ISC Class 12 Biology Question Paper Solution 2015

The ISC Class 12 Biology – Paper I (Theory) was conducted on 25th March 2015. The exam started at 2 PM. The paper was of 3 hours of time duration and 70 Marks. Students can find the ISC Class 12 Biology Solved Question Paper 2015 along with the marking scheme on this page. The answers provided in the solution pdf are in the same way as expected from the students to write during the exam. Also, step wise marking is provided for each answer. This will help students to understand how to express answers in the exam to score more marks. So, after solving the ISC Class 12 Biology Previous Year Question Papers, students must go through the solution pdf. They can download the ISC Class 12 Biology Question Paper and Solution pdf 2015 from the link below.

Science Stream students can also access the Solved ISC Class 12 Previous Year Question Papers with Solutions for all subjects compiled at one place. They can have a look at the ISC Class 12 Biology Question Paper Solution 2015 below.

Difficult Topics of ISC Class 12 Biology Paper 2015

Topics which students found difficult while solving the Biology 2015 paper are mentioned below:

• Water Potential and its components.
• DNA finger printing technique.
• Action of T cells to antigens.
• Placentation in angiosperms.
• Cannabinoids and Barbiturates
• Mass flow hypothesis.
• Characteristics of Dryopithecus

Confusing ISC Class 12 Biology Questions 2015

Biology concepts between which students got confused during the exam are mentioned below.

• Gene flow and genetic drift/genetic erosion
• Types of placenta and types of ovules.
• DNA finger printing and finger printing
• Pisciculture and Ichthyology
• Cannabinoids and Cannibalism
• Biotic potential and carrying capacity
• Chromosomal and anatomical features of apes and man.
• Infertility in humans and infertility in soil
• Features of pea plant and laws of Mendel

ISC Class 12 Biology Question Paper Solution 2015

Question 1:

(a Give a brief answer for each of the following:

(i) What is heterosis?

(ii) Why is non-cyclic photo phosphorylation considered as a non-cyclic pathway?

(iii) Define test cross.

(iv) What are introns?

(b) Each of the following question(s)/statement(s) has four suggested answers. Choose the correct option in each case.

1. Triple Fusion involves:

(i) Fusion of one male gamete with female gamete

(ii) Fusion of tube nucleus with generative nucleus

(iii) Fusion of two polar nuclei

(iv) Fusion of second male gamete with two polar nuclei

2. An EEG represents spontaneous electrical activity of the:

(i) Kidney

(ii) Spinal cord

(iii) Heart

(iv) Brain

3. The genotype of a person with Turner’s syndrome will be:

(i) 44+XXY

(ii) 44+XYY

(iii) 44+XO

(iv) 44+XXYY

4. Transcription is the transfer of genetic code from a DNA molecule to:

(i) RNA molecule

(ii) Second DNA molecule

(iii) Ribosomal sub unit

(iv) Sequence of amino acids in a protein molecule

(c) Give a scientific term for each of the following:

(i) The first formed category of photosynthetic organisms.

(ii) The surgical removal of a section of fallopian tube.

(iii) An animal behaviour which benefits others but is of no advantage to itself.

(iv) The hydrostatic pressure developed inside the cell on the cell wall due to endosmosis.

(d) Expand the following abbreviations:

(i) STD

(ii) NADP

(iii) MRI

(iv) DDT

(e) Name the scientists who are associated with the following:

(i) Discovered the fossil of Australopithecus

(ii) Microspheres

(iii) Coined the term Diffusion Pressure Deficit

(iv) Invented the CT scan

Answer:

(a) (i) Heterosis or hybrid vigour is the exhibition of superiority of the hybrid/F1/offspring over both of its parents in one or more traits such as the ability to give higher yield or disease or pest resistance (or explained).

(ii) In non-cyclic photophosphorylation the electrons ejected from PS II do not return back, but instead are used by PS I (or explained diagrammatically).

(iii) Test cross: A test cross is when the F1 hybrid is crossed with the double recessive parent/homozygous recessive/Tt x tt

(iv) Introns are the non-coding segments of eukaryotic chromosomes./DNA/gene/Non coding sequences/sequences removed/spliced/intervening sequences/segments not expressed.

