ISC Biology 2018 Class-12 Previous Year Question Papers

ISC Biology 2018 Class-12 Previous Year Question Paper Solved for practice. Step by step Solutions with Part-I, and II (Section-A,B and C). By the practice of ISC Biology 2018 Class-12 Solved Previous Year Question Paper you can get the idea of solving.

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ISC Biology 2018 Class-12 Previous Year Question Papers Solved

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Sections-A, Part- II,

Sections-B, Part- II,

Sections-C, 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, B and C.
  • 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 [ ].

Part-I (20 MARKS)

Previous Year Question Papers of ISC Biology 2018 Class-12 Papers Solved

Question 1.

(a) Answer the following questions briefly and to the point: [8 × 1]
(i) Give a significant point of difference between Oestrous and Menstrual cycle.
(ii) Give the biological name of the organism causing typhoid.
(iii) If the haploid number of chromosomes in a plant species is 20, how many chromosomes will be present in the cells of the shoot tip?
(iv) Name a plant which flowers every twelve years.
(v) Name the diagnostic test for AIDS.
(vi) Name the terminal stage of ageing in the life cycle of plants.
(vii) Which organisms constitute the last trophic level?
(viii) What is emasculation?

(b) Each of the following questions has four choices. Choose the best option in each case :
(i) Length of the DNA with 23 base pairs is:
(1) 78.4 A
(2) 78.2 A
(3) 78 A
(4) 74.8 A

(ii) Opium is obtained from:
(1) Papaver somniferum
(2) Cannabis sativa
(3) Erythroxylum coca
(4) Datura metel

(iii) According to Abiogenesis, life originated from:
(1) Non-living matter
(2) Pre-existing life
(3) Oxygen
(4) Extra-terrestrial matter

(iv) The largest unit in which gene flow is possible is:
(1) Organism
(2) Population
(3) Species
(4) Genes

(c) Give one significant contribution of each of the following scientists: [4 × 1]
(i) P. Maheshwari
(ii) E. Wilson
(iii) M. S. Swaminathan
(iv) H. Boyer

(d) Define the following: [2 × 1]
(i) Biopatent
(ii) Parthenocarpy

(e) Give a reason for each of the following: [2 × 1]
(i) Pollen grains of wind pollinated flowers are produced in large quantities.
(ii) Equilibrium of a forest ecosystem can be disturbed by uncontrolled hunting of big predators.
Answer 1.


Oestrous cycle Menstrual cycle
Occurs in non-primate mammals There is no blood flow.
Female allows mating only during heat
Occurs in primate mammals. Blood flow occurs.
Female always allow mating.

(ii) Salmonella typhi
(iii) 40 chromosomes
(iv) Strobilanthes kunthiana / Neelakuranji
(v) ELISA test/ Western blot test
(vi) Senescence
(vii) Decomposers
(viii) Emasculation is removal of anthers from bisexual flower before maturing for cross-fertilization.


(i) 78.2 A / 74.8 A
(ii) Papaver somniferum
(iii) Non-living matter
(iv) Species / population


(i) P. Maheshwari : Plant tissue culture
(ii) E. Wilson: Coined the term Biodiversity
(iii) M. S. Swaminathan: Green revolution in India
(iv) H. Boyer: Discovered restriction enzyme endonuclease.


(i) Biopatent: It is a legal right given to the owner for a biological discovery.
(ii) Parthenocarpy: Development of fruits without fertilization.


(i) Pollen grains of wind pollinated flowers are produced in large quantities because large number of pollen grains may be wasted on the way. It ensures pollination.
(ii) Equilibrium of a forest ecosystem can be disturbed by uncontrolled hunting of big predators as herbivore population will increase and disturb the food chain.

Section-A (14 Marks)

Previous Year Question Papers of ISC Biology 2018 Class-12 Papers Solved

Question 2.
(a) A woman with blood group O married a man with blood group AB. Show the possible blood groups of the progeny. List the alleles involved in this inheritance. [2]
(b) If the mother is a carrier of colour blindness and the father is normal, show the possible genotype and phenotype of the offspring of the next generation, with the help of a Punnet square.
Answer 2.
ISC Class 12 Biology Previous Year Question Papers Solved 2018 1
(i) Possible blood groups of the progeny: A and B
(ii) Alleles involved – IA, IB, 1O

ISC Class 12 Biology Previous Year Question Papers Solved 2018 2

(i) Phenotype: 25% of males colour blind, 75% progeny normal
(ii) Genotype: 50% Normal progeny, 25% Carrier female and 25% colour blind male

Question 3.
Define life span. Give the life span of an elephant. [2]
Answer 3.
Life span is the time period from birth to death of an organism.
Life span of Elephant is between 20 – 90 years.

