Download Biology special study material

STUDENT SPECIAL STUDY MATERIAL
Class XII
BIOLOGY
Session 2016-17
Kendriya Vidyalaya Sangathan
Regional Office
Guwahati
OUR SOURCE OF INSPIRATION
CHIEF PATRON
SHRI SANTOSH KUMAR MALL, IAS
COMMISSIONER
KENDRIYA VIDYALAYA SANGATHAN
NEW DELHI
PATRONS
SHRI CHANDRA P NEELAP
DEPUTY COMMISSIONER
KENDRIYA VIDYALAYA SANGATHAN
GUWAHATI REGION
SMT. ANJANA HAZARIKA
&
SHRI D. PATLE
ASSISTANT COMMISSIONERS
KENDRIYA VIDYALAYA SANGATHAN
GUWAHATI REGION
CONVENOR
SHRI DHIRENDRA KUMAR JHA
PRINCIPAL
KV AFS BORJHAR
GUWAHATI
PREPARED BY:
SMT. NG. SARJUBALA DEVI
PGT(BIOLOGY), KV AFS BORJHAR
This study module is aimed at ensuring at least pass mark in the board exams and is prepared using the
available study materials of KVS but in a concise manner. This also includes previous years CBSE questions
and marking scheme so that students will have idea on what type of questions can come from a particular
chapter and what points need to be in their answers to get marks. Hope this module will boost your
confidence both during the preparatory stage as well as during the examinations.
Students can follow these steps:
8, &10 carrying weightage of 42 marks, for complete and comprehensive revision
Make a target to learn 2 or 3 concepts a day and follow it up with weekly slip tests.
checking answers with the scoring key.
Practise these diagrams and flow charts in NCERT Textbook:
(i) 2.3 a-TS of young anther
(ii) 2.5- enlarged view of pollen grain
(iii) 2.7(d)-Anatropous ovule
(iv) 2.8 c-mature embryo sac
(v) 2.13 b-stages of embryo development (vi) 2.14-typical dicot, monocot embryo
(vii) 3.3b-Human female reproductive system (viii) 3.5: Sectional view of seminiferous tubule
(ix)3.6-human sperm
(x) 3.7-human ovary section
(xi) 3.8-spermatogenesis & oogenesis
(xii) 3.9-menstrual cycle
(xiii) 3.11-transport of embryo through fallopian tube
(xiv) 6.4a-nucleosome
(xv) 6.8-DNA replication
(xvi) 6.9-transcription unit
(xvii) 6.14-lac operon
(xviii) 7.1-Millers experiment
(xix) 8.1- life cycle of Plasmodium
(xx) 8.4-Antibody molecule
(xxi) 8.6-Life cycle of HIV
(xxii) 10.8-Biogas plant
(xxiii) 11.2-rDNA technology
(xxiv) 11.6-PCR
(xxv) 11.7-Bioreactor
(xxvi) 13.3-organismic response representation
(xxvii) 13.4-human age pyramids
(xxviii) 13.5-growth curves
(xxix) 14.4-ecological pyramids
(xxx) 14.5-Primary succession
(xxxi) 14.6-carbon cycle
(xxxii) 16.7-greenhouse gases
UNIT VI – REPORODUCTION
CHAPTER -1 REPRODUCTION IN ORGANISMS

Life Span: Period from birth till natural death.
 Special Flowering: Bamboo- once in life, generally after 50-100 years.
 Strobilanthus kunthiana (Neelakuranji) – flowers once in 12 years.
 Dioecious: Only one type of reproductive structure in a plant. Eg. Papaya
 Monoecious : Reproductive organs at different positions in same plant eg. Cucurbits, Maize.
 Hermophrodite : Reproductive organs at different positions in same animal eg. Earthworm.
Cell division during gamete formation:





Haploid-parent (n) produces haploid gametes (n) by mitotic division, eg. Monera, fungi, algae and
bryophytes.
Diploid parent (2n) produces haploid gametes (n) by meiosis division (possess only one set of chromosomes)
and such specialized parent cell is called meiocyte or gamete mother cell (2n).
Parthenogenesis: Female gamete develops into new organism without fertilization. eg- Honey bee, turkey,
lizard, rotifers (Protozoans).
Seedless fruits formed by parthenocarpy.
Clone: A group of individuals of the same species that are morphologically and genetically similar to each
other & their parents.
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. A male honey bee has 16 chromosomes whereas its female has 32 chromosomes. Give reason.1
Ans: Because male honey bee is produced by parthenogenesis, ie., it is developed from female gamete without
fertilization.
2. No organism is immortal, then why do we say there is no natural death in single – celled organisms?
Ans: Parent cell divides to give rise to new individuals
3. How many chromosomes do drones of honeybee possess? Name the type of cell division involved in
the production of sperms by them.
Ans. 16, Mitosis = ½ + ½
4. Meiosis is an essential event in the sexual life cycle of any organism. Give two reasons.
Ans. (i) Meiosis helps in formation of gametes by reductional division & maintains number of chromosomes
constant / maintains ploidy = ½
(ii) Recombination of genes in offsprings / brings variation = ½
5. Name any two organisms and the phenomenon involved where the female gamete
undergoes development to form new organisms without fertilization .
Ans. Rotifers / honeybees / some lizards / turkey (Any two) = ½ + ½; Parthenogenesis = 1
--------------------------------------------------------------------------------
CHAPTER : 2 SEXUAL REPRODUCTION IN FLOWERING PLANTS
FLOWERS : modified shoot, Site of sexual Reproduction.
POLLINATION: Transfer of pollen from anther to stigma
MICROSPOROGENESIS: The process of formation of micro spores from pollen mother cell (2n) through meiosis.
MEGASPOROGENESIS
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. Explain the process of pollination in Vallisneria.
