ISC Class 12 Biology Question Paper Solution 2016

The ISC Class 12 Biology – Paper I (Theory) was conducted on 18th March 2016. The exam started at 2 PM. The paper was of 3 hours of time duration and 70 Marks. Here, the answers of Biology 2016 theory paper is provided. Students who were looking for the detailed answers can refer to the ISC Class 12 Biology Question Paper Solution 2016. Going through the solutions will help in accessing the strong and weak areas of students in Biology. So, after solving the previous years ISC Class 12 Biology Question Papers they must give time for evaluating their analysing their answer sheet. Students can download the ISC Class 12 Biology Question Paper and Solution pdf 2016 from the link provided below.

For students convenience, we have also compiled the solved ISC Class 12 Previous Year Question Papers of Maths, Physics, Chemistry and Biology subjects at one place. Students can download and access them for free. They can have a look at the ISC Class 12 Biology Question Paper Solution 2016 below.

Difficult Topics of ISC Class 12 Biology Paper 2016

Topics which students found difficult while solving the Biology 2016 paper are mentioned below:

– Chemiosmotic hypothesis

– Example of cross showing complete linkage

– Artificial chromosomes

– RNA interference

– Integrated Pest Management

Confusing ISC Class 12 Biology Questions 2016

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

– Action and absorption spectrum

– Living and non-living with reference to the levels of organization

– Role of hormones and their function during the menstrual cycle

– Multiple Ovulation Embryo transfer technology and artificial insemination

ISC Class 12 Biology Question Paper Solution 2016

Question 1:

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

(i) What is central dogma?

(ii) Define cryopreservation.

(iii) What is symbiosis?

(iv) Explain the term perianth.

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

1. The curve showing the amount of light absorbed at each wavelength is:

(i) Action spectrum

(ii) Absorption spectrum

(iii) Quantum yield

(iv) Quantum requirement

2. After fertilisation, the integument of an ovule develops into:

(i) Seed

(ii) Seed coat

(iii) Fruit

(iv) Fruit wall

3. Meselson and Stahl’s Experiment proved:

(i) Transduction

(ii) Transformation

(iii) DNA is the genetic material

(iv) Disruptive DNA replication

4. The act of expelling the full term foetus from the uterus is termed as:

(i) Gestation

(ii) Implantation

(iii) Parturition

(iv) Capacitation

(c) Give scientific terms for each of the following:

(i) The smallest unit of DNA which can mutate.

(ii) Type of water absorption by roots where metabolic energy is required.

(iii) Statistical study of human population.

(iv) Multiple effects of a gene on the phenotype of an organism.

(d) Expand the following abbreviations:

(i) rDT

(ii) BAC

(iii) SSBP

(iv) IUCD

(e) Name the scientists who have contributed to the following:

(i) Reverse transcription

(ii) Photorespiration

(iii) Principle of limiting factors

(iv) Photolysis of water

Answer:

(a) (i)

1. Unilateral flow of information from the DNA to the mRNA to the ribosomes.
2. The sequence of nucleotide on DNA determines the sequence of nucleotide on mRNAwhich determines the sequence of amino acids on the polypeptide chain.
3. DNA (Replication) DNA (Transcription) RNA (Translation) Protein
4. DNA transcribed into RNA which is translated into proteins.
5. Instructions in DNA are converted into a functional product/ protein.
6. Flow of genetic information in living organisms

(ii) Cryopreservation is a process where cells/tissues/germplasm are preserved by cooling to low/sub-zero temperatures (such as –196OC) /liquid nitrogen/biological/organic/living material

(iii)

1. Symbiosis is an interaction between two species which is mutually beneficial to both species. (E.g.: Rhizobium in root nodules of legumes, lichens.)
2. Living together of two living organisms (of different species).

