ISC Biology 2012 Class-12 Previous Year Question Papers

ISC Biology 2012 Class-12 Previous Year Question Paper Solved for practice. Step by step Solutions with Part-I, and II (Section-A,B). By the practice of ISC Biology 2012 Class-12 Solved Previous Year Question Paper you can get the idea of solving.

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ISC Biology 2012 Class-12 Previous Year Question Papers Solved

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Sections-A, Part- II,

Sections-B, Part- II,

Maximum Marks: 70

Time allowed: Three hours

(Candidates are allowed additional 15 minutes for only reading the paper. They must not start writing during this time)

  • This paper comprises TWO PARTS – Part I and Part II.
  • Part I contains one question of 20 marks having four sub parts.
  • Part II consists of Sections A, and B.
  • Section A contains seven questions of two marks each Section B contains seven questions of three marks each, and
  • Section C contains three questions of five marks each.
  • Internal choices have been provided in two questions in Section A, two questions in Section B and in all three questions of Section C.
  • The intended marks for questions or parts of questions are given in brackets [ ].

(Attempt All Questions)

ISC Biology 2012 Class-12 Previous Year Question Papers Solved

Question 1.
(a) Give one difference between each of the following :
(i) Ventricular systole and Ventricular diastole.
(ii) Sertoli cells and Spermatids.
(iii) Dwarfism and Cretinism.
(iv) Antibodies and Interferons.
(v) Glucocorticoids and Mineralocorticoids.

(b) Give reasons for the following :
(i) Nerve impulse travels in one direction.
(ii) Jamming of wooden doors and windows takes place during rainy season.
(iii) A cut plant wilts fast even if its cut end is dipped in water.
(iv) Urine excreted during summer months is hypertonic.
(v) A person has difficulty in focusing on nearer objects, as the age increases.

(c) Give a scientific term for each of the following:
(i) A single isolated contraction of muscle fibre.
(ii) Inhibition of lateral bud growth by terminal bud.
(iii) Specialised structure through which guttation occurs.
(iv) Development of embryo from the egg without the process of fertilization.
(v) Process of splitting of water molecules during photosynthesis.
(vi) Passing out of urine.

(d) Mention the most significant function of each of the following :
(i) Tapetum cells
(ii) Serotonin
(iii) Lenticels
(iv) Cerebrospinal fluid
(v) Islets of Langerhans
(vi) Bundle sheath

(e) State the most significant contribution of the following scientists :
(i) Hans Berger
(ii) Dixon and Jolly
(iii) J.B. Lamarck
(iv) William Harvey

(e) State the most significant contribution of the following scientists: [2]

(f) Expand the following: [2]
(1) OP
(ii) RuBP
(iii) IBA
(iv) PEP


Ventricular systole Ventricular diastole
Ventricles contract and the blood flows from ventricles to arteries. Ventricles are relaxed and the blood flows from auricles to the ventricles.


Sertoli cell Spermatids
Present in the germinal epithelium of the seminiferous tubules which provides nour­ishment to the developing sperms. Few cells of the germinal epithelium of the seminiferous tubules undergo meiosis to produce spermatids.


Dwarfism Cretinism
It is caused due to hyposecretion of growth hormone during childhood. Caused due to malfunctioning of the thyroid gland in infant.


Antibodies Interferons
Acts slowly and long-lasting. The action is quick but temporary.


Glucocorticoids Mineralocorticoids
Influences the metabolism of carbohy­drates, proteins and fats. Affects the transport of electrolyte and absorption of sodium ions by the various parts of the uriniferous tubules.


(i) Nerve impulse travels in one direction Since the neurotransmitter is present only in the axon terminal so nerve impulse always travels from the axon terminal of one neurone to the dendrite or cell body of the next neurone.

(ii) Jamming of wooden doors and windows takes place during rainy season The cellulose of the wood absorbs water and swells up. This causes increase in size of doors and windows resulting in jam.

(iii) A cut plant wilts fast even if the cut end is dipped in water As the rate of transpiration is higher than that of the absorption by the cut end, the wilting is fast.

(iv) Urine excreted during summer months is hypertonic During the summer, excessive water is lost as sweating. Urine becomes hypertonic to reduce the loss of water in urine and maintains the osmotic concentration of the blood constant.

(v) A person has difficulty in focusing on nearer objects as the age increases With the advancement of age, the elasticity of the lens decreases. It causes the person difficult to focus the nearer object.

