labelled diagram showing the ultrastructure of a plant cell.
Answer:

Plant Cell – Ultrastructure

Question 34.
Classify simple epithelium and mention the structural modifications of cells in them.
Answer:
Simple epithelium: Simple epithelium is further classified into:

1. Squamous epithelium.
2. Columnar epithelium.
3. Cuboidal epithelium.
4. Ciliated epithelium.
5. Glandular epithelium.

1. Squamous epithelium :
1. It occurs in the skin of vertebrates, alveoli of lungs, the mouth cavity, etc.
2. The cells are nearly hexagonal, flat and plate-like.

3. A spherical or oval nucleus is found at the cell centre.
4. The cells are closely fitted without intercellular space. Hence look like a mosaic pavement. Because of this squamous epithelium is also called pavement epithelium.

Function:
1. It forms a continuous cover and protects the inner tissues.
2. It helps in ultrafiltration in Bowman’s capsule.
3. It helps in the exchange of gases in alveoli.

Note: Squamous epithelium lining the blood vessels is called endothelium( derived from endoderm). Squamous epithelium lining internal organs such as the stomach, the intestine is called mesothelium (derived from mesoderm).

Question 35.
Explain the pressure-flow hypothesis of the translocation of organic solutes.
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 fonned 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 Tc roots.

5. In the roots the food ¡s 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 the 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.
(a) Explain the mechanism of clotting of blood. (3M)
(b) Differentiate pulmonary and systemic circulations. (2M)
Answer:
Blood has a unique property. As long as circulating inside the body it retains the fluid state however as soon as blood comes out of the body as a result of a cut or an injury, blood is transformed from fluid state to gel state, preventing further loss. This self-regulating mechanism of the blood is known as clotting of the blood.

The mechanism of blood clotting is explained by various theories out of these ‘Best and Taylor’s theory is one. According to this theory. four factors are responsible for blood clotting.

1. Prothrombin: Produced by the liver and present in plasma.
2. Thmmboplastin: An enzyme released from damaged tissues, present in the body tissues.
3. Calcium ions: Present in the plasma of the blood.
4. Fibrinogen: Produced by the liver and present in plasma.

Chemical events leading to the formation of b od clot as suggested by ‘Best and Taylor’s theory can be represented as follows.

The reaction shown above indicates that the inactive Prothrombin is converted into active thrombin by the catalytic activity of the enzyme Thromboplastin in the presence of Ca₂. The activated thrombin reacts with soluble fibrinogen, resulting in the formation of insoluble fibrin.

The fibrin threads form a mesh-like network in which the blood components get entangled, and results in blood clots and thus prevents bleeding.

Best and Taylor’s theory doesn’t consider the role of platelets in the blood clotting mechanism.

Question 37.
Draw a labelled diagram of the human brain in the sagittal section.
Answer:

Students can Download 1st PUC Biology Model Question Paper 8 with Answers, 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 Model Question Paper 8 with Answers

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 a herbarium?
Answer:
A herbarium is defined as a collection of plants, that have been dried, pressed, and preserved on a sheet.

Question 2.
Give an example for palmately compound leaf.
Answer:
Silk cotton tree.

Question 3.
Collenchyma is called a simple tissue. Why?
Answer:
Because it is made up of cells that are similar in structure and function.

Question 4.
What are mesosomes?
Answer:
Mesosomes are balloon or tubular ingrowths of plasma membrane formed in prokaryotes and contain enzymes for respiration.

Question 5.
Mention the significance of meiosis.
Answer:
Meiosis maintains the chromosome number constant in sexually reproducing organisms. It is essential since the chromosome number is doubled after fertilization.

Question 6.
Define root pressure.
Answer:
The hydrostatic pressure is created inside the xylem of the root due to the osmotic entry of water.

Question 7.
Name the oxygen scavenger molecule that protects nitrogenase in nodules.
Answer:
Leg – hemoglobin.

Question 8.
What is emphysema?
Answer:
The loss of elasticity of walls of alveoli resulting in their expansion is called emphysema. It is induced by cigarette smoke and other irritants.

Question 9.
Which blood group is called a universal donor?
Answer:
‘O’ blood group.

Question 10.
Why the filtration of blood in Bowman’s capsule is referred to as ultrafiltration?
Answer:
Because almost all the constituents of plasma except proteins are filtered into the lumen of the bow marks capsule due to the presence of micropores in the epithelial layer of Bowman’s capsule.

PART-B

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

Question 11.
Write any four characters of fungi.
Answer:
Fungi are eukaryotic, achlorophyllous organisms that are but, generally multicellular a few are unicellular (yeast).

1. The fungal body consists of long, slender, thread-like structures called hyphae, which form a network called mycelium.
2. The hyphae have a cell wall made up of chitin.
3. They may be aseptate and multi-nucleate i.e., coenocytic or septate.
4. All fungi are heterotrophs -they are saprotrophs or parasites or live as symbionts in the roots of higher plants (mycorrhizae) and in lichens (mycobiont).
5. They reproduce vegetatively by fragmentation, fission, or budding.