(iv)

1. When petals and sepals cannot be distinguished from each other, they are collectively called perianth.
2. Petals and sepals are collectively called perianth.
3. Collection of tepals
4. Collection of accessory/non-essential whorls

(b)

1. (i) Absorption spectrum
2. (ii) Seed coat
3. (iv) Disruptive DNA replication
4. (iii) Parturition

(c) (i) Muton

(ii) Active (non-osmotic) (Absorption) (transport)

(iii) Demography

(iv) Pleiotropy/Pleiotropism/Pleiotropic (genes)

(d) (i) Recombinant DNA Technology

(ii) Bacterial Artificial Chromosome

(iii) Single Strand Binding Protein

(iv) Intra Uterine Contraceptive Device

(e) (i) Temin, Baltimore (and Dulbecco)/ Temin/ Baltimore/ Dulbecco (any one)

(ii) Decker

(iii) Blackman

(iv) Robert Hill (Scarisbrick)/ Hill

Question 2:

(a) Mention three features of the Neanderthal Man.

(b) Differentiate between connecting link and missing link.

(c) What is adaptive radiation?

Answer: (a) Important features of the Neanderthal man:

1. Heavily built/muscular body/ broad shoulder
2. Outwardly curved thigh bones/ legs bent.
3. Short limbs
4. Semi-erect/ stooping posture
5. Rounded cranium
6. Thick skull bones
7. Low and slanting (flat and receding) forehead
8. Thick / heavy brow ridges
9. Orthognathous/ Prognathous/ Deep jaws/ no chin
10. Large/ rounded orbits (eye sockets)
11. Large and broad nose
12. Cranial capacity around 1300 – 1600 CC.
13. Height 5 – 5½ feet /1.5 to 1.6 meters
14. Used stone tools, weapons and fire.
15. Intelligent
16. Good hunters
17. Used animal skin for clothing
18. Buried the dead
19. Performed ceremonies/rituals
20. Constructed huts/ dwelling structures/ lived in caves
21. Communicated by mono-syllabiclanguage
22. Developed primitive social life/lived in families.

(b)

(c) It is the process of evolution of different species in a given geographical area from a common ancestor/ divergent evolution/ Darwin’s finches explained/ Australian marsupials/ branching descent.

Question 3:

(a) Give an account of Lederberg’s replica plating experiment to show the genetic basis of evolution.

(b) Define phylogeny.

(c) What is Founder’s effect?

Answer: (a)

1. Obtained a master plate of several bacterial colonies without antibiotic
2. Created replica plates with antibiotic (streptomycin) of colonies/penicillin
3. Colonies which could grow were resistant to antibiotic.
4. Supports natural selection/ pre-adaptive mutations/ variations. (Any three)

(b) Phylogeny: The evolutionary (ancestral) history of an organism/ group/ race/ species.

(c)

1. Founder effect: When a few individuals move away from the parent population to establish an isolated population, they will not have all the genes of the parent population and so can cause a change in the allelic frequency leading to genetic drift.
2. An example of genetic drift, when a new colony is started by a few members of the original population in a new area.
3. An example of genetic drift due to non-random sampling of genes of the original population.
4. Genetic drift due to migration.

Question 4:

(a) With reference to the levels of organisation, differentiate between living organisms and non-living objects.

(b) Mention one cause for variation in nature.

(c) What is the difference between the teeth of apes and the teeth of man?

Answer: (a)

(b)

1. Genetic recombinations

(i) Random fusion of gametes during sexual reproduction/ Independent assortment

(ii) Crossing over

2. Mutation/types/chromosomal aberrations

3. Migration

4. Genetic drift/ gene flow

5. Selection

6. Hybridisation/Polyploidy/Ploidy

7. Any environmental factor

(c)

Question 5:

(a) Give a graphic representation of the C₃ cycle.

(b) Discuss the role of cambium in secondary growth of dicot stems.

(c) State two advantages of vegetative propagation.