2
Ans. The female flower reaches the surface of water by long stalk , male flower releases the pollen grains on
surface of water,pollen grains are carried by water currents,some of them reach the stigma and achieve
pollination. ½ x 4=2
2. Name the type of fruit apple is categorised under and why? Mention two other examples which belong to the
same category as apple.
2
Ans. False fruit , thalamus contributes to fruit formation : Strawberry, Cashew ½ x 4=2
3. Double fertilization is reported in plants of both, castor and groundnut. However, the mature seeds
of groundnut are non-albuminous and castor are albuminous. Explain the post fertilization events
that are responsible for it.
3
Ans. Development of endosperm (preceding the embryo) takes place in both , developing embryo derives
nutrition from endosperm = ½ + ½; Endosperm is retained / persists / not fully consumed in castor ,
endosperm is consumed in groundnut = 1 + 1
4. A flower of tomato plant following the process of sexual reproduction produces 240 viable
seeds. Answer the following questions giving reasons : [1 ×5 = 5 Marks]
(a) What is the minimum number of pollen grains that must have been involved in the pollination of its pistil ?
(b) What would have been the minimum number of ovules present in the ovary ?
(c) How many megaspore mother cells were involved ?
(d) What is the minimum number of microspore mother cells involved in the above case ?
(e) How many male gametes were involved in this case ?
Ans. (a) 240 , one pollen grain participates in fertilisation of one ovule = ½ + ½
(b) 240 , one ovule after fertilisation forms one seed = ½ + ½
(c) 240 , each MMC forms four megaspores out of which only one remain functional= ½ + ½
(d) 60, each microspore mother cell meiotically divides to form four pollen grains (240/4 = 60) ½, ½
(e) 480 , each pollen grain carries two male gametes (which participate in double
fertilisation) (240 × 2 = 480) = ½ + ½
-----------------------------------------------------------------------------------
CHAPTER 3: HUMAN REPRODUCTION
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. Testes normally remain suspended in scrotum in mammals. Why? 1
Ans: Scrotum helps in maintaining low temperature , necessary for spermatogenesis
2. How many spermatozoa will be produced from 100 primary spermatocytes and how many ova will be
produced from 100 primary oocytes?
1
Ans: 400 spermatozoa , 100 eggs
3.Draw a sectional view of the ovary showing the different follicular stages of a humanfemale in her
preovulatory phase of menstrual cycle.
2
Ans: Primary, secondary, tertiary and graafian follicles(4 part to be labeled) ½X4=2
4. Draw the microscopic structure of human sperm and relate its different parts with their functions. 5
Ans: Plasma membrane-envelops the whole body of sperm; Acrosome – filled with enzymes that help in fertilization
if the ovum; Middle piece – Contains numerous mitochondria which produce energy for the movement of tail; Tailfacilitates sperm motility essential for fertilization; Mitochondria – produce energy; Nucleus- Carries haploid set (n) of
genes (1 mark for diagram, 4 marks for function of any four parts)
5. During the reproductive cycle of a human female, when, where and how does a placenta develop? What is
the function of placenta during pregnancy and embryo development?
5
Ans. After implantation , uterus , chorionic villi and uterine tissue become interdigitated (physically fused) = 1 + 1 + 1
Placenta facilitates supply of oxygen / nutrients to the embryo = ½ Removal of carbon dioxide / waste material /
excretory material produced by the embryo = ½ Production of hCG / hPL / estrogens / progestogens (Any two) = ½ ×
2
-----------------------------------------------------------------------------------------
Chapter - 4: REPRODUCTIVE HEALTH
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. Explain how IUDs act as contraceptives. 3
A: IUDs . increase phagocytosis , suppress sperm motility fertilizing capacity of sperm uterus unsuitable for
implantation cervix hostile to sperms
2. (i) Expand MTP. (ii) Give two situations when MTP is advised. (iii) Write when amniocentesis and MTP can
be misused.
3
Ans. (i) Medical Termination of Pregnancy
1
(ii) (a) to get rid of unwanted pregnancies, (b)when continuation of pregnancy could be harmful or
fatal to mother/foetus/ both. ½ x2=1
(iii) for identifying the sex of the foetus 1
3. (a)HIV and Hepatitis –B are STDs. Mention the two other ways by which they can be transmitted to a
healthy person. (b)Why is early detection of STD essential? What can it lead to otherwise? Explain . 3
Ans. HIV and Hepatitis B can also be spread by- Infected needle, From infected mother to the child, Transfusion of
infected Blood (any 2) 1
b. Leads to complication in life later; pelvic inflammatory diseases /abortion/infertility/cancer (any 2)
4. Mrs. X was blamed for being childless though the problem was due to low sperm counts in the ejaculate of
her husband. Suggest a technique which could help the couple to have a child.4
Ans: IVF and its detail process and importance
----------------------------------------------------------------------------------------------
UNIT VII: GENETICS
Chapter 5: PRINCIPLES OF INHERITANCE AND VARIATION
MENDEL’S LAW OF INHERITANCE
 Mendel's first law (Law of dominance ) :
(i)Characters are controlled by discrete units called factors (genes).
(ii)Factors occur in pairs. (iii)In a dissimilar pair of factors one member of the pair dominates (dominant) the
other (recessive).
 Mendel's Law of segregation (Purity of Gametes): The two alleles received, one from each parent,
segregate independently in gamete formation, so that each gamete receives one or the other with equal
probability. (Can be explained by monohybrid cross).
 Mendel's law of Independent Assortment : Two characters determined by two unlinked genes are
recombined at random in gamete formation, so that they segregate independently of each other, each
according to the first law (note that recombination here is not used to mean crossing-over in meiosis). (Can be
explained by dihybrid cross).
Test Cross : Individual with dominant phenotype is crossed with homozygous recessive individuals to find the
homozygosity/heterozygosity .
