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Karnataka 1st PUC Biology Previous Year Question Paper March 2014 (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 museum?
Museums are those places that have collections of preserved animals and plants for taxonomic studies.

Question 2.
Where do you find bulliform cells?
The upper epidermis of monocot leaf.

Question 3.
Mention the function of gizzard in cockroaches.
Gizzard helps in grinding the food particles and mixing them with digestive enzymes.

Question 4.
Who proposed the Omnis cellula – e cellula?

Question 5.
Expand PPLO.
Pleuropneumonia like organism.

KSEEB Solutions 1st PUC Biology Previous Year Question Paper March 2014 (North)

Question 6.
What is Ammonification?
The process by which organic nitrogenous compounds are decomposed to produce ammonia is known as ammonification.

Question 7.
What is plasmolysis?
Plasmolysis is a phenomenon in which the cell shrinks due to exosmosis when it is surrounded by a hypertonic solution.

Question 8.
Write the dental formula of man.
\(\frac{2123}{2123}\) × 2 in adults

Question 9.
What is emphysema?
Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased it occurs due to cigarette smoking.

Question 10.
State the function of fibrinogen in the blood.
Soluble fibrinogen results in the formation of insoluble fibrin which forms a mesh-like network in which the blood components get entangled and results in a blood clot.


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

Question 11.
Write the important rules of Binomial nomenclature Give one example.
(a) 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, Mangifera and Homo are generic names, while Indica and sapiens are the names of the species belonging to Mangifera 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.

(b) 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 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.

KSEEB Solutions 1st PUC Biology Previous Year Question Paper March 2014 (North)

Question 12.
How are viroids different from viruses.
Viroids are naked single-stranded infectious RNA molecules causing plant diseases.

Viruses are obligate parasites made up of a protein coat called capsid which encloses genetic material either DNA or RNA. They cause diseases in plants, animals, and bacteria.

Question 13.
Sketch and label cardiac muscle tissue.
1st PUC Biology Previous Year Question Paper March 2014 (North) 1

Question 14.
Write any two differences between mitosis and meiosis.

  1. Mitosis consists of single nuclear divisions.
    Meiosis consists of two nuclear divisions.
  2. In Mitosis, crossing over is absent.
    In Meiosis, crossing over is present.

Question 15.
What is hydroponics? Give one application of this technique.
The technique of growing plants in a nutrient solution is known as hydroponics. The application of this technique is that the plants can be grown in a soil-free, defined mineral solution.

Question 16.
Mention the role of hydrochloric acid indigestion.
(a) It activates inactive pepsinogen into active pepsin.
(b) It converts inactive prorenin into active renin.
(c) It provides an acidic medium (pH 1.5 – 2.5) for the action of both pepsin, and renin.
(d) HCl kills harmful bacteria and loosens fibrous food.
(e) It inactivates the ptyalin.

Question 17.
Hemoglobin is an oxygen carrier Explain.
Hemoglobin is called an oxygen carrier because about 97% of the oxygen is transported by RBCs. RBCs contain hemoglobin. Hemoglobin has an affinity towards oxygen. Hence hemoglobin bonds with oxygen molecules to form oxyhemoglobin.
1st PUC Biology Previous Year Question Paper March 2014 (North) 2
Oxyhaeinoglobin is a very unstable compound. Hence it dissociates into hemoglobin, and molecular oxygen, when it reaches tissues.

Question 18.
Name anyone gonadotropin hormones, giving a function of each.
(a) FSH (Follicle-stimulating hormone).

  1. In females, it promotes the growth and maturation of the Graafian folliçie.
  2. In females, it also stimulates the follicle cells of the ovary to secrete female sex hormone or estrogen.

(b) LH (Luteinizing hormone).

  1. In females stimulates final maturation and release of the ovum from the Graafian follicle.
  2. It in females helps in the conversion of the Graafian follicle into the corpus luteum. It stimulates the corpus luteum to secrete progesterone.
  3. In males, it stimulates the interstitial cells of the testis called Leydig cells to secrete the male sex hormones called testosterone.


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

Question 19.
Assign the following to their respective phyla.
(a) Sycon (b) Aurelia (c) Liver fluke
(a) Phylum Porifera
(b) Phylum Coelenterata
(c) Phylum Platyhelminthes.

KSEEB Solutions 1st PUC Biology Previous Year Question Paper March 2014 (North)

Question 20.
Mention the salient features of the family Solanaceae
Vegetative characters: Plants mostly herbs, shrubs, and rarely small trees.
Stem: Herbaceous rarely woody, aerial, erect, cylindrical, branched, solid or hollow, hairy or glabrous, underground stem in potato (Solanum tuberosum).
1st PUC Biology Previous Year Question Paper March 2014 (North) 3
Solanum nigrum (makoi) plant: (a) Flowering twig (b) Flower (c) L.S. of flower (d) Stamens (e) Carpel (f) Floral diagram