(b) 1. (iv) fusion of second male gamete with two polar nuclei

2. (iv) brain

3. (iii) 44+XO

4. (i) RNA molecule

(c) (i) Protobionts/photoautotrophs/cyanobacteria/blue green algae/ chemosynthetic/ Archaebacteria/sulphur bacteria

(ii) Tubectomy

(iii) Altruism/Commensalism

(iv) Turgor pressure

(d) (i) STD – Sexually Transmitted Disease

(ii) NADP – Nicotinamide Adenine Dinucleotide Phosphate

(iii) MRI – Magnetic Resonance Imaging

(iv) DDT – Di-chloro Diphenyl Tri-chloro ethane

(e) (i) Raymond Dart

(ii) Sydney Fox

(iii) Meyer

(iv) Godfrey Hounsfield

Question 2:

(a) Give any three characters that have developed during human evolution.

(b) Explain the term chemogeny.

(c) Give any two distinctive features of Dryopithecus.

Answer:

(a)

1. Distinct lumber curve

2. Hind limbs to support weight/Hind limbs longer than fore limbs.

3. Forelimbs for grasping / opposable thumb

4. Pelvis broadened to balance trunk / Development of broad basin-shaped iliac bones in the pelvic girdle.

5. Skull shifted on upper end

6. Stereoscopic vision/binocular vision/3D vision.

7. Large size of brain/High cranial capacity/1400 to 1500 cc.

8. Ability to learn

9. Unspecialized teeth

10. Low fertility rate

11. Bipedal locomotion

12. Straight posture

13. Acetabular cavities shifted inward to give straight posture/erect

14. Flattening of face/orthognathus

15. Loss of supraorbital ridges/brow ridges not distinct.

16. Straightening of forehead/flattening/prominent

17. Formation of chin/jutting out

18. Sparse body hair

19. Narrowing of nose/elevated nose/nose bridge.

20. Thinning of jawbones

21. Reduction in the size of canines/small canines.

22. Increase in intelligence

23. Social and cultural organization/communication.

24. Foramen magnum shifted downward.

25. Simian gap/Diastema absent.

26. Parabolic denture

27. Loss of tail.

28. Curvature of sole/plantigrade/sub plantigrade locomotion/heel formation.

29. Flattening of sternum.

(b) Chemogeny: chemical origin of life/ abiotic synthesis of macromolecules/chemical evolution

Formation of various simple and complex organic molecules from ammonia, methane, water and vapours/ Formation of chemicals/biomolecules (during origin of life) in the primitive

atmosphere.

(c) Distinctive features of Dryopithecus:

1. Their arms and legs were of almost equal length

2. They had a semi erect posture

3. Large canines like those of modern apes

4. Frontally broadened jaws/large jaws.

5. No brow ridges

6. 4 ft. tall

7. Cranial capacity (500 to 700cc)

Question 3:

(a) Explain the evolution of giraffe’s neck according to Lamarck’s theory of evolution.

(b) Give two chromosomal similarities between man and apes.

(c) Name any two temporary embryonic structures in vertebrates which provide evidence for evolution.

Answer:

(a)

• Giraffes have evolved from deer-like ancestors/goat like/small height
• Had short neck and forelimbs
• Grazed on grass
• Ground vegetation disappeared /grass disappeared/ replaced by high trees
• Stretched their neck continuously to feed upon branches/leaves of trees
• Resulted in gradual elongation of neck and forelimbs
• Increase was transmitted to the next generation/acquired by descendants.

(b) Chromosomal similarities between man and apes:

• Close similarities in chromosome number. (apes = 24 pairs, humans = 23 pairs.
• Banding patterns of chromosome 3 and 6 are similar
• Similarity in DNA sequences./amount/content.

(c) Temporary embryonic structures:

• Visceral clefts or gill clefts develop in all land vertebrates, but are not present in the adult.
• They are only useful to fish. / Gill slits functional in fishes only but of no use for land vertebrates.
• Tooth buds develop in embryos of toothless whales and birds, which are absent in adults.
• Embryos of all vertebrates develop a notochord which is replaced by a vertebral column in adults.
• Post anal tail

Question 4:

(a) Persons suffering from sickle cell anaemia are at an advantage in Malaria infested areas. Explain.

(b) Define the term gene flow.

(c) What are analogous organs? Describe with one example from the plant kingdom.