(i) Single muscle twitch
(ii) Apical dominance
(iii) Hydathodes
(iv) Parthenocarpy
(v) Photolysis
(vi) Micturition

(i) Tapetum cells: Rich in food materials, surrounds the microspore mother cell and supplies food to the developing spores.
(ii) Serotonin : Acts as a neurotransmitter.
(iii) Lenticels : Allows exchange of gases between the atmosphere and the interior of the cells.
(iv) Cerebrospinal fluid : Forms the protective cushion over the brain and spinal cord against shock and mechanical injury.
(v) Islets of Langerhans: Secrete two hormones –

(a) insulin and
(b) glucagon
(vi) Bundle sheath : Surrounds the vascular bundles in monocot leaf, provides mechanical support.

(i) hans Berger : First to record EEG
(ii) Dixon and Jolly: Theory of ascent of sap
(iii) J. B Lamarck: Theory of inhentance of acquired characters.
(iv) William Harvey : Discovered closed circulatory system

(i) OP – Osmotic pressure/potentiaL
(ii) RuBP — Ribulose biphosphate
(iii) IBA – Indole – 3 – butyric acid
(iv) PEP — Phosphoenol pyruvic acid

Section – A Part-II 

(Attempt any three questions)

ISC Biology 2012 Class-12 Previous Year Question Papers Solved

Question 2.
(a) Give any four anatomical differences between monocotyledonous and dicotyledonous leaf. [4]
(b) Explain the phases of growth in the meristem of plants. [3]
(c) Draw a neat labelled diagram of a matured anatropous ovule before fertilization. [3]
(a) MonocoG’ledonous/Isobilateral Leaf

  1. Both the surfaces are alike.
  2. Stomata are equally distributed on both sides. Mesophyll undifferentiated.
  3. Vascular bundles are partially or completely surrounded by a sclerenchymatous sheath.
  4. Vascular bundles have phloem on the upper side and xylem on the lower side.

Dicotyledonous/Dorsiven frai leaf

  1. Distinct upper and lower surface.
  2. Stomata are mostly present on the lower side. Mesophyll differentiated.
  3. Vascular bundles has sclerenchymatous patches on the upper side.
  4. Vascular bundles have downward phloem and upward xylem.

(b) Phases of growth in meristem of plant:
(i) Cell formation phase – During this phase meristematic cell divides to form new cells. The newly formed cells are thin wailed.
(ii) Cell enlargement phase – During this phase, the newly formed cells absorb water by osmosis resulting in the increase in turgidity and expansion and dialation of the elastic cell wall.
(iii) Cell differentiation phase – This occurs below the zone of elongation. The thin cell wall grows in thickness and the cells gradually undergoes structural and physiological changes.

This occurs below the zone of elongation. The thin cell wall grows in thickness and the cells gradually undergoes structural and physiological changes.

Question 3.
(a) Give an account of activity of cambium in the secondary growth of the stem. ‘
(b) Write three differences between C3 and C4 cycles.
(c) Mention two advantages each of the following :
(i) Hydroponics’
(ii) Turgidity to plants
(iii) Cross-pollination
(a) Activity of cambium in the secondary growth of the stem –
1. F ormation of cambium ring – The vascular bundles of dicot stem have strips of cambium in between xylem and phloem. During secondary growth, the cells of medullary rays in a line with this cambium develop meristematic activity and forms strips of cambium. The intra and interfascicular cambium unites to form cambium ring or phellogen.

2. Formation of secondary tissue – The cambium ring becomes active as a whole and starts cutting off new cells. The cut off cells of the outer side get differentiated into phloem and are called secondary phloem. The cut off cells of the innerside are modified into secondary xylem. The activity of the cambium ring is more on the innerside than on the outerside which results in the increase of the xylem more rapidly than the phloem.

3. Secondary medullary ray – Certain cells of the cambium form some narrow bands of living parenchyma cells-passing through secondary xylem and secondary phloem and are called secondary medullary rays. This provides radial conduction of food from the phloem and water and mineral salts from the xylem.

4. Annual Ring – Activity of cambium is not uniform in those plants which grows in the regions where favourable climate conditions regularly alternates with the unfavourable conditions. The cambium is more active in spring and less active in winter. The wood formed in the spring is known as spring wood and that formed in summer and winter is known as autumn wood. The two woods appear together in concentric rings in the trunk and is known as annual ring. The age of the plant is approximately calculated by counting the number of annual rings.

(b) C3 cycle

  1. RuBP is the CO2 acceptor.
  2. PGA is the first stable product.
  3. The process runs at a optimum temperature of 1O°C—25°C.