Question 4.
Give two characteristic features of each of the following: [2]
(a) Ramapithecus
(b) Cro-Magnon man
Answer 4.
(a) Ramapithecus : Cranial capacity 600 – 700 cc and about 4 feet tall.
Bipedal locomotion with canines small, short face and small brain.
(b) Cro Magnon man: 1.8 m tall with erect posture.
Body less hairy, chin prominent, omnivores and cave dwellers.

Question 5.
(a) List any four effects of global warming. [2]
(b) State any four measures to control noise pollution.
Answer 5.
(a) Four effects of global warming:

– Mean temperature of earth increases
– Melting of ice caps and rise in sea level
– Coastal floods
– Extinction of large number of plant and animal species

(b) Four measures to control noise pollution:

– Use of better designed, quieter machine industries.
– Noisy industries should be away from residential areas.
– Enforce restriction on the use of loudspeakers and amplifiers.
– Plantation of trees on highways.

Question 6.
Define BOD. What is its significance in an aquatic ecosystem? [2]
Answer 6.
BOD: (Biological Oxygen Demand) It is the amount of dissolved oxygen needed by aerobic organisms to breakdown organic material present in a given water sample.

Its significance in an aquatic ecosystem is that it indicates water pollution.

Question 7.
Give one significant difference between each of the following pairs : [2]
(a) Humoral immunity and cell-mediated immunity.
(b) Benign tumour and malignant tumour.
Answer 7.
(a) Humoral immunity:
Consists of B-lymphocytes and is active against viruses and bacterias.

Cell-mediated immunity:
Consists of T-lymphocytes and is active against all pathogens including fungi and protozoa.

(b) Benign tumour:
Non-cancerous and growth rate is slow.

Malignant tumour:
Cancerous and growth rate is rapid.

Question 8.
Give four causes of infertility in males.
Answer 8.
Four causes of infertility in males:

– Cryptorchidism
– Low sperm count
– Vas deferens blockage
– Smoking and alcoholism

Section-B (21 Marks)

Previous Year Question Papers of ISC Biology 2018 Class-12 Papers Solved

Question 9.
(a) Draw a labelled diagram of L.S. of human testis.
(b) Draw a labelled diagram of the mature embryo sac of angiosperms.
Answer 9.
ISC Class 12 Biology Previous Year Question Papers Solved 2018 3

Question 10.
Explain gene therapy, with reference to treatment of SCID. [3]
Answer 10.
Gene therapy is the correction of genetic defect through genetic engineering or rDT.

Deficiency of adenosine deaminase causes SCID in which patient lacks T cells and B cells. The cells from the bone marrow of the patient are cultured outside the body. The functional ADA gene is transferred into these cells through retrovirus vector.

The transformed cells are transferred to the patient back. The T cells and B cells appear in the blood. It is effective in embryonic condition. After birth it has to be done repeatedly.

Question 11.
Study the table given below. Do not copy the table, but write the answers in the correct order. [3]

Scientific Name Commercial Product Use
(a) _____________

Monascus purpureus

(e) _____________


(c) _____________

Lactic acid

(b) _____________

(d) _____________

(f) _____________

Answer 11.
(a) Streptococcus
(b) Clot dissolving or clot buster
(c) Statins
(d) Blood cholesterol lowering agent
(e) Lactobacillus
(f) Curd production

Question 12.
Explain industrial melanism. [3]
Answer 12.
Two varieties of peppered moth were seen. Biston betularia grey coloured and Biston carbonaria black coloured. Before industrialization number of grey variety was more. After industrialization number of black variety increased. Due to pollution lichens disappeared and the tree became dark. The black variety got advantage through camouflage and were saved by the attack of predator birds. This showed natural selection of the black coloured moths.

Question 13.
Describe the tissue culture technique in plants. [3]
Answer 13.
Tissue culture technique in plants is as follows:

The explants are selected and sterilised. These are then placed in culture medium. Incubation under ideal, sterile and aseptic condition is done. Callus formation takes place. Auxins are added for cell differentiation. Organogenesis occurs and plantlets are generated. Hardening of plantlets are done and then in the end they are transferred in the soil.

Question 14.
Define the following: [3]
(a) Spermiogenesis
(b) Reproductive health
(c) Amenorrhea
Answer 14.
(a) Spermiogenesis: Changing of spermatids into sperms.
(b) Reproductive health: Ability of a person to produce healthy offsprings.
(c) Amenorrhea: Stoppage or absence of menstrual cycle.

