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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.

KSEEB Solutions 1st PUC Biology Model Question Paper 7 with Answers

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.

KSEEB Solutions 1st PUC Biology Model Question Paper 7 with Answers

Question 12.
Sketch and label the Areolar connective tissue?
Answer:
1st PUC Biology Model Question Paper 7 with Answers 1

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:
1st PUC Biology Model Question Paper 7 with Answers 2

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.

KSEEB Solutions 1st PUC Biology Model Question Paper 7 with Answers

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.
1st PUC Biology Model Question Paper 7 with Answers 3
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.

KSEEB Solutions 1st PUC Biology Model Question Paper 7 with Answers

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:
1st PUC Biology Model Question Paper 7 with Answers 4

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.

KSEEB Solutions 1st PUC Biology Model Question Paper 7 with Answers

Question 26.
Draw a neat labeled diagram of V.S. of maize seed.
Answer:
1st PUC Biology Model Question Paper 7 with Answers 5

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:
1st PUC Biology Model Question Paper 7 with Answers 6
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.

KSEEB Solutions 1st PUC Biology Model Question Paper 7 with Answers

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.
1st PUC Biology Model Question Paper 7 with Answers 7
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.

KSEEB Solutions 1st PUC Biology Model Question Paper 7 with Answers

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.
1st PUC Biology Model Question Paper 7 with Answers 8

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).
1st PUC Biology Model Question Paper 7 with Answers 9
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.
1st PUC Biology Model Question Paper 7 with Answers 10
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.
1st PUC Biology Model Question Paper 7 with Answers 11
3. Carboxypeptidase: It converts polypeptides into Tripeptides, Dipeptides, and Amino acids.
1st PUC Biology Model Question Paper 7 with Answers 12

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.
1st PUC Biology Model Question Paper 7 with Answers 13
2. Tripeptidase: It acts on the Tripeptide units and converts them into amino acids.
1st PUC Biology Model Question Paper 7 with Answers 14
3. Dipeptidase: It acts on Dipeptide units and converts them into Amino acids.
1st PUC Biology Model Question Paper 7 with Answers 15
4. Erepsin: It s a powerful protease of the intestine. It converts polypeptides and other short-chain Amino acids into free Amino acids.
1st PUC Biology Model Question Paper 7 with Answers 16

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:
1st PUC Biology Model Question Paper 7 with Answers 17
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.

KSEEB Solutions 1st PUC Biology Model Question Paper 7 with Answers

Question 34.
Draw a neatly labeled diagram of the digestive system of cockroaches.
Answer:
1st PUC Biology Model Question Paper 7 with Answers 18
Cockroach

Question 35.
Explain the structure of a multipolar myelinated neuron.
Answer:
1st PUC Biology Model Question Paper 7 with Answers 19

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.
1st PUC Biology Model Question Paper 7 with Answers 20
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.

KSEEB Solutions 1st PUC Biology Model Question Paper 7 with Answers

Question 37.
Write the schematic representation of Kreb’s cycle.
Answer:
1st PUC Biology Model Question Paper 7 with Answers 21
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.