C4 cycle

  1. PEP is the first carbon dioxide acceptor.
  2. Oxaloacetic acid is the first stable product.
  3. The process runs at a optimum temperature of 30°C to 45°C.

(c) Two advantages of the following –
(i) Hydroponics :
(a) No soil required for growing the plants hence useful in infertile and dry soil.
(b) Plants are free from soil pathogens and weeds.

(ii) Ttirgidity to plants:
(a) Provides mechanical supports to non-woody parts of the plant.
(b) Regulates the opening and closing of the stomata.

(iii) Cross-pollination:
(a) New varieties with useful characters are produced.
(b) Results in healthy and stronger offsprings.

Question 4.
(a) Explain the movement of water cell to cell across the root from the soil to the xylem. [4]
(b) Draw a labelled diagram of T.S of hyaline cartilage. Write a brief note on its functions. [3]
(c) What is the full form of ADH? How does ADH control osmoregulation in human kidney? [3]
(a) Root hairs are in contact with soil particles. These provide a large surface area for water absorption. Water diffuses into the root hairs as a result of diffusion pressure deficit gradient. Water enters as long as the DPD of the cell sap is greater. The DPD falls following an increase in TP. The DPD of the adjoining cortical cells being higher and water moves from root hairs into cortical cells.

The DPD falls following an increase in TP. The DPD of the adjoining cortical cells being higher and water moves from root hairs into cortical cells.

The DPD of the next inner cells being higher, water moves into them. In this way, water moves from one cortical cell to the other along a DPD gradient till it reaches the passage cells in the endodermis. Through the radial and inner tangential walls of the passage cell water enters the pericycle and then to the protoxylem element.

(b) Hyaline cartilage – The matrix is homogeneous, translucent and fibreless and some what elastic. It forms the embryonic skeleton invertebrates and the skeleton of elasmobranch fishes.

Fig. Hyaline cartilage

(c) ADH- Antidiuretic hormone It is released from the posterior part of the pituitary gland. It acts on distal convoluted tubules and collecting ducts of nephron and increase their permeability for water. The secretion of ADH varies with the intake of water. If less water is taken more ADH is produced to reduce the urine volume and when more water is taken less ADH is produced so that only small amount of water is reabsorbed and majority goes out in the urine.

Question 5.
(a) Explain the process of oogenesis in humans. [4]
(b) State three differences between red muscle fibre and white muscle fibre. [3]
(c) Mention a cause and symptom of each of the following : [3]
(i) Emphysema
(ii) Renal calculi
(iii) Diarrhoea. [2]
(a) Oogenesis occurs in the germinal epithelium cells of the ovary resulting in the formation of a mature ovum.
The process is divided into –
(i) Multiplication phase – Oogonia multiply by mitotic division forming primary oocytes.
(ii) Growth phase – Long phase, extend over many years primary oocyte grows into large size.
(iii) Maturation phase – Primary oocyte undergoes two maturation divisions to form a single ovum and two polar bodies.

Primary oocyte undergoes two maturation divisions to form a single ovum and two polar bodies.

(b) Red muscle fibre:

  • Thin muscle fibres.
  • Contains pigment myoglobin hence dark red in colour.
  • Gets energy from aerobic respiration.

White Muscle fibre:

  • Thick muscle fibres.
  • Light in colour due to lack of myoglobin.
  • Gets energy from anaerobic respiration.

(c) Disease:



  • Air pollution and smoking
  • Precipitates (ppt.) of uric acid and accumulation of oxalate crystals
  • Flagellate protozoa Giardia intestinalis


  1. Wall of alveoli contracts and becomes thin so the capacity of gaseous exchange is lowered.
  2. Severe pam or blockade of the ureter
  3. Loose motion with pain in the stomach

Question 6.     (ISC Biology 2012 Class-12 Solved Previous Year Question)
(a) Describe the structure of an artery and a vein. Explain how their structure helps in their functioning. [4]
(b) Write three differences between short day and long-day plants. [3]
(c) Name the 2nd, 3rd and 8th cranial nerves in man and write a function of each. [3]
Answers :
(a) Structure of Artery:

  • Wall is thick and elastic.
  • Lumen is narrow.
  • Valves are absent.
  • Endothe liai cells oftunica are bore elongated. Tunica media is more inuscular and tunica externa is less developed.

Structure of vein:

  • Wall is thin and less elastic.
  • Lumen is wide.
  • Valves are present.
  • Endothelial cells of tunica arc lcss elongated. Tunica media is less muscular and tunica externa morc developed.