Question 15.
(a) Define the following: [3]
(i) Hotspots
(ii) Ramsar Sites
(iii) Red data book
(b) Define the following:
(i) Biodiversity
(ii) Eutrophication
(iii) PAR
Answer 15.

(i) Hotspots : A region with high level of endemic species. An area rich in biodiversity.
(ii) Ramsar sites: These are the wetland areas of biodiversity.
(iii) Red Data book: A book containing a list of endangered species.

(i) Biodiversity: It is the degree of variation of life forms.
(ii) Eutrophication: The process of accumulation of inorganic nutrients in an aquatic ecosystem.
(iii) PAR : Photosynthetically Active Radiation is the wavelength of light available for photosynthesis.

Section-C (15 Marks)

Previous Year Question Papers of ISC Biology 2018 Class-12 Papers Solved

Question 16.
(a) Describe post transcriptional processing of RNA in eukaryotes. [5]
(b) Describe Avery, McLeod and McCarty’s experiment. State its significance.
Answer 16.
(a) Post transcriptional processing of RNA in eukaryotes:
It is the changes in the primary transcript (hnRNA) to mature or functional mRNA.
The processing is as follows:
Capping : It is done with methylguanosine.
Tailing / Poly-A tail: It is the addition of adenylate residues at 3′ end.
Splicing : In this removing of introns and joining of exons take place.
Capping and tailing : It prevents degradation of mRNA.
(b) Avery, McLeod and McCarty’s experiment:
From heat-killed S strain bacteria, extraction and purification of proteins, DNA and RNA, etc., is done.

The extract was treated with protease, RNAse and DNAse. DNAs treated extract did not kill mice. Protease killed mice. RNAse killed the mice.

This proved that DNA is the genetic material.

Hereditary material can be extracted from one organism and introduced into another (transformation).

Hence, DNA is transforming principle.

Question 17.
(a) Write a short note on Chipko Movement. [5]
(b) Write a short note on Joint forest management.
Answer 17.
(a) Chipko movement is related to forest conservation. The villages in Gharwal started this movement. Sunderlal Bahuguna, Amrita Devi etc., were the leaders for this movement and it gained momentum under their guidance. The aim was to protect the trees from cutting and felling. The people in great number hugged the trees so as to protect them from cutting. They embraced the trees. Chipko means embrace. This movement prevented contractors from cutting the trees. This became a turning point in the history of eco development. Many people lost their lives in this movement. This movement encouraged planting of trees for timber, food, shelter and making villages self-sufficient.
(b) Joint forest management was established by Government of India. It involves both the state forest department and local rural communities. Currently a large forest areas are under them. The rural communities protect the forest from fire, grazing, illegal harvesting and poaching. In return forest department provides non-timber forest products and share of revenue from the sale. It gives employment to the villagers. The products offered gums, resins and medicines to the villagers. This management avoids human tribal conflict.

Question 18.
(a) What does PCR stand for? Describe the different steps of PCR. [5]
(b) Give an account of the Blue-White Method of selection of recombinants.
Answer 18.
(a) PCR stands for Polymerase Chain Reaction or gene amplification is the method of making multiple copies of the gene of interest using oligonucleotides and DNA polymerase.

Different steps of PCR are as follows:
(i) Denaturation : The target DNA is heated to a high temperature of 94° — 96°C. This results in the separation of two strands. Each strand serves as a template for DNA synthesis.
(ii) Annealing: The two oligonucleotide primers are hybridized or annealed to each single-stranded DNA following rule of complementarity. It is caused at low temperature of 40° – 60°C.
(iii) Extension (Polymerisation) : With the help of Tag DNA polymerase, the DNA is synthesized the primers using dNTPs and magnesium ions. The temperature is 72°C.
(iv) The newly synthesized DNA is subjected to same procedure to double the DNA content from the previous cycle.


(b) The Blue-White Method of selection of recombinants:
Based on insertional inactivation the foreign gene is inserted into Lac Z gene coding for Beta-galactosidase enzyme which are used as selectable markers. The Beta-galactosidase converts colourless substrate (X gel) into blue coloured product. In transformed hosts, Lac Z is activated due to insertion of foreign DNA. Non-transformed colonies convert colourless X gel (chromogenic substrate) into blue coloured product. The transformed colonies remain white or colourless. It helps to differentiate between transformant or recombinant and non-transformant or non-recombinant.

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