ISC Class 12 Biology Question Paper Solution 2017

The ISC Class 12 Biology – Paper I (Theory) was conducted on 24th March 2017. The exam started at 2 PM. The paper was of 3 hours and consisted of a maximum of 70 marks. Here, we have made available the ISC Class 12 Biology Question Paper Solution 2017 for students exam preparation. Solving previous years ISC Class 12 Biology Question Papers and analysing the mistakes made by students will help in improving their performance in the board exam. They can download the ISC Class 12 Biology Question Paper and Solution pdf 2017 from the link below.

ISC Class 12 Biology Question Paper 2017

ISC Class 12 Biology Question Paper Solution 2017 PDF

Students can also access the solved ISC Class 12 Previous Year Question Papers for Maths, Biology, Biology and Biology subjects compiled at one place. They can have a look at the ISC Class 12 Biology Question Paper Solution 2017 below.

 

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Difficult Topics of ISC Class 12 Biology Paper 2017

Topics which students found difficult while solving the Biology 2017 paper are mentioned below:

  • Photorespiration and photooxidation
  • Apomixis
  • Pleiotropy
  • Transcription in prokaryotic cell
  • Tissue culture
  • Lac – operon
  • Transgenic animals
  • Electroporation

Confusing ISC Class 12 Biology Questions 2017

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

  • Multiple alleles and polygenic inheritance
  • Protobionts and Eobionts
  • Transgenic and transcription
  • In situ and Ex situ strategies for conservation of biodiversity
  • Electroporation and electrophoresis

ISC Class 12 Biology Question Paper Solution 2017

Question 1:

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

(i) Why do Green plants start evolving CO2 instead of O2, at high temperatures?

(ii) Define Apomixis.

(iii) What is a Recon?

(iv) Why are the spores of Bacillus thuringiensis used as bio insecticide?

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

1. Initiation codon of protein synthesis in Eukaryotes:

(i) GUA

(ii) GGA

(iii) CCA

(iv) AUG

2. Type of Interaction where an individual sacrifices its own welfare (life) for the benefit of another animal of its own species:

(i) Altruism

(ii) Scavenging

(iii) Protocooperation

(iv) Commensalism

3. Wings of Insect and Birds are examples of:

(i) Analogous

(ii) Homologous

(iii) Vestigial

(iv) Atavism

4. The pressure of the cell contents on the cell wall is known as:

(i) Wall pressure

(ii) Osmotic pressure

(iii) Turgor pressure

(iv) Diffusion pressure

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

(i) An act of expelling the full-term foetus from mother’s uterus at the end of gestation.

(ii) Entry of pollen tube into an ovule through integuments.

(iii) An alternative form of the single gene which influences the same character and produces different expressions in different individuals of a species.

(iv) The study of human population covering all aspects and parameters.

(d) Expand the following abbreviations:

(i) MTP

(ii) IVF

(iii) HIV

(iv) DPD

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

(i) Discovered the fossil of Cro-Magnon man.

(ii) Classified active and passive absorption of water by roots.

(iii) Reported Haemophilia.

(iv) Discovered double fertilization.

Answer:

(a) (i) Green plants start evolving CO2 instead of O2, at high temperatures due to photorespiration, where RubisCO functions as RuBPoxygenase. More photosynthetically fixed carbon is lost by photorespiration.

(ii) The development of an embryo without the occurrence of fertilization, especially in plants. /Substitution of sexual reproduction by asexual reproduction/Agamospermy/Formation of seed without fertilization/reproduction without formation of zygote/without fertilization.

(iii) Recon is the smallest unit of DNA capable of undergoing crossing over and recombinations.

(iv) Spores of Bacillus thuringiensis used as bio insecticide:

The spores produce insecticidal cryoprotein / endotoxins which kills larvae of certain insects/crystalline protein/toxic protein/insecticidal protein/BT toxin

(b)

  1. (iv)AUG
  2. (i)Altruism
  3. (i)Analogous
  4. (iii)Turgor pressure

(c) (i) Parturition

(ii) Mesogamy

(iii) Multiple alleles.

(iv) Demography

(d) (i) MTP – Medical Termination of Pregnancy

(ii) IVF – In Vitro Fertilization

(iii) HIV – Human Immuno Deficiency Virus

(iv) DPD – Diffusion Pressure Deficit

(e) (i) Mac Gregor

(ii) Kramer

(iii) John Otto

(iv) G.Nawaschin

Question 2:

(a) Differentiate between Apes and Man with respect to the following characteristics:

(i) Posture

(ii) Brow ridges

(iii) Cranial capacity

(b) Define protobionts.