Function of artery: Arteries do not collapse due to the fast flow of blood under pressure as they have thick walls.
Function of vein : In veins the blood flows smoothly because of large lumen and the valves prevent backward flow of blood.

(b) Short day plants:

  • They flower when exposed to day lengths shorter than a certain critical miniminm
  • Do not flower if the dark period is interrupted by a flash of light.
  • They normally flower in the early spring or autumn.

Long day plants

  • They flower when exposed to day lengths longer than a certain critical minimwn.
  • Flowering is stinmiated if the dark period is interrupted by a flash of light.
  • They normally flower in the late spring or early summer

(c) 2nd — Optic — Vision
3rd — Oculomotor — Movement of eyeball, accommodation
8th — Auditory — Hearing and body balance

Section – B Part-II  

(Attempt any three questions)

ISC Biology 2012 Class-12 Previous Year Question Papers Solved

Question 7.
(a) Explain mass selection and pure line selection. How is pure line selection a better method for crop Improvement? (4]
(b) Write short notes on [3]
(i) Atavism
(ii) Protoplast fusion
(iii) Rh factor
(a) Mass selection – Most common and old method of crop selection. In this selection, large num-ber of similarly appearing plants are selected for the desired trait and their seeds are mixed together which is then sown to raise new crops.

Pure line selection : It is the progeny of a single homozygous self-pollinated plant. The single plant of the desired trait is selected out of the variable populate in the field. Seeds from the selected plants are sown in different rows to produce progeny by self-pollination.
Pure line selection is better method because the selected plants retain desirable characters for several generation.

(b) Write notes on:
(i) Atavism: Reappearence of certain ancestral characters which had either disappeared or were reduced.
(ii) Protoplast Fusion : It is also called somatic fusion. It is the genetic modification in plants by which protoplasts of two distinct species fuses to form somatic hybrid.

(iii) Rh Factor: The erythrocytes of most people have an agglutinogen named the Rh factor. It was first found in rhesus monkey. Such peoples are called Rh-positive. The minority with no such agglutinogen are known as Rh-negative. When the donor’s blood is incompatible with the recipients blood in Rh-factor, transfusion of such blood results in destruction of erythrocytes.


(i) Turner’s syndrome : Individual has 45 chromosomes (44+X) i.e. one chromosome less than normal and hence sterile female.
(ii) Klinefelter’s Syndrome : This disorder is due to XXY genotype. Such an individual has 47 chromosomes (2n +1) and the male is sterile with underdeveloped testes.
(iii) Down’s syndrome : This disorder is associated with an extra chromosome 21. Such indi-viduals have 47 chromosomes (45 + XX in female and 45 + XY in male)

Question 8.
(a) How does the human body protect itself from infections? [4]
(b) Write short notes on the following : [3]
(i) Biomedical Engineering
(ii) Stem cells
(iii) Cryopreservation
(c) Give an account of Darwin’s finches. [3]
(a) Human body has two lines of defence against pathogens :
1. Non-specific mechanism
2. Specific mechanism (Immune System)

1. Non-specific defence mechanism – It is of two types
(a) external defence or 1st line of defence and
(b) internal defence or 2nd line of defence
(a) Skin, mucus membrane lying the respiratory tract, throat, digestive tract and urinogenital system is the first line of defence. It secretes certain chemicals which inhibits the growth of pathogens.
(b) Body’s second line of defence is carried on by white blood corpuscles, macrophages, mast cells, inflammatory reactions and fever.

2. Specific mechanism – This mechanism is also called immune system which has two com-ponents 1. humoral immune system (Antibody-mediated immune system) and 2. cell-mediated immune system (defends the body against pathogens-protists and fungi)


(i) Biomedical Engineering – It is the branch of science which deals with the instruments
used in diagnosis and the treatment of human diseases. The instruments are classified into following three categories :
1. Diagnostic instrument
2. Imaging instrument
3. Therapeutic instrument.

(ii) Stem cell – These cells can divide and differentiate into diverse specialised cell type and self.
(iii) Cryopreservation – It is the storage of living organisms in ultra-low temperature such that it can be revived and restored into the same living state as before it was stored.

(c) Darwin’s finches : The islands situated near the mainlands usually possess flora and fauna related to those of mainland. However, they have a unique diversity of species despite being related. One such example is Galapagos islands (Spanish Galapago – Giant Tortoise). Darwin visited the island in 1835 and described them as living laboratory of evolution. The islands have 26 species of birds and 11 species of tortoises. Out of 26 species of birds, 23 are endemic or found only in the islands. Amongst them were species of sparrow-like small black birds called finches. They are called Galapagos finches or Darwin’s finches (Lack, 1947).