Answer:

(a) Graphic representation of the C₃ cycle:

(b) Secondary growth in a dicot stem:

I.Role of vascular cambium/ intrastellar/ stellar

(i) Intra-fascicular cambium

(ii) Fascicular cambium

(iii) Cambium ring

(iv) Fusiform initials

(v) Secondary xylem on outer side

(vi) Secondary phloem on the inner side

(vii) Ray initials form medullary/ pith rays

(viii) Secondary growth takes place due to activity of cambium

II.Role of cork cambium /Exrastellar

(i) Ruptured epidermis

(ii) Outer layer of cortex or epidermal cells form cork cambium/ phellogen

(iii) Outer cork/ phellem/ dead cells

(iv) Inner side secondary cortex/ phelloderm

(v) Annual rings/ Autumn & Spring wood

(vi) Heart wood/ sap wood

(vii) Bark/ periderm

(c) Advantages of vegetative propagation

1. Rapid method
2. Sure
3. Easy/simple
4. Less expensive
5. Useful in plants that cannot produce viable seeds/ seedless/grow only vegetatively
6. Clones with superior traits of the parent/ Maintains purity of race
7. Suitable for plants which produce less seeds.
8. Useful for plants with long seed dormancy
9. Breed improvement
10. Pest free

Question 6:

(a) Explain the role of hormones during the menstrual cycle.

(b) Give four adaptations shown by flowers pollinated by wind.

(c) Give two differences between heart wood and sap wood.

Answer: (a)

• Whole cycle starts again as without progesterone lining peels off

(Any 4 hormones and one function of the corresponding hormone.)

(b)

1. Large number of pollen grains
2. Light, dry or non-sticky/ winged pollen grains
3. Long and feathery stigma that hangs out of the flower.
4. Versatile anther
5. Inconspicuous flowers
6. Dull/ colourless flowers.
7. No fragrance
8. No nectar
9. If flower is unisexual more number of staminate flowers
10. Male flowers are placed at a higher level.

(c) Differences between heart wood and sap wood.

Question 7:

(a) Explain chemiosmotic hypothesis for ATP synthesis.

(b) Draw a neat labelled diagram of the vertical section of a monocot leaf.

(c) Mention any two functions of the human placenta.

Answer: (a)

1. Photolysis at PSII releases protons (H+) inside the lumen of thylakoids
2. Protons accumulate inside the thylakoids
3. Reduction of NADP removes protons from the stroma.
4. This creates a high concentration of protons inside the lumen creating a gradient.
5. Protons flow through the channel of F₀ component of the ATPase located in the thylakoid membrane.
6. This provides enough energy to cause a conformational change in the F₁ component of the ATPase that protrudes into stroma.
7. Formation of ATP by ADP + P_i

• Peter Mitchell

(b) Labelled diagram of the vertical section of a monocot leaf:

1. Stoma
2. Spongy mesophyll
3. Sub stomatal air cavity
4. Border parenchyma/ bundle sheath
5. Sclerenchyma
6. Xylem
7. Phloem
8. Lower epidermis
9. Upper epidermis
10. Guard cells
11. Cuticle
12. Bulliform cells/ Motor cells (Any eight labels)

(c) Functions of the human placenta:

1. Elimination of nitrogenous waste from foetus
2. Exchange of respiratory gases
3. Diffusion of digested material from maternal blood to foetal blood
4. Endocrine function
5. Placental barrier
6. Immunity
7. Placenta stores fat/glycogen/iron

Question 8:

(a) Explain the process of sex determination in honey bees.

(b) Define complete linkage. Give an example of a cross, showing complete linkage.

(c) Write a short note on Multiple Ovulation Embryo Transfer Technology.

Answer: (a) The process of sex determination in honey bees:

• Haploid-diploid method of sex determination/chromosomal method
• Fertilised eggs produce females (2n = 32)
• Females (workers and queen) are diploid
• Unfertilized eggs develop into males (drones) by parthenogenesis
• Males are haploid (n = 16)

(b) Complete linkage without breakage of chromosome:

Genes are closely located in the chromosome. They are transmitted together in their parental combination to the same gamete.

e.g.

1. Genes for grey body and vestigial wings in male Drosophila
2. Cross between red-eyed and normal winged female Drosophila.

grey body and long wings X black body and vestigial wings

• Only parental combinations are formed in F₂ generation
• No appearance of any new combinations.
• The results indicate that genes of grey body character and long wings are completely linked.

(c) Give a brief account of Multiple Ovulation Embryo Transfer technology.