Leaves: alternate, simple, rarely pinnately compound, exstipulate, venation reticulate. Floral Characters:
Inflorescence: Solitary, axillary, or cymose as in Solanum.
Flower: bisexual, actinomorphic.
Calyx: sepals five, united, persistent, valvate aestivation.
Corolla: petals five, united: valvate aestivation.
Androecium: stamens five, epipetalous.
Gynoecium: carpellary, syncarpous, ovary superior, bilocular, placenta swollen with many ovules.
Fruits: berry or capsule.
Seeds: many, endosperms.
1st PUC Biology Previous Year Question Paper March 2014 (North) 4
Class: Monocotyledonae Order: Liliflorae Family: Liliaceae
Vegetative characters: Perennial herbs with underground bulbs/corms/ rhizomes. Leaves: mostly basal, alternate, linear, exstipulate with parallel venation.
Floral characters:
Inflorescence: solitary / cymose; often umbellate clusters.
Flower: bisexual; actinomorphic.
Perianth: tepal six (3 + 3), often united into a tube; valvate aestivation.
Androecium: stamen six (3 + 3).
Gynoecium: Tricarpellary, syncarpous, ovary superior, trilocular with many ovules; axile placentation.
Fruit: capsule, rarely a berry.
Seed: endosperms
1st PUC Biology Previous Year Question Paper March 2014 (North) 5

Question 21.
What is a vascular bundle? Explain the types of vascular bundles.
The Vascular Tissue System: Vascular bundles carried defined as a compact strand of xylem and phloem with or without cambium. Vascular tissues derived from the procambium, during the primary growth of the plant, are called primary vascular tissues. They include primary xylem and primary phloem. Vascular tissues derived from vascular cambium., during secondary growth of the plants, are called secondary vascular tissues. They include secondary xylem and secondary phloem.

  • Protoxylem: It is the first-formed xylem, in which the elements are much elongated and have narrow lumen.
  • Metaxylem: It is the later formed xylem, in which the elements are shorter and have wider lumen.
  • Exarch xylem: A condition of the xylem in which the protoxylem is facing away from the center is called Exarch xylem.
  • Endarch xylem: A condition of the xylem, in which the protoxylem is facing towards the center is called endarch xylem.

Types of Vascular bundles:
1. Radial vascular bundles: The vascular bundles in which xylem and phloem are present in separate bundles and lie on different radii are called radial vascular bundles.

2. Conjoint vascular bundles: The vascular bundles in which xylem and phloem are present in the same bundle on the same radius are called conjoint vascular bundles.

Question 22.
Describe the following:
(a) Synapsis (b) bivalent (c) chiasmata
(a) The process of the pairing of homologous chromosomes is called Synapsis.
(b) A chromosome containing two chromatids is called a bivalent.
(c) The regions where segments of non-sister chromatids of the bivalents are exchanged are called chiasmata.

Question 23.
How are the minerals absorbed by the plants?

  1. The process of absorption occurs in two distinct phases.
  2. In the first phase, there is a rapid uptake of irons into the ‘free space’ or outer space of the cell called the apoplast. it is a passive process and takes place through non-channels.
  3. In the second phase, the ions are taken slowly into the inner space – the symplast of the cells. This is an active process and requires the expenditure of metabolic energy.

KSEEB Solutions 1st PUC Biology Previous Year Question Paper March 2014 (North)

Question 24.
Write a note on ultrafiltration.
Blood brought by the efferent arteriole is filtered in the glomerular capillaries because of the net effective pressure of 25mm of Hg. Net effective pressure is due to the differences in glomerular hydrostatic pressure, blood colloidal osmotic pressure, and capsular hydrostatic pressure.

This can be represented as,
Ghp – (BCOP + CHP) = NEP
i.e., 75mm Hg – (30mm Hg + 20mm Hg) = 25mm Hg.

During filtration, the blood cells and large organic molecules are retained in the capillaries. The fluid containing dissolved substances is passed into the capsular lumen. This is called glomerular filtrate. It contains a number of useful substances like glucose. amino acids, minerals, etc., as well as waste products.

Question 25.
(a) What is a pacemaker?
SAN is also known as the pacemaker of the heart because it initiates the electrical impulses for the heartbeat.

(b) Define a cardiac cycle and the cardiac output.
Cardiac cycle (Heart Beat):
The sequential cyclic events of systole (contraction state) and diastole (relaxation state) together are known as one cardiac cycle or heartbeat. In a healthy person, the heart completes 72bcats/min. On average, each complete heartbeat requires 0.8 sec in which systole of ventricles requires 0.3 sec, systole of auricles require 0.1 sec, diastole of the heart requires 0.4 sec.

Cardiac Output:
The amount of blood pumped by the right and left Ventricles per minute is known as cardiac output. It is determined by multiplying the stroke volume with the number of heartbeats/mm.

Cardiac Output = stroke volume × number of heart beats / minute
= 70 ml × 72 beats / min
= 5040 ml of blood.

Question 26.
Describe the different types of movements exhibited by cells of the human body.
Types of Movement:
Cells of the human body show three basic types of movement:
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.

3. Muscular Movement:
The contractile property of muscles is responsible for muscular movement.

  • Movements of jaws, limbs, eyelids, etc. are muscular movements involving striated muscles.
  • Movement of food ¡n the alimentary canal and movement of urine in ureters are movements involving smooth muscles.