Incomplete Dominance : Dominant gene is not fully expressed on recessive gene. So, the phenotype of hybrid do not
resemble with any of the parents. Eg- Antirrhinum majus (snapdragon), Mirabilis jalapa
Genotypic & phenotypic ratio- 1:2:1
Co-dominance : Both parental genes expressed in F1 progeny so the offspring shows resemblance with both the
parents. Eg- ABO blood group types in human.
Blood group shows 3 different alleles (IA, IB, i) and 6 different possible genotypes.
Cross between IA i x I B i shows the law of dominance, co-dominance & multiple alleles.
Mutation : Sudden changes in DNA. Mutagens : Chemicals/agents that caused mutation
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.1 Ans.
Round/Wrinkled = ½, Yellow/ Green = ½
2. Give an example of a sex - linked recessive disorder in humans.
1
Ans. Colour blindness /any other correct example
3. How does the gene ‘I’ control ABO blood groups in humans ? Write the effect the gene has on the structure
of red blood cells.
2
Ans. – Gene ‘I’ has three different alleles IA , IB , i = ½
– IA produces A type of sugar / Antigen A group; IB produces B type of sugar / Antigen B group
– i - No sugar - O group = ½
– Structure - sugar polymers protrude from the surface of plasma membrane of RBCs = ½
4. Write the types of sex-determination mechanisms the following crosses show. Give an example of each type.
(i) Female XX with Male XO (ii) Female ZW with Male ZZ 2
Ans. (i) Male heterogamety , Grasshopper = ½ + ½ (ii) Female heterogamety , Birds = ½ + ½
5. In Snapdragon, A cross between true breeding red flower (RR) plants and true breeding white flower (rr)
plants showed a Progeny of plants with all pink flowers. (a) The appearance of pink flowers is not known as
blending. Why? (b) What is the phenomenon known as?
2
Ans. (a) R (dominant allele red colour) is not completely dominant over r (recessive allele white
colour) / r maintains its originality and reappear in F2 generation. = 1 (b) Incomplete dominance = 1
6. A colourblind child is born to a normal couple. Work out a cross to show how it is
possible. Mention the sex of this child.
Ans:
:, Male child-1.
---------------------------------------------------------------------------------------
CHAPTER: 6 MOLECULAR BASIS OF INHERITANCE
Transcription in Eukaryotes:
Protein Synthesis:
PREVIOUS YEARS QUESTIONS & MARKING SCHEME:
1. A DNA segment has a total of 1000 nucleotides, out of which 240 of them are adenine containing nucleotides.
How many pyrimidine bases this DNA segment possesses? 1
2. Why does hnRNA need to undergo splicing ? Where does splicing occur in the cell?
2
Ans. hn RNA has both exons and introns, Introns are non-coding regions , which are
removed by the process called splicing ,splicing occurs in the nucleus. ½ x 4=2
3. (a) How many codons code for amino acids and how many are unable to do so? (b) Why are codes said to be
(i) degenerate and (ii) unambiguous? 2
Ans.(a) 61, 3 (½ + ½) (b) i. Degenerate-One amino acid may be coded by several codons 1
ii. unambiguous or specific- each codon codes for a specific amino acid. 1
4. Following the collision of two trains a large number of passengers are killed. A majority of them are beyond
recognition. Authorities want to hand over the dead to their relatives. Name a modern scientific method and
write the procedure that would help in the identification of kinship. 3
Ans. DNA fingerprinting (analysis) = ½
- Isolation and digestion of DNA by restriction endonuclease
- Separation of DNA fragments by electrophoresis and transferring them to synthetic membranes
/ nitrocellulose / nylon
- Hybridisation using labelled VNTR probe
- Detection of hybridised DNA fragments by autoradiography
- Matching banding pattern of DNA / DNA fingerprints / autoradiograms of the passengers
killed and that of relatives = ½ × 5
5. (a) Explain the role of regulatory gene, operator, promoter and structural genes in lac operon when E.Coli is
growing in a culture medium with the source of energy as lactose.
(b) Mention what would happen if lactose is withdrawn from the culture medium.
5
Ans. (a) Regulatory gene, codes for repressor of lac operon 1
Operator ; provides site for binding of repressor protein to prevent transcription 1
Promoter ; provides site for binding of RNA polymerase 1
Structural Genes; codes for enzymes / gene products required for metabolism of lactose 1
(b) If Lactose is withdrawn from the culture medium the operon is not induced or expressed
1
---------------------------------------------------------------------------------------------------
CHAPTER - 7: EVOLUTION
Evolution: Process that results in heritable changes in a population spread over many generations (change in allele
frequencies over time) leading to diversity of organisms on earth.
Theory of Chemical Evolution:
hat the first form of life could have come from pre-existing
non-living organic molecules (e.g. RNA, protein, etc.) and that formation of life was preceded by chemical evolution.
onditions on earth – high temperature, volcanic storms, reducing atmosphere containing CH4, NH3, etc.
0
C.
ds. In similar experiments others observed, formation of sugars, nitrogen bases,
pigment and fats.
Evidences for Organic Evolution:
Palaeontological evidences : fossils founds in rock and support the evolution.
Embryological Evidences
Comparative anatomy & morphology :
Anthropogenic Action: Industrial Melanism
Moths are able to camouflage themselves, i.e. hide in the background, survived. This type of evolution is due to
anthropogenic action. Lichen act as a industrial pollution indicator.
scale of months or years and not centuries.
Adaptive radiation
The evolutionary process which produces new species from single point origin and spread to other geographical areas
(habitat) is called adaptive radiation.
Eg. Darwin finches found in Galapagos Island and Australian Marsupials.
MECHANISM OF EVOLUTION:
Hardy-Weinberg principle:
stable and is constant from generation to generation- genetic equilibriumSum total of all the allelic frequencies is 1. p2 + 2pq + q2 = 1 or, (p + q)2 = 1
Five factors are known to affect Hardy-Weinberg equilibrium:
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. State a reason for the increased population of dark coloured moths coinciding with the loss of lichens (on tree
barks) during industrialization period in England.