(c) What is cognogeny?

Answer: (a)

Characteristics

Apes

Man

Posture

Semi-erect/stooping/bent

Erect/upright/straight

Brow ridges

Prominent/ heavy/protruded /conspicuous/broad thick/dense/projecting

Are reduced/inconspicuous

Cranial capacity

390 – 650 cc

1350 – 1600 cc

(b) Protobionts: aggregate of Biomolecules

A protobiont is an aggregate of abiotically produced organic molecules surrounded by a membrane-like structure and are considered to have possibly been the precursors to first cells.\Pre-cell like structures/Protocells/Precursor of cell/cell like

(c) Cognogeny: Diversification and evolution of the prokaryotic and eukaryotic cells /Development of complexities/The process of evolution of complex life forms/ development of senses of perception/ communication/modification. /Cognition/expression/evolution.

Question 3:

(a) Explain any three molecular (genetic) evidence in favour of organic evolution.

(b) Define biogenesis.

(c) Define fossils.

Answer:

(a) Evidences from molecular biology: (Any three points)

  1. Cytology:
  • Basic structure of cell is same in all organism
  • Mitochondria are same in plants and animals
  • Photosynthetic machinery/plastids/chloroplasts same. /chlorophyll/ribosomes same
  1. Molecular biology:
  • Genetic code is universal … Anything related to codon
  • DNA/Nucleotide sequence is almost similar.
  • Similar chromosome bands.
  • Similarity in chromosome number.
  • Structure of actin and tubulin proteins is also same.
  • Homology occurs in amino acid sequence of cytochrome C/Haemoglobin. Enzymes, hormones
  1. Metabolic process:
  • Same biochemical reactions are found from bacteria to humans. /Glycolytic pathway/Krebs cycle/ETS
  • CO2 is released by oxidation of glucose/oxygen consumed
  • Energy released is stored in ATP molecules.
  • Nitrogenous wastes in all living organism is produced in the form of ammonia.
  1. Serological test:
  • Shows similarities in blood groups.
  • Similar antigen-antibody profile (Precipitin test) in man and apes.

(b) Biogenesis: Formation of first living form/Eobionts/Biological evolution

The production of living organisms from other living organisms / Origin of life.

(c) Fossils: The remains / impression of a prehistoric plant or animal embedded in a rock / preserved in petrified form.

Question 4:

(a) List any three drawbacks of Darwinism.

(b) State Hardy Weinberg’s principle.

(c) Differentiate between Directional natural selection and Disruptive natural selection.

Answer:

(a)

  1. Could not explain – characters are transmitted from generation to generation
  2. Theory of pangenesis
  3. Inheritance of small variations / laid too much stress on large variation
  4. Ignored mutation
  5. Existence of vestigial organs
  6. Overspecialization
  7. Arrival of the fittest.
  8. Could not explain causes and sources of variations
  9. No role of genes/Theory of germplasm
  10. Considered individual (not population) as unit of evolution
  11. Could not explain the connecting link.

(b) The relationship of gene frequencies / genotype frequencies of alleles in the gene pool of a population. gene (allele) frequencies remain constant in a large, randomly breeding population/ genetic equilibrium in population

OR

(p+q)2 = 1 OR

p2 + 2 pq + q2 = 1

(c)

Directional Natural Selection

Relatively frequent

Disruptive Natural Selection

Rare in nature

A mode of natural selection in which a single phenotype is favoured, causing the allele frequency to continuously shift in one direction

A mode of natural selection in which extreme values for a trait are favoured over intermediate values

Results in relatively uniform phenotype

Polymorphism/more than one distinct forms

Frequency of any one extreme phenotype is very high

One side of mean of average frequency

Frequencies of both extreme phenotypes are high (that of the average phenotype is very low)

Example – With explanation

Example – Both sides of mean

ISC Class 12 Biology Question Paper Solution 2017-1

Question 5:

(a) Give four anatomical differences between a dicot leaf and monocot leaf.

(b) Briefly describe the secretory phase of the menstrual cycle.

(c) Define:

(i) Menarche

(ii) Actinomorphic symmetry

Answer: (a)

Monocot Leaf

Dicot Leaf

Mesophyll Compact/small intercellular spaces

Large intercellular spaces

At both surfaces (abaxial and adaxial) sclerenchymatous bundle sheath is present

At abaxial surface sclerenchymatous bundle sheath is present.