4. Flagella movement: It is caused by a long whip-like process called flagella. It is seen in spermatozoa, euglena, and sponges.

Part-D (Section – I).

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

Question 27.
Describe the important characteristics of gymnosperms.
Gymnosperms represent primitive phanerogams (flowering plants) or spermatophytes. They are generally regarded as ‘naked seeded plants’ due to the presence of exposed seeds without being protected by the ovary wall as there is no well-defined carpel.

Gymnosperms are distributed throughout the world, represented in tropical and temperate forests, and dominating mountainous ranges. The group comprises nearly seven hundred species. They exhibit diverse habits showing the presence of woody climbers (lianas), shrubs, and trees.

General features:
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 beating microspores and megaspores within microsporangia and megasporangia respectively. These structures occur on leaf-like ¡iicrosporophylls and megasporophylls. These are further organized into fertile structures called strobili or cones.

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

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

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

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

Question 28.
Draw a labeled diagram of the Alimentary canal system of cockroaches.
1st PUC Biology Previous Year Question Paper March 2014 (North) 6

Question 29.
Write any five differences between cartilage and bony fishes.

  1. Chondrichthyes (cartilaginous fishes).
  2. Osteichthyes (bony fishes).
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 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: Sciiodon (Shark)
Narcine (Electric ray)
Trygon (Stingray).
8. Claspers are absent.
eg: Anguilla (Freshwater Eel)
Carias (Catfish)
Hippocampus (Sea horse).

Question 30.
A neat labeled diagram explains the structure of mitochondria.
1st PUC Biology Previous Year Question Paper March 2014 (North) 7

KSEEB Solutions 1st PUC Biology Previous Year Question Paper March 2014 (North)

Question 31.
Explain the mode of enzyme action.
Enzymes are specific in their substrate molecules.
When enzyme (E) and its substrate (S) come together, they make an intermediary complex called the E-S complex, like a key coming with a lock and making a complex. This is the primary step for enzyme action.

So the enzymes are highly specific for the type of reaction they catalyze.
Over a fraction of a time, the enzyme cleaves the substrate into its product.

Since the product is different from the substrate, it cannot remain in contact with the enzyme, the enzyme becomes free.

The free enzyme can take up another substrate molecule.
E + S = ES → E + P,
1st PUC Biology Previous Year Question Paper March 2014 (North) 8

Question 32.
Describe the transpiration pull model of water transport in plants.
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 adhesin.

Due to the adhesive and cohesive properties of water, a continuous column of water is present 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.


  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.


  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

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

Question 33.
Explain the cyclic photophosphorylation with schematic representation.
1st PUC Biology Previous Year Question Paper March 2014 (North) 9
Cyclic Photophosphorylation :
It is a cyclic path of electrons expelled from chlorophyll through a series of substrates that are arranged in a suitable oxidation-reduction potential. The energy in the electrons is used for the phosphorylation of ADP to ATP.

In PS I the absorbed photons of light excite chlorophylL a 700 to eject energized electrons on makes it positively charged and unstable. Electrons pass through the sequence → FRS → FD → Cyt b6 → Cytf → PC and generate ATP at two places. Finally, an electron from PC returns to chi-a restoring its stability.

Question 34.
Draw a labeled diagram of a multipolar neuron.
1st PUC Biology Previous Year Question Paper March 2014 (North) 10

Question 35.
Give the schematic representation of Kreb’s cycle?
1st PUC Biology Previous Year Question Paper March 2014 (North) 11
Note: To account for two molecules of acetyl CoA produced from ¡ molecule of glucose the entire reaction has to be multiplied by two.


  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.

KSEEB Solutions 1st PUC Biology Previous Year Question Paper March 2014 (North)

Question 36.
(a) What is Photoperiodism? What is its significance?
(b) Explain the sigmoid growth curve.
Plants, in order to flower, required a particular day length or light period called photoperiod, and the response of the plants to photoperiod ¡n 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. eg: Wheat, oats, etc.
  2. Short Day Plants: These flower ¡n 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 duration of light for their flowering. e.g: Tomato, cucumber, cotton, etc.

Growth curve:
The curve obtained when the growth in size or weight or on entire plant or its individual parts is plotted against time is called growth curve ft is always ‘S’ or sigmoid in shape shows free distinct phases namely.

  1. Lag phase.
  2. Log phase.
  3. Steady phase.

1. Lag phase: This is the initial stage where the growth rate is very slow But it gradually increases with time.

2. lag phase: In this phase, the rate of growth is very rapid. The plant shows a high rate of growth and reaches maximum height. Hence this phase is called the exponential phase or grand of growth.

3. Steady phase or stationary phase: This is the last phase in which plants show no growth. Therefore the curve becomes almost orbital.
1st PUC Biology Previous Year Question Paper March 2014 (North) 12

Question 37.
Diagrammatically indicates the location of the various endocrine glands in our body.
1st PUC Biology Previous Year Question Paper March 2014 (North) 13