Answer: (a)

• Sickle cell anaemia is a genetic disorder/autosomal/recessive where normal haemoglobin HbA is replaced by HbS, reducing the oxygen carrying capacity.
• RBCs become sickle shaped/malaria parasite cannot multiply and rupture at low concentration of oxygen/hypoxia leading to severe haaemolytic anaemia.
• Homozygous individuals die (Hb^S/Hb^S), but heterozygotes (HbA/HbS) remain normal.
• Natural selection has allowed this condition to remain in Africa, where the heterozygous individuals are able to cope with malarial infection, better than the persons with normal RBCs.

(b) Movement/ transfer/ migration of alleles/ genes from one population to another as a result of interbreeding between members of the two populations./Transfer of genes from generation to generation/transfer of genes among members of the same species/population.

(c) Analogous organs:

Analogous structures are those structures which perform the same function but have different

origin/internal structure/basic plan/. Organs which reflect or show convergent evolution/ adaptive convergence.

e.g. Plant leaves and ruscus cladode/asparagus both photosynthesise but are of different origin.

Sweet potato and potato have same function but different origin.

Tendrils help in climbing but have different origins like modified stipules in smilax petioles in Nepenthes, leaflets in pea and axillary bud in Passiflora.

Question 5:

(a) With the help of diagrams, name and describe the different types of placentation seen in angiosperms.

(b) Give four points of anatomical differences between a monocot stem and a dicot stem.

(c) Define the following terms:

(i) Racemose inflorescence

(ii) Osmotic pressure

Answer: (a) Types of placentation:

1. Marginal: In monocarpellary, unilocular ovary, placenta is borne on the fused margins of the same carpel. The ovules are present along the ventral suture of the carpel./Placenta forms a ridge along the ventral suture of the ovary and ovules are attached on this ridge side by side in two alternating rows e.g. Pea
2. Axile: Multicarpellary, syncarpous multilocular gynoecium. Placenta is borne on fused margins of the same carpel. The ovules are borne on confluent margins which meet on the central axis./ovary is divided into several chambers or locules and placentae are borne along the septa of the ovary e.g. Tomato/China rose
3. Parietal: Multicarpellary, syncarpous unilocular ovary. Placenta is borne on the fused margin of the same carpel./ovules develop on the inner wall of the ovary or on the peripheral part e.g. Cruciferae (mustard)
4. Free central: Multicarpellary, syncarpous, unilocular ovary. Ovules appear to arise from the central column/Ovules are borne on the central axis and septa are absent e.g. Carnation/Dianthus.
5. Basal: Unilocular ovary, with a solitary ovule which appears to arise from the base of the ovary./The placenta with one ovule attached to it, lies at the base of the ovary e.g. Sunflower
6. Superficial: multicarpellary, sycarpous ovary. Most of the internal surface of the ovary wall is covered with ovules/Any portion of the inner wall of the ovary may serve as a placenta e.g. Water Lily
7. Pendulous: Placenta at the top of the ovary and ovule hanging down.
8. Lamellar: Placenta enlarges considerably and extends towards the centre.

(b)

(c) (i) Racemose inflorescence: The floral axis shows indeterminate growth, main axis is elongated and unbranched with older flowers at the bottom and younger flowers at the top (acropetal succession)./Main axis does not end in flower/unlimited growth

E.g. Gladiolus, Triticum, etc. (diagram accepted but tip should not have flower)

(ii) Osmotic pressure: Maximum pressure developed in a solution when separated from pure water by a semipermeable membrane/pressure required to prevent osmosis.

Question 6:

(a) Draw a diagram of the internal structure of the human ovary.

(b) Define the term water potential. What are its components? Explain.

(c) Give definition and importance of:

(i) Imbibition

(ii) Parturition

Answer: (a)

Labelling:

• Germinal epithelium/ epithelium/ peritoneum
• Cortex
• Medulla
• Primary follicles/primordial follicle
• Secondary follicle
• Tertiary follicle
• Graphian follicle
• Ovum released
• Corpus luteum
• Corpus albicans
• Corpus hemorrhagium
• Blood clot
• Stroma/ connective tissue
• Antrum
• Egg nest
• Atretic follicle

(b) Water potential: The difference in the free energy/kinetic energy of water molecules in the solution and that of pure water at the same temperature and pressure. Kinetic energy per mole of water/ the tendency of water to leave the system/sum of matrix potential, solute (osmotic) potential & pressure potential. (equation accepted)

Components:

1. Matrix potential: the hydrophillic colloidal particles to which water is adsorbed/ component of WP affected due to presence of hydrophilic substances/decrease in the WP due to the presence of matrix.
2. Solute potential/Osmotic potential: the decrease in chemical potential of pure water or solvent due to the presence of solute particles/ the component of WP affected due to presence of solutes.
3. Pressure potential: Pressure which governs the movement of water into a cell, developed due to turgor and wall pressure/ the hydrostatic pressure applied by the cell contents on the cell wall in a turgid cell.