Question 12.
List any four salient features of phylum Coelenterata.
Answer:

1. Coelenterates are all aquatic, most of them are marine and a few are freshwater forms.
2. Members are multicellular and the cells are compactly arranged to form definite tissues.
3. Coelenterates exhibit radial symmetry.
4. These organisms have two germ layers viz. outer ectoderm and inner endoderm, and thus they are diploblastic. Between these two cellular germ layers, is a thick non-cellular gelatinous mesoglea.
5. Two distinct types of individuals are seen in poly or hydra-like (hydroid) type and the medusoid type.
6. Some forms exhibit polymorphism (eg: Obelia).
7. Digestion is both extracellular and intracellular.
8. Reproduction takes place by both sexual and asexual methods.

Question 13.
Give reasons for the following:
(a) The blood vascular system in cockroaches is considered as an open type.
(b) The vision in cockroaches is referred to as mosaic vision.
Answer:
(a) As blood flows in blood sinusoids which are open channels and in the hemocoel.
(b) As the eye contain several ommatidia to receive light signals instead of a single lens.

Question 14.
Draw a labeled diagram of a section of the chloroplast.
Answer:

Electron microscopic structure of the chloroplast

Question 15.
What is vernalization? Mention any one of its importance.
Answer:
Some plants require low-temperature treatment for flowering. This treatment is known as vernalization.

Vernalization prevents precocious reproductive development late in the growing season and enables the plant to have sufficient time to reach maturity.

Significance: Subjecting the biennial plants like sugar best cabbage, carrot, etc to low temperature or cold treatment stimulates subsequent flowering.

Question 16.
Explain how the exchange of O₂ and CO₂ is achieved between alveoli and deoxygenated blood.
Answer:
Exchange of O₂ and CO₂ between alveoli and deoxygenated blood occurs by simple diffusion based on pressure/concentration gradient.

The PO₂ in alveoli is 104 mm Hg and PO₂ in deoxygenated blood is 40 mm Hg.

Due to the difference in partial pressure, the O₂ from Alveoli diffuses into deoxygenated blood. The PCO₂ in deoxygenated blood is 45 mm Hg and in the alveoli is 40 mm Hg. Due to this difference in partial pressure, the CO₂ from deoxygenated blood diffuses into alveoli.

Question 17.
Explain the role of the Atrial Natriuretic factor in the regulation of kidney function.
Answer:
ANF is a peptide hormone secreted by the atrial wall. An increase in the blood flow to the atria of the heart can cause the release of ANF. ANF can cause vasodilation of blood vessels and thereby decrease blood pressure. ANF mechanism acts as a cheek on the renin-angiotensin mechanism.

Question 18.
What is osteoporosis? Mention the common cause that leads to osteoporosis.
Answer:
Osteoporosis is an age-related disorder characterized by decreased bone mass and increased chances of fractures.
The most common cause is the decreased levels of estrogen.

PART-C

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

Question 19.
List the important rules of binomial nomenclature. Write the scientific name of a housefly.
Answer:

1. Scientific names are, generally, written in Latin or derived from Latin, irrespective of their origin.
2. The scientific names are written in italics or underlined. The first word denotes the name of the genus, and the second word denotes the species.
3. The generic name starts with a capital letter, while the specific name starts with a small letter.
4. The name of the author is written in an abbreviated form after the specific name e.g, Mangifera indica Linn.-indicates that this species was first described by Linnaeus.
5. The name should be short, precise, and easy to pronounce.
   The scientific name of housefly: Musca domestica.

Question 20.
defies the terms: (i) Monodelphous condition (ii) Apocarpous condition (iii) Zygomorphic flower.
Answer:
(i) Monoadelphous: The stamens are united into one bunch or one bundle.
(ii) Apocarpous condition: The carpels in an ovary are free.
(iii) Zygomorphic flower: When a flower can be divided into two similar halves only in one particular vertical plane, it is referred to as a zygomorphic flower.

Question 21.
Write any three anatomical differences between the dicot leaf and the monocot leaf.
Answer:

Isobilateral (Monocot)	Dorsiventral (Dicot)
1. The upper epidermis has large bulliform cells.	1. There are no bulliform cells.
2. Stomata generally occur on both the epidermal layers.	2. Stomata are generally found only on the lower epidermis.
3. The mesophyll is not differentiated into palisade and spongy tissues.	3. Mesophyll is well differentiated.
4. The bundle sheath extensions are made up of sclerenchyma cells.	4. The extensions are made up of collenchyma and parenchyma cells.

Question 22.
Write a note on the G₀ phase of the cell cycle.
Answer:
G₀ phase: Some cells in the adult animals exit the G₁ phase and enter an inactive phase called quiescent state or G₀ phase of the cell cycle.

Cells in this stage remain metabolically active. But these cells -do not exhibit cell division, unless called on to do so depending upon the requirement of an organism.

Therefore, they divide occasionally, as needed to replace cells that have been lost due to injury or cell death.

Question 23.
Draw the sigmoid growth curve. Write the formula to express, exponential growth.
Answer:

Formula to express exponential growth: W¹ = WOe^rt.