• Used to improving the breed of cattle/increase herd size in the short time
• FSH like hormones injected to cause superovulation
• Ova fertilized by artificial insemination
• 4-10 embryos are produced at a time.
• 8-32 cell stage embryo is transplanted into a surrogate mother
• who may not have superior quality who carries it to full term.
• So a number of super milch cows or high quality meat producing bulls can be produced.
• By deep freezing it is possible to preserve 7 day old foetus for years.

Question 9:

(a) Give an account of artificial chromosomes in transfer of genetic material.

(b) Mention any four methods involved in the treatment of cancer.

(c) What is RNA interference? Give any one application of RNA interference.

Answer: (a) Artificial chromosomes in transfer of genetic material:

Bacterial Artificial Chromosome (BAC)

• Those which are constructed from bacterial genomes are called BAC/Prokaryotes.
• The construction of BAC is based on F-plasmid.
• BAC can accept DNA inserts of the range of 300-350 kb.
• It is more stable than Yeast Artificial Chromosome (YAC).
• BAC has many restriction sites and therefore is also used in genome projects.

Yeast Artificial Chromosomes (YAC)

• Constructed from yeast genome/eukaryotes.
• It can accept large fragments of foreign DNA (upto 1 million bases).
• It has centromeric and telomeric regions, ori sites, and selectable markers along with autonomously replication sequences.
• Both BAC and YAC are used as vectors for HGP.(Any four)

(b) Methods involved in the treatment of cancer:

• Surgery
• Radiation
• Chemotherapy
• Immunotherapy
• Alternative medicine (Naturopathy, Acupuncture, Ayurveda etc.)/homeopathy
• Stem cell therapy
• Bone marrow transplantation
• Hormonal treatment

(c) RNA interference is silencing a specific mRNA by introducing a gene that produces the sense and antisense strand of mRNA so that a double stranded mRNA is produced which cannot be translated into the specific protein. Inhibiting gene expression.

This technique is used to create nematode resistant plants, delaying ripening of fruits, to know gene functions, production of vaccines/treatment of papilloma virus/ Hepatitis B/ genetic defects etc.

Question 10:

(a) What is Integrated Pest Management?

(b) Explain the structure of a typical antibody molecule.

(c) Why are bio-fertilizers preferred over chemical fertilizers?

Answer: (a) Integrated pest management is an ecosystem-based strategy that focuses on control of pests through a combination of techniques (such as biological control, habitat manipulation, modification of cultural practices, and use of resistant varieties)/selective breeding. Use of bio pesticides. Selects and integrates different methods of pest control without excessive use of pesticides.

IPM is based on integration of the following methods:

• Cultural method – crop rotation, mulching, pheromones etc.
• Mechanical and physical control- catch and kill, burning, confusion technique, sterilization of males etc.
• Use of resistant varieties
• Biological control – use of parasites and predators of pests to control them

Advantages/benefits

• Economic
• Sociological
• Ecological
• Environmental

(b) The structure of a typical antibody molecule:

• Each antibody molecule is Y-shaped.
• Formed of four peptide chains – 2 heavy (longer), 2. light (small)/ (H₂L₂)
• Inter- and intra-chain disulphide bonds (– S – S –)
• Stem of ‘Y’ is formed by heavy chains
• The arm (fork) is formed by both light and heavy chains.
• Variable region (V-region) or antigen binding fragment / Fab.
• Constant region or crystalline fragment / FC.
• Glycoprotien
• Immunoglobulin

(c) Bio-fertilizers are preferred over chemical fertilizers because:

1. They do not pollute environment/eco-friendly/do not change the natural quality of soil
2. They are cheap
3. They are easy to produce
4. No energy is utilised in their production
5. Enrich the soil with nitrogenous nutrients/organic nutrients/humus/water holding capacity/increases aeration

The answers provided in ISC Class 12 Biology Question Paper Solution 2016 must have helped students in their exam preparation. So, be regular in your studies and keep working hard. Stay tuned to BYJU’S for the latest update on ICSE/CBSE/State Boards/Competitive exams. Also, download the BYJU’S App for interactive study videos.