Stomata present on the upper and lower epidermis /isobilateral/ equifacial / amphistomatic

Stomata mainly present on the lower epidermis/dorsiventral/bifacial/hypostomatic

Some epidermal cells are modified into bulliform cells

Epidermis does not have bulliform cells.

Mesophyll is undifferentiated/only spongy

Mesophyll is differentiated into palisade and spongy parenchyma

Bundle sheaths double layered

Bundle sheath single layered

Guard cells dumbbell shaped

Kidney /bean shape

Equally thick cuticle on both sides

Thicker on upper (ventral) side.

All veins of equal size/ parallel venation

One main vein/Prominent vein/reticulate or net venation

(b) Luteal or secretory phase:

  • Glycogen content of endometrium increases.
  • Graafian follicle ruptures to release the ovum (ovulation) due to LH surge.
  • Uterine movement/contraction decreases
  • cervical mucus becomes thick
  • formation of a yellow colour body called corpus luteum.
  • The corpus luteum secretes large amounts of progesterone.
  • Thickness/ Vascularization of the endometrium increases.
  • Uterine glands secrete “uterine milk”/corkscrew shaped/active/tortuous/coiled
  • In case ovum is not fertilized corpus luteum degenerates into corpus albicans
  • Longest phase of the cycle /14 days duration/extends from days 14 to 25
  • Decline of progesterone level starts menstrual phase.

(c) (i) Menarche: Beginning / Onset of the menstrual cycle in young females.

(ii) Actinomorphic symmetry:

Radial symmetry / flower can be divided into two equal halves in any plane passing through the centre.

Question 6:

(a) Give a graphic representation of the Hatch Slack or C4 cycle.

(b) Give two significant differences between:

(i) Transpiration and Guttation

(ii) Chlorophyll ‘a’ and Chlorophyll ‘b’

(c) Define the following:

(i) Amniocentesis

(ii) Polyembryony

Answer: (a) Graphic representation of the Hatch Slack or C4 cycle:

ISC Class 12 Biology Question Paper Solution 2017-2

(b) Difference between:

(i)

Transpiration

Guttation

During the day

Night and early morning

Water loss in the form of water vapour

Water loss in the form of liquid droplets

Through stomata, lenticels, cuticle/ Leaf surface

Through hydathodes/ leaf margin

Under dry conditions

Under humid conditions

Only water is lost

Both water and minerals are lost

Due to evaporation

Due to root pressure

Cooling effect

Helps to get rid of excess water

(ii)

Chlorophyll ‘a’

Chlorophyll ‘b’

The empirical formula of Chlorophyll ‘a’ is C55H72O5N4Mg

The empirical formula of Chlorophyll ‘b’ is C55H70O6N4Mg

Blue green in colour

Yellow green in colour/olive green

Primary pigment

Accessory pigment

It has a methyl group attached to third carbon

It has a aldehyde group attached to third carbon

Soluble in petroleum ether

Soluble in methyl alcohol

Molecular weight is 893

Molecular weight is 907

Maximum absorption of violet-blue/red and orange-red light (430 and 662 nm)/Red 430-450 & 660-690

Maximum absorption of blue light (453 and 642 nm) 450-480 & 640-650

More abundant/All green plants 2 parts/ more

Less abundant/ Not found in diatoms etc 1 part

(c) (i) Amniocentesis: Prenatal diagnostic technique: A procedure in which a small sample of amniotic fluid is drawn out of the uterus through a needle inserted in the abdomen. The fluid is then analysed to detect genetic abnormalities in the foetus.

(ii) Polyembryony: The phenomenon of development of more than one embryo in a seed.

Question 7:

(a) Describe K+ transport stomatal mechanism.

(b) Draw a neat-labelled diagram of L.S. of anatropous ovule.

(c) Differentiate between the following:

(i) Spermatogenesis and oogenesis

(ii) Apocarpous ovary and syncarpous ovary.

Answer: (a) K+ transport stomatal mechanism:

In light

In dark

Starch converted to malic acid.

Photosynthesis stops, respiration continues, increases CO2 conc.

Malic acid– malate & H+ ions

High CO2, K ion transport stops.

H+ ions move to epidermal cells, K+ ions enter guard cells (ion exchange)

Abscisic acid is formed, reverses H – K pump.

K+ ions balanced by organic ions (malate)

Malate ions produce malic acid with H ions lowering its synthesis by PEP carboxylase.

Cl- ions are also taken in to balance K ions

These changes induce reversal of K ions.