Question 24.
Give a brief account of the absorption of fatty acids and glycerol in the small intestine.
Answer:
Give a brief account of the absorption of fatty acids and glycerol in the small intestine.

Fatty acid and glycerol are insoluble, hence cannot be absorbed into the blood. They are first incorporated into small droplets called micelles. Micelles move to the intestinal mucosa, where they are reformed into very small protein-coated globules called chylomicrons.

Chylomicrons are transported into lymph vessels in villi and lymph vessels release these absorbed substances into the bloodstream.

Question 25.
Explain the mechanism of coagulation of blood.
Answer:
Blood has a unique property. As long as circulating inside the body it retains the fluid state however as soon as blood comes out of the body as a result of a cut or an injury, blood is transformed from fluid state to gel state, preventing further loss. This self-regulating mechanism of the blood ¡s known as clotting of the blood.

The mechanism of blood clotting is explained by various theories out of these ‘Best and Taylor’s theory is one. According to this theory. four factors are responsible for blood clotting.

1. Prothrombin: Produced by the liver and present in plasma.
2. Thromboplastin: An enzyme released from damaged tissues, present ¡n the body tissues.
3. Calcium ions: Present in the plasma of the blood.
4. Fibrinogen: Produced by the liver and present in plasma.

Chemical events leading to the formation of b; od clot as suggested by ‘Best and Taylor’s theory can be represented as follows.

The reaction shown above indicates that the inactive Prothrombin is converted into active thrombin by the catalytic activity of the enzyme Thromboplastin in the presence of Ca2. The activated thrombin reacts with soluble fibrinogen, resulting in the formation of insoluble fibrin.

The fibrin threads form a mesh-like network in which the blood components get entangled and results in blood clots and thus prevents bleeding.

Best and Taylor’s theory doesn’t consider the role of platelets in the blood clotting mechanism.

Question 26.
Write a note on myosin protein.
Answer:

1. Myosin (Thick) Filament: length is about 1.6 nm and breadth the 15nm.
2. Myosin ¡s a polymeric protein, whose monomeric proteins are called meromyosins.
3. Each meromyosin has two important parts, a globular head with a short arm and a tail.
4. The head with the short arm is called the heavy meromyosin (HMM) and the tail is called light meromyosin (LMM).
5. 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.
6. The globular head functions as an ATPase enzyme and has the binding sites for ATP and active sites of actin.

PART-D (Section I)

Answer any FOUR of the following questions in 200-250 words each wherever applicable: (4 x 5 = 20)

Question 27.
Write the general characters of angiosperms.
Answer:
General characters:

1. Plant body ¡s a diploid sporophyte and is well-differentiated into the underground root system and aerial shoot system.
2. Plants produce special sexual reproductive structures called flowers.
3. Flowers contain a male reproductive organ androecium and a female reproductive organ gynoecium.
4. The androecium is composed of stamens which produce pollen grains. The gynoecium is composed of carpels that contain ovules.
5. Pollen grain represents the male gametophyte and produces male gametes. The embryo sac in the ovule represents the female gametophyte and bears female gamete eggs and secondary nucleus.
6. Fertilization is double fertilization where two male gametes involve in the process, one unit with the egg and the other with the secondary nucleus of the female gametophyte.
7. After fertilization fertilized egg develops into an embryo. Ovule develops into a seed that is enclosed in the fruit structure developed from the ovary. Seed on germination gives rise to the sporophyte.
8. The life cycle shows distinct heteromorphic alternation of generation between independent predominant diploid sporophytic phase and dependent reduced haploid gametophytic phase.

Question 28.
Differentiate between chordates and non-chordates.
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 gillslits.	(c) Absesnce of gilislits.
(d) Presence of ventral heart.	(d) Presence of dorsal heart.

Question 29.
Draw a labeled diagram of the complete digestive system of the frog showing internal organs.
Answer:

Question 30.
List the functions of the plasma membrane.
Answer:
The functions of the plasma membrane are:

1. It maintains the size and shape of the cell.
2. Osmosis: Osmosis is the process by which water molecules pass through a semi-permeable membrane (here it ¡s the plasma membrane) from the region of its higher concentration- to one of lower concentration.
3. Active transport: It is energy (from ATP) dependent transport of molecules or ions across a semi-permeable membrane against the concentration (electrochemical gradient).
4. It can also be called “metabolically linked transport”.
   e.g: Sodium – Potassium Pump: (Revolving door model) :
5. The moving machinery of Na & Kt through active transport ¡s called ‘Sodium – Potassium pump’.

Question 31.
Explain how the p^H and concentration of substrate affect enzyme activity with graphical representation.
Answer:
Effect of P^H on enzyme activity: Enzymes generally function in a narrow range of P^H. Each enzyme shows its highest activity at a particular P^H called optimum P^H. Activity decreases both below and above optimum value.

Effect of concentration of substrate on enzyme activity with the increase in substrate concentration, the velocity of enzymatic reaction raises at first. The reaction ultimately reaches a maximum velocity (Vmax) which is not exceeded by further raise in the concentrate of the substrate.