They resemble mainland finches in plumage, body plan and short tails but differ amongst themselves as well as from mainland finches in shape and size of beaks, food habits, colour of feathers and body size. Thirteen species of finches occur in Galapagos island and one species in nearby Cocos island. Six species are of ground finches (Geospiza species), six species of tree finches (Camarhyncus species) and two types of warbler finches (Certhidia and Pinarolaxis species). The ancestors of Galapagos finches must have come from mainland.

They were seed eating. However, environment and availability of food on the various islands were different from those of the mainland. As a result variations appeared in them to suit their habitats. They ultimately produced the different species of finches, some of which evolved insect eating patterns. Divergence of organisms of a common stock due to adaptive changes to suit new environmental conditions is called adaptive radiation.

Question 9.  (ISC Biology 2012 Class-12 Solved Previous Year Question)
(a) Explain the convergent and divergent evolution with suitable examples. [4]
(b) What is manure ? How does green manure differ from biofertilizers ? [3]
(c) What is IPM ? Give an example of bioinsecticides and bioherbicides and how do they help in pest control. [3]
(a) When basically dissimilar parts of different animals are modified for the same purpose it is ‘ called convergent evolution, e.g. wing of bat-modification of forelimb and wings of insect are the extension of body wall and when the same basic organ becomes adapted by specialisation to different functions, it is known as divergent evolution e.g. limbs of vertebrates.

(b) Manures are the organic wastes which after partial decay is added to the soil to increase the crop productivity. It supplies all the essential mineral requirement for the crop plant.

The green manure crop supplies the organic as well as inorganic components to the soil. It also provides the protective action against erosion and leaching where, as biofertilizers are micro-organisms which enrich the soil in nutrients by enhancing the availability of nutrients.

(c) IPM – Integrated Pest Management.
Bioinsecticide – Bacillus thuringiensis, the spores of soil bacteria-produce a protein which kills larvae of certain insect pests.
Bioherbicides – Cochineal insect – Used to reduce the over growth of cacti.

Question 10.
(a) What is mental illness ? Explain any three methods of treatment of mental illness. [4]
(b) What are Koch’s postulates ? Why are they not applicable to viruses ? [3]
(c) Name the Causative agent and the main symptom of each of the following diseases : [3]
(i) Filariasis
(ii) Rabies
(iii) Chickenpox
(a) Mental illness
(i) It is the abnormal change in thinking, feeling, memory leading to change in behaviour and in the manner of talking. The patient may have partial or total loss of memory, self-de-structive behaviour, hallucination etc.
Treatements – Psychotherapy, Chemotherap, Electric Shock Treatment (ECT)

(i) Psychotherapy (psychological treatment) : It can help the patient to adjust to his sur-roundings (home, place of work, society). Social therapy is aimed at rehabilitation of the victim. Re-creational activities, involvement in family life, removal of maladjustments can give relief, if not complete cure to chronic mental patients.
(ii) Chemotherapy : Drugs can cure psychoses fully if started early and used regularly e.g. amphetamines, barbiturates etc.
(iii) Electric shock treatment (ECT) : It is a crude method, used to relieve severe depression. The introduction of the psychotropic drugs has considerably reduced this method.

(b) Koch’s Postulates In 1876, a German microbiologist, Robert Koch discovered that a large number of microorganisms such as viruses, bacteria, fungi, protozoans, etc. cause diseases in human body. He stated that certain requirements should be fulfilled if the disease causing character of any organism is to be proved. These requirements are Koch’s postulates which are given below.

  1. The organisms (pathogen) must be regularly found in the body of the animal that is suffering from a disease.
  2. The organism must be isolated that grow in pure culture on artificial media.
  3. The same disease must be produced when the cultured organisms are injected into other healthy animals.
  4. The same organism must be recovered from the injected animals.

These postulates originally were applied for animal diseases but are equally applicable for human diseases. However, Koch s postulates are not applicable to viral diseases because they cannot be cultured on artificial media. Koch’s postulates are also not applicable to the bacteria of leprosy.


  • Filariasis
  • Rabies
  • Chickenpox

Causative agent:

  • Wuchereria bancrofti
  • rabies virus
  • Vpricella zoster


  • the inflammatory thickness of the wall of lymphatic vessel.
  • Severe headache, high fever, severe and painful spasm of muscles
  • Skin eruption occurs as small red papules which grows out into pustules

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