H – K ion exchange requires energy supplied by photosynthesis, respiration

Thus, decreases the OP of guard cells

Increased K ions & malate ions increase the OP thus water enters the guard cells and stomata open

Guard cells – flaccid – stomata close

(b) Diagram of L.S. of anatropous ovule.

ISC Class 12 Biology Question Paper Solution 2017-3

(c) (i) Spermatogenesis and oogenesis:

Spermatogenesis

oogenesis

Starts at puberty

Before birth

Process of formation of haploid spermatids from diploid germinal cells

Process of formation of ovum.

Four spermatids are produced from a single spermatogonium

One functional ovum and three non-functional polar bodies.

No formation of polar body

(Two/three) polar bodies formed

Spermiogenesis (transformation/tail formation)

Direct formation

Continuous process

Meiotic arrest during formation of oocytes

Regulated by androgens/ testosterone

Regulated by oestrogens

Growth phase not significant

No synthesis of yolk/no vitellogenesis

Significant increase in the size of Pr. Oocyte

Synthesis of yolk/ vitellogenesis

In testes

In ovary

(ii)

Apocarpous Ovary

Syncarpous Ovary

Aggregate inflorescence

Multiple inflorescence

Have carpels that are free from one another

Consist of united carpels

Used of a single flower with two or more separate pistils

Flowers bear single pistil

Ovary is unilocular

Ovary can be unilocular or multilocular.

Question 8:

(a) Explain Pleiotropy with reference to phenylketonuria.

(b) Explain the mechanism of transcription in a prokaryotic cell.

(c) Explain Rh factor incompatibility during pregnancy.

Answer:

(a) Pleiotropy: One gene with many effects.

Phenylketonuria:

  • Associated with chromosome 12
  • Caused by deficiency of phenylalanine hydroxylase
  • Affected individuals fail to convert phenyl pyruvic acid into P-hydroxyphenyl pyruvic acid.
  • Autosomal recessive/genetic disorder-Accumulation in blood.
  • Can cause mental retardation
  • Skin pigmentation
  • loss of hair

(b) The mechanism of transcription in prokaryotic cell:

Transcription proceeds in the following general steps:

  1. Initiation: RNA polymerase, together with one or more general transcription factor (rho factor), binds to promoter DNA.
  2. RNA polymerase creates a transcription bubble, which separates the two strands of the DNA helix. This is done by breaking the hydrogen bonds between complementary DNAnucleotides.
  3. Elongation: RNA polymerase adds matching RNA nucleotides to the complementary nucleotides of one DNA strand.
  4. Various ribonucleotide triphosphate is converted to ribonucleo monophosphate on binding to DNA template chain.
  5. RNA polymerase can cause polymerization only in 5’ – 3’direction. using energy
  6. Termination: when polymerase reaches the termination signal it leaves the DNA
  7. Hydrogen bonds of the RNA–DNA helix break, freeing the newly synthesized RNA strand.
  8. Bacteria use two different strategies for transcription termination – Rho-independent termination and Rho-dependent termination. /Poly A- tail.

OR

(i) Initiation

  1. Transcription starts at a specific sequence called promoter sequence
  2. RNA polymerase binds at promoter sequence
  3. Sigma factor helps in recognising promoter sequence
  4. RNA polymerase creates a transcription bubble, by breaking the hydrogen bonds between complementary DNA nucleotides.

Elongation

  1. 3’ – 5’ strand of DNA acts as template.
  2. RNA polymerase adds complementary nucleotides in 5’ – 3’ direction.

Termination

  1. Rho-dependent
  2. Rho-independent
  3. Newly synthesised RNA dissociates from DNA – RNA helix.

ISC Class 12 Biology Question Paper Solution 2017-4

(c) Rh factor incompatibility during pregnancy:

  • Father Rh positive and mother Rh negative
  • Rh –ve woman bears Rh +ve child
  • Rh antibody formation in mother due to exposure to Rh antigen
  • Antibodies less during first conception so first child normal
  • Second child – Erythoblastosis foetalis.

Question 9:

(a) Discuss the various In-situ and Ex-situ strategies for conservation of biodiversity.

(b) List any four applications of tissue culture.

(c) Mention the causative agent and the preventive measures for each of the following:

(i) Gonorrhoea

(ii) Pneumonia

Answer: (a) In-situ and Ex-situ strategies for conservation of biodiversity:

In-situ conservation is also known as “on-site conservation”. This technique is more applicable for conserving wild species in its natural habitats. It includes protected areas like: Hotspots, wetland/ Ramsar sites, sacred groves.

National Parks: is an area strictly reserved for betterment of wildlife where no human activities are permitted and no private ownership right is allowed.