Question 32.
Explain the physical properties of water that govern the transpiration-driven ascent of sap. Explain how these properties help in the ascent of sap or transpiration pull and how a “pull” is achieved.
Answer:
It is one of the most successful physical theories of the ascent of sap. Dixon and Jolly proposed this theory. According to this theory, the ascent of sap is due to three factors namely,
(a) Cohesive force of water
(b) Adhesion of water molecules to the walls of the xylem vessel
(c) Transpiration pull

A strong intermolecular force of attraction exists between water molecules. Thus water molecules are bound to each other forming a continuous column of water.

There is a force of attraction between water molecules on the inner walls of xylem elements.

Therefore water molecules are attached to the wall of the xylem. It is called adhesion.

Due to the adhesive and cohesive properties of water, a continuous column of water is preserved in the xylem.

Transpiration at the leaf surface forces the mesophyll cells of the leaves to draw water from the neighboring xylem elements. This creates a suction force known as Transpiration pull.

This suction force pulls the water column in the xylem upwards. This ascent of sap is due to the combined effect of adhesive, and cohesive properties of water and transpiration pull.

Merits:

1. The ascent of sap is directly proportional to the rate of transpiration.
2. It is a physical process and does not need energy.
3. Strong cohesive, and adhesive forces are sufficient to prevent the rupture of the water
   columns in xylem vessels.

Demerits:

1. The presence of an air bubble breaks the continuity of the water column.
2. Ascent of sap continues even in the absence of transpiration, as at night times.
3. The strength of the water column is thoughtful against two opposing forces such as gravitation force and transpiration tension.

Section-II

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

Question 33.
What are macronutrients? Describe the roles played by calcium and magnesium in plants.
Answer:
Macronutrients: Macronutrients are those elements, which are generally present in large amounts in the plant tissues, i.e., in excess of 10 m. mole kg-I of dry matter.
e.g., Carbon, hydrogen, oxygen, nitrogen, phosphorus, sulfur, potassium, and magnesium.

Calcium:
Calcium is are obtained as calcium ions (Cat) from the soil.

Functions:

1. It is necessary for the selective permeability of cell membranes.
2. It occurs as calcium pectate in the middle lamella of the cell wall hence is necessary for cell enlargement.
3. It is used in the formation of mitotic spindles during cell division, hence is required in meristematic, and differentiating tissues (root apex, and shoot apex).
4. it activates certain enzymes in the metabolism.

Deficiency symptoms:
(i) Stunted growth.
(ii) Necrosis of meristematic regions.

Question 34.
Explain the events of the C₄ pathway. Mention any two special features of C₄ plants.
Answer:

1. Pathway of CO₂ fixation in which the first stable compound formed is a 4-carbon compound oxalic acetic acid in the presence of PEP carboxylase is called the C₄ pathway.
2. It was first observed by Hatch and Slack.
   e.g: Monocots like maize, sugarcane, ragi, and dicots like euphorbia.

Mechanism:

• In mesophyll cells, CO₂ is fixed by phosphoenolpyruvate (PEP) to form the first stable carbon compound called oxalic acetic acid, a four-carbon sugar in the presence of PEP carboxylase.
• Oxalic acetic acid converts into malic acid.
• Malic acid is transferred to bundle sheath cells, where it is converted into pyruvic acid and CO₂
• The released CO₂ is used in the Calvin cycle and gets reduced to carbohydrates by the rubisco.
• Pyruvic acid is transported back to mesophyll cells.

Question 35.
Write the schematic representation of the overall view of the Citric Acid cycle.
Answer:

Note: To account for two molecules of acetyl COA produced from 1 molecule of glucose the entire reaction has to be multiplied by two.

Biocncrgetics:
1. Number of ÄTPs produced = 2
2. Number of NADH2 produced = 6 (6 × 3 = 18ATPs)
3. Number of FADH2 produced 2 (2 × 2 = 4 ATPs)
The total yield of ATPs = 24
Note:- Efficiency of Krebs cycle along with its predatory reaction is 30 ATPs.

Question 36.
Describe the events of reflex action with a diagrammatic representation of the knee jerk reflex.
Answer:
The reflex action pathway comprises the following events.

The stimulus is received by a receptor (eg. muscle spindle). An afferent neuron receives signals from this sensory organ and transmits the impulse via a dorsal nerve root into the CNS (at the level of the spinal cord). The efferent neuron then carries the signals from CNS to the effectors.

The stimulus and response thus form a reflex arc.

Question 37.
What are hormones? Mention one function each for (i) ACTH (ii) Melatonin (iii) Parathyroid hormone (iv) Thymosin.
Answer:
Hormones are non-nutrient chemicals that act as intercellular messengers and are produced in trace amounts.
1. Adrenocorticotrophic hormone: ACTH stimulates the synthesis and secretion of steroid hormones called glucocorticoids from the adrenal cortex.

2. Melatonin: Melatonin plays a very important role in the regulation of the 24-hour (diurnal) rhythm of our body.
Melatonin helps in maintaining the normal rhythm of the sleep-wake cycle.

3. Parathyroid hormone: Parathyroid hormone increases the Ca²⁺ levels in the blood. Parathyroid hormone acts on bones and stimulates the process of bone resorption/ dissolution.demineralization.