1
Ans. Natural selection / survival of fittest / escaped predators due to camouflage
2. How can evolution by natural selection be explained by melanised moths before and after industrialisation in
England?
3
Ans. Before industrialization – mostly light moths (peppered moth) as they merged with the background of trees,
having lichen on their bark ,fewer black moths due to predation by birds as they were clearly visible against the grey
background ½ x 3= 1 ½
After Industrilisation – deposition of soot due to emission from factories made the tree barks black and the black
moths got camouflaged in the background , as the grey moths were now visible , they were predated upon by the birds
and their number got reduced . thus black moths selected over the grey ones. ½ x 3= 1 ½
3. How did S.L. Miller provide an experimental evidence in favour of Oparin and Haldane’s hypothesis?
Explain.
3
Ans. Miller created electric discharge in a closed flask, containing methane hydrogen ammonia and water vapour at
8000c, he observed formation of amino acids. 1x3=3
4. Explain adaptive radiation with the help of a suitable example.
3
Ans. Evolution of different species in a given geographical area starting from a point and literally radiating to other
geographical areas / habitat is called adaptive radiation = 1
A number of marsupials each different from other / Tasmanian wolf / Tiger Cat / Banded anteater /
Marsupial rat / Kangaroo/ Wombat / Bandicoot / Koala / Marsupial mole / Sugar glider (any two
or more) , evolved from an ancestral stock , but all within Australian continent = 1 + ½ + ½ = 2
// Darwin’s finches , from original seed eating features many other forms with altered beaks arose , enabling them to
become insectivorous / vegetarian finches on the same (Galapagos) islands = 1 + ½ + ½ = 2
5. (i) List any four evidences of evolution.
(ii) Explain any one of the evidences that helps to understand the concept of evolution.
5
i. Four evidence= Fossils/ comparative anatomy/ homologous organs / Analogous organs/ Bio- Chemical evidences /
embryological evidences (any four) ½ x 4=2
ii. Any one evidence explained, Definition/ concept 1, Example 1, how it explains evolution 1
----------------------------------------------------------------------------------
UNIT VIII: BIOLOGY IN HUMAN WELFARE
CHAPTER 8: HUMAN HEALTH AND DISEASES
Health- physical, mental and social well being
Common Human Diseases:
TYPES OF DRUGS
SOURCE PLANT
OPIOIDS
CANNABINOIDS
Poppy plant Papaver Cannabis sativa
somniferum
BODY CNS & GI Tract
Cardiovascular system
COCA ALKALOIDS
Erythroxylum coca
AFFECTED
PART
EXAMPLES
CNS
Heroin, morphine
Ganja,
marijuana
hashish, Cocaine, coke
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. Name any two secondary lymphoid organs in a human body and state the function of any of them. 2
Ans. Spleen, lymph nodes, tonsils, Peyer’s patches of small intestine, (vermiform) appendix (any two)½+ ½=1
They act as sites for interaction of lymphocytes with the antigen and cause immune response//
Function: Spleen: Trap blood-borne microorganisms thus filters blood/
Lymph nodes: Trap the microorganisms / antigens (which happen to get into the lymph and tissue fluid)
2. How are oncogenic viruses different from proto-oncogenes?
2
Oncogenic viruses are cancer causing viruses / external cancer causing factor
Proto-oncogenes- Identified in normal cells which when activated (under certain condition) could lead
to oncogenic transformation of cells / internal cancer causing factor 1+1=2
3. Why are adolescents especially advised not to smoke? How does smoking affect the functioning of the body?
2
Ans. Because smoking paves the way to hard drugs, causes increased chances of cancer, cause oxygen deficiency in
the body (any two); Nicotine (in cigarette) stimulates adrenal gland , which raises blood pressure / increased heart rate
. ½X4=2
4.(i) Which organ of the human body is initially affected when bitten by an infected female Anopheles? Name
the stage of the parasite that infects this organ. (ii)Explain the events that are responsible for chill and high
fever in the patient. 3
Ans.i. Liver cells, sporozoites ½+1/2=1
ii. Parasites reproduce asexually in RBC/multiply, Rapture of RBCs ,is associated with release of
toxic substance, haemozoin ½X4=2
5 (a) Why is mother’s milk considered very essential for the healthy growth of infants?
(b)What is the milk called that is produced in the initial days of lactation?
Ans.(a) The milk secreted has nutrients and contain antibodies Ig A , provide passives immunity 1
(b) Colostrum 1
6. Modern life style in big cities and towns is surely making the life more easy and comfortable for people. On
the contrary many more health issues and problems are on the rise and one of them is allergic reactions.
(a) Write any four steps you would suggest to minimise the cause of the above allergic responses.
(b) List any two allergens. How does the human body respond to them? Explain.
Ans. (a) Reduce air pollution , improve exposure /sensitivity of the children to the environment to
reduce vulnerability , improve resistance, improve food habits resulting in good health , introduce
physical exercise (any other appropriate measure ) (any four) ½X 4=2
(b) Mites in dust ,pollen, animal dander (any two) ½+ ½ =1
Immune system respond by producing antibodies of IgE type, inducing Mast cells release chemicals
like histamine/ serotonin in response to allergens 2
------------------------------------------------------------------------------------
CHAPTER-9: STRATEGIES FOR ENHANCEMENT IN FOOD PRODUCTION
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. How has mutation breeding helped in improving the production of mung bean crop ?