(c) (i) Imbibition: Surface adsorption of water by non-living hydrophilic substances like cellulose/colloids due to surface attraction

Initial stages of absorption of water.

Initial stages of germination of seeds.

(ii) Parturition – Act of expelling the full term foetus from mother’s uterus at the end of gestation.

Question 7:

(a) Give four adaptations in flowers pollinated by insects.

(b) Describe the mass flow hypothesis for translocation of organic solutes (food) in plants.

(c) Write a brief note on the causes of infertility.

Answer: (a)

• Large conspicuous
• Brightly coloured
• Sweet smell/Fragrant
• Nectar secreting
• Pollen grains are rough and sticky/spiny
• Stigma is sticky
• Guidelines on petals
• Arranged in inflorescence
• Lever mechanism or mimics the female

(b) The transport of food along the conc. gradient/TP gradient/enmass movement.)

– Munch

– Sugar prepared in the mesophyll cells of leaf increases the osmotic pressure

– Water from xylem elements and neighbouring cells increases the TP. This forces some of the dissolved food into sieve tube

– The cells of the root and storage organ have low osmotic and turgor pressure due to low food concentration.

– This creates TP gradient between leaf and phloem

– As a result of this mass flow of water containing dissolved organic food takes place from the upper end to the lower end of the plant through phloem

– The source of supply is the leaf and the storage organ is the root.

– This theory could not explain the bidirectional movement of metabolites.

(i) Sugar added

(ii) Entry of sugar into Bulb A

(iii) Movement to Bulb B through P

(iv) Exit from Bulb B

(v) Removal of sugar

(vi) Movement of water from X vessel to Y vessel through T

(c) Causes of infertility:

• Cryptorchidism: failure of testes to descend into the scrotum
• Hyperthermia: higher temperature of the scrotal sac
• Blockage of vas deferens /sperm duct Blockage of the fallopian tube
• Age related
• DNA damage
• Genetic factors
• Diabetes mellitus / thyroid disorders
• Hypothalamic pituitary factors (hyperprolactinemia and hypopituitarism)
• Low sperm count/oligospermia/ azospermia/ abnormal sperm structure/ poor sperm motility
• Irregular or no ovulation/ less egg production
• Defect in the genital tract
• Defective endometrium, cervix, vaginal growth
• Deficiency of sex hormones/hormonal imbalance
• Hostile response to sperm by the production of antibodies by the woman’s blood.
• Overweight / underweight females.
• Improper fertilization
• Inability of meeting of sperm and egg
• Polycystic ovary

Question 8:

(a) Give any four reasons for Mendel’s success.

(b) Briefly describe the technique employed in DNA fingerprinting.

(c) Give any two features of Genetic Code.

Answer: (a) Mendel selected the pea plant in which:

• Concentrated at a time only on one particular trait/one character at a time
• Maintained accurate record of observations/ used statistical/ mathematical analysis
• Several varieties available
• Easy to cultivate
• Artificial cross breeding between varieties was possible so hybrids were totally fertile.
• Genes coding for seven pairs of contrasting characters were on different chromosomes. /
• Fortunate in choosing seven pairs of characters
• The characters he chose did not show any interaction / linkage.
• Used pure varieties
• Bisexuality.
• Short life span
• Large sample size/high yield

(b) DNA fingerprinting technique: Alec Jeffery

• Isolation of DNA by high speed centrifugation.
• DNA amplification by polymerase chain reaction in case the sample is very small.
• Fragmentation of amplified DNA into segments of variable lengths by digesting with
• restriction endonuclease enzymes.
• Separation of DNA fragments by electrophoresis over agarose gel. The separated segments are called restriction fragment length polymorphism.
• Denaturation of DNA fragments by alkali treatment.
• Transfer of single stranded DNA fragments from gel onto a synthetic membrane such as nitrocellulose or nylon by southern blotting method.
• Fixation of separated DNA fragments to the membrane by exposing to UV light.
• Hybridisation of single stranded DNA with radio labeled VNTR probes.
• Exposure of membrane containing hybrids of radioactive DNA probes and VNTR to X rays, so that the hybridized VNTRs appear as dark bands. The film provides DNA profile and is called autoradiogram.