Parathyroid hormone also stimulates the reabsorption of Ca²⁺ by the renal tubules and increases Ca²⁺ absorption from the digested food.

4. Thymosin: Thymosin plays a major role in the differentiation of T-lymphocytes which provide cell-mediated immunity.

Students can Download 1st PUC Biology Model Question Paper 7 with Answers, 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 Model Question Paper 7 with Answers

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 illustration do not attract any marks.

Part – A

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

Question 1.
Write the Botanical name of wheat?
Answer:
Triticum aestivum.

Question 2.
What is a Manual?
Answer:
Manual contains compiled or complete information about the particular area, key, descriptions of families, genus, and species.

Question 3.
What is a Holoenzyme?
Answer:
A complete enzyme consisting of Apoenzyme and prosthetic group.

Question 4.
Name the chemical present in viroid?
Answer:
RNA.

Question 5.
What are spicules?
Answer:
Spicules from the skeleton in sponges. They may be made up of calcium, silica, or spongin protein.

Question 6.
What are hydathodes?
Answer:
Hydathodes are the water stomata through which Guttation takes place.

Question 7.
What is bolting?
Answer:
Internode elongation and early flowering induced by gibberellins are called bolting.

Question 8.
What are flame ceils?
Answer:
Flame cells are the excretory organs of the phylum Platyhelminthes. They are known as solenocytes.

Question 9.
What is tidal volume?
Answer:
It is the volume of air inhaled or exhaled with each inhalation and exhalation without any extra effort. Its volume is 500 ml.

Question 10.
Name the Instrument used to measure B.P?
Answer:
Sphygmomanometer.

PART-B

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

Question 11.
Mention any two functions of the Golgi apparatus?
Answer:

1. It secretes various enzymes and hormones.
2. It forms secretory vesicles which store enzymes, proteins, carbohydrates, etc.
3. It takes part in the formation of the cell plates and cell walls along with ER at the time of telophase in plant cells.
4. It helps in the formation of primary Lysosomes.

Question 12.
Sketch and label the Areolar connective tissue?
Answer:

Question 13.
Write any four functions of parenchyma tissue.
Answer:
(1) Parenchyma is chiefly storage in function.
(2) It forms the basic packing tissue.
(3) Parenchyma when modified as aerenchyma, provides buoyancy to aquatic plants.
(4) Chlorenchyma performs the function of photosynthesis.

Question 14.
Draw a neatly labelled diagram of the walking leg of a cockroach?
Answer:

Question 15.
What is bilateral symmetry? Give an example?
Answer:
The body is divisible into two halves in only one lateral plane, eg: Platyhelminthes to chordates.

Question 16.
Classify phylum Coelenterata with an example for each class?
Answer:
The phylum Coelenterata is divided into 3 classes:-

1. Class: Hydrozoa: eg: Hydra.
2. Class: Scyphozoa: eg: Aurelia or Jelly Fish.
3. Class: Anthozoa: eg: Corals.

Question 17.
List any two physiological effects of Gibberellins on the plants.
Answer:

1. Gibberellins break genetic dwarfism in dwarf varieties of pea and beans.
2. Can induce internode elongation, and early flowering in rosette plants like cabbage and cauliflower. This is called bolting.
3. Can break seed dormancy, and bud dormancy.
4. Can induce parthenocarpic fruits.
5. Gibberellins can substitute cold treatment i.e., can replace vernalization.
6. Can be used in inducing flowers even in off-seasons in long-day plants.

Question 18.
Why earthworms are called “friends of farmers”?
Answer:
Earthworms are called friends of farmers because they make burrows in the soil and make it porous which helps in respiration and penetration of developing plant roots. The fertility of the soil is also increased by their vermicompost.

PART-C

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

Question 19.
Write the salient features of the class Reptilia?
Answer:
Reptilians were the dominant group of vertebrates during ‘the Mesozoic period which was known as the ‘golden age of reptiles’. They are the first group of vertebrates to have developed completely for land life (terrestrial life).
1. Reptiles are poikilothermic (cold-blooded) animals, which are completelÿ adapted for a terrestrial mode of life; but some have secondarily adapted for an aquatic mode of life.

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

3. The skin is dry and covered by hard, horny, epidermal scales or. scutes, which prötect the animal from desiccation and from predators. The epidemic scales from a sort of exoskeleton to dry skin. Skin ¡s devoid of skin glands.

4. The head region possesses a mouth guarded by jaws with teeth, a pair of compound eyes, a pair of external nares, and a pair of external ears (tympanic membrane). In snakes, the tympanic membrane is poorly developed.

5. Trunk region possesses two pairs of pentadactyl limbs, whose digits terminate with horny claws.

6. The endoskeleton is bony.

7. Digestive system is complete and it is associated with digestive glands (pancreas and liver).

8. Respiration is pulmonaiy. The respiratory system comprises external nostrils, internal flares, pharynx, and lungs. Gills are absent.

9. Circulatory system includes a 3- chambered heart (4 chambered heart in crocodiles), and blood vessels. ¡t is a closed-type circulatory system.