2
Ans. Produce disease resistant varieties, against yellow mosaic virus / powdery mildew = 1 + 1
2. (i) Why is inbreeding necessary? Give two reasons. (ii) What does continued inbreeding lead to? 3
Ans. Inbreedingis necessary if we want to evolve a pure line in any animal. 1
Inbreeding exposes harmful recessive genes that are eliminated by selection leading to accumulation of superior
genes, continued inbreeding reduces fertility and productivity( Inbreeding depression) 1+1=2
3. Give two reasons for keeping beehives in crop fields during flowering period.
3
Ans. Keeping beehives in crop fields during flowering period increases pollination efficiency, it improves yield of
crops and honey. 1+1=2
4. Explain how and why controlled breeding experiment is carried out in cattle.
3
Ans. Controlled breeding experiment are carried out using artificial insemination. Semen is collected from the male chosen as parent, injected into the reproduction tract of the selected female (cow), stored Semen may be used at a later
date and desirable matings are carried out. ½x4 =2
Helps overcome several problems of normal mating; improve the quality and quantity of desired yield 1
5. Enlist the steps involved in inbreeding of cattle. Suggest two disadvantages of this practice.
Ans. Inbreeding involves mating of closely related individuals within the same breed for 4-6 generations = ½
Superior males and superior females are identified and mated in pairs , the progeny are evaluated ,
superior males and females among them are selected for further mating = ½ × 3
Disadvantages : Inbreeding depression , reduction in fertility , reduction in productivity (any 2) = ½ × 2
------------------------------------------------------------------------------------
CHAPTER: 10 MICROBES IN HUMAN WELFARE
Microbes in household products-
Microbes in industrial products-
Enzymes and other bioactive molecules-
Enzymes and their actions-
Microbes -- production of Bio-Gas- methanogens (e.g.) Methanobacterium.
Act on cellular material to produce methane, seen in anaerobic sludge digesters, rumen of cattle, flooded rice field &
in cattle dung
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. Name a free living and a symbiotic bacterium that serve as Biocontrol agents. Why are they so called?2
Ans. Free living-Azospirillum/Azotobacter ½ , Symbiotic- Rhizobium ½
Because they enrich the nutrient quality of the soil.1
2. Mention a product of human welfare obtained with the help of each one of the following microbes: 2
(a) LAB (b) Saccharomyces cerevisiae (c) Propionibacterium sharmanii (d) Aspergillus niger.
Ans. a) Milk to curd = ½ b) Bread / ethanol / alcoholic drinks / whiskey / brandy / beer / rum = ½
c) Swiss cheese = ½ d) Citric acid = ½
3. Name a genus of baculovirus. Why are they considered good biocontrol agents. 2
Ans. Nucleopolyhedrovirus 1, Species specific & narrow spectrum insecticidal applications
1
4. Choose any three microbes, from the following which are suited for organic farming which is in great
demand these days for various reasons. Mention one application of each one chosen. Mycorrhiza ; Monascus ;
Anabaena ; Rhizobium ; Methanobacterium ; Trichoderma. 3
Ans. Mycorrhiza : (Fungal symbiont of the association) Abosrb phosphorus from soil
Anabaena : Fix atmospheric nitrogen / Adds organic matter to the soil
Rhizobium : Fix atmospheric nitrogen (in leguminous plants)
Methanobacterium : They digest cellulosic material and the product / spent slurry can be used as fertiliser
Trichoderma : Biocontrol agent for several plant pathogens (Any 3 microbes = ½ × 3 = 1½)
(Any 3 corresponding roles = ½ × 3 = 1½)
5. Make a list of three household products along with the names of the micro-organism producing them. Ans.
Lactic acid bacteria – curd; Sacharomyces cerevisiae- bread; Propionibacterium sharmanii- swiss cheese
6x1/2=3
--------------------------------------------------------------------------------------
UNIT IX: BIOTECHNOLOGY
CHAPTER -11: BIO TECHNOLOGY PRINCIPLES & PROCESSES
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. Mention the use of gel electrophoresis in biotechnology experiments.
1
Ans. Cut fragments of DNA can be segregated/separated
2. Name the technique that is used to alter the chemistry of genetic material (DNA, RNA) to obtain desired
result. 1
Ans. Genetic Engineering / Biochemical Engineering / Biotechnology
3. Write the functions of the following n biotechnology.
3
(a) Polymerase chain reaction technique (b) Restriction endonucleases (c) Thermus aquaticus
Ans a) Multiple copies of gene of interest can be obtained.
(b) They can cut DNA molecule at a particular point by recognizing a specific sequence of base pairs. Thus they are
useful in forming recombinant DNA.
(c) Thermus aquaticus – is the source of Taq- polymerase which remains active during high temperature induced
denaturation of DNA in PCR technique and therefore allows chain reaction to proceed .
4. (i) Why was a bacterium used in the first instance of the construction of an artificial recombinant DNA
molecule? (ii) Name the scientists who accomplished this and how?
Ans. (i) Bacterium has a plasmid in which the desired gene is introduced / the gene to betransferred is from a
bacterium –the anti-biotic resistance gene/ the host cell(bacterial cell) is required for gene cloning / bacteria produce
restriction endonucleases ½ + ½ (ii) Herbert Boyer and Stanley Cohen ½ + ½
Antibiotic resistant gene was isolated using restriction enzyme and introduced into the plasmid of bacterium –
Salmonella typhimurium ½ Later the recombinant plasmid was introduced into the bacterium – E.coli so that it could
make copies of gene. ½
5. Explain the mode of action of Eco RI.
Ans. EcoRI first inspects the length of a DNA sequence, then it binds with specific recognition sequence of DNA,
EcoRI cut each of the two strands of double helix at specific points in their sugar – phosphate backbones // diagram
with same value points to be accepted (1x3=3 Marks)
----------------------------------------------------------------------------------
CHAPTER- 12: BIO TECHNOLOGY & ITS APPLICATION
Application in agriculture : Genetically modified organisms (GMO)
Transgenic crops (GMO) -Crops contain or express one or more useful foreign genes.
Advantages -i) More tolerant to stresses (heat, cold, draught).
ii) Pest resistants GM crops, reduce the use of Chemical pesticides. Eg- BT-Cotton
iii) Reduced post harvest losses. iv) Enhance nutritional value of food.