Wildlife sanctuaries: is an area where protection is given to fauna and certain operations like harvesting of timber and collection of forest products is permitted and also private ownership rights are allowed.

Biosphere reserves: is divided into three zones. Its aim is to conserve the gene pool of flora and fauna and traditional lifestyle of tribals.it provides area for ecological research and training. Ex-situ conservation “is also known as “Off-site conservation. In this technique, the conservation of biodiversity components is done outside of their natural habitats.

Home gardens: useful plants grown at home

Botanical gardens: is a garden dedicated to the collection, cultivation and display of a wide range of plants labelled with their botanical names.

Zoo: Is a facility in which animals are confined within enclosures, displayed to the public, and in which they may also be bred. Aquarium is a clear-sided container in which water-dwelling plants and animals are kept.

Cryopreservation: The storage of seeds, pollen, tissue, or embryos in liquid nitrogen at -196 °C. Arboreta seed bank/gene bank/ germplasm bank/ pollen bank Safari parks

(b)

  • Propagation of a large number of plants in a short duration / micropropagation.
  • Plants formed are genetically identical to the original plant / somaclonal propagation.
  • Variations appearing during tissue culture / somaclonal variations.
  • Production of disease free plants
  • Anther culture and formation of androgenic haploids
  • Embryo culture of successful hybridization
  • Induction and selection of mutants.
  • Production of transgenic plants
  • Production of weedicide resist plants
  • Production of abiotic stress resist plants
  • Production of high yielding varieties
  • Production of secondary metabolites
  • Conservation of germplasm/biodiversity
  • Production of pest or insect resistant plants
  • Protoplast culture

(c)

Disease

Causative agent

Preventive measure

Gonorrhoea

Nisseria gonorrhoeae

Avoid sexual contact with infected person.


Avoid Homosexuality

Pneumonia

Streptococcus phenumoniae / Haemophilus influenzae

Maintain personal and public hygiene

Avoid sharing glasses, utensils, food or water with infected persons.

Question 10:

(a) Name the components of lac operon and discuss their role.

(b) Give the significance of transgenic animals.

(c) Give one significant difference between:

(i) Electroporation and Gene Gun.

(ii) ECG and EEG

Answer:

(a)

  • Lac operon consists of three structural genes, promoter, termination and an operator.
  • Lac Z encodes B-Galactosidase – hydrolyses lactose into glucose and galactose
  • Lac Y codes for enzyme permease
  • Pumps B-galactosides into the cell
  • Lac A – codes for transacetylase

Only lac Z and lac Y are necessary for lactose catabolism.

  • Lac I/R/Rep/Reg codes for repressor protein
  • Promoter gene/Lac P – binding of RNA polymerase/Initiation of transcription
  • Operator gene/Lac O – Place for binding of repressor protein/switch
  • Structural genes – codes for enzymes required for metabolising lactose.

(b) Any four points Genetically modified animals currently being developed can be placed into six different broad classes based on the intended purpose of the genetic modification:

  1. To research human diseases (for example, to develop animal models for these diseases);
  2. To produce industrial or consumer products (fibres for multiple uses);
  3. To produce products intended for human therapeutic use (pharmaceutical products or tissue for implantation);
  4. To enrich or enhance the animals’ interactions with humans (hypo-allergenic pets);
  5. Enhance production or food quality traits (faster growing fish, pigs that digest food more efficiently); fishes are used for scientific research and as pets, and are being considered for use as food and as aquatic pollution sensors.
  6. Improve animal health: genetically modified viruses to deliver genes that can cure disease in humans. It has been used to treat genetic disorders (disease resistance)
  7. Improvement of quality of human race/eugenics
  8. To study normal physiological process
  9. Testing safety of vaccine
  10. To study chemical toxicity/ teratogenesity.

(c) (i)

Electroporation

Gene Gun

Electroporation is a method that uses short pulses of high voltage to carry DNA across the cell membrane. This shock is thought to cause temporary formation of pores in the cell membrane, allowing DNA molecules to pass through.

In this technique, DNA is coated onto gold particles/platinum/tungsten and loaded into a device which generates a force to achieve penetration of the DNA into the cells.

Can be used only for naked protoplast

Can be used for cells with cell wall as well

(ii)

ECG

EEG

Used for recording rate and rhythm of atria and ventricles.

For examination of Brain

We hope, solving ISC Class 12 Biology Question Paper Solution 2017 has boosted students’ exam preparation. Be continuous in your studies and keep practising more questions. 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.

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