10. Urinogenital system comprises paired metanephric kidneys which excrete nitrogenous waste ¡n the form of uric acid.

11. Sexes are separate. Gonads are paired, Males have either single or double copulatory organs. Fertilization is internal. Most of the forms are oviparous, while a few are ovoviviparous.

12. Fertilized egg is Cleidoic and is laid on land.

13. Development is direct.

14. Nervous system comprises a brain, 12 pairs of cranial nerves, and a peripheral nervous system.
eg: Garden lizard (calories)
Wall lizard (gecko)
Flying lizard (Draco)
Chameleon
Snakes – Cobra, Kraft, Viper, etc.,

Question 20.
What is a flower? Describe the parts of a typical angiosperm flower.
Answer:
Flower:
The flower is modified that night for sexual reproduction in phanerogams.
Velurnba Nucifera (Kelumbiwn, Creciosuni) commonly known as sacred lotus, is the National flower of India.
However, the flowers are present only in angiosperms.

• The flower of Rafflesia is the largest flower.
• The flower of Woiffia microscope is the smallest flower.

A typical flower consists of the following parts:
1. Bract: A foliaceous or scaly or variously modified structure in the axil of which the flower ¡s developed. Such a flower with a bract is called a bracteate flower. A flower without a bract is known as an ebracteate flower.

2. Pedicel: The stalk of the flower. A flower with a pedicel is called pedicellate, while one lacking the pedicel is called pedicellate.

Lotus is a sessile flower.

3. Thalamus: The condensed floral axis on which the floral whorls are developed is called thalamus. It may be conical, disc-shaped, or cup-shaped. In certain plants, the floral axis may be elongated and differentiated into nodes and internodes.

4. Floral whorls: A typical flower possesses the following four whorls:
(a) Calyx: Outermost whorl composed of sepals.
(b) Corolla: The second whorl composed of petals.
(c) Androecium: The third whorl composed of stamens.
(d) Gynoecium: The innermost and central whorl composed of one or more carpels.

Of the four whorls, calyx and corolla are called accessory whorls, while androecium and gynoecium are called essential whorls.

Question 21.
What is heterospory? Briefly comment on its significance? Give two examples.
Answer:

1. Liverwort: Capsule part of sporophyte at the time of spore formation.
2. Moss: Capsule part of sporophyte active time of spore formation.
3. Fern: Spoatigia at the time of spore formation.
4. Gymnosperm: Microsporangium at the time of microspore or pollen formation and ovule or megasporangium at the time of megaspore formation.
5. Angiosperm: Microsporangium and ovule or megasporangium at the time of microspore (pollen grain) and megaspore formation.

Question 22.
(a) Mention the functions of the following.
(i) Trachea
(ii) Flame cells
(iii) Suckers
(iv) Contractile vacuole
Answer:
(i) Respiration.
(ii) Excretion.
(iii) for mechanical support in Hirudinea.
(iv) Osmoregulation and excretion.

(b) Name the phyla in which you come across the following:
(i) Pseudocoel
(ii) Jointed legs
(iii) Flame cells
(iv) Gastro vascular cavity
Answer:
(i) Aschelminthes
(ii) Arthropoda
(iii) Platyhelminthes
(iv) Coelenterata

Question 23.
Give an account of the distinguishing characters of fungi?
Answer:
Fungi are eukaryotic, achlorophyllous organisms that are but, generally multicellular a few are unicellular (yeast).

1. The fungal body consists of long, slender, thread-like structures called hyphae, which form a network called mycelium.
2. The hyphae have a cell wall made up of chitin.
3. They may be aseptate and multi-nucleate ¡e., coenocytic or septate.
4. All fungi are heterotrophs -they are saprotrophs or parasites or live as symbionts in the roots of higher plants (mycorrhizae) and in lichens (mycobiont).
5. . They reproduce vegetatively by fragmentation, fission, or budding.
6. Asexual reproduction is by the formation of zoospores as in Saprolegnia and Pythium by Aplanospores /sporangiospores formed endogenously (within the sporangium) as in Rhizopus and Mucor or by conidia, which are produced exogenously on the swollen tip of the conidiophore as in Penicillium and Aspergillus.

Question 24.
Write a neatly labeled diagram of V.S. of the human heart.
Answer:

Question 25.
Name the organs of respiration in the following organisms:
(a) cockroach
(b) birds
(c) frog
Answer:
(a) Respiratory trachea
(b) lungs
(c) skin, lungs, and buccopharyngeal lining.

Question 26.
Draw a neat labeled diagram of V.S. of maize seed.
Answer:

PART-D (Section I)

Answer any FOUR of the following questions in 200-250 words each wherever applicable: (4 x 5 = 20)

Question 27.
Describe the structure of the nucleus with a neat labeled 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 in a cell, but some -cells also contain two or more nuclei.

Question 28.
Briefly explain the stages in Prophase I of meiosis?
Answer:
1. Prophase I: It lasts for quite a long period and is studied under five substages.