PEST RESISTANT PLANTS: Bt- cotton
Protection of plants against nematodes –, Meloidogyne incognita infects tobacco plants & reduces yield. Specific
genes (DNA) from nematodes introduced into the plants using Agrobacterium tumifecians (soil bacteria). Genes
produce sense and antisense complementary RNA. Act as dsRNA and initiates RNAi ( RNA interference) and silences
the specific mRNA. Complementary RNA neutralizes the specific RNA of nematodes by a process called RNA
Interference and parasite cannot live in transgenic host.
In medicine:
Molecular diagnosis -accurate detection of diseases can be done through : PCR (Polymerase chain reaction): Short stretches of
pathogenic genome is amplified for detection of suspected AIDS, Cancer or genetic disorder.
ELISA (Enzyme Linked Immunosorbent Assay) used to detect AIDS based on detection of antibodies produced
against antigen of pathogen.
Transgenic Animals
1. To know how genes contribute to development of disease.
2. To get biologically useful products . Eg. The first transgenic cow Rosie produced human protein enriched milk 3.
To verify vaccine and chemical safety.
Biropiracy -- Some organizations and multinational companies exploit or patent bioresources of other nations without
proper authorization. Indian patent bill is there to prevent such unauthorized exploitation.
GEAC- For validity of GM research and the safety of introducing GM organism.
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. People are quite apprehensive to use GM crops. Give three arguments in support of GM
crops so as to convince the people in favour of such crops. 3
Ans. i. Made crops more tolerant to abiotic stresses (cold, drought, salt, heat)
ii. Reduced reliance on chemical pesticides (pest- resistance crops)
iii. Helped to reduce post harvest losses.
iv. Increased efficiency of mineral usage by plants/this prevent early exhaustions of fertility of soil.
v. Enhanced nutritional value of food /vitamin ‘A’ enriched rice.
vi Create tailor made plants to supply alternative resources to industry (any three) 1X3=3
2. Explain how the company Eli Lilly was able to produce human insulin using recombinant DNA technique.
3
Ans.Eli lilly prepared two DNA sequences A and B, and introduced them in plasmids of E. Coli to produce insulin
chains , later extracted and combined by creating disulphide bonds. 1X3=3
4. How has the study of biotechnology helped in developing pest resistant cotton crop? Explain.3
Ans. Some strains of Bacillus thuringiensis produce proteins that kill insects (pests), these crystals contain a toxic
insecticidal protein, once the insect ingests this (inactive) toxin it is converted into an active form, due to alkaline pH
of the gut , activated toxin binds to surface of midgut epithelial cells and creates pores , causing swelling and lysis
leading to death of pest ½ x 6=3
5. Why is molecular diagnosis preferred over conventional methods? Name any two techniques giving one use
of each.
3
Ans. To allow early detection 1 Example –rDNA technology/ PCR / ELISA/ Probe (any two) ½ + ½
PCR- to detect low concentration of bacteria/ virus (HIV), ELISA- to detect antigens / to detect antibodies produce by
those antigens / to detect HIV, Probe- to detect a mutated gene (from a normal one) (any two corresponding functions)
½ + ½= 1
------------------------------------------------------------------------------------------
UNIT - X ECOLOGY
CHAPTER: 13 ORGANISMS AND POPULATIONS
Ecology: deals with the interaction among organisms between organisms & physical environment.
Biome: the largest ecological regions distinguishable by characteristic plants and animals.
There are six: tundra, conifer, deciduous forest, grassland, tropical, and desert.
RESPONSE TO ABIOTIC FACTORS
 Regulation: Organisms maintain homeostasis achieved by physiological and behavioral means Thermo
regulation and osmoregulation.
 Conformation Cannot maintain constant internal Environment # Body temperature and osmotic concentration
of body changes with ambient temperature and concentration of medium.-Thermo confirmer and osmo
confirmer
 Migration : Organism moves away temporarily to another habitat in stressful condition
 Suspension: Organisms suspend their metabolic activities during stressful condition
Resume their function at the return of favorable conditions.E.g. Hibernation (winter sleep) of Frog, Reptiles,
Polar Bear etc , Aestivation (summer sleep) in Snail and Fish.
Adaptation
 Morphological, physiological and behavioral changes that enable organisms to adjust to the ever changing
environment . E.g. Kangaroo rat survives in desert conditions through internal oxidation of fat, removing
concentrated urine of limited quantity.
# Allen‘s rule-cold climate mammals have shorter ears and limbs to minimize heat loss.
# Polar mammals like seals have blubber to prevent heat loss.
# Burrowing habit to escape from heat
# Higher count of RBC, Hb(haemoglobin) at high altitudes.
Population attributes
*Birth Rate/ NATALITY – Number of individuals born per thousand per year.
*Death Rate/MORTALITY – Number of individuals die per thousand per year.
*Sex Ratio – Ratio of male-female in the population.
Population density. - the number of individual organisms per unit area (appropriate measure – total numbersometimes difficult to determine or meaningless because 4 factors N+I-M+E are concerned w.r.t habitat concerned
Age pyramids
# Three ecological ages: Pre-reproductive, Reproductive and Post-Reproductive , High proportion pre-reproductive
individuals occur in Expanding population , Pre-reproductive and reproductive individuals are uniform in Stable
population and Pre-reproductive individuals are less in Declining population.
Population growth : Growth Models :
(i) Exponential growth: When resources in the habitat are unlimited.
(ii) Logistic growth: When resources are limited.
Density of population at any time at a given place depends on Natality, Mortality, Emigration Immigration
POPULATION INTERACTION
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. What is mutualism ? Mention any two examples where the organisms involved are commercially exploited in
agriculture.
2
Ans. Interaction between two species in which both are benefitted =1
i. Rhizobium in the roots (nodules) of legumes = ½ ii. Mycorrhiza / Glomus with the roots of higher plants = ½
2. How do snails, seeds, bears, zooplanktons, fungi and bacteria adapt to conditions unfavourable for their
survival ?