(c)

• It is always triplet
• Non overlapping
• Universal
• No punctuations
• Collinear
• Degenerative / Redundant
• Initiator codon AUG
• Non sense codons / termination codon

Question 9:

(a) Explain the mechanism of action of T cells to antigens.

(b) Explain how insulin can be produced using recombinant DNA technology.

(c) What is pisiculture? Give one advantage.

Answer: (a) Mechanism of action of T cells to antigens:

• T cells provide cell mediated immunity – recognize specific antigens.
• The T lymphocyte divides rapidly/differentiate to form a clone of T cells of 4 types:
• Killer cells/CT cells: destroy infected cells having the foreign antigen attached to their surface.
• Memory T cells: are sensitized by antigens and retain their sensitization for the future/remember the nature of antigen for future (secondary) response.
• Suppressor T cells: inhibit immune response by releasing cytokines that suppress activity of other T and B cells.
• Helper T cells: secrete substances that enhance or activate immune response./stimulate antibody production of B – cells.

(b) Using recombinant DNA to make Insulin:

Two polypeptide chains (Chain – A with 21 amino acids, Chain – B with 30 amino acids

interlinked through disulphide bridges)

Restriction enzymes used to produce nicks in insulin gene in E.coli plasmid at the same

restriction sites producing sticky ends.

Mutant strains of E.coli used to avoid bacteria attacking “foreign” genes.

Insert insulin gene next to E.coli galactosidase gene which controls transcription.

Bacterial cells replicate and make copies of insulin gene.

Insulin protein is purified (beta-galactosidase removed).

Chains are mixed and disulphide bridges formed.

Final product insulin is chemically identical to human insulin.

The Rhesus Factor

(c) Pisiculture: The process of fish farming in isolated water bodies./rearing of fish.

Advantage:

• Provide income and employment to fisherman/economic, helps to enhance food production/nutrition/fish oil (cod liver oil), leather (shagreen).
• Increase in organic fertilisation by fish excreta.
• Better tilling of rice seedlings.
• Reduction in number of harmful insects whose larvae are eaten by fish.

Question 10:

(a) Name the causative organism and preventive measures for each of the following:

(i) Swine flu

(ii) Typhoid

(iii) Filariasis

(iv) Syphilis

(b) State four causes and four consequences of population growth.

(c) Differentiate between:

(i) Cannabinoids and Barbiturates

(ii) Biotic potential and Carrying capacity

Answer: (a) Name the causative organism and preventive measures for each of the following:

(b) Causes:

• Advancement in agriculture
• Control of famines
• Better public health
• Control of diseases
• Better storage facilities
• More children reach reproductive age
• Low mortality rate
• Illiteracy/lack of education/lack of awareness.
• Desire for a male child
• Better socio – economic conditions/more children more money
• Religious beliefs
• Early marriage
• Lack of use of contraceptives/lack of family planning.
• Poverty/lack of recreation.

Consequences:

• Food crisis/economic crises.
• Acute clothing shortage/shelter/poverty/housing.
• Shortage of drinking water
• Pollution
• Danger of epidemics
• Unemployment
• Lack of educational facilities
• Acute shortage of natural resources/deforestation.
• Poor health of mother and child

(c) Difference:

(i) Cannabinoids:

• Obtained from Cannabis sativa/Natural
• Intoxicating and hallucinogenic/mood swings/loss of memory/loss of motor
• coordination/narcosis
• Interact with cannabinoid receptors present principally in the brain.
• Generally inhaled or ingested orally

Barbiturates:

• Derivatives of barbituric acid/synthetic
• Used as sedative/induce sleep/narcotic effect
• Moderate doses produce relaxing effect and relieve stress
• Larger doses impair one’s physical, psychological or psychological functions/cause drowsiness/confusion
• Taken orally

(ii) Biotic potential

Biotic potential refers to the inherent power of a population to increase in numbers when all environmental conditions are favourable/physiological capacity to produce offsprings under ideal conditions

Carrying capacity

Carrying capacity refers to the maximum population size that a given environment can support/sustain.

ISC Class 12 Biology Question Paper Solution 2015 must have helped students in analysing their current level of exam preparation. Keep practising more questions and stay tuned to BYJU’S for the latest update on ICSE/CBSE/State Boards/Competitive exams. Also, download the BYJU’S App to get interactive study related videos.