1. Leptotene
2. Zygotene
3. Pachytene
4. Diplotene
5. Diakinesis.

1. Leptotene: (Gr leptons slender; one – thread)
The chromosomes are thin, long, and uncoiled. Each is a double chromosome consisting of two chromatids. These chromatids are held firmly.
Each chromosome appears as a string of beads, the beads are the chromomeres.

2. Zygotene: (Gr: Zygon-male; tene- thread)
During this stage, the pairing of homologous chromosomes (half of the maternal and half of them paternal). This pairing is called synapsis. The pairs so formed are called bivalents.

Each bivalent consists of four chromatids and is therefore called a tetrad.
The two chromatids of the same chromošomes are called sister chromatids and the belonging to two different chromosomes of a homologous pair are termed non-sister chromatids.

3. Pachytene: (Gr Pachus – thick tene – thread)
Crossing over takes place by breakage and reunion of chromatid segments. after crossing over, the two chromatids of a chromosome become dissimilar.
The points of interchange are X-shaped and are called chiasmata. (sing. chiasma)

4. Diplotene: (Gr: Diplo – double; tene – thread)

• Repulsion between homologous chromosomes begins.
• Each tetrad now appears in different shapes j.c, X-shaped, ‘8’ shaped, or O’-shaped.
• Repulsion results in the criminalization of chiasmata (sliding of chiasmata towards the ends of chromosomes.

5. Diakinesis: (Gr: dia-cross; kinesis- movement).

• The nucleolus breaks down and disappears.
• Spindle fibers appêar.
• The nuclear membrane breaks down.
• Chromosomes are released into the cytoplasm.

Question 29.
List the hormones produced from the adenohypophysis and mention any one of their function?
Answer:
It accounts for nearly 75% of the total weight of the gland. It is derived from the buccal cavity in the form of a projection called Rathke’s pouch.

Adenohypophysis is further divided into three regions: Pars distaìis, Pars intermedia, and Pars tubular. Pars intermedia degenerates during fetal development and occurs only as a small strip in adults.

Pars tubular has no functional significance. Adenohypophysis cells secrete seven major hormones.
1. Somatotrophic hormone or Human growth hormone (STH or HGH): Promotes the growth of muscles. It stimulates the uptake of amino acids by tissues and their synthesis into proteins.

• Hypo secretion in childhood causes Pituitary dwarfism. Such an individual will be abnormally dwarf and is called a midget.
• Hypersecretion in childhood causes Piluirary gigantism. Such an individual will be abnormally tall.
• Hypersecretion in adolescence causes acromegaly which is characterized by the formation of disproportionately large hands, feet, cheekbones, jaws, etc.,

2. Thyroid-stimulating Hormone (TSH): Controls secretion of thyroid hormones by the thyroid gland and also regulates iodine intake by the thyroid gland.

• Hypo secretion leads to goiter, cretinism, and myxoedema.
• Hypo secretion leads to Hyperthyroidism.

3. Adrenocorticotrophic hormone (ACTH): It controls the1 secretion of hormones (Cortisone) by the adrenal cortex.

• Hypersecretion of ACTH causes Cushing’s syndrome.
• Hypo secretion causes Addison’s disease.

4. Follicle Stimulating Hormone (FSH):. In females the induces growth and maturation of Graffian follicle and stimulates follicular secretion of estrogen. In males, it stimulates the testis to produce sperm.

5. Luteinizing Hormone or Interstitial Cell Stimulating hormone (LH or ICSH): In females, it stimulates ovulation and the formation of the corpus luteum. In males, it stimulates interstitial cells in the testis to secrete testosterone.

6. Prolactin or Lactogenic or Luteotropic Hormone (LTH): In females, it causes growth and development of breasts during pregnancy. It stimulates milk production and secretion after childbirth. it inhibits ovulation during pregnancy and breastfeeding.

7. Melanocyte Stimulating Hormone (MSH): Increases skin pigmentation by stimulating dispersion of melanin in amphibians but its exact role in humans is unknown.

Question 30.
Explain the Mass flow hypothesis of organic translocation?
Answer:
According to Munch, food materials are translocated from the source (leaves) to sink (roots) en mass, 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 ¡n 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 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 31.
Write short notes on.
(a) Hypertension
Answer:
Whenever the Blood Pressure values are less than the normal values then the condition is termed as hypotension and Blood Pressure is called Low Blood Pressure.

Low blood Pressure ⇒ \(\frac{60-80 \mathrm{~mm} \text { of } \mathrm{Hg}}{55-45 \mathrm{~mm} \text { of } \mathrm{Hg}}\)

Causes :

1. Endocrine disorders.
2. Steroids deficiency.
3. Severe loss of blood due to accidents and severe burns.
4. Excessive loss of body fluids due to vomiting or diarrhea.

Effects:
Tiredness, giddiness, feeling cold, and low metabolic activities.

(b) Renal Calculi
Answer:
The accumulation of salts and glycoproteins in the urinary tract (Kidney tubules) is known is Kidney stones. Kidney stones contain calcium oxalate salts.

Causes:

1. High level of calcium in the blood.
2. Infection of the urinary tract.
3. Tumors in the kidney.
4. Less intake of water.
5. Intake of more milk (when milk proteins are not adequately utilized by body cells and when it enters to Kidney tubules).
6. Consumption of food increases the alkalinity of the urine.