2
Ans. Snail-aestivation = ½; Seeds-dormancy/suspended metabolic activities = ½; Bear-Hibernation = ½; Zooplanktondiapause/suspended development = ½; Fungi-Spore/Zygospore = ½ Bacteria-Cyst/spore = ½
3. Many fresh water animals can not survive in marine environment. Explain.
2
Ans. High salt concentration outside / hypertonic surroundings = 1
Loss of water from body / exosmosis from animal body / animal suffers osmotic problems = 1
4. (a) List the different attributes that a population has and not an individual organism.
(b) What is population density ? Explain any three different ways the population density can be measured, with
the help of an example each.
5
(a) Attributes of population: Birth rate , Death Rate , sex ratio,age pyramids/age distribution (any two) = ½ × 2
(b) Population density - Number of individuals per unit area at a given time / period = 1
1. Biomass / % Cover , e.g Hundred Parthenium plants and 1 huge banayan tree = ½×2
2. Relative Density , e.g Number of fish caught per trap from a lake =½×2
3. Numbers , e.g Human population = ½ × 2
4. Indirect estimation , e.g without actually counting/seeing them e.g tiger census based on
pugmarks and fecal pellets = ½ × 2
----------------------------------------------------------------------------------------
Chapter - 14: ECOSYSTEM
The components of the ecosystem are seen to function as a unit: Productivity, Decomposition, Energy flow and
Nutrient cycle.
PRODUCTIVITY:
Primary productivity:
o The amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis.
Gross primary productivity: (GPP) is the rate of production of organic matter during photosynthesis.
Net primary productivity: GPP – R = NPP.
Secondary productivity: is defined as the rate of formation of new organic matter by the consumer.
DECOMPOSITION:The process of decomposition completed in following steps:
o Fragmentation : Break down of detritus into smaller particles by detritivore (earthworm).
o Leaching: Water soluble inorganic nutrients go down into the soil horizon and get precipitated as unavailable salts.
o Catabolism : Bacterial and fungal enzymes degrade detritus into simple inorganic substances.
o Humification: Accumulation of dark coloured amorphous substances called humus.
o Mineralization: The humus is further degraded by some microbes and release of inorganic nutrients occur.
ENERGY FLOW IN ECOSYSTEM:
Only 10% of energy transferred from one trophic level to other.
Food chain:
Grazing food chain (GFC): it extends from producers through herbivore to carnivore.
Grass---- Grass hopper---- Frog----- Snake------- Hawk
Detritus food chain (DFC): Begins with dead organic matter (detritus) and pass through detritus feeding organism
in soil to organisms feeding on detritus-feeders.
Detritus--- Earthworm--- Bacteria/ Fungi ---- Plants--- Animals
Standing crop: each trophic level has a certain mass of living material at a particular time called as the standing
crop.
ECOLOGICAL PYRAMID:
Three types : number, energy or biomass. In most ecosystems, all the pyramids, of number, of energy and biomass are
upright.
number in a tree ecosystem is inverted.
biomass in sea also inverted because the biomass of fishes is far exceeds that of phytoplankton.
energy is always upright, can never be inverted, because when energy flows from a particular trophic
level to the next, some energy is always lost as heat at each step.
ECOLOGICAL SUCCESSION:
ecological succession.
the changes lead finally to a community that is in near equilibrium with the environment and that is called
climax community.
Primary succession: succession that starts where no living organisms are there- these could be areas where no
living organism ever existed may be a bare rock or new water body.
Secondary succession: succession that starts in areas that somehow, lost all the living organisms that existed there.
TYPES : Based on the nature of habitat – whether it is water or it is on very dry areas- succession of plants is called
hydrarch or xerarch.
Xerarch succession: Succession in bare rock:
Lichen—Mosses—herbaceous plants--- shrubs--- trees
Hydrarch (succession in aquatic environment)
Phytoplankton--- Zooplanktons --- rooted hydrophytes---- Sub merged and free-floating plant stage----- Reedswamp stage---- Marsh-meadow stage--- Shrub stage--- Trees--- the forest
NUTRIENT CYCLING: Of two types: Gaseous cycle & Sedimentary cycle.
Carbon Cycle
rbon is fixed in the biosphere by photosynthesis, annually.
of Carbon is lost to sediments and removed from circulation.
releasing CO2 to atmosphere.
Phosphorus cycle:
horus is the rock, which contain phosphorus in the form of phosphates.
Difference between Carbon and Phosphorus cycle:
1. No respiratory release of phosphorus
2. Reservoir for Carbon is atmosphere but for Phosphorus it is rocks.
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. ‘‘Man can be a primary as well as a secondary consumer.’’ Justify this statement.
1
Ans. Vegetarian diet - Primary consumer = ½ Non vegetarian diet - Secondary consumer = ½
2. How are productivity, gross productivity, net primary productivity and secondary productivity interrelated
?
Ans. Productivity is rate of biomass production = ½
GPP - R = NPP = 1
NPP - is biomass available to consumers for secondary productivity = ½
3. ‘‘It is often said that the pyramid of energy is always upright. On the other hand, the pyramid of biomass
can be both upright and inverted.’’ Explain with the help of examples and sketches. 5
Ans. Upright Pyramid of Energy diagram: e.g of any Grassland food chain depicting energy transfer at each trophic
level = 1+1
Upright Pyramid of Biomass: e.g grassland food chain(Any other relevant example) =
1 for Diagram + ½ for example
Inverted Pyramid of Biomass: e.g aquatic ecosystem where small standing crop of phytoplanktons supports large
standing crop of zooplanktons =1 for Diagram + ½ for example
-------------------------------------------------------------------------------------
Chapter - 15: BIODIVERSITY AND CONSERVATION
Levels of biodiversity: Genetic diversity, Species diversity, Ecological diversity
Pattern of Biodiversity:
Latitudinal gradients: Species diversity decreases as we move away from the equator towards the pole.