Effects :

1. Severe pain in the abdomen, pelvis, and legs.
2. Pain during urination.
3. Frequent urination with blood.
4. Nausea, and vomiting
5. Severe pain in the Kidney region.

Treatment :

1. It can be treated by passing LASER beam shock waves to pulverize the stones. This technique is called extra Corporeal Shock Wave Lithotripsy (ESWL).
2. Endoscopic removal of stones.
3. Surgical removal of stones.

Question 32.
Explain the process of protein digestion in the human digestive system.
Answer:
Digestion in Stomach:
Gastric juice: Pepsin, and Renin are the proteases present in gastric juice
1. Pepsin: It is an enzyme produced in an inactive form called pepsinogen. Pepsinogen is activated by HCI. Pepsin is an endopeptidase and it breaks inner peptide bonds of the protein molecule to form proteoses, peptones, and polypeptides.

2. Renin: Renin is also secreted in an inactive form called prorenin. It ¡s activated by HCI. Renin acts on the milk protein called casein and converts ¡tinto paracasein. Paracasein in turn combines with calcium salts and forms calcium paracaseinate (curdy precipitate).

This process of conversion into calcium Paracaseinate is called the curdling of milk.

Pancreatic juice: Trypsin, Chymotrypsin, and Carboxypeptidase are the proteases secreted by the pancreas.
1. Trypsin: It is secreted as inactive trypsinogen. Later gets activated by enterokinase of the intestinal juice. Trypsin is an endopeptidase. it acts on proteins and converts them into Proteoses, Peptones, and Polypeptides.

2. Chymotrypsin: It is secreted as an inactive chymotrypsin and later gets activated by trypsin. Chymotrypsin is also an endopeptidase and converts Protein into Polypeptides.

3. Carboxypeptidase: It converts polypeptides into Tripeptides, Dipeptides, and Amino acids.

Intestinal juice: The four proteases forming the intestinal juice are Aminopeptidase.
Dipeptidase, Tripeptidase, and Erepsin.
1. Aminopeptidase: It is an exopeptidase that acts on the terminal ends of the polypeptides and converts them into Tripeptides, Dipeptides, and Amino acids.

2. Tripeptidase: It acts on the Tripeptide units and converts them into amino acids.

3. Dipeptidase: It acts on Dipeptide units and converts them into Amino acids.

4. Erepsin: It s a powerful protease of the intestine. It converts polypeptides and other short-chain Amino acids into free Amino acids.

Section II

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

Question 33.
Explain the structure of the Dicot leaf with a labeled diagram.
Answer:

Helianthus lear

e.g. Helianthus Annus (sunflower).
A thin transverse section of a dorsiventral leaf shows the following features.

1. Epidermis
2. Mesophyll tissue
3. Vascular bundles.

1. Epidermis:
The epidermis covers the upper and lower surface of a leaf and is known respectively as.
(a) Upper epidermis: It consists of a single layer of parenchyma cells arranged without intercellular space. The outer walls of epidermal cells bear a thick waxy layer called the cuticle. Multicellular epidermal hairs are present. Stomata are usually absent on the upper epidermis.

(b) Lower epidemic: It is almost similar to the upper epidermis but numerous stoma are found to be distributed throughout the lower surface ( Hypostomatic condition). These are pores concerned with the gaseous exchange. Each stomatal pore is surrounded by two specialized cells called guard cells containing chloroplasts. The other epidermal cells do not possess chloroplasts. The epidermis protects internal tissues.

Question 34.
Draw a neatly labeled diagram of the digestive system of cockroaches.
Answer:

Cockroach

Question 35.
Explain the structure of a multipolar myelinated neuron.
Answer:

Question 36.
Explain the histology of the blood.
Answer:
Blood: It is a fluid tissue consisting of two components.
(a) Blood: It is a fluid tissue consisting of,

1. Plasma. ,
2. Corpuscles or cells.

1. Plasma: It is a straw yellow (almost colorless) homogenous matrix containing a large amount of water (about 90%). A number of gases (O₂, CO₂) and substances are dissolved in it. Plasma also carries metabolic wastes, hormones, antibodies, etc. Plasma is the matrix and blood corpuscles are suspended in it.

2. Blood corpuscles: They are of 3 types. namely

1. Erythrocytes or RBCs (Red Blood corpuscles).
2. leucocytes or WBCs (White Blood corpuscles).
3. Platelets.

Question 37.
Write the schematic representation of Kreb’s cycle.
Answer:

Steps involved in Krebs Cycle
Note:- To account for two molecules of acetyl COA produced from ¡ molecule of glucose the entire reaction has to be multiplied by two.

Bioenergetics:

1. Number of ÄTPs produced =2
2. Nuriber of NADH₂ produced 6 (6 × 3 = 18 ATPs)
3. Number of FADH₂ produced 2 (2 × 2 = 4 ATPs)

The total yield of ATPs = 24
Note:- Efficiency of Krebs cycle along with its predatory reaction.is 30 ATPs.