Reasons why tropical rain forest has greater biodiversity:
environments. Unlike temperate ones, are less seasonal, relatively more constant and predictable, promotes niche
specialization and lead to greater species diversity.
Species area relationship: Within a region species richness increased with increasing explored area but only up to a
limit.
Causes of biodiversity loss: Four major causes “The Evil Quartet”:
1) Habitat loss and fragmentation
2) Over-exploitation:
3) Alien species invasion: Nile perch introduced into Lake Victoria in east Africa led to extinction of 200 species of
cichlid fish in the lake.
Parthenium, (carrot grass), Lantana, and water hyacinth (Eichornia) posed a thread to
Clarias gariepinus for aquaculture purposed is posing a threat to indigenous
catfishes in our rivers.
4) Co-extinction
BIODIVERSITY CONSERVATION:
Reasons: Narrowly utilitarian, Broadly utilitarian and Ethical
Methods:
1. In situ conservation: When we conserve and protect the whole ecosystem, its biodiversity at all level is protected
– we save the entire forest to save the tiger
Biodiversity hot spot: regions with very high levels of species richness and high degree of endemism.(species
confined to that region and not found anywhere else)
Sacred groves: tract of forest were set aside, and all the trees and wildlife within were venerated and given total
protection.
2. Ex situ conservation: threatened animals and plants are taken out from their natural habitat and placed in special
setting where they can be protected and given special care.
cryopreservation, Genetic strains
are preserved in seed bank.
Convention on Biodiversity: “The Earth Summit” held in Rio de Janeiro, World Summit on Sustainable
development held in Johannesburg, South Africa
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. List any four techniques where the principle of ex-situ conservation of biodiversity has been employed. 2
Ans. Cryopreservation, in vitro fertilisation, micro propagation / tissue culture , sperm bank/ seed bank / gene bank ½
×4
2. Why are sacred groves highly protected? 2
Ans. Because they are the last refuges for a large number of rare and threatened plants.
3. What is meant by ‘alien species’ invasion? Name one plant and one animal alien species that are threat to
our Indian native species.
2
Ans. When alien species are introduced unintentionally or deliberately for whatever purpose, some of them turn
invasive, and cause decline or extinction of indigenous species. 1
Plant- Carrot grass/ lantana ½ ; Animal- African catfish Clarias gariepinus ½
4. Since the origin of life on the earth, there were five episodes of mass extinction of species. (i) How is the 'Sixth
Extinction', presently in progress, different from the previous episodes? (ii) Who is mainly responsible for the
'Sixth Extinction''? (iii) List any four points that can help to overcome this disaster.3
Ans. (i) The rates are faster / accelerated / current species extinction rate are estimated to be 100-1000 times faster
than in the pre-human times. = ½ (ii) Human activities. = ½ (iii) a. Preventing habitat loss and fragmentation b.
Checking overexploitation c. Preventing alien species invasion d. Preventing co-extinction e. Conservation /
Preservation of species.(any four) = ½×4 = 2 [1+2= 3 marks]
5. (a) Why is there a need to conserve biodiversity?
(b) Name and explain any two ways that are responsible for the loss of biodiversity. 5
Ans. (a) 1. to continue to get the products of human consumption
2. plays a major role in many eco system services that nature provides and that is invaluable
3. moral duty to pass on biological legacy in good order to future generations (Any two) = (1×2 = 2)
(b) 1. Habitat loss and fragmentation- large habitats when broken lead to loss of habitat for
animals needing large territories (are badly affected) – population decline
2. Overexploitation- leading to extinction of many, especially commercially important species
3. Alien species invasion - alien species when introduced may turn invasive causing decline
and extinction of indigenous species // explain with an example.
4. Coextinction- when one species become extinct , any other organism intimately associated
also becomes extinct. (any two) (1½ × 2) [2 + 3 = 5 marks]
----------------------------------------------------------------------------------------
CHAPTER-16: ENVIRONMENTAL ISSUES
GREEN HOUSE EFFECT AND GLOBAL WARMING: The greenhouse effect is a naturally occurring
phenomenon that is responsible for heating of Earth’s surface and atmosphere.
global warming or
enhanced green house effect.
OZONE DEPLETION IN THE STRATOSPHERE: Thickness of ozone layer measured in Dobson units (DU)
Effects of UV rays:
– B radiation and high dose of UV – B causes inflammation of cornea called
snow-blindness, cataract etc.
Prevention:
Montreal Protocol was signed at Montreal (Canada) in 1987 to control emission of ozone depleting substances.
PREVIOUS YEARS QUESTIONS & MARKING SCHEME
1. List two advantages of the use of unleaded petrol in automobiles as fuel. 1
Ans. (i) Allows the catalytic convertor to remain active = ½
(ii) Reduces air pollution = ½
2. State the cause of Accelerated Eutrophication
1
Ans. Pollutants from human activities /effluents from industries / effluents from home / sewage /
agricultural (chemical) wastes radically accelerate the ageing process
3. What is joint forest management? How can it help in conservation of forests ?
2
Ans. JFM - A programme (initiated by Govt. of India in 1980) where govt. works closely with local
communities for protecting & managing forests = 1
Forests are conserved by locals in a sustainable manner as locals are also benefitted with forest
products / (fruits / gum / rubber / medicines etc) = 1
4.“Determination of Biological Oxygen Demand (BOD) can help in suggesting the quality of a water body.”
Explain.
3
Ans. High BOD of a water body indicates more number of micro-organisms in water , resulting in bad
quality of water / death of aquatic creatures , more polluting potential = 1 × 3
// Lower BOD of water body indicates less number of micro-organisms in water , good quality of
water / aquatic life flourishes , less polluting potential = 1 × 3
5. With the help of a flow chart, show the phenomenon of biomagnification of DDT in an aquatic food chain.
BEST